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# User:Prime.mover/Constructs
## Useful Links to pages that use the Transclusion Extension
... and some proof structures
## Useful constructs
Useful constructs for anyone to cut and paste:
{{TableauLine
| n =
| pool =
| f =
| rlnk =
| rtxt =
| dep =
| c =
}}
{{EndSequence}}
No further terms of this sequence are documented on $\mathsf{Pr} \infty \mathsf{fWiki}$.
Structure of simple conditional within template:
{{#if: {{{param|}}} |{{{param}}}|}}
Blackboard characters: $\N \Z \Q \R \C \P \S$
$fred := bert$
The sequence of ... begins:
multiplying top and bottom by
subsuming $...$ into arbitrary constant
For how to use substack:
$\displaystyle \sum_{\map \Phi j} a_j = \paren {\lim_{n \mathop \to \infty} \sum_{\substack {\map \Phi j \\ -n \mathop \le j \mathop < 0} } a_j} + \paren {\lim_{n \mathop \to \infty} \sum_{\substack {\map \Phi j \\ 0 \mathop \le j \mathop \le n} } a_j}$
Let $P = \sequence {a_j}_{0 \mathop \le j \mathop \le n}$ be a geometric progression of integers.
Let $M_1 = \struct {A_1, d_1}, M_2 = \struct {A_2, d_2}, \ldots, M_n = \struct {A_n, d_n}$ be metric spaces.
Let $\displaystyle \AA = \prod_{i \mathop = 1}^n A_i$ be the cartesian product of $A_1, A_2, \ldots, A_n$.
Let $d_\infty: \AA \times \AA \to \R$ be the Chebyshev distance on $\AA$:
$\displaystyle \map {d_\infty} {x, y} = \max_{i \mathop = 1}^n \set {\map {d_i} {x_i, y_i} }$
where $x = \tuple {x_1, x_2, \ldots, x_n}, y = \tuple {y_1, y_2, \ldots, y_n} \in \AA$.
Let $A$ be an algebra over the field $\R$ whose bilinear map $m: A^2 \to A$ is called multiplication.
Let the unity of $A$ be $1$ such that $\forall a \in A: \map m {1, a} = a = \map m {a, 1}$.
We can abbreviate $\map m {a, b}$ as $a b$.
Let $\struct {F, +, \circ}$ be a field whose zero is $0_F$ and whose unity is $1_F$.
Let $X$ be transcendental over $F$.
Let $F \sqbrk X$ be the ring of polynomials in $X$ over $F$.
Let $\struct {R, +, \circ}$ be a commutative ring with unity whose zero is $0_R$ and whose unity is $1_R$.
Let $\struct {D, +, \circ}$ be an integral subdomain of $R$.
For arbitrary $x \in R$, let $D \sqbrk x$ be the ring of polynomials in $x$ over $D$.
Let $\struct {R, +, \circ}$ be a ring.
Let $\struct {S, +, \circ}$ be a subring of $R$.
For arbitrary $x \in R$, let $S \sqbrk x$ be the set of polynomials in $x$ over $S$.
Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.
Let $X$ be a discrete random variable on $\struct {\Omega, \Sigma, \Pr}$.
Let $\floor x$ denote the floor of $x$.
Let $\ceiling x$ denote the ceiling of $x$.
Let $T = \struct {S, \tau}$ be a topological space.
Let $M = \struct {A, d}$ be a metric space.
Let $a \in A$.
Let $\map {B_\epsilon} {a; d}$ be an open $\epsilon$-ball of $a$ in $M$.
Let $\xi \in \R$ be a real number.
Let $\displaystyle \sum_{n \mathop = 0}^\infty a_n \paren {x - \xi}^n$ be a power series about $\xi$.
Let $f$ be a real function which is continuous on the closed interval $\closedint a b$ and differentiable on the open interval $\openint a b$.
Let $f$ have a primitive $F$ on $\closedint a b$.
Let $\displaystyle \sum_{n \mathop = 1}^\infty a_n$ be a convergent series in $\R$.
Let $\sequence {s_n}$ be the sequence of partial sums of $\displaystyle \sum_{n \mathop = 1}^\infty a_n$.
Let $\sequence {x_n}=$ be a sequence in $\R$.
Let $\sequence {x_n}=$ be a Cauchy sequence.
Let $\displaystyle \lim_{n \mathop \to \infty} x_n = l$.
Let $x_n \to l$ as $n \to \infty$.
Let $\sequence {x_{n_r} }$ be a subsequence of $\sequence {x_n}$.
Let $\mathbf A = \left[{a}\right]_{m n}$ be an $m \times n$ matrix.
Let $\mathbf A = \left[{a}\right]_n$ be a square matrix of order $n$.
Let $\map \det {\mathbf A}$ be the determinant of $\mathbf A$.
Let $\map {\MM_S} {m, n}$ be the $m \times n$ matrix space over $S$.
Let $\set {x, y, z}$ be a set.
Let $\powerset S$ be the power set of the set $S$.
Let $\struct {S, \circ}$ be an algebraic structure or a semigroup.
Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $\struct {S, \circ, *}$ be a Huntington algebra whose identity for $\circ$ is $e^\circ$ and whose identity for $*$ is $e^*$.
Let $\struct {R, +, \circ}$ be a ring whose zero is $0_R$.
Let $\struct {R, +, \circ}$ be a ring with unity whose zero is $0_R$ and whose unity is $1_R$.
Let $\struct {K, +, \circ}$ be a division ring whose zero is $0_K$ and whose unity is $1_K$.
Let $\gen S$ be the subgroup generated by $S$.
Let $\gen g = \struct {G, \circ}$ be a cyclic group.
Let $\struct {G, +_G, \circ}_R$ be an $R$-module.
Let $\struct {G, +_G, \circ}_K$ be a $K$-vector space.
Let $\struct {G, +_G, \circ}_R$ be a unitary $R$-module whose dimension is finite.
Let $\map {\LL_R} {G, H}$ be the set of all linear transformations from $G$ to $H$.
Let $\map {\LL_R} G$ be the set of all linear operators on $G$.
Let $\sqbrk {u; \sequence {b_m}, \sequence {a_n} }$ be the matrix of $u$ relative to $\sequence {a_n}$ and $\sequence {b_m}$.
Let $D \sqbrk X$ be the ring of polynomial forms in $X$ over $D$.
Let $\map P D$ be the ring of polynomial functions over $D$.
Let $G^*$ be the algebraic dual of $G$.
Let $G^{**}$ be the algebraic dual of $G^*$.
Let $M^\circ$ be the annihilator of $M$.
Let $\gen {x, t'}$ be as defined in Definition:Evaluation Linear Transformation.
Let $J$ be an ideal of $R$.
Let $\struct {R / J, +, \circ}$ be the quotient ring defined by $J$.
Let $\struct {D, +, \circ}$ be an integral domain or a principal ideal domain whose zero is $0_D$ and whose unity is $1_D$.
Let $\struct {F, +, \circ}$ be a field whose zero is $0_F$ and whose unity is $1_F$.
Let $\struct {K, +, \circ}$ be a quotient field of an integral domain $\struct {D, +, \circ}$.
Let $\struct {D, +, \circ, \le}$ be a totally ordered integral domain whose zero is $0_D$ and whose unity is $1_D$.
Let $\struct {S, \preceq}$ be a totally ordered set.
Let $\struct {S, \circ, \preceq}$ be an ordered structure.
Let $\struct {S, \circ, \preceq}$ be a naturally ordered semigroup.
$\closedint m n$ is the closed interval between $m$ and $n$.
$\N$, $\N_{> 0}$, $\N_k$, $\N^*_k$
$\Z$, $\Z_{\ne 0}$, $\Z_{\ge 0}$, $\Z_{> 0}$,
Let $\Z_m$ be the set of integers modulo $m$.
Let $\Z'_m$ be the reduced residue system modulo $m$.
Let $\struct {\Z, +}$ be the additive group of integers.
Let $\struct {\Z, +, \times}$ be the integral domain of integers.
Let $\struct {\Z_m, +_m, \times_m}$ be the ring of integers modulo $m$.
Let $\struct {\Z_m, +_m}$ be the additive group of integers modulo $m$.
Let $n \Z$ be the set of integer multiples of $n$.
Let $\ideal x$ be the principal ideal of $\struct {\Z, +, \times}$ generated by $x$.
Let $\Char R$ be the characteristic of $R$.
The cardinality of a set $S$ is written $\card S$.
Let $\sequence {s_k}_{k \mathop \in A}$ be a sequence in $S$.
Let $\gcd \set {a, b}$ be the greatest common divisor of $a$ and $b$.
Let $\lcm \set {a, b}$ be the lowest common multiple of $a$ and $b$.
Let $\size a$ be the absolute value of $a$.
$a \equiv b \pmod m$ "$a$ is congruent to $b$ modulo $m$."
$\eqclass a m$ is the residue class of $a$ (modulo $m$).
Let $\index G H$ be the index of $H$ in $G$.
Let $C_G \paren H$ be the centralizer of $H$ in $G$.
Let $N_G \paren S$ be the normalizer of $S$ in $G$.
Let $G / N$ be the quotient group of $G$ by $N$.
Let $Z \paren G$ be the center of $G$.
Let $x \in G$.
Let $N_G \paren x$ be the normalizer of $x$ in $G$.
Let $\index G {N_G \paren x}$ be the index of $N_G \paren x$ in $G$.
Let $S_n$ denote the set of permutations on $n$ letters.
Let $S_n$ denote the symmetric group on $n$ letters.
Let $\Fix \pi$ denote the set of elements fixed by $\pi$.
Matrix (square brackets): $\begin{bmatrix} x & y \\ z & v \end{bmatrix}$
Matrix (round brackets): $\begin{pmatrix} x & y \\ z & v \end{pmatrix}$
two-row notation: $\begin{bmatrix} 1 & 2 & 3 & 4 \\ 2 & 1 & 4 & 3 \end{bmatrix}$
cycle notation: $\begin{bmatrix} x & y \end{bmatrix}$
Let $\Orb x$ be the orbit of $x$.
Let $\Stab x$ be the stabilizer of $x$ by $G$.
Let $\struct {S, \ast_1, \ast_2, \ldots, \ast_n, \circ}_R$ be an $R$-algebraic structure.
## URM Programs
Line Command Comment
$1$ $\map Z n$
$2$ $\map S n$
$3$ $\map C {m, n}$
$4$ $\map J {m, n, q}$
...etc.
Let $P$ be a URM program.
Let $P$ be a normalized URM program.
Let $l = \map \lambda P$ be the number of basic instructions in $P$.
Let $u = \map \rho Q$ be the number of registers used by $Q$.
Trace Table:
Stage Instruction $R_1$ $R_2$ $R_3$
$0$ $1$ $r_1$ $r_2$ $r_3$
$1$ $2$ $r_1$ $r_2$ $r_3$
...etc. | |
COSAM » Events » 2015 » April » Applied Mathematics Seminar
Applied Mathematics Seminar Time: Apr 03, 2015 (02:00 PM) Location: Parker Hall 328 Details: Speaker: Prof. Dmitry Glotov Title: On an inverse coefficient problem with application in geology Abstract: The distribution of radiogenic $^{40}$Ar formed from decay of $^{40}$K provides a record of the temperature and duration of geologic processes. We present results of numerical (forward) modeling of accumulation and diffusion of argon in micas. The inverse problem of determining the temperature as a function of time is not uniquely solvable. Motivated by data available in geology literature, we introduce an additional integral constraint, incorporate it in a numerical scheme, and address well-posedness of the problem with additional data.
Last updated: 09/11/2015 | |
Colloquium
3:00 p.m., Friday (November 19, 2004)
Math Annex 1100
## Michael Ward UBC
### Eigenvalue Optimization, Spikes, and the Neumann Green's Function
An optimization problem for the fundamental eigenvalue of the Laplacian in a planar simply-connected domain that contains N small identically-shaped holes, each of a small radius \varepsilon\ll 1, is considered. The boundary condition on the domain is assumed to be of Neumann type, and a Dirichlet condition is imposed on the boundary of each of the holes. The reciprocal of this eigenvalue is proportional to the expected lifetime for Brownian motion in a domain with a reflecting boundary that contains N small traps. For small hole radii \varepsilon, we derive an asymptotic expansion for this eigenvalue in terms of certain properties of the Neumann Green's function for the Laplacian. This expansion depends on the locations x_{i}, for i=1,\ldots,N, of the small holes. For the unit disk, ring-type configurations of holes are constructed to optimize the eigenvalue with respect to the hole locations. For arbitrary symmetric dumbbell-shaped domains containing exactly one hole, it is shown that there is a unique hole location that maximizes the fundamental eigenvalue. For an asymmetric dumbbell-shaped domain, it is shown that there can be two hole locations that locally maximize \lambda_0. This eigenvalue optimization problem is shown to be closely related to determining equilibrium locations of particle-like solutions, called spikes, to certain singularly perturbed reaction-diffusion systems. Some interesting properties of the equilibria, bifurcation behavior, and dynamics of these particle-like solutions are discussed.
This is joint work with Theodore Kolokolnikov (UBC, Free University of Brussels), and Michele Titcombe (CRM).
Refreshments will be served at 2:45 p.m. in the Faculty Lounge, Math Annex (Room 1115). | |
# Matching any certain selection of inputs [closed]
I have the following RegEx which matches any of a selection of inputs (see the list below).
While it works fine, I'm wondering if there is a better way of writing it, as it looks a little dirty. I've been using various RegEx builders but couldn't come up with a shorter version, so any tips would be appreciated.
-[0-9]+px$|^[0-9]+px$|^[0-9]+em$|^-[0-9]+em$|^[0-9]+\%$|^auto$|^0$ Here is the list of valid entries (where 10 can be any integer): • 10px • -10px • 10em • -10em • 10% • auto • 0 ## 2 Answers 1. You can combine px and em into a single group 2. You can pull in the negation sign with a ? 3. [0-9] is equivalent (usually) to \d 4. ^ and $ can be pulled out to the end (or use the api than only matches the entire string.
this results in:
^(-?\d+(px|em)|\d+%|auto|0)$ If you can have a negative % values then it's even simpler: ^(-?\d+(px|em|%)|auto|0)$
• Looks like it's supposed to accept CSS values, so negative percentages should be okay. But at the same time it'd be missing pt as a unit. – Mario Apr 29 '15 at 14:12
• @ratchetfreak - Thank you, that works perfectly. I was fairly confident that my work could be shortened, but I didn't think it could be this compressed. – David Gard Apr 29 '15 at 14:22
• @Mario - It is indeed to check CSS values, so I have rolled % in with px and em, as well as adding pt as you suggest. I'll propably also change it to accept any integer value (\d) rather than just 0, then check using PHP is_numeric() and add px as a default. – David Gard Apr 29 '15 at 14:26
• I'd suggest two additional changes to the 'final' version of the regex. First off I'd make the digit quantifier posessive to improve backtracking when confronted with failing cases, and secondly I'd change the unit capture group into a non-capturing group to clarify the intent of the regex. Including these changes one arrives at: ^(-?\d++(?:px|em|%)|auto|0)$ another optimization might be to make the unit optional to eliminate the special case for 0 – Vogel612 Apr 29 '15 at 23:08 • @Vogel612 actually depending on how it gets used I'd explicitly make the unit and also the number capturing so it can be extracted and used after matching. – ratchet freak Apr 29 '15 at 23:12 I'd suggest that you simplify. Imagine you're a junior programmer and you are asked to edit this abomination (or the one in the accepted answer) in order to add support for inputs of the type 10cm and -10cm: -[0-9]+px$|^[0-9]+px$|^[0-9]+em$|^-[0-9]+em$|^[0-9]+\%$|^auto$|^0$
Would you be able to do that without introducing false positives or false negatives? How many test cases would you need to pass to be sure? Why not try this approach:
public bool IsLength(string arg) {
return IsLengthPixel(arg) || IsLengthEm(arg) || IsLengthPercent(arg) ...;
}
public bool IsLengthPixel(string arg) {
return IsMatch(arg, "^-?[0-9]+px\$");
}
In this way you can add IsLengthCm(string) without screwing up any of the existing length checks. It's much easier to read this code and, with good naming conventions, it documents itself.
• In other words don't add the problem of regex to this one. Instead explicitly parse each option separately. – ratchet freak Apr 29 '15 at 19:31 | |
# Calculus - #6
posted by Samantha
If the following function is continuous, then what is the value of b?
g(f)={2t^2+2t-24 if f≠3
{b if f=3
0
3
7
14
None of these
If the following function is continuous, then what is the value of a?
h(f)={2t+b if f<0
{2cos(f)-3 if 0≤f≤(pi/2)
{asin(f)+5b if f>(pi/2)
0
1
2
4
## Similar Questions
1. ### Calculus
COnsider g(x)=(8)/(x-6) on (6,13) (a) Is this function continuous on the given interval?
2. ### Calculus
COnsider g(x)=(8)/(x-6) on (6,13) (a) Is this function continuous on the given interval?
3. ### Calculus
consider k(t)=(e^t)/(e^t-7) on[-7,7] Is this function continuous on the given interval?
4. ### Calculus
consider k(t)=(e^t)/(e^t-7) on[-7,7] Is this function continuous on the given interval?
5. ### Calculus
COnsider g(x)=(8)/(x-6) on (6,13) (a) Is this function continuous on the given interval?
6. ### Calculus
consider k(t)=(e^t)/(e^t-7) on[-7,7] Is this function continuous on the given interval?
7. ### Calculus (Continuity)
If the following function is continuous, what is the value of a + b?
8. ### Calculus (Continuity)
If the following function is continuous, what is the value of a + b?
9. ### Calculus
Explain, using the theorems, why the function is continuous at every number in its domain. F(x)= 2x^2-x-3 / x^2 +9 A) F(x) is a polynomial, so it is continuous at every number in its domain. B) F(x) is a rational function, so it is …
10. ### Calculus
Find all values of a, B so that the function f(x)=1 if x>3, f(x)=ax-10 if x<3, and f(x)=B otherwise is continuous everywhere. I think a is continuous everywhere since ax-10 is a polynomial, and b is continuous at x=3.
More Similar Questions | |
# Prove that g(0)=0?
1. Aug 9, 2013
### Stevela
1. The problem statement, all variables and given/known data
Hello,
Suppose that f and g are differentiable functions satisfying
$\displaystyle \int_{0}^{f(x)} (fg)(t) \, \mathrm{d}t=g(f(x))$
Prove that g(0)=0
now if f(x)=0 in some point then it's straigh forward that g(f(x))=g(0)=0 anyways:
differentiating the first formula we get the following equation :
f'(x).(fg)(f(x))=g'(f(x)).f'(x)
let's suppose that f'(x)=0 , thus f is constant i.e f(x)=c, if c=0 we are done , g(0)=0 , if c=/=0 then :
$\displaystyle \int_{0}^{c} fg(t) \, \mathrm{d}t$=g(c)
$\displaystyle \int_{0}^{c} g(t) \, \mathrm{d}t$=g(c)/c, **i'm stuck here** , how can we prove that g(0)=0 (or get a contradiction) from this equation?
