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The RSA private-key operation (used for decryption and signature generation) amounts to solving for $x$ the equation $y\equiv x^e\pmod N$, knowing $y$, the factorization of the public modulus $N$ into $k\ge2$ distinct primes $N=r_1\dots r_k$, public exponent $e$ such that $\gcd(e,r_i-1)\ne1$, and that $0\le x<N$.
For an efficient implementation, we can solve this equation modulo each of the $r_i$; then use the CRT to combine solutions between products of moduli for which we already have a solution, until reaching a solution modulo $N$. The common way, implicit in PKCS#1v2.2, is:
precompute the following quantities $d_i$ (the CRT exponents) and $t_i$ (the CRT inverses/coefficients), e.g. at key generation time, including the results in the private key: for $i\in\{1,\dots,k\}$ $d_i\gets e^{-1}\bmod(r_i-1)$, or equivalently $d_i\gets d\bmod(r_i-1)$ $m\gets r_1$ for $i$ from $2$ to $k$ $t_i\gets m^{-1}\bmod r_i$ $m\gets m\cdot r_i$ when needing to use the private key and solve $y\equiv x^e\pmod N$ for $i\in\{1,\dots,k\}$ [note: should be parallelized if possible] $x_i\gets(y\bmod r_i)^{d_i}\bmod r_i$ $x\gets x_1$, $m\gets r_1$ for $i$ from $2$ to $k$ [loop invariant: $0\le x<m$, $y\equiv x^e\pmod m$ ] $x\gets x+m\cdot((x_i-x\bmod r_i)\cdot t_i\bmod r_i)$ $m\gets m\cdot r_i$
Correctness follows from the loop invariant. See this question for attribution. See this other one for how the bitsize of $N$ relates to a maximum reasonable number of primes.
Artificially small example with 3 primes:
e=5
r1=931164518537359 r2=944727352543879 r3=982273258722607
N=864102436520313334659779717201860718296307527
d1=558698711122415 d2=566836411526327 d3=785818606978085
t2=360227672914825 t3=882117903741868
y=529481440313141057262802385309623737292746309
x1=436496882968258 x2=903092574358267 x3=806961802724
x=710532117316769399313215266414 (when i=2)
x=111222333444555666777888999000000000000000042
The effort saved compared to a standard (non-CRT) implementation is by a factor at most (and near) $k^2$, if modular multiplication has cost $\mathcal O(n^2)$ for arguments of $n$ bits. The time saved can be higher, up to a factor at most (and near) $k^3$ if parallelization is used on $k$ independent modexp units.
It is critical to make a final check that $y\equiv x^e\pmod N$, and not disclose $x$ otherwise. If this precaution was not taken, the implementation would be vulnerable to the cardinal "Bellcore" fault attack: D. Boneh, R. A. DeMillo, R. Lipton;
On the Importance of Eliminating Errors in Cryptographic Computations (in Journal of Cryptology 14(2), 2001; or there).
Implementations should be adequately protected from a variety of other attacks, including timing, power analysis, and other side-channel attacks.
The question also mentions encryption, where only the public key $(N,e)$ is known, not the factorization of $N$. Hence, for that RSA public-key operation (also used for signature verification), there is no similar shortcut applying to the computation $y\gets x^e\bmod N$. However, typically, that remains of low cost compared to the RSA private-key operation, because $e$ is small. |
I wanted to check a couple of these changing rate problems, since these are kind of tricky to me still:
1.
(a) If A is the area of a circle with radius r and the circle expands as time passes, find $\frac{dA}{dT}$ in terms of $\frac{dR}{dT}$.
(b) Suppose oil spills from a ruptured tanker and spreads in a circular pattern. If the radius of the oil spill increases at a constant rate of 1 m/s, how fast is the area of the spill increasing when the radius is 30m?
Answer Attempt
(a) $$A = \pi \cdot r^2$$ so: $$\frac{dA}{dT} = 2 \cdot \pi r \cdot \frac{dR}{dT}$$ (b) if r = 30m and $\frac{dR}{dT} = 1,$ then:
$$ \frac{dA}{dT} = 2 \cdot \pi \cdot (30m) \cdot 1 \frac{m}{s}$$ $$ = 60 \frac{m^2}{s}$$
2.
If $x^2 + y^2 + z^2 = 9$, $\frac{dx}{dt} = 5$, and $\frac{dy}{dt} = 4$, find $\frac{dz}{dt}$ when (x, y, z) = (2, 2, 1)
Answer Attempt:
$$2x\frac{dx}{dt} + 2y\frac{dy}{dt} + 2x\frac{dz}{dt} = 0$$
$$4 \cdot 5 + 4 \cdot 4 + 2 \cdot \frac{dz}{dt} = 0$$
$$2\frac{dz}{dt} = 36$$
$$\frac{dz}{dt} = 18$$ |
I have come across an exercise which asks to prove that the series of functions$$\sum\frac{x^n}{1+x^n}$$ is convergent for $x\in [0,1)$. It also asks us to prove that the series converges uniformly for each $a: 0<a<1$. I think I know how to answer the second question (proving the uniform convergence of the series): $$\sum\frac{x^n}{1+x^n}\leq\sum x^n$$ We can then use the Weierstrass M-test and let $\sum M_k$= $\sum x^n$, which then converges to $\frac{1}{1-x}$ for $\vert x\vert<1$. Hence $\sum\frac{x^n}{1+x^n}$ is uniformly convergent. However, I am not sure how to answer the first question. I think that I could just say that since the series is uniformly convergent, it must converge. Or I could use the same inequality as above and use the comparison test to show that it converges. However, I am not sure. What am I supposed to do? Is it asking whether the series of functions is pointwise convergent? If not what is the difference between proving that a series of functions converges pointwise vs. proving that a series of functions converges? Am I missing something? Also, is my approach in proving uniform convergence correct?
We can then use the Weierstrass $M$-test and let $\sum M_k= \sum x^n$
No. The whole point of $M$-test is that $M$-numbers do not involve $x$. What should be done instead: on the interval $[0,a]$ we have $$\frac{x^n}{1+x^n}\le x^n \le a^n$$ Let $M_n=a^n$ and apply the test.
I think that I could just say that since the series is uniformly convergent, it must converge.
Yes. More precisely: every point $x\in [0,1)$ is contained in some interval $[0,a]$ with $a<1$ (take any $a$ with $x<a<1$), and we already know that the series converges, even uniformly, on $[0,a]$. Thus it converges at $x$; and this is all we need to show to demonstrate pointwise convergence on $[0,1)$.
Apparently, you were not asked to show that uniform convergence fails on $[0,1)$. For the same of completeness, here is a sketch: for every $n$ there is $x\in [0,1)$ such that $x^n=1/2$. Hence $\frac{x^n}{1+x^n}=1/3$. So, the terms of the series do not tend to $0$ uniformly, thus failing a necessary condition for uniform convergence. |
Let $A$ be an $n \times n$ matrix. Let $C_{ij}$ be the $ij^{th}$
cofactor of $A$, defined by $C_{ij} = (-1)^{i+j} \det A^{ij}$ and let$$C = \begin{bmatrix} C_{11} & C_{12} & \cdots & C_{1n} \\ C_{21} & C_{22} & \cdots & C_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ C_{n1} & C_{n2} & \cdots & C_{nn}\end{bmatrix}$$be the cofactor matrix. Also let $M$ be the matrix of minors; that is, the matrix that appears in the left hand side of the equation in the question. We will show that $\det C = (\det A)^{n-1}$, and $\det M = \det C$,
which together establish the claim.
Claim 1. From this wikipedia page, we know that $$C^T A = A C^T = (\det A) I_n.$$
If $A$ is invertible, then taking determinants on both sides, we get:$$\det (C^T) \cdot \det A = (\det A)^n \cdot 1.$$Since $\det (C^T) = \det C$ and $\det A \neq 0$, we get $\det C = (\det A)^{n-1}$, and we are done.
On the other hand, if $A$ is noninvertible, then $\det A = 0$ and $ C^T A = AC^T = 0$. In this case, (WLOG assuming $n \geq 1$) it suffices to show that $\det C = 0$, which is equivalent to showing that $C$ is noninvertible as well. Here, we have two subcases:
If $A$ is the zero matrix, then $C$ is also the zero matrix, and hence non-invertible.
Suppose $A$ is nonzero. Then if $C$ is invertible, then $C^T A = 0$ implies that $$A = \Big((C^{-1})^T C^T \Big) A = (C^{-1})^T \cdot 0 = 0 ,$$ which is a contradiction. Hence $C$ is noninvertible, and we are done.
Claim 2. We now prove that $\det M = \det C$. We have the relation $C_{ij} = (-1)^{i+j} M_{ij} = (-1)^{i+j} \det A^{ij}$. So, if we take the matrix $M$, and multiply the $i^{\rm th}$ row by $(-1)^i$ and $j^{\rm th}$ column by $(-1)^j$, then we end up with the matrix $C$. In this process, the determinant gets multiplied by $$\left(\prod_{i=1}^n (-1)^i \right) \left(\prod_{j=1}^n (-1)^j \right) = \left(\prod_{i=1}^n (-1)^i \right)^2 = \prod_{i=1}^n (-1)^{2i} = 1.$$In other words, $\det C = \det M$. |
Problem 676
Let $V$ be the vector space of $2 \times 2$ matrices with real entries, and $\mathrm{P}_3$ the vector space of real polynomials of degree 3 or less. Define the linear transformation $T : V \rightarrow \mathrm{P}_3$ by
\[T \left( \begin{bmatrix} a & b \\ c & d \end{bmatrix} \right) = 2a + (b-d)x – (a+c)x^2 + (a+b-c-d)x^3.\]
Find the rank and nullity of $T$.Add to solve later
Problem 675
The space $C^{\infty} (\mathbb{R})$ is the vector space of real functions which are infinitely differentiable. Let $T : C^{\infty} (\mathbb{R}) \rightarrow \mathrm{P}_3$ be the map which takes $f \in C^{\infty}(\mathbb{R})$ to its third order Taylor polynomial, specifically defined by
\[ T(f)(x) = f(0) + f'(0) x + \frac{f^{\prime\prime}(0)}{2} x^2 + \frac{f^{\prime \prime \prime}(0)}{6} x^3.\] Here, $f’, f^{\prime\prime}$ and $f^{\prime \prime \prime}$ denote the first, second, and third derivatives of $f$, respectively.
Prove that $T$ is a linear transformation.Add to solve later
Problem 674
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 $T : \mathrm{P}_4 \rightarrow \mathrm{P}_{4}$ be the map defined by, for $f \in \mathrm{P}_4$,
\[ T (f) (x) = f(x) – x – 1.\]
Determine if $T(x)$ is a linear transformation. If it is, find the matrix representation of $T$ relative to the standard basis of $\mathrm{P}_4$.Add to solve later
Problem 673
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 $T : \mathrm{P}_3 \rightarrow \mathrm{P}_{5}$ be the map defined by, for $f \in \mathrm{P}_3$,
\[T (f) (x) = ( x^2 – 2) f(x).\]
Determine if $T(x)$ is a linear transformation. If it is, find the matrix representation of $T$ relative to the standard basis of $\mathrm{P}_3$ and $\mathrm{P}_{5}$.Add to solve later
Problem 672
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.Add to solve later
Problem 669 (a) Suppose that a $3\times 3$ system of linear equations is inconsistent. Is the coefficient matrix of the system nonsingular? (b) Suppose that a $3\times 3$ homogeneous system of linear equations has a solution $x_1=0, x_2=-3, x_3=5$. Is the coefficient matrix of the system nonsingular?
Add to solve later
(c) Let $A$ be a $4\times 4$ matrix and let \[\mathbf{v}=\begin{bmatrix} 1 \\ 2 \\ 3 \\ 4 \end{bmatrix} \text{ and } \mathbf{w}=\begin{bmatrix} 4 \\ 3 \\ 2 \\ 1 \end{bmatrix}.\] Suppose that we have $A\mathbf{v}=A\mathbf{w}$. Is the matrix $A$ nonsingular? Problem 668
Consider the system of differential equations
\begin{align*} \frac{\mathrm{d} x_1(t)}{\mathrm{d}t} & = 2 x_1(t) -x_2(t) -x_3(t)\\ \frac{\mathrm{d}x_2(t)}{\mathrm{d}t} & = -x_1(t)+2x_2(t) -x_3(t)\\ \frac{\mathrm{d}x_3(t)}{\mathrm{d}t} & = -x_1(t) -x_2(t) +2x_3(t) \end{align*} (a) Express the system in the matrix form. (b) Find the general solution of the system.
Add to solve later
(c) Find the solution of the system with the initial value $x_1=0, x_2=1, x_3=5$. Solve the Linear Dynamical System $\frac{\mathrm{d}\mathbf{x}}{\mathrm{d}t} =A\mathbf{x}$ by Diagonalization Problem 667 (a) Find all solutions of the linear dynamical system \[\frac{\mathrm{d}\mathbf{x}}{\mathrm{d}t} =\begin{bmatrix} 1 & 0\\ 0& 3 \end{bmatrix}\mathbf{x},\] where $\mathbf{x}(t)=\mathbf{x}=\begin{bmatrix} x_1 \\ x_2 \end{bmatrix}$ is a function of the variable $t$.
Add to solve later
(b) Solve the linear dynamical system \[\frac{\mathrm{d}\mathbf{x}}{\mathrm{d}t}=\begin{bmatrix} 2 & -1\\ -1& 2 \end{bmatrix}\mathbf{x}\] with the initial value $\mathbf{x}(0)=\begin{bmatrix} 1 \\ 3 \end{bmatrix}$. Prove that $\{ 1 , 1 + x , (1 + x)^2 \}$ is a Basis for the Vector Space of Polynomials of Degree $2$ or Less Problem 665
Let $\mathbf{P}_2$ be the vector space of polynomials of degree $2$ or less.
(a) Prove that the set $\{ 1 , 1 + x , (1 + x)^2 \}$ is a basis for $\mathbf{P}_2$.
Add to solve later
(b) Write the polynomial $f(x) = 2 + 3x – x^2$ as a linear combination of the basis $\{ 1 , 1+x , (1+x)^2 \}$. Problem 663
Let $\R^2$ be the $x$-$y$-plane. Then $\R^2$ is a vector space. A line $\ell \subset \mathbb{R}^2$ with slope $m$ and $y$-intercept $b$ is defined by
\[ \ell = \{ (x, y) \in \mathbb{R}^2 \mid y = mx + b \} .\]
Prove that $\ell$ is a subspace of $\mathbb{R}^2$ if and only if $b = 0$.Add to solve later
Problem 659
Fix the row vector $\mathbf{b} = \begin{bmatrix} -1 & 3 & -1 \end{bmatrix}$, and let $\R^3$ be the vector space of $3 \times 1$ column vectors. Define
\[W = \{ \mathbf{v} \in \R^3 \mid \mathbf{b} \mathbf{v} = 0 \}.\] Prove that $W$ is a vector subspace of $\R^3$. Problem 658
Let $V$ be the vector space of $n \times n$ matrices with real coefficients, and define
\[ W = \{ \mathbf{v} \in V \mid \mathbf{v} \mathbf{w} = \mathbf{w} \mathbf{v} \mbox{ for all } \mathbf{w} \in V \}.\] The set $W$ is called the center of $V$.
Prove that $W$ is a subspace of $V$.Add to solve later |
States of Matter: Gases and Liquids Gas Laws and Ideal Gas Equation and Kinetic Molecular Theory of Gases Gas Laws: \tt V\propto\frac{1}{P} (Boyle's Law) PV = constant
Charles' Law: V ∝ T (or) \tt \frac{V_{1}}{V_{2}}=\frac{T_{1}}{T_{2}} Gay-Lussac Law: \tt \frac{P_{1}}{T_{1}}=\frac{P_{2}}{T_{2}}
Avagadro's Law: V ∝ n Ideal Gas Equation: \tt V \propto \frac{1}{P} V ∝ T V ∝ n \tt \therefore V \propto \frac{nT}{P} PV ∝ nT PV = nRT Graham Law of Diffusion of Gases: \tt r\propto\frac{1}{\sqrt{d}}\propto\frac{1}{\sqrt{M.wt}}\propto\frac{1}{\sqrt{V.D}} (Constant T and P) \tt \frac{r_{1}}{r_{2}}=\frac{\sqrt{d_{2}}}{\sqrt{d_{1}}}=\sqrt{\frac{M_{2}}{M_{1}}}=\sqrt{\frac{V.D_{2}}{V.D_{1}}}=\frac{V_{1}t_{2}}{V_{2}t_{1}} \tt \frac{r_{H_{2}}}{r_{CH_{4}}}=\frac{\sqrt{M_{CH_{4}}}}{\sqrt{M_{H_2}}}=\sqrt{\frac{16}{2}}=2\sqrt{2} Dalton's Law of Partial Pressure: P total = P 1 + P 2 + P 3 + .... P.P = M.F × T.P = Xi × T.P [Xi = mole fraction] \tt Xi=\frac{P.P}{T.P} P total = P A + P B + P C = (n 1 + n 2 + n 3) \tt \frac{RT}{V} Kinetic Gas Equation: U rms = \tt \sqrt{\frac{u_1^2+u_2^2+u_3^2...+u_n^2}{n}} K.E equation = \tt PV=\frac{1}{3}MC^2 \tt K.E=\frac{3}{2}RT \tt K.E=\frac{3}{2}kT (PV=RT) Part1: View the Topic in this Video from 3:47 to 54:04 Part2: View the Topic in this Video from 1:43 to 53:00 Part3: View the Topic in this Video from 0:56 to 1:00:00
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1.
Boyle's law PV = K or \tt P_{1}V_{1} = P_{2}V_{2} = K (For two or more gases)
2. Boyle's law can also be expressed as \tt \left(\frac{dP}{dV}\right)_{T} = \frac{-k}{V^{2}} \ or \ \frac{dV}{V} = \frac{-dP}{P}
3. \tt K = \frac{V}{T} \ or \ \frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}} = K \left(For \ two \ or \ more \ gases\right)
4. Charle's law can be represented as \tt \left(\frac{dV}{dt}\right)_{P} = K
5. \tt K = \frac{P}{T} \ or \ \frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}} = K \left(For \ two \ or \ more \ gases\right)
6. Avogadro's Law: \tt \frac{V_{1}}{n_{1}} = \frac{V_{2}}{n_{2}} = ............ = K
7. PV = nRT, This is called ideal gas equation.
8. If a number of gases having volume V
1, V 2, V 3 ...... at pressure P 1, P 2, P 3 ....... are mixed together in container of volume V, then, \tt P_{Total} = \frac{P_{1}V_{1} + P_{2}V_{2} + P_{3}V_{3}......}{V} or = \tt \left(n_{1} + n_{2} + n_{3}.......\right)\frac{RT}{V} \ \left(\because PV = nRT\right) or = \tt n\frac{RT}{V} \ \left(\because n = n_{1} + n_{2} + n_{3}.......\right)
9. \tt \frac{r_{1}}{r_{2}} = \sqrt{\frac{d_{2}}{d_{1}}} = \sqrt{\frac{d_{2} \times 2}{d_{1} \times 2}} = \sqrt{\frac{M_{2}}{M_{1}}}
Where, M 1 and M 2 are the molecular weights of the two gases.
10. For gases \tt \frac{V_{1}}{V_{2}} = {\frac{n_{1}}{n_{2}}}
\tt {\frac{n_{1}}{n_{2}}}\times{\frac{t_{2}}{t_{1}}}= \sqrt{\frac{M_{2}}{M_{1}}}
11. When equal volume of the two gases diffuse, i.e. V
1 = V 2 then, \tt {\frac{r_{1}}{r_{2}}} = {\frac{t_{2}}{t_{1}}}= \sqrt{\frac{d_{2}}{d_{1}}}
12. When volumes of the two gases diffuse in the same time, i.e. t
1 = t 2 then, \tt {\frac{r_{1}}{r_{2}}} = {\frac{V_{1}}{V_{2}}}= \sqrt{\frac{d_{2}}{d_{1}}}
13.
Kinetic gas equation : On the basis of above postulates, the following gas equation was derived, \tt PV = \frac{1}{3}mnu_{rms}^2 where, P = pressure exerted by the gas V = Volume of the gas m = average mass of a molecule n = number of molecules u rms = root mean square (RMS) velocity of the gas. |
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 wizard is
Leonhard Euler
and his "union" is
the famous equation $e^{i\pi}+1=0$.
The two "helpers" are
the + and = symbols
and the five others are
the actual numbers involved.
Here are my best guesses at how the clues work. Some of them are not very confident guesses.
An unpredictable guy, the wizard's adopted son, who was a prominent & important figure in the court of an English warlord!
This must be $e$, named of course for the "wizard" Euler and "unpredictable" because irrational. (Either because the digits look random, though actually that isn't implied by irrationality, or because irrational
people may be unpredictable.) In comments, OP indicates that the English warlord is in fact another mathematician. Well, pretty much any mathematican in the last 300 years will have made much use of $e$. Newton? (Co-discoverer of the calculus, from which $e$ emerges very naturally.) Taylor? (The exponential function is strongly associated with its nice simple Taylor series.) I hope it isn't Napier, pioneer of logarithms, because Napier was Scottish and not English. Nor, because he too was Scottish, Maxwell (nothing to do with logarithms but plenty of "E"s in his work for entirely different reasons.) Perhaps the word "warlord" is a hint, but I have completely failed to figure out how. (Lots of mathematicians have done military work; hard to see how to pick out a particular one.) ... Or maybe "from the same profession" doesn't mean the warlord was a mathematician. Perhaps e.g. it refers to a "profession of faith"; Euler was a Christian, as we know from the silly story about him and Diderot, so perhaps our English chap was notably religious. I dunno.
This was actually an imaginary guy & as per our wizard he came from ancient Greece.
Presumably this is $i$ ("imaginary"). I've no idea in what sense "per our wizard he came from ancient Greece": I think Euler consistently wrote $\sqrt{-1}$ explicitly rather than (as some occasionally do) $\iota$, and the ancient Greeks certainly didn't have complex numbers. The letter $i$ is, I guess, derived from Greek $\iota$, but again I don't think Euler used it.
Another crazy guy, from somewhere in Asia. It was believed that he came from some sort of spiritual realm.
It seems this must be $\pi$, but I don't understand how this comes from Asia. (Though some of the earliest numerical approximations to it are Chinese.) $\pi$ is, though, a
transcendental number, which is close enough to "spiritual" for our purposes here.
This guy is a representative of his people as well as this union. He was an old & ancient fellow, and nobody knew where he came from, though some people believe he attained this current status in India.
This must be either 0 or 1. Probably 1, perhaps with a suggestion of "one man, one vote" ("representative of his people"). Our numerals are of Indian origin.
Another old guy from India, who had the look of an ancient Himalayan yogi & claimed to hold the truth about whole cosmos.
If the previous one is 1 then this must be 0 (which definitely comes from India; more specifically, from Brahmagupta in what we parochially call the seventh century). The last bit might fit better if this one is 1 ("all is one"...) but it looks to me as if the numbers are being listed in the order of their appearance in the usual form $e^{i\pi}+1=0$ of the formula, in which case this must be 0. I suppose some people have claimed that our whole world is some sort of illusion, which would make 0 kinda-sorta hold the truth about the whole cosmos.
In comments below, Silenus suggests
putting #4 and #5 the other way around, 0 being "representative" because both sides of the equation are 0 and 1 for the same "all is one" reason I mentioned above -- and also because the numeral 1 is thin, like a Himalayan yogi! This could well be right; I can't work out whether I find these reasons more or less persuasive than the fact that otherwise the numbers seem to be "in order of appearance".
Things I'm most conspicuously confused by:
Who's the "English warlord", of the same profession as Euler, whose "court" prominently includes $e$? In what sense is $i$ Greek or $\pi$ Asian? If 0 and 1 appear in that order, in what sense is 0 "representative of his people"? If in the other order, why does 0 claim "to hold the truth about whole cosmos"? |
I know there are questions that may look similiar at first glance, but they ask about different aspects.
In a book I read there is a certain result proven($[\sigma]$ denotes the conjugacy class of $\sigma$):
Let $n \geq 2$, and let $\sigma \in A_n$. Then $[\sigma]_{A_n} = [\sigma]_{S_n}$ or the size of $[\sigma]_{A_n}$ is half the size of $[\sigma]_{S_n}$, according to whether the centralizer $Z_{S_n}(\sigma)$ is not or is contained in $A_n$.
Then it says that in case if $Z_{S_n}(\sigma)$ lies in $A_n$(that is, $|[\sigma]_{S_n}| = 2|[\sigma]_{A_n}|$ ) then $[\sigma]_{S_n}$ actually splits into two conjugacy classes in $A_n$(one of which is, obviously, $[\sigma]_{A_n}$).
But how come $[\sigma]_{S_n} \setminus [\sigma]_{A_n}$ is even a conjugacy class in $A_n$?
What I know is that $g \in [\sigma]_{S_n} \setminus [\sigma]_{A_n}$ iff $g = \tau \sigma \tau^{-1}$ for some odd permutation such that for all $\rho \in A_n \ \ \ \rho^{-1} \tau \notin Z_{S_n}( \sigma ) \Leftrightarrow \rho^{-1} \tau \sigma \tau^{-1} \rho \neq \sigma \Leftrightarrow \tau \sigma \tau^{-1} \neq \rho \sigma \rho^{-1}$.
It's obvious that $g \in A_n$. But why all such $g$'s are conjugates to some even permutation? |
The Annals of Probability Ann. Probab. Volume 17, Number 3 (1989), 1255-1263. Central Limit Theorems for Infinite Urn Models Abstract
An urn model is defined as follows: $n$ balls are independently placed in an infinite set of urns and each ball has probability $p_k > 0$ of being assigned to the $k$th urn. We assume that $p_k \geq p_{k + 1}$ for all $k$ and that $\sum^\infty_{k = 1} p_k = 1$. A random variable $Z_n$ is defined to be the number of occupied urns after $n$ balls have been thrown. The main result is that $Z_n$, when normalized, converges in distribution to the standard normal distribution. Convergence to $N(0, 1)$ holds for all sequences $\{p_k\}$ such that $\lim_{n \rightarrow \infty} \operatorname{Var}Z_{N(n)} = \infty$, where $N(n)$ is a Poisson random variable with mean $n$. This generalizes a result of Karlin.
Article information Source Ann. Probab., Volume 17, Number 3 (1989), 1255-1263. Dates First available in Project Euclid: 19 April 2007 Permanent link to this document https://projecteuclid.org/euclid.aop/1176991268 Digital Object Identifier doi:10.1214/aop/1176991268 Mathematical Reviews number (MathSciNet) MR1009456 Zentralblatt MATH identifier 0685.60023 JSTOR links.jstor.org Citation
Dutko, Michael. Central Limit Theorems for Infinite Urn Models. Ann. Probab. 17 (1989), no. 3, 1255--1263. doi:10.1214/aop/1176991268. https://projecteuclid.org/euclid.aop/1176991268 |
Particles will move parallel to the magnetic field at the sound speed.
$v_s = \sqrt{\frac{Z T_e + 3 T_i}{M}}$
For a typical tokamak, the plasma is usually mostly deuterium, so $M=2$ u and $Z=1$. Let's say $T_e\sim500$ eV and $T_i\approx T_e$. This gets us $v_s = 3.1\times10^5$ m/s.
The safety factor $q = \frac{d\phi}{d\theta}$ measures how the magnetic field is twisted. For a quick estimate, let's pick a typical value of $q=3$. So for every poloidal turn, there will be three toroidal turns. Taking the circular approximation and $R=2$ m, $r=0.5$ m, an ion has to go a distance of roughly $2\pi R q + 2\pi r = 41$ m to get to around the poloidal circumference once.
Traveling 41 m at $v_s$ would take about 130 $\mathrm{\mu}$s. During this time, particles would have some opportunity to mix energy with neighboring field lines. If you only want one turn toroidally, then it's just one third the distance (since we assumed $q=3$), or 43 $\mathrm{\mu}$s.
So, spreading of heat and particles across a flux surface generally happens on the scale of microseconds. |
Category: Ring theory Problem 624
Let $R$ and $R’$ be commutative rings and let $f:R\to R’$ be a ring homomorphism.
Let $I$ and $I’$ be ideals of $R$ and $R’$, respectively. (a) Prove that $f(\sqrt{I}\,) \subset \sqrt{f(I)}$. (b) Prove that $\sqrt{f^{-1}(I’)}=f^{-1}(\sqrt{I’})$
Add to solve later
(c) Suppose that $f$ is surjective and $\ker(f)\subset I$. Then prove that $f(\sqrt{I}\,) =\sqrt{f(I)}$ Problem 618
Let $R$ be a commutative ring with $1$ such that every element $x$ in $R$ is idempotent, that is, $x^2=x$. (Such a ring is called a
Boolean ring.) (a) Prove that $x^n=x$ for any positive integer $n$.
Add to solve later
(b) Prove that $R$ does not have a nonzero nilpotent element. Problem 543
Let $R$ be a ring with $1$.
Suppose that $a, b$ are elements in $R$ such that \[ab=1 \text{ and } ba\neq 1.\] (a) Prove that $1-ba$ is idempotent. (b) Prove that $b^n(1-ba)$ is nilpotent for each positive integer $n$.
Add to solve later
(c) Prove that the ring $R$ has infinitely many nilpotent elements. |
$T^{ij}$ is nothing more or less than the flow of $i$-momentum across surfaces of constant $j$.
1 As a result, the force exerted across a surface $S$ with unit normal one-form components $n_j$ has components$$ F_{(S)}^i = \int\limits_S T^{ij} n_j \,\mathrm{d}^{d-1}x $$in $d$ dimensions.
The argument for symmetry is not that the cube is static. The argument is that the cube cannot have infinite angular acceleration as its size shrinks. That is, because we are dealing with a continuous fluid, it should be well behaved as we take our region to be arbitrarily small.
Consider a two-dimensional example of a square covering the area $x_0-L/2 < x < x_0+L/2$, $y_0-L/2 < y < y_0+L/2$. The left surface $x = x_0-L/2$ has $n^\mathrm{left}_j = \delta_j^x$ (sign chosen to correspond to flow of momentum into the square), and so the force on our square due to interactions across the left face has components $F_\mathrm{left}^i \approx L T_\mathrm{left}^{ix}$, where $T_\mathrm{left}^{ix}$ is $T^{ix}$ evaluated at the midpoint $(x_0-L/2,y_0)$. On the opposite surface, $n^\mathrm{right}_j = -\delta_j^x$, so $F_\mathrm{right}^i \approx -L T_\mathrm{right}^{ix}$. Similarly, $F_\mathrm{bottom}^i \approx L T_\mathrm{bottom}^{iy}$ and $F_\mathrm{top}^i \approx -L T_\mathrm{top}^{iy}$.
Torque is a $(d-2)$-form: $\tau = {}^*(\tilde{r} \wedge \tilde{F})$, with $\tilde{r}$ and $\tilde{F}$ the one-forms corresponding to displacement $\vec{r}$ and force $\vec{F}$. In 2D, $\tau = \epsilon_{ij} (r^i F^j - r^j F^i)$. If force components $F_\mathrm{left}^i$ are applied at $(x_0-L/2,y_0)$, then $r_\mathrm{left}^i = -(L/2) \delta^i_x$ and $\tau_\mathrm{left} \approx -(1/2) L^2 T^{yx}$. You can also check $\tau_\mathrm{right} \approx -(1/2) L^2 T^{yx}$ and $\tau_\mathrm{bottom} \approx \tau_\mathrm{top} \approx (1/2) L^2 T^{xy}$.
As we shrink the square down, the midpoints at which we evaluate $T^{ij}$ approach one another and we find the total torque is $\tau \approx L^2 (T_\mathrm{center}^{xy} - T_\mathrm{center}^{yx})$. However, the moment of inertia for a square of surface density $\sigma$ is $\sigma L^4/6$. Thus angular acceleration is$$ \alpha = \lim_{L\to0} \frac{6(T^{xy}-T^{yx})}{\sigma L^2} $$at any point in the fluid. Thus we must have $T^{xy} = T^{yx}$, in order to avoid $\alpha \to \infty$. Note that this argument holds in higher dimensions, in more general settings than fluids, and for more general geometries/spacetimes.
2
The argument does not however hold in the linear acceleration case. For example, the net $x$-force will have terms like $L (T_\mathrm{left}^{xx} - T_\mathrm{right}^{xx})$ and $L (T_\mathrm{bottom}^{xy} - T_\mathrm{top}^{xy})$. Even though mass is $\sigma L^2$, which would seem to imply linear acceleration goes as $1/L$, the fact is the pairs of stress tensor components naturally cancel as they are evaluated at the same point ($T_\mathrm{left}, T_\mathrm{right}, T_\mathrm{bottom}, T_\mathrm{top} \to T_\mathrm{center}$). No constraints are imposed from this consideration.
1I avoid saying "in the $j$-direction, since we are really interested in surfaces, and these are characterized by one-forms, not vectors. This is more apparent in non-Cartesian (better still non-diagonal) coordinate systems.
2This symmetry always holds, for any material or field, as long as momentum is conserved. You occasionally see reference to the antisymmetric part of the stress tensor, but this comes from splitting the physics into separate domains, and pretending that momentum is lost when going from one to another (e.g. torques can transfer angular momentum from bulk flow into particle spins, and we choose to treat the latter as some momentum-conservation-violating sink as far as the continuum-modeled fluid is concerned). |
I read about black holes, about the Schwarzschild metric, Einstein field equations and their solution in the vacuum for a spherical body.
I understood black holes are object whose gravity is enormous, and I also understood that there does exist the so called Schwarzschild radius, which is
$$R_s = \frac{2GM}{c^2}$$
which is the radius of a sphere such that, if all the mass of an object were to be compressed within that sphere, the escape velocity from the surface of the sphere would equal the speed of light.
So I understood of course why General Relativity is more than essential to describe those objects.
But now, if we make an example with our Sun (which won't ever become a black hole, but let's just play), we would get
$$R_s \approx 2.95\cdot 10^3\ m$$
Now this sphere seems everything but infinitesimal to me, so my question is: why black holes do need also quantum mechanics to be completely understood? Or is it referred only to the "inside" of a Black Hole (namely the region beyond the Events Horizon)?
I'm sorry, this question may be unclear or stupid but don't down vote it. I'm sure you all got some doubts in your past, and the most beautiful thing is someone who explains them to you. |
title: “Introduction to ‘eva’ and its capabilities” author: Brian Bader date: “2018-10-04” output: rmarkdown::html_vignette vignette: > % %
The ‘eva’ package, short for ‘Extreme Value Analysis with Goodness-of-Fit Testing’, provides functionality that allows data analysis of extremes from beginning to end, with model fitting and a slew of newly available tests for diagnostics. In particular, some highlights are:
Implementation of the \(r\) largest order statistics (GEV\(_r\)) model - data generation, fitting, and return levels.
Efficient handling of the near-zero shape parameter.
Maximum product spacings (MPS) estimation for parameters in the block maxima (GEV\(_1\)) and generalized pareto distributions.
Sequential tests for the choice of \(r\) in the GEV\(_r\) model, as well as tests for the selection of threshold in the peaks-over-threshold (POT) approach. For the boostrap based tests, the option to run in parallel is provided.
P-value adjustments to control for the false discover rate (FDR) and family-wise error rate (FWER) in the sequential testing setting.