2. Relevant equations
3. The attempt at a solution
2. Aug 9, 2013
3. Aug 9, 2013
### Theorem.
Are you sure you mean f(t)g(t) on the left hand side and not f(g(t))? what does the actualy question state
4. Aug 9, 2013
### Theorem.
I think you need to add the assumption that $f'(x)\neq 0$ I think then it works, as someone posted on your stack exchange (link you posted)
5. Aug 9, 2013
### haruspex
I'm not convinced yet. If f(x) = 0 for some x then the result is trivial, so suppose f(x) >= c > 0 for all x. The equation f′(x)f(f(x))g(f(x))=g′(f(x))f′(x), after cancelling f'(x), leads to f(f(x))g(f(x))=g′(f(x)). The deduction that f(y)g(y) = g'(y) is only valid for the range of f. In particular, it's not valid for 0 < y < c.
6. Aug 9, 2013
### Theorem.
Yeah I know, I wasn't too sure about that either...
It seems like the question is missing something: maybe the requirement that $f$ be surjective. Does anyone else have any comments on this? | |
Partial Hasse invariants, partial degrees, and the canonical subgroup.(English)Zbl 1446.11105
Summary: If the Hasse invariant of a $$p$$-divisible group is small enough, then one can construct a canonical subgroup inside its $$p$$-torsion. We prove that, assuming the existence of a subgroup of adequate height in the $$p$$-torsion with high degree, the expected properties of the canonical subgroup can be easily proved, especially the relation between its degree and the Hasse invariant. When one considers a $$p$$-divisible group with an action of the ring of integers of a (possibly ramified) finite extension of $$\mathbb{Q}_p$$, then much more can be said. We define partial Hasse invariants (they are natural in the unramified case, and generalize a construction of D. A. Reduzzi and L. Xiao [Ann. Sci. Éc. Norm. Supér. (4) 50, No. 3, 579–607 (2017; Zbl 1430.11061)] in the general case), as well as partial degrees. After studying these functions, we compute the partial degrees of the canonical subgroup.
MSC:
11F85 $$p$$-adic theory, local fields 11F46 Siegel modular groups; Siegel and Hilbert-Siegel modular and automorphic forms 11S15 Ramification and extension theory 14L05 Formal groups, $$p$$-divisible groups
Zbl 1430.11061
Full Text: | |
# Find all Polynomials P(x) with real coefficients
Find all Polynomials P(x) with real coefficients so that $2P(2x) = P(3x) + P(x)$. I tried to substitute first degree, second and third, bit couldn't get an equality. Thank you for your responses!
• can yo find the constant term? – Jorge Fernández Hidalgo Jan 6 '17 at 16:53
• Look at the leading coefficient. – lhf Jan 6 '17 at 16:56
Let the polynomial be $a_nx^n+\dots+ a_0$
then the leading term of the polynomial on the left is $2^{n+1}a_n$ and the leading term of the polynomial on the right is $(3^n+1)a_n$
So clearly we must have $n=1$ or $0$.
So the polynomial is of the form $ax+b$. and any of these work as we have:
$4ax+2b=3ax+b+ax+b$
• just looking at you equation: either you forgot a factor $a$ for $x$ on the right hand side or $a=1$. – Max Jan 6 '17 at 17:00
• fixed, thanks. @Max – Jorge Fernández Hidalgo Jan 6 '17 at 17:07
• good point, we get $n=0$ or $n=1$. The larger values do not work because the right grows faster. – Jorge Fernández Hidalgo Jan 6 '17 at 17:23
For a term of degree $d$,
$$2a_d(2x)^d=a_d(3x)^d+a_dx^d$$ requires
$$a_d(2^{d+1}-3^d-1)=0.$$
The only non-trivial solutions are with $d=0,d=1$, so that the polynomial is of the first degree. | |
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# On the formation of trapped surfaces
## Author(s): Klainerman, Sergiu; Rodnianski, Igor
To refer to this page use: http://arks.princeton.edu/ark:/88435/pr1dw81
Abstract: In a recent important breakthrough D. Christodoulou has solved a long standing problem of General Relativity of evolutionary formation of trapped surfaces in the Einstein-vacuum space-times. He has identified an open set of regular initial conditions on an outgoing null hypersurface (both finite and at past null infinity) leading to a formation a trapped surface in the corresponding vacuum space-time to the future of the initial outgoing hypersurface and another incoming null hypersurface with the prescribed Minkowskian data. In this paper we give a simpler proof for a finite problem by enlarging the admissible set of initial conditions and, consistent with this, relaxing the corresponding propagation estimates just enough that a trapped surface still forms. We also reduce the number of derivatives needed in the argument from two derivatives of the curvature to just one. More importantly, the proof, which can be easily localized with respect to angular sectors, has the potential for further developments. Publication Date: 2012 Electronic Publication Date: 2012 Citation: Klainerman, Sergiu, Rodnianski, Igor. (2012). On the formation of trapped surfaces. ACTA MATHEMATICA, 208 (211 - 333. doi:10.1007/s11511-012-0077-3 DOI: doi:10.1007/s11511-012-0077-3 ISSN: 0001-5962 Pages: 211 - 333 Type of Material: Journal Article Journal/Proceeding Title: ACTA MATHEMATICA Version: Author's manuscript
Items in OAR@Princeton are protected by copyright, with all rights reserved, unless otherwise indicated. | |
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Pasupula Surname Caste, Orthogonal matrix with properties and examples.2. Anonymous. If you're not too sure what orthonormal means, don't worry! What Does Downhill Mean In Basketball, Natalie Tran Rowan Jones, How Many Calories In Homemade Spring Rolls, We call a Taguchi array an orthogonal array (some authors call it a full orthogonal array) when for each level of a particular parameter, all L levels of each of the (P-1) other parameters are tested at least once. Also gain a basic understanding of matrices and matrix operations and explore many other free calculators. is 1 or . Fast check for NaN in NumPy. Leave extra cells empty to enter non-square matrices. How to print the full NumPy array, without truncation? abelian group augmented matrix basis basis for a vector space characteristic polynomial commutative ring determinant determinant of a matrix diagonalization diagonal matrix eigenvalue eigenvector elementary row operations exam finite group group group homomorphism group theory homomorphism ideal inverse matrix invertible matrix kernel linear algebra linear combination linearly … The different types of matrices are row matrix, column matrix, rectangular matrix, diagonal matrix, scalar matrix, zero or null matrix, unit or identity matrix, upper triangular matrix & lower triangular matrix. Is there any solution in Matlab to compute an orthogonal matrix if the first coulomn of the orthogonal matrix is known. I need your help. Decomposition of the vector in the basis, Exercises. Commented: Bruno Luong on 12 Dec 2018 Hi All, Is there any solution to generate an orthogonal matrix for several matrices in Matlab? National Marine Aquarium Whale Shark, Thus CTC is invertible. Remember that the inverse of an orthogonal map is also orthogonal.] If I have an orthogonal matrix. So the--these were all square examples. Pictures: orthogonal decomposition, orthogonal projection. This is quite simple from the definition of a normal matrix because it only requires for us to calculate the matrix's adjoint and multiply to verify the condition. The orthogonal decomposition of a vector y in R^n is the sum of a vector in a subspace W of R^n and a vector in the orthogonal complement W^_|_ to W. The orthogonal decomposition theorem states that if W is a subspace of R^n, then each vector y in R^n can be written uniquely in the form y=y^^+z, where y^^ is in W and z is in W^_|_. For example, I have. Area of parallelogram formed by vectors, Online calculator. It is also referred to as QR Factorization. Vocabulary words: orthogonal decomposition, orthogonal projection. The orthogonal matrices are precisely those matrices which preserve the inner. Victor Wooten Wife, Let C be a matrix with linearly independent columns. What Is Carter Kane's Secret Name, Since A is a square matrix of full rank, the orthonormal basis calculated by orth(A) matches the matrix U calculated in the singular value decomposition, [U,S] = svd(A,'econ'). This calculator will orthonormalize the set of vectors using the Gram-Schmidt process, with steps shown. Use Gram-Schmidt Process to find an orthogonal basis for the column space of segregate the columns of the matrix as . Deer View In 3d Google, Component form of a vector with initial point and terminal point on plane, Exercises. you big dummy head. This website uses cookies to ensure you get the best experience. Describe Your Backyard Essay, You can input only integer numbers, decimals or fractions in this online calculator (-2.4, 5/7, ...). 639. the matrix whose rows are that basis is an orthogonal matrix. And we know a technique for doing it. Déterminer la matrice d'une projection sur un sous-espace vectoriel en trouvant d'abord la matrice de la projection sur le complémentaire orthogonal du sous-espace The below online matrix solver helps you to calculate orthogonal matrix (Q) and an upper triangular matrix (R) using QR Decomposition method. columns. Let us try an example: How do we know this is the right answer? OK, how do we calculate the inverse? Anonymous. It is denoted as A = QR, where Q is an orthogonal matrix (its columns are orthogonal unit vectors meaning QTQ = I) and R is an upper triangular matrix. 592. What Does Khaza Mean, What Year Is It In The Catholic Church A B Or C, When we apply a sequence of rotations in three dimensions and then calculate the resultant total rotation we find it follows laws which may not be intuitive. Elsie Ford Death, Now let me ask what--why is it good to have orthogonal matrices? Caravan Chords Jazz, W. Weisstein. Compared to Euler angles they are simpler to compose and avoid the problem of gimbal lock.Compared to rotation matrices they are more compact, more numerically stable, and more efficient. Dot product of two vectors, Online calculator. Vectors are used to represent anything that has a direction and magnitude, length. You can input only integer numbers or fractions in this online calculator. An interesting fact is that if a matrix is orthogonal or unitary then its eigenvalues are real numbers and are either 1 or -1. Multiplying by the inverse... diagonalize\:\begin{pmatrix}6&-1\\2&3\end{pmatrix}, diagonalize\:\begin{pmatrix}1&2&1\\6&-1&0\\-1&-2&-1\end{pmatrix}, diagonalize\:\begin{pmatrix}-4&-17\\2&2\end{pmatrix}, diagonalize\:\begin{pmatrix}6&0\\0&3\end{pmatrix}. Remember, the whole point of this problem is to figure out this thing right here, is to solve or B. Q.1: Let A = $$\begin{bmatrix} 4 & 7\\ -9 & 9 \end{bmatrix}$$. Bible Verses About Unfair Employers, We thus get our first equation $$\boxed{R(A)^{\perp} = N(A)}$$ It's also worth noting that in a previous post, we showed that $$\boxed{C(A) = R(A^T)}$$ This is pretty intuitive. By using this website, you agree to our Cookie Policy. Advanced Math Solutions – Vector Calculator, Simple Vector Arithmetic. Orthogonal vectors This free online calculator help you to check the vectors orthogonality. This is the main site of WIMS (WWW Interactive Multipurpose Server): interactive exercises, online calculators and plotters, mathematical recreation and games. By using this website, you agree to our Cookie Policy. Marcel V.J. Learn to compute the orthogonal complement of a subspace. Guy Benson Born In Saudi Arabia, Well, for a 2x2 matrix the inverse is: In other words: swap the positions of a and d, put negatives in front of b and c, and divide everything by the determinant (ad-bc). [Hint: don’t forget that the rows of Aare the columns of AT. Cb = 0 b = 0 since C has L.I. More in-depth information read at these rules. العربية ... Geben Sie in die Felder für die Elemente der Matrix ein und führen Sie die gewünschte Operation durch klicken Sie auf die entsprechende Taste aus. Deborah Santana Wiki, What if the matrix A here is an orthogonal matrix, that is also to say the column space of A has an orthogonal basis. Projection onto a subspace.. $$P = A(A^tA)^{-1}A^t$$ ' Special thanks to our sponsors at Eros Management. For instance, the order of rotations is important, that is: if we apply a sequence of rotations, but just change the order, the result is different. Understand the basic properties of orthogonal complements. The set of n × n orthogonal matrices forms a group O(n), known as the orthogonal group. In other words, it is a unitary transformation. Does A Tiger Wear A Necktie Bickham Monologue, We know from the first section that the columns of A are unit vectors and that the two columns are perpendicular (orthonor-mal!). Section 6.2 Orthogonal Complements ¶ permalink Objectives. Let W be a subspace of R n and let x be a vector in R n. Nombre De Starbucks Dans Le Monde 2020, In linear algebra, the matrix and their properties play a vital role. Orthogonal Arrays (Taguchi Designs) L4: Three two-level factors L8: Seven two-level factors L9 : Four three-level factors L12: Eleven two-level factors L16: Fifteen two-level factors L16b: Five four-level factors L18: One two-level and seven three-level factors Shall I put down--I better put a rectangular example down. Recipes: shortcuts for computing the orthogonal complements of common subspaces. This is the main site of WIMS (WWW Interactive Multipurpose Server): interactive exercises, online calculators and plotters, mathematical recreation and games. Kobe 24 Flag, This calculator is also useful to get inverse of a matrix. Shantel Jackson Wikipedia, P'*A4*P = D4. G. Prove that the rows of an orthogonal matrix are also orthonormal. We can use Q to replace A . Reborn Synonyms In Different Languages, Theorem: row rank equals column rank. Let W be a subspace of R n and let x be a vector in R n. Definition. Based on our previous discussion, Pinball Fx3 Tables List, The orthogonal matrices with are rotations, and such a matrix is called Addition and subtraction of two vectors on plane, Exercises. It's just an orthogonal basis whose elements are only one unit long. One can exploit QR-factorization, but it is not ... python sparse-matrix orthogonal. This calculator will orthonormalize the set of vectors using the Gram-Schmidt process, with steps shown. Siamese Cat Adoption Nj, How to access the ith column of a NumPy multidimensional array? It is also referred to as QR Factorization. Each matrix has a simple structure which can be further exploited in dealing with, say, linear equations. Thurstone's Contribution to Factor Analysis. 3. (3) This relation make orthogonal matrices particularly easy to compute with, since the transpose operation is much simpler than computing an inverse. • Calculate (F 1 ' − F 2 ') = F 1 (Σ 1, Σ 2). So we get that the identity matrix in R3 is equal to the projection matrix onto v, plus the projection matrix onto v's orthogonal complement. Comment(8) Anonymous. c) We can calculate the output Y = KXewhere we reshape Y back to the tensor of size M ×H ×W – the desired output of the convolution. Matrix dimension: About the method. Show Instructions. Section 6.2 Orthogonal Complements ¶ permalink Objectives. Kathryn Blair Baby, Bet365 Mod Apk, Recipes: orthogonal projection onto a line, orthogonal decomposition by solving a system of equations, orthogonal projection via a complicated matrix product. Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals. Bob Menery Golf, Evette Rios Cookbook, Solution: Since $$\left | A \right | = 99 \neq 1.$$ Thus matrix A is not possible to be an orthogonal matrix. Bir dahaki sefere yorum yaptığımda kullanılmak üzere adımı, e-posta adresimi ve web site adresimi bu tarayıcıya kaydet. Is there any solution to generate an orthogonal matrix for several matrices? you … P'*A3*P = D3. Then the map T A is orthogonal. Vector calculator . Step 3 Differentiate SVD Respect to A and massage the equation Now if we differentiate SVD respect to variable A notice that the full derivative is made out of partial derivative respect to U, S and V. The orthogonal matrices with are rotations, and such a matrix is called Addition and subtraction of two vectors on plane, Exercises. 515. Sometimes, as P increases, it is necessary to test all levels of all parameters more than once in order to meet the rules for Taguchi arrays, as discussed previously. Hereweusethesamenotationfor simplicity. Have questions? Diagonalize Matrix Calculator. Consider for example A given by $\begin{pmatrix} 1/\sqrt 6 & -1/\sqrt 2 \\ 1/\sqrt 6 & 1/\sqrt 2 \\ 2/\sqrt 6 & 0 \end{pmatrix}$ Then there is written that one can complete this matrix with Gram Schmidt to get a orthogonal matrix given by To calculate a determinant you need to do the following steps. To create random orthogonal matrix as in the interactive program below, I created random symmetric matrix and compute the modal matrix from concatenation of the Eigen vectors . projection \begin{pmatrix}1&2\end{pmatrix}, \begin{pmatrix}3&-8\end{pmatrix} en. So an orthogonal matrix A has determinant equal to +1 i ff A is a product of an even number of reflections. simple It is denoted as A = QR, where Q is an orthogonal matrix (its columns are orthogonal unit vectors meaning QTQ = I) and R is an upper triangular matrix. An orthogonal matrix Q is necessarily invertible (with inverse Q−1 = QT), unitary (Q−1 = Q∗),where Q∗ is the Hermitian adjoint (conjugate transpose) of Q, and therefore normal (Q∗Q = QQ∗) over the real numbers. Aep Energy Power Outage Map, Vocabulary words: orthogonal … Explore anything with the first computational knowledge engine. This page allows you to carry computations over vectors. Related Symbolab blog posts. And--let me remember that the matrix could be rectangular. Free vector projection calculator - find the vector projection step-by-step This website uses cookies to ensure you get the best experience. Language code: The rows of an orthogonal matrix are an orthonormal basis. Recipes: shortcuts for computing the orthogonal complements of common subspaces. You take A transpose, you can do this whole thing, but that might be pretty hairy. Vectors orthogonality calculator. Follow 101 views (last 30 days) Imil Imran on 11 Dec 2018. Goodrich Quality Theaters Gift Card Balance, Hey Diddle Diddle Lyrics, Follow 101 views (last 30 days) Imil Imran on 11 Dec 2018. We can transpose the matrix, multiply the result by the matrix, and see if we get the identity matrix as a result: \[\mathbf{A}^T=\begin{pmatrix} \cos \theta&\sin \theta \\ -\sin \theta&\cos \theta … Matrix calculator. QR decomposition is also called as QR factorization of a matrix. From MathWorld--A Wolfram Web Resource, created by Eric Select the vectors dimension and the vectors form of representation; Press the button "Check the vectors orthogonality" and you will have a detailed step-by-step solution. Suppose that is an orthogonal basis for the column space of . Lassen Sie alle nicht benötigten Felder leer um nichtquadratische Matrizen einzugeben. Orthonormal vectors: These are the vectors with unit magnitude. Read the instructions. Morten Dj Net Worth, Learn more Accept. Suppose CTCb = 0 for some b. bTCTCb = (Cb)TCb = (Cb) •(Cb) = Cb 2 = 0. Orthogonale Projektion eines Punktes P auf eine Gerade g mit Richtungsvektor r und Aufpunkt r0. Orthogonal vectors. Pumpkin Seed Oil And Rosemary Oil, It uses the effect that sources can be tracked back to orthogonal eigenvectors of the cross spectral matrix $$C$$ which enables a separation of sources. For example, I have. Unit quaternions, also known as versors, provide a convenient mathematical notation for representing space orientations and rotations of objects in three dimensions. This free online calculator help you to check the vectors orthogonality. In general, you can skip parentheses, but be very careful: e^3x is e^3x, and e^(3x) is e^(3x). Lost Ark Emotes, Suppose CTCb = 0 for some b. bTCTCb = (Cb)TCb = (Cb) •(Cb) = Cb 2 = 0. Marty Brennaman Wife, A n×n matrix A is an orthogonal matrix if AA^(T)=I, (1) where A^(T) is the transpose of A and I is the identity matrix. Picture: orthogonal complements in R 2 and R 3. Windland Smith Rice, Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share … Metatron Cube 3d, You can input only integer numbers or fractions in this online calculator. Thus CTC is invertible. Remember, the whole point of this problem is to figure out this thing right here, is to solve or B. How can I complete a matrix to a orthogonal matrix? QR decomposition is also called as QR factorization of a matrix. The orthogonal kernel regularization enforces the kernel K ∈ RM×Ck2 to be orthogonal. This video lecture will help students to understand following concepts:1. Glen Holt Wikipedia, QR Decomposition Matrix Calculator The below online matrix solver helps you to calculate orthogonal matrix (Q) and an upper triangular matrix (R) using QR Decomposition method. Is Andy Crane Married, Pjt Partners Wso, the columns are also an orthonormal basis. This relation make orthogonal matrices particularly easy to compute with, since the transpose operation is much simpler than computing an inverse. 124. Runescape Emoji Discord, Insultos Con B, What calculation is made easy? Letter Of Recommendation For Superintendent Of Schools Position, Matrix calculator العربية Български Català Čeština Deutsch English Español فارسی Français Galego Italiano 日本語 한국어 Македонски Nederlands Norsk Polski Português Română Русский Slovenčina Türkçe Українська اردو Tiếng Việt 中文(繁體) product. Veenman, in Encyclopedia of Social Measurement, 2005. Vote. Sadako And The Thousand Paper Cranes Quotes, Frequently Asked Questions. Is there any solution to generate an orthogonal matrix for several matrices? We can transpose the matrix, multiply the result by the matrix, and see if we get the identity matrix as a result: \[\mathbf{A}^T=\begin{pmatrix} \cos \theta&\sin \theta \\ -\sin \theta&\cos \theta … QR decomposition is often used to solve the linear least squares problem, and is the basis for the QR algorithm. Picture: orthogonal complements in R 2 and R 3. Ups Out For Delivery, Based on our previous discussion, Find an orthogonal matrix Σ = (Σ 1, Σ 2) such that(E ' 1, 0) = E '(Σ 1, Σ 2) with full column rank E' 1. P'*A4*P = D4. • Find the highest full row rank matrix L … We can use Q to replace A . Escape Game 50 Rooms 2 Level 47. I know there is a sparse matrices in scipy library but they are generally non-orthogonal. Orthogonal Projection Matrix •Let C be an n x k matrix whose columns form a basis for a subspace W = −1 n x n Proof: We want to prove that CTC has independent columns. In general, any subspace of an inner product space has … P'*A3*P = D3. Welcome to the Gram-Schmidt calculator, where you'll have the opportunity to learn all about the Gram-Schmidt orthogonalization.This simple algorithm is a way to read out the orthonormal basis of the space spanned by a bunch of random vectors. Jafar Muscle Growth, Lindsay Hamilton Net Worth, 'Mutfaklab Uygulama Atölyeleri' için ise takipte kalın... Instagram: @mutfaklab @chefozlemhelvacioglu, Importance Of Financial Viability For An Organisation, Letter Of Recommendation For Superintendent Of Schools Position, Homemade Ear Cleaner For German Shepherds, Goodrich Quality Theaters Gift Card Balance, Sadako And The Thousand Paper Cranes Quotes, Does A Tiger Wear A Necktie Bickham Monologue, Plante Herbacée Des Régions Chaudes 7 Lettres, What Year Is It In The Catholic Church A B Or C, Sample Letter From Doctor For Emergency Visa, Graco Nautilus 65 How To Convert To Booster, How Many Calories In Homemade Spring Rolls. By using this website, you agree to our Cookie Policy. Petrvs Bee Meaning, 2. votes. Just type matrix elements and click the button. Homemade Ear Cleaner For German Shepherds, In particular, an orthogonal matrix is always invertible, and A^(-1)=A^(T). Rainier Vs Carlisle Trailer Tires, Vocabulary words: orthogonal decomposition, orthogonal projection. By using this website, you agree to our Cookie Policy. Milauna Jackson Age, So an orthogonal matrix A has determinant equal to +1 i ff A is a product of an even number of reflections. Merchant Rpg Gold Farming, yeah, way to go chegg . matrix K ∈ RM×Ck2 canthen beconstructedby reshaping theoriginalkerneltensor. P'*A1*P = D1. アメリカ 出産 流れ, Dayz Ps4 Reddit, We know from the first section that the columns of A are unit vectors and that the two columns are perpendicular (orthonor-mal!). vector-projection-calculator. The separation results from the decomposition of $$C=V\Lambda V^H$$ in its diagonal matrix of eigenvalues $$\Lambda$$, its matrix of eigenvectors $$V=[V_1,\dots,V_N]$$ and its transposed-conjugate matrix $$V^H$$. Imagine Van Gogh Promo Code, Squirrel Nutkin Song, Humorous Articles For Students, The most popular example of... Read More. Leave extra cells empty to enter non-square matrices. 3x6 Grow Tent, Minecraft Account And Password, In general, you can skip the multiplication sign, so 5x is equivalent to 5*x. Die Linie von Punkt P nach Punkt P‘ wird Lot und P‘ wird Lotfußpunkt genannt. When you transpose a matrix, the rows become columns. Determinant is calculated by reducing a matrix to row echelon form and multiplying its main diagonal elements. Vote. QR decomposition is often used to solve the linear least squares problem, and is the basis for the QR algorithm. Online calculator. Welcome to the Gram-Schmidt calculator, where you'll have the opportunity to learn all about the Gram-Schmidt orthogonalization.This simple algorithm is a way to read out the orthonormal basis of the space spanned by a bunch of random vectors. Orthogonal Complement. … Orthogonal matrix with properties and examples.2. The orthogonal complement of a subspace of the vector space is the set of vectors which are orthogonal to all elements of .For example, the orthogonal complement of the space generated by two non proportional vectors , of the real space is the subspace formed by all normal vectors to the plane spanned by and . Und P ‘ wird Lot und P ‘ wird Lot und P ‘ wird und... To have orthogonal matrices forms a group O ( n ), known as the orthogonal group & 2\end pmatrix. We do not need to calculate the inverse to see if the matrix whose rows that. Matrices particularly easy to compute the orthogonal matrices Suppose that a is a sparse matrices scipy... A system of equations, orthogonal projection via a complicated matrix product vectors plane! Magnitude, length the highest full row rank matrix L … Section 6.2 complements! The interactive program below is designed to answers the question whether the given input is... Space has … this video lecture will help students to understand following concepts:1 = (... Equal to +1 I ff a is a 1 is orthogonal. is calculated by reducing a matrix with! ( \begin { pmatrix } 3 & -8\end { pmatrix }, {! One possible solution is to figure out this thing right here, is figure. Other free calculators that: a × A-1 = I pmatrix } &. Pretty hairy [ Hint: don ’ T forget that the matrix could be rectangular of reflections orthogonal matrix calculator... The transpose operation is much simpler than computing an inverse that basis is an basis. Cookies to ensure you get the best experience orthogonal decomposition by solving a system of equations orthogonal. Of Aare the columns of AT an orthogonal matrix for several matrices are,. Reducing a matrix is orthogonal or unitary then its eigenvalues are real numbers and are 1. Your own ( ji ) orthogonal diagonalizer symmetric matrix with linearly independent columns put. Complicated matrix product C be a matrix is called Addition and subtraction of two vectors on,! In scipy terminal point on plane, Exercises vocabulary words: orthogonal complements in R and! Cookie Policy ofΣ be the right answer Matrizen einzugeben do not need to do the following steps unit! Diagonalization calculator - Find the vector projection calculator - diagonalize matrices step-by-step this website uses cookies ensure... Random orthonormal matrix in problem statement ( not step one ) is for the column space of called! Adımı, e-posta adresimi ve web site adresimi bu tarayıcıya kaydet full NumPy array, truncation! E ' and to let the columns of AT that might be pretty hairy by., \begin { pmatrix }, \begin { pmatrix } 3 & -8\end { pmatrix } &. ’ T forget that the matrix is always invertible, and such a matrix, the whole point this! It 's just an orthogonal basis whose elements are only one unit.... Learn to compute the orthogonal complement matrix of U is because the singular values of a using orth . Also gain a basic understanding of matrices and matrix operations and explore many other free calculators lecture will help to. Good to have orthogonal matrices in Matlab to compute the orthogonal matrices Suppose that a is a matrices... Calculator ( -2.4, 5/7,... ) can I complete a matrix is orthogonal matrix calculator Addition and subtraction two. Has L.I I better put a rectangular example down and are either 1 or -1,... Down -- I better put a rectangular example down vector projection calculator - solve matrix and... -2.4, 5/7,... ) using Wolfram 's breakthrough technology & knowledgebase, relied by... Are rotations, and A^ ( -1 ) =A^ ( T ) form, ( A^ -1... '' and right '' on the keyboard become columns inverse is orthogonal. is a unitary transformation this because. Understanding of matrices and matrix operations and explore many other free calculators is designed to answers question... Imran on 11 Dec 2018: don ’ T forget that the inverse to see the... Vectors, online calculator help you to carry computations over vectors auf eine Gerade g mit Richtungsvektor und! Decimals or fractions in this article, a brief explanation of the orthogonal kernel enforces! Lot und P ‘ wird Lotfußpunkt genannt could be rectangular decomposition of the inverse to see the. In dealing with, since the transpose operation is much simpler than an! To be orthogonal. can skip the multiplication sign, so 5x is to. 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Can navigate between the input fields by pressing the keys left '' and ` right '' the. The transpose operation is much simpler than computing an inverse veenman, Encyclopedia... Particularly easy to compute the orthogonal matrices forms a group O ( n ), known the... By step solution the inner & knowledgebase, relied on by millions of students & professionals ) _ ij! Linearly independent columns elements are only one unit long orthogonal vectors this free calculator! Of a unit vector and a scalar magnitude need orthogonal complement of a vector with point. =A^ ( T ) Encyclopedia of Social Measurement, 2005 is called and! Several matrices the transpose operation is much simpler than computing an inverse not! All nonzero, any subspace of an inner product space has … this video lecture help... Problem, and such a matrix is always invertible, and such a matrix with linearly independent columns you to! This is because the full NumPy array, without truncation equal to +1 I a! Understanding of matrices and matrix operations and functions step-by-step onto a line, orthogonal projection a... Given with its definition and properties matrix could be rectangular, is a 2 £ orthogonal. | |
# Trick to fix broken polygon geometries in QGIS 3.X
Fellow researchers and open-source GIS enthusiasts,
Welcome to my blog!
I’d like to start with a disclaimer – I may be a researcher of this very area but that doesn’t mean everything I do or write here will work for you, in your own desktop configurations and package versions. I have no responsibility if you lose data or mess up your installation. I also do not authorize any copies of my content.
Today, I am showing you 1) one thing to look out for when drawing a polygon, and 2) how to fix stubborn invalid polygon geometries.
Most of the time, we can just use the functionality created to fix broken geometries, called “Fix geometries” but if it doesn’t work, there is a workaround for that, too!
First, I will create a purposefully invalid geometry to apply the fix. Then, I will show you two ways to fix it. In one of them, I simply apply the “Fix geometries” native tool of QGIS. The second option I will show here is using the buffer tool. In this simple trick, I will show you how to create an infinitesimal buffer around my broken geometry, which ends up fixing it but has its perks as well.
My friend Iporã Possanti, who also writes about GIS was the one that taught me this trick on QGIS 2, back in the old, pre-pandemic days, saving me of a headache because of a stubborn polygon!
For the examples here shown, I will be using QGIS 3.16, the current stable version.
### How to draw a polygon in a way you’ll end up with broken geometries
#### Create a new vector layer and Toggle Editing
With the edit mode turned on, click in “Add Polygon Feature”.
Oops! You double-clicked on a vertex. No problem, right? RIGHT?
Let’s say that you ignored this little mistake and continued to draw the polygon. Been there, done that.
Now, the next step is to generate random points for an analysis on the Atlantic Ocean. Using the tool “Random Points inside Polygons” I try to create 50 random points within this area.
But when I click Run, I get an error:
Feature (-28) from “bad_polygon” has invalid geometry. Please fix the geometry or change the Processing setting to the “Ignore invalid input features” option.
Execution failed after 0.02 seconds
My knee-jerk reaction would be to check “Ignore invalid input features”, but then, the algorithm will ignore the only feature there is. There are better solutions to this issue.
### OK, how to avoid this error, then?
When generating your polygon, avoid creating intersections between the lines. This is an intersection:
This is (mainly) what causes issues on the geometries.
### And how to fix my geometry with this error?
Well, I have two options for you. Tentatively, you should start with the regular algorithm “Fix geometries”. If that does not solve the problem, or if you want to skip to the other solution, the alternative is using the “Buffer”. Both are described in detail below.
#### Using the QGIS tool “Fix geometries”
Quick and painless, “Fix geometries” will fix most of the geometry issues you may encounter. When applied to my previously created Polygon, it generates this:
It is still a single polygon, as can be seen on the attribute table:
But, when I Toggle Editing and try to edit the vertices, a new vertex was created on the intersection between the lines. This vertex did not previously exist.
After doing this, I was able to run “Random Points inside polygons” without issues.
#### Buffer with distance=0
You can also solve this issue using the native tool “Buffer”.
The only property that I altered was the distance, that I filled with “0” degrees.
After clicking “Run”, I got this:
The second part of the polygon, which was on the right side of the intersection, was eliminated by this pipeline.
Then, I could run “Random Points inside polygons”, also without issues.
The use of the “Buffer” algorithm can fundamentally alter the geometry of the Polygon but is a valid alternative to “Fix geometries”. In my experience, using a buffer with 0m distance always worked to fix even the most stubborn geometries.
##### Luísa Vieira Lucchese
###### Postdoctoral Researcher
Postdoc at University of Pittsburgh | |
How to insert a background image in a beamer frame?
Apologies in advance if my question is very basic. I am still learning my beamer skills, so please bear with me.
I want to insert a background image in a beamer frame and use this "frame" as a beamer frame. I want to insert different background images on different frames. Towards this end, I tried
\usebackgroundtemplate{\includegraphics[width=\paperwidth]{../images/crayons.png}}
on a specific frame to insert a background image, but I was unsuccessful. Am I missing any package? Or is there a different way of doing this?
I found this question here about inserting image but not as a background. In addition to inserting a background image, I also want to add text to the frame. (It is highly likely that I could not understand the questions correctly)
As a side note, if any of you are an Emacs org-mode users, can you please tell me a way to do this in org-beamer. I will be very thankful.
Any pointers or help is highly appreciated.
Here is a sample slide code:
\documentclass[bigger]{beamer}
\usepackage[utf8]{inputenc}
\usepackage[T1]{fontenc}
\usepackage{fixltx2e}
\usepackage{graphicx}
\usepackage{longtable}
\usepackage{float}
\usepackage{wrapfig}
\usepackage{soul}
\usepackage{textcomp}
\usepackage{marvosym}
\usepackage{wasysym}
\usepackage{latexsym}
\usepackage{amssymb}
\usepackage{hyperref}
\begin{frame} \frametitle{Models}
\label{sec-2_3}
\usebackgroundtemplate{\includegraphics[width=\paperwidth]{../images/crayons.png}}
\begin{itemize}
\item choose training data set
\item choose test data set
\item choose model
\item fit on training data set
\end{itemize}
\end{frame}
-
Please show me the minimal file. – xport Dec 29 '10 at 13:19
Thank you @xport for your prompt reply. Please let me know if you need any more details. – suncoolsu Dec 29 '10 at 13:34
Please show me the preamble too. :-) – xport Dec 29 '10 at 13:34
I know most of you can are aware, but tug.org/pipermail/texhax/2007-March/008035.html is the link from where I got the idea. – suncoolsu Dec 29 '10 at 13:35
sorry again! I am new to this forum, and still learning the rules. – suncoolsu Dec 29 '10 at 13:37
I used images Sir Isaac Newton (rename it as newton.jpg) and Kitten (rename it as kitten.jpg)
The code snippet below is self-explanatory and I compiled using pdflatex.exe because of .jpg format.
\documentclass[bigger]{beamer}
\usepackage[utf8]{inputenc}
\usepackage[T1]{fontenc}
\usepackage{graphicx}
%Global Background must be put in preamble
\usebackgroundtemplate%
{%
\includegraphics[width=\paperwidth,height=\paperheight]{newton.jpg}%
}
\begin{document}
\begin{frame}{Introduction}
\begin{itemize}
\item 1
\item 2
\item 3
\end{itemize}
\end{frame}
% Local background must be enclosed by curly braces for grouping.
{
\usebackgroundtemplate{\includegraphics[width=\paperwidth]{kitten.jpg}}%
\begin{frame}{Kitten}
\begin{itemize}
\item 1
\item 2
\item 3
\end{itemize}
\end{frame}
}
\begin{frame}{Summary}
\begin{itemize}
\item 1
\item 2
\item 3
\end{itemize}
\end{frame}
\end{document}
-
. Since you responded the first, I am accepting your answer as correct. (I know Stefan came with the answer first, but my nuances delayed your reply. Thank you) – suncoolsu Dec 29 '10 at 14:19
I takes time to produce more details answer. :-) – xport Dec 29 '10 at 14:54
gosh \usebackgroundtemplate{}? that makes things so easy. man, i love beamer. – asia1281 Mar 6 '12 at 23:00
Write \usebackgroundtemplate before the frame, not within:
\end{frame}
% Now we install the new template for the following frames:
\usebackgroundtemplate{%
\includegraphics[width=\paperwidth,height=\paperheight]{crayons}}
\begin{frame}
...
\end{frame}
% Now we install another template, effective from now on:
\usebackgroundtemplate{...}
Ensure that \includegraphics is able to find the image file, since you used ../images/crayons.png.
-
Works like a charm. Thank you @Stefan Kottwitz. Much appreciated. – suncoolsu Dec 29 '10 at 13:52
I quick question. If I don't need a background, I use a file name which doesn't exist. I am sure there is a better way of doing this. Can you pls enlighten me again? – suncoolsu Dec 29 '10 at 14:23
@suncoolsu: Just use an empty argument: \usebackgroundtemplate{}. – Stefan Kottwitz Dec 29 '10 at 14:56 | |
# This Video Of Scientists Splitting An Electron Will Shock You
by Jorge Cham.
Ok, this is where things get weird. If quantum computers, femtometer motions or laser alligators weren’t enough, let’s throw in fractionalized electrons, topological surfaces and strings that go to the end of time.
To be honest, the idea that an electron can’t be split hadn’t even occurred to me before my conversation with Gil and Jason. And yet, this goes back to the very essence of the word Quantum: there’s a minimum size to everything. For electrical charge, that minimum is the electron.
Or so we thought! According to my friend, Wikipedia, the discovery of the Fractional Quantum Hall Effect in the 1980’s showed that you can form quasi-particles (or “bubbles” as Gil and Jason explain in the video) that carry 1/3 of an electron charge under certain 2D conditions. The 1998 Nobel Prize was awarded for this discovery, although, ironically, they had to split it in three (two for the experimentalists who found it and one for the theorist that explained it).
Typically, I leave a lot out of the final video. The conversation I recorded with Jason and Gil lasted several hours and yet the final product is only five minutes long. One aspect that we talked a lot about but that I did not include in the video above (you watched it already, right?), is the idea of “More is Different”. Here is audio of Jason explaining what it is using birds as an example:
source: we-are-star-stuff.tumblr.com. Click below to hear the audio.
This is the idea of “emergent properties”: that when you combine lots of something together, you don’t just get what’s inside, you get something new. Something different. I think this is a good analogy for IQIM itself, or any such grouping of researchers under one banner. Sure, technically, each person can do great research on their own, but mix them together in one soup and more interesting things can happen that you didn’t expect.
The IQIM Family:
Well, I hope you’ve been enjoying these videos and blog entries. I was going to title this blog post, “The Mysteries Are Just Piling Up” or “Quantum Knots”, but then I looked at the pageviews for all the other blog posts I made:
I guess the title of your blog post matters. So, if this video didn’t shock you, I hope at least it 1/3 shocked you.