# load package
library(eva)
# A naive implementation of the GEV cumulative density function
pgev_naive <- function(q, loc = 0, scale = 1, shape = 1) {
exp(-(1 + (shape * (q - loc))/scale)^(-1/shape))
}
curve(pgev_naive(1, 0, 1, x), 1e-20, .01, log = "x", n = 1025,
xlab = "Shape", ylab = "GEV CDF", col = "red", lty = 1, lwd = 1)
curve(pgev(1, 0, 1, x), 1e-20, .01, log = "x", n = 1025, col = "blue", lty = 6, lwd = 3, add = TRUE)
# Similarly for the GPD cdf
pgpd_naive <- function(q, loc = 0, scale = 1, shape = 1) {
(1 - (1 + (shape * (q - loc))/scale)^(-1/shape))
}
curve(pgpd_naive(1, 0, 1, x), 1e-20, .01, log = "x", n = 1025,
xlab = "Shape", ylab = "GPD CDF", col = "red", lty = 1, lwd = 1)
curve(pgpd(1, 0, 1, x), 1e-20, .01, log = "x", n = 1025, col = "blue", lty = 6, lwd = 3, add = TRUE)
The GEV\(_r\) distribution has the density function \[f_r (x_1, x_2, ..., x_r | \mu, \sigma, \xi) = \sigma^{-r}\exp\left\{-(1+\xi z_r)^{-\frac{1}{\xi}} - \left(\frac{1}{\xi}+1\right)\sum_{j=1}^{r}\log(1+\xi z_j)\right\}\] for some location parameter \(\mu\), scale parameter \(\sigma > 0\) and shape parameter \(\xi\), where \(x_1 > \cdots> x_r\), \(z_j = (x_j - \mu) / \sigma\), and \(1 + \xi z_j > 0\) for \(j=1, \ldots, r\). When \(r = 1\), this distribution is exactly the GEV distribution or block maxima.
This package includes data generation (rgevr), density function (dgevr), fitting (gevrFit), and return levels (gevrRl) for this distribution. If one wants to choose \(r > 1\), goodness-of-fit must be tested. This can be done using function gevrSeqTests. Take, for example, the dataset Lowestoft, which includes the top ten sea levels at Lowestoft harbor from 1964 - 2014. Two tests are available to run in sequence - the entropy difference and score test.
data(lowestoft)
gevrSeqTests(lowestoft, method = "ed")
## r p.values ForwardStop StrongStop statistic est.loc est.scale
## 1 2 0.9774687 1.455825 0.9974711 0.02824258 3.431792 0.2346591
## 2 3 0.7747741 1.163697 1.0869254 -0.28613586 3.434097 0.2397408
## 3 4 0.5830318 1.116989 1.1500564 0.54896156 3.447928 0.2404563
## 4 5 0.6445475 1.157363 1.2470248 0.46135009 3.452449 0.2376723
## 5 6 0.4361569 1.181963 1.2676077 -0.77869930 3.455478 0.2396332
## 6 7 0.6329943 1.334209 1.4133341 0.47751655 3.454680 0.2372572
## 7 8 0.4835074 1.444819 1.4790559 -0.70067260 3.455901 0.2376215
## 8 9 0.8390270 1.836882 2.0321878 -0.20313820 3.458135 0.2356543
## 9 10 0.8423291 1.847245 3.4235418 0.19891516 3.459470 0.2342272
## est.shape
## 1 0.10049739
## 2 0.09172687
## 3 0.06802070
## 4 0.05451138
## 5 0.04709329
## 6 0.04555449
## 7 0.03838020
## 8 0.02536685
## 9 0.01964612
The entropy difference test fails to reject for any value of \(r\) from 1 to 10. A common quantity of interest in extreme value analysis are the \(m\)-year return levels, which can be thought of as the average maximum value that will be exceeded over a period of \(m\) years. For the Lowestoft data, the 250 year sea level return levels, with 95% confidence intervals are plotted for \(r\) from 1 to 10. The advantage of using more top order statistics can be seen in the plots below. The width of the intervals decrease by over two-thirds from \(r=1\) to \(r=10\). Similarly decreases can be seen in the parameter intervals.
# Make 250 year return level plot using gevr for r = 1 to 10 with the LoweStoft data
data(lowestoft)
result <- matrix(0, 20, 4)
period <- 250
for(i in 1:10) {
z <- gevrFit(as.matrix(lowestoft[, 1:i]))
y1 <- gevrRl(z, period, conf = 0.95, method = "delta")
y2 <- gevrRl(z, period, conf = 0.95, method = "profile", plot = FALSE)
result[i, 1] <- i
result[i, 2] <- y1$Estimate
result[i, 3:4] <- y1$CI
result[(i + 10), 1] <- i
result[(i + 10), 2] <- y2$Estimate
result[(i + 10), 3:4] <- y2$CI
}
result <- cbind.data.frame(result, c(rep("Delta", 10), rep("Profile", 10)))
colnames(result) <- c("r", "Est", "Lower", "Upper", "Method")
result <- as.data.frame(result)
prof <- subset(result, Method == "Profile")
del <- subset(result, Method == "Delta")
par(mfrow = c(2, 1))
plot(prof$r, prof$Est, main = "Profile Likelihood",
xlab = "r", ylab = "250 Year Return Level",
xlim = c(1, 10), ylim = c(4, 7))
polygon(c(rev(prof$r), prof$r), c(rev(prof$Lower), prof$Upper), col = 'grey80', border = NA)
points(prof$r, prof$Est, pch = 19, col = 'black')
lines(prof$r, prof$Est, lty = 'solid', col = 'black')
lines(prof$r, prof$Lower, lty = 'dashed', col = 'red')
lines(prof$r, prof$Upper, lty = 'dashed', col = 'red')
plot(del$r, del$Est, main = "Delta Method",
xlab = "r", ylab = "250 Year Return Level",
xlim = c(1, 10), ylim = c(4, 7))
polygon(c(rev(del$r), del$r), c(rev(del$Lower), del$Upper), col = 'grey80', border = NA)
points(del$r, del$Est, pch = 19, col = 'black')
lines(del$r, del$Est, lty = 'solid', col = 'black')
lines(del$r, del$Lower, lty = 'dashed', col = 'red')
lines(del$r, del$Upper, lty = 'dashed', col = 'red')
par(mfrow = c(1, 1))
In addition, the profile likelihood confidence intervals are compared with the delta method intervals. The advantage of using profile likelihood over the delta method is the allowance for asymmetric intervals. This is especially useful at high quantiles, or large return level periods. In the Lowestoft plots directly above, the asymmetry can be seen in the stable lower bound across values of \(r\), while the upper bound decreases.
set.seed(7)
n <- 100
r <- 10
x <- rgevr(n, r, loc = 100 + 1:n / 50, scale = 1 + 1:n / 100, shape = 0)
## Plot the largest order statistic
plot(x[, 1])
## Creating covariates (linear trend first)
covs <- as.data.frame(seq(1, n, 1))
names(covs) <- c("Trend1")
## Create some unrelated covariates
covs$Trend2 <- rnorm(n)
covs$Trend3 <- 30 * runif(n)
## Use full data
fit_full <- gevrFit(data = x, method = "mle", locvars = covs, locform = ~ Trend1 + Trend2*Trend3,
scalevars = covs, scaleform = ~ Trend1)
## Only use r = 1
fit_top_only <- gevrFit(data = x[, 1], method = "mle", locvars = covs, locform = ~ Trend1 + Trend2*Trend3,
scalevars = covs, scaleform = ~ Trend1)
In the previous chunk of code, we look at non-stationary fitting in the GEV\(_r\) distribution. The ‘eva’ package allows generalized linear modeling in each parameter (location, scale, and shape), as well as specifying specific link functions. As opposed to some other packages, one can use formulas when specifying the models, so it is quite user friendly. Additionally, to benefit optimization, there is efficient handling of the near-zero shape parameter in the likelihood and covariates are automatically centered and scaled when appropriate (they are transformed back to the original scale in the output).
The previous chunk of code demonstrates the benefit of using a larger \(r\) than just the block maxima. Data is generated with sample size 100 and \(r=10\), with a linear trend in both the location and scale parameter. From a visual assessment of the largest order statistic, it is difficult to see either trend in the data. We include two erroneous covariates, named ‘Trend2’ and ‘Trend3’, and fit the nonstationary distribution with the full data (\(r=10\)) and only the block maxima (\(r=1\)). The results are summarized in the gevrFit function.
## Show summary of estimates
fit_full
## Summary of fit:
## Estimate Std. Error z value Pr(>|z|)
## Location (Intercept) 100.0856115 0.2276723 439.60378 0.0000e+00 ***
## Location Trend1 0.0193838 0.0042366 4.57535 4.7543e-06 ***
## Location Trend2 -0.0716415 0.0949454 -0.75455 4.5052e-01
## Location Trend3 0.0031478 0.0049866 0.63124 5.2788e-01
## Location Trend2:Trend3 0.0032205 0.0053048 0.60710 5.4378e-01
## Scale (Intercept) 1.0831201 0.0922684 11.73879 8.0627e-32 ***
## Scale Trend1 0.0097441 0.0016675 5.84352 5.1108e-09 ***
## Shape (Intercept) 0.0393603 0.0293665 1.34031 1.8014e-01
## ---
## Signif. codes: 0 '***' 0.001 '*' 0.01 '*' 0.05 '.' 0.1 ' ' 1
fit_top_only
## Summary of fit:
## Estimate Std. Error z value Pr(>|z|)
## Location (Intercept) 99.5896128 0.4373915 227.68985 0.0000e+00 ***
## Location Trend1 0.0235401 0.0052891 4.45072 8.5582e-06 ***
## Location Trend2 -0.3567545 0.2858703 -1.24796 2.1205e-01
## Location Trend3 0.0170264 0.0170313 0.99971 3.1745e-01
## Location Trend2:Trend3 0.0021764 0.0194109 0.11212 9.1073e-01
## Scale (Intercept) 1.1098238 0.2654124 4.18151 2.8958e-05 ***
## Scale Trend1 0.0036369 0.0042099 0.86390 3.8764e-01
## Shape (Intercept) 0.1375594 0.1063641 1.29329 1.9591e-01
## ---
## Signif. codes: 0 '***' 0.001 '*' 0.01 '*' 0.05 '.' 0.1 ' ' 1
From the output, one can see from the fit summary that using \(r=10\), both the correct trend in location and scale are deemed significant. However, using \(r=1\) a test for significance in the scale trend fails.
Next, we fit another two models (using \(r = 10\)). The first, labeled ‘fit_reduced1’ is a GEV\(_{10}\) fit incorporating only the true linear trends as covariates in the location and scale parameters. The second, ‘fit_reduced2’ fits the Gumbel equivalent of this model (\(\xi = 0\)). Note that ‘fit_reduced1’ is nested within ‘fit_full’ and ‘fit_reduced2’ is further nested within ‘fit_reduced1’.
One way to compare these models is using the Akaike information criterion (AIC) and choose the model with the smallest value. This metric can be extracted by using AIC(\(\cdot\)) on a gevrFit object. By this metric, the chosen model is ‘fit_reduced2’, which agrees with the true model (Gumbel, with linear trends in the location and scale parameters). We can also test between nested models using the likelihood ratio test (LRT). For example, the test of \(H_0\): \(\xi = 0\) between ‘fit_reduced1’ and ‘fit_reduced2’ is approximately distributed as chi-square with one degree of freedom and its p-value is found to be 0.95. Thus, the further reduced (Gumbel) model ‘fit_reduced2’ is favored.
## Compare AIC of three models
fit_reduced1 <- gevrFit(data = x, method = "mle", locvars = covs, locform = ~ Trend1,
scalevars = covs, scaleform = ~ Trend1)
fit_reduced2 <- gevrFit(data = x, method = "mle", locvars = covs, locform = ~ Trend1,
scalevars = covs, scaleform = ~ Trend1, gumbel = TRUE)
AIC(fit_full)
## [1] 127.0764
AIC(fit_reduced1)
## [1] 120.4856
AIC(fit_reduced2)
## [1] 118.4895
LRT <- as.numeric(2 * (logLik(fit_reduced1) - logLik(fit_reduced2)))
pval <- pchisq(LRT, df = 1, ncp = 0, lower.tail = FALSE, log.p = FALSE)
round(pval, digits = 3)
## [1] 0.95
One can also use the ‘gevrFit’ function to estimate the parameters of a multivariate model with dependence between GEV marginal distributions using independence likelihood. Here we generate correlation using the multivariate normal distribution and then transform the marginal distributions into GEV.
set.seed(7)
require(SpatialExtremes)
## Loading required package: SpatialExtremes
## Warning: package 'SpatialExtremes' was built under R version 3.5.1
##
## Attaching package: 'SpatialExtremes'
## The following objects are masked from 'package:eva':
##
## dgpd, pgev, pgpd, qgev, qgpd, rgpd
n.site <- 4
n.obs <- 15
## Simulate a max-stable random field
locations <- matrix(runif(2 * n.site, 0, 10), ncol = 2)
colnames(locations) <- c("lon", "lat")
## Smith model
U <- rmaxstab(n.obs, locations, "gauss", cov11 = 16, cov12 = 0, cov22 = 16)
cor(U, method = "spearman")
## [,1] [,2] [,3] [,4]
## [1,] 1.0000000 0.2857143 0.1250000 0.2714286
## [2,] 0.2857143 1.0000000 0.6964286 0.7714286
## [3,] 0.1250000 0.6964286 1.0000000 0.6500000
## [4,] 0.2714286 0.7714286 0.6500000 1.0000000
The previous chunk of code generates data from a multivariate max stable random variable on the Frechet scale. We can see the correlation is somewhat large. Next, we transform the uniform data into GEV margins with site specific location means [8, 10, 12, 9] and shared scale, trend parameters. The design matrix labels each of the four sites as 1, 2, 3, or 4. The estimated means can be constructed from the output. Site 1 is chosen as the reference level, so its location parameter is estimated by the location intercept – 8.12. The other site specific location parameter estimates can be obtained by adding the intercept and the corresponding site value. For example, site 3 has a location estimate of 8.12 + 4.02 = 12.14.
## Transform to GEV margins
locations <- c(8, 10, 12, 9)
out <- frech2gev(U, loc = 0, scale = 1, shape = 0.2)
out <- out + t(matrix(rep(locations, nrow(out)), ncol = nrow(out)))
out <- as.vector(out)
## Create design matrix for the location parameters
loc <- cbind.data.frame(as.factor(sort(rep(seq(1, n.site, 1), n.obs))))
colnames(loc) <- c("Site")
z <- gevrFit(out, locvars = loc, locform = ~ Site)
z
## Summary of fit:
## Estimate Std. Error z value Pr(>|z|)
## Location (Intercept) 8.1197181 0.32172 25.238373 1.5196e-140 ***
## Location Site2 1.9309253 0.37440 5.157381 2.5043e-07 ***
## Location Site3 4.0150971 0.39637 10.129764 4.0763e-24 ***
## Location Site4 0.9416902 0.39483 2.385056 1.7077e-02 *
## Scale (Intercept) 0.9465077 0.10922 8.666143 4.4700e-18 ***
## Shape (Intercept) 0.0095865 0.12631 0.075896 9.3950e-01
## ---
## Signif. codes: 0 '***' 0.001 '*' 0.01 '*' 0.05 '.' 0.1 ' ' 1 |
In mathematical logic and model theory, one considers
interpretations of syntactic expressions: terms without free variables are interpreted as elements of some structure, formulas without free variables have truth values, formulas with free variables can be interpreted as relations.
Multiple expressions may have identical interpretations. For example, $\ulcorner 1 + 1\urcorner$ and $\ulcorner 2\urcorner$ are both interpreted as $2$.
Question: does anyone ever consider formal languages where terms can have multiple interpretations? Is there some standard approach or framework?
I am thinking about this because i am trying to understand the $\omicron$ and $O$ notation in analysis, like in $$ \ln(x) =\omicron(x),\quad x\to +\infty. $$ Also, when calculating an indefinite integral, on often writes $$ \int 2x\,dx = x^2 + C. $$
Update.I understand that when the equality sign is used with $\omicron$/$O$ notation, it does not represent an equivalence relation.I also know that $\omicron(g(x))$ can be viewed as a set of functions.However, this interpretation does not fit my intuition well.When i write $\sin x = x + \omicron(x^2)$, $x→0$, i do not think about sets of functions, i think that i am replacing an anonymous implicitly understood function with a placeholder.In other words, the designated object does not change (it is still a function or a number, not a set of functions or set of numbers), only the notation is abbreviated and made less explicit, a bit like when i write "$1 + 2 + 3$" instead of "$((1 + 2) + 3)$". |
Wendy Testaburger and Sally Turner play a game with two unfair coins. A coin flip with Wendy's coin shows head with probability $\frac {1} {100}$. A coin flip with Sally's coin shows head with probability $p$. Wendy does the first coin flip. Wendy and Sally flip their coins alternately until a coin shows heads. The girl with the coin showing head wins the game. What value of $p$ makes this game a fair game?
Wendy wins on the first flip with probability $1/100$. Otherwise, the game keeps going and Sally has probability $p$ to win on the next flip, which has overall probability $99/100 \times p$. If not, the game returns to the start, which has no effect on fairness. So, to be fair, these two probabilities should be equal, which happens for $p=1/99$.
The answer is
$p = \frac{1}{99}$
Proof
Let $p(W,k)$ be the probability of Wendy winning on her $k$th turn and $p(S,k)$ be the probability of Sally winning on her $k$th turn. Then,
$p(W,1) = \frac{1}{100}, p(W,2) = \frac{99}{100}(1-p)\frac{1}{100}$ and, by simple induction $p(W, k) = \left(\frac{99}{100}(1-p) \right)^{k-1} \left(\frac{1}{100} \right)$. Also, by a simple induction it can be shown that $p(S, k) = \left(\frac{99}{100}(1-p) \right)^{k-1}\left(\frac{99p}{100} \right)$. We must then enforce $\sum_{k=0}^{\infty} p(W,k) = \sum_{k=0}^{\infty} p(S,k)$ which implies $\frac{1}{100} = \frac{99p}{100}$ as required
The game is fair for
p = $\frac1{99}$
Proof:
If Wendy's coin has odds $\frac1n$, a fair game is achieved when Sally's coin has odds $\frac1{n-1}$. It's easy to see that if Wendy's coin has $\frac12$ odds, Sally needs a coin that always comes up heads. If Wendy's coin comes up heads $\frac13$ of the time, and Sally's comes up heads $\frac12$ of the time, there's a $\frac13$ chance Wendy wins on turn 1, and a $\frac12 \times \frac23=\frac13$ chance Sally wins on turn 2 ($\frac12$ as her odds of getting heads, and $\frac23$ as the odds of the game lasting long enough for there to be a turn 2). Then there's a $\frac13 \times \frac13 = \frac19$ chance Wendy wins on turn 3, a $\frac12 \times \frac29=\frac19$ Sally wins on turn 4... for any pair of turns, Wendy and Sally will have equal odds. We can see this trend holds for any $\frac1n$, leading us to the above answer of $\frac1{99}$.
The answer is 0.0101010101...
Proof
Say that $w$ is the chance of Wendy’s coin coming up heads; $s$ is the chance for Sally’s coin. Wendy will win with probability:
$$p(W)= w + w's'w + w's'w's'w + w's'w's'w's'w ... = w ( 1 + (w's')^2 + (w's')^3 ... )$$
Sally wins with probability:
$$p(S) = w's + w's'w's + w's'w's'w's ... = s ( w' + w'^2s'^1 + w'^3s'^2 + ....) = sw'(1 + (w's')^2 + (w's')^3 ... $$
Set them equal to each other:
$$sw'(1 + (w's')^2 + (w's')^3 ... = w ( 1 + (w's')^2 + (w's')^3$$ $$sw' = w$$ $$w= 0.01, w'=0.99$$
Therefore, $s = 0.01 / 0.99 = 1/99 = 0.0101010101 ...$.
The key simplification is that:
since the game repeats from the beginning if no-one wins after they each have a turn, we only have to make sure that they chance of them each winning on their first toss is equal.
Therefore it must hold that:
1/100 = (1 - 1/100)p
So
p = 1/99
First instinct, but I may be wrong on this:
1/100. Then both of them have the same chance of winning.
[EDIT]
Second instinct :
1/99
Why my second instinct?
let's limit to 1 toss each.
Wendy's chance. $1/100$. Sally's chance $99 * p/100 $. For this to be equal chances $p = 1/99$. Note: will try my third instinct for 2 tosses each.
Let $f(n)$ be Wendy's chance of winning at her $n$th turn. $f(1)=1/100$ with $f(n+1) = f(n)*(1-p)*99/100$ in general. Her total chance of winning is the sum of the infinite series consisting of all $f(n)$ values. If we let it be $A$, we get $A*(1-p)*99/100 + 1/100 = A$. Since $A$ is supposed to equal 1/2, $p=1/99$.
You can't make this a fair game because if you give 1/100 to one person and 1/99 to another, after like 200 chances the 1/99 person would be in advantage. If the chances are the same(1/100), the person starting first have the advantage. |
where $L$ is the length of the wheel base and $h$ is the distance to the centre of gravity.
The idea is to find what the maximum angle of the slope is before gravity will overcome the friction between the tires and road, and supposedly, in this limiting case, $F=\mu N_{2}$. I'm not quite sure how to work out what $N_{2}$ is.
Don't worry so much about the numerical answer right at the end, he's just subbed in the values, I'm more interested in the derivation.
I don't get where $\frac{h}{L}\sin(\theta)+ \frac{1}{2}\cos(\theta)$ came from and I also don't understand why were using torques? Basically I don't understand any of it. |
How to Perform a Nonlinear Distortion Analysis of a Loudspeaker Driver
A thorough analysis of a loudspeaker driver is not limited to a frequency-domain study. Some desirable and undesirable (but nonetheless exciting) effects can only be caught by a nonlinear time-domain study. Here, we will discuss how system nonlinearities affect the generated sound and how to use the COMSOL Multiphysics® software to perform a nonlinear distortion analysis of a loudspeaker driver.
Understanding Linear and Nonlinear Distortions
A transducer converts a signal of one energy form (input signal) to a signal of another energy form (output signal). In regard to a loudspeaker, which is an electroacoustic transducer, the input signal is the electric voltage that, in the case of a moving coil loudspeaker, drives its voice coil. The output signal is the acoustic pressure that the human ear perceives as a sound. A distortion occurs when the output signal quantitatively and/or qualitatively differs from the input signal.
Schematic representation of a moving coil loudspeaker.
The distortion can be divided into two principal parts:
Linear distortion Nonlinear distortion
The term
linear distortion, which might sound rather confusing, implies that the output signal has the same frequency content as the input signal. In this distortion, it is the amplitude and/or phase of the output signal that is distorted. In contrast, the term nonlinear distortion suggests that the output signal contains frequency components that are absent in the input signal. This means that the energy is transferred from one frequency at the input to several frequencies at the output. Input and output signals in linear and nonlinear transducers.
Let the input sinusoidal signal, A_\text{in} \sin \left( 2\pi f t \right), be applied to a transducer with a nonlinear transfer function. The frequency content of the output signal will then have more than one frequency. Apart from the fundamental portion, which corresponds to the frequency f, there will be a distorted portion. Its spectrum usually (but not always) consists of the frequencies f^{(2)}, f^{(3)}, f^{(4)}, \ldots, which are multiples of the fundamental frequency f^{(n)} = n f, in which n \geq 2. These frequencies, called
overtones, are present in the sound, and it is the overtones that make musical instruments sound different: A note played on a violin sounds different from the same note played on a guitar. The same happens with the sound emitted from a loudspeaker.
The distortion is a relative quantity that can be described by the value of the
total harmonic distortion (THD). This value is calculated as the ratio of the amplitude of the distorted portion of the signal to that of the fundamental part:
The profile of a signal with a higher THD visibly differs from the pure sinusoidal.
Unfortunately, the value of the THD of the output signal itself might not be enough to judge the quality of the loudspeaker. A signal with a lower THD may sound worse than a signal with a higher THD. The reason is that the human ear perceives various overtones differently.
The distortion can be represented as a set of individual even-order, 2nf, and odd-order, (2n-1)f, components. The former are due to asymmetric nonlinearities of the transducer, while the latter are due to symmetric nonlinearities. The thing is that the sound containing even-order harmonics is perceived as “sweet” and “warm”. This can be explained by the fact that there are octave multiples of the fundamental frequency among them. The odd-order harmonics sound “harsh” and “gritty”. That is quite alright for a guitar distortion pedal, but not for a loudspeaker. What matters is, of course, not just the presence of those harmonics, but rather their level in the output signal.
Another interesting effect, called
intermodulation, occurs when the input signal contains more than one frequency component. The corresponding output signals start to interact with each other, producing frequency components absent in the input signal. In practice, if a two-tone sine wave such as A_\text{in} \sin \left( 2\pi f_1 t \right) + B_\text{in} \sin \left( 2\pi f_2 t \right) (in which f_2 > f_1) is applied to the input, the system nonlinearities will result in the modulation of the higher-frequency component by the lower one. That is, the frequencies f_2 \pm f_1, f_2 \pm 2f_1, and so on will appear in the frequency spectrum of the output signal. The quantitative measure of the intermodulation that corresponds to the frequency, f_2 \pm (n-1) f in which n \geq 2, is the n th-order intermodulation distortion (IMD) coefficient. It is defined as:
In practice, using an input signal containing three or more frequencies for the IMD analysis is not advisable, as the results become harder to interpret.
Transient Nonlinear Analysis of a Loudspeaker Driver
To summarize, the linear analysis of the loudspeaker, though a powerful tool for a designer, might not be sufficient. The loudspeaker can only be completely described if an additional nonlinear analysis is carried out. The nonlinear analysis is supposed to answer the following questions:
How does the nonlinear behavior of the loudspeaker affect the output signal? What are the limits of the input signal that ensure the loudspeaker functions acceptably? How should I compensate for the undesired distortion of the loudspeaker?
From the simulation point of view, there is both bad and good news. The bad news is that the full nonlinear analysis cannot be performed in the frequency domain. It requires the transient simulation of the loudspeaker, which is more demanding and time consuming than the frequency-domain analysis. The good news is that the effect of certain nonlinearities is only significant at low frequencies.
For example, the voice coil displacement is greater at lower frequencies and therefore the finite strain theory must be used to model the mechanical parts of the motor. Using the finite strain theory is redundant at higher frequencies, where the infinitesimal strain theory is applicable. The figures below show the results for the transient loudspeaker tutorial, driven by the same amplitude (V_0 = 10 V) of input voltage:
Voice coil motion in the air gap of the loudspeaker driver for a single-tone input voltage signal: 70 Hz on the left and 140 Hz on the right. Acoustic pressure at the listening point for a single-tone input voltage. The blue curves correspond to the nonlinear time-domain analysis, while the red curves correspond to the frequency-domain analysis: 70 Hz on the left and 140 Hz on the right.
The animations above depict the magnetic field in the voice coil gap and the motion of the former and the spider (both in pink) as well as the voice coil (in orange). As expected, the displacements, as well as the spider deformation, are higher at the lower frequency. The spider deformation obeys the geometrically nonlinear analysis and therefore the linear approximation is inaccurate in this case. This is confirmed by the output signal plots. These plots depict the acoustic pressure at the listening point located about 14.5 cm in front of the speaker dust cap tip.
The acoustic pressure profile obtained from the nonlinear time-domain modeling for the 70-Hz input signal deviates from the sinusoidal shape to a certain extent, which means that higher-order harmonics start playing a definite role. This is not visible for the input signal at 140 Hz: There’s only a slight difference in the amplitude between the linear frequency-domain and nonlinear time-domain simulation results. The THD value of the output signal drops from 4.3% in the first case to 0.9% in the second case. The plots below show how the harmonics contribute to the sound pressure level (SPL) at the listening point.
Frequency spectra of the SPL at the listening point: single-tone input voltage (70 Hz on the left and 140 Hz on the right).
The IMD analysis of the loudspeaker is carried out in a similar way. What’s different is the input signal applied to the voice coil, which contains two harmonics parts:
whose amplitudes, V_1 and V_2, usually correlate as 4 : 1, which corresponds to 12 dB.
The example below studies the IMD of the same test loudspeaker driver. The dual-frequency input voltage, in which f_1 = 70 Hz and f_2 = 700 Hz, serves as the input signal. The SPL plot on the left shows how the second- and third-order harmonics arising in the low-frequency part of the output signal generate a considerable level of the corresponding order IMDs in the high-frequency part. The IMD level becomes sufficiently lower if the signal frequency f_1 is increased to 140 Hz. This is seen in the right plot below.
Frequency spectra of the SPL at the listening point for a two-tone input voltage. Modeling Tips for Analyzing a Loudspeaker Driver
Since transient nonlinear simulations tend to be demanding, the loudspeaker driver model should not be overcomplicated. The 2D axisymmetric formulation is a good starting approach and was used for the tutorial examples in the previous section. After that, it’s important to estimate which effects are more important than others. This will help you set up an adequate multiphysics model of a loudspeaker.
The system nonlinearities include, but are not limited to, the following:
Nonlinear behavior of the magnetic field in the loudspeaker pole piece made of high-permeability metal Geometric nonlinearities in the moving parts of the motor Topology change as the voice coil moves up and down in the air gap
Speaking the lumped parameters’ language, this means that they are no longer constants like the Thiele-Small parameters, but functions of the voice coil position, x, and the input voltage, V. The above-mentioned nonlinearities will be reflected in the nonlinear inductance, L \left( x, V \right); compliance, C \left( x, V \right); and dynamic force factor, Bl \left( x, V \right). For instance, the tutorial example shows that the nonlinear behavior of the force factor is more distinct at 70 Hz, whereas it is almost flat (that is, closer to linear) at 140 Hz.
Nonlinear (left) and almost linear (right) behavior of the dynamic force factor: 70 Hz on the left and 140 Hz on the right.
With the following steps, the discussed nonlinearities can be incorporated into the model. First, the nonlinear magnetic effects are taken into account through the constitutive relation for the corresponding material. In the test example, the BH curve option is chosen for the iron pole piece. Next, the
Include geometric nonlinearity option available under the Study Settings section forces the structural parts of the model to obey the finite strain theory. Lastly, the topology change is captured by the Moving Mesh feature. Whenever applied, the feature ensures that the mesh element nodes move together with the moving parts of the system. Since the displacements can be quite high, it is likely that the mesh element distortion reaches extreme levels and the numerical model becomes unstable. The Automatic Remeshing option is used as a remedy against highly distorted mesh elements.
All in all, the nonlinear time-domain analysis of the loudspeaker requires much more effort and patience than the linear frequency-domain study. This is especially relevant when the model includes the
Moving Mesh feature with the Automatic Remeshing option activated. Investing some time in the geometry and mesh preprocessing will pay off, as the moving mesh is very sensitive to the mesh quality. That is, highly distorted mesh elements and near-zero angles between the geometric entities have to be avoided. A proper choice of the Condition for Remeshing option may also require some trial and error.
The loudspeaker design discussed here might not be considered “good” by most standards. The odd-order harmonics prevail in the frequency content of the output signal.
Next Steps
To perform your own nonlinear distortion analysis of a loudspeaker, click on the button below. This will take you to the Application Gallery, where you can find the MPH-files for this model together with detailed modeling instructions. (Note: You must have a COMSOL Access account and valid software license.)
Additional Resources Check out other examples of modeling loudspeakers in these tutorials: Further reading: L.L. Beranek and T.J. Mellow, Acoustics: Sound Fields and Transducers, Academic Press, 2012. Brüel & Kjær, “Audio Distortion Measurements,” Application Note BO0385, 1993. W. Marshall Leach, Jr., Introduction to Electroacoustics and Audio Amplifier Design, Kendall Hunt, 2010. L.L. Beranek and T.J. Mellow, Comentários (4) CATEGORIAS Chemical COMSOL Now Electrical Fluid General Interfacing Mechanical Today in Science |
I'm trying to understand the expression $Cov(\hat y,\hat \epsilon)$ in regards to the usual linear regression model/assumptions $y = \beta_0 + \beta_1 x_1 + \beta_2 x_2 + ... \beta_n x_n +\epsilon$. Also, $x_i$ are not considered to be random variables. Rather, they are considered to be fixed measurements with negligible error.
Each $\hat y_i$ is the sum of the $x_i$s weighted by the $\hat \beta_i$s (with $x_0=1)$. And each $\hat \epsilon_i$ is the difference ($y_i - \hat y_i$).
Are $\hat y,\hat \epsilon$, and for that matter, $\hat \beta,$ random vectors?
If u,v are random vectors, then $Cov(u,v)$ is the matrix of elements $Cov(u_i,v_j)$
If $u,v$ are not random vectors, then $Cov(u,v)$ is the scalar $\Sigma u_i v_i$. |
Continuity and Differentiability Derivatives of Functions in Parametric Forms
Sometimes the relation between two variables is neither explicit nor implicit, but some link of a third variable with each of the two variables, separately, establishes a relation between the first two variables. In such a situation, we say that the relation between them is expressed via a third variable. The third variable is called the parameter. More precisely, a relation expressed between two variables x and y in the form x =
f( t), y = g( t) is said to be parametric form with t as a parameter. In order to find derivative of function in such form, we have by chain rule. \frac{dy}{dt}= \frac{dy}{dx}\frac{dx}{dt}
\frac{dy}{dx}= \frac{\frac{dy}{dt}}{\frac{dx}{dt}}\tt \left(whenever\ \frac{dx}{dt}\neq 0\right)
Thus \frac{dy}{dx}=\frac{g'(t)}{f'(t)}\ \left(as\ \frac{dy}{dt}=g'(t)\ and\ \frac{dx}{dt}=f'(t)\right) [provided f'(t) ≠ 0]
Part1: View the Topic in this video From 49:18 To 53:58 Part2: View the Topic in this video From 00:40 To 05:53
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If y=\sqrt{f(x)+y} then \frac{dy}{dx}=\frac{f'(x)}{2y-1} If y=f(x)+\frac{1}{y} then \frac{dy}{dx}=\frac{y^{2}f'(x)}{y^{2}+1} If f(x + y) = f(x) f(y) ∀ x, y ∈ R and f(x) ≠ 0 f(a) = k, f'(0) exists, then f'(a) = kf'(0). \frac{d}{dx}\left[\tan^{-1}\left(\frac{a\cos f(x)+b \sin f(x)}{a\cos f(x)-a \sin f(x)}\right)\right]=f'(x) \frac{d}{dx}\left[\tan^{-1}\left(\frac{a\cos f(x)-b \sin f(x)}{b\cos f(x)+a \sin f(x)}\right)\right]=f'(x) If \sqrt{1-x^{2}}+\sqrt{1-y^{2}}=a(x-y) then \frac{dy}{dx}=\sqrt{\frac{1-y^{2}}{1-x^{2}}} If \sqrt{1-x^{6}}+\sqrt{1-y^{6}}=a(x^3-y^3) then \frac{dy}{dx}=\frac{x^{2}}{y^{2}}\sqrt{\frac{1-y^{6}}{1-x^{6}}} If x\sqrt{1-y^{2}}+y\sqrt{1-x^{2}}=1, then \frac{dy}{dx}=-\sqrt{\frac{1-y^{2}}{1-x^{2}}} If x m. y n= a m + nthen \frac{dy}{dx}=\frac{-my}{nx} If f(x, y) = 0 is a homogeneous equation in x and y then \frac{dy}{dx}=\frac{y}{x}. If f(x + y) = f(x) f(y) ∀ x, y ∈ R then f'(x) = f(x) + f'(0) If ax 2+ 2hxy + by 2+ 2gx + 2fy + c = 0 then \frac{dy}{dx}=\frac{-(ax+hy+g)}{(hx+by+f)} |
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 $M$. |
If I would try to treat macroscopic systems consisting of a super-large number of particles (also when environment is included), I have to compute $2N$-point correlation functions with very large particle number $N$. These are given by
$$A(x_1',\dots,x_N';x_1,\dots,x_N) = \langle\prod_{i=1}^N\psi^\dagger(x_i') \psi(x_i)\rangle$$
The primed variables denote final states and $\psi(x)$ generate one fermion at point $x$. Transition probabilities are then proportional to $|A|^2$. For a classical behavior, these transition probabilities have to be deterministic, i.e. must be in the form of a delta distribution
$$\propto \prod_i \delta(x_i'-f_i(x_1,\dots,x_N))$$
for the (differential equation) solution function $f_i$. How can I derive that deterministic transition amplitudes arise in everyday life scenarios from the path integral average?