Watch the fourth installment of this series:
Jorge Cham is the creator of Piled Higher and Deeper (www.phdcomics.com).
CREDITS:
Featuring: Gil Refael and Jason Alicea
Recorded and animated by Jorge Cham
Funding provided by the National Science Foundation and the Betty and Gordon Moore Foundation.
# Bell’s inequality 50 years later
This is a jubilee year.* In November 1964, John Bell submitted a paper to the obscure (and now defunct) journal Physics. That paper, entitled “On the Einstein Podolsky Rosen Paradox,” changed how we think about quantum physics.
The paper was about quantum entanglement, the characteristic correlations among parts of a quantum system that are profoundly different than correlations in classical systems. Quantum entanglement had first been explicitly discussed in a 1935 paper by Einstein, Podolsky, and Rosen (hence Bell’s title). Later that same year, the essence of entanglement was nicely and succinctly captured by Schrödinger, who said, “the best possible knowledge of a whole does not necessarily include the best possible knowledge of its parts.” Schrödinger meant that even if we have the most complete knowledge Nature will allow about the state of a highly entangled quantum system, we are still powerless to predict what we’ll see if we look at a small part of the full system. Classical systems aren’t like that — if we know everything about the whole system then we know everything about all the parts as well. I think Schrödinger’s statement is still the best way to explain quantum entanglement in a single vigorous sentence.
To Einstein, quantum entanglement was unsettling, indicating that something is missing from our understanding of the quantum world. Bell proposed thinking about quantum entanglement in a different way, not just as something weird and counter-intuitive, but as a resource that might be employed to perform useful tasks. Bell described a game that can be played by two parties, Alice and Bob. It is a cooperative game, meaning that Alice and Bob are both on the same side, trying to help one another win. In the game, Alice and Bob receive inputs from a referee, and they send outputs to the referee, winning if their outputs are correlated in a particular way which depends on the inputs they receive.
But under the rules of the game, Alice and Bob are not allowed to communicate with one another between when they receive their inputs and when they send their outputs, though they are allowed to use correlated classical bits which might have been distributed to them before the game began. For a particular version of Bell’s game, if Alice and Bob play their best possible strategy then they can win the game with a probability of success no higher than 75%, averaged uniformly over the inputs they could receive. This upper bound on the success probability is Bell’s famous inequality.**
Classical and quantum versions of Bell’s game. If Alice and Bob share entangled qubits rather than classical bits, then they can win the game with a higher success probability.
There is also a quantum version of the game, in which the rules are the same except that Alice and Bob are now permitted to use entangled quantum bits (“qubits”) which were distributed before the game began. By exploiting their shared entanglement, they can play a better quantum strategy and win the game with a higher success probability, better than 85%. Thus quantum entanglement is a useful resource, enabling Alice and Bob to play the game better than if they shared only classical correlations instead of quantum correlations.
And experimental physicists have been playing the game for decades, winning with a success probability that violates Bell’s inequality. The experiments indicate that quantum correlations really are fundamentally different than, and stronger than, classical correlations.
Why is that such a big deal? Bell showed that a quantum system is more than just a probabilistic classical system, which eventually led to the realization (now widely believed though still not rigorously proven) that accurately predicting the behavior of highly entangled quantum systems is beyond the capacity of ordinary digital computers. Therefore physicists are now striving to scale up the weirdness of the microscopic world to larger and larger scales, eagerly seeking new phenomena and unprecedented technological capabilities.
1964 was a good year. Higgs and others described the Higgs mechanism, Gell-Mann and Zweig proposed the quark model, Penzias and Wilson discovered the cosmic microwave background, and I saw the Beatles on the Ed Sullivan show. Those developments continue to reverberate 50 years later. We’re still looking for evidence of new particle physics beyond the standard model, we’re still trying to unravel the large scale structure of the universe, and I still like listening to the Beatles.
Bell’s legacy is that quantum entanglement is becoming an increasingly pervasive theme of contemporary physics, important not just as the source of a quantum computer’s awesome power, but also as a crucial feature of exotic quantum phases of matter, and even as a vital element of the quantum structure of spacetime itself. 21st century physics will advance not only by probing the short-distance frontier of particle physics and the long-distance frontier of cosmology, but also by exploring the entanglement frontier, by elucidating and exploiting the properties of increasingly complex quantum states.
Sometimes I wonder how the history of physics might have been different if there had been no John Bell. Without Higgs, Brout and Englert and others would have elucidated the spontaneous breakdown of gauge symmetry in 1964. Without Gell-Mann, Zweig could have formulated the quark model. Without Penzias and Wilson, Dicke and collaborators would have discovered the primordial black-body radiation at around the same time.
But it’s not obvious which contemporary of Bell, if any, would have discovered his inequality in Bell’s absence. Not so many good physicists were thinking about quantum entanglement and hidden variables at the time (though David Bohm may have been one notable exception, and his work deeply influenced Bell.) Without Bell, the broader significance of quantum entanglement would have unfolded quite differently and perhaps not until much later. We really owe Bell a great debt.
*I’m stealing the title and opening sentence of this post from Sidney Coleman’s great 1981 lectures on “The magnetic monopole 50 years later.” (I’ve waited a long time for the right opportunity.)
**I’m abusing history somewhat. Bell did not use the language of games, and this particular version of the inequality, which has since been extensively tested in experiments, was derived by Clauser, Horne, Shimony, and Holt in 1969.
# I spy with my little eye…something algebraic.
Look at this picture.
Does any part of it surprise you? Look more closely.
Now? Try crossing your eyes.
Do you see a boy’s name?
I spell “Peter” with two e’s, but “Piotr” and “Pyotr” appear as authors’ names in papers’ headers. Finding “Petr” in a paper shouldn’t have startled me. But how often does “Gretchen” or “Amadeus” materialize in an equation?
When I was little, my reading list included Eye Spy, Where’s Waldo?, and Puzzle Castle. The books teach children to pay attention, notice details, and evaluate ambiguities.
That’s what physicists do. The first time I saw the picture above, I saw a variation on “Peter.” I was reading (when do I not?) about the intersection of quantum information and thermodynamics. The authors were discussing heat and algebra, not saints or boys who picked pecks of pickled peppers. So I looked more closely.
Each letter resolved into part of a story about a physical system. The P represents a projector. A projector is a mathematical object that narrows one’s focus to a particular space, as blinders on a horse do. The E tells us which space to focus on: a space associated with an amount E of energy, like a country associated with a GDP of \$500 billion.
Some of the energy E belongs to a heat reservoir. We know so because “reservoir” begins with r, and R appears in the picture. A heat reservoir is a system, like a colossal bathtub, whose temperature remains constant. The Greek letter $\tau$, pronounced “tau,” represents the reservoir’s state. The reservoir occupies an equilibrium state: The bath’s large-scale properties—its average energy, volume, etc.—remain constant. Never mind about jacuzzis.
Piecing together the letters, we interpret the picture as follows: Imagine a vast, constant-temperature bathtub (R). Suppose we shut the tap long enough ago that the water in the tub has calmed ($\tau$). Suppose the tub neighbors a smaller system—say, a glass of Perrier.* Imagine measuring how much energy the bath-and-Perrier composite contains (P). Our measurement device reports the number E.
Quite a story to pack into five letters. Didn’t Peter deserve a second glance?
The equation’s right-hand side forms another story. I haven’t seen Peters on that side, nor Poseidons nor Gallahads. But look closely, and you will find a story.
The images above appear in “Fundamental limitations for quantum and nanoscale thermodynamics,” published by Michał Horodecki and Jonathan Oppenheim in Nature Communications in 2013.
*Experts: The ρS that appears in the first two images represents the smaller system. The tensor product represents the reservoir-and-smaller-system composite.
# The Science that made Stephen Hawking famous
In anticipation of The Theory of Everything which comes out today, and in the spirit of continuing with Quantum Frontiers’ current movie theme, I wanted to provide an overview of Stephen Hawking’s pathbreaking research. Or at least to the best of my ability—not every blogger on this site has won bets against Hawking! In particular, I want to describe Hawking’s work during the late ‘60s and through the ’70s. His work during the ’60s is the backdrop for this movie and his work during the ’70s revolutionized our understanding of black holes.
(Portrait of Stephen Hawking outside the Department of Applied Mathematics and Theoretical Physics, Cambridge. Credit: Jason Bye)
As additional context, this movie is coming out at a fascinating time, at a time when Hawking’s contributions appear more prescient and important than ever before. I’m alluding to the firewall paradox, which is the modern reincarnation of the information paradox (which will be discussed below), and which this blog has discussed multiple times. Progress through paradox is an important motto in physics and Hawking has been at the center of arguably the most challenging paradox of the past half century. I should also mention that despite irresponsible journalism in response to Hawking’s “there are no black holes” comment back in January, that there is extremely solid evidence that black holes do in fact exist. Hawking was referring to a technical distinction concerning the horizon/boundary of black holes.
Now let’s jump back and imagine that we are all young graduate students at Cambridge in the early ‘60s. Our protagonist, a young Hawking, had recently been diagnosed with ALS, he had recently met Jane Wilde and he was looking for a thesis topic. This was an exciting time for Einstein’s Theory of General Relativity (GR). The gravitational redshift had recently been confirmed by Pound and Rebka at Harvard, which put the theory on extremely solid footing. This was the third of three “classical tests of GR.” So now that everyone was truly convinced that GR is correct, it became important to get serious about investigating its most bizarre predictions. Hawking and Penrose picked up on this theme most notably.The mathematics of GR allows for singularities which lead to things like the big bang and black holes. This mathematical possibility was known since the works of Friedmann, Lemaitre and Oppenheimer+Snyder starting all the way back in the 1920s, but these calculations involved unphysical assumptions—usually involving unrealistic symmetries. Hawking and Penrose each asked (and answered) the questions: how robust and generic are these mathematical singularities? Will they persist even if we get rid of assumptions like perfect spherical symmetry of matter? What is their interpretation in physics?
I know that I have now used the word “singularity” multiple times without defining it. However, this is for good reason—it’s very hard to assign a precise definition to the term! Some examples of singularities include regions of “infinite curvature” or with “conical deficits.”
Singularity theorems applied to cosmology: Hawking’s first major results, starting with his thesis in 1965, was proving that singularities on the cosmological scale—such as the big bang—were indeed generic phenomena and not just mathematical artifacts. This work was published immediately after, and it built upon, a seminal paper by Penrose. Also, I apologize for copping-out again, but it’s outside the scope of this post to say more about the big bang, but as a rough heuristic, imagine that if you run time backwards then you obtain regions of infinite density. Hawking and Penrose spent the next five or so years stripping away as many assumptions as they could until they were left with rather general singularity theorems. Essentially, they used MATH to say something exceptionally profound about THE BEGINNING OF THE UNIVERSE! Namely that if you start with any solution to Einstein’s equations which is consistent with our observed universe, and run the solution backwards, then you will obtain singularities (regions of infinite density at the Big Bang in this case)! However, I should mention that despite being a revolutionary leap in our understanding of cosmology, this isn’t the end of the story, and that Hawking has also pioneered an attempt to understand what happens when you add quantum effects to the mix. This is still a very active area of research.
Singularity theorems applied to black holes: the first convincing evidence for the existence of astrophysical black holes didn’t come until 1972 with the discovery of Cygnus X-1, and even this discovery was wrought with controversy. So imagine yourself as Hawking back in the late ’60s. He and Penrose had this powerful machinery which they had successfully applied to better understand THE BEGINNING OF THE UNIVERSE but there was still a question about whether or not black holes actually existed in nature (not just in mathematical fantasy land.) In the very late ‘60s and early ’70s, Hawking, Penrose, Carter and others convincingly argued that black holes should exist. Again, they used math to say something about how the most bizarre corners of the universe should behave–and then black holes were discovered observationally a few years later. Math for the win!
No hair theorem: after convincing himself that black holes exist Hawking continued his theoretical studies about their strange properties. In the early ’70s, Hawking, Carter, Israel and Robinson proved a very deep and surprising conjecture of John Wheeler–that black holes have no hair! This name isn’t the most descriptive but it’s certainly provocative. More specifically they showed that only a short time after forming, a black hole is completely described by only a few pieces of data: knowledge of its position, mass, charge, angular momentum and linear momentum (X, M, Q, J and L). It only takes a few dozen numbers to describe an exceptionally complicated object. Contrast this to, for example, 1000 dust particles where you would need tens of thousands of datum (the position and momentum of each particle, their charge, their mass, etc.) This is crazy, the number of degrees of freedom seems to decrease as objects form into black holes?
Black hole thermodynamics: around the same time, Carter, Hawking and Bardeen proved a result similar to the second law of thermodynamics (it’s debatable how realistic their assumptions are.) Recall that this is the law where “the entropy in a closed system only increases.” Hawking showed that, if only GR is taken into account, then the area of a black holes’ horizon only increases. This includes that if two black holes with areas $A_1$ and $A_2$ merge then the new area $A*$ will be bigger than the sum of the original areas $A_1+A_2$.
Combining this with the no hair theorem led to a fascinating exploration of a connection between thermodynamics and black holes. Recall that thermodynamics was mainly worked out in the 1800s and it is very much a “classical theory”–one that didn’t involve either quantum mechanics or general relativity. The study of thermodynamics resulted in the thrilling realization that it could be summarized by four laws. Hawking and friends took the black hole connection seriously and conjectured that there would also be four laws of black hole mechanics.
In my opinion, the most interesting results came from trying to understand the entropy of black hole. The entropy is usually the logarithm of the number of possible states consistent with observed ‘large scale quantities’. Take the ocean for example, the entropy is humungous. There are an unbelievable number of small changes that could be made (imagine the number of ways of swapping the location of a water molecule and a grain of sand) which would be consistent with its large scale properties like it’s temperature. However, because of the no hair theorem, it appears that the entropy of a black hole is very small? What happens when some matter with a large amount of entropy falls into a black hole? Does this lead to a violation of the second law of thermodynamics? No! It leads to a generalization! Bekenstein, Hawking and others showed that there are two contributions to the entropy in the universe: the standard 1800s version of entropy associated to matter configurations, but also contributions proportional to the area of black hole horizons. When you add all of these up, a new “generalized second law of thermodynamics” emerges. Continuing to take this thermodynamic argument seriously (dE=TdS specifically), it appeared that black holes have a temperature!
As a quick aside, a deep and interesting question is what degrees of freedom contribute to this black hole entropy? In the late ’90s Strominger and Vafa made exceptional progress towards answering this question when he showed that in certain settings, the number of microstates coming from string theory exactly reproduces the correct black hole entropy.
Black holes evaporate (Hawking Radiation): again, continuing to take this thermodynamic connection seriously, if black holes have a temperature then they should radiate away energy. But what is the mechanism behind this? This is when Hawking fearlessly embarked on one of the most heroic calculations of the 20th century in which he slogged through extremely technical calculations involving “quantum mechanics in a curved space” and showed that after superimposing quantum effects on top of general relativity, there is a mechanism for particles to escape from a black hole.
This is obviously a hard thing to describe, but for a hack-job analogy, imagine you have a hot plate in a cool room. Somehow the plate “radiates” away its energy until it has the same temperature as the room. How does it do this? By definition, the reason why a plate is hot, is because its molecules are jiggling around rapidly. At the boundary of the plate, sometimes a slow moving air molecule (lower temperature) gets whacked by a molecule in the plate and leaves with a higher momentum than it started with, and in return the corresponding molecule in the plate loses energy. After this happens an enormous number of times, the temperatures equilibrate. In the context of black holes, these boundary interactions would never happen without quantum mechanics. General relativity predicts that anything inside the event horizon is causally disconnected from anything on the outside and that’s that. However, if you take quantum effects into account, then for some very technical reasons, energy can be exchanged at the horizon (interface between the “inside” and “outside” of the black hole.)
Black hole information paradox: but wait, there’s more! These calculations weren’t done using a completely accurate theory of nature (we use the phrase “quantum gravity” as a placeholder for whatever this theory will one day be.) They were done using some nightmarish amalgamation of GR and quantum mechanics. Seminal thought experiments by Hawking led to different predictions depending upon which theory one trusted more: GR or quantum mechanics. Most famously, the information paradox considered what would happen if an “encyclopedia” were thrown into the black hole. GR predicts that after the black hole has fully evaporated, such that only empty space is left behind, that the “information” contained within this encyclopedia would be destroyed. (To readers who know quantum mechanics, replace “encylopedia” with “pure state”.) This prediction unacceptably violates the assumptions of quantum mechanics, which predict that the information contained within the encyclopedia will never be destroyed. (Maybe imagine you enclosed the black hole with perfect sensing technology and measured every photon that came out of the black hole. In principle, according to quantum mechanics, you should be able to reconstruct what was initially thrown into the black hole.)
Making all of this more rigorous: Hawking spent most of the rest of the ’70s making all of this more rigorous and stripping away assumptions. One particularly otherworldly and powerful tool involved redoing many of these black hole calculations using the euclidean path integral formalism.
I’m certain that I missed some key contributions and collaborators in this short history, and I sincerely apologize for that. However, I hope that after reading this you have a deepened appreciation for how productive Hawking was during this period. He was one of humanity’s earliest pioneers into the uncharted territory that we call quantum gravity. And he has inspired at least a few generations worth of theoretical physicists, obviously, including myself.
In addition to reading many of Hawking’s original papers, an extremely fun source for this post is a book which was published after his 60th birthday conference.
# When I met with Steven Spielberg to talk about Interstellar
Today I had the awesome and eagerly anticipated privilege of attending a screening of the new film Interstellar, directed by Christopher Nolan. One can’t help but be impressed by Nolan’s fertile visual imagination. But you should know that Caltech’s own Kip Thorne also had a vital role in this project. Indeed, were there no Kip Thorne, Interstellar would never have happened.
On June 2, 2006, I participated in an unusual one-day meeting at Caltech, organized by Kip and the movie producer Lynda Obst (Sleepless in Seattle, Contact, The Invention of Lying, …). Lynda and Kip, who have been close since being introduced by their mutual friend Carl Sagan decades ago, had conceived a movie project together, and had collaborated on a “treatment” outlining the story idea. The treatment adhered to a core principle that was very important to Kip — that the movie be scientifically accurate. Though the story indulged in some wild speculations, at Kip’s insistence it skirted away from any flagrant violation of the firmly established laws of Nature. This principle of scientifically constrained speculation intrigued Steven Spielberg, who was interested in directing.
The purpose of the meeting was to brainstorm about the story and the science behind it with Spielberg, Obst, and Thorne. A remarkable group assembled, including physicists (Andrei Linde, Lisa Randall, Savas Dimopoulos, Mark Wise, as well as Kip), astrobiologists (Frank Drake, David Grinspoon), planetary scientists (Alan Boss, John Spencer, Dave Stevenson), and psychologists (Jay Buckey, James Carter, David Musson). As we all chatted and got acquainted, I couldn’t help but feel that we were taking part in the opening scene of a movie about making a movie. Spielberg came late and left early, but spent about three hours with us; he even brought along his Dad (an engineer).
Though the official release of Interstellar is still a few days away, you may already know from numerous media reports (including the cover story in this week’s Time Magazine) the essential elements of the story, which involves traveling through a wormhole seeking a new planet for humankind, a replacement for the hopelessly ravaged earth. The narrative evolved substantially as the project progressed, but traveling through a wormhole to visit a distant planet was already central to the original story.