I know that I can make a decomposition $\psi(x) = \psi_0(x)+\psi'(x)$ with classical solutions of the equation of motion $\psi_0$ and quantum fluctuations $\psi'$ that obey $\langle\psi'\rangle=0$. Taylor expansion until second order in fluctuations of the action functional leads to the Gaussian distribution similar to
$$\exp\left(\frac{i}{2}\frac{\delta^2}{\delta \psi^2}S|_{\psi_0}\psi'^2\right)$$
This distribution becomes sharper peaked as the second functional derivative of the action functional gets larger.
I can introduce dimensionless variables and see that this Gaussian standard deviation depends on characteristic length scales and with higher length scales, standard deviations will get less.
That is what I know.
But by considering above multiparticle scattering amplitudes, making the expansion of quantum fields around classical solution I will pick up also LOTS of quantum corrections $\langle\psi'^2\rangle,\langle\psi'^3\rangle,...$. There will be $\frac{N(N-1)}{2}$ quadratic fluctuation terms which is a huge number. But why these are irrelevant in macroscopic systems?
Is there a detailed answer? |
Apologies if this is a duplicate. I searched for some time and the closest I could find was the Knapsack Problem, which does not fit because I have a fixed team size & a different value function.
I have $N \approx 50$ Pokémon, from which I need to choose a team of six to attack an enemy. Each pokemon can survive for a certain time "$t$" against the enemy, and each pokemon does a certain amount of damage per second "$DPS$" to the enemy. I may or may not defeat the enemy, but I would like to optimize the damage I can do to it within the time limit of 300 seconds without my entire team dying.
These data are sufficient to determine the optimal team. The damage done by a team, which fights in order of highest > lowest $DPS$, is the sum $$Damage(DPS_i, t_i) = \sum_{i=1} DPS_i * \hat{t_i} $$ where $0<\hat{t_i}<t_i$ is the amount of time each pokemon battled for. Typically the first few pokemon will battle until they die ($\hat{t_i}=t_i$), one pokemon will win with a fraction of its health remaining ($0<\hat{t_i}<t_i$), and the rest may never get to battle ($\hat{t_i}=0$).
The optimization problem can be phrased as follows:
Maximize the function $Damage(DPS_1, ST_1, ..., DPS_6, ST_6)$ Given the constraint that $\sum_i t_i >300$
It is clear that this can be brute-forced by checking the damage and survival time of all team combinations. This is not an option because it takes too long.
Another possibility was to start with the highest-DPS team and then make "trades" in order of highest gain in $t$ per sacrifice of $DPS$. However I'm not convinced this will converge to the optimal solution.
I would be very thankful for any suggestions or references to analogous problems. |
I'm currently studying atomic term symbols. I wanted to try it on a simple atomic carbon with the electron configuration $1s^22s^22p^2$.
I know, that only open-shell electrons are involved in the term symbol classification, so that leaves us working with solely 2 electrons in $2p$ sub-shell.
First of all, we have ${6}\choose{2}$$= \frac{6!}{2!(6-2)!}=15$ possible microstates for the $1s^22s^22p^2$ configuration.
And, if I understand it correctly, we can categorize them into several "subsets" labeled with the corresponding term symbols.
The atomic term symbol is defined as ${}^{2S+1}L_J$.
In our case, the possible values of S, L and J are following: $$\max(M_L) = \max(L) = +2 \Rightarrow L = 0,1,2$$ $$\max(M_S) = \max(S) = +1 \Rightarrow S = 0, 1$$ $$J = L+S, L+S-1, \ldots, |L-S| = 3,2,1,0$$
Considering possible values of $S$, multiplicity can be $1$ or $3$, i.e. singlet or triplet. Values of $L$ enable symbols $S,P,D$.
So, our presumed set of states is ${}^3D, {}^1D, {}^3P, {}^1P, {}^3S, {}^1S$.
${}^3D$ "contains" $S=1$ and $L=2$. In that situation, both electrons would have to be spin-up and positioned together in the orbital with $m_l = +1$. That is impossible, as it contradicts Pauli exlusion principle. I.e. the state ${}^3D$ does not exist.
${}^1D$ is possible ($L=2,1,0, S=0$), so we can compute the number of corresponding microstates $N$ with the formula $N = (2L+1)(2S+1) = 5\cdot 1 = 5$.
Using the same formula, ${}^3P$ is going to contain 9 microstates.
With the "substraction method" described in this video, we can arrive to the conclusion, that both ${}^1P$ and ${}^3S$ won't exist.
And finally, ${}^1S$ will contain the last one state.
Question Now I know the possible term symbols, but I'm not sure, which microstates belong to them specifically.
It's clear, that microstates 13 and 15 will belong to ${}^1D$, but how can I determine it for the other states? ${}^1D$ should contain 3 more microstates with $S=0$ and $m_L$ being equal to -1, 0 and +1, but there are multiple candidates for every configuration.
So how could I distinguish between, e.g. microstates 7 and 9 or 8, 11 and 14?
I don't understand this point, as even $J=2$ for ${}^1D$, so it can't be used to distinguish among the "similar" microstates and to assign them properly. |
Let $p$ and $q$ be distinct prime numbers and $n=pq$. Show that $HK=\mathbb{Z}_n^\times$ for the subgroups $H=\{[x]\in\mathbb{Z}_n^\times\mid x\equiv 1\pmod{p}\}$ and $K=\{[y]\in\mathbb Z_n^\times \mid y\equiv 1\pmod{q} \}$.
What I have so far is that I know that $\phi(pq)=(p-1)(q-1)$. My book says that I should use a counting argument or CRT.
I chose a counting argument. So would it be correct if I said that there are $p-1$ congruence classes relatively prime to $p$ and $q-1$ congruences classes relatively prime to $q$. Thus set $HK$ has a total of $\phi(pq)$ distinct equivalence classes. Thus $HK=\mathbb Z_n^\times$.
By the chinese remainder theorem way. Suppose $$x \equiv 1 \pmod{p}$$ and $$y \equiv 1 \pmod{q}$$ where $\gcd(p,q)=1$. By the chinese remainder theorem we know that there exists a unique solution call it $a$. It follws that $p|(a-1)$ and $q|(a-1)$ From here im stuck now. |
The question is related to the question: detecting weak equivalences in a simplicial model category
Suppose that we have a simplicial model category $M$ and denote by $M^{f}$ the full simplicial subcategory of fibrant objects. Suppose that $R$ is a subcategory of $M^{f}$ such that for any object $m\in M^{f}$ there exists an object $r\in R$ such that $r$ is (zigzag) equivalent to $m$ i.e. $r$ and $m$ are isomorphic in $Ho(M)$ the homotopy category of $M$. Let $w: a\rightarrow b$ be a morphism in $M$ such that for any object $r\in R$ the induced map of simplicial sets $w^{\ast}:Map_{M}(b,r)\rightarrow Map_{M}(a,r)$ is a weak homotopy equivalence of simplicial sets. Can we conclude that $w$ is a weak equivalence in the model category $M$ ?
If $R=M^{f}$ this is true and is proved in Hirschhorn's book.
EDIT : The question is very general, and it should have a formal answer in case it is true. After trying all suggestions (in the comments and deleted answer) and reading Hirshorn's book I had the impression that maybe the answer to my question is no, and there should be a counterexample. In the Hirshorn's book the fact we test for all fibrant objects seems to be essential. |
I struggle to find how to estimate the volume occupied by one atom in a compound. As an example let's take $\beta$-Ga$_2$O$_3$, since that's the material on which I work the most. The crystal structure is monoclinic (space group C2/m) with lattice constants $a=1.2214$nm, $b=0.30371$nm, $c=0.57981$nm and $\beta = 103.83^\circ$ (see attached picture). Additionally gallium oxide has a molar mass $M_{\text{Ga}_2\text{O}_3}=187.444$g$\cdot$mol$^{-1}$ and a density $\rho=5.88$g$\cdot$cm$^{-3}$.
I know how to calculate the volume occupied by each molecule of gallium oxide, but is there a way to know the mean volume occupied by each atom of gallium or of oxygen individually?
Here is my effort to try to answer the question: the unit cell has a volume $V=|\vec{a} \times \vec{b}| \cdot \vec{c}=0.20885$nm$^3$ and contains 12 atoms of oxygen and 8 atoms of gallium. The density of atoms/molecules can be calculated with:
$\rho = \frac{M}{N_\text{A} \cdot V}$
where $M$ is the molar mass of the atoms/molecules of interest and $N_\text{A}$ is the Avogadro constant. This formula seems to work approximately when I calculate the density for Ga$_2$O$_3$ molecules:
$\rho_{\text{Ga}_2\text{O}_3} = 5.9607$g$\cdot$cm$^{-3}$
If I do the same with the Ga and O atoms individually (that is, by inserting $M_\text{Ga}=69.723$g/mol and $M_\text{O}=15.999$g/mol) I get:
$\rho_\text{Ga} = 2.217196$g$\cdot$cm$^{-3}$
$\rho_\text{O} = 0.508769$g$\cdot$cm$^{-3}$
Does that make sense or should I proceed differently?
Thank you very much in advance. |
The simple Newton-like explanation of dipole gravitational radiation unexistence is following.
The gravitational analog of
electric dipole moment is$$\mathbf d = \sum_{\text{particles}}m_{p}\mathbf r_{p}$$The first time derivative$$\dot{\mathbf d} =\sum_{\text{particles}}\mathbf p_{p}, $$while the second one is$$\ddot{\mathbf d} = \sum_{\text{particles}}\dot{\mathbf p}_{p} = 0,$$indeed due to momentum conservation.
"Magnetic" dipole gravitational radiation is analogically impossible due to conservation law of angular momentum. Indeed, since by the definition it is the sum of cross products of position of point on corresponding current: $$\mathbf{M} = \sum_{\text{particles}}\mathbf r_{p}\times m_{p}\mathbf{p}_{p} = \sum_{\text{particles}}\mathbf{J}_{p} \Rightarrow \dot{\mathbf M} = \ddot{\mathbf M} = 0$$
What's about general relativity? As you know, the propagation of gravitational waves is described by linearized Einstein equations for perturbed metric $h_{\mu \nu}$, and in this limit they coincide with EOM for helicity 2 massless particles in the presence of stress-energy pseudotensor $\tau_{\mu \nu}$:$$\square h_{\mu \nu} = -16 \pi \tau_{\mu \nu}, \quad \partial_{\mu}h^{\mu \nu} = 0, \quad \partial_{\mu}\tau^{\mu \nu} = 0$$Since $\tau^{\mu \nu}$ is conserved, this protects $h_{\mu \nu}$ from the contributions from monopole or dipole moments of sources as well as from additional helicities.
Formally the deep difference between gravitational and EM radiations is that we associate General relativity symmetry $g_{\mu \nu} \to g_{\mu \nu} + D_{(\mu}\epsilon_{\nu )}$ (it is infinitesimal version of $g_{\mu \nu}(x)$ transformation under $x \to x + \epsilon$ transformation) with covariant stress-energy tensor conservation (indeed,
tensor current conservation, from which we can extract conservation of 4-momentum vector current), while EM gauge symmetry is associated with vector current conservation (from which we can extract the conservation of electrical charge scalar quantity). So that corresponding conservation laws affect on different quantities; the nature of radiation in EM and GR cases are different, and the first one rules primarily by Maxwell equations (and hence conservation of charge plays the huge role), while the second one rules by linearized Einstein equations (and hence the momentum conservation is genuine).
For example, heuristically speaking, due to conservation of EM charge EM monopole radiation is impossible (it is expressed therough the time derivative of charge), but nothing restricts dipole moment radiation. In GR due to conservation of momentum vector, which is related to metric (an so to gravitational waves, in the sense I've shown above), dipole moment radiation is impossible.
This, indeed as anna v said in comments, is connected with the fact that EM field represents helicity-1 particles, while linearized gravitational field coincides with the field which represents helicity-2 particles. As you see, such thinking doesn't require presence of plus minus masses. |
This question already has an answer here:
This is the question-
Let $A[1,....,n]$ ba an array storing a bit $(1\,\,or\,\,0)$ at each location, and $f(m)$ is a function whose time complexity is $\theta(m)$. Consider the following program -
counter = 0;for( i = 1; i <= n; i++){ if(A[i] == 1) counter++; else { f(counter); counter = 0; }}
Options are :
(A)$\Omega(n^2)$ (B)$\Omega(n\log n)$ and $O(n^2)$ (C)$\Theta(n)$ (D)$\mathcal{O}(n)$
The solution says - "Since it contains only one loop, so compute with linear complexity, hence the complexity will be $\Theta(n)$".
My doubt is, what if the time complexity of the function $f(m)$ was given to be $\theta(2^n)$ or of the order that grows in exponential time? Would the solution given in the book be correct in case of exponential time? |
Recall that a connected semisimple algebraic group $G$ over an algebraically closed field $K$ of arbitrary characteristic was defined by Chevalley to be
simply connected if the character group $X(T)$ of a maximal torus $T$ in $G$ is "as large as possible": equal to the abstract weight lattice associated to the root lattice $\mathbb{Z} \Phi \subset X(T)$ (here $\Phi$ denotes the root system of $G$ relative to $T$). Since all maximal tori are conjugate, this is independent of the choice of $T$. Fortunately this notion of "simply connected" agrees with the usual topological notion when $K = \mathbb{C}$. A typical example is $\mathrm{SL}_n$.
The closed reductive subgroups $H$ of $G$ containing $T$ have been well studied, but one loose end still bothers me. The connected component of the identity $H^\circ$ is generated by $T$ and certain pairs of opposite root groups for roots forming a subsystem $\Psi$ of $\Phi$ (Borel-Tits). Typical examples are the derived groups of
Levi subgroups in parabolic subgroups of $G$ (complementary there to the unipoent radical), which are always connected. Imitating work of Borel and de Siebenthal (1949) on compact Lie groups, one gets these and sometimes more examples by taking root subsystems corresponding to proper subsets of vertices in the extended Dynkin diagram.
Equally natural are the groups $H = C_G(s)$, centralizers of semisimple elements (not necessarily connected, unless $G$ is simply connected). These have especially been studied in prime characteristic and have identity components which turn out to be of the type described above: work of Springer and Steinberg, Carter and Deriziotis, McNinch and Sommers.
Assume for convenience that $G$ is both simply connected and almost-simple, thus of Lie type $A - G$.
Is the semisimple derived group $H'$ of a connected reductive subgroup of $G$ including $T$ always simply connected? (If so, is there a uniform proof?)
When $H$ is a Levi subgroup of some parabolic subgroup of $G$, this was proved by Borel-Tits (in the more complicated context of a field of definition). Some experimental work with subgroups of Borel and deSiebenthal type suggests that the derived groups are also simply connected, but if so the reason is obscure. Maybe the topological case in characteristic 0 is suggestive?
Example: Take $G$ to be of type $G_2$ (so it is both simply connected and adjoint), with short simple root $\alpha$ and long simple root $\beta$ (picture here). The six long roots form a subsystem of type $A_2$. Since the original highest root is $3\alpha +2\beta$, its negative (call it $\gamma$) along with $\delta := \beta$ correspond to two vertices of the extended Dynkin diagram and span a subsystem $\Psi$ of type $A_2$ belonging to a subgroup $H$. Here $H$ is already connected and almost-simple. In fact it is simply connected: direct calculation shows that the two fundamental weights for $H$ are $-2\alpha -\beta$ and $-\alpha$, thus lie in the weight lattice (= root lattice) of $G$.
ADDED: My formulation (and example) probably oversimplify the computations involved in most cases. When $H'$, with a maximal torus $S := T \cap H'$, has lower rank than $G$ or has a root system with multiple irreducible components, the study of $X(S)$ tends to be more complicated. This already shows up in the usual Levi subgroups. For instance, the group $G_2$ has a standard Levi subgroup with derived group of type $\mathrm{SL}_2$ corresponding to either $\alpha$ or $\beta$, but the fundamental weight in $X(S)$ won't lie in $X(T)$. I hoped one might see whether or not a uniform pattern exists without too much case-by-case work, but that may be unrealistic.
UPDATE: The comments (especially by Paul) help to clarify what is going on, though I'm left with the partial question: Is there any predictable pattern to the occurrence or non-occurrence of simply connected groups $H'$? I was motivated to look more closely at an extreme case involving the simple (both adjoint and simply connected) group $G$ of type $E_8$. By removing the vertex of the extended Dynkin diagram corresponding to $\alpha_2$ (Bourbaki numbering), one gets a subgroup $H= H'$ of type $A_8$ with the same maximal torus $T$ and character group $X(T)$. Using Chevalley's classification, $H$ will be one of three nontrivial quotients of $\mathrm{SL}_9$ (as a group scheme). Some tedious bookkeeping with roots and weights shows that $H$ is the "intermediate" group, whose group of rational points has a center of order 3 (when the characteristic is not 3). Thus $H$ fails to be simply connected or adjoint, the former not obvious
a priori but certainly consistent with results of McNinch-Sommers.. |
This can be calculated by dividing $f(e,s)/(e_k+s)$ (assuming all $e_i$'s and $s$ are known to me) and raising $g$ to it.
First, if the prover knows $s$, it doesn't need to know the $e_i$'s to create membership witnesses. It can simply raise the accumulator $\mathsf{acc}$ to $1/(e_k + s)$:\begin{align*}\mathsf{wit}_{e_k} &= \mathsf{acc}^{\frac{1}{e_k + s}} = {g^{f(e,s)/{(e_k + s)}}} = g^{\prod_{i\ne k}{(e_i+s)}}\end{align*}
This brings me to my second point: knowledge of $s$ allows the prover to create
fake membership witnesses for any $e$ that was not accumulated. This is because the prover can compute $\frac{1}{e + s}$ for any $e$.
So, yes, computing $f(e,s)/(e_k+s)$ and then raising $g$ to it is one way to calculate the witness, but it's a bad idea: the
prover can easily create fake witnesses for any non-accumulated $e$ if it knows $s$. Computing witnesses in this way only makes sense when there's a trusted party that knows $s$.This trusted party computes and outsources witnesses to an untrusted "prover" who merely relays these witnesses to verifiers (i.e., the 2- or 3- party models of authenticated data structures).
Requiring a trusted party that knows $s$ is unreasonable in many applications.In practice, we want
to prevent the (untrusted) prover from faking witnesses, by ensuring it does not know $s$.We can do this via a trusted setup phase that gives the prover public parameters $\left(g^{s^i}\right)_{0\le i\le q}$ but does not give him $s$.This trusted setup phase is run by some trusted "guy in the sky," which in practice can be implemented via MPC protocols (see Scalable Multi-party Computation for zk-SNARK Parameters in the Random Beacon Model).Here $q$ is a strict upper bound on the number of elements you will ever accumulate.
Now, in order to commit to a set $A = \{e_1, \dots, e_n\}, n\le q$ of items, you first have to compute the
coefficients of the polynomial $f(e, x)$ using Lagrange interpolation in $O(n^2)$ time, Newton interpolation in $O(n^2)$ time or FFT interpolation in $O(n\log^2{n})$ time.
Note that you
don't need to know $s$ to interpolate. Consider $A = \{1,2,3\}$. Its polynomial $f_A(x) = (x-1)(x-2)(x-3) = x^3 - 6 x^2 + 11 x - 6$, so its coefficients are $(1, -6, 11, -6)$.
Once you have the coefficients $\{c_1, \dots, c_n\}$ you can compute the accumulator $acc$ by raising the $\left(g^{s^i}\right)_{0\le i\le n}$ public parameters to the coefficients:\begin{align*}acc &= \left(g^{s^n}\right)^{c_n} \left(g^{s^{n-1}}\right)^{c_{n-1}} \dots (g^s)^{c_1} \left(g\right)^{c_0}\\ &= g^{c_n s^n} g^{c_{n-1} s^{n-1}} \dots g^{c_1 s} g^{c_0}\\ &= g^{c_n s^n+ c_{n-1} s^{n-1} + \dots + c_1 s + c_0}\\ &= g^{f(e,s)}\end{align*}
For example, for our set $A$ above, the prover computes its accumulator as $(g^{s^3})^1 (g^{s^2})^{-6} (g^{s})^{11} g^{-6} = g^{s^3 - 6 s^2 + 11 s - 6} = g^{f_A(s)}$.Note that the prover does not need $s$ to compute $g^{f_A(s)}$ as long as it has the public parameters.In my experience, most accumulator papers do not explain this step, partly because they only look at the two- or three-party models of authenticated data structures where there actually exists a trusted party that knows $s$.(In other words, the "Hey, let's play with fire!" model.)
To answer your other questions,
While computing $f(e,s)$, do I have to do modular arithmetic on $P$ where $P$ is a prime number representing the order of group $\mathbb{G}$?
If you know the trapdoor $s$ (i.e., insecure mode), then yes, computing $f(e,s)$ involves adding and multiplying things modulo $P$ as per its definition.
If you don't know the trapdoor $s$, then computing $f(e,s)$ involves adding, multiplying and inverting numbers (for Lagrange and Newton) modulo $P$. If you're using FFT interpolation, it also involves
roots of unity in the field of order $P$ (i.e., the field "in the exponent").
Also, one paper stated that it is impossible to create a witness for an element which is not accumulated into it. Will anybody please elaborate on it?
As I described before, the prover cannot fake witnesses only when the prover
does not know the trapdoor $s$. I can't find the security proof in Lan Nguyen's original paper on accumulators, but it's a reduction to $q$-Strong Diffie-Hellman ($q$-SDH). The intuition behind why membership witnesses are secure is that without the trapdoor, nobody can compute any pair of the form $(e, g^\frac{1}{e + s})$.
By dividing $f(e,s)$ by $(e_k+s)$ one can calculate a witness value.
Yes, but remember that this requires the prover to know the trapdoor, which is not secure (unless you're in the "Hey, let's play with fire!" model).
Is there any way to distinguish between the witness (let me call it) values that where computed when $e_k$ is accumulated in $f(e,s)$ and when it is not accumulated $f(e,s)$?
Before I can answer this question, you have to clarify what element(s) these witnesses are for and what does the adversary know? Do they know the elements accumulated in the two accumulators? |
Let's work from the outside in.
for (int i = 0; i < length - 1; i++) {
Clearly, this loop is executed $n = \mathtt{length}-1$ times, so we get $\sum_{i=0}^{n-1} \dots$ where $\dots$ stands for the time needed by the loop's body (for iteration $i$). Inside, we have
for (int j = i + 1; j < length; j++) {
which we can translate similarly, obtaining $\sum_{i=0}^{n-1} \sum_{j = i+1}^{n} \dots$. Last but not least, the innermost operation
System.out.println(input.substring(i,j));
is apparently assumed to take $j-i$ steps (one operation per character).
Putting everything together, we get
$\qquad \begin{align} T(n) &= \sum_{i=0}^{n-1} \sum_{j = i+1}^{n} j - i \\ &= \sum_{i=0}^{n-1} \left[\left(\sum_{j = i+1}^{n} j\right) - (n-i)i\right] \\ &= \sum_{i=0}^{n-1} \left[\left(\sum_{j = 1}^{n} j - \sum_{j = 1}^{i} j\right) - (n-i)i\right] \\ &= \sum_{i=0}^{n-1} \left[\left(\frac{n(n+1)}{2} - \frac{i(i+1)}{2}\right) - (n-i)i\right] \\ &= \sum_{i=0}^{n-1} \left[ \frac{n^2 - (2i - 1)n + i^2 - i}{2}\right]\end{align}$
The term in brackets is what you are looking for.
The whole sum can be evaluated using Gauss' formula and its sibling for summand $i^2$:
$\qquad \begin{align} 2T(n) &= \sum_{i=0}^{n-1} n^2 - \sum_{i=0}^{n-1} 2in + \sum_{i=0}^{n-1} n + \sum_{i=0}^{n-1} i^2 - \sum_{i=0}^{n-1} i \\ &= n^3 - 2n \cdot \sum_{i=0}^{n-1} i + n^2 + \sum_{i=0}^{n-1} i^2 - \sum_{i=0}^{n-1} i \\ &= n^3 - n^2(n-1) + n^2 + \frac{(n-1)n(2n - 1)}{6} - \frac{n(n-1)}{2} \\ &= \frac{12n^2 + 2n^3 - 3n^2 + n - 3n^2 + 3n}{6} \\ &= \frac{2n^3 + 6n^2 + 4n}{6}\end{align}$
which immediately yields
$\qquad \displaystyle T(n) = \frac{n^3 + 3n^2 + 2n}{6}$. |
Let $X/k$ be a scheme of finite type over a field $k$, $G/k$ be a finite group scheme and suppose $G$ acts on $X$, i.e we have a $k$-morphism $ \mu : G \times_k X \rightarrow X$ satisfies some conditions, see Mumford's book "Abelian Varieties", p. 108.
Suppose that $k$ is algebraically closed and consider the map $\mu$ induced on closed points $G(k) \times X(k) \rightarrow X(k)$, which gives $G(k) \rightarrow \mathrm{Aut}(X(k))$. Under the assumption that each $x \in X$ has an open affine neighborhood $U$ such that $U$ is invariant under $G$, one can form the quotient $X/G$. In particular, quasi-projective varieties have this property. This assumption reduces the construction from general variety to affine case.
For a general field $k$, if one can cover $X$ by open affine subset $U_i$ such that the image of $ G \times_k U_i$ under $\mu$ is contained in $U$, then one reduces the construction to the affine case. But I couldn't figure out if a quasi-projective variety $X$ over $k$ always has such a covering. By working with base change to the algebraic closure of $k$, one gets a $G$-invariant open affine covering of $X_{\overline{k}}$. The images of these open affine subsets of $X_{\overline{k}}$ under the projection to $X$ is still $G$-invariant, but not necessary affine. So for quasi-projective varieties over $k$, do we have the existence of the quotient $X/G$?
Another question is that if it exists, then if it is also quasi-projective? What I know is that, in the classical case ( over algebracially closed field and giving finite group action $H \rightarrow \mathrm{Aut}_k (X))$, if the quotient exists and $X$ is complete, then $X/H$ is also complete. |
Building a Magnetohydrodynamic Multiphysics Model in COMSOL®
The COMSOL Multiphysics® software is built from the ground up to be multiphysics capable so that the user can easily combine models representing different physics phenomena in whichever way they wish. Sometimes this can be achieved simply by using built-in features of the software, but in other cases, the user will need to do a little bit of extra work. Let’s look at such a workflow in the context of building a magnetohydrodynamic (MHD) model.
Modeling the Multiphysics of Magnetohydrodynamics
The modeling of MHD phenomena is intrinsically a multiphysics problem; one must numerically solve for the coupling between fluid flow, current flow, and magnetic fields. These different fields are all described by partial differential equations, which can be solved for via the finite element method.
An MHD problem of a conductive fluid in a channel between two magnets with applied current.
Let’s look at how to do this in the context of a relatively simple-looking problem, as sketched above, of an incompressible conductive fluid within an insulative rectangular channel that connects two infinite reservoirs (not modeled) of equal hydrostatic pressure. There are two electrodes protruding through the flow channel on either side that drive an electrical current through the fluid by applying a potential difference. In addition, two circular magnets are placed above and below. The magnets set up a static magnetic field, \mathbf{B}, such that the fluid with conductivity, \sigma, moving with a velocity, \mathbf{v}, through this field will experience induced currents of \mathbf{J} = \sigma \left( \mathbf{v \times B}\right) . In addition to these induced currents, there are currents that arise as a consequence of the boundary conditions on the electric potential field, V, so that the total current in the fluid becomes:
This current, flowing through the magnetic field, will lead to a volume force on the fluid of \mathbf{F = J \times B} , and this will act to pump the fluid from one reservoir to the other. We will assume that the system operates at steady state.
Coupling the Electric, Magnetic, and Flow Fields
For this problem, we need to solve a system of partial differential equations within the fluid to describe the electric and magnetic fields. The equations are the following:
and
This set of equations is solved via the
Magnetic and Electric Fields interface, which is part of the AC/DC Module, using the Ampère’s Law and Current Conservation feature along with the separate Velocity (Lorentz Term) feature.
In the space surrounding the moving fluid, there is no current flow, so we only need to solve the single vector equation:
where \mathbf{B_r} is the remanent flux density, which is only nonzero in the magnet domains. When solving solely the above equation, use the
Ampère’s Law feature available within the Magnetic and Electric Fields interface.
We assume that the properties of the channel walls do not affect the fields and thus omit them from the model. A set of material properties and boundary conditions are used that will give illustrative results. The magnetic field boundary conditions everywhere are the
Magnetic Insulation condition, except for on the xy-plane, where the Perfect Magnetic Conductor condition is used to exploit the symmetry of the system. The domains representing the electrodes must extend all the way to the boundary of the modeling domain, touching the Magnetic Insulation boundaries, to provide a current return path. The Ground and Terminal of type Voltage are applied on these exterior faces, while Electric Insulation conditions are applied on all other applicable boundaries.
In addition, we also need to solve for the flow field in the channel. We will assume that the flow is laminar and thus solve the Navier–Stokes equations in the channel domain. If the flow were turbulent, we could have added a turbulence model. The
Open Boundary condition is applied at either end of the channel, with a gauge pressure of zero. The Symmetry condition is applied on the xy-plane. The computational domain is shown in the image below. The computational domains and boundary conditions.
The flow will be driven by the volume force that arises due to the interaction of the electric currents in the fluid and the magnetic fields, which is \mathbf{F = J \times B} . The expression for this force is not built into the software, so here we will need to do a little bit of manual work. We need to find the built-in expressions for the components of the current flow and magnetic field, which we can do by looking at the
Equation View and generating a report, as described in a Knowledge Base entry on implementing user-defined multiphysics couplings. These built-in expressions are used to define the volume force on the fluid, as shown in the screenshot below. Screenshot showing the variables that compute the force components.
Lastly, to couple the computed velocity field back to the electromagnetic problem, use the
Velocity (Lorentz Term) feature within the Magnetic and Electric Fields interface, as shown in the screenshot below. Note that the software automatically recognizes the fluid velocity field as an input to this feature. And that is all there is to it! The coupling between the two physics is now completely implemented. Screenshot showing how the velocity is coupled into the Magnetic and Electric Fields interface. Meshing and Solving the MHD Problem
As far as the element meshing and element order, a significant concern here is the computational size of the model. Solving for the magnetic and electric fields in the fluid and surrounding domains is the most computationally expensive part of the model, and we would thus like to keep the total number of mesh elements in the whole model at a minimum. Based on some rules of thumb for linear static problems, we can say that having at least second-order elements is a good starting point. Thus, we will switch the discretization of the fluid flow to a P2+P2 discretization, meaning that both velocity and pressure are described with second-order base functions. The magnetic and electric fields are both described using second-order discretization. Since all fields are discretized to at least second order, the geometry shape order will thus automatically also be second order. A complete investigation of alternative mesh orders and mesh size is left as an exercise to the motivated reader.
When solving, the software will automatically use a so-called segregated approach that switches back and forth between determining the electromagnetic fields and the velocity field and calculates the linear subsystem for those fields, each with its own optimized iterative solver. As this multiphysics problem is inherently nonlinear, it is also generally helpful to be aware of the issues that might arise when solving such problems and how to address them, as discussed in this Knowledge Base entry on improving the convergence of nonlinear stationary models.
The results of this multiphysics analysis are shown in the plots below. We observe a distinct pumping effect: The applied voltage causes current to flow through the fluid, and as these charges move through the magnetic field, they experience a force, which is imparted to the fluid.
Results showing a fluid pumping due to the MHD multiphysics couplings. Simplifying the MHD Model
What we’ve built up to this point is a model involving magnetic fields, electric currents, and fluid flow, and we have considered bidirectional couplings between all physics equations. That is, every physics phenomenon can affect all other physics phenomena. As it turns out, though, we do not need to do so for this particular case. Let’s next look at why this is and how it makes our model much simpler.
If we go back and look at all of the governing equations from earlier, we can see that there are only two equations that introduce couplings between the physics phenomena. There is the equation \mathbf{F = J \times B} that imposes a force on the fluid due to the current and magnetic field, and there is the equation for the total current in the fluid \mathbf{J} = \sigma \left( – \nabla V + \mathbf{v \times B}\right). The latter equation says that the current arises due to the imposed voltage boundary conditions as well as due to the movement of the conductive fluid through the magnetic field. However, if we assume that the former term is much larger than the latter (that is, – \nabla V \gg \mathbf{v \times B}), then we simplify the current equation to: \mathbf{J} = \sigma \left( – \nabla V \right) . This means that the fluid flow problem does not affect the current, which means that the flow equations can be solved entirely separately from the electromagnetic field equations. That is, we can first solve for the electromagnetic fields, and once those are known, use those fields as an input to the flow problem, which makes the problem unidirectionally coupled.
An additional simplification can be made. Strictly speaking, the magnetic field arises due to the magnets as well as the current flow. However, for the boundary conditions and material properties that we consider here, the magnetic field that arises due to the current flow is much smaller than the magnetic field due to the magnets. Thus, we can make the simplifying assumption that the magnetic field arises solely due to the magnets; that is, the currents do not produce a significant magnetic field. As a consequence, we can solve for the magnetic fields under the assumption of no currents and solve for the electric currents separately using the
Magnetic Fields, No Currents and Electric Currents interfaces, respectively. These physics interfaces have a similar set of boundary and domain conditions to the ones previously discussed.
The
Magnetic Fields, No Currents interface defines the equation \nabla \cdot \left( \mu_0 \mu_r \mathbf{H + B_r} \right) = 0, which is much less computationally expensive than the set of equations defined in the Magnetic and Electric Fields interface. Also, this equation can be solved independently of the electric currents. Screenshot showing the setup of the simplified model.
The screenshot above shows the setup of a new model after considering these simplifications. The expression for the volume force on the fluid will use different variable names, but otherwise, the model is very similar to before. Note that three different physics interfaces are solved in three separate study steps. The
Magnetic Fields, No Currents and the Electric Currents interface equations can be solved separately and both must be solved prior to the Laminar Flow interface equations. Results of the simplified MHD model.
The solution time will be greatly reduced when solving this simplified case, as compared to the fully coupled case, since the physics equations are solved separately and the software does not need to iterate between them. We can see from the results shown above that the solutions are almost identical to the previous, unsimplified case. Of course, these assumptions and simplifications that we’ve made do have their limitations, so it never hurts to check against a full model, but the power and flexibility of the COMSOL Multiphysics platform let us easily build both simplified and complete models, compare them, and modify them in whatever way we desire. Are you ready to get started with your own multiphysics modeling? Contact COMSOL here!
If you want to download the example model presented, it is available by clicking the button below (note that you must log into a COMSOL Access account and have a valid software license):
Comentários (0) CATEGORIAS Chemical COMSOL Now Electrical Fluid General Interfacing Mechanical Today in Science |
156 10
Hi, I'm aware of a typical example of injective immersion that is not a topological embedding: figure 8
##\beta: (-\pi, \pi) \to \mathbb R^2##, with ##\beta(t)=(\sin 2t,\sin t)## As explained here an-injective-immersion-that-is-not-a-topological-embedding the image of ##\beta## is compact in ##\mathbb R^2## subspace topology while the domain open interval is not, thus ##\beta## is not a smooth embedding. Consider it from the point of view of "homeomorphism onto its image" definition, I was trying to find out an instance of image subset open in the subspace topology that actually is not open in the domain topology or the other way around. Can you help me ? Thanks.