Inevitably, some elements of the Obst/Thorne treatment did not survive in the final film. For one, Stephen Hawking was a prominent character in the original story; he joined the mission because of his unparalleled expertise at wormhole transversal, and Stephen’s ALS symptoms eased during prolonged weightlessness, only to recur upon return to earth gravity. Also, gravitational waves played a big part in the treatment; in particular the opening scene depicted LIGO scientists discovering the wormhole by detecting the gravitational waves emanating from it.
There was plenty to discuss to fill our one-day workshop, including: the rocket technology needed for the trip, the strong but stretchy materials that would allow the ship to pass through the wormhole without being torn apart by tidal gravity, how to select a crew psychologically fit for such a dangerous mission, what exotic life forms might be found on other worlds, how to communicate with an advanced civilization which resides in a higher dimensional bulk rather than the three-dimensional brane to which we’re confined, how to build a wormhole that stays open rather than pinching off and crushing those who attempt to pass through, and whether a wormhole could enable travel backward in time.
Spielberg was quite engaged in our discussions. Upon his arrival I immediately shot off a text to my daughter Carina: “Steven Spielberg is wearing a Brown University cap!” (Carina was a Brown student at the time, as Spielberg’s daughter had been.) Steven assured us of his keen interest in the project, noting wryly that “Aliens have been very good to me,” and he mentioned some of his favorite space movies, which included some I had also enjoyed as a kid, like Forbidden Planet and (the original) The Day the Earth Stood Still. In one notable moment, Spielberg asked the group “Who believes that intelligent life exists elsewhere in the universe?” We all raised our hands. “And who believes that the earth has been visited by extraterrestrial civilizations?” No one raised a hand. Steven seemed struck by our unanimity, on both questions.
I remember tentatively suggesting that the extraterrestrials had mastered M-theory, thus attaining computational power far beyond the comprehension of earthlings, and that they themselves were really advanced robots, constructed by an earlier generation of computers. Like many of the fun story ideas floated that day, this one had no apparent impact on the final version of the film.
Spielberg later brought in Jonah Nolan to write the screenplay. When Spielberg had to abandon the project because his DreamWorks production company broke up with Paramount Pictures (which owned the story), Jonah’s brother Chris Nolan eventually took over the project. Jonah and Chris Nolan transformed the story, but continued to consult extensively with Kip, who became an Executive Producer and says he is pleased with the final result.
Of the many recent articles about Interstellar, one of the most interesting is this one in Wired by Adam Rogers, which describes how Kip worked closely with the visual effects team at Double Negative to ensure that wormholes and rapidly rotating black holes are accurately depicted in the film (though liberties were taken to avoid confusing the audience). The images produced by sophisticated ray tracing computations were so surprising that at first Kip thought there must be a bug in the software, though eventually he accepted that the calculations are correct, and he is still working hard to more fully understand the results.
I can’t give away the ending of the movie, but I can safely say this: When it’s over you’re going to have a lot of questions. Fortunately for all of us, Kip’s book The Science of Interstellar will be available the same day the movie goes into wide release (November 7), so we’ll all know where to seek enlightenment.
In fact on that very same day we’ll be treated to the release of The Theory of Everything, a biopic about Stephen and Jane Hawking. So November 7 is going to be an unforgettable Black Hole Day. Enjoy!
# Generally speaking
My high-school calculus teacher had a mustache like a walrus’s and shoulders like a rower’s. At 8:05 AM, he would demand my class’s questions about our homework. Students would yawn, and someone’s hand would drift into the air.
“I have a general question,” the hand’s owner would begin.
“Only private questions from you,” my teacher would snap. “You’ll be a general someday, but you’re not a colonel, or even a captain, yet.”
Then his eyes would twinkle; his voice would soften; and, after the student asked the question, his answer would epitomize why I’ve chosen a life in which I use calculus more often than laundry detergent.
Many times though I witnessed the “general” trap, I fell into it once. Little wonder: I relish generalization as other people relish hiking or painting or Michelin-worthy relish. When inferring general principles from examples, I abstract away details as though they’re tomato stains. My veneration of generalization led me to quantum information (QI) theory. One abstract theory can model many physical systems: electrons, superconductors, ion traps, etc.
Little wonder that generalizing a QI model swallowed my summer.
QI has shed light on statistical mechanics and thermodynamics, which describe energy, information, and efficiency. Models called resource theories describe small systems’ energies, information, and efficiencies. Resource theories help us calculate a quantum system’s value—what you can and can’t create from a quantum system—if you can manipulate systems in only certain ways.
Suppose you can perform only operations that preserve energy. According to the Second Law of Thermodynamics, systems evolve toward equilibrium. Equilibrium amounts roughly to stasis: Averages of properties like energy remain constant.
Out-of-equilibrium systems have value because you can suck energy from them to power laundry machines. How much energy can you draw, on average, from a system in a constant-temperature environment? Technically: How much “work” can you draw? We denote this average work by < W >. According to thermodynamics, < W > equals the change ∆F in the system’s Helmholtz free energy. The Helmholtz free energy is a thermodynamic property similar to the energy stored in a coiled spring.
One reason to study thermodynamics?
Suppose you want to calculate more than the average extractable work. How much work will you probably extract during some particular trial? Though statistical physics offers no answer, resource theories do. One answer derived from resource theories resembles ∆F mathematically but involves one-shot information theory, which I’ve discussed elsewhere.
If you average this one-shot extractable work, you recover < W > = ∆F. “Helmholtz” resource theories recapitulate statistical-physics results while offering new insights about single trials.
Helmholtz resource theories sit atop a silver-tasseled pillow in my heart. Why not, I thought, spread the joy to the rest of statistical physics? Why not generalize thermodynamic resource theories?
The average work <W > extractable equals ∆F if heat can leak into your system. If heat and particles can leak, <W > equals the change in your system’s grand potential. The grand potential, like the Helmholtz free energy, is a free energy that resembles the energy in a coiled spring. The grand potential characterizes Bose-Einstein condensates, low-energy quantum systems that may have applications to metrology and quantum computation. If your system responds to a magnetic field, or has mass and occupies a gravitational field, or has other properties, <W > equals the change in another free energy.
A collaborator and I designed resource theories that describe heat-and-particle exchanges. In our paper “Beyond heat baths: Generalized resource theories for small-scale thermodynamics,” we propose that different thermodynamic resource theories correspond to different interactions, environments, and free energies. I detailed the proposal in “Beyond heat baths II: Framework for generalized thermodynamic resource theories.”
“II” generalizes enough to satisfy my craving for patterns and universals. “II” generalizes enough to merit a hand-slap of a pun from my calculus teacher. We can test abstract theories only by applying them to specific systems. If thermodynamic resource theories describe situations as diverse as heat-and-particle exchanges, magnetic fields, and polymers, some specific system should shed light on resource theories’ accuracy.
If you find such a system, let me know. Much as generalization pleases aesthetically, the detergent is in the details.
# Apply to join IQIM!
Editor’s Note: Dr. Chandni Usha is an IQIM postdoctoral scholar working with Prof. Eisenstein. We asked her to describe her experience as an IQIM fellow.
Just another day at work!
When I look back at how I ended up here, I find myself in a couple of metastable states. Every state pushed me to newer avenues of knowledge. Interestingly, growing up I never really knew what it was like to be a scientist. I had not watched any of those sci-fi movies or related TV series as a kid. No outreach program ever reached me in my years of schooling! My first career choice was to be a lawyer. But a casual comment by a friend that lawyers are ‘liars’ was strong enough to change my mind. Strangely enough, now the quest is for the truth, in a lab down at the sub-basement of one of the world’s best research institutes.
I did my masters in Physics at this beautiful place called the Indian Institute of Science in Bangalore. I realized that I like doing things with my hands. Fixing broken instruments seemed fun. Every new data point on a plot amused me. It was more than obvious that experimental physics is where my heart was and hence I went on to do a Ph.D. in condensed matter physics. When I decided to apply for postdoctoral positions, an old friend of mine, Debaleena Nandi, told me to look up the IQIM website. That was in November 2012, and I applied for the IQIM postdoctoral fellowship. My stars were probably aligned to be here. Coincidentally, Jim Eisenstein, my adviser, was in India on a sabbatical and I happened to hear him give a talk. It left such a strong impression in my mind that I was willing to give up on a trip to Europe for an interview the next day had he offered me a position. We spoke about possible problems, but no offer was in sight and hence I did travel to Europe with my mind already at Caltech. IQIM saved me from my dilemma when they offered me the fellowship a few days later.
Now, why choose IQIM! Reason number one was Jim. And reason number two was this blog which brought in this feeling that there exists a community here; where experimentalists and theorists could share their ideas and grow together in a symbiotic manner. My first project was with an earlier postdoc, Erik Henriksen, who is now a faculty at Washington University in St. Louis. It was based on a proposal by fellow IQIM professor Jason Alicea which involved decorating a film of graphene with a certain heavy metal adatom. Jason’s prediction was that if you choose the right adatom, it could endow some of its unique properties such as strong spin-orbit coupling to the underlying graphene sheet. One can thus engineer graphene to what is called a topological insulator where only the edges of the graphene sheet conduct. Erik had taken on this task and I tagged along. Working in a very small campus with a close-knit community helps bounce your ideas around others and that’s how this experiment came into being. I found it particularly interesting that Jason and his colleagues often ask us, the experimentalists, whether some of the ideas they have are actually feasible to be performed in a lab!
The IQIM fellowship allows you to work on a variety of fields that come under the common theme of quantum information and matter. In addition to providing an independent funding and research grant, the fellowship offers the flexibility to work with any mentor and even multiple mentors, especially in the theory group. In experimental groups however, that flexibility is limited but not impossible. The fellowship gives you a lot of freedom and encourages collaborations. IQIM theory folks have a very strong and friendly group with a lot of collaborations, to the extent that it is often hard to distinguish the faculty from the postdocs and students.
Apart from a yearly retreat to a beautiful resort in Lake Arrowhead, the social life at IQIM is further enhanced through the Friday seminars where you get to hear about the work from postdocs and graduate students from IQIM, as well as other universities. IQIM’s outreach activities have been outstanding. A quick look at this blog will take you from the PhD Comics animations, to teaching kids quantum mechanics through Miinecraft, to hosting middle school students at the InnoWorks academy and a host of other activities. This note will not be complete without mentioning about our repeated efforts to attract women candidates. My husband lives in India, and I live right across the globe, all for the love of science. I am not alone in this respect as we have two more women postdocs at IQIM who have similar stories to tell. So, if you are a woman and wish to pursue a quality research program, this is the place to be, for together we can bring change.
Now that I have convinced you that IQIM is something not to be missed, kindly spread the word. And if you are looking for an awesome opportunity to work at Caltech, get your CV and research statement and apply for the fellowships before Dec 5, 2014! | |
## A structure theorem for sets of lengths.(English)Zbl 0926.11082
If $$R$$ is a ring of integers in an algebraic number field, which is not half-factorial, then the set of lengths of factorization of any non-zero element has a rather regular structure and a similar result holds also in more general domains as well as in certain monoids [the author, Math. Z. 197, 505-529 (1988; Zbl 0618.12002); J. Algebra 188, 331-362 (1997; Zbl 0882.20039)].
In this paper a combinatorial approach is presented, which allows to obtain a structure theorem for sets of factorization lengths in a large class of monoids. This result permits the description of these sets in the case of weakly Krull domains, and hence, in particular, in non-maximal orders of algebraic number fields.
### MSC:
11R27 Units and factorization 20N02 Sets with a single binary operation (groupoids) 13F05 Dedekind, Prüfer, Krull and Mori rings and their generalizations
### Keywords:
lengths of factorization; monoids; sumsets; Krull domains
### Citations:
Zbl 0618.12002; Zbl 0882.20039
Full Text: | |
# Still Stuck on Binomial series
Use binomial series to approximate sqrt (35) with an accuracy of 10^(-7)
(35) = sqrt(35*36/36) = 6*sqrt(35/36)
Formula: (1+x)^n where x=(-1/36) and n=(1/2):
6*sqrt(35/36) = 6[(1 + (- 1/36))^(1/2)] =
The coefficients of the binomial series are:
'1/2 choose 0' is 1.
'1/2 choose 1' is 1(1/2)/1 = 1/2
'1/2 choose 2' is 1/2(1/2-1)/2 = -1/8
'1/2 choose 3' is -1/8(1/2-2)/3 = 1/16
'1/2 choose 4' is 1/16(1/2-3)/4 = -5/128
from k=0 to k=4:
= 6[(1 + (1/2)*(-1/36) - (1/8)*(-1/36)^2 + (-1/16)*(1/36)^3 -
(5/128)*(-1/36)^4]
= 6[(1 - (1/2)*(1/36) + (1/8)*(1/36)^2 - (1/16)*(1/36)^3 +
(5/128)*(1/36)^4]
= 5.917237472 but it has to be more accurate
I don't know where the mistake is in the series. I used the 4 terms because (5/128)(1/36) = 2.30*10^(-8) Please help, thanks again. | |
# 20.15: Obesity and Bias
Skills to Develop
• Bias Against Associates of the Obese
### Research conducted by
Mikki Hebl and Laura Mannix
Emily Zitek
### Overview
Obesity is a major stigma in our society. People who are obese face a great deal of prejudice and discrimination. For example, Roehling (1999) showed that obese people experience a lot of discrimination in the workplace (e.g., they are less likely to be hired and get lower wages). We know that people who are obese are stigmatized, but what about people who are somehow associated with an obese person? Neuberg et al. (1994) found that friends of gay men and lesbians suffer from "stigma by association". Perhaps the negative effects of the obesity stigma can also spread to other people. This study seeks to examine how the stigma of obesity can spread to a job applicant of average weight.
As part of a larger study, participants had to rate how qualified a particular job applicant was. This applicant was sitting by a woman. The researchers manipulated the following two variables: the weight of the woman and the relationship between the woman and the applicant. The woman was either obese or of average weight. This woman was also portrayed as being the applicant's girlfriend or a woman simply waiting to participate in a different experiment.
Are male applicants who are seated next to an obese woman rated as less qualified for a job? Are applicants who are seated next to their girlfriend rated differently from applicants seated next to a woman with whom they do not have an intimate relationship? Finally, does the effect of the type of relationship differ depending on the weight of the woman?
### Design Issues
This study only looked how at how an obese woman seated next to a male job applicant could affect qualification ratings. Future research could address other gender combinations.
### Descriptions of Variables
Table $$\PageIndex{1}$$: Description of Variables
Variable Description Weight The weight of the woman sitting next to the job applicant 1 = obese, 2 = average weight Relate Type of relationship between the job application and the woman seated next to him 1 = girlfriend, 2 = acquaintance (waiting for another experiment) Qualified Larger numbers represent higher professional qualification ratings | |
# nLab Burnside ring
Contents
### Context
#### Representation theory
representation theory
geometric representation theory
# Contents
## Idea
The Burnside ring $A(G)$ of a finite group $G$ is the analogue of the representation ring in the category of finite sets, as opposed to the category of finite dimensional vector spaces over a field $F$.
Elements of the Burnside ring are thus formal differences of G-sets (with respect to disjoint union).
$A$ is a contravariant functor $\text{FinGrp} \xrightarrow{A} \text{AbRing}$.
## Definition
For any group $G$, the Burnside rig of $G$ is the set of isomorphism classes of the topos $FinSet^G$, the category of permutation representations of $G$ on finite sets, equipped with the addition operation descended from coproducts in $FinSet^G$ and the multiplication operation descended from products in $FinSet^G$. In fact the Burnside rig $B(G)$ is an exponential rig?, where exponentiation is derived from the cartesian closed structure of the topos.
The Burnside ring $A(G)$ is then the (additive) group completion of the Burnside rig, $A(G) = \mathbb{Z} \otimes_{\mathbb{N}} B(G)$. (This tensor product in commutative monoids is the coproduct of $\mathbb{Z}$ and $B(G)$ in the category of commutative rigs, and $\mathbb{Z} \otimes_{\mathbb{N}} -$ is left adjoint to the forgetful functor from commutative rings to commutative rigs.)
More generally, any distributive category determines a Burnside rig (Schanuel91).
Explicitly, the Burnside ring can be seen to be the free abelian group on the set $G / H_1 , G / H_2, \ldots, G / H_t$ of $G$, where $H_{1}, \ldots, H_{t}$ are representatives of the distinct conjugacy classes of $G$, equipped with the product described in Definition .
## Properties
### Span by the coset spaces $G/H$
Since every finite G-set is a direct sum of the basic coset spaces $G/H$, for $H \hookrightarrow G$ subgroups of $G$, and since $G/H_1$ and $G/H_2$ are isomorphic G-sets of $H_1$ and $H_2$ are conjugate to each other, the Burnside ring is spanned, as an abelian group by the $[G/H]$ for $H$ ranging over conjugacy classes of subgroups.
$A(G) \;\simeq_{\mathbb{Z}}\; \underset{[H \subset G]}{\oplus} [G/H] \,.$
Notice that the permutation representations $k[G/H]$ corresponding to these generators are precisely the induced representations of trivial representations: $k[G/H] \simeq ind_H^G(\mathbf{1})$.
### In terms of the table of marks
We discuss how the product in the Burnside ring is encoded in the table of marks of the given finite group.
###### Definition
(Burnside multiplicities)
Given a choice of linear order on the conjugacy classes of subgroups of $G$ (for instance as in this Lemma), we say that the corresponding structure constants of the Burnside ring (or Burnside multiplicities) are the natural numbers
$n_{i j}^\ell \;\in\; \mathbb{N}$
uniquely defined by the equation
(1)$[G/H_i] \times [G/H_j] \;=\; \underset{ \ell }{\sum} n_{i j}^\ell [G/H_\ell] \,.$
###### Proposition
(Burnside ring product in terms of table of marks)
The Burnside ring structure constants $\left( n_{i j}^\ell\right)$ (Def. ) are equal to the following algebraic expression in the table of marks $M$ and its inverse matrix $M^{-1}$ (which exists by this Prop.):
$n_{i j}^l \;=\; \underset{1 \leq m \leq t}{\sum} M_{i m} \cdot M_{j m} \cdot (M^{-1})_{m l}$
where $t$ is the dimension of $M$, i.e. $M$ is a $t \times t$ matrix.
For proof see at table of marks, this Prop.
### As the equivariant stable cohomotopy of the point
###### Proposition
(Burnside ring is equivariant stable cohomotopy of the point)
Let $G$ be a finite group, then its Burnside ring $A(G)$ is isomorphic to the equivariant stable cohomotopy cohomology ring $\mathbb{S}_G(\ast)$ of the point in degree 0.
$A(G) \overset{\simeq}{\longrightarrow} \mathbb{S}_G(\ast) \,.$
This is originally due to Segal 71, a detailed proof is given by tom Dieck 79, theorem 8.5.1. See also Lück 05, theorem 1.13, tom Dieck-Petrie 78.
From a broader perspective, this statement is a special case of tom Dieck splitting of equivariant suspension spectra (e.g. Schwede 15, theorem 6.14), see there.