##\beta: (-\pi, \pi) \to \mathbb R^2##, with ##\beta(t)=(\sin 2t,\sin t)##
As explained here an-injective-immersion-that-is-not-a-topological-embedding the image of ##\beta## is compact in ##\mathbb R^2## subspace topology while the domain open interval is not, thus ##\beta## is not a smooth embedding.
Consider it from the point of view of "homeomorphism onto its image" definition, I was trying to find out an instance of image subset open in the subspace topology that actually is not open in the domain topology or the other way around.
Can you help me ? Thanks. |
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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I would like to "reopen" the previous post regarding Modus ponens because, frankly speaking, I'm not satisfied with some (most of ?) answers by the mathematicians community.
Disclaim: I'm not aiming to "unravel the mystery", but I'm not convincd either that mathematicians and philosophers speaks completely different languages.
This is my argument, in two steps : a "mental experiment" followed by some considerations about formalization and natural language.
The experiment I'm trying is based on a reformulation of McGee's first example (see Vann McGee,
A Counterexample to Modus Ponens (1985)), regarding the US presidential election of 1980.
I'll neglect the aspects regarding "belief" and the nuances connected to verbal tense (see the paper of Robert Fogelin & W.Sinnott-Armstrong,
A defense of Modus Ponens (1986)), also because I'm not a native english speaker.
I assume as
domain of the problem a non-empty universe (call it $US$) where there are only two mutually exclusive subsets : $rep$ and $dem$ (so that : $rep \cap dem = \emptyset$).
I assume that the set $rep$ has only two elements $R$ and $A$ (i.e. $rep = \{ R, A \}$, and $A \ne R$).
I assume only one "obvious" axioms, translating the "rule of the game", using a single predicate $win$ :
$win(dem) \lor win(rep)$.
The first consideration - we will discuss it later - is that the above condition is really a "XOR": "a republican will win or a democrat will win, but not both".
We have also :
$\lnot win(rep) \equiv win(dem)$.
So we have the "tirvial" :
$\lnot win(rep) \lor win(rep)$.
But due to the fact that the only republican candidates are $R$ and $A$, the last amount to :
$\lnot win(rep) \lor [win(R) \lor win(A)]$ ---
(A). Note : we are not using $\rightarrow$ in this argument; if we would use it, with the classical truth-functional semantics, the sub-formula between the square brackets would amount to : $\lnot win(R) \rightarrow win(A)$.
I introduce now what I'll call
Shoenfield rule (from Joseph Shoenfield, Mathematical Logic (1967), page 28 :
if $\vdash A$ and $\vdash \lnot A \lor B$, then $\vdash B$.
The above rule is proved in Shoenfield's system using three of the four "propositional" primitive rules [page 21 : the last one, the
Associative Rule, is not used in the proof below] :
Expansion Rule: infer $B \lor A$ from $A$
Contraction Rule: infer $A$ from $A \lor A$
Cut Rule: infer $B \lor C$ from $A \lor B$ and $\lnot A \lor C$.
With the
Cut Rule and the (only) propositional axiom : $\lnot A \lor A$, we can derive the Lemma 1 : if $\vdash A \lor B$, then $\vdash B \lor A$.
Now we prove
Shoenfield rule : (1) --- $\vdash A$ (2) --- $\vdash B \lor A$ --- from (1) by Expansion (3) --- $\vdash A \lor B$ --- from (2) by Lemma 1 (4) --- $\vdash \lnot A \lor B$ (5) --- $\vdash B \lor B$ --- from (3) and (4) by Cut (6) --- $\vdash B$ --- from (5) by Contraction. Disclaim: nothing new; all is trivial (classical) propositional logic.
Now, we go back to
(A) :
$\lnot win(rep) \lor (win(R) \lor win(A))$
and add the premise :
$win(rep)$;
by
Shoenfield rule we conclude the "obvious" :
$win(R) \lor win(A)$.
Nothing has gone wrong ... We only has used standard rules for propositional connectives in a
classical framework, with the use of $\lor$ in a situation where the alternative are mutually exclusive.
Question: Is the previous argument "sound" ?
The above argument, assuming it is "sound" suggests to me some considerations about formalization and natural language.
The "regimentation" that symbolic logic - from Frege on - has deliberately imposed on natural language has been greatly fruitful; this does not imply that the richness of natural language can be wholly "explained away" with formalization.
The dissatisfaction of McGee about the
modus ponens seems to me the "old" dissatisfactions about the translation of "if ... then" in term of the truth-functional connective $\rightarrow$.
This one is blind about the nuances of natural language (that
relevant logic try to recover). In the same way, when I use $\lor$ in a context where the alternatives are mutually exclusive, I "loose" some presuppositions (some implicit information that the speaker aware of the context knows).
This does not means that the rule of logic are "wrong"; neither that philosopher does not know logic. Aristotle and Leibniz and Peirce and Frege and Russell were all philosophers.
In conclusion, I think that there is no "contradiction" between the way mathematical logic formalize truth-functional connectives and natural language. |
I'm trying to define a dashed
\Uparrow: my first try, using
tikz, consists simply in drawing two white rectangle over the normal
\Uparrow, manually trying to find a good positioning of the rectangles.
\documentclass{article}\usepackage[utf8]{inputenc}\usepackage[T1]{fontenc}\usepackage{tikz}\newcommand{\Dasheduparrow}{\begin{tikzpicture}[x=1, y=1]\node (0,0) [inner sep=0] {$\Uparrow$};\fill[fill=white] (-3,0.5) rectangle (2,-1);\fill[fill=white] (-3,-2) rectangle (2,-3.5);\end{tikzpicture}}
By typing now
\begin{document}Text $\Dasheduparrow$ $\Downarrow$ $\Uparrow$ text.\end{document}
I get this
and I'm quite satisfied with the output. However, I've got three questions:
How to get the same alignment of the normal
\Downarrowand
\Uparrow? My new symbol is aligned with the baseline of the text, while the two default ones are a little down-shifted.
How to make this new symbol correctly stretchable? By doing
\LARGE Text $\Dasheduparrow$ $\Downarrow$ $\Uparrow$ text.the output is where it's evident that the underlying
\Uparrowhas been enlarged, while the white rectangles have not.
Is there a better solution than the one I proposed? |
Absolute Values
We start by defining absolute values on $\mathbb Q$, but much of this extends naturally to absolute values on other fields. An
absolute value on $\mathbb Q$ is a function $\mathbb Q \rightarrow [0,\infty)$ such that: positive definiteness: $|x| = 0$ iff $x = 0$, multiplicativity: $|x y| = |x| |y|$ for all $x, y \in \mathbb Q$ triangle inequality: $|x + y| \leq |x| + |y|$ for all $x,y \in \mathbb Q$.
If, in place of (3), $|\cdot|$ satisfies the
strong triangle inequality: $|x + y| \leq \max\{|x|, |y|\}$, then we say $| \cdot |$ is a non-archimedean absolute value. Example
Let $p$ be a prime, and define the standard $p$-adic absolute on $\mathbb Q$ by
$$ \left|p^{\ell} \frac{a}b\right| = p^{-\ell}. $$ Here we assume that $a$ and $b$ are rational integers not divisible by $p$. $| \cdot|_p$ is a non-archimedean absolute value. Note that if $n$ is a rational integer, then $|n| \leq 1$. Completions
Given an absolute value $|\cdot|$ on $\mathbb Q$, we say a sequence $(x_n)$ is
Cauchy if given $\epsilon > 0$, there exists $N \in \mathbb N$ so that $n, m > N$ implies $|x_n – x_m| < \epsilon$. Similarly, we say that $(x_n)$ converges to zero if given $\epsilon > 0$ there exists $N \in \mathbb N$ so that $n > N$ implies $|x_n| < \epsilon$. Using the standard operations on sequences, the set of Cauchy sequences $\mathcal C$ is a (commutative) ring, and the set of sequences which converge to zero, $\mathcal Z$ is a maximal ideal. It follows that $\mathcal C / \mathcal Z$ is a field, and we call this the completion of $\mathbb Q$ with respect to the absolute value $| \cdot |$. We may embed $\mathbb Q$ into $\mathcal C / \mathcal Z$ by the diagonal embedding $x \mapsto (x,x,\cdots)$. The fact that $\mathbb Q$ is dense in $\mathcal C / \mathcal Z$ allows us to extend $| \cdot |$ to a unique absolute value on $\mathcal C / \mathcal Z$.
The completion of $\mathbb Q$ with respect to the usual absolute value is $\mathbb R$.
The completion of $\mathbb Q$ with respect to $| \cdot |_p$ is called the field of $p$-adic numbers and denoted $\mathbb Q_p$. The Topology of $\mathbb Q_p$
A basic neighborhood in $\mathbb Q_p$ is a set of the form $B(x, \epsilon) = \{y \in \mathbb Q_p : |y – x|_p \leq \epsilon\}$. Since $| \cdot |_p$ takes values in the discrete set $\{p^{-n} : n \in \mathbb Z\}$, every basic neighborhood is both open and closed, since for instance
$$ \{y : |y|_p \leq p^{-n} \} = \{y : |y|_p < 2p^{-n} \}. $$
Let’s investigate the ball $B(1, p^{-1})$ by looking at what integers it contains. Clearly $1 \in B(1, p^{-1})$. The condition $|y – 1| \leq p^{-1}$ is equivalent to $y \equiv 1 \bmod p$. Similarly, $B(1, p^{-n})$ contains all the integers $y$ such that $y \equiv 1 \bmod p^n$. For instance, when $p = 7$, the integers $1$ and $344$ are close, a distance of $7^{-3}$ from each other, since $344 – 1 = 343 = 7^3$.
For a slightly more complicated ball, consider $B(1/2, p^{-2})$. This is by definition
$$ \{y \in \mathbb Q_p : |y – 1/2|_p \leq p^{-2} \}. $$ From the definition of $| \cdot |_p$, if $y$ is a rational number so that $|y – 1/2|_p = p^2$, then there exist integers $a$ and $b$, both relatively prime to $p$, so that $y – 1/2 = p^2 a/b$. More generally, if $|y – 1/2|_p \leq p^2$ then there exist integers $a$ and $b$, with $b$ relatively prime to $p$, so that $y – 1/2 = p^2 a/b$. It follows that the rational numbers in $B(1/2, p^{-2})$ are of the form $1/2 + p^2 r$ where $r$ is a rational number which contains no factors of $p$ in the denominator when written in lowest terms. The General Setup
Suppose $K$ is a number field of degree $d$ over $\mathbb Q$ and suppose $\mf O$ is the ring of integers of $K$. Given a prime ideal $\mf p \subseteq \mf O$ we can define an absolute value $| \cdot |_{\mf p}$ by fixing $q > 0$ and setting
\[ |x|_{\mf p} = q^{-n} \qquad \mbox{where} \qquad x \in \mf p^n \setminus \mf p^{n+1}. \] The completion of $K$ with respect to $| \cdot |_{\mf p}$ is denoted $K_{\mf p}$.
The ring of integers in $K_{\mf p}$ is defined to be $\mf o_{\mf p} = \{x \in K_{\mf p} : |x|_{\mf p} \leq 1\}$. The ideal $\mf m_{\mf p} \subseteq \mf o_{\mf p}$ given by $\mf m_{\mf p} = \{x \in \mf o_{\mf p} : |x| < 1 \}$. The units in $\mf o_{\mf p}$ are given by $U_{\mf p} := \{x \in \mf o_{\mf p} : |x|_{\mf p} = 1 \}$.
The field $\mf o_{\mf p} / \mf m_{\mf p}$ is a finite field of size $p^f$ for some rational prime $p$ and positive integer $f.$ It is convenient to take $q$ in the definition of $| \cdot |_{\mf p}$ to be $q := p^f.$
Often times, when we are working with a fixed non-archimedean absolute value, we drop the subscript $\mf p$ from various quantities. When in this situation we assume $(K, |\cdot|)$ is a field complete with respect to $|\cdot|$ and with integers $\mf o$, maximal ideal $\mf m$ and units $U$. Unless specifically indicated otherwise we take $q = [\mf o : \mf m]$. The maximal ideal is principal, and we set $\varpi$ to be a generator or
uniformizer for the ideal: $\mf m = \varpi \mf o$. Clearly then $|\varpi| = 1/q$. |
(a) If $AB=B$, then $B$ is the identity matrix. (b) If the coefficient matrix $A$ of the system $A\mathbf{x}=\mathbf{b}$ is invertible, then the system has infinitely many solutions. (c) If $A$ is invertible, then $ABA^{-1}=B$. (d) If $A$ is an idempotent nonsingular matrix, then $A$ must be the identity matrix. (e) If $x_1=0, x_2=0, x_3=1$ is a solution to a homogeneous system of linear equation, then the system has infinitely many solutions.
Let $A$ and $B$ be $3\times 3$ matrices and let $C=A-2B$.If\[A\begin{bmatrix}1 \\3 \\5\end{bmatrix}=B\begin{bmatrix}2 \\6 \\10\end{bmatrix},\]then is the matrix $C$ nonsingular? If so, prove it. Otherwise, explain why not.
Let $A, B, C$ be $n\times n$ invertible matrices. When you simplify the expression\[C^{-1}(AB^{-1})^{-1}(CA^{-1})^{-1}C^2,\]which matrix do you get?(a) $A$(b) $C^{-1}A^{-1}BC^{-1}AC^2$(c) $B$(d) $C^2$(e) $C^{-1}BC$(f) $C$
Let $\calP_3$ be the vector space of all polynomials of degree $3$ or less.Let\[S=\{p_1(x), p_2(x), p_3(x), p_4(x)\},\]where\begin{align*}p_1(x)&=1+3x+2x^2-x^3 & p_2(x)&=x+x^3\\p_3(x)&=x+x^2-x^3 & p_4(x)&=3+8x+8x^3.\end{align*}
(a) Find a basis $Q$ of the span $\Span(S)$ consisting of polynomials in $S$.
(b) For each polynomial in $S$ that is not in $Q$, find the coordinate vector with respect to the basis $Q$.
(The Ohio State University, Linear Algebra Midterm)
Let $V$ be a vector space and $B$ be a basis for $V$.Let $\mathbf{w}_1, \mathbf{w}_2, \mathbf{w}_3, \mathbf{w}_4, \mathbf{w}_5$ be vectors in $V$.Suppose that $A$ is the matrix whose columns are the coordinate vectors of $\mathbf{w}_1, \mathbf{w}_2, \mathbf{w}_3, \mathbf{w}_4, \mathbf{w}_5$ with respect to the basis $B$.
After applying the elementary row operations to $A$, we obtain the following matrix in reduced row echelon form\[\begin{bmatrix}1 & 0 & 2 & 1 & 0 \\0 & 1 & 3 & 0 & 1 \\0 & 0 & 0 & 0 & 0 \\0 & 0 & 0 & 0 & 0\end{bmatrix}.\]
(a) What is the dimension of $V$?
(b) What is the dimension of $\Span\{\mathbf{w}_1, \mathbf{w}_2, \mathbf{w}_3, \mathbf{w}_4, \mathbf{w}_5\}$?
(The Ohio State University, Linear Algebra Midterm)
Let $V$ be the vector space of all $2\times 2$ matrices whose entries are real numbers.Let\[W=\left\{\, A\in V \quad \middle | \quad A=\begin{bmatrix}a & b\\c& -a\end{bmatrix} \text{ for any } a, b, c\in \R \,\right\}.\]
(a) Show that $W$ is a subspace of $V$.
(b) Find a basis of $W$.
(c) Find the dimension of $W$.
(The Ohio State University, Linear Algebra Midterm)
The following problems are Midterm 1 problems of Linear Algebra (Math 2568) at the Ohio State University in Autumn 2017.There were 9 problems that covered Chapter 1 of our textbook (Johnson, Riess, Arnold).The time limit was 55 minutes.
This post is Part 3 and contains Problem 7, 8, and 9.Check out Part 1 and Part 2 for the rest of the exam problems.
Problem 7. Let $A=\begin{bmatrix}-3 & -4\\8& 9\end{bmatrix}$ and $\mathbf{v}=\begin{bmatrix}-1 \\2\end{bmatrix}$.
(a) Calculate $A\mathbf{v}$ and find the number $\lambda$ such that $A\mathbf{v}=\lambda \mathbf{v}$.
(b) Without forming $A^3$, calculate the vector $A^3\mathbf{v}$.
Problem 8. Prove that if $A$ and $B$ are $n\times n$ nonsingular matrices, then the product $AB$ is also nonsingular.
Problem 9.Determine whether each of the following sentences is true or false.
(a) There is a $3\times 3$ homogeneous system that has exactly three solutions.
(b) If $A$ and $B$ are $n\times n$ symmetric matrices, then the sum $A+B$ is also symmetric.
(c) If $n$-dimensional vectors $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3$ are linearly dependent, then the vectors $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3, \mathbf{v}_4$ is also linearly dependent for any $n$-dimensional vector $\mathbf{v}_4$.
(d) If the coefficient matrix of a system of linear equations is singular, then the system is inconsistent.
The following problems are Midterm 1 problems of Linear Algebra (Math 2568) at the Ohio State University in Autumn 2017.There were 9 problems that covered Chapter 1 of our textbook (Johnson, Riess, Arnold).The time limit was 55 minutes.
This post is Part 2 and contains Problem 4, 5, and 6.Check out Part 1 and Part 3 for the rest of the exam problems.
Problem 4. Let\[\mathbf{a}_1=\begin{bmatrix}1 \\2 \\3\end{bmatrix}, \mathbf{a}_2=\begin{bmatrix}2 \\-1 \\4\end{bmatrix}, \mathbf{b}=\begin{bmatrix}0 \\a \\2\end{bmatrix}.\]
Find all the values for $a$ so that the vector $\mathbf{b}$ is a linear combination of vectors $\mathbf{a}_1$ and $\mathbf{a}_2$.
Problem 5.Find the inverse matrix of\[A=\begin{bmatrix}0 & 0 & 2 & 0 \\0 &1 & 0 & 0 \\1 & 0 & 0 & 0 \\1 & 0 & 0 & 1\end{bmatrix}\]if it exists. If you think there is no inverse matrix of $A$, then give a reason.
Problem 6.Consider the system of linear equations\begin{align*}3x_1+2x_2&=1\\5x_1+3x_2&=2.\end{align*}
(a) Find the coefficient matrix $A$ of the system.
(b) Find the inverse matrix of the coefficient matrix $A$.
(c) Using the inverse matrix of $A$, find the solution of the system.
(Linear Algebra Midterm Exam 1, the Ohio State University) |
Prove or disprove:
if $\sum_{n=0}^\infty a_n x^n$ converges at $x=x_1$ , then $\sum_{n=0}^\infty n\cdot a_n \cdot x^{n-1}$ converges at $x=x_1$
I am quite new to this material (and taylor series especially).
I am pretty sure, that if I differentiate a power series, the radius of convergence stays the same, but:
I'm not sure why.
if $R=x_1$ (The radius of convergence), and it converges in the original series, I don't think it still holds for the differentiate.
Would love some guidelines. |
On the law of logarithm of the recurrence time
1.
Department of Mathematics, Korea Advanced Institute of Science and Technology, Daejeon, 305-701, South Korea
2.
School of Mathematics, Korea Institute for Advanced Study, Seoul, 130-722, South Korea
$ \lim_{n\to\infty}\frac{\log K_n(x)}{n}=1 \quad$and$\quad \lim_{n\to\infty}\frac{\log K_n(x,y)}{n}=1 \quad $a.e.
Mathematics Subject Classification:Primary: 28D05, 37E0. Citation:Chihurn Kim, Dong Han Kim. On the law of logarithm of the recurrence time. Discrete & Continuous Dynamical Systems - A, 2004, 10 (3) : 581-587. doi: 10.3934/dcds.2004.10.581
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Alternating Current Power in AC Circuit: The Power Factor
The power is defined as the rate at which work is being in the circuit.
The average power in an AC circuit,
P av = V rms i rms cos θ
= \frac{V}{\sqrt{2}} \frac{i}{\sqrt{2}} \cos \theta = \frac{Vi}{2} \cos \theta
Where \cos \theta = \tt \frac{Resistance (R)}{Impedance (Z)} is called the power factor of AC Circuit.
View the Topic in this video From 31:38 To 59:00
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A voltage
v = v m sin ω tapplied to a series RLCcircuit drives a current in the circuit given by i= isin (ω m t+ Φ) where
i_{m} = \frac{v_{m}}{Z} \ {\tt and} \ \phi = \tan^{-1} \frac{X_{C} - X_{L}}{R} |
Meleda, or Chinese rings, is a game of Chinese origin which, so the story goes, was invented by the soldier hero Hung Ming (181-234 A.D.) who gave it to his wife when he went to war.
Counting is not always as simple as $1,2,3,\cdots $, but, as I hope to show in these articles, it can be a lot more interesting.
The harmonic series:
$$S=1^{-1} + 2^{-1} + 3^{-1} + 4^{-1}+\cdots $$ is divergent, that is if we add enough terms together, we can produce a partial sum which is as large as we like.
Is a degree in mathematics any use? I mathematics at university interesting? Is it fun?
Algernon announced that on his birthday this year his age would be equal to the sum of the digits of the year in which he was born. When was he born?
We shall consider some problems involving the roots $\alpha, \beta$ and $\gamma$, say, of the cubic equation
$$x^3 + qx + r = 0$$
Dear Sir, Do you think that there would be two people in your class whose birthdays fall on the same day and month?
Q.455 The rule for leap years runs as follows: A year which is divisible by 4 is a leap year except that years which are divisible by 100 are not leap years unless they are divisible by 400. |
Eigenvalues of Squared Matrix and Upper Triangular Matrix
Problem 184
Suppose that $A$ and $P$ are $3 \times 3$ matrices and $P$ is invertible matrix.If\[P^{-1}AP=\begin{bmatrix}1 & 2 & 3 \\0 &4 &5 \\0 & 0 & 6\end{bmatrix},\]then find all the eigenvalues of the matrix $A^2$.
We give two proofs. The first version is a short proof and uses some facts without proving.The second proof explains more details and give proofs of the facts which are not proved in the first proof.
Proof (short version).
Let $B=P^{-1}AP$. Since $B$ is an upper triangular matrix, its eigenvalues are diagonal entries $1, 4, 6$.
Since $A$ and $B=P^{-1}AP$ have the same eigenvalues, the eigenvalues of $A$ are $1, 4, 6$. Note that these are all the eigenvalues of $A$ since $A$ is a $3\times 3$ matrix.
It follows that all the eigenvalues of $A^2$ are $1, 4^2, 6^2$, that is, $1, 16, 36$.
Proof (long version.)
Let us put $B:=P^{-1}AP$.The eigenvalues of $B$ are $1, 4, 6$ since $B$ is an upper triangular matrix and eigenvalues of an upper triangular matrix are diagonal entries.We claim that the eigenvalues of $A$ and $B$ are the same.
To prove this claim, we show that their characteristic polynomials are equal.Let $p_A(t), p_B(t)$ be the characteristic polynomials of $A, B$, respectively.Then we have\begin{align*}p_B(t)&=\det(B-tI)=\det(P^{-1}AP-tI)\\&= \det(P^{-1}(A-tI)P)\\&=\det(P^{-1})\det(A-tI)\det(P)\\&=\det(P)^{-1}p_A(t)\det(P)\\&=p_A(t).\end{align*}Therefore, the characteristic polynomials of $A$ and $B$ are the same, and hence the matrices $A$ and $B$ have the same eigenvalues.Thus $1, 4, 6$ are eigenvalues of the matrix $A$ and these are all the eigenvalues of $A$ since a $3\times 3$ matrix has at most three eigenvalues.Now we claim that in general if $\lambda$ is an eigenvalue of $A$, then $\lambda^2$ is an eigenvalue of $A^2$. We prove this statement below. Assuming this claim for the moment, we finish the problem.By the claim, the matrix $A^2$ has eigenvalues $1^2, 4^2, 6^2$. Thus all the eigenvalues of $A^2$ are\[1, 16, 36.\]To complete the solution, let us prove the last claim.Since $\lambda$ is an eigenvalue of $A$, we have an eigenvector $\mathbf{x}$ such that\[A\mathbf{x}=\lambda \mathbf{x}.\]
Multiplying by $A$ we obtain\begin{align*}A^2T\mathbf{x}&=\lambda (A\mathbf{x})\\&=\lambda (\lambda \mathbf{x})\\&=\lambda^2 \mathbf{x}\end{align*}and thus $\lambda^2$ is an eigenvalue of $A^2$.
The Inverse Matrix of an Upper Triangular Matrix with VariablesLet $A$ be the following $3\times 3$ upper triangular matrix.\[A=\begin{bmatrix}1 & x & y \\0 &1 &z \\0 & 0 & 1\end{bmatrix},\]where $x, y, z$ are some real numbers.Determine whether the matrix $A$ is invertible or not. If it is invertible, then find […]
Determinant/Trace and Eigenvalues of a MatrixLet $A$ be an $n\times n$ matrix and let $\lambda_1, \dots, \lambda_n$ be its eigenvalues.Show that(1) $$\det(A)=\prod_{i=1}^n \lambda_i$$(2) $$\tr(A)=\sum_{i=1}^n \lambda_i$$Here $\det(A)$ is the determinant of the matrix $A$ and $\tr(A)$ is the trace of the matrix […]
Find All the Eigenvalues of $A^k$ from Eigenvalues of $A$Let $A$ be $n\times n$ matrix and let $\lambda_1, \lambda_2, \dots, \lambda_n$ be all the eigenvalues of $A$. (Some of them may be the same.)For each positive integer $k$, prove that $\lambda_1^k, \lambda_2^k, \dots, \lambda_n^k$ are all the eigenvalues of […] |
The hint by Yuval in his comment is the right one. An approach to writing context-sensitive grammars is to design them like a machine. Send messages over the string. This is very close to writing a linear bounded automaton (or linear space Turing machine). In such a machine the reading/writing head scans over the tape while updating the string. This head (with the machine state) can be modelled by a production, like $ap\to qb$ for "in state $p$ while reading $a$, write $b$ and move left, changing state to $q$", where we assume the state is written to the right of the position of the head.
There are differences. The CSG is more flexible as one can insert letters in the middle of the string, and moreover there can be parallel processes going on at the same time. The latter can be a nuisance when arguing the approach is correct.
Lets see how the messages work in an example:$L = \{ww\mid w\in\{a,b\}^*, |w|\ge 1 \}$, or squares.
Start with left and right sides of the string with the rule $S\to LR$. As markers and messengers cannot disappear in a CSG they must represent a letter in the final string. This letter can be $a$ or $b$ so we have two rules instead: $S\to L_aR_a\mid L_bR_b$. Now the left marker generates new letters and a messenger that becomes a copy at the other half of the string. $L_\sigma\to L_\sigma aM_a\mid L_\sigma bM_b$. Everywhere in these rules $\sigma,\tau$ may take the values $a,b$.Messengers move over other letters: $M_\sigma\tau\to \tau M_\sigma$. At the beginning of the second half write the letter: $M_\sigma R_\tau \to R_\tau \sigma$.
We should end the derivation by removing the boundary markers $L_\sigma\to\sigma$, $R_\sigma\to \sigma$. We have a problem when the right marker is gone, while there is still a messenger under way. This can be solved by extra symchronization. However, no terminal string will be generated this way; the drivation is lost, so there is no real problem.
I tried a similar approach here for the language $\{a^ib^jc^{ij} \mid i,j\ge 1\}$. Here a hint for squares (numbers) $\{a^{n^2} \mid n\ge 1\}$. As you see, I practise a lot.
PS. I hope you mean a monotonic grammar rather than a "real" context-sensitive one. Those are harder, but there is a standard (boring) transformation. |
The title is pretty self explanatory. If I have $V$ nodes and $E$ edges in a connected undirected graph, is there a formula to determine an upper bound on the maximum possible diameter? The exact graph is unknown, but the number of edges and the number of vertices is. I do know that when $E=V(V-1)/2$ (complete graph), the maximum possible diameter is $1$, and when $E=V-1$ (line graph), the maximum possible diameter is $V-1$, but I have no idea about anything in between.
We assume that $v \geq n-1 $ and $v \leq \frac{n(n-1)}{2}$
Given $v$ edges and $n$ nodes, let's compute the minimal number of nodes $u$ needed to spend the excess of edges in a spending-hole $SH$:
We know that the saturation of $u$ nodes needs $\frac{n(n-1)}{2}$ edges and it remains $w = n-u$ edges to constitute a linear graph generator of long distances.
First, let's try to minimize $u$ and maximize $w$ in :
$n = w + u$ $v = w-1 + 1 + \frac{u(u-1)}{2}$
where $w-1$ is for the long distance subgraph , $\frac{u(u-1)}{2}$ for the spending-hole $SH$ and $1$ edge to join them.
$=> u' = \frac{(3+\sqrt{(8(v-n)+9)})}{2}$ $=> u = \lceil {u'} \rceil = \lceil {\frac{(3+\sqrt{(8(v-n)+9)})}{2} }\rceil$ ,
$ceil(u')$ because we deal with integers and $u'$ was just a computed bound.
Then we compute the remaining nodes $n-u$ , we take from the $SH$ as many edges needed to reach $n-u$ and we may compute the distance $d = n-u +1$ which must be added $1$ if the $SH$ is saturated, ie if $\frac{u(u-1)}{2}-u= v-n$ .
Numerical application :
/*main(5,true) :5 nodes :for v=4 : 2 nodes will consume 1 edges from 1 ; it remains 3 nodes 3 edges,d =4 5 : 3 3 3 ; 2 2 3 6 : 4 5 6 ; 1 1 3 7 : 4 6 6 ; 1 1 2 8 : 5 8 10 ; 0 0 2 9 : 5 9 10 ; 0 0 2 10 : 5 10 10 ; 0 0 1
Note how the distance $d$ changes with the edges in excess and how it decreases by $1$ when the $SH$ is saturated. I recall that the number of edges is bounded by the question and we cannot have double edges, no edges or edges without nodes.
function main(n,all){ var u , spent , spentmax , v , V = n*(n-1)/2 , res = n+" nodes :\n" , exm = -1,d , firstline = true ; for ( v = n-1 ; v <= V ; v ++ ) { u = Math.ceil( (3+Math.sqrt(8*(v-n)+9))/2 ) ; spentmax = (u*(u-1)/2) ; spent = (v-n+u) ; if(u!=exm || all ) { if(u!=exm ) { if( all ) res += "\n" ; exm = u ; } d = 1 + n-u + ( spentmax == spent ? 0 : 1 ); if( firstline ) res += ( "for v="+v+" : "+u +" nodes will consume "+spent+" edges from "+spentmax+" ; it remains "+(n-u)+" nodes " + (v-spent) + " edges,d ="+d+"\n" ) ; else res += ( " "+v+" : "+u +" "+spent+" "+spentmax+" ; "+(n-u)+" " + (v-spent) + " "+d+"\n" ) ; firstline = false ; } } return res ;}// scratchpad formalism to get the result by typing CTRL L at the end of the scriptvar z1 , z2 = main(5,false) ; // number of nodes and true to get all the intermediate edges stepsz1=z2;/*main(12,false) :12 nodes :for v=11 : 2 nodes will consume 1 edges from 1 ; it remains 10 nodes 10 edges,d =11 12 : 3 3 3 ; 9 9 10 13 : 4 5 6 ; 8 8 10 15 : 5 8 10 ; 7 7 9 18 : 6 12 15 ; 6 6 8 22 : 7 17 21 ; 5 5 7 27 : 8 23 28 ; 4 4 6 33 : 9 30 36 ; 3 3 5 40 : 10 38 45 ; 2 2 4 48 : 11 47 55 ; 1 1 3 57 : 12 57 66 ; 0 0 2 */
Even if the proof is not fundamentally detailed, one can see that the construction is minimal, starting from a linear graph with $v = n-1$ and adding the edges in excess in a Spending hole ( or a ball of wool if one prefers ). When the latter is saturated, we "sacrify" a new node until a new saturation. When all the edges in excess have used a minimum of nodes, it remains a piece of linear graph which is joined to the $SH$ by one edge to one node.
The same question with the possibility of not connection is interesting too ... This kind of problems has a lot of applications when the algorithm may add nodes at its convienience ( Steiner tree problems family ).
ps : feel free to edit and correct obscure translations, TY
This is not a complete answer but rather an observation which leads to good results in some special cases. An interesting family of graphs to consider is the following. Take a complete graph $K_k$ and draw a simple path od length $v-k$ from one of its vertices. The thing you obtain has $v$ vertices, $v-(k^2-3k)/2$ edges, and diameter $v-k+1$. (The diameter is attained at a furthest point on the path and any vertex in $K_k$ that is not on the path.) A straightforward computation shows that you can get a graph with diameter $$V-\left\lceil\sqrt{2E-2V+\frac94}+\frac12\right\rceil,$$ $V$ vertices, and $E$ edges. Note that this bound is asymptotically best possible when there are not too many edges, that is, when $E=o(V^2)$, because then you get a graph with diameter $V+o(V)$. My bound is not supposed to be good for large $E$ though. |
5.2 Least squares estimation
In practice, of course, we have a collection of observations but we do not know the values of the coefficients \(\beta_0,\beta_1, \dots, \beta_k\). These need to be estimated from the data.
The least squares principle provides a way of choosing the coefficients effectively by minimising the sum of the squared errors. That is, we choose the values of \(\beta_0, \beta_1, \dots, \beta_k\) that minimise \[ \sum_{t=1}^T \varepsilon_t^2 = \sum_{t=1}^T (y_t - \beta_{0} - \beta_{1} x_{1,t} - \beta_{2} x_{2,t} - \cdots - \beta_{k} x_{k,t})^2. \]
This is called
least squares estimation because it gives the least value for the sum of squared errors. Finding the best estimates of the coefficients is often called “fitting” the model to the data, or sometimes “learning” or “training” the model. The line shown in Figure 5.3 was obtained in this way.
When we refer to the
estimated coefficients, we will use the notation \(\hat\beta_0, \dots, \hat\beta_k\). The equations for these will be given in Section 5.7.
The
tslm() function fits a linear regression model to time series data. It is similar to the
lm() function which is widely used for linear models, but
tslm() provides additional facilities for handling time series.
Example: US consumption expenditure
A multiple linear regression model for US consumption is \[ y_t=\beta_0 + \beta_1 x_{1,t}+ \beta_2 x_{2,t}+ \beta_3 x_{3,t}+ \beta_4 x_{4,t}+\varepsilon_t, \] where \(y\) is the percentage change in real personal consumption expenditure, \(x_1\) is the percentage change in real personal disposable income, \(x_2\) is the percentage change in industrial production, \(x_3\) is the percentage change in personal savings and \(x_4\) is the change in the unemployment rate.
The following output provides information about the fitted model. The first column of
Coefficients gives an estimate of each \(\beta\) coefficient and the second column gives its standard error (i.e., the standard deviation which would be obtained from repeatedly estimating the \(\beta\) coefficients on similar data sets). The standard error gives a measure of the uncertainty in the estimated \(\beta\) coefficient.
fit.consMR <- tslm( Consumption ~ Income + Production + Unemployment + Savings, data=uschange)summary(fit.consMR)#> #> Call:#> tslm(formula = Consumption ~ Income + Production + Unemployment + #> Savings, data = uschange)#> #> Residuals:#> Min 1Q Median 3Q Max #> -0.8830 -0.1764 -0.0368 0.1525 1.2055 #> #> Coefficients:#> Estimate Std. Error t value Pr(>|t|) #> (Intercept) 0.26729 0.03721 7.18 1.7e-11 ***#> Income 0.71448 0.04219 16.93 < 2e-16 ***#> Production 0.04589 0.02588 1.77 0.078 . #> Unemployment -0.20477 0.10550 -1.94 0.054 . #> Savings -0.04527 0.00278 -16.29 < 2e-16 ***#> ---#> Signif. codes: #> 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1#> #> Residual standard error: 0.329 on 182 degrees of freedom#> Multiple R-squared: 0.754, Adjusted R-squared: 0.749 #> F-statistic: 139 on 4 and 182 DF, p-value: <2e-16
For forecasting purposes, the final two columns are of limited interest. The “t value” is the ratio of an estimated \(\beta\) coefficient to its standard error and the last column gives the p-value: the probability of the estimated \(\beta\) coefficient being as large as it is if there was no real relationship between consumption and the corresponding predictor. This is useful when studying the effect of each predictor, but is not particularly useful for forecasting.
Fitted values
Predictions of \(y\) can be obtained by using the estimated coefficients in the regression equation and setting the error term to zero. In general we write,\[\begin{equation} \hat{y}_t = \hat\beta_{0} + \hat\beta_{1} x_{1,t} + \hat\beta_{2} x_{2,t} + \cdots + \hat\beta_{k} x_{k,t}. \tag{5.2}\end{equation}\]Plugging in the values of \(x_{1,t},\dots,x_{k,t}\) for \(t=1,\dots,T\) returns predictions of \(y_t\) within the training-sample, referred to as
fitted values. Note that these are predictions of the data used to estimate the model, not genuine forecasts of future values of \(y\).
The following plots show the actual values compared to the fitted values for the percentage change in the US consumption expenditure series. The time plot in Figure 5.6 shows that the fitted values follow the actual data fairly closely. This is verified by the strong positive relationship shown by the scatterplot in Figure 5.7.
Goodness-of-fit
A common way to summarise how well a linear regression model fits the data is via the coefficient of determination, or \(R^2\). This can be calculated as the square of the correlation between the observed \(y\) values and the predicted \(\hat{y}\) values. Alternatively, it can also be calculated as, \[ R^2 = \frac{\sum(\hat{y}_{t} - \bar{y})^2}{\sum(y_{t}-\bar{y})^2}, \] where the summations are over all observations. Thus, it reflects the proportion of variation in the forecast variable that is accounted for (or explained) by the regression model.
In simple linear regression, the value of \(R^2\) is also equal to the square of the correlation between \(y\) and \(x\) (provided an intercept has been included).
If the predictions are close to the actual values, we would expect \(R^2\) to be close to 1. On the other hand, if the predictions are unrelated to the actual values, then \(R^2=0\) (again, assuming there is an intercept). In all cases, \(R^2\) lies between 0 and 1.
The \(R^2\) value is used frequently, though often incorrectly, in forecasting. The value of \(R^2\) will never decrease when adding an extra predictor to the model and this can lead to over-fitting. There are no set rules for what is a good \(R^2\) value, and typical values of \(R^2\) depend on the type of data used. Validating a model’s forecasting performance on the test data is much better than measuring the \(R^2\) value on the training data.
Example: US consumption expenditure
Figure 5.7 plots the actual consumption expenditure values versus the fitted values. The correlation between these variables is \(r=0.868\) hence \(R^2= 0.754\) (shown in the output above). In this case model does an excellent job as it explains 75.4% of the variation in the consumption data. Compare that to the \(R^2\) value of 0.16 obtained from the simple regression with the same data set in Section 5.1. Adding the three extra predictors has allowed a lot more of the variation in the consumption data to be explained.
Standard error of the regression
Another measure of how well the model has fitted the data is the standard deviation of the residuals, which is often known as the “residual standard error”. This is shown in the above output with the value 0.329.
It is calculated using \[\begin{equation} \hat{\sigma}_e=\sqrt{\frac{1}{T-k-1}\sum_{t=1}^{T}{e_t^2}}, \tag{5.3} \end{equation}\] where \(k\) is the number of predictors in the model. Notice that we divide by \(T-k-1\) because we have estimated \(k+1\) parameters (the intercept and a coefficient for each predictor variable) in computing the residuals.
The standard error is related to the size of the average error that the model produces. We can compare this error to the sample mean of \(y\) or with the standard deviation of \(y\) to gain some perspective on the accuracy of the model.
The standard error will be used when generating prediction intervals, discussed in Section 5.6. |
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.
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(d) Determine the group structure of the kernel of $\phi$. Problem 612
Let $C[-2\pi, 2\pi]$ be the vector space of all real-valued continuous functions defined on the interval $[-2\pi, 2\pi]$.
Consider the subspace $W=\Span\{\sin^2(x), \cos^2(x)\}$ spanned by functions $\sin^2(x)$ and $\cos^2(x)$. (a) Prove that the set $B=\{\sin^2(x), \cos^2(x)\}$ is a basis for $W$.
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(b) Prove that the set $\{\sin^2(x)-\cos^2(x), 1\}$ is a basis for $W$. Problem 611
An $n\times n$ matrix $A$ is called
orthogonal if $A^{\trans}A=I$. Let $V$ be the vector space of all real $2\times 2$ matrices.
Consider the subset
\[W:=\{A\in V \mid \text{$A$ is an orthogonal matrix}\}.\] Prove or disprove that $W$ is a subspace of $V$. Problem 607
Let $\calP_3$ be the vector space of all polynomials of degree $3$ or less.
Let \[S=\{p_1(x), p_2(x), p_3(x), p_4(x)\},\] where \begin{align*} p_1(x)&=1+3x+2x^2-x^3 & p_2(x)&=x+x^3\\ p_3(x)&=x+x^2-x^3 & p_4(x)&=3+8x+8x^3. \end{align*} (a) Find a basis $Q$ of the span $\Span(S)$ consisting of polynomials in $S$. (b) For each polynomial in $S$ that is not in $Q$, find the coordinate vector with respect to the basis $Q$.
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(The Ohio State University, Linear Algebra Midterm) Read solution Problem 606
Let $V$ be a vector space and $B$ be a basis for $V$.
Let $\mathbf{w}_1, \mathbf{w}_2, \mathbf{w}_3, \mathbf{w}_4, \mathbf{w}_5$ be vectors in $V$. Suppose that $A$ is the matrix whose columns are the coordinate vectors of $\mathbf{w}_1, \mathbf{w}_2, \mathbf{w}_3, \mathbf{w}_4, \mathbf{w}_5$ with respect to the basis $B$.
After applying the elementary row operations to $A$, we obtain the following matrix in reduced row echelon form
\[\begin{bmatrix} 1 & 0 & 2 & 1 & 0 \\ 0 & 1 & 3 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{bmatrix}.\] (a) What is the dimension of $V$? (b) What is the dimension of $\Span\{\mathbf{w}_1, \mathbf{w}_2, \mathbf{w}_3, \mathbf{w}_4, \mathbf{w}_5\}$?
Add to solve later
(The Ohio State University, Linear Algebra Midterm) Read solution Problem 605
Let $T:\R^2 \to \R^3$ be a linear transformation such that
\[T\left(\, \begin{bmatrix} 3 \\ 2 \end{bmatrix} \,\right) =\begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} \text{ and } T\left(\, \begin{bmatrix} 4\\ 3 \end{bmatrix} \,\right) =\begin{bmatrix} 0 \\ -5 \\ 1 \end{bmatrix}.\] (a) Find the matrix representation of $T$ (with respect to the standard basis for $\R^2$). (b) Determine the rank and nullity of $T$.
Add to solve later
(The Ohio State University, Linear Algebra Midterm) Read solution Problem 604
Let
\[A=\begin{bmatrix} 1 & -1 & 0 & 0 \\ 0 &1 & 1 & 1 \\ 1 & -1 & 0 & 0 \\ 0 & 2 & 2 & 2\\ 0 & 0 & 0 & 0 \end{bmatrix}.\] (a) Find a basis for the null space $\calN(A)$. (b) Find a basis of the range $\calR(A)$. (c) Find a basis of the row space for $A$.
Add to solve later
(The Ohio State University, Linear Algebra Midterm) Read solution Problem 603
Let $C[-2\pi, 2\pi]$ be the vector space of all continuous functions defined on the interval $[-2\pi, 2\pi]$.
Consider the functions \[f(x)=\sin^2(x) \text{ and } g(x)=\cos^2(x)\] in $C[-2\pi, 2\pi]$.
Prove or disprove that the functions $f(x)$ and $g(x)$ are linearly independent.
Add to solve later
(The Ohio State University, Linear Algebra Midterm) Read solution Problem 601
Let $V$ be the vector space of all $2\times 2$ matrices whose entries are real numbers.
Let \[W=\left\{\, A\in V \quad \middle | \quad A=\begin{bmatrix} a & b\\ c& -a \end{bmatrix} \text{ for any } a, b, c\in \R \,\right\}.\] (a) Show that $W$ is a subspace of $V$. (b) Find a basis of $W$. (c) Find the dimension of $W$.
Add to solve later
(The Ohio State University, Linear Algebra Midterm) Read solution |
Until recently, all my knowledge of measure theory and Lebesgue integration are from Rudin's book, which focuses solely on the Lebesgue measure, its construction and nothing else. I have just put my hands on a nice book "Measure and Integration Theory" by Heinz Bauer and I'm currently enjoying it. I have encountered the definition of a Dynkin system $\mathcal D$, which is a family of subsets of a set $\Omega$ satisfying
1.) $\Omega\in\mathcal D$.
2.) If $A\in\mathcal D$, then $A^c\in\mathcal D$.
3.) For $n\in\Bbb N$, if $A_n\in\mathcal D$ are pairwise disjoint then $\bigcup_{n=1}^{\infty}A_n\in\mathcal D$.
I have some idea about what a $\sigma$-algebra is, but not about a Dynkin system. I would really appreciate if someone could give me an intuition about Dynkin systems or what they're supposed to represent. What is the characteristics of a Dynkin system that let you recognize it once you see it?
I know the $\pi$-$\lambda$ theorem and facts like a Dynkin system $\mathcal D$ is a $\sigma$-algebra if it is closed under intersection, it would be also nice if anyone could explain to me why should we expect such a result. Thank you in advance. |
How to find the $m^\text{th}$ term for the following expression:
$$ \left.\frac{\partial^m}{\partial s^m}e^{a s^2}\right|_{s=0}$$
Is there any analytical approach?
I computed first few terms and used mathematica "FindSequenceFunction", which yielded the $m^\text{th}$ term as (
edited expression):
$$ \frac{2^{m-1} \left(1+(-1)^{m}\right) a^{\frac{m}{2}} \Gamma \left(\frac{m+1}{2}\right)}{\sqrt{\pi }}$$
The first few terms:
$1,0,2 a,0,12 a^2,0,120 a^3,0,1680 a^4,0,30240 a^5,0,665280 a^6,0,17297280 a^7,0,518918400 a^8,0,17643225600 a^9,0,670442572800 a^{10}$
Any help will be appreciated. |
The complete elliptic integral of the first kind is defined as $$K(k)=\int_0^{\pi/2} \frac{dx}{\sqrt{1-k^2\sin^2{x}}}$$ and the complete elliptic integral of the second kind is defined as $$E(k)=\int_0^{\pi/2}\sqrt{1-k^2\sin^2{x}}~dx$$ for $0\leq k<1$.
I'm supposed to prove the following relation $$K'(k)=\frac{E(k)}{k(1-k^2)}-\frac{K(k)}{k}.$$
What I tried so far
Without much thought about the exchange of integration and differentiation I tried to compute \begin{align}K'(k)&=\int_0^{\pi/2} \frac{k\sin^2{x}}{(1-k^2\sin^2{x})^{3/2}}dx=-\frac{1}{k}\int_0^{\pi/2}\left(\frac{1-k^2\sin^2{x}}{(1-k^2\sin^2{x})^{3/2}}-\frac{1}{(1-k^2\sin^2{x})^{3/2}}\right)dx\\ &=-\frac{K(k)}{k}+\frac{1}{k}\int_0^{\pi/2}\frac{1}{(1-k^2\sin^2{x})^{3/2}} dx.\end{align} Comparing this with the result I'm supposed to obtain it would remain to show $$\int_0^{\pi/2}\frac{1}{(1-k^2\sin^2{x})^{3/2}} dx=\int_0^{\pi/2}\frac{\sqrt{1-k^2\sin^2{x}}}{1-k^2}dx.$$ Some numerical computations suggest that this identity is correct but I have know idea how to show it. Any hints or solutions would be appreciated! |
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 $\phi$. Problem 612
Let $C[-2\pi, 2\pi]$ be the vector space of all real-valued continuous functions defined on the interval $[-2\pi, 2\pi]$.
Consider the subspace $W=\Span\{\sin^2(x), \cos^2(x)\}$ spanned by functions $\sin^2(x)$ and $\cos^2(x)$. (a) Prove that the set $B=\{\sin^2(x), \cos^2(x)\}$ is a basis for $W$.
Add to solve later
(b) Prove that the set $\{\sin^2(x)-\cos^2(x), 1\}$ is a basis for $W$. Problem 611
An $n\times n$ matrix $A$ is called
orthogonal if $A^{\trans}A=I$. Let $V$ be the vector space of all real $2\times 2$ matrices.
Consider the subset
\[W:=\{A\in V \mid \text{$A$ is an orthogonal matrix}\}.\] Prove or disprove that $W$ is a subspace of $V$. Problem 607
Let $\calP_3$ be the vector space of all polynomials of degree $3$ or less.
Let \[S=\{p_1(x), p_2(x), p_3(x), p_4(x)\},\] where \begin{align*} p_1(x)&=1+3x+2x^2-x^3 & p_2(x)&=x+x^3\\ p_3(x)&=x+x^2-x^3 & p_4(x)&=3+8x+8x^3. \end{align*} (a) Find a basis $Q$ of the span $\Span(S)$ consisting of polynomials in $S$. (b) For each polynomial in $S$ that is not in $Q$, find the coordinate vector with respect to the basis $Q$.
Add to solve later
(The Ohio State University, Linear Algebra Midterm) Read solution Problem 606
Let $V$ be a vector space and $B$ be a basis for $V$.
Let $\mathbf{w}_1, \mathbf{w}_2, \mathbf{w}_3, \mathbf{w}_4, \mathbf{w}_5$ be vectors in $V$. Suppose that $A$ is the matrix whose columns are the coordinate vectors of $\mathbf{w}_1, \mathbf{w}_2, \mathbf{w}_3, \mathbf{w}_4, \mathbf{w}_5$ with respect to the basis $B$.
After applying the elementary row operations to $A$, we obtain the following matrix in reduced row echelon form
\[\begin{bmatrix} 1 & 0 & 2 & 1 & 0 \\ 0 & 1 & 3 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{bmatrix}.\] (a) What is the dimension of $V$? (b) What is the dimension of $\Span\{\mathbf{w}_1, \mathbf{w}_2, \mathbf{w}_3, \mathbf{w}_4, \mathbf{w}_5\}$?
Add to solve later
(The Ohio State University, Linear Algebra Midterm) Read solution Problem 605
Let $T:\R^2 \to \R^3$ be a linear transformation such that
\[T\left(\, \begin{bmatrix} 3 \\ 2 \end{bmatrix} \,\right) =\begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} \text{ and } T\left(\, \begin{bmatrix} 4\\ 3 \end{bmatrix} \,\right) =\begin{bmatrix} 0 \\ -5 \\ 1 \end{bmatrix}.\] (a) Find the matrix representation of $T$ (with respect to the standard basis for $\R^2$). (b) Determine the rank and nullity of $T$.
Add to solve later
(The Ohio State University, Linear Algebra Midterm) Read solution Problem 604
Let
\[A=\begin{bmatrix} 1 & -1 & 0 & 0 \\ 0 &1 & 1 & 1 \\ 1 & -1 & 0 & 0 \\ 0 & 2 & 2 & 2\\ 0 & 0 & 0 & 0 \end{bmatrix}.\] (a) Find a basis for the null space $\calN(A)$. (b) Find a basis of the range $\calR(A)$. (c) Find a basis of the row space for $A$.
Add to solve later
(The Ohio State University, Linear Algebra Midterm) Read solution Problem 603
Let $C[-2\pi, 2\pi]$ be the vector space of all continuous functions defined on the interval $[-2\pi, 2\pi]$.
Consider the functions \[f(x)=\sin^2(x) \text{ and } g(x)=\cos^2(x)\] in $C[-2\pi, 2\pi]$.
Prove or disprove that the functions $f(x)$ and $g(x)$ are linearly independent.
Add to solve later
(The Ohio State University, Linear Algebra Midterm) Read solution Problem 601
Let $V$ be the vector space of all $2\times 2$ matrices whose entries are real numbers.
Let \[W=\left\{\, A\in V \quad \middle | \quad A=\begin{bmatrix} a & b\\ c& -a \end{bmatrix} \text{ for any } a, b, c\in \R \,\right\}.\] (a) Show that $W$ is a subspace of $V$. (b) Find a basis of $W$. (c) Find the dimension of $W$.
Add to solve later
(The Ohio State University, Linear Algebra Midterm) Read solution |
Tagged: subspace Problem 709
Let $S=\{\mathbf{v}_{1},\mathbf{v}_{2},\mathbf{v}_{3},\mathbf{v}_{4},\mathbf{v}_{5}\}$ where
\[ \mathbf{v}_{1}= \begin{bmatrix} 1 \\ 2 \\ 2 \\ -1 \end{bmatrix} ,\;\mathbf{v}_{2}= \begin{bmatrix} 1 \\ 3 \\ 1 \\ 1 \end{bmatrix} ,\;\mathbf{v}_{3}= \begin{bmatrix} 1 \\ 5 \\ -1 \\ 5 \end{bmatrix} ,\;\mathbf{v}_{4}= \begin{bmatrix} 1 \\ 1 \\ 4 \\ -1 \end{bmatrix} ,\;\mathbf{v}_{5}= \begin{bmatrix} 2 \\ 7 \\ 0 \\ 2 \end{bmatrix} .\] Find a basis for the span $\Span(S)$. Problem 706
Suppose that a set of vectors $S_1=\{\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3\}$ is a spanning set of a subspace $V$ in $\R^5$. If $\mathbf{v}_4$ is another vector in $V$, then is the set
\[S_2=\{\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3, \mathbf{v}_4\}\] still a spanning set for $V$? If so, prove it. Otherwise, give a counterexample. Problem 663
Let $\R^2$ be the $x$-$y$-plane. Then $\R^2$ is a vector space. A line $\ell \subset \mathbb{R}^2$ with slope $m$ and $y$-intercept $b$ is defined by
\[ \ell = \{ (x, y) \in \mathbb{R}^2 \mid y = mx + b \} .\]
Prove that $\ell$ is a subspace of $\mathbb{R}^2$ if and only if $b = 0$.Add to solve later
Problem 659
Fix the row vector $\mathbf{b} = \begin{bmatrix} -1 & 3 & -1 \end{bmatrix}$, and let $\R^3$ be the vector space of $3 \times 1$ column vectors. Define
\[W = \{ \mathbf{v} \in \R^3 \mid \mathbf{b} \mathbf{v} = 0 \}.\] Prove that $W$ is a vector subspace of $\R^3$. Problem 658
Let $V$ be the vector space of $n \times n$ matrices with real coefficients, and define
\[ W = \{ \mathbf{v} \in V \mid \mathbf{v} \mathbf{w} = \mathbf{w} \mathbf{v} \mbox{ for all } \mathbf{w} \in V \}.\] The set $W$ is called the center of $V$.
Prove that $W$ is a subspace of $V$.Add to solve later
Problem 612
Let $C[-2\pi, 2\pi]$ be the vector space of all real-valued continuous functions defined on the interval $[-2\pi, 2\pi]$.
Consider the subspace $W=\Span\{\sin^2(x), \cos^2(x)\}$ spanned by functions $\sin^2(x)$ and $\cos^2(x)$. (a) Prove that the set $B=\{\sin^2(x), \cos^2(x)\}$ is a basis for $W$.
Add to solve later
(b) Prove that the set $\{\sin^2(x)-\cos^2(x), 1\}$ is a basis for $W$. Problem 611
An $n\times n$ matrix $A$ is called
orthogonal if $A^{\trans}A=I$. Let $V$ be the vector space of all real $2\times 2$ matrices.
Consider the subset
\[W:=\{A\in V \mid \text{$A$ is an orthogonal matrix}\}.\] Prove or disprove that $W$ is a subspace of $V$. Problem 604
Let
\[A=\begin{bmatrix} 1 & -1 & 0 & 0 \\ 0 &1 & 1 & 1 \\ 1 & -1 & 0 & 0 \\ 0 & 2 & 2 & 2\\ 0 & 0 & 0 & 0 \end{bmatrix}.\] (a) Find a basis for the null space $\calN(A)$. (b) Find a basis of the range $\calR(A)$. (c) Find a basis of the row space for $A$.
Add to solve later
(The Ohio State University, Linear Algebra Midterm) Read solution Problem 601
Let $V$ be the vector space of all $2\times 2$ matrices whose entries are real numbers.
Let \[W=\left\{\, A\in V \quad \middle | \quad A=\begin{bmatrix} a & b\\ c& -a \end{bmatrix} \text{ for any } a, b, c\in \R \,\right\}.\] (a) Show that $W$ is a subspace of $V$. (b) Find a basis of $W$. (c) Find the dimension of $W$.
Add to solve later
(The Ohio State University, Linear Algebra Midterm) Read solution |
Actually, a more direct answer to the original question explains why there is nothing like Kähler potential theory in the hypercomplex and quaternionic cases.
Recall what a Kähler potential does: You start with a complex $n$-manifold, i.e., a $2n$-manifold $M^{2n}$ endowed with a torsion-free $\mathrm{GL}(n,\mathbb{C})$-structure $B$, represented by an integrable complex structure tensor $J:TM\to TM$. Now, $\mathrm{U}(n)$ is a maximal compact in $\mathrm{GL}(n,\mathbb{C})$, and we can select a torsion-free $\mathrm{U}(n)$-structure $B'\subset B$ over an open set $U\subset M$ by choosing a smooth function $u\in C^\infty(U)$ such that $\omega = i\partial\bar\partial u$ is a positive $(1,1)$-form. Locally, this is how all of the torsion-free (i.e., Kähler) structures subordinate to the $\mathrm{GL}(n,\mathbb{C})$-structure $B$ are described. Of course, every torsion-free $\mathrm{U}(n)$-structure $B'$ on $M$ underlies a unique torsion-free $\mathrm{GL}(n,\mathbb{C})$-structure $B = B'{\cdot}\mathrm{GL}(n,\mathbb{C})$. Now, the nice thing in this case is that, modulo diffeomorphism, locally at least, integrable $\mathrm{GL}(n,\mathbb{C})$-structures are all the same, i.e., they have no local geometry, whereas the torsion-free $\mathrm{U}(n)$-structures depend locally on $1$ function of $2n$-variables (which is represented exactly by the Kähler potential as above).
Consider what you would need to be true in either the hypercomplex case or the quaternionic case in order to have such an analogue:
First, consider the hypercomplex case: Now, the group is $\mathrm{GL}(n,\mathbb{H})\subset \mathrm{GL}(4n,\mathbb{R})$, with maximal compact $\mathrm{Sp}(n)\subset \mathrm{GL}(n,\mathbb{H})$. Thus, on a $4n$-manifold $M^{4n}$ you'd want a way to start with a torsion-free $\mathrm{GL}(n,\mathbb{H})$-structure $B$ on $M$ (i.e., a hypercomplex structure) and specify the torsion-free $\mathrm{Sp}(n)$-structures (i.e., hyperKähler structures) that it contains. As in the complex/Kähler case, the reverse operation is automatic: Each torsion-free $\mathrm{Sp}(n)$-structure $B'$ lies in a unique torsion-free $\mathrm{GL}(n,\mathbb{H})$-structure $B= B'{\cdot}\mathrm{GL}(n,\mathbb{H})$. However, you immediately see the problem: Modulo diffeomorphism, the torsion-free $\mathrm{GL}(n,\mathbb{H})$-structures in dimension $4n$ depend on $4n^2$ functions of $2n{+}1$ variables, while the the torsion-free $\mathrm{Sp}(n)$-structures in dimension $4n$ depend on only $2n$ functions of $2n{+}1$ variables, so the 'generic' torsion-free $\mathrm{GL}(n,\mathbb{H})$-structure in this dimension does not contain
any torsion-free $\mathrm{Sp}(n)$-structures, so there can't be a potential theory like the complex/Kähler potential theory in this case. You might think that you could get around this by considering the underlying $\mathrm{SL}(n,\mathbb{H})$-structures instead, but, modulo diffeomorphism, the torsion-free $\mathrm{SL}(n,\mathbb{H})$-structures in dimension $4n$ depend on $4n^2{-}2n$ functions of $2n{+}1$ variables, which is still too high (except when $n=1$, when $\mathrm{SL}(1,\mathbb{H}) = \mathrm{Sp}(1)$, so there is nothing to do). (This is why they punt and go for HKT structures instead in this case.)
Second, in the quaternionic case, the group is the group is $\mathrm{GL}(n,\mathbb{H}){\cdot}\mathrm{Sp}(1)\subset \mathrm{GL}(4n,\mathbb{R})$, with maximal compact $\mathrm{Sp}(n){\cdot}\mathrm{Sp}(1)\subset \mathrm{GL}(n,\mathbb{H}){\cdot}\mathrm{Sp}(1)$. Thus, on a $4n$-manifold $M^{4n}$ you'd want a way to start with a torsion-free $\mathrm{GL}(n,\mathbb{H}){\cdot}\mathrm{Sp}(1)$-structure $B$ on $M$ (i.e., a quaternionic structure) and specify the torsion-free $\mathrm{Sp}(n){\cdot}\mathrm{Sp}(1)$-structures (i.e., quaternion-Kähler structures) that it contains. As in the complex/Kähler case, the reverse operation is automatic: Each torsion-free $\mathrm{Sp}(n){\cdot}\mathrm{Sp}(1)$-structure $B'$ lies in a unique torsion-free $\mathrm{GL}(n,\mathbb{H}){\cdot}\mathrm{Sp}(1)$-structure $B= B'{\cdot}\mathrm{GL}(n,\mathbb{H}){\cdot}\mathrm{Sp}(1)$. However, again, there's a problem: Modulo diffeomorphism, the torsion-free $\mathrm{GL}(n,\mathbb{H}){\cdot}\mathrm{Sp}(1)$-structures in dimension $4n$ depend on $2n(2n{+}1)$ functions of $2n{+}1$ variables, while the the torsion-free $\mathrm{Sp}(n){\cdot}\mathrm{Sp}(1)$-structures in dimension $4n$ depend on only $2n$ functions of $2n{+}1$ variables, so, again, the 'generic' torsion-free $\mathrm{GL}(n,\mathbb{H}){\cdot}\mathrm{Sp}(1)$-structure in this dimension does not contain
any torsion-free $\mathrm{Sp}(n){\cdot}\mathrm{Sp}(1)$-structures, so, again, there can't be a potential theory like the complex/Kähler potential theory in this case.
For these function counts for the generality of torsion-free structures modulo diffeomorphism, in case you are curious, you might consult my paper: R. Bryant -
Classical, exceptional, and exotic holonomies: a status report, in Actes de la Table Ronde de Géométrie Différentielle en l’Honneur de Marcel Berger, Soc. Math. France, 1996, 93–166. |
> > MathModePlugin <--
Contributions to this plugin are appreciated. Please update the plugin page at
http://twiki.org/cgi-bin/view/Plugins/MathModePlugin or provide feedback at
http://twiki.org/cgi-bin/view/Plugins/MathModePluginDev.
If you are a TWiki contributor please update the plugin in the SVN repository.
-->Add math formulas to TWiki topics using LaTeX markup language Description This plugin allows you to include mathematics in a TWiki page, with a format very similar to LaTeX. The external program
latex2html is used to generate
gif (or
png) images from the math markup, and the image is then included in the page. The first time a particular expression is rendered, you will notice a lag as
latex2html is being run on the server. Once rendered, the image is saved as an attached file for the page, so subsequent viewings will not require re-renders. When you remove a math expression from a page, its image is deleted. Note that this plugin is called MathModePlugin, not LaTeXPlugin, because the only piece of LaTeX implemented is rendering of images of mathematics.
Syntax Rules <latex [attr="value"]* > generates an image from the contained
formula </latex>
formula. In addition attribute-value pairs may be specified that are passed to the resulting
img html tag. The only exeptions are the following attributes which take effect in the latex rendering pipeline:
The formula will be displayed using a
size: the latex font size; possible values are tiny, scriptsize, footnotesize, small, normalsize, large, Large, LARGE, huge or Huge; defaults to %LATEXFONTSIZE%
color: the foreground color of the formula; defaults to %LATEXFGCOLOR%
bgcolor: the background color; defaults to %LATEXBGCOLOR%
math latex environment by default. If the formula contains a latex linebreak (
\\) then a
multline environment of amsmath is used instead. If the formula contains an alignment sequence (
& = &) then an
eqnarray environment is used.Note that the old notation using
%$formula$% and
%\[formula\]% is still supported but are deprecated.If you might want to recompute the images cached for the current page then append
?refresh=on to its url, e.g. click
here to refresh the formulas in the examples below. Examples The following will only display correctly if this plugin is installed and configured correctly.
<latex title="this is an example">
\int_{-\infty}^\infty e^{-\alpha x^2} dx = \sqrt{\frac{\pi}{\alpha}}
</latex>
<latex>
{\cal P} & = & \{f_1, f_2, \ldots, f_m\} \\
{\cal C} & = & \{c_1, c_2, \ldots, c_m\} \\
{\cal N} & = & \{n_1, n_2, \ldots, n_m\}
</latex>
<latex title="Calligraphics" color="orange">
\cal
A, B, C, D, E, F, G, H, I, J, K, L, M, \\
\cal
N, O, P, Q, R, S, T, U, V, W, X, Y, Z
</latex>
<latex>
\sum_{i_1, i_2, \ldots, i_n} \pi * i + \sigma
</latex>
This is new inline test.
Greek letters
\alpha
\theta
\beta
\iota
\gamma
\kappa
\delta
\lambda
\epsilon
\mu
\zeta
\nu
\eta
\xi
Plugin Installation Instructions Download the ZIP file Unzip
in your twiki installation directory. Content:
MathModePlugin.zip
File: <-- --> Description:
data/TWiki/MathModePlugin.txt
lib/TWiki/Plugins/MathModePlugin.pm
lib/TWiki/Plugins/MathModePlugin/Core.pm
pub/TWiki/MathModePlugin/latex2img This plugin makes use of three additional tools that are used to convert latex formulas to images. These are Make sure they are installed and check the paths to the programs
latex,
dvipng and
convert in the latex2img shiped with this plugin
Edit the file
<path-to-twiki>/pub/TWiki/MathModePlugin/latex2img accordingly and set execute permission for your webserver on it
Visit
configure in your TWiki installation, and enable the plugin in the {Plugins} section.
Troubleshooting If you get error like
"fmtutil: [some-dir]/latex.fmt does not exist", run
fmtutil-sys --all on your server to recreate all latex formatstyles.
If your generated image of the latex formula does not show up, then you probably have encoding issues. Look into the source of the <img>-tag in your page's source code. Non-ASCII characters in file names might cause troubles. Check the localization in the TWiki configure page. Configuration There are a set of configuration variables that an be set in different places. All of the below variables can be set in your
LocalSite.cfg file like this:
$TWiki::cfg{MathModePlugin}{<Name>} = <value>;
Some of the below variables can only be set this way, some of the may be overridden by defining the respective prefrence variable.
Name Preference Variable Default
HashCodeLength
32 length of the hash code. If you switch to a different hash function, you will likely have to change this
ImagePrefix
'_MathModePlugin_' string to be prepended to any auto-generated image
ImageType
%LATEXIMAGETYPE% 'png' extension of the image type; possible values are 'gif' and 'png'
Latex2Img
'.../TWiki/MathModePlugin/latex2img' the script to convert a latex formula to an image
LatexPreamble
%LATEXPREAMBLE% '\usepackage{latexsym}' latex preamble to include additional packages (e.g. \usepackage{mathptmx} to change the math font) ; note, that the packages
amsmath and
color are loaded too as they are obligatory
ScaleFactor
%LATEXSCALEFACTOR% 1.2 factor to scale images
LatexFGColor
%LATEXFGCOLOR% black default text color
LatexBGColor
%LATEXBGCOLOR% white default background color
LatexFontSize
%LATEXFONTSIZE% normalsize default font size Plugin Info <-- provided for compatibility only --> Set SHORTDESCRIPTION = Add math formulas to TWiki topics using LaTeX markup language
Plugin Author: TWiki:Main.GraemeLufkin, TWiki:Main.MichaelDaum Copyright: © 2002 Graeme Lufkin gwl@u.washington.edu; © 2006-2007 Michael Daum http://wikiring.de; © 2007-2011 TWiki:TWiki.TWikiContributor License: GPL (GNU General Public License) Plugin Version: 2.98 Change History: <-- versions below in reverse order --> 2012-12-10 TWikibug:Item7071: MathModePlugin doesn't support multiple disks. 2011-10-17 TWikibug:Item6799: Add troubleshoot info -- TWiki:Main.MatthiasRella 2010-04-25 TWikibug:Item6433: Doc improvements -- TWiki:Main.PeterThoeny 07 Dec 2007: replaced
templfile with
mktemp in the latex2img helper script
13 Nov 2007: fixed plugin on 4.2 18 Dec 2006: only use one bgcolor 02 Oct 2006: don't fail on hierarchical webs; backwards compatible tempfile cleanup 31 Aug 2006: added NO_PREFS_IN_TOPIC; using
xcolor instead of
color latex package now to be able to specify colors in html typical codes; default preamble uses latexsym now
07 Aug 2006: switched from latex2html to latex+dvipng+convert; added
size,
color,
bgcolor to <latex> tag; rendering pngs by default now; reworked plugin settings; added a latex2img shell script; returning full latex error report
04 Aug 2006: major rewrite; fixed security issues by using TWiki's sandbox feature and creating tempfiles properly; added new <latex>...</latex> tag to support multiline formulas; better configurability; better error reporting; fixed issues where images have not been cleaned up regularly; speedup don't clean orphaned images during
view but during
save; speedup by adding lazy compilation and initialization; implemented a
postRenderingHandler for TWiki V4; prevent auto-generated images stored in pub from being auto-attached using TWiki V4
03 Apr 2002: Initial version CPAN Dependencies: none Other Dependencies: LaTeX, dvipng, ImageMagick Perl Version: 5.8 TWiki:Plugins/Benchmark: GoodStyle nn%, FormattedSearch nn%, MathModePlugin nn% Plugin Home: TWiki:Plugins/MathModePlugin Feedback: TWiki:Plugins/MathModePluginDev Appraisal: TWiki:Plugins/MathModePluginAppraisal
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Hi Guys, I just want to say that JOMO 5 has started! Please reshare and tell your friends about it. You can check out the paper here. Please read the rules before starting
Note by Yan Yau Cheng 5 years, 4 months ago
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Great problems and solutions in JOMO5! Looking forward to the next one.
Log in to reply
@Bogdan Simeonov @Sharky Kesa After Triple-checking, there IS a problem in question 2, we are terribly sorry and we have edited the question.@Finn Hulse You might want to resubmit for question 2 as well since you already submitted
Could you please tell me what the edit to the question is? I am not able to access the new file.