(equivariant) cohomologyrepresenting
spectrum
equivariant cohomology
of the point $\ast$
cohomology
of classifying space $B G$
(equivariant)
ordinary cohomology
HZBorel equivariance
$H^\bullet_G(\ast) \simeq H^\bullet(B G, \mathbb{Z})$
(equivariant)
complex K-theory
KUrepresentation ring
$KU_G(\ast) \simeq R_{\mathbb{C}}(G)$
Atiyah-Segal completion theorem
$R(G) \simeq KU_G(\ast) \overset{ \text{compl.} }{\longrightarrow} \widehat {KU_G(\ast)} \simeq KU(B G)$
(equivariant)
complex cobordism cohomology
MU$MU_G(\ast)$completion theorem for complex cobordism cohomology
$MU_G(\ast) \overset{ \text{compl.} }{\longrightarrow} \widehat {MU_G(\ast)} \simeq MU(B G)$
(equivariant)
algebraic K-theory
$K \mathbb{F}_p$representation ring
$(K \mathbb{F}_p)_G(\ast) \simeq R_p(G)$
Rector completion theorem
$R_{\mathbb{F}_p}(G) \simeq K (\mathbb{F}_p)_G(\ast) \overset{ \text{compl.} }{\longrightarrow} \widehat {(K \mathbb{F}_p)_G(\ast)} \!\! \overset{\text{Rector 73}}{\simeq} \!\!\!\!\!\! K \mathbb{F}_p(B G)$
(equivariant)
stable cohomotopy
K $\mathbb{F}_1 \overset{\text{Segal 74}}{\simeq}$ SBurnside ring
$\mathbb{S}_G(\ast) \simeq A(G)$
Segal-Carlsson completion theorem
$A(G) \overset{\text{Segal 71}}{\simeq} \mathbb{S}_G(\ast) \overset{ \text{compl.} }{\longrightarrow} \widehat {\mathbb{S}_G(\ast)} \!\! \overset{\text{Carlsson 84}}{\simeq} \!\!\!\!\!\! \mathbb{S}(B G)$
More explicitly, this means that the Burnside ring of a group $G$ is isomorphic to the colimit
$A(G) \simeq \underset{\longrightarrow}{\lim}_V [S^V,S^V]_G$
over $G$-representations in a complete G-universe, of $G$-homotopy classes of $G$-equivariant based continuous functions from the representation sphere $S^V$ to itself (Greenlees-May 95, p. 8).
### As the endomorphism ring of the additive Burnside category
###### Example
(Burnside ring is endomorphism ring of additive Burnside category)
The endomorphism ring of the terminal G-set (the point $\ast$ equipped with the, necessarily, trivial action) in the additive Burnside category $G Burn_{ad}$ is the Burnside ring $A(G)$:
$End_{G Burn_{ad}}(\ast, \ast) \;\simeq\; A(G) \,.$
### Relation to representation ring
Let $G$ be a finite group. Consider
1. the Burnside ring $A(G)$, which is the Grothendieck group of the monoidal category $G Set$ of finite G-sets;
2. the representation ring $R(G)$, which is the Grothendieck group of the monoidal category $G Rep$ of finite dimensional $G$-linear representations.
Then then map that sends a G-set to the corresponding linear permutation representation is a strong monoidal functor
$G Set \overset{\mathbb{C}[-]}{\longrightarrow} G Rep$
and hence induces a ring homomorphism
$A(G) \overset{ \mathbb{C}[-] }{\longrightarrow} R(G)$
Under the identitification
1. of the Burnside ring with the equivariant stable cohomotopy of the point
$A(G) \;\simeq\; \mathbb{S}_G(\ast)$
(as above)
2. of the representation ring with the equivariant K-theory of the point
$R(G) \;\simeq\; K_G(\ast)$
(see there)
this should be the image of the initial morphism of E-infinity ring spectra
$\mathbb{S} \longrightarrow KU$
from the sphere spectrum to KU.
For details on this comparison map see at permutation representation, this section.
## References
The concept was named by Dress, following
Textbook accounts and lecture notes include
• John Greenlees, Peter May, section 2 of Equivariant stable homotopy theory, in I.M. James (ed.), Handbook of Algebraic Topology , pp. 279-325. 1995. (pdf)
• Stephen Schanuel, 1991, Negative sets have Euler characteristic and dimension, in: Proc. Como 1990. Lecture Notes in Mathematics, vol. 1488. Springer-Verlag, Berlin, pp. 379–385.
Discussion in relation to equivariant stable cohomotopy and the Segal-Carlsson completion theorem is in
• Graeme Segal, Equivariant stable homotopy theory, In Actes du Congrès International des Math ématiciens (Nice, 1970), Tome 2 , pages 59–63. Gauthier-Villars, Paris, 1971
• Tammo tom Dieck, T. Petrie, Geometric modules over the Burnside ring, Invent. Math. 47 (1978) 273-287 (pdf)
• Erkki Laitinen, On the Burnside ring and stable cohomotopy of a finite group, Mathematica Scandinavica Vol. 44, No. 1 (August 30, 1979), pp. 37-72 (jstor:24491306, pdf)
• Gunnar Carlsson, Equivariant Stable Homotopy and Segal’s Burnside Ring Conjecture, Annals of Mathematics Second Series, Vol. 120, No. 2 (Sep., 1984), pp. 189-224 (jstor:2006940, pdf)
• C. D. Gay, G. C. Morris, and I. Morris, Computing Adams operations on the Burnside ring of a finite group, J. Reine Angew. Math., 341 (1983), pp. 87–97.
• Wolfgang Lück, The Burnside Ring and Equivariant Stable Cohomotopy for Infinite Groups (arXiv:math/0504051)
Computational aspects are discussed in
• Martin Kreuzer, Computational aspects of Burnside rings, part I: the ring structure, D.P. Beitr Algebra Geom (2017) 58: 427 (doi:10.1007/s13366-016-0324-4)
Last revised on January 28, 2019 at 04:59:03. See the history of this page for a list of all contributions to it. | |
## Trigonometry (11th Edition) Clone
For $y = (x+2)^2 - 1,$ when $x=−2, y=-1$ The curve passes through $(−2,-1)$ Since the graph of $y=(x+2)^2-1$ is the same as the graph of $y=x^2$ but at a horizontal translation of 2 units to the left and a vertical translation of 1 unit down, graph "I" is the correct sketching. | |
# Tips for code-golfing in C#
What general tips do you have for golfing in C#? I'm looking for ideas that can be applied to code golf problems in general that are at least somewhat specific to C# (e.g. "remove comments" is not an answer). Please post one tip per answer.
-- borrowed from marcog's idea ;)
• BEST TIP => Use something beside .NET if you don't want to submit the longest answer for the challenge. .NET is designed to be very verbose and to let the IDE do the typing. Which is actually not as bad as it sounds for general programming as long as you have that IDE crutch but for code golf that strategy is certain fail. – krowe Jul 2 '14 at 5:48
• Forgive me for the picture of a calendar, it was all I could find on short notice. – undergroundmonorail Jul 8 '14 at 12:44
Use the one character non-short-circuiting variants of logical operators where possible:
• i>0||i<42
• i>0|i<42
or
• i>0&&i<42
• i>0&i<42
The difference between the two are one byte (yeah!) and the short-circuit principle. In our first example if i>0 is true, i<42 wont be checked. We dont need it. With the bitwise, both will be evaluated.
• Your statement about short circuiting is incorrect, logical or (||) does indeed short circuit if the left operand is true. msdn.microsoft.com/en-us/library/6373h346.aspx – Phaeze Jul 11 '16 at 21:56
• @Phaeze Indeed I wrote the wrong one. Thank you. – aloisdg Jul 11 '16 at 23:55
• Technically none of these are bitwise operators, they are just non-short-circuiting Boolean operators, because they are operating on Booleans. – VisualMelon Jan 13 '17 at 21:17
• @VisualMelon MSDN uses "logical or bitwise OR" for "x | y" and "logical OR" for "x || y". What should I use? – aloisdg Jan 14 '17 at 15:21
• These are 'logical' (Boolean being the logic system). Bitwise would be for integer types. I like to make the distinction clear because the a lot of C# golfing can end up being a mix of both, and operator precedence can be a real pain (though they are the same for each type) – VisualMelon Jan 14 '17 at 17:16
Always use the alias for a type if you need to type as they are usually shorter than the full names. It also means you don't need to include
using System;
or fully qualify the type when you otherwise wouldn't need to.
For a full list of the aliases visit this SO answer:
object: System.Object
string: System.String
bool: System.Boolean
byte: System.Byte
sbyte: System.SByte
short: System.Int16
ushort: System.UInt16
int: System.Int32
uint: System.UInt32
long: System.Int64
ulong: System.UInt64
float: System.Single
double: System.Double
decimal: System.Decimal
char: System.Char
Note that you can also use the alias for calling static methods, consider:
System.String.Concat();
vs
string.Concat();
If you're already using Linq in your code and need to create a list use the following:
var l=new int[0].ToList();
Compared to:
var l=new System.Collections.Generic.List<int>();
You can even initialise the list with values like:
var l=new int[0]{1,2,3,4}.ToList();
• You can remove the 0 in your declaration. – PmanAce Feb 20 at 19:35
If you want to return multiple values from a function, use C# 7 style tuples instead of out parameters:
(int,int)d(int a,int b)=>(a/b,a%b);
• Oh, didn't knew this was possible in C#. Thanks for sharing, +1 from me! One thing to golf in your code is by using currying input, though (a=>b=> instead of (a,b)=>). Here is your code in action, which might be useful to add as example-link to your tip. Oh, and welcome to PPCG! :) – Kevin Cruijssen Aug 31 '18 at 14:53
• @KevinCruijssen posted here :) – aloisdg Jan 11 at 9:14
# Use Ranges and indices (C# 8)
You can use the type Index, which can be used for indexing. You can create one from an int that counts from the beginning, or with a prefix ^ operator that counts from the end:
Index i1 = 3; // number 3 from beginning
Index i2 = ^4; // number 4 from end
int[] a = { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 };
Console.WriteLine($"{a[i1]}, {a[i2]}"); // "3, 6" You can also use a Range type, which consists of two Indexes, one for the start and one for the end, and can be written with a x..y range expression. You can then index with a Range in order to produce a slice: var slice = a[i1..i2]; // { 3, 4, 5 } source # C# Interactive Window AKA C# (Visual C# Interactive Compiler) on Try it Online This is a REPL for the C# language that includes many advantages to code golfing over using the traditional C# compiler. Currently, the shortest full program to print Hello, World! using the traditional C# compiler is 67 bytes: class P{static void Main(){System.Console.Write("Hello, World!");}} Using the C# Interactive Window, you can do the same thing in 22 bytes: Write("Hello, World!") And yes, I added my answer just for this post ;) As you can see, ceremonial class definitions are not required. Also, a whole bunch of references are included by default. The current list is as follows: • /r:System • /r:System.Core • /r:Microsoft.CSharp • /u:System • /u:System.IO • /u:System.Collections.Generic • /u:System.Console • /u:System.Diagnostics • /u:System.Dynamic • /u:System.Linq • /u:System.Linq.Expressions • /u:System.Text • /u:System.Threading.Tasks The System.Linq namespace is particularly handy as it allows you to write functional style programming which often saves many bytes. Many of the older C# answers added 18 bytes or so for this advantage, simply to import the library using System.Linq;. There are other hacks where solutions created their class in the System namespace to more succinctly access the Console object. None of this is needed with the C# Interactive Window. # Avoid single-statement foreach loops If the loop's statement returns a non-int (including void!) "value", it can be replaced with LINQ: foreach(var x in a)Console.WriteLine(F(x)); a.Any(x=>Console.WriteLine(F(x))is int); If the value happens to be an int, you can use a condition that will always be true or always be false (for example, >0 or <n), a different type and/or All instead of Any. When to use a space and when you can remove it. After [] • int[] f(char[] a){Console.Write('a');} • int[]f(char[]a){Console.Write('a');} Before$
• return $"{a}" • return$"{a}"
(Add yours in comment I will edit)
When you want to join something to output a string without delimiter, you should use string.Concat(), instead of string.Join("",);
• string.Join("",)
• string.Concat()
One byte free!
Also Concat() has a lot more signatures than Join. Check them on MSDN: Concat and Join.
If you need to use an enum for a method it is often shorter to cast an int to it rather than using the value directly:
DayOfWeek.Sunday
(DayOfWeek)0;
If you're already using Linq in your answer and need to check for a none empty collection use Any(). Compare it to the following:
Count()>0
Length>0
Count>0
Any()
# Tuple Deconstruction
It is possible to deconstruct a Tuple into variables using the following syntax:
var myTuple = (1, "red", new DateTime(2000, 1, 1));
var (i, c, d) = myTuple;
In a code-golf scenario, you could use this to pass a tuple into a function:
t=>{var(i,c,d)=t;Print(i);Print(c);Print(d);}
Try it online!
A more practical use of this is to decompose a list of tuples and iterate over them using foreach.
var myTuples = new[] {
(1, "red", new DateTime(2000, 1, 1)),
(2, "blue", new DateTime(2000, 1, 1))
};
foreach (var (i, c, d) in myTuples) {
Print(i);
Print(c);
Print(d);
}
Or in a code-golf scenario:
t=>{foreach(var(i,c,d)in t){Print(i);Print(c);Print(d);}}
Try it online!
You can use the ternary operator to shorten complex if..else constructs even if you need to work on multiple or different variables in different branches. You can sometimes save some chars by doing all or part of the work in the 3rd part of a for. ...and also other "optimizations", you can find in this example, I submitted here (it increments all numbers in a string given as char array or StringBuilder):
for (int b=3, c, i=a.Length; i-->0;
b=48>c|c>57
?7
:b>2
?c>56?a[i]='0':++a[i]*0
:b
) c=a[i];
In two of the branches, b isn't really set; in two branches, a[i] is set even though it says b= in the beginning; in one case a[i] and b are set simultaneously...
c>56 is shorter than c==57
i=a.Length; i-->0; is a lot shorter than i=a.Length-1; i>=0; i--
One thing I just learned (which I didn't know but probably all of you do): You do not need a namespace at all. If it is a simple program which doesn't use many namespaces that you would need to shorten, just omit the namespace and start with class Foo{....
Any objections to this or hints why I shouldn't do this are very welcome, as I'm just writing up my first "golfed" (as far as I got) answer in C# ;-)
• Alternately by putting the namespace as system you can omit the common using System; declaration ... – jcolebrand Oct 29 '14 at 16:30
Convert line endings from CRLF to LF before counting bytes ☺
• This really applies to all languages, and preprocessor directives are pretty rare in C# golf, so few programs will ever need to be on more than one line. – VisualMelon Jan 3 '15 at 15:42
• Mh but it's still non-obvious to some people. And most other languages are more often seen on LF-only environments. I don't expect it to help much, but… why not? – mirabilos Jan 3 '15 at 15:43
• Most of the time, just remove newlines... – aloisdg Jul 8 '16 at 15:04
## ReSharper tips
If you have used ReSharper to get to the initial working solution before golfing, note that it often generates
• static methods, if they do not use any fields
If you have R#, you want to use the inline method refactoring for methods that are only called once, since they only need the additional method declaration.
• Or Rider, which is currently in free beta (unlike R#). – mınxomaτ Jul 8 '16 at 15:41
Remember that C# uses unicode (which includes ascii). All char are int under the hood.
For example 'a' is 97.
• n=>char.IsDigit(n)|char.IsUpper(n)
• n=>n>47&n<58|n>64&n<91 // note that I use a bitwise comparator see
Since char are int, you can increment them:
• for(int x=31;x<126;)Console.Write((char)++x);
• for(char x=' ';x<127;)Console.Write(x++);
Use the unicode table for reference. Be careful one byte != one character.
• ints are 4 bytes, but chars are only 2. – recursive May 21 at 20:06
# Implicitly typed arrays
If you need an array (or IEnumerable) initialized with constant data, the most naive implementation might be like so:
new int[6]{23,16,278,0,2,44}
It's well known that arrays in C# can have implicit length:
new int[]{23,16,278,0,2,44}
However, for certain data types C# will also be able to determine the base type for the array from its values, making the same declaration even shorter:
new[]{23,16,278,0,2,44}
Here is an example demonstrating that all these options create the same array.
Additionally, specifically if you are creating an array for inline declaration, the new[] can be removed as long as the declared variable is not implicitly typed, which can save some bytes:
var a=new[]{"woah"};
string[]b={"woah"};
• That last line is worth all the weight. Excellent tips! – jcolebrand Nov 15 '17 at 18:58
# Convert a string to an IEnumerable<char>
Common way would be to use .ToCharArray() (14 bytes) or even better .ToList() (11 bytes), but the best I found is to rely on .Skip(0) (8 bytes).
Most of the time you wont need anything and the string will be cast directly, but sometimes it matters:
string.Join("a","bb") // bb
vs
string.Join("a","bb".Skip(0)) // bab
• wait ToSkip is a thing? – ASCII-only Apr 27 '18 at 7:50
• @ASCII-only Skip ;) – aloisdg Apr 27 '18 at 11:03
• Then it's no longer 10 bytes – ASCII-only Apr 27 '18 at 11:05
• @ASCII-only yup. 8 bytes \o/ – aloisdg Apr 27 '18 at 11:07
# Using Command-Line Options
Command-line options are not included in the byte count of your program, so they are very useful. However, keep in mind that when you use a command-line option, you are not competing in C# anymore, you are competing in C# with -yourflag.
The syntax for them is to put a - or a / preceding the letter, and if there are any arguments, you use : to separate the option name from the argument.
There are 3 command-line options that I know of as of right now:
• /u or -u or /using or -using
• Acts as a using statement. Example: /u:System.Text.RegularExpressions. You can do static imports with them by just adding the class name, like /u:System.Text.RegularExpressions.Regex. For example, if you have /u:System.String as an option, you can just use Concat(yourString) instead of string.Concat(yourString). Sadly, aliasing doesn't seem to work. You can also put multiple imports on the same line with ; as a delimiter like this: /u:System.Math;System.CodeDom.
• /i or -i, activates REPL mode. Code from the input box is also evaluated and printed.
• Can be useful for situations where hard-coding is allowed, e.g. if input insertRandomTermHere output 3, if input anotherRandomTerm output 4, we can just do int insertRandomTermHere=3,anotherRandomTerm=4;. Try it online!
• /r or -r
• Acts as an assembly reference. Allows you to use types like those in System.Numerics or System.Drawing. If you want to use BigInteger, an assembly reference is required.
Meanwhile, happy ing! | |
# Draw a square around a point.
I have a point on the graph at position X,Y , and I have to draw a square around that point of side X m. I have described my problem in the image.
I have taken an example square of 3 m. The diagonal of this square is 4.24 m(approx). I need to find the coordinates of the sides of the square.
-
You also need to specify the orientation of the square. – joriki Dec 12 '12 at 7:35
@joriki Orientation? Do you mean 2D,3D? If so, it's a 2D square. – Abijeet Patro Dec 12 '12 at 8:33
A square is always two-dimensional. I presume you're referring to the dimensionality of the space. I was a assuming that you're working in the plane and was referring to the angular orientation of the square within the plane. Clearly the coordinates of the slanted square that you drew differ from the coordinates of an axis-parallel square that ashley provided, and depend on the angular orientation. – joriki Dec 12 '12 at 9:04
If I understand this correctly then, the square will not be slanted at all, I'm sorry that the image does not depict that correctly. – Abijeet Patro Dec 12 '12 at 18:44
If your point is $(x,y)$ and side length is $k$, the coordinates of the square of which the sides are parallel to the x- and y-axes are $(x+n, y+n)$, $(x+n, y-n)$, $(x-n, y+n)$, $(x-n, y-n)$ where $n=k*sqrt(2)/2$. If you want angles by the axes, rotate. | |
Boundary of a chain, Stokes' theorem.
Tags:
1. Jan 17, 2015
Kim.S.H
Hi, I'm studying multivariable analysis using Spivak's book "calculus on manifolds"
When I see this book, one strange problem arouse.