Oh yes I did! Thanks. :D
I cannot find any non-negative solutions to Qn 2.Is there an issue in the problem?
Happened to me as well.
We are sorry Bogdan and all others for the inconvenience but there IS a problem in question 2 .... We have edited the file and soon , the correct version will be uploaded on the website ..... Apologies for your time that it consumed unnecessarily .....
Why is JOMO and Proofathon always together? Can you please colobrate time so that I can participate in both.
I think this one is pretty easy ,, Nice job, All the best for all participants
The proof problems were a bit strange. No casework, no induction, not really much of anything! :P
Really? Last weeks for me was the easiest, this one for me was moderately tough.
I am a problem writer at JOMO but i also feel that this one is easier than previous JOMO (of course feeling easy/difficult may differ among people) .... Go on , all the best ! Progress expected this time .......
I didnt attempt any of the contest. If u are saying of the previous combinatorics contest, then I admit my worst prepared topic is combinatorics, while my strongest topic is inequalities.. But, previous contest was also easy, but I didn't get a few. As for this contest, I got all ques but one till now.. R u taking in it? all the best then
Arrgghh... I cant do this, the problems arent bad but it just takes too much to convert everything to PDF 's and everything, i barely have that much time. sorry!
Don't worry for that now , we'll now allow word documents .... You may submit it as microsoft office Word document too !
Dude I finished in 40 minutes. :P
For question 2, I can't find any non-negative integer solutions for xxx, yyy and zzz. Did yours have any negative numbers?
@Sharky Kesa – We are sorry Sharky and all others for the inconvenience but there IS a problem in question 2 .... We have edited the file and soon , the correct version will be uploaded on the website ..... Apologies for your time that it consumed unnecessarily .....
@Sharky Kesa – there are no problems to question 2, I double checked
@Yan Yau Cheng – I got a non-integral answer and I'm sure it's right...
@Finn Hulse – It asked for nonnegative integers. I got two solutions for the equation but both of them had negative integers.
@Sharky Kesa – Yeah I know.
@Yan Yau Cheng – I'll check again.
I've told all of my friends! This is going to be really fun but @Yan Yau Cheng could you go a little easy on the trig? After all, this is Junior. I just finished. :D
@Aditya Raut
What ? I think the correct file(ques 2 problem fixed) is now uploaded on the website ..... All participants , we're sorry for that inconvenience , really!!
@Aditya Raut – I am not able to access the file. Could you please post the difference in the question?
@Nanayaranaraknas Vahdam – The 1st equation was previously xy+2x+3y=2 but it was supposed to be xy+3x+2y=2 so we've edited it now
@Yan Yau Cheng – Thank you!
totally agreed, especially since I just learnt basic trig only .-.
Haha, yeah but I'm in Algebra 1 so... sucks for me. :P
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What is transverse energy? Why we use transverse total energy instead of energy and transverse momentum in place of Total momentum in the particle detectors?
Transverse momentum, $\vec{p}_T$, is the momentum of an object transverse to the beam. Transverse energy is defined as $E_T = \sqrt{m^2+p_T^2}$ for an object with mass $m$ and transverse momentum $p_T$.
The initial longitudinal momentum in a parton collision is unknown, because the partons that make up a proton share the momentum. We do know, however, that the initial transverse momentum was zero. So we look for missing transverse momentum, defined $E_T^\textrm{miss} = -\sum_i \vec{p}_T(i)$ for visible particles $i$. Finding missing transverse momentum would indicate that new, unaccounted for particle(s) had escaped the detector.
Confusingly, $E_T^\textrm{miss} = -\sum_i \vec{p}_T(i)$ is commonly called missing transverse energy or MET. Missing transverse energy is equivalent to missing transverse momentum only if the missing particle(s) were massless.
Also, events in which the products have large transverse momentum are more likely to be genuine, interesting events. |
Science Advisor
Homework Helper
2,559 3 The elements of [itex]SL_2(\mathbb{Z})[/itex], the group of 2 x 2 matrices which have integer entries and determinant +1, act on the upper half of the complex plane as Möbius transformations. Let [itex]\alpha[/itex] denote the arc of the unite circle which joins [itex]\exp (i\pi /3)[/itex] to [itex]\exp (i\pi /2)[/itex] in the upper half plane and define [itex]\Gamma[/itex] to be the union of all segments [itex]g(\alpha )[/itex] where [itex]g \in SL_2 (\mathbb{Z})[/itex]. Draw a picture of [itex]\Gamma[/itex] and check that it represents a tree. Show that the action of [itex]SL_2(\mathbb{Z})[/itex] on this tree is not a free action.
The Möbius transformation induced by the 2 x 2 matrix with entries a, b, c, and d will send the complex number z to
(az + b)/(cz + d) in
CU {∞}.
Is there any easy way to see what the orbit is? If we treat the plane as
R² in the natural way, the point (x,y) is sent to:
(a + c + bd + (ad + bc)x, y)/(c² + 2cdx + d²)
= (a + c + bd + (ad + bc)x, y)/(c² + 2cdx + d²)
This doesn't seem to help too much. Does the fact that the determinant is 1 tell me anything special, in terms of preserving the shape in some way? Anyways, assuming that we can tell what the image of the original arc will be if we can determine the image of it's endpoints, we have two endpoints, (0,1) and (sqrt(3)/2, 0.5). In the first case, we get:
(a + c + bd, 1)/(c² + d²)
Knowing that ad - bc = 1, does that place any nice restrictions on a + c + bd? I don't even want to touch the other case right now. |
My question is a very basic one. It seems feasible to believe that $\mathsf{P = NP}$, because there is some "pathological" good algorithm for SAT, yet it is impossible to prove that the algorithm is actually correct and/or that it runs in polynomial time, in a standard axiomatic system like ZFC.
Let's fix a logic $L$ that is strong enough to encode statements about Turing machines. (Edit: I should have clarified that by this I mean the same requirements as Godel's second incompleteness theorem. For the purposes of this question, assume that
$L$ proves the peano axioms PA.)
Then define \begin{align*} \mathsf{ProvableP} &:= \{A \mid L \text{ proves } [A \in P]\} \\ \mathsf{ProvableNP} &:= \{A \mid L \text{ proves } [A \in NP] \} \end{align*}
Now we have that $\mathsf{ProvableP} \subseteq \mathsf{ProvableNP} \subseteq \mathsf{NP}$. But does $\mathsf{ProvableP} = \mathsf{ProvableNP}$? How does this relate to the original P vs NP question?
Partial answer: The SAT problem is $\mathsf{ProvableNP}$-complete, but $\mathsf{ProvableNP}$ is closed only under provably polynomial-time reductions, not necessarily all polynomial-time reductions. However, this does seem to show that if $\mathsf{ProvableP} = \mathsf{ProvableNP}$, then since SAT is in $\mathsf{ProvableNP}$, $\mathsf{P} = \mathsf{NP}$.
So what is remaining is the converse: if $\mathsf{P} = \mathsf{NP}$, does $\mathsf{ProvableP} = \mathsf{ProvableNP}$?
I picked the P vs. NP question because it's the "canonical" unsolved problem in complexity. But, a similar question could be asked about any two complexity classes. |
Importance of differential forms is obvious to any geometer and some analysts dealing with manifolds, partly because so many results in modern geometry and related areas cannot even be formulated without them: for example if you want to learn the definition of symplectic manifold, you must first learn what is differential form.
However I have heard another opinion from people with non-geometric background (say, functional analysis, probability) that differential forms are probably used only in problems where they already appear in basic definitions (like symplectic geometry). When I taught an undergraduate course "Analysis on manifold" I had a feeling that my students might have got the same impression at the end: in the second half of the semester I developed calculus of differential forms with rather few concrete applications. Personally I am completely convinced in usefulness of differential forms, but find it not so easy to explain this to people who are not used to them.
I would like to make a list of concrete applications of differential forms. Also it would be interesting to know historical reasons to introduce them. To start the list, below are some examples which come to my mind.
1) The general Stokes formula $\int_M d\omega=\pm \int_{\partial M}\omega$ generalizes all classical formulas of Green, Gauss, Stokes.
2) The Haar measure on a Lie group can be easily constructed as a translation invariant top differential form. (Some experts often prefer to consider this as a special case of a more general and more difficult fact of existence of Haar measure on locally compact groups. In that generality there is no such a simple and relatively explicit description of Haar measure as there is no language similar to differential forms.)
3) The cohomology of the de Rham complex of differential forms on a manifold is canonically isomorphic to singular cohomology (with real coefficients) of the manifold. One use of this isomorphism is that even Betti numbers of a symplectic manifold are non-zero. Another non-trivial use of this fact is the Hodge decomposition of the cohomology (with complex coefficients) of a compact Kahler manifold which makes it, in particular, a bi-graded algebra (rather than just graded) and provides new information on the Betti numbers, say $\beta_{2k+1}$ must be even. |
Amrit bought an article for Rs. 2,200 and sold it for Rs. 2,500. Here, his selling price is greater than the cost price. Hence, he got a profit of Rs. 2,500 - Rs. 2,200 = Rs. 300. If he had sold the article for Rs. 2000, he would have a loss of Rs. 2,200 - Rs. 2,000 = Rs. 200. The price for which an article is bought is known as the cost price (C.P.). The price for which it is sold is known as selling price (S.P.). If the selling price is greater than cost price, there is profit or gain. On the other hand, if the selling price is less than the cost price, there is a loss.
So, Profit = Selling price (S.P) - Cost price (C.P)
P = SP - CP and Loss =Cost price (C.P) -Selling price (S.P) L = CP - SP
The percentage profit or loss can be calculated using the following formula.
Actual profit = profit% of cost price
\(\text {Profit percentage} = \frac {Profit} {C.P}\times 100\)
Actual loss = loss% of Cost price.
\(\text {Loss percentage} = \frac {Loss} {C.P}\times 100\)
If S.P and profit or loss percent are given then
\(C.P = \frac {S.P \times 100} {100 + P\%} \: {or}\: C.P = \frac {S.P \times 100} {100 - L\%}\)
If C.P. and profit or loss percentage are given then
\(S.P = \frac {C.P \times (100 + P\%)} {100}\: {or}\: S.P = \frac {C.P \times (100 - L\%)} {100}\)
The seller may deduct a certain amount from the price of goods. The deduction is known as discount. The price from which the discount is deducted is called the marked price or labeled price. The price obtained by deducting the discount from marked price is called selling price
i.e. Selling price (S.P) = Market price (M.P) - Discount S.P = M.P - D or, M.P = S.P + D or, D = M.P - S.P
If there is no discount, selling price = marked price [ S.P = M.P ]
\(\text {Discount percentage} = \frac {Discount} {M.P} ×100%\)
Value Added Tax is a tax imposed by the government based on goods and services in each step of production and distribution. VAT is levied in the amount after allowing the discount (if there is) from the market price. In general, VAT is expressed in terms percentage which is called the rate of the VAT and it is fixed by the government. The cost of goods is determined by adding the VAT.
S.P = Orginal cost + VAT
\(\text {Rate of VAT} = \frac {VAT \;Amount} {Cost \; after \; discount (S.P)} \times 100\%\)
VAT amount = Rate of VAT (in%) \(\times\) discounted price.
Solution:
Profit% = 12 %
Profit = Rs 60 Selling price (SP) =?
If profit Rs 12 then SP is Rs 112
If profit Rs 1 then SP is Rs \(\frac{112}{12}\) If profit Rs 60 then SP is \(\frac{112}{12} \times 60 = Rs \: 560\)
\(\therefore SP\) = Rs 560
Ans.
Solution:
Cost price (CP) = Rs 4500
Profit % (P) = 30% Selling price (SP) =?
\begin{align*} SP &= \left( \frac{100 + P%}{100} \right) \times CP \\ &= \frac{100 + 30}{100} \times 4500 \\ &= \frac{4500 \times 130}{100}\\ &= Rs \: 5850\end{align*}
\(\therefore SP\) = 5850
Ans.
Solution:
Cost price (CP) = Rs 3405.50
Gain (G) = Rs 120 Selling price (SP) = ?
We know that,
\begin{align*} SP &= CP + profit \\ &= Rs 3405.0 + Rs 120 \\ &= Rs 3525.50_{ANS.} \end{align*}
Solution:
Cost price (CP) = Rs 220,000 + Rs 83500 = Rs 303,500
Selling price (SP) = Rs 300,000
Loss% =?
\begin{align*} Loss\% &= \frac {CP -SP} {CP} \times 100\% \\ &= \frac{303,500 - 300,000}{303500} \times 100\% \\ &= 1.15\% \end{align*}
\(\therefore \) Loss = 1.15%
Ans.
Solution:
Selling price (SP) = Rs 2700
Loss% = 10%
\begin{align*} Cost \: price (CP) &= \frac{SP \times 100}{100 - L\%} \\ &= \frac {2700 \times 100}{100 - 10 }\\ &= \frac{270000}{90} \\ &= Rs \: 3000 \end{align*}
Again,
CP = Rs 3000
Profit % = 7.5% SP = ?
\begin{align*} SP &= \frac{1100 + P\%}{100 } \times CP \\ &= \frac{100 + 7.5\times 3000}{100}\\ &= Rs \: 3225 \end{align*}
\( \therefore \) selling price = 3325
Ans
Solution:
The price of doll before discount = Rs 180
The price of doll after discount = Rs 160 Amount of discount = Rs 180 - Rs 160 = Rs 20
\begin{align*} Discount \% &= \frac{Amount \: of \: discount }{Initial \: price} \times 100\% \\ &= \frac{20}{180} \times 100\% \\ &= 11.11\% _{Ans.} \end{align*}
Solution:
Selling price (SP) = Rs. 164
Loss = 18%
\begin{align*} Cost \: price (C.P.) &= \frac{S.P. \times 100}{100 - Loss \%} \\ &= \frac{164 \times 100}{100 - 18}\\ &= \frac{16400}{82}\\ &= Rs. 200 _{Ans}\end{align*}
Solution:
Market price (MP) = Rs. 1000
Discount % = 10%
\begin{align*}Payment\: amount &= MP - discount\% of MP \\ &= Rs. \: 1000 - \frac{10}{100}\times 1000\\ &= Rs. 1000 - Rs. 100 \\ &= Rs. 900_{Ans} \end{align*}
Solution:
Marked price (MP) = Rs. 150
Selling price after discount (SP) = Rs. 130
\begin{align*} Discount\% &= \frac{MP - SP}{MP} \times 100\% \\&= \frac {150 - 130}{150} \times 100\% \\ &= 13\frac{1}{3} \% \: \: _{Ans}\end{align*}
Solution:
Marked price (P) = Rs 2700
VAT = 13%
\begin{align*} Selling \: price \: (SP) &= MP + VAT\% of MP\\ &= 2700 + \frac{13}{100} \times 2700 \\ &= 2700 + 351 \\ &= Rs. \: 3051 \: \: \: _{Ans.} \end{align*}
Solution:
Let, cost price of calculator (CP
1) = Rs x
Cost price of the watch (CP
2) = RS (4000 - x)
\begin{align*} SP \: of \: calculator \: (SP_1) &= CP + profit \\ &= x + x \: of \: 10\% \\ &= x + x \times \frac{10}{100}\\ &= \frac{11x}{10}\end{align*}
\begin{align*}SP \: of\: watch \: (SP_2) &=CP - loss\\ &= (4000 - x) -20\% \: of \: (4000 + x)\\ &= (4000 -x) - \frac{20}{100} \times (4000 - x)\\ &=\frac{32000 - 5x -4000 + x}{5}\\ &= 3200 - \frac{4x}{5} \end{align*}
\begin{align*} Total \: SP &= SP_1 + SP_2 \\ &= \frac{11x}{10} + 3200 - \frac{4x}{5}\\ &= \frac{3x}{10} + 3200 \end{align*}
Total CP = 4000
Profit = 1%
\begin{align*}SP &= CP + Profit\\ or, \frac{3x}{10} + 3200 &= 4000 + 1\% of 4000\\ or, \frac{3x}{10} + 3200 &= 4000 + \frac{1}{100} \times 4000\\ or, \frac{3x}{10} &= 4000 + 40 - 3200\\ x &= 840 \times \frac{10}{3}\\ &= Rs \: 2800 \end{align*}
\begin{align*}\text{CP of watch = Rs} \: 4000 -x \\ &= 4000 - 2800 \\ &= 1200 \end{align*}
\( \therefore \) CP of calculator = Rs 2800
\(\therefore\) CP of watch = Rs 1200 Ans.
Solution:
Marked price (MP) = Rs 1350
Selling Price (SP) = Rs 1282.50
\begin{align*} Discount &= MP -SP \\&= 1350 - 1282.50 \\ &= Rs \: 67.50 \end{align*}
\begin{align*} Discount\% &= \frac{Discount}{MP} \times 100\% \\ &= \frac{67.50}{1350} \times 100 \\ &= 5\% \: \: _{Ans.} \end{align*}
Solution:
Selling price (SP) = Rs 29660
VAT % = 10 %
\begin{align*} \text {Amount of VAT} &= 29660 \times \frac{10}{100} \\ &= Rs \: 2966 \end{align*}
Solution:
Let, MP = Rs x,
VAT = 10%
\begin{align*} x + x \: of \: 10\% &= 17050 \\ or, x + x \times \frac{10}{100} &= 17050 \\ or, \frac{10x + x}{10} &= 17050 \\ or, x &= \frac{17050 \times 10}{11} \\ \therefore x &= Rs \: 15500 \end{align*}
\begin{align*} \text{Amount of VAT } &= Rs 17050 - Rs 15500 \\ &= Rs 1550 \: _{Ans.} \end{align*}
Solution:
Let, cost price (CP) = Rs x
VAT = 10%
\begin{align*} x + x \: of \: 10\% &= 650 \\ or, x + x \times \frac{10}{100} &= 650\\ or, \frac{11x}{10} &= 650 \\ or, x &= \frac{650 \times 10}{11}\\ \therefore x &= Rs \: 590.90 \end{align*}
Return money for 1 set = Rs 650 - Rs 590.90 = Rs 59.10
Return money for 5 sets = 5 \(\times\) 59.10 = Rs 295.50
Solution:
Price of TV = Rs 24,000
Amount of discount = Rs 1200 Discount % = ?
\begin{align*} Discount\% &= \frac{Discount \: Amount}{Price \: of \: TV} \times 100\% \\ &= \frac{1200}{24000} \times 100\%\\ &= 5\% \end{align*}
\(\therefore \) Discount = 5%
Solution:
Marked price (MP) = Rs 260
Discount % = 5% Selling price (SP) = ?
\begin{align*} SP &=MP - MP \: of \: discount\% \\ &= 260 - 260 \times \frac{5}{100} \\ &= Rs \:260 -13 \\&= Rs \: 247 \end{align*}
\(\therefore \) SP = Rs 247 \(_{Ans}\)
Solution:
Let marked price (MP) = Rs x
The price of the article with VAT = Rs 690 VAT =15%
We know that,
The price of the article with \begin{align*} VAT &= x + x \: of \: 15\% \\ 690 &= x + x \times \frac{15}{100} \\ or, 690 &= \frac{23x}{20}\\ or,x &= \frac{690 \times 20}{23} \\ x &= Rs \: 600 \end{align*}
The price excluding VAT is Rs 600.
Solution:
Marked price (MP) = Rs 80,000
Discount = 5%
\begin{align*}Selling\: price \:(SP) &= MP - MP \: of \: discount\% \\ &= Rs 80000 - 80000 \times \frac{5}{100}\\ &= Rs \: 80,000 - 4000 \\&= Rs \: 76,ooo \: \: _{Ans.} \: \end{align*}
Solution:
The price of computer before VAT = Rs x, VAT = 15%
Cost of computer after adding VAT = Rs 46000
\begin{align*} x + x \: of \: 15 &= Rs \: 46000\\ or, x + x \times \frac{15}{100} &= Rs \: 46000\\ or, \frac{20x + 3x}{20} &= 46000 \\ or, x &= \frac{46000 \times 20}{23}\\ \therefore x &= Rs \: 40,000 \end{align*}
\(\therefore \) The price of exclusive of the VAT = Rs 40,000 \(_{Ans.}\)
An article is bought for Rs 800 & sold for 5/4 of the cost price. What is the profit percentage?
20%
25%
15%
10%
If a dozen is bought for Rs. 48 and sold for Rs 5 per piece, what percent is the profit?
25%
30%
22%
32%
Raman sold a camera for Rs. 2520 at 5% profit. Find the cost price.
Rs 2400
Rs2100
Rs 2200
Rs2000
Shila brought 5 books for Rs 1500 and sold them at 20% loss, Find the selling price of the book.
Rs 20
Rs 20
Rs 15
Rs 25
If the cost price of 10 chairs is equal to the selling price of 16 chairs, find the loss or gain percentage.
Loss:30%
loss: 37.5%
Loss:20 %
Loss:25 %
By selling a watch for Rs 4500 a dealer got 10% loss. At what price should he sell it so as to gain 10%?
Rs 5200
Rs 5100
Rs 5500
Rs 4500
Sohan bought a radio for Rs. 800 and sold it to Rohan at profit of 20% Rohan sold it to Mohan at a loss of 10%. For how much did Mohan buy it?
Rs 800
Rs 864
Rs 750
Rs 850
Amisha sold a cycle to Amir at a profit of 10%. Amir sold the same cycle to Abhishek at a profit of 20%. If Abhishek has sold it for Rs 3300 thereby earning a profit of 25% find the cost price of Amisha .
Rs 1550
Rs 2000
Rs 2400
Rs 2100
Salman sold two computer for RS 2400 each. On one he gained 20% and on the other he lost 20% and on the other he lost 20%. Find his gain or loss percentage in the whole transaction.
P=5%
P = 4 %
P= 3%
P=7%
Lata bought two watches for Rs. 800.She sold them to gain 20% on one and lose 20% the other. Calculate her final gain or loss percent if the selling price of both the watches is the same.
5% loss
4% loss
7% loss
9% loss
A man bought two books for Rs 1040. He sold one at a loss of 15% and the other at a profit of 36% then he found that each book was sold for the same price. Find the cost price of each book.
Rs 600, Rs 350
Rs 500, Rs 300
Rs 640, Rs 400
Rs 700,Rs 500
Rambilash bought two radio sets for Rs 500. He sold one at a loss of 12% and the other at a gain of 8%. He neither gained nor lost on his transaction.Find the cost price of each radio.
Rs 200, Rs 300
Rs 350, Rs 400
Rs 275, Rs 375
Rs 100, Rs 150
A man bought a hen and a duck for Rs. 370 and sold them for Rs.402, thereby gaining 20% on the former and losing 15% on the later. Find the cost price of the duck.
Rs 100
Rs 150
Rs 130
Rs 120
A person sold an article at a profit of 15% .If he sold it for Rs 81 less, his loss would have been 12% Find the cost price of the article.
Rs 300
Rs 240
Rs 250
Rs 350
If the selling price of a sofa is increased by Rs 7920; the loss of 15%converts into a profit of 18%.Find the cost price.
Rs 2200
Rs 1500
Rs 24000
Rs 1850 |
The aim of this test case is to validate the following functionalities:
The simulation results of SimScale were compared to the analytical results in [NAFEMS_R27]. The tetrahedral meshes used were created with the parametrized-tetrahedralization-tool on the SimScale platform.
Tool Type : Code_Aster Analysis Type : Static – nonlinear Mesh and Element types :
Case Mesh type Number of nodes Element type (A) second order hexahedral 20 3D isoparametric (B) second order hexahedral 20 3D reduced integration (C) second order tetrahedral 21 3D isoparametric (D) second order tetrahedral 21 3D reduced integration
Material:
Important Information
The material Data of the reference was given in MPa, mm and h units and were converted to SI units using the following relation: $$A_{SI} = rac{1}{3600} cdot rac{1}{3600^m} cdot rac{1}{10^{6n}} cdot A_{MPa;h}$$ for the parameters as given in followig Time Hardening formulation: $$dot{epsilon} = A cdot sigma^n cdot t^m$$
Loads:
Advanced Automatic Time Stepping:
$$\epsilon_{xx}^c = – \epsilon_{zz}^c = \frac{0.004218}{60} \sqrt{t}$$
$$\epsilon_{eff}^c = \frac{0.004871}{60} \sqrt{t}$$ $$\epsilon_{yy}^c = 0.0$$
The equations used to solve the problem are derived in [NAFEMS_R27]. As the SimScale solution is calculated using SI units, the reference solution was adopted to a time unit of
seconds instead of hours.
Comparison of the average creep strain $$\epsilon_{xx}^c$$
of the cube after a creep time of 3.6e6 s (1000 h) with the analytical solution presented in [NAFEMS_R27]:
Case time [s] [NAFEMS_R27] SimScale Error (%) (A) 3600000 0.13338 0.133107 0.205 (B) 3600000 0.13338 0.133107 0.205 (C) 3600000 0.13338 0.133107 0.205 (D) 3600000 0.13338 0.133107 0.205
[NAFEMS_R27] (1, 2, 3, 4) (1993) “NAFEMS Fundamental Tests of Creep Behaviour”, Test 11, NAFEMS Publication Ref: R0027 |
Let the function $T$ be a mapping from $\mathbb{R^3} \rightarrow \mathbb{R^4}$ given by
$$T\left(\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}\right) = \begin{bmatrix}1\\2\\-1\\-2\end{bmatrix}$$
Is this function a linear transformation?
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Let the function $T$ be a mapping from $\mathbb{R^3} \rightarrow \mathbb{R^4}$ given by
$$T\left(\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}\right) = \begin{bmatrix}1\\2\\-1\\-2\end{bmatrix}$$
Is this function a linear transformation?
Every linear transformation applies $$T(0)=0$$ The given transformation : $$T(0,0,0)=(1,2,-1,-2) \neq (0,0,0,0)$$ Hence, Not a linear transformation.
It might also be helpful to consider the proof of why T(0) must be 0 from the axioms of linearity
0 in a field with (second operation multiplication, which is what we consider here in this scalar multiplication) is defined to be the element at which $\forall f \in F$, we have $0 + f = f + 0 = f$ (note that I made this definition two-sided, because in any ring [all fields are rings] the additive operation forms an abelian group). It can be shown using the distributive axiom of rings that $0 * f = f* 0 = 0$. This is also true for rings in general, not just fields, but we use fields with vector spaces for more useful axioms. https://en.wikipedia.org/wiki/Proofs_of_elementary_ring_properties may also be helpful for general ring/field properties.
By linearity we have $T(0 * v) = 0 * T(v)$. This is not finished, but that tangent on fields was to really show that vector spaces have very similar structure, except that scalar multiplication is NOT binary operation. However, we can use the distributive property in similar manner as in Workaholic's answer. Vector spaces - Multiplying by zero scalar yields zero vector
Then we can see that this transformation will not be linear we the transformation applied to $\vec{0}$ is not also $\vec{0}$. |
The question is to find the variation in induced electromotive force w.r.t to time or put simply $\varepsilon(t)$ for a circular loop being removed from the region of the magnetic field at a constant velocity $v$.
Obviously $$\varepsilon= \frac{d\phi}{dt}$$ Now $$\phi = BA\cos \theta=B(\pi r^2)$$ Now, $$\varepsilon= \frac{d\phi}{dt} = 2\pi rB \frac{dr}{dt}$$ But there's no clear way to determine $\frac{dr}{dt}$. So is there any other way to determine the rate of change of area (how much of the circle is leaving the region per $dt$ time). Any help would be very good right now.
The question is to find the variation in induced electromotive force w.r.t to time or put simply $\varepsilon(t)$ for a circular loop being removed from the region of the magnetic field at a constant velocity $v$.
As I mentioned in a comment, a magnetic field abruptly decreasing to zero is unphysical, unless you have currents or time-varying E-fields around it to confine it. However I will ignore this concern for what follows.
If the magnetic field is confined to a large rectangular region in space, when the loop is leaving this region the magnetic field is passing through only a portion of the loop. This portion has the shape of the intersection of a circle and a half-plane. Let the edge of this half-plane be located a distance $x$ from the center of the circle. The secant line defined by the boundary of the region with magnetic field has length $2\sqrt{r^2-x^2}$. The rate of change of the area enclosing the magnetic field with $x$ is this secant length. If the loop is halfway outside the magnetic field at $t=0$, we can write $x=-vt$ where $v$ is the speed. Putting everything together,
$$\mathcal{E} = -\frac{d\Phi}{dt} = -B \cos\theta \frac{dA}{dt} = -B \cos\theta \frac{dx}{dt} \frac{dA}{dx} = 2B v \sqrt{r^2 - (vt)^2} \cos\theta $$
for $|t| < r/v$, and $0$ for all other times.
What you've derived is a formula for the emf generated in a single turn coil whose radius is changing at a rate $\frac{dr}{dt}$. The coil's axis is parallel to a constant uniform field of magnitude
B.
But you're trying to find an emf in a coil of fixed radius? You can't, I'm afraid, find how the emf varies for "a circular loop being removed from the region of the magnetic field at a constant velocity 𝑣" unless you know how the field varies from point to point, and the path along which you move the loop. Even if you do know these things it may still be hard to do what you want. |
Can there be a large cardinal $\kappa$ and a forcing of size $\kappa$ that makes $\kappa$ a singular cardinal? The motivation is that the standard Prikry forcing does not have a dense set of size $\kappa$.
Edit:
In response to some attempts at a positive answer, let me explain something that does not work. If $\mathbb{P}$ is the Prikry forcing and $\mathbb{Q}$ is something like $Coll(\kappa,2^\kappa)$, one may expect under suitable indestructibility hypotheses, $\mathbb{P}$ works in $V^\mathbb{Q}$. But this never works.
The following lemma is based on an exercise in Kunen's book: Suppose $\kappa$ is a singular cardinal and $\mathbb{R} = \{ f : f$ is a partial function from $\kappa$ to $2$ with domain bounded below $\kappa \}$, ordered by extension. Then $\mathbb{R}$ collapses $\kappa$ to $cf(\kappa)$.
Proof: Suppose for simplicity $cf(\kappa) = \omega$, and let $\langle \kappa_n : n \in \omega \rangle$ be an increasing cofinal sequence. If $G \subseteq \mathbb{R}$ is generic, define in $V[G]$ the function $f : \omega \to \kappa$ by $f(n) = \beta$ where $\beta < \kappa_n$ and for some $\delta$, the ordinal $\kappa_n \cdot \delta + \beta$ is the $\kappa_n$-th element of $\{ \alpha : \bigcup G(\alpha) = 1 \}$. A simple density argument shows that $f$ is surjective.
Now let $\kappa$ be our large cardinal. It follows from a general folklore fact that there is a dense embedding $e : Add(\kappa,1) \times Coll(\kappa,2^\kappa) \to Coll(\kappa,2^\kappa)$. After forcing with $\mathbb{P}$, the $Add(\kappa,1)$ of the ground model becomes the forcing with bounded functions from the lemma, and the map $e$ is still a dense embedding. So if $G \times H$ is $\mathbb{P} \times \mathbb{Q}$-generic, then by the lemma, $\kappa$ is collapsed to $\omega$. Therefore in $V^\mathbb{Q}$, $\mathbb{P}$ collapses $\kappa$ to $\omega$.
I suspect that if a positive answer is possible, the forcing must be significantly different from the standard Prikry forcing or some combination of it with simple forcings. |
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Change to browse by: References & Citations Bookmark(what is this?) Condensed Matter > Statistical Mechanics Title: Sequential disruption of the shortest path in critical percolation
(Submitted on 22 Jun 2019)
Abstract: We investigate the effect of sequentiallydisrupting the shortest path of percolation clusters at criticality by comparing it with the shortest alternative path. We measure the difference in length and the enclosed area between the two paths. The sequential approach allows to study spatial correlations. We find the lengths of the segments of successively constant differences in length to be uncorrelated. Simultaneously, we study the distance between red bonds. We find the probability distributions for the enclosed areas A, the differences in length $\Delta l$, and the lengths between the redbonds $l_r$ to follow power law distributions. Using maximum likelihood estimation and extrapolation we find the exponents $\beta$ = 1.38 $\pm$ 0.03 for $\Delta l$, $\alpha$ = 1.186 $\pm$ 0.008 for A and $\delta$ = 1.64 $\pm$ 0.025 for thedistribution of $l_r$. Submission historyFrom: Oliver Gschwend [view email] [v1]Sat, 22 Jun 2019 20:10:52 GMT (164kb,D) |
Can someone help me finding the expected value of the solution to Merton's jump diffusion model:
\begin{align} S_t &= S_0 \exp \left( \left(r - \frac{\sigma^2}{2} - \lambda k \right) t + \sigma W_t \right) \prod_{j=1}^{N_t} (1+\epsilon_i) \end{align}
where $W_t$ is a BM and $N_t$ is a Poisson process with intensity $\lambda$ and $k$ is the expectation of $\epsilon_i$. The Brownian Motion and Poisson Process are independent.
I know that
\begin{align} E \left[ \exp \left( \left(r - \frac{\sigma^2}{2} \right) t + \sigma W_t \right) \right] = \exp(rt) \end{align}
but what is
\begin{align} E \left[ \prod_{j=1}^{N_t} (1+\epsilon_i) \right] = ? \end{align} |
Answer
See the answer below.
Work Step by Step
a) $W=-P\Delta V$ $W=-101\ kPa\times -22.4\ L\div 1000\ L/m^3=2.26\ kJ$ b)$\Delta H= \Delta U +P\Delta V=\Delta U-W$ $\Delta U=-483.6\ kJ+2.26\ kJ$ $\Delta U=-481.3\ kJ$
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Since version 0.9.45.1 of the ‘mkin’ package, a function for calculating time weighted average concentrations for decline kinetics (
i.e. only for the compound applied in the experiment) is included. Strictly speaking, they are maximum moving window time weighted average concentrations, i.e. the maximum time weighted average concentration that can be found when moving a time window of a specified width over the decline curve.
Time weighted average concentrations for the SFO, FOMC and the DFOP model are calculated using the formulas given in the FOCUS kinetics guidance (FOCUS Work Group on Degradation Kinetics 2014, 251):
SFO:
\[c_\textrm{twa} = c_0 \frac{\left( 1 - e^{- k t} \right)}{ k t} \]
FOMC:
\[c_\textrm{twa} = c_0 \frac{\beta}{t (1 - \alpha)} \left( \left(\frac{t}{\beta} + 1 \right)^{1 - \alpha} - 1 \right) \]
DFOP:
\[c_\textrm{twa} = \frac{c_0}{t} \left( \frac{g}{k_1} \left( 1 - e^{- k_1 t} \right) + \frac{1-g}{k_2} \left( 1 - e^{- k_2 t} \right) \right) \]
Often, the ratio between the time weighted average concentration \(c_\textrm{twa}\) and the initial concentration \(c_0\)
\[f_\textrm{twa} = \frac{c_\textrm{twa}}{c_0}\]
is needed. This can be calculated from the fitted initial concentration \(c_0\) and the time weighted average concentration \(c_\textrm{twa}\), or directly from the model parameters using the following formulas:
SFO:
\[f_\textrm{twa} = \frac{\left( 1 - e^{- k t} \right)}{k t} \]
FOMC:
\[f_\textrm{twa} = \frac{\beta}{t (1 - \alpha)} \left( \left(\frac{t}{\beta} + 1 \right)^{1 - \alpha} - 1 \right) \]
DFOP:
\[f_\textrm{twa} = \frac{1}{t} \left( \frac{g}{k_1} \left( 1 - e^{- k_1 t} \right) + \frac{1-g}{k_2} \left( 1 - e^{- k_2 t} \right) \right) \]
Note that a method for calculating maximum moving window time weighted average concentrations for any model fitted by ‘mkinfit’, and also for metabolites in such models, can be found in the ‘pfm’ package, which is currently not published on CRAN, but available from github and documented at my website.
FOCUS Work Group on Degradation Kinetics. 2014.
Generic Guidance for Estimating Persistence and Degradation Kinetics from Environmental Fate Studies on Pesticides in Eu Registration. 1.1 ed. http://esdac.jrc.ec.europa.eu/projects/degradation-kinetics. |
Advances in Differential Equations Adv. Differential Equations Volume 6, Number 12 (2001), 1517-1540. Radial solutions for a quasilinear equation via Hardy inequalities Abstract
We establish an analogue of the Sobolev critical exponent for the inclusion $V^p(a)\hookrightarrow L^q(b)$, where $a$ and $b$ are weight functions, $V^p(a)$ is a weighted Sobolev space, and $L^q(b)$ is a weighted Lebesgue space. We use this result to study existence of radial solutions to the problem with weights $$(D_w)\quad\quad\begin{cases} -{\rm div}(\tilde{a}(|x|)|\nabla u|^{p-2}\nabla u)= \tilde{b}(|x|)|u|^{q-2}u \qquad \mbox{in} \quad \Omega\subset \mathbb R^N ,\\~~u=0\qquad \mbox{on}\quad \partial\Omega, \end{cases} $$ where $\Omega$ is a ball, $1 <p <q,$ and $\tilde{a}(|x|)=|x|^{1-N}a(|x|),$ $\tilde{b}(|x|)=|x|^{1-N}b(|x|)$. We are interested in the interplay between $q$ and a suitable critical exponent and its consequences for the existence and nonexistence of positive solutions of problem $(D_w).$
Article information Source Adv. Differential Equations, Volume 6, Number 12 (2001), 1517-1540. Dates First available in Project Euclid: 2 January 2013 Permanent link to this document https://projecteuclid.org/euclid.ade/1357139957 Mathematical Reviews number (MathSciNet) MR1858431 Zentralblatt MATH identifier 1140.35441 Subjects Primary: 35J60: Nonlinear elliptic equations Secondary: 34B16: Singular nonlinear boundary value problems 35B33: Critical exponents 35J25: Boundary value problems for second-order elliptic equations Citation
García-Huidobro, M.; Kufner, A.; Manásevich, R.; Yarur, C. S. Radial solutions for a quasilinear equation via Hardy inequalities. Adv. Differential Equations 6 (2001), no. 12, 1517--1540. https://projecteuclid.org/euclid.ade/1357139957 |
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 $\Fun(S, V)$. |
Note that indeed the ratio test is inconclusive, as $$\lim_{n\to\infty} \frac{a_{n+1}}{a_n} = 1\,.$$
If you know Stirling's approximation of the factorial,$$n! \operatorname*{\sim}_{n\to\infty} \sqrt{2\pi n}\left(\frac{n}{e}\right)^n$$then you can show that $a_n \operatorname*{\sim}_{n\to\infty} \frac{1}{\sqrt{\pi n}}$ and the series $\sum_n a_n$ thus diverges by theorems of comparison (for series with positive terms).
If you do not but like probabilities, recall that this is exactly $$a_n = \mathbb{P}\{ X = n \}$$where $X\sim \mathrm{Bin}(2n,1/2)$ is a Binomial random variable with parameters $2n$ and $1/2$. By standard concentration and anticoncentration results, the Binomial distribution is "roughly uniform" (i.e., its probability mass constant is within constant factors) within $\pm \sqrt{n}$ (that is, more or less the order of a standard deviation) of its expectation, and a constant fraction of its probability mass is on this interval. This implies that $$a_n = \Theta(1/\sqrt{n})$$ leading to the same result.
I consider the second method the most fun, but the first is quite useful (if you do not know Stirling's appproximation but want to, here is a great occasion to do so.) And, of course, there are other ways. |
[some formatting tweaked, and the question copied from the title to the main body, by YC]
Hi,
I've been struggling a lot to calculate this integral.
$$ \int_0^\infty \frac{k^{n-1}}{\prod_{i=1}^n (k^2+ x_i^2)}\; dk $$ where $x_i$ are constants and $n\geq 1$.
I did the calculation for n=1,2,3,4, with the hope of identifying some form and then find the result by induction. But here is what I got:
n=1: I= (pi/2) * abs(x1)
n=2: I= (1/2) * 1/(x2ˆ(2)-x1ˆ(2)) * log(x2ˆ(2) / x1ˆ(2))
n=3: I= (pi/2) * [abs(x1) (x2ˆ(2)-x3ˆ(2)) +abs(x2) (x3ˆ(2)-x1ˆ(2))+ abs(x3) (x1ˆ(2)-x2ˆ(2))] / [(x2ˆ(2)-x3ˆ(2) (x3ˆ(2)-x1ˆ(2)) (x1ˆ(2)-x2ˆ(2)]
n=4: I= (1/2) * [ A1 log(x1ˆ(2)) + A2 log(x2ˆ(2)) +... A4 log(x4ˆ(2))), where Ai= xiˆ(2) / [ prod (xjˆ(2)-xiˆ(2))]
-->> This makes me think that the result depends on whether n is even or uneven; that is, we would have a form in log( ) for n even, and something in pi/2 for n uneven?
Could you please help me here? What is the correct result and how to get it?
Your help is so much appreciated, many many thanks in advance! Elise |
Mathematically one way to see it is that the (combined)
step functions become Dirac pulses i.e $\theta(t) \to \delta(t)$, which diverge. When $t \to t'$, the fields have nearly identical values, thus the time-ordered product involving step functions degenerates into dirac pulse (a dirac pulse is the derivative of the step function)
Physically another way to see this is: This post imported from StackExchange Physics at 2014-06-27 11:27 (UCT), posted by SE-user Nikos M.
S-matrix describes interactions between states and particles, as such a matrix element for $t \to t'$, requires an interaction to take place instanteneously, thus it would require infinite energy at that point (also related to time-energy unceratinty). |
Problem 27
Solve the following system of linear equations using Gauss-Jordan elimination.
\begin{align*} 6x+8y+6z+3w &=-3 \\ 6x-8y+6z-3w &=3\\ 8y \,\,\,\,\,\,\,\,\,\,\,- 6w &=6 \end{align*}
Add to solve later
We use the following notation.
Elementary row operations.
The three elementary row operations on a matrix are defined as follows.
(1) Interchanging two rows:
$R_i \leftrightarrow R_j$ interchanges rows $i$ and $j$.
(2) Multiplying a row by a non-zero scalar (a number):
$tR_i$ multiplies row $i$ by the non-zero scalar (number) $t$.
(3) Adding a multiple of one row to another row:
$R_j+tR_i$ adds $t$ times row $i$ to row $j$.
Solution.
The augmented matrix of the system is
\[A=\left[ \begin{array}{rrrr|r} 6 & 8 & 6 & 3 & -3 \\ 6 & -8 & 6 & -3 & 3\\ 0 & 8 & 0 & -6 & 6 \end{array} \right].\] We apply elementary row operations as follows to reduce the system to a matrix in reduced row echelon form.
\[A \xrightarrow{R_2 – R_1}
\left[\begin{array}{rrrr|r} 6 & 8 & 6 & 3 & -3 \\ 0 & -16 & 0 & -6 & 6\\ 0 & 8 & 0 & -6 & 6 \end{array}\right] \xrightarrow[R_2+2R_3]{R_1-R_3} \left[\begin{array}{rrrr|r} 6 & 0 & 6 & 9 & -9 \\ 0 & 0 & 0 & -18 & 18\\ 0 & 8 & 0 & -6 & 6 \end{array}\right] \] \[\xrightarrow[\frac{1}{2}R_3]{\frac{1}{3}R_1, \frac{-1}{18}R_2} \left[\begin{array}{rrrr|r} 2 & 0 & 2 & 3 & -3 \\ 0 & 0 & 0 & 1 & -1\\ 0 & 4 & 0 & -3 & 3 \end{array}\right] \xrightarrow[R_3+3R_2]{R_1-3R_2} \left[\begin{array}{rrrr|r} 2 & 0 & 2 & 0 & 0 \\ 0 & 0 & 0 & 1 & -1\\ 0 & 4 & 0 & 0 & 0 \end{array}\right] \]
\[
\xrightarrow[\frac{1}{4}R_3]{\frac{1}{2}R_1} \left[\begin{array}{rrrr|r} 1 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & -1\\ 0 & 1 & 0 & 0 & 0\end{array}\right] \xrightarrow{R_2 \leftrightarrow R_3} \left[\begin{array}{rrrr|r} 1 & 0 & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & -1 \end{array}\right] \] The last matrix is in reduced row echelon form. The corresponding system of linear equations is \begin{align*} x+z &=0\\ y&=0 \\ w&=-1 \end{align*}
Let $z=t$ be a free variable. Then the solution is $(x,y,z,w)=(-t,0,t,-1)$ for any number $t$.
Similar Problem.
Another similar problem is
Solving a system of linear equations using Gaussian elimination. The instruction of the problem says to use Gaussian elimination, but try to solve it using Gauss-Jordan elimination as well. Click here if solved
Add to solve later
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More from my site Solving a System of Linear Equations Using Gaussian Elimination
Solve the following system of linear equations using Gaussian elimination.
\begin{align*}
x+2y+3z &=4 \\
5x+6y+7z &=8\\
9x+10y+11z &=12
\end{align*}
Elementary row operations
The three elementary row operations on a matrix are defined as […] Give a Formula for a Linear Transformation if the Values on Basis Vectors are Known
Let $T: \R^2 \to \R^2$ be a linear transformation.
Let
\[
\mathbf{u}=\begin{bmatrix}
1 \\
2
\end{bmatrix}, \mathbf{v}=\begin{bmatrix}
3 \\
5
\end{bmatrix}\]
be 2-dimensional vectors.
Suppose that
\begin{align*}
T(\mathbf{u})&=T\left( \begin{bmatrix}
1 \\
[…] Find a Polynomial Satisfying the Given Conditions on Derivatives
Find a cubic polynomial
\[p(x)=a+bx+cx^2+dx^3\]
such that $p(1)=1, p'(1)=5, p(-1)=3$, and $ p'(-1)=1$.
Solution.
By differentiating $p(x)$, we obtain
\[p'(x)=b+2cx+3dx^2.\]
Thus the given conditions are
[…] Express a Vector as a Linear Combination of Other Vectors
Express the vector $\mathbf{b}=\begin{bmatrix}
2 \\
13 \\
6
\end{bmatrix}$ as a linear combination of the vectors
\[\mathbf{v}_1=\begin{bmatrix}
1 \\
5 \\
-1
\end{bmatrix},
\mathbf{v}_2=
\begin{bmatrix}
1 \\
2 \\
1
[…] Vector Space of Polynomials and Coordinate Vectors
Let $P_2$ be the vector space of all polynomials of degree two or less.
Consider the subset in $P_2$
\[Q=\{ p_1(x), p_2(x), p_3(x), p_4(x)\},\]
where
\begin{align*}
&p_1(x)=x^2+2x+1, &p_2(x)=2x^2+3x+1, \\
&p_3(x)=2x^2, &p_4(x)=2x^2+x+1.
\end{align*}
(a) Use the basis […] Vector Form for the General Solution of a System of Linear Equations
Solve the following system of linear equations by transforming its augmented matrix to reduced echelon form (Gauss-Jordan elimination).
Find the vector form for the general […] Find Values of $a$ so that Augmented Matrix Represents a Consistent System
Suppose that the following matrix $A$ is the augmented matrix for a system of linear equations.
\[A= \left[\begin{array}{rrr|r}
1 & 2 & 3 & 4 \\
2 &-1 & -2 & a^2 \\
-1 & -7 & -11 & a
\end{array} \right],\]
where $a$ is a real number. Determine all the […] Determine Whether Matrices are in Reduced Row Echelon Form, and Find Solutions of Systems
Determine whether the following augmented matrices are in reduced row echelon form, and calculate the solution sets of their associated systems of linear equations.
(a) $\left[\begin{array}{rrr|r} 1 & 0 & 0 & 2 \\ 0 & 1 & 0 & -3 \\ 0 & 0 & 1 & 6 \end{array} \right]$.
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J/Ψ production and nuclear effects in p-Pb collisions at √sNN=5.02 TeV
(Springer, 2014-02)
Inclusive J/ψ production has been studied with the ALICE detector in p-Pb collisions at the nucleon–nucleon center of mass energy √sNN = 5.02TeV at the CERN LHC. The measurement is performed in the center of mass rapidity ...
Suppression of ψ(2S) production in p-Pb collisions at √sNN=5.02 TeV
(Springer, 2014-12)
The ALICE Collaboration has studied the inclusive production of the charmonium state ψ(2S) in proton-lead (p-Pb) collisions at the nucleon-nucleon centre of mass energy √sNN = 5.02TeV at the CERN LHC. The measurement was ...
Event-by-event mean pT fluctuations in pp and Pb–Pb collisions at the LHC
(Springer, 2014-10)
Event-by-event fluctuations of the mean transverse momentum of charged particles produced in pp collisions at s√ = 0.9, 2.76 and 7 TeV, and Pb–Pb collisions at √sNN = 2.76 TeV are studied as a function of the ...
Centrality, rapidity and transverse momentum dependence of the J/$\psi$ suppression in Pb-Pb collisions at $\sqrt{s_{NN}}$=2.76 TeV
(Elsevier, 2014-06)
The inclusive J/$\psi$ nuclear modification factor ($R_{AA}$) in Pb-Pb collisions at $\sqrt{s_{NN}}$=2.76TeV has been measured by ALICE as a function of centrality in the $e^+e^-$ decay channel at mid-rapidity |y| < 0.8 ...
Multiplicity Dependence of Pion, Kaon, Proton and Lambda Production in p-Pb Collisions at $\sqrt{s_{NN}}$ = 5.02 TeV
(Elsevier, 2014-01)
In this Letter, comprehensive results on $\pi^{\pm}, K^{\pm}, K^0_S$, $p(\bar{p})$ and $\Lambda (\bar{\Lambda})$ production at mid-rapidity (0 < $y_{CMS}$ < 0.5) in p-Pb collisions at $\sqrt{s_{NN}}$ = 5.02 TeV, measured ...
Multi-strange baryon production at mid-rapidity in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV
(Elsevier, 2014-01)
The production of ${\rm\Xi}^-$ and ${\rm\Omega}^-$ baryons and their anti-particles in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV has been measured using the ALICE detector. The transverse momentum spectra at ...
Measurement of charged jet suppression in Pb-Pb collisions at √sNN = 2.76 TeV
(Springer, 2014-03)
A measurement of the transverse momentum spectra of jets in Pb–Pb collisions at √sNN = 2.76TeV is reported. Jets are reconstructed from charged particles using the anti-kT jet algorithm with jet resolution parameters R ...
Two- and three-pion quantum statistics correlations in Pb-Pb collisions at √sNN = 2.76 TeV at the CERN Large Hadron Collider
(American Physical Society, 2014-02-26)
Correlations induced by quantum statistics are sensitive to the spatiotemporal extent as well as dynamics of particle-emitting sources in heavy-ion collisions. In addition, such correlations can be used to search for the ...
Exclusive J /ψ photoproduction off protons in ultraperipheral p-Pb collisions at √sNN = 5.02TeV
(American Physical Society, 2014-12-05)
We present the first measurement at the LHC of exclusive J/ψ photoproduction off protons, in ultraperipheral proton-lead collisions at √sNN=5.02 TeV. Events are selected with a dimuon pair produced either in the rapidity ...
Measurement of quarkonium production at forward rapidity in pp collisions at √s=7 TeV
(Springer, 2014-08)
The inclusive production cross sections at forward rapidity of J/ψ , ψ(2S) , Υ (1S) and Υ (2S) are measured in pp collisions at s√=7 TeV with the ALICE detector at the LHC. The analysis is based on a data sample corresponding ... |
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.
Offline
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.
Offline
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.
Offline
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.
Offline
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.
Offline |
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. |
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 |
In fact the statement "The frobenius method states that for repeated roots or roots that differ by an integer, an alternative method must be used to find the second solution once one is found." is always true for multiple roots cases but not always true for the cases that some of the roots are differed by integers. Since for the cases that some of the roots from the indical equations are differed by integers, in some situations some of the roots from the indical equations can find more than one group of the linearly independent solutions:
$1.$ For the cases that having more than one indical equations, when some of the roots from the indical equations are common, as taking the common ones can eliminate more than one indical equations at the same time, so we can find more than one group of the linearly independent solutions at the same time. For example the question no. $5$ in http://tw.knowledge.yahoo.com/question/question?qid=1511121805995 and Can I use Frobenius method here? excluding the case of $n=3$ .
$2.$ When some of the roots from the indical equations can obtain the self-fitful recurrence relations. For example for the recurrence relations of the form $f(n)a_n=g(n)a_{n-k}$ where $k\in\mathbb{N}$ , when there exist some non-negative integers $n$ such that $f(n)=0$ , they are self-fitful. So we can find more than one group of the linearly independent solutions from each of the recurrence relations. For example the question no. $1$ in http://tw.knowledge.yahoo.com/question/question?qid=1512050807030 and http://faculty.pccu.edu.tw/~meng/Math3.pdf#page=6.
Now for this problem,
Let $y=\sum\limits_{n=0}^\infty a_nz^{n+\sigma}$ ,
Then $y'=\sum\limits_{n=0}^\infty(n+\sigma)a_nz^{n+\sigma-1}$
$y'=\sum\limits_{n=0}^\infty(n+\sigma)(n+\sigma-1)a_nz^{n+\sigma-2}$
$\therefore\sum\limits_{n=0}^\infty(n+\sigma)(n+\sigma-1)a_nz^{n+\sigma-2}-\dfrac{6}{z^2}\sum\limits_{n=0}^\infty a_nz^{n+\sigma}=0$
$\sum\limits_{n=0}^\infty(n+\sigma)(n+\sigma-1)a_nz^{n+\sigma-2}-\sum\limits_{n=0}^\infty6a_nz^{n+\sigma-2}=0$
$\sum\limits_{n=0}^\infty((n+\sigma)(n+\sigma-1)-6)a_nz^{n+\sigma-2}=0$
Since there are not any indical equations present, that means $\sigma$ can be chosen as any complex number.
However, in fact, take $\sigma=3$ or $\sigma=-2$ will bring the above equation most simplified.
Moreover, in fact, we can find all groups of the linearly independent solutions by just taking $\sigma=-2$ :
$\sum\limits_{n=0}^\infty((n-2)(n-3)-6)a_nz^{n-4}=0$
$\sum\limits_{n=0}^\infty n(n-5)a_nz^{n-4}=0$
$\therefore n(n-5)a_n=0$
$\therefore\begin{cases}a_0=a_0\\a_5=a_5\\a_n=0~\forall n\in\mathbb{N}\setminus\{5\}\end{cases}$
Hence $y=c_1z^{-2}+c_2z^3$
Therefore in this case the alternative method for a second solution of course is not required. |
I do not think that there is an exact value and believe that it would depend on the content. It is really hard to precisely align what is on the margins to what is in the main body because
\marginnote command will usually prevent the ability of aligning contents that is provided in LaTeX environments or commands. This is usually due to nesting incompatibility or simply because primitive TeX instructions used to define environment may not apply equally to both main body contents and marginpar contents. However, here is a good way that is simple but also does not rely merely on trial and error to align the contents.
To help align them systematically, we must add virtual contents to the smaller one of them, which is what is in the main body. Use
\vphantom with the argument being the smaller contents (whether the one in the main text or in the margin)
$\displaystyle \frac{\left(2a\right)^3 3a^\frac{1}{3}}{a^{-\frac{2}{3}}6a^2} = 2^2 \cdot a^2 \vphantom{\displaystyle \begin{aligned}[c] & \frac{A^m}{A^n} = A^{m-n} \\ & A^0 = 1 \text{ where } A \ne 0 \end{aligned}}$
The output is then
What is left now is to do some tuning. It is best to determine the offset in terms of the
\baselineskip length. After some tuning, it was obvious that
-0.1\baselineskip is the best offset. The content is now
This approach works even with different number of lines for the
aligned environment.
What you need to do is to install virtual height to the shorter content using
\vphantom where the argument of
\vphantom is the larger content. By doing that, you will establish virtual boxes of equal heights and what would be left is to do some fine-tuning
\documentclass[twoside,12pt]{article}
\usepackage[a4paper,marginparwidth=8cm,includemp,innermargin=4cm,outermargin=4cm]{geometry}
\usepackage{amsmath}
\usepackage{adjustbox}
\usepackage{marginnote}
\begin{document}
\begin{enumerate}
\item
\marginnote{
$\displaystyle
\begin{aligned}[c]
& \frac{A^m}{A^n} = A^{m-n} \\
& A^0 = 1 \text{ where } A \ne 0
\end{aligned}$}[-0.1\baselineskip]
$\displaystyle \frac{\left(2a\right)^3 3a^\frac{1}{3}}{a^{-\frac{2}{3}}6a^2} = 2^2 \cdot a^2
\vphantom{\displaystyle
\begin{aligned}[c]
& \frac{A^m}{A^n} = A^{m-n} \\
& A^0 = 1 \text{ where } A \ne 0
\end{aligned}}$
\end{enumerate}
\end{document} |
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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Science Advisor
Homework Helper
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I've been reviewing Euler's proof for [itex]\zeta(2)[/itex] and though some of you might find it interesting too. We wish to find:
[tex]\zeta(2)=\sum_{n=1}^{\infty}\frac{1}{n^2}[/tex]
First a lemma:
If a polynomial P(x), has non-zero roots [itex]r_i[/itex], and P(0)=1, then:
[tex]P(x)=\left(1-\frac{x}{r_1}\right) \left(1-\frac{x}{r_2}\right) \left(1-\frac{x}{r_3}\right)...\left(1-\frac{x}{r_n}\right)[/tex]
I found that interesting to prove and will leave it for the reader if they wish to do so.
Now consider the polynomial:
[tex]P(x)=1-\frac{x^2}{3!}+\frac{x^4}{5!}-\frac{x^6}{7!}+ . . .[/tex]
Note that P(0)=1 but we don't know anything about it's roots yet.
Also, consider the power series for Sin(x):
[tex]Sin(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+...[/tex]
Note that:
[tex]xP(x)=Sin(x)[/tex]
Now, since the Sin(x) has roots of 0, and [itex]\pm \pi[/tex], and the 'x' accounts for the zero root on the left, we are left with P(x) containing the remaining roots. Thus P(x) has non-zero roots and we can thus use the lemma above and state:
[tex] \begin{align*} 1-\frac{x^2}{3!}+\frac{x^4}{5!}-\frac{x^6}{7!}+ . . .&=\left(1-\frac{x}{\pi}\right)\left(1+\frac{x}{\pi}\right)\left(1-\frac{x}{2\pi}\right)\left(1+\frac{x}{2\pi}\right)...\\ &=\left(1-\frac{x^2}{\pi^2}\right)\left(1-\frac{x^2}{4\pi^2}\right)\left(1-\frac{x^2}{9\pi^2}\right)\left(1-\frac{x^2}{16\pi^2}\right)... \end{align} [/tex]
Now I tried multiplying four of those together by hand and with extreme difficulty was able to do so in some manner of order. Apparently Euler was able to do many, many more since he calculated by hand [itex]\zeta(26)[/itex]!
Expanding this product and equating the coefficients to those of P(x) is the key to solving this problem . . .
[tex]\zeta(2)=\sum_{n=1}^{\infty}\frac{1}{n^2}[/tex]
First a lemma:
If a polynomial P(x), has non-zero roots [itex]r_i[/itex], and P(0)=1, then:
[tex]P(x)=\left(1-\frac{x}{r_1}\right) \left(1-\frac{x}{r_2}\right) \left(1-\frac{x}{r_3}\right)...\left(1-\frac{x}{r_n}\right)[/tex]
I found that interesting to prove and will leave it for the reader if they wish to do so.
Now consider the polynomial:
[tex]P(x)=1-\frac{x^2}{3!}+\frac{x^4}{5!}-\frac{x^6}{7!}+ . . .[/tex]
Note that P(0)=1 but we don't know anything about it's roots yet.
Also, consider the power series for Sin(x):
[tex]Sin(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+...[/tex]
Note that:
[tex]xP(x)=Sin(x)[/tex]
Now, since the Sin(x) has roots of 0, and [itex]\pm \pi[/tex], and the 'x' accounts for the zero root on the left, we are left with P(x) containing the remaining roots. Thus P(x) has non-zero roots and we can thus use the lemma above and state:
[tex]
\begin{align*}
1-\frac{x^2}{3!}+\frac{x^4}{5!}-\frac{x^6}{7!}+ . . .&=\left(1-\frac{x}{\pi}\right)\left(1+\frac{x}{\pi}\right)\left(1-\frac{x}{2\pi}\right)\left(1+\frac{x}{2\pi}\right)...\\
&=\left(1-\frac{x^2}{\pi^2}\right)\left(1-\frac{x^2}{4\pi^2}\right)\left(1-\frac{x^2}{9\pi^2}\right)\left(1-\frac{x^2}{16\pi^2}\right)...
\end{align}
[/tex]
Now I tried multiplying four of those together by hand and with extreme difficulty was able to do so in some manner of order. Apparently Euler was able to do many, many more since he calculated by hand [itex]\zeta(26)[/itex]!
Expanding this product and equating the coefficients to those of P(x) is the key to solving this problem . . . |
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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 ... |
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. |
$\mathsf{DFS(G, u) \text{}}$, $G = (V,E)$
Input : A Directed graph $G$ and a source vertex $u$. Find : Is $v$ reachable from vertex $u$ for all $v \in V$ ? Model of computation : Word RAM , one can represent a vector $A[1,\cdots ,n]$ of elements from a finite alphabet $\Sigma$ using $n \lg |\Sigma| + O(\lg^2 n)$ bits, such that any element of vector can be read or written in constant time.
$\mathsf{DFS \text{}}$ (Depth first search) takes $O(m + n)$ time and $O(n\log n)$ bits of space . Other known results for DFS is $O((m + n ) \log n)$ time and space $O(n)$ bits ( see this paper) (please note that DFS in this paper have defined as two phase process 1) Forward phase 2) Backtack phase )
It is a open question that $\mathsf{DFS \text{}}$ can be done in $O(m + n)$ time and $O(n)$ bits of space. (see this paper)
How to prove the lower bound that for $\mathsf{DFS \text{}}$ if we want to keep the runtime $O(m + n)$, we cannot do better than $O(n)$ in terms of space ?
Thanks in advance. |
מר מגניב, מר פינקל, ומר מצטרף לשאלה, שלום
מצורפת בזאת סקיצה של פתרון השאלה
Cain got $z^{10} \pmod {N_1}$ and $z^{22} \pmod {N_1}$. Using extended gcd to compute inverses modulo $N_1$, he can compute $z^{2} \pmod {N_1}$ (using the fact that $22-2\cdot 10 = 2$). Likewise, Cain got $z^{16} \pmod {N_2}$ and $z^{6} \pmod {N_2}$. Using extended gcd (and the fact that $3\cdot 6- 16 = 2$), he can compute $z^{2} \pmod {N_2}$.
Note that typically $z^{2} > N_1$, $z^{2} > {N_2}$ (over the integers), and taking square roots of $z^{2} \pmod {N_1}$, $z^{2} \pmod {N_2}$ modulo any of these moduli (individually) is as hard as factoring.
Using the Chinese remainder theorem, Cain can efficiently find an integer $r, \ \ 0 \leq r < N_1N_2$ such that $r = z^{2} \pmod {N_1}$ and $r = z^{2} \pmod {N_2}$. Since $z < N_1$ and $z< {N_2}$, we have $z^{2} < N_1N_2$, and therefore such $r$ will satisfy $r = z^2$ (without any modular reductions). So we got $z^{2}$, and we can now take its square root
over the integers, which can be efficiently computed. |
Problem 676
Let $V$ be the vector space of $2 \times 2$ matrices with real entries, and $\mathrm{P}_3$ the vector space of real polynomials of degree 3 or less. Define the linear transformation $T : V \rightarrow \mathrm{P}_3$ by
\[T \left( \begin{bmatrix} a & b \\ c & d \end{bmatrix} \right) = 2a + (b-d)x – (a+c)x^2 + (a+b-c-d)x^3.\]
Find the rank and nullity of $T$.Add to solve later
Problem 675
The space $C^{\infty} (\mathbb{R})$ is the vector space of real functions which are infinitely differentiable. Let $T : C^{\infty} (\mathbb{R}) \rightarrow \mathrm{P}_3$ be the map which takes $f \in C^{\infty}(\mathbb{R})$ to its third order Taylor polynomial, specifically defined by
\[ T(f)(x) = f(0) + f'(0) x + \frac{f^{\prime\prime}(0)}{2} x^2 + \frac{f^{\prime \prime \prime}(0)}{6} x^3.\] Here, $f’, f^{\prime\prime}$ and $f^{\prime \prime \prime}$ denote the first, second, and third derivatives of $f$, respectively.
Prove that $T$ is a linear transformation.Add to solve later
Problem 674
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 $T : \mathrm{P}_4 \rightarrow \mathrm{P}_{4}$ be the map defined by, for $f \in \mathrm{P}_4$,
\[ T (f) (x) = f(x) – x – 1.\]
Determine if $T(x)$ is a linear transformation. If it is, find the matrix representation of $T$ relative to the standard basis of $\mathrm{P}_4$.Add to solve later
Problem 673
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 $T : \mathrm{P}_3 \rightarrow \mathrm{P}_{5}$ be the map defined by, for $f \in \mathrm{P}_3$,
\[T (f) (x) = ( x^2 – 2) f(x).\]
Determine if $T(x)$ is a linear transformation. If it is, find the matrix representation of $T$ relative to the standard basis of $\mathrm{P}_3$ and $\mathrm{P}_{5}$.Add to solve later
Problem 672
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.Add to solve later
Problem 669 (a) Suppose that a $3\times 3$ system of linear equations is inconsistent. Is the coefficient matrix of the system nonsingular? (b) Suppose that a $3\times 3$ homogeneous system of linear equations has a solution $x_1=0, x_2=-3, x_3=5$. Is the coefficient matrix of the system nonsingular?
Add to solve later
(c) Let $A$ be a $4\times 4$ matrix and let \[\mathbf{v}=\begin{bmatrix} 1 \\ 2 \\ 3 \\ 4 \end{bmatrix} \text{ and } \mathbf{w}=\begin{bmatrix} 4 \\ 3 \\ 2 \\ 1 \end{bmatrix}.\] Suppose that we have $A\mathbf{v}=A\mathbf{w}$. Is the matrix $A$ nonsingular? Problem 668
Consider the system of differential equations
\begin{align*} \frac{\mathrm{d} x_1(t)}{\mathrm{d}t} & = 2 x_1(t) -x_2(t) -x_3(t)\\ \frac{\mathrm{d}x_2(t)}{\mathrm{d}t} & = -x_1(t)+2x_2(t) -x_3(t)\\ \frac{\mathrm{d}x_3(t)}{\mathrm{d}t} & = -x_1(t) -x_2(t) +2x_3(t) \end{align*} (a) Express the system in the matrix form. (b) Find the general solution of the system.
Add to solve later
(c) Find the solution of the system with the initial value $x_1=0, x_2=1, x_3=5$. Solve the Linear Dynamical System $\frac{\mathrm{d}\mathbf{x}}{\mathrm{d}t} =A\mathbf{x}$ by Diagonalization Problem 667 (a) Find all solutions of the linear dynamical system \[\frac{\mathrm{d}\mathbf{x}}{\mathrm{d}t} =\begin{bmatrix} 1 & 0\\ 0& 3 \end{bmatrix}\mathbf{x},\] where $\mathbf{x}(t)=\mathbf{x}=\begin{bmatrix} x_1 \\ x_2 \end{bmatrix}$ is a function of the variable $t$.
Add to solve later
(b) Solve the linear dynamical system \[\frac{\mathrm{d}\mathbf{x}}{\mathrm{d}t}=\begin{bmatrix} 2 & -1\\ -1& 2 \end{bmatrix}\mathbf{x}\] with the initial value $\mathbf{x}(0)=\begin{bmatrix} 1 \\ 3 \end{bmatrix}$. Prove that $\{ 1 , 1 + x , (1 + x)^2 \}$ is a Basis for the Vector Space of Polynomials of Degree $2$ or Less Problem 665
Let $\mathbf{P}_2$ be the vector space of polynomials of degree $2$ or less.
(a) Prove that the set $\{ 1 , 1 + x , (1 + x)^2 \}$ is a basis for $\mathbf{P}_2$.
Add to solve later
(b) Write the polynomial $f(x) = 2 + 3x – x^2$ as a linear combination of the basis $\{ 1 , 1+x , (1+x)^2 \}$. Problem 663
Let $\R^2$ be the $x$-$y$-plane. Then $\R^2$ is a vector space. A line $\ell \subset \mathbb{R}^2$ with slope $m$ and $y$-intercept $b$ is defined by
\[ \ell = \{ (x, y) \in \mathbb{R}^2 \mid y = mx + b \} .\]
Prove that $\ell$ is a subspace of $\mathbb{R}^2$ if and only if $b = 0$.Add to solve later
Problem 659
Fix the row vector $\mathbf{b} = \begin{bmatrix} -1 & 3 & -1 \end{bmatrix}$, and let $\R^3$ be the vector space of $3 \times 1$ column vectors. Define
\[W = \{ \mathbf{v} \in \R^3 \mid \mathbf{b} \mathbf{v} = 0 \}.\] Prove that $W$ is a vector subspace of $\R^3$. Problem 658
Let $V$ be the vector space of $n \times n$ matrices with real coefficients, and define
\[ W = \{ \mathbf{v} \in V \mid \mathbf{v} \mathbf{w} = \mathbf{w} \mathbf{v} \mbox{ for all } \mathbf{w} \in V \}.\] The set $W$ is called the center of $V$.
Prove that $W$ is a subspace of $V$.Add to solve later |
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. |
Hydrogen Physical and Chemical properties of water and heavy water Water : Water has high thermal capacity, high heat of vaporization, and hydrogen bond.
→ Angular / V -shape / bent shape
→ sp 3 hybridization → 2 lone pair and 2 bond pairs. Ice : At atmospheric pressure, has hexagonal shape at very low temperature, it has cubical structure. In polar region, in rivers, the aquatic life is surviving due to thermal isolation property of water. Physical properties :
H 2O
D 2O max. density 4°C 116°C Boiling point 100°C 101.4°C Melting point 0°C 3.8°C molecular weight 18 20 Dielectric constant 78.39 78.06 Viscosity less more Amphoteric nature : NH_{3} + H_{2}O \rightleftharpoons NH_4^+ + OH^{-} HCl + H_{2}O \rightleftharpoons H_{3}O^{+} + OH^{-} Autoprotolysis : H_{2}O + H_{2}O \rightleftharpoons H_{3}O^{+} + OH^{-} D_{2}O + D_{2}O \rightleftharpoons D_{3}O^{+} + OD^{-} Hydrates formation :
H - bonded water (hydrates)
interstitial water CuSO
4.5H 2O BaCl
2.2H 2O [Cu(H_{2}O)_{4}]^{+2}SO_4^{-2}H_{2}O BaCl
2.2D 2O Co - ordinated water : CrCl 3.6H 2O [Cr(H 2O) 6]Cl 3 [Cr(H_{2}O)_{6}]^{+3}3Cl^{-} Hard water : Which does not give lather (foam) with soap
Soft water : Which gives lather with soap. Hardness : Permanent hardness : due to Cl − and SO_4^{-2} of Ca and Mg Temporary hardness : due to bicarbonates of Ca and Mg. By using washing soda (Na 2CO 3.10H 2O) both temporary and permanent hardness can be removed. Temporary hardness : Na_{2}CO_{3} + Ca(HCO_{3})_{2} \rightarrow CaCO_{3}\downarrow + \ 2NaHCO_{3} Permanent hardness: Na_{2}CO_{3} + MCl_{2} \rightarrow MCO_{3}\downarrow + \ 2NaCl \ (M = Ca^{+2} \ or \ Mg^{+2} ) Na_{2}CO_{3} +MSO_{4} \rightarrow + 2Na_{2}SO_{4} + MCO_{3}\downarrow Calgon method : (sodium hexa meta phosphate)
Mg^{+2} + Na_{2}\left[Na_{4}(PO_{3})_{6}\right]\rightarrow Na_{2}\left[Na_{2}M(PO_{3})_{6}\right] + 2Na^{+}
Ca^{+2} + Na_{2}\left[Na_{4}(PO_{3})_{6}\right]\rightarrow Na_{2}\left[Na_{2}Ca(PO_{3})_{6}\right] + 2Na^{+} Synthetic Resin Method : Ion Exchange Method : ↓ Cation Exchange Resin :
Ca^{+2} + 2RCOOH \rightarrow (RCOO)_{2}Ca + 2H^{+}
Mg^{+2} + 2RSO_{3}H \rightarrow (RSO_{3})_{2}Ca + 2H^{+} Anion Exchange Resin : can be revived by using dil NaOH / Na 2CO 3 Cl^{-} + RNH_{3}OH \rightarrow RNH_{3}Cl + OH^{-} SO_4^{-2} + 2RNH_{3}OH \rightarrow R(NH_{3})_{2}SO_{4} + 2OH^{-} Calculation of degree of hardness of water : Equivalents of salts of Ca and Mg are given below.
CaCO
3 = 100 g Mg(HCO
3) 2 = 146 g Ca(HCO
3) 2 = 162 g CaSO
4 = 136 g MgSO
4 = 120 g CaCl
2 = 111 g MgCl
2 = 95 g
\tt Hardness \ in \ ppm = \frac{weight \ of \ salt}{mol \ wt \ of \ salt} \times \frac{100}{wt \ of \ H_{2}O \ in \ gram} \times 10^{6}
Uses of D 2O (Deuterium oxide) used a moderator (or) coolant in nuclear reaction As a traces used in reaction mechanism used as coolant in nuclear power plants. Part1: View the Topic in this Video from 0:05 to 6:26 Part2: View the Topic in this Video from 0:06 to 3:55
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Measurement of $J/\psi$ production as a function of event multiplicity in pp collisions at $\sqrt{s} = 13\,\mathrm{TeV}$ with ALICE
(Elsevier, 2017-11)
The availability at the LHC of the largest collision energy in pp collisions allows a significant advance in the measurement of $J/\psi$ production as function of event multiplicity. The interesting relative increase ...
Multiplicity dependence of jet-like two-particle correlations in pp collisions at $\sqrt s$ =7 and 13 TeV with ALICE
(Elsevier, 2017-11)
Two-particle correlations in relative azimuthal angle (Δ ϕ ) and pseudorapidity (Δ η ) have been used to study heavy-ion collision dynamics, including medium-induced jet modification. Further investigations also showed the ...
The new Inner Tracking System of the ALICE experiment
(Elsevier, 2017-11)
The ALICE experiment will undergo a major upgrade during the next LHC Long Shutdown scheduled in 2019–20 that will enable a detailed study of the properties of the QGP, exploiting the increased Pb-Pb luminosity ...
Azimuthally differential pion femtoscopy relative to the second and thrid harmonic in Pb-Pb 2.76 TeV collisions from ALICE
(Elsevier, 2017-11)
Azimuthally differential femtoscopic measurements, being sensitive to spatio-temporal characteristics of the source as well as to the collective velocity fields at freeze-out, provide very important information on the ...
Charmonium production in Pb–Pb and p–Pb collisions at forward rapidity measured with ALICE
(Elsevier, 2017-11)
The ALICE collaboration has measured the inclusive charmonium production at forward rapidity in Pb–Pb and p–Pb collisions at sNN=5.02TeV and sNN=8.16TeV , respectively. In Pb–Pb collisions, the J/ ψ and ψ (2S) nuclear ...
Investigations of anisotropic collectivity using multi-particle correlations in pp, p-Pb and Pb-Pb collisions
(Elsevier, 2017-11)
Two- and multi-particle azimuthal correlations have proven to be an excellent tool to probe the properties of the strongly interacting matter created in heavy-ion collisions. Recently, the results obtained for multi-particle ...
Jet-hadron correlations relative to the event plane at the LHC with ALICE
(Elsevier, 2017-11)
In ultra relativistic heavy-ion collisions at the Large Hadron Collider (LHC), conditions are met to produce a hot, dense and strongly interacting medium known as the Quark Gluon Plasma (QGP). Quarks and gluons from incoming ...
Measurements of the nuclear modification factor and elliptic flow of leptons from heavy-flavour hadron decays in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 and 5.02 TeV with ALICE
(Elsevier, 2017-11)
We present the ALICE results on the nuclear modification factor and elliptic flow of electrons and muons from open heavy-flavour hadron decays at mid-rapidity and forward rapidity in Pb--Pb collisions at $\sqrt{s_{\rm NN}}$ ...
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 ... |
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The ALICE Transition Radiation Detector: Construction, operation, and performance
(Elsevier, 2018-02)
The Transition Radiation Detector (TRD) was designed and built to enhance the capabilities of the ALICE detector at the Large Hadron Collider (LHC). While aimed at providing electron identification and triggering, the TRD ...
Constraining the magnitude of the Chiral Magnetic Effect with Event Shape Engineering in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV
(Elsevier, 2018-02)
In ultrarelativistic heavy-ion collisions, the event-by-event variation of the elliptic flow $v_2$ reflects fluctuations in the shape of the initial state of the system. This allows to select events with the same centrality ...
First measurement of jet mass in Pb–Pb and p–Pb collisions at the LHC
(Elsevier, 2018-01)
This letter presents the first measurement of jet mass in Pb-Pb and p-Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV and 5.02 TeV, respectively. Both the jet energy and the jet mass are expected to be sensitive to jet ...
First measurement of $\Xi_{\rm c}^0$ production in pp collisions at $\mathbf{\sqrt{s}}$ = 7 TeV
(Elsevier, 2018-06)
The production of the charm-strange baryon $\Xi_{\rm c}^0$ is measured for the first time at the LHC via its semileptonic decay into e$^+\Xi^-\nu_{\rm e}$ in pp collisions at $\sqrt{s}=7$ TeV with the ALICE detector. The ...
D-meson azimuthal anisotropy in mid-central Pb-Pb collisions at $\mathbf{\sqrt{s_{\rm NN}}=5.02}$ TeV
(American Physical Society, 2018-03)
The azimuthal anisotropy coefficient $v_2$ of prompt D$^0$, D$^+$, D$^{*+}$ and D$_s^+$ mesons was measured in mid-central (30-50% centrality class) Pb-Pb collisions at a centre-of-mass energy per nucleon pair $\sqrt{s_{\rm ...
Search for collectivity with azimuthal J/$\psi$-hadron correlations in high multiplicity p-Pb collisions at $\sqrt{s_{\rm NN}}$ = 5.02 and 8.16 TeV
(Elsevier, 2018-05)
We present a measurement of azimuthal correlations between inclusive J/$\psi$ and charged hadrons in p-Pb collisions recorded with the ALICE detector at the CERN LHC. The J/$\psi$ are reconstructed at forward (p-going, ...
Systematic studies of correlations between different order flow harmonics in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV
(American Physical Society, 2018-02)
The correlations between event-by-event fluctuations of anisotropic flow harmonic amplitudes have been measured in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV with the ALICE detector at the LHC. The results are ...
$\pi^0$ and $\eta$ meson production in proton-proton collisions at $\sqrt{s}=8$ TeV
(Springer, 2018-03)
An invariant differential cross section measurement of inclusive $\pi^{0}$ and $\eta$ meson production at mid-rapidity in pp collisions at $\sqrt{s}=8$ TeV was carried out by the ALICE experiment at the LHC. The spectra ...
J/$\psi$ production as a function of charged-particle pseudorapidity density in p-Pb collisions at $\sqrt{s_{\rm NN}} = 5.02$ TeV
(Elsevier, 2018-01)
We report measurements of the inclusive J/$\psi$ yield and average transverse momentum as a function of charged-particle pseudorapidity density ${\rm d}N_{\rm ch}/{\rm d}\eta$ in p-Pb collisions at $\sqrt{s_{\rm NN}}= 5.02$ ...
Energy dependence and fluctuations of anisotropic flow in Pb-Pb collisions at √sNN=5.02 and 2.76 TeV
(Springer Berlin Heidelberg, 2018-07-16)
Measurements of anisotropic flow coefficients with two- and multi-particle cumulants for inclusive charged particles in Pb-Pb collisions at 𝑠NN‾‾‾‾√=5.02 and 2.76 TeV are reported in the pseudorapidity range |η| < 0.8 ... |
I'm a little stuck on this question, any help would be appreciated!
Given that the Hamiltonian Path (HP) and the Hamiltonian Circuit/Cycles (HC) problems are known to be NP-complete, show that HCE is NP-complete.
HCE: Given an undirected graph G and an edge e of G, does G have a Hamiltonian circuit/cycle that uses e?
I've tried to approach this by showing that HC $\leq$ HCE, but I'm wondering if my approach is too convoluted.
EDIT:
I think I have a solution. Consider a graph $G=(V,E)$ where $V$ is the set of vertices in $G$, and $E$ is the set of edges in $G$. Let $f(G)=G'=(V',E')$ where
\begin{alignat*}{1} V'= & V\cup\{v_{\alpha},v_{\beta},v_{\gamma}\},v_{\alpha},v_{\beta},v_{\gamma}\notin V\\ E'= & E\cup\{(v_{\alpha},v_{\beta}),(v_{\beta},v_{\gamma})\}\cup\{\bigcup_{i\in\{\alpha,\gamma\},v\in V}(v,v_{i})\} \end{alignat*}
Let the edge $e=(v_{\alpha},v_{\beta})$.
$G'$is the graph $G$ with three additional vertices. $v_{\alpha}$ and $v_{\gamma}$ are connected to all the vertices in $G$ and $v_{\beta}$. $v_{\beta}$ has a degree of 2, and is only connected to $v_{\alpha}$and $v_{\gamma}$. $f$ can be computed in p-time.
Consider some $G$ that has a HP along vertices $v_{1},v_{2},...,v_{n}$. Then $G'$ will also have a path $v_{1},v_{2},...,v_{n}$ with each vertex only appearing once in the path. In order to turn this path into a HC, the three additional vertices will have to be included. In order to do so, the path has to be extended in either $v_{n},v_{\alpha},v_{\beta},v_{\gamma},v_{1}$ or $v_{n},v_{\gamma},v_{\beta},v_{\alpha},v_{1}$. $G'$ thus have a HC that will always include the edge $e$.
$\therefore$ G $\in$ HP $\implies$$f(G)=G'\in$ HCE
Consider some $G'$ with a HCE along some path $v_{1},v_{2},...,v_{n},v_{\alpha},v_{\beta},v_{\gamma},v_{1}$. Since $G$ has vertices $V=V'\backslash{v_{\alpha},v_{\beta},v_{\gamma}}$, $G$ has a HP along vertices $v_{1},v_{2},...,v_{n}$.
$\therefore$ $G'\in$ HCE $\implies G\in$ HP
And thus $G\in$ HP iff $f(G)=G'\in$ HCE. Since $f$ can run in p-time, HP $\leq$ HCE.
$\therefore$ HCE is NP-complete. |
Original research Open Access Published: A group from Hackenbush game Journal of the Egyptian Mathematical Society volume 27, Article number: 13 (2019) Article metrics
147 Accesses
Abstract
This research presents the reader some algebraic operations related to combinatorial games and gives a detailed outlook of a special game called Hackenbush game. We will deduce a group with a special feature with the help of some basic algebraic concepts. A fresh outlook to some combinatorial mathematical algebraic operations and concepts through the evaluation of a deduced group from this game.
Introduction
As a result of the many mathematical research done by Lagrange, Abel, Galois, and others in the fields of geometry, number’s theory, and algebraic equations at the end of the eighteenth century, early nineteenth century, the mathematical theory of “groups” was discovered [1,2,3]. That theory has led researchers to the discovery of many essential mathematical concepts such as “sub-groups,” “graphs and groups,” and plenty others. By building on previous credited works and researches, we can express mathematical groups as graphs. By analyzing the structures of graphs, we can deduce the properties of their related groups, which can be done by exploiting the notion of “identity” in “group theory.”
By expressing groups by their associated graphs, or in other words “identity graphs,” we can utilize the “identity graphs” theory’s equations and concepts to conclude the following.
Two elements x, y in a group are adjacent or can be joined by an edge if
x. y = e ( e, identity element of G).
Since we know that
x. y = y. x = e, there is no need to utilize the property of commutatively. It is by convention; every element is adjoined with the identity of the group G. This is the convention we use when trying to represent a group by a graph. The vertices correspond to the elements of the group, hence the order of the group G corresponds to the number of vertices in the identity graph [4, 5].
In this research, we will utilize the concepts of “graph theory” to analyze some combinatorial mathematical algebraic operations and concepts through the evaluation of a deduced group from a combinatorial game called “Hackenbush game.”
Combinatorial games theory
“Combinatorial game theory” describes the study of sequential games with perfect information. When playing “Hackenbush,” all players know all the possible outcomes from a given position with no randomness [6]. The game is defined with the following attributes:
(A)
There are two players, left and right.
(B)
There is always a finite number of positions, in addition to a starting position.
The game is played by the following rules:
(C)
Left and right alternate making moves.
(D)
Both players have access to all information always.
(E)
There is no randomness to moves made, such as rolling a dice.
(F)
A player loses when he/she can no longer make any legal moves.
(G)
The game ends when the ending condition is met.
If we have a game
G and this game includes options for player L named G and options for player R named L G . R
Now, we will try to explain the manner of dealing with the games, as well as showing its rules.
(A) G= { G | L G } R
Numbers represent the number of free moves or the moves possible to a particular player. By convention, positive numbers represent an advantage for left, while negative numbers represent an advantage for right. They are defined recursively with 0 being the base case.
(B)
0 = { | } we will explain Zero game in detail in the section “Hackenbush game”.
(C) G= 0 means the first player loses (D)
1 = {0 | }, 2 = {1 | }, 3 = {2 | }
(E)
−1 = { | 0}, − 2 = { | −1}, − 3 = { | −2 }
Star, written as * or {0|0}, is the first player’s wins since either player must move to a Zero game, which will be defined later in the research, therefore wins.
* + * = 0, because the first player must turn one copy of * to a 0, and the other player will have to turn the other copy of * to a 0 as well, at this point, the first player would lose, since 0 + 0 admits no moves.
Like all other games, the game * is neither positive nor negative, in which the first player wins.
(F)
∗ = {0 | 0 }
(G)
∗ = − ∗
(H) G║ 0 means the second player loses (I) G> 0 means left wins (J) G< 0 means right wins
Up, written as ↑, is a position in combinatorial game theory [7]. In standard notation,
(K)
↑ = {0| ∗}
Down, written as ↓, is a position in combinatorial game theory [7]. In standard notation,
(L)
↓ = {∗| 0 }
(M)
↓ = − ↑
We will explain the last two definitions in detail in the next section.
Hackenbush game
In Hackenbush, the drawn figure uses vertices and line segments, which are connected to a final line called the ground. Players take turns deleting one of their lines. Therefore, left and right take on the colors blue and red respectively. If there is, at any time, a path that cannot be drawn from the ground to a line segment, the segment is deleted. This allows for more strategic plays as a player can delete an opponent’s move during their turn [7, 8]. Figure 3 is an example of the Hackenbush game.
Now, we will discuss how the game is played. In a game represented by Fig. 3. We will pick left to make the first move. Left has two legal moves. He can either remove the line on the right or the line on the top of the red line. However, the latter move is best, because it allows right to take his middle piece, and effectively remove left’s piece with it. Likewise, right should take his middle piece if he moves first for that same exact reason. That would be the optimal plays in that current example. Mathematically, after every move, the resulting game’s board becomes a sub-game of the original game’s board, or in other words, a new game with new independent possibilities [9, 10].
Assuming optimal plays are always being made, finite values can be assigned to each game or sub-game. These values are determined by observing the advantages the left player has, after each move. For example, after Left makes a move, he has a moves advantage over right, but after right moves, left has
b moves advantage over right.
Those values could be written in the form {
a| b}. This form does not make any quantifiable sense at the moment, because of the actual value determined by what “ a” and “ b” are.
Before explaining how to find those values, we will first consider the case in which there are no legal moves for neither players, left or right. Therefore, the player’s (with no legal moves) score is left blank in the notation. If both players have no legal moves, the result is a Zero game.
Definition 3.1. A Zero-game
A Zero-game is a game that scores { | } = 0, or in other words, the player who moves next loose, assuming all moves made are optimal.
For Hackenbush, the simplest form of a Zero game equates to an empty board at the beginning. Thus, it is obvious the first player to move has no legal move, therefore loses. In another scenario, if we were to add one blue line, left would have 0 moves after he removes the only possible line, and right would have no legal move giving left a clear 1 move advantage, written {0| } = 1. This trend continues in such a manner that {
n| } = n + 1, as n equals the number of remaining moves left has, after making the optimal move [11].
However, what will happen if we add the red line instead? As done before, these values are applied with respect to the advantage of the left player. Thus, adding one line for right results in putting left at a one move disadvantage, or a (−1) advantage. So, adding one red line results in { | 0} = − 1. Adding two red lines would then be { | −1} = − 2. And so on to a general form of { | −
n} = − ( n + 1), as n equals the number of remaining moves right has after the optimal move.
Now with the groundwork out of the way, we can start using the values of sub-games to determine the value of an overall game, for example, see Fig. 4.
After left moves, right has one legal move. In that case, we clearly have a { | 0 } = − 1 situation. Conversely, if right moves first, we have { 0 | } = 1. This results in the overall game having a value of {−1| 1} = 0. It makes sense because we equally added one independent move for both players to an empty board [12], which means that the advantage does not change. But what about the game in Fig. 5?
Now both players have been given one line each. But if left moves, right’s only legal move is eliminated, but if right moves, left will still have one legal move. It is not so clear what the given value of this game by implementing the “scoring rules” discussed earlier. That is why we must utilize a new rule called the “simplicity rule,” to assign values to the previous example [13].
Definition 3.2. A * game
A * game (pronounced star game) is an infinitesimal game that scores.
{0| 0} = ∗ resulting in the first player to move wins, assuming all moves made are optimal [14].
For example, Hackenbush’s game has another line type that is green, which is claimable by either player. Then, we get the following game which results in a value of {0| 0} = ∗ (see Fig. 6). Building on the previous concept, we can see results like {
n| − n} = n∗, as n ∗ = n + ∗. It is also worth noting that ∗ has the property such that ∗ + ∗ = 0, see Fig. 6.
Furthermore, there are more two infinitesimal games.
Definition 3.3. An ↑ game
The negative version of an up game is called a down game and defined as follows.
Definition 3.4. An ↓game
A ↓ game (pronounced “down game”) is a negative infinitesimal game as the score is {∗| 0} which favors the right player [15].
The relation between “up” and “down” games represents the relation among the inverses of games. In every game, there is a way to reverse every move, which results in negating the value of the game originally obtained. In Hackenbush, this is obtained by replacing every red line with a blue line and vice versa. See Fig. 8.
Property 3.1. (A)
\( -\frac{1}{2^{\mathrm{n}}}<\downarrow <0<\uparrow <\frac{1}{2^{\mathrm{n}}} \)
for any integer n > 0.
(B)
↓ = − ↑ .
(C)
Both ↑ and ↓ are fuzz to ∗.
(D)
\( {\displaystyle \begin{array}{l}\left\{\uparrow |\downarrow \right\}=\left\{\uparrow |\ 0\right\}=\\ {}=\left\{0\ |\downarrow \right\}=\left\{0\ |\ 0\right\}=\ast .\end{array}} \)
Hackenbush group
For every Hackenbush’s game, a class can be defined. For example, the class of Zero game, see Fig. 9.
Let us define the binary addition operation to a set (
H) of unique classes defined above Theorem 4.1. Addition is commutative. Proof
Base case:
G + 0 = 0 + G for all games G. Theorem 4.2. Addition is associative
(x +
y) + z = x + ( y + z) The proof is clear.
We get negate by replacing the red (
R) sides with blue ( L) and vice versa and keep the green ( R, L) as it is, see Fig. 10.
The effect here is that all moves of left and right are switched.
G + (− G) is a second player wins. For example, right first:
Right to move, so left wins.
We can classify the games by its outcomes (Table 1).
Definition 4.1. Hackenbush group
The set
H is defined as a group because the following conditions are met.
if
h, h 1, h 2, h 3 ∈ H (A)
Closure: if
h 1, h 2belong to H, h 1+ h 2is also in H. (B)
Associative
h 1+ ( h 2+ h 3) = ( h 1+ h 2) + h 3for all h 1, h 2, h 3in H (C)
Identity element: there is a class 0 in
Hsuch that h+ 0 = 0 + h= hfor all hin H (D)
Inverse element: for each
hin H, there is an element − hin Hsuch that h+ (− h) = (− h) + h= 0 see Fig. 11 (E)
Commutative
h 1+ h 2= h 2+ h 1for all h 1, h 2in H Definition 4.2. The group under addition modulo (n)
We can create one to one function as shown in Table 2.
If we take
H = {0, 1, 2, … , n n− 1}; 0 is the CO (class of) Zero games, 1 is the class of One games and so on. Let us define the operation ⊕:
for every
h 1, h 2 ∈ H ; n
It is clear from the definition that the operation ⊕ is closure, associative, commutative, and has an identity element (0). For every element
h ∈ H there is an inverse element n n− h, or − h= n− h. Therefore, H is a commutative group under addition modulo n n. For example, H 6= {0, 1, 2, , 3, 4, 5} the group under addition modulo 6 is shown in Table 3.
The inverse of the element 1 is −1 = 6 − 1 = 5 and the inverse of 2 is −2 = 6 − 2 = 4
Example 4.1. This group (퐻 2 × 퐻 2, +) consists of two tuples with addition defined
element-by-element modulo 2. An addition to the group table is shown in Table 4.
What is special in Hackenbush’s group?
Hackenbush group (
H) is an infinite commutative group that has an element (not the identity) when added to itself, we get the identity element, see Fig. 12. Conclusion
In this research, we have utilized combinatorial mathematical algebraic operations and concepts, through the evaluation of a deduced group from a combinatorial game called Hackenbush game to define the unique Hackenbush group.
Now, what happens if we considered the multiplication’s operation on what we have analyzed above.
Or the division’s operation that is done in terms of reciprocal and multiplication
Would that lead to the creation of Rings, Fields, etc.… and would such research be applied to the world of Algebra? Only further research will tell how far Combinatorial Game Theory can go.
Abbreviations ↑:
Up game
↓:
Down game
CO:
Class of
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Acknowledgements
The authors are very thankful to the editor.
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About this article Received Accepted Published DOI Keywords Combinatorial games Hackenbush game Groups Graph theory Hackenbush group Moving to graph AMS Subject Classification 91Axx 20Nxx 05Cxx |
The exam practice question is as follows:
We call a relation $R$ anti-reflexiveiff $\forall a \in A : (a,a) \notin R$ and anti-symmetriciff $\forall a,b \in A : (a,b) \in R \rightarrow (b,a) \notin R$. Prove or refute that any anti-reflexive and transitive relation is also anti-symmetric.
I know it works because: $A = {(1,2),(2,3),(1,3)}$ is anti-reflexive, transitive and anti-symmetric - but I am struggling to prove this using
only the definitions of the set relations. |
We can multiply $a$ and $n$ by adding $a$ a total of $n$ times.
$$ n \times a = a + a + a + \cdots +a$$
Can we define division similarly using only addition or subtraction?
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It only takes a minute to sign up.Sign up to join this community
We can multiply $a$ and $n$ by adding $a$ a total of $n$ times.
$$ n \times a = a + a + a + \cdots +a$$
Can we define division similarly using only addition or subtraction?
To divide $60$ by $12$ using subtraction:
$$\begin{align*} &60-12=48\qquad\text{count }1\\ &48-12=36\qquad\text{count }2\\ &36-12=24\qquad\text{count }3\\ &24-12=12\qquad\text{count }4\\ &12-12=0\qquad\;\text{ count }5\;. \end{align*}$$
Thus, $60\div 12=5$.
You can even handle remainders:
$$\begin{align*} &64-12=52\qquad\text{count }1\\ &52-12=40\qquad\text{count }2\\ &40-12=28\qquad\text{count }3\\ &28-12=16\qquad\text{count }4\\ &16-12=4\qquad\;\text{ count }5\;. \end{align*}$$
$4<12$, so $64\div 12$ is $5$ with a remainder of $4$.
If $n$ is divisible by $b$ ($\frac{n}{b}$ is a whole number), then keep doing $n - b - b - b - b - b - \cdots - b$ until the value of that is $0$. The number of times you subtract $b$ is the answer. For example, $\frac{20}{4} \rightarrow 20 - 4 - 4 - 4 - 4 - 4$. We subtracted '$4$' five times, so the answer is $5$.
You can also use additions. One should use results from intermediate calculations to speed up.
Let us divide 63 by 12. $$ \begin{split} 12+12=24,&\qquad\textrm{count }1+1=2\\ 24+24=48,&\qquad\textrm{count }2+2=4\\ 48+24=72,&\qquad\textrm{count }4+2=6\textrm{ (exceeded 63)}\\ 48+12=60,&\qquad\textrm{count }4+1=5\textrm{ (so we try adding less)}\\ 63-60=3,&\qquad\textrm{(calculation of the remainder)}\\ \end{split} $$
You can define division as repeated subtraction:$${72\over 9}=72-9-9-9-9-9-9-9-9$$Subtracting by $9$ eight times is the same as subtracting by $72$ since $9\cdot8=72$. So, the answer is $8$. Also, this is why ${n\over a}=n-a-a-a-a\cdots$ for whatever whole number $a$ is other than zero.
If you have a remainder, then you just do this:$${13\over 2}=13-2-2-2-2-2-2-1$$as you just saw, subtracting by $2$ six times is the same as subtracting by $12$ since $2\cdot6=12$, but there's a remainder of $1$ being sutracted, so it's the same as subtracting by $13$ since $2\cdot6+1=13$, so the answer is $6$ R$1$ or $6.5$.
Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).
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According to Brahmagupta-Fibonacci Identity, for $p=q\cdot r$ we can prove if any two of the integers $p,q,r$ are of the form $a^2+n\cdot b^2,$ the third must of the same form
Now, I want to determine the $n$ such that if $M=r\cdot s=a^2+n\cdot b^2$ where $(r,s)=1, r$ and $s$ must be of the
same form for any integer pair $a,b$
Using Program, it seems that $n=1,2,3,7$ satisfy this
I think the proof for $2$ will be required here
We can safely assume $n$ to be square-free, as its square part(if any) can easily be merged with $b$
Now, if $(a,b)=d$ and $\frac aA=\frac bB=d\implies (A,B)=1$ and $ a^2+n\cdot b^2=d^2(A^2+n\cdot B^2)$
As $n$ is square-free, $(A^2,n)=(A,n)=D$(say) and $\frac A{A_1}=\frac nN=D\implies (A_1,N)=1$
Subsequently, $A^2+n\cdot B^2$ becomes $D^2\cdot A_1^2+N\cdot D\cdot B^2=D(D\cdot A^2_1+N\cdot B^2)$ which is not of the form $a^2+n\cdot b^2$ unless $D$ or $N=1$
So, we can focus on $a^2+n\cdot b^2,$ where $(a,n\cdot b)=1$
Some observations can be made:
$(1):$ If $p^c$ divides $M=a^2+n\cdot b^2,$ where integer $c\ge1$ and $p$ is prime,
$a^2\equiv-n\cdot b^2\pmod {p^c}\iff (a\cdot b^{-1})^2\equiv-n\pmod {p^c}$ $\implies -n$ must be a Quadratic residue of $p^c$
This is a necessary condition for $p^c$ to be of the form $a^2+n\cdot b^2$
So, if $2^c$ (where $c\ge3,$) divides $M,n\equiv-1\pmod 8$ as $x^2\equiv e\pmod {2^c}$ is solvable with exactly $4$ solutions $\iff e\equiv1\pmod 8$
$(2):$ Generalizing the solution of this problem,
Let's consider $2^x=a^2+n\cdot b^2$
As $n$ is odd and $(a,b)=1, a\cdot b$ must be odd
One value of $x$ is $y,$ i..e, $2^y=a_1^2+nb_1^2$
and if the smallest value of $x$ is $x_\text{min},$ i.e., $2^{x_\text{min}}=a_2^2+n\cdot b_2^2$
$2^{x_\text{min}+y}=(a_1^2+n\cdot b_1^2)(a_2^2+n\cdot b_2^2)=(a_1a_2\pm n\cdot b_1b_2)^2+n(a_1b_2\mp a_2b_1)^2$
Observe that $a_1a_2\pm n\cdot b_1b_2,a_1b_2\mp a_2b_1$ are even and the highest powers of $2$ that divides each will be same $=2^k$(say).
So, $2^{x_\text{min}+y-2k}=\left(\frac{a_1a_2\pm n\cdot b_1b_2}{2^k}\right)^2+n\left(\frac{a_1b_2\mp a_2b_1}{2^k}\right)^2$
Following this line, we can be prove
$4^k,k\ge 2$ can be represented as $a^2+15b^2$
$2^{3k+2},k\ge 1$ can be represented as $a^2+31b^2$ |
Tagged: determinant of a matrix Problem 718
Let
\[ A= \begin{bmatrix} 8 & 1 & 6 \\ 3 & 5 & 7 \\ 4 & 9 & 2 \end{bmatrix} . \] Notice that $A$ contains every integer from $1$ to $9$ and that the sums of each row, column, and diagonal of $A$ are equal. Such a grid is sometimes called a magic square.
Compute the determinant of $A$.Add to solve later
Problem 686
In each of the following cases, can we conclude that $A$ is invertible? If so, find an expression for $A^{-1}$ as a linear combination of positive powers of $A$. If $A$ is not invertible, explain why not.
(a) The matrix $A$ is a $3 \times 3$ matrix with eigenvalues $\lambda=i , \lambda=-i$, and $\lambda=0$.
Add to solve later
(b) The matrix $A$ is a $3 \times 3$ matrix with eigenvalues $\lambda=i , \lambda=-i$, and $\lambda=-1$. Problem 582
A square matrix $A$ is called
nilpotent if some power of $A$ is the zero matrix. Namely, $A$ is nilpotent if there exists a positive integer $k$ such that $A^k=O$, where $O$ is the zero matrix.
Suppose that $A$ is a nilpotent matrix and let $B$ be an invertible matrix of the same size as $A$.
Is the matrix $B-A$ invertible? If so prove it. Otherwise, give a counterexample. Problem 571
The following problems are Midterm 1 problems of Linear Algebra (Math 2568) at the Ohio State University in Autumn 2017.
There were 9 problems that covered Chapter 1 of our textbook (Johnson, Riess, Arnold). The time limit was 55 minutes.
This post is Part 2 and contains Problem 4, 5, and 6.
Check out Part 1 and Part 3 for the rest of the exam problems.
Problem 4. Let \[\mathbf{a}_1=\begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}, \mathbf{a}_2=\begin{bmatrix} 2 \\ -1 \\ 4 \end{bmatrix}, \mathbf{b}=\begin{bmatrix} 0 \\ a \\ 2 \end{bmatrix}.\]
Find all the values for $a$ so that the vector $\mathbf{b}$ is a linear combination of vectors $\mathbf{a}_1$ and $\mathbf{a}_2$.
Problem 5. Find the inverse matrix of \[A=\begin{bmatrix} 0 & 0 & 2 & 0 \\ 0 &1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 1 & 0 & 0 & 1 \end{bmatrix}\] if it exists. If you think there is no inverse matrix of $A$, then give a reason. Problem 6. Consider the system of linear equations \begin{align*} 3x_1+2x_2&=1\\ 5x_1+3x_2&=2. \end{align*} (a) Find the coefficient matrix $A$ of the system. (b) Find the inverse matrix of the coefficient matrix $A$. (c) Using the inverse matrix of $A$, find the solution of the system.
(
Linear Algebra Midterm Exam 1, the Ohio State University) Read solution Problem 546
Let $A$ be an $n\times n$ matrix.
The $(i, j)$
cofactor $C_{ij}$ of $A$ is defined to be \[C_{ij}=(-1)^{ij}\det(M_{ij}),\] where $M_{ij}$ is the $(i,j)$ minor matrix obtained from $A$ removing the $i$-th row and $j$-th column.
Then consider the $n\times n$ matrix $C=(C_{ij})$, and define the $n\times n$ matrix $\Adj(A)=C^{\trans}$.
The matrix $\Adj(A)$ is called the adjoint matrix of $A$.
When $A$ is invertible, then its inverse can be obtained by the formula
For each of the following matrices, determine whether it is invertible, and if so, then find the invertible matrix using the above formula.
(a) $A=\begin{bmatrix} 1 & 5 & 2 \\ 0 &-1 &2 \\ 0 & 0 & 1 \end{bmatrix}$. (b) $B=\begin{bmatrix} 1 & 0 & 2 \\ 0 &1 &4 \\ 3 & 0 & 1 \end{bmatrix}$. Problem 509
Using the numbers appearing in
\[\pi=3.1415926535897932384626433832795028841971693993751058209749\dots\] we construct the matrix \[A=\begin{bmatrix} 3 & 14 &1592& 65358\\ 97932& 38462643& 38& 32\\ 7950& 2& 8841& 9716\\ 939937510& 5820& 974& 9 \end{bmatrix}.\]
Prove that the matrix $A$ is nonsingular.Add to solve later
Problem 505
Let $A$ be a singular $2\times 2$ matrix such that $\tr(A)\neq -1$ and let $I$ be the $2\times 2$ identity matrix.
Then prove that the inverse matrix of the matrix $I+A$ is given by the following formula: \[(I+A)^{-1}=I-\frac{1}{1+\tr(A)}A.\]
Using the formula, calculate the inverse matrix of $\begin{bmatrix}
2 & 1\\ 1& 2 \end{bmatrix}$. Problem 486
Determine whether there exists a nonsingular matrix $A$ if
\[A^4=ABA^2+2A^3,\] where $B$ is the following matrix. \[B=\begin{bmatrix} -1 & 1 & -1 \\ 0 &-1 &0 \\ 2 & 1 & -4 \end{bmatrix}.\]
If such a nonsingular matrix $A$ exists, find the inverse matrix $A^{-1}$.
(
The Ohio State University, Linear Algebra Final Exam Problem) Read solution Eigenvalues of Orthogonal Matrices Have Length 1. Every $3\times 3$ Orthogonal Matrix Has 1 as an Eigenvalue Problem 419 (a) Let $A$ be a real orthogonal $n\times n$ matrix. Prove that the length (magnitude) of each eigenvalue of $A$ is $1$.
Add to solve later
(b) Let $A$ be a real orthogonal $3\times 3$ matrix and suppose that the determinant of $A$ is $1$. Then prove that $A$ has $1$ as an eigenvalue. |
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