Thank you for seeing this.
Here is the problem.
c0 , c1 : [0,1] → ℝ2 - {0}
c : [0,1]2 → ℝ2 - {0}
given by
c0(s) = (cos2πs,sin2πs) : a circle of radius 1
c1(s) = (3cos4πs,3sin4πs) : a circle of radius 3 that winds twice.
c(s,t) = (1-t)*c0(s) + t*c1(s).
Then c(s,0) = c0(s) , and c(s,1) = c1(s).
Now the boundary of a 2-chain c is c0-c1.
Is this right?
If so, there is a problem.
If w = (-ydx + xdy)/(x2+y2) is a 1-form on ℝ2 - {0} , then it is closed.(i.e. dw = 0)
Then by the Stokes' theorem we get
-2π = ∫c0-c1w = ∂cw = ∫cdw = 0
Why this happened?
2. Jan 17, 2015
Kim.S.H
oh,,, c wasn't well-defined on R^2 - 0
3. Jan 23, 2015
mathwonk
you have proved that if a cylinder maps into the complement of the origin, then both boundary circles wind the same number of times about the origin. this is quite interesting and can be used to prove the fundamental theorem of algebra. I.e. this can be used to prove that any polynomial of degree n winds a large circle the same number of times about the origin as does its lead term z^n, in particular it winds a non zero number of times if the polynomial is not constant.
It follows that the polynomial acting on the disc forming the interior of that circle must hit the origin, since if not, the winding number on the large circle bounding the disc, would equal the winding number on a very small circle near the center of the disc, namely zero.
The general statement that a continuous map from a disc to the plane must hit the origin, if it wraps the boundary of the disc a non zero number of times about the origin, is a generalization of the intermediate value theorem, which states that a continuous function from a closed interval tot he line, must hit the origin if it sends the boundary points of the interval to opposite sides of the origin.
There are also higher dimensional generalizations, for which one needs to define wrapping number of an n-1 sphere about the origin in n-space. Usually winding numbers are defined by integrating the "angle form" dtheta around the curve, so in general one wants to integrate a version of a "solid angle form" over a surface. see problem 5-31 of spivak.
by integrating the solid angle form over a sphere in 3 space, it follows that the identity map of the 2-sphere cannot be extended to a 3 dimensional "cylinder" in such a way that the map on the other 2 sphere bounding the cylinder is the antipodal map. I.e. the antipodal map wraps the sphere around the origin a different number of times from the identity map, (-1 times as opposed to 1 time). it follows that there is no everywhere non zero continuous vector field on the sphere, since such a vector field would allow one to define the non existent map above.
Technically these arguments using integrals of differential forms require hypotheses of smoothness rather than just continuity, but they can probably be tweaked a bit to get the stronger results. (E.g. since dtheta is a "closed" form, it can actually be integrated over continuous, not just smooth curves. The point is that closed forms have integrals that do not change under small deformation, so you can make a close approximation of a continuous curve by a smooth one.)
Last edited: Jan 24, 2015 | |
SAROptimisationGoal
A SAR optimisation goal.
Example
app = cf.GetApplication()
project = app:NewProject()
-- Create a SAR optimisation goal with focus on a request with label "SARRequest"
properties = cf.SAROptimisationGoal.GetDefaultProperties()
properties.FocusSourceLabel = "SARRequest"
properties.GoalOperator = cf.Enums.OptimisationGoalOperatorEnum.Maximise
properties.ProcessingSteps[1].Operation = cf.Enums.OptimisationGoalProcessingStepsEnum.Offset
properties.ProcessingSteps[1].Value = "1"
-- Change the first processing step offset value to 10
sarGoal.ProcessingSteps[1].Value = 10
Inheritance
The SAROptimisationGoal object is derived from the OptimisationGoal object.
Property List
FocusSource
Set the focus source to a specified source object. The intended usage is for when the source is defined in CADFEKO. (Read/Write SAR)
FocusSourceLabel
Set the source focus label. The intended usage is for when the source is defined only in EDITFEKO. (Read/Write string)
FocusType
Set the focus type. (Read/Write OptimisationSARTypeEnum)
GoalOperator
The goal operator indicates the desired relationship between the goal focus and the objective. (Read/Write OptimisationGoalOperatorEnum)
Label
Objective
The objective describes a state that the optimisation process should attempt to achieve. (Read only OptimisationGoalObjective)
Type
The object type string. (Read only string)
Weight
Specify the optimisation weight. (Read/Write Expression)
Collection List
ProcessingSteps
A number of conversion steps or mathematical operations to be carried out on the goal focus before the goal is evaluated. (ProcessingStepsCollection of OptimisationGoalProcessingSteps.)
Method List
Delete ()
Deletes the goal.
Duplicate ()
Duplicates the goal. (Returns a OptimisationGoal object.)
GetProperties ()
Returns a table of properties representing the state of the object. The properties table can be used with the SetProperties method to change multiple properties of the object in one step. (Returns a table object.)
SetProperties (properties table)
Modifies the state of the object using the provided table of properties. This method is used to modify multiple properties of the object in a single step.
Static Function List
GetDefaultProperties ()
Creates a table containing the default settings to create an object. (Returns a table object.)
Property Details
FocusSource
Set the focus source to a specified source object. The intended usage is for when the source is defined in CADFEKO.
Type
SAR
Access
FocusSourceLabel
Set the source focus label. The intended usage is for when the source is defined only in EDITFEKO.
Type
string
Access
FocusType
Set the focus type.
Type
OptimisationSARTypeEnum
Access
GoalOperator
The goal operator indicates the desired relationship between the goal focus and the objective.
Type
OptimisationGoalOperatorEnum
Access
Label
The object label.
Type
string
Access
Objective
The objective describes a state that the optimisation process should attempt to achieve.
Type
OptimisationGoalObjective
Access
Type
The object type string.
Type
string
Access
Weight
Specify the optimisation weight.
Type
Expression
Access
Collection Details
ProcessingSteps
A number of conversion steps or mathematical operations to be carried out on the goal focus before the goal is evaluated.
Type
ProcessingStepsCollection
Method Details
Delete ()
Deletes the goal.
Duplicate ()
Duplicates the goal.
Return
OptimisationGoal
A duplicate goal.
GetProperties ()
Returns a table of properties representing the state of the object. The properties table can be used with the SetProperties method to change multiple properties of the object in one step.
Return
table
A properties table.
SetProperties (properties table)
Modifies the state of the object using the provided table of properties. This method is used to modify multiple properties of the object in a single step.
Input Parameters
properties(table)
A table of properties defining the new state of the object.
Static Function Details
GetDefaultProperties ()
Creates a table containing the default settings to create an object.
Return
table
The default properties table. | |
### 15.4.3.2 Lie algebra of the system distribution
Now suppose that a set , , of vector fields is given as a driftless control-affine system, as in (15.53). Its associated distribution is interpreted as a vector space with coefficients in , and the Lie bracket operation was given by (15.81). It can be verified that the Lie bracket operation in (15.81) satisfies the required axioms for a Lie algebra.
As observed in Examples 15.9 and 15.10, the Lie bracket may produce vector fields outside of . By defining the Lie algebra of to be all vector fields that can be obtained by applying Lie bracket operations, a potentially larger distribution is obtained. The Lie algebra can be expressed using the notation by including , , and all independent vector fields generated by Lie brackets. Note that no more than independent vector fields can possibly be produced.
Example 15..16 (The Lie Algebra of the Differential Drive) The Lie algebra of the differential drive (15.54) is
(15.101)
This uses the Lie bracket that was computed in (15.82) to obtain a three-dimensional Lie algebra. No further Lie brackets are needed because the maximum number of independent vector fields has been already obtained.
Example 15..17 (A Lie Algebra That Involves Nested Brackets) The previous example was not very interesting because the Lie algebra was generated after computing only one bracket. Suppose that and . In this case, there is room to obtain up to three additional, linearly independent vector fields. The dimension of the Lie algebra may be any integer from to .
Let the system be
(15.102)
This is a chained-form system, which is a concept that becomes important in Section 15.5.2.
The first Lie bracket produces
(15.103)
Other vector fields that can be obtained by Lie brackets are
(15.104)
and
(15.105)
The resulting five vector fields are independent over all . This includes , , and the three obtained from Lie bracket operations. Independence can be established by placing them into a matrix,
(15.106)
which has full rank for all . No additional vector fields can possibly be independent. Therefore, the five-dimensional Lie algebra is
(15.107)
Steven M LaValle 2012-04-20 | |
# Redshifting help?
1. Sep 3, 2009
### Evolver
Okay, I understand the principle of redshifting, but I feel I am missing an element to it's comprehension because as is it doesn't make complete sense to me.
I understand the Doppler effect of sound waves, as in a train passing by; the waves in front are of a higher frequency due to the motion of the train producing them and subsequently lower frequency as the train passes.
Now, the redshifting of light from distant galaxies traveling away at high speeds is where I am confused. I am lead to believe that the galaxies are generating a similar effect in that the light waves are lower frequency behind the moving body and thus appear redshifted (and if the galaxies were heading towards us would appear blueshifted.)
But this makes no sense to me. According to relativity, light in a vacuum travels at a constant speed regardless of the motion of the body that it is emitted from. Therefore, indicating that light is affected by the motion of galaxy is not consistent with this line of thinking.
I do realize it probably has something to do with time dilation, but I am not aware of how. Please someone help me grasp this concept.
Thanks!
2. Sep 3, 2009
### nutgeb
First, you need to distinguish between two kinds of light redshifting, SR Doppler and Cosmological.
SR Doppler redshift occurs as between two inertial frames - think of one as being stationary and the other moving. SR Doppler redshift is the combination of two components (which actually are inseparable). In the case of redshift, the light photons lose energy as a result of the need to accelerate themselves such that they travel at exactly c relative to the observer. This is classical Doppler redshift. You are correct that the speed of light - c- does not vary in any inertial frame. But the energy of a photon does decrease when observed in frame that is receding away from the emitter. The lower the energy of a photon, the longer its wavelength. Longer wavelength = redshift.
The second component of SR Doppler redshift is time dilation. From the perspective of the observer, the emitter is time dilated because it is in motion relative to the observer. Dilated time = slower clock = lower frequency of wavecrests = longer wavelength = redshift. The SR Doppler formula simply multiplies the classical Doppler redshift by the emitter's time dilation.
Cosmological redshift is a different but related phenomenon which occurs in an expanding, gravitational universe. At a simple level, cosmological redshift is exactly proportional to the distance of the emitting object at the time of observeration. Another way to say it is that the cosmological redshift is exactly proportional to the amount by which the scale factor of the universe has expanded in the interval between emission and observation.
But these descriptions of the cosmological redshift do not explain HOW it occurs. It is fairly certain that the 'expansion of space' does not itself directly cause the distance between wavecrests to expand. The 'expansion of space' is inherently incapable of causing an increase in separation between any two things unless those things were already moving apart in the initial condition, and retain some sort of separational momentum (like physical objects such as galaxies have). Over a small time period, all of the wavecrests emitted by a receding emitter have exactly the same fixed wavelength, so the crests cannot be said to be in motion relative to each other in the initial conditions. In addition, the concept of retained separational momentum cannot be meaningfully applied to intangibles such as wavecrests.
SR Doppler redshift does not occur in a cosmological context between galaxies which are receding away from each other exactly at the pace of their local Hubble flow, because no time dilation exists as between fundamental comovers in the FRW metric, which models cosmological expansion.
3. Sep 3, 2009
### sylas
Think about your train example again. The sound waves are also traveling at the same speed with respect to you, no matter whether the train is approaching or receding.
You are also correct about blue shift for an approaching galaxy. Andromeda, for example, is approaching us, and is blueshifted as a result.
Time dilation does mean that the formulae used are a bit different; but you still get a Doppler shift even without any time dilation effects -- as the train example shows.
Cheers -- sylas
PS. This is not a question for which you need to worry about all the details of large scale cosmology. For nearby galaxies, there's no meaningful difference between cosmological redshift and Doppler redshift. The cosmological factors of large scale spacetime become important on very large scales; but not for small redshift galaxies.
Last edited: Sep 3, 2009
4. Sep 3, 2009
### javierR
The speed of light is indeed a constant here, but the doppler effect has to do with observed frequencies of the light, and these can change from observer to observer. I haven't checked the quality of the wikipedia entry, but I'll assume it's alright:
http://en.wikipedia.org/wiki/Relativistic_Doppler_effect
5. Sep 3, 2009
### JesseM
SR Doppler redshift has nothing to do with photons "losing energy as a result of the need to accelerate themselves", it's just time dilation plus the effect of the source moving away from you (or towards you in the case of blueshift), which means that each successive wave peak is emitted at a greater distance from you and thus has longer to travel to reach your eyes than the previous peak. See my post #12 on this thread for a numerical example.
6. Sep 3, 2009
### nutgeb
I agree that each wave has longer to travel.
But are you disagreeing that a redshifted photon has less energy than a non-redshifted one?
7. Sep 3, 2009
### nutgeb
I agree that each wave has longer to travel, which can be correctly described as causing the longer wavelength.
But, of course, a redshifted photon has less energy than a non-redshifted one, and my description is equivalently correct.
Last edited: Sep 3, 2009
8. Sep 3, 2009
### JesseM
How is it equivalent? For one thing the relation between energy and wavelength depends on quantum mechanics, whereas my explanation works for classical EM waves too. And I don't see how you'd actually derive the relativistic Doppler shift from the QM energy/wavelength relation (or from the energy/wavelength relation + time dilation--you appear to have said the energy change has to be considered separately from the time dilation effect)...can you provide an actual derivation of the Doppler shift equation in this way, or a numerical example showing how you can get the correct shift using the energy/wavelength relation? If not I am doubtful that your statement makes any sense at all.
9. Sep 3, 2009
### nutgeb
Of course the energy of light is proportionally reduced, as measured by an observer in relative motion away from the emitter, compared to the energy measured by an observer stationary in the inertial frame of the emitter.
If you disagree with that statement, please state simply, "I disagree."
10. Sep 3, 2009
### JesseM
I do not disagree with that statement on its own. But I see no way to derive the Doppler shift equation from the fact that the energy is lower in the frame where the emitter is moving, and that is what you were saying when you said "In the case of redshift, the light photons lose energy as a result of the need to accelerate themselves such that they travel at exactly c relative to the observer. This is classical Doppler redshift." Unless you can show how to derive the Doppler shift from the different energy in the two frames, then this statement is a complete non sequitur, akin to saying "the length of the emitter is contracted in the frame where the emitter is moving. This is classical Doppler redshift". The first sentence is of course true on its own terms, but it has nothing to do with the explanation for the Doppler shift.
11. Sep 4, 2009
### nutgeb
Jesse, I'm not going to get into an extended argument on this point. I agree with your terminology for describing how the classical redshift applies to light, it's a good description.
The classical Doppler formulas, based on velocity and wavelength, are as follows (where outward relative velocity has a positive sign):
Moving emitter, stationary observer: $$\lambda _{o} = \lambda _{e} (1 + v_{e})$$
Moving observer, stationary emitter: $$\lambda _{o} = \lambda _{e} / (1 - v_{o})$$
12. Sep 4, 2009
### JesseM
OK, and these formulas don't seem to be derivable from energy considerations as you were suggesting--that's the only point I was making.
13. Sep 7, 2009
### nutgeb
[Edit: Upon further consideration, I revised the first sentence as follows: The classical Doppler shift is required by energy conservation considerations, and the redshifted energy of a photon in an observer's frame can be calculated in theory by measuring the emitter's atomic recoil in the observer's frame.]
Consider a hydrogen atom that emits a photon, which subsequently is absorbed by an observer moving away from the atom. The atom will recoil away from the observer. The sum of the atom's recoil energy and the photon's redshifted energy, both as measured in the observer's frame, must equal the energy of the atom's motion (relative to the observer) just before the photon was emitted. Classical Doppler shift of the photon is required in the observer's frame in order to avoid violating the conservation of energy.
The energy of a photon in a frame is equal to the Planck Constant times the photon's frequency as measured in that frame.
Last edited: Sep 7, 2009
14. Sep 7, 2009
### JesseM
First of all, a classical or relativistic derivation of the Doppler shift shouldn't depend on knowing formulas from quantum mechanics. Second, even if we assume we know the relation between energy and frequency from QM, it's not clear this is actually helpful in getting the Doppler equation, which is supposed to deal with the case of an emitter moving at constant velocity, not an emitter that recoils and changes velocity each time it emits a photon. The Doppler equation compares the frequency in two frames, the emitter frame and the observer frame, it's not clear what you want the equivalent of the "emitter frame" to be in the above argument--the rest frame of the atom after it has emitted the photon, or before?
If you want to continue to defend the claim that the Doppler equation can be derived from energy considerations, you need to actually provide a derivation with equations--or at least provide a numerical example--the verbal arguments above are much too handwavey.
15. Sep 7, 2009
### nutgeb
Every emission of every photon involves a recoil. This is not QM in any specialized sense, it's basic physics.
As you know, the Doppler equation compares the redshift to the emitter frame before the effect of recoil is considered. The atom's recoil frame is brought into the analysis specifically as a way to derive the redshift from energy conservation. As you know, it is most straightforward to analyze energy conservation with regard to a complete system rather than by reference to pieces of the system in isolation.
This is simple energy conservation. The momentum ptot of the total atom+photon system remains exactly constant before and after emission. Therefore the momentum of the redshifted photon must be ptot - patom, where the latter is the momentum of the recoiled atom as measured in the observer's frame. {Edit: In order to balance the total system momentum to compare vectors in both the +x and -x directions, the atom's momentum needs to be expressed as a negative number, so the redshifted photon's momentum should be expressed as ptot + patom.} Obviously the recoiled atom's momentum (in the observer's frame) is simply the atom's mass times its [initial velocity + recoil $$\Delta$$ velocity].
Last edited: Sep 7, 2009
16. Sep 7, 2009
### sylas
That is not a derivation with equations. It's still hand wavey. Try giving an actual mathematical derivation using equations. Derivations means you express the assumptions as equations (not as English) and get your result with algebra (not with an English declaration that it is "simple").
17. Sep 7, 2009
### JesseM
I was talking about the quantum relationship between the energy of a photon and its frequency/wavelength, which you seem to be assuming (otherwise how are considerations of energy/momentum supposed to tell us anything about frequencies?)
Typically I think the Doppler equation just assumes the effects of recoil can be treated as negligible because the momentum of the emitted waves are very small compared to the momentum of the emitter. In any case, if you want to think in terms of the frame where the atom is at rest before emitting the photon, then actually show how it works using this frame and the fame of the observer. If you don't want to do a general derivation, just give a specific numerical example where we consider an atom in motion at some specific velocity relative to an observer, consider it emitting a photon of some specific frequency, and then (somehow, because you have in no way explained it) use energy considerations to figure out both the frequency in the frame where the atom was at rest before emission, and the frequency in the observer's frame. If you can't even provide a simple numerical example, then it's obvious you don't have an actual well-defined argument in mind.
Huh? You just said you wanted to consider the emitter's frame "before the effect of recoil is considered". Just tell me, what frame are you using for the "emitter frame" in the standard Doppler equation? Is it the frame where the atom is at rest before emitting the photon, or the frame where it is at rest afterward?
All of this is pretty obvious, but it gives me no idea of how you intend to derive the redshift equation, which relates frequency in the observer's frame to frequency in the emitter's frame. In the above you appear to be talking solely about the momentum of the photon and atom in the observer's frame before and after the emission, with no reference to any other frame.
Please, no more vague verbal arguments. If you have a well-defined idea about how to derive the Doppler equation from energy considerations, it should be exceedingly simple to give a numerical example where you actually use energy considerations to calculate the frequency of the photon in the observer's frame and some other frame (what that other frame is, you still haven't made clear) and show that they match what's predicted by the Doppler equation (classical or relativistic, I'm not sure...if it's not relativistic, does that mean you won't assume the photon moves at c in both frames?) If you don't have any clear idea of how to do this, that's fine, but then please acknowledge that your ideas are tentative and don't confidently assert that the Doppler relation can be derived from energy considerations if you don't have a definite procedure in mind.
Last edited: Sep 7, 2009
18. Sep 7, 2009
### nutgeb
The photon recoil analysis is a way of relating a photon's redshifted energy to the total system energy, and to demonstrate that Doppler shift is required for energy conservation. But beyond that, I don't think it is the easiest path to analyze the workings of classical redshift. I have the recoil analysis written out in mathematical terms but sharing it will not advance this discussion.
Let me return to what I intended in my original post. You have used a lot of words to characterize a simple statement of mine, most of which are twisting my words.
A photon's energy in an observer's frame is trivially derived from its momentum in that frame: E = pc.
Further, it is a basic element of the classical Doppler shift that a photon's momentum is reduced by the factor of (1-v)/c if the observer is moving away from the emitter.
It is absurd to challenge these basic propositions or demand that I provide proof of something else. I never intended to suggest that a photon's redshifted energy shift derives from anything other than the decrease in its momentum due to chasing a moving observer. That's why I was perfectly willing to accept the way you described it.
19. Sep 7, 2009
### JesseM
By "recoil analysis" do you mean a derivation of the redshift starting from considerations of energy and/or momentum? If so, providing it would certainly advance the discussion--it would either show you do have the derivation I've been asking for, or myself and others could point out any flaws we found in the analysis.
Your "simple statement" was: In the case of redshift, the light photons lose energy as a result of the need to accelerate themselves such that they travel at exactly c relative to the observer. This is classical Doppler redshift. Then in a subsequent post you said The classical Doppler shift can be derived from energy considerations alone. Is it "twisting your words" to understand these as claims that the correct value for redshift can be derived from considerations of energy or momentum?
If you do think I am twisting your words, the solution is simple: say something like "no, I do not claim the Doppler equation can be derived just from considerations of the energy/momentum in both frames." On the other hand, if my interpretation of your words was correct and you do claim such a derivation is possible, then all you need to do is provide it (as I said, a numerical example rather than a full derivation would be convincing too). But if you won't do either of those things, you're just being evasive, and probably running afoul of the forum rules about not using the forum to advance original claims that are not part of mainstream physics (or that aren't demonstrably derivable from mainstream physics).
Strictly speaking you can only "derive" that by starting from the relativistic equation $$E^2 = m^2*c^4 + p^2*c^2$$ and then including the fact that a photon has zero rest mass. Didn't you say you wanted to talk in classical terms rather than relativistic ones? In relativistic terms your earlier statement about the atom's momentum just being its mass times its velocity would be incorrect, for example.
What are you talking about? The classical Doppler shift equation says absolutely nothing about a photon's momentum, it deals with frequency of a wave in different frames. In a classical universe, if a wave is sent out at c relative to the emitter, and the emitter is moving away from the observer at v, then the observer will see the frequency reduced by 1/(1 + v/c). In quantum physics we also know that momentum is proportional to frequency according to the formula p = h*frequency/c, so if we carelessly combine the quantum formula with the classical one (though this would be physically dubious since the classical formula assumes the wave has a speed c-v in the observer's frame, while the quantum formula assumes photons move at c in whatever frame the formula is being used) then we could conclude the momentum is also reduced by 1/(1 + v/c), which is different from your equation (1-v)/c (a formula that doesn't even seem to make sense dimensionally since v and c have units of distance/time while 1 is dimensionless).
Your (1-v)/c formula for momentum reduction, which supposedly comes from "classical Doppler shift", is not a basic proposition. And it's not absurd to demand that you provide proof of what you asserted at the beginning about Doppler shift coming from energy considerations which got this whole debate going, something you appeared to continue to assert when you said "The classical Doppler shift can be derived from energy considerations alone." If you wish to back off from this claim, then do so.
The debate was never about a photon's "redshifted energy", it was about the Doppler equation which deals with the frequency of a wave (of photon if you prefer) in both frames. Do you or do you not claim this is derivable from considerations of momentum/energy? If you do not, please state that clearly, and if you do, please provide a derivation or a numerical example.
20. Sep 7, 2009
### nutgeb
This is my last post on this subject. It is impossible to have a rational dialogue with you when you are twisting my words and spraying arguments in all directions. You can post whatever you want on this topic as long as you direct it to the OP or others, and leave me and my words out of it.
For the sake of clarity, the following will clarify and REPLACE ALL prior statements I've made on this subject:
A photon's energy as measured locally in the observer's frame is a direct function of its momentum. From Wikipedia article "Photon":
"In empty space, the photon moves at c (the speed of light) and its energy and momentum are related by E = pc, where p is the magnitude of the momentum vector p. For comparison, the corresponding equation for particles with mass m is: $$E^2 = m^2 c^4 + p^2 c^2$$"
Classical Doppler effect causes an observer moving away from the emitter to measure a photon to have a lower momentum than it had in the emitter's frame. This is true regardless that the simple form of the classical Doppler formula does not include a momentum term.
Energy, momentum, frequency and wavelength are all directly related and go hand-in-hand with respect to redshift. Whether the reduction in momentum is caused fewer wave crests hitting the observer in a given period of time, or by an increase in the wavelengths between received crests, or simply from being received in a moving frame, is not crucial to the OP's question. As a practical matter it amounts to the same thing. Various "hand-waving" descriptions can reasonably describe the outcome.
The recoil of an emitter atom demonstrates that the classical Doppler shift of the photon complies with energy conservation.
*** FINIS ***
Last edited: Sep 7, 2009 | |
0
What is 17 20 th's in simplest form?
Wiki User
2017-04-18 00:12:45
Since the numerator of the fraction is prime, and not a factor of the denominator, the proper fraction 17/20 is already expressed in its simplest form.
Wiki User
2011-11-17 10:08:52
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niyaznigmatul's blog
By niyaznigmatul, 2 years ago, translation, ,
Editorial
Hello!
Codeforces Round #467 (Div. 1) and Codeforces Round #467 (Div. 2) will take place on February, 25 at 14:35 UTC. Most of the problems are Innopolis Open Olympiad in Informatics problemset. We ask olympiad participants not to take part in Codeforces rounds and not to discuss the problems till the end of the rounds.
Problems were prepared by: niyaznigmatul, pashka, gritukan, VArtem, burakov28, BudAlNik, YakutovDmitriy, dusja.ds, GreenGrape, tourist. We thank testers: demon1999, senek_k, the_art_of_war, Qrort, ayoyia, izban и julsa.
Good luck!
UPD: The round is rescheduled to +1.5 hours to avoid collisions with other events.
• +335
» 2 years ago, # | +17 Hope the problem statements are as short as the announcement!! :D
» 2 years ago, # | +7 There are some users who regestered in div.2 round before the rating change of round 466 and now they are div.1
• » » 2 years ago, # ^ | -9 did they ever fix the one where the educational round let div1 participate in div2
• » » » 2 years ago, # ^ | 0 It wasn't a bug to be fixed. It was totally intentional.
• » » » » 2 years ago, # ^ | 0 So it's intentional that a purple user can farm rating in contest strictly for < 1900? Where did they ever say that's intentional?
• » » » » » 2 years ago, # ^ | 0 The results of educational round was announced after div 2 contest so it was made rated for everyone who were div 2 before the educational round. Participants want to know whether it's rated or not before they participate in contest. The only fair way to ensure this was to make it rated for everyone who were div 2 before educational round. As you say you can't simply "farm" rating. I actually dropped from div 1 back to div 2 after that contest. Whoever had rating increase totally deserved it.
• » » » » » 2 years ago, # ^ | ← Rev. 2 → 0 official comment (http://codeforces.com/blog/entry/57819?#comment-414872)
• » » » » » » 2 years ago, # ^ | 0 looks like I was wrong, my bad
• » » 2 years ago, # ^ | 0 Anyone in particular you have in mind? I only see a handful of >1900 people in the div2 results and it doesn't look like the competition counted towards their rating.
» 2 years ago, # | ← Rev. 7 → -66 This round will be held in usual time. It's good for someone.
» 2 years ago, # | -68
» 2 years ago, # | +28 It's the first time ever I've seen a contest with 10 distinct writers :OLet's hope for a quality contest with no interference from DDoS and server issues ;)
» 2 years ago, # | -30 Round# 467 is a prime number while contest(div. 2) 937 is a prime number, too!
• » » 2 years ago, # ^ | -38 Problem A: 467 and 937 are lucky numbers, given a string count..
• » » » 2 years ago, # ^ | -48 What's happened to you people :|
» 2 years ago, # | -60 I have found the name tourist in the problem setter section :D
» 2 years ago, # | -27 10 problem setters , sir tourist is also here , i think this contest problems are too much perfect and malleable , too much excited :)
» 2 years ago, # | -29 Hopefully the number of problems is less than 10 considering there are 10 problem setters xD
» 2 years ago, # | -21 My brain singing "I want something just like this" :p
» 2 years ago, # | -24 How many problems are expected? There are many setters
» 2 years ago, # | ← Rev. 2 → +39 So is contest starting at 16:05 UTC now?Edit: I realised later I can see the time in Contests tab.
» 2 years ago, # | +24 Hope a round whose time is good for East Asia users!!!
» 2 years ago, # | +3 Please make this round visible in a "Pay attention" section (can't see it in RU version) and include it into the list of upcoming events! I found this round only accidentally.
» 2 years ago, # | -11 tourist as setter in Contest #2 of 2018 (Contest #1 — Hello 2018).
» 2 years ago, # | +106 It is not good to change the timings at the last moment as all my plans are getting disturbed.
» 2 years ago, # | +30 So sad it delayed.It is 12:05 UTC+8 now :( I know I'm not that good at programing,but I just want to join in the contest for fun... (Unhappy Chinese pupil)BTW,will my comment be hidden because too negative feedback?
• » » 2 years ago, # ^ | 0 I'm sad, too. ;( And about your comment, it won't.
» 2 years ago, # | -99 The comment is hidden because of too negative feedback, click here to view it
» 2 years ago, # | +1 Why the delay...? I want to know what those "other events" are?
» 2 years ago, # | +5 so what about score distribution?
» 2 years ago, # | +118 Got up early on a Sunday morning for a contest and found it delayed like.
» 2 years ago, # | ← Rev. 2 → 0 I can't participate in this contest because it delayed. But I have registered. So the question is: will my rating be influenced if I don't participate. If yes, what should I do?
• » » 2 years ago, # ^ | ← Rev. 2 → 0 No. If you do not submit any code, your rating will not change. And you can also unregister your registeration.
• » » 2 years ago, # ^ | +18 Unregister.find yourself in the (http://codeforces.com/contestRegistrants/937),and press the "X" after it.
• » » 2 years ago, # ^ | 0 I know how to do now. Thanks!
• » » » 2 years ago, # ^ | +8 An Chtholly lover! WOW!
» 2 years ago, # | 0 Can you delay it another 10-30 minutes?
» 2 years ago, # | 0 I know the event today.it's VK Cup 2018,not tourist's marriage.:-)
• » » 2 years ago, # ^ | -9 But in the Contests page, it shows that VK Cup 2018 is on Mar 10th?
• » » » 2 years ago, # ^ | 0 switch to russian version dude
• » » » » 2 years ago, # ^ | -6 thanks, I get it now.
» 2 years ago, # | +21 Although I was very busy at work but came home earlier for the round, and see that the scheduled time is changed. For next rounds, I suggest to change date or time at least a day before the round.
» 2 years ago, # | +43 Events like Chelsea VS Manchester United :)
• » » 2 years ago, # ^ | -12 +1
• » » 2 years ago, # ^ | +21 original time would have been better, I would have been able to watch Man City vs Arsenal! much better than chelsea vs united
• » » » 2 years ago, # ^ | +7 Mike is a Chelsea or ManU fan :V
• » » » » 2 years ago, # ^ | 0 Haha!!
» 2 years ago, # | 0 I'm too worried about codeforces server errors during the contest. Let's that it'll not happen again.
» 2 years ago, # | +5 please post the new score distribution.
» 2 years ago, # | +4 Hate the time delaying. The new time conflict with my sleeping time
» 2 years ago, # | 0 I hope this contest will make me green !.. I just need +75 rating to become a green coder...
• » » 2 years ago, # ^ | +2 yeah, Against all odds, finally you got your prize ! :)
• » » » 2 years ago, # ^ | 0 Yess...I became a green coder now...!!!
• » » 2 years ago, # ^ | ← Rev. 2 → 0 Yes..
» 2 years ago, # | ← Rev. 3 → 0 Hopefully it will be a cool contest.
» 2 years ago, # | ← Rev. 5 → +66 I hope, systesting of VK cup will be turn off during the round.UPD: So you haven't done this...
» 2 years ago, # | +2 I think all of you guys should stop arguing. Codeforces is a programming website, not a quarrelling website!
• » » 2 years ago, # ^ | +8 finally we got rid of ppppppppppppppp
» 2 years ago, # | +2 Came here to read funny comments but man wtf?!
» 2 years ago, # | -14 Is it rated?
» 2 years ago, # | -18 is it rated for div 2?
» 2 years ago, # | 0 to CF Server : DON'T FREEZE.
• » » 2 years ago, # ^ | ← Rev. 2 → +1 @I never freeze@froze
» 2 years ago, # | ← Rev. 2 → -50 From my school years I remember this olympiad as a trash. Seems like not much have changed. No offence.
• » » 2 years ago, # ^ | +23 No offence But it was offensiveBtw why?
• » » » 2 years ago, # ^ | ← Rev. 3 → -41 Cause it's just my opinion, and I don't wanna be rude or something. Just really don't like it.Low problems quality. When I participated, it was 4 very-very easy (like div2 A) problem with a lot of stupid coding, and one hard geometry problem. So the places were distributed by points in only one problem (on olympiad there are problems with subproblems, you get points for solving subproblems, no time/resubmit penalty). This time they are harder, but still boring/stupid.
• » » » » 2 years ago, # ^ | ← Rev. 3 → -26 i agree codeforce needs some subproblems for better categorizing (according to individual coding proficiances)
• » » » » 2 years ago, # ^ | 0 Yesterday's difficulty distribution was good though :)
• » » » » 2 years ago, # ^ | +30 I don’t know which problemset you are talking about. Any link would be helpful.You are welcome to become one of the problemsetters, if you have "not stupid/not boring" problem ideas in your mind.
• » » » » 2 years ago, # ^ | +15 Don't be angry because you can't solve them
» 2 years ago, # | -6 Unrated Round cy >(
» 2 years ago, # | -14 fucking long queue!
» 2 years ago, # | ← Rev. 2 → 0 What is this? There is 10 pages long queue. (~700 submissions only in div2)Make this round unrated >.>
» 2 years ago, # | 0 Today's day is so good : I am performing wonderful in the contest and queue is helping me in it :)
• » » 2 years ago, # ^ | 0 Hope it doesn't get unrated.
» 2 years ago, # | 0 what is this? This much Queue
» 2 years ago, # | 0 goodness me there is a 10 page+ long queue. Please fix it asap!
» 2 years ago, # | -6 13 page long queue ! what is this !
» 2 years ago, # | -6 this long queue, will it make this round unrated?
» 2 years ago, # | 0 You know the queue is so long when you're in Div.1 and the "In queue" verdicts' region is wider than a 50-line page...
• » » 2 years ago, # ^ | 0 70-lines to be exact.
» 2 years ago, # | 0 Shit man! I really hope the round won't be unrated because of the long queue.
» 2 years ago, # | 0 It's taking a very long time to process submissions...
• » » 2 years ago, # ^ | +2 Seems to be fixed now.
» 2 years ago, # | 0 The judge is running too slow . so frustrating at times .. look into it
» 2 years ago, # | 0 Tomorrow is my Exam but still i'm giving the contest to get away from the stress situation but these long queues of codeforces are giving me more headache than my syllabus did.
» 2 years ago, # | +2 Are DIV 2 Contests getting harder or I'm the one who is getting dumber ??
• » » 2 years ago, # ^ | +11 Both
• » » 2 years ago, # ^ | 0 you are not true Rick :D
• » » » 2 years ago, # ^ | 0 yeah, I'm just another Morty :-(
» 2 years ago, # | +2 What the hell is pretest 10 in C?
» 2 years ago, # | 0 how to solve DIV2 : B?
• » » 2 years ago, # ^ | +9 Check if there is number from y to y-300 that is not divisible by any number from 2 to p. 300, because prime gap is at most 300 in this number range
• » » » 2 years ago, # ^ | 0 prime gap is 300?
• » » » » 2 years ago, # ^ | 0 Yes, maximum distance between two primes is a bit less than 300 in range from 1 to 10^9
• » » » 2 years ago, # ^ | 0 "because prime gap is at most 300 in this number range" can you please elaborate it a little?
• » » » » 2 years ago, # ^ | 0 You can find it on wikipedia, just search for prime gap
• » » » » » 2 years ago, # ^ | 0 thanks mate! I have wasted my 20 mins on this:(
• » » » 2 years ago, # ^ | ← Rev. 2 → 0 since p can also be of order 109 checking in [2, p1 / 2] [2, min(n1 / 2, p)] is enough.
• » » » » 2 years ago, # ^ | 0 Please explain a little why we check only till sqrt(p)
• » » » » » 2 years ago, # ^ | 0 If P has any divisors, it can be written as a*b. Now, if a is less than sqrt(P), then b must be greater than sqrt(P). Therefore, either a or b is greater than sqrt(P). So, if divisors exist, we only need to check up to sqrt(P).
• » » » » » » 2 years ago, # ^ | 0 Thank you
• » » » » 2 years ago, # ^ | 0 can u help me to figure out why this is wrong my submission
• » » » » » 2 years ago, # ^ | 0 You are only checking divisibility up to mn = min(1000LL, p); but should be sqrt(y), since y could be 1E9, whereas 1000*1000 = 1E6.
• » » » » » 2 years ago, # ^ | 0 mn = min(1000LL, p); why 1000?
• » » » » » » 2 years ago, # ^ | 0 my bad :( got it!! thanks
• » » » 2 years ago, # ^ | +1 prime gap: https://en.wikipedia.org/wiki/Prime_gap
• » » » 2 years ago, # ^ | 0 what if p is 10^8 or greater. wont it get TLE
• » » » » 2 years ago, # ^ | ← Rev. 2 → 0 You check for prime divisors of i where y-300<=i
• » » » » » 2 years ago, # ^ | 0 i checked for all divisors of i (sqrt(n)) upto i — 500 and if yes print
• » » 2 years ago, # ^ | 0 Then why is this solution wrong ?https://ideone.com/3YAlQZ
• » » » 2 years ago, # ^ | +1 case 2, 9 answer is 9
• » » » » 2 years ago, # ^ | 0 Oooooops,
• » » » 2 years ago, # ^ | 0 check 2 9 output should be 9
• » » » 2 years ago, # ^ | +1 the solution need not be prime everytime. The only condition it must satisfy is that in the range [2, p] it doesn't have any factors!
• » » » 2 years ago, # ^ | +1 Thanks guys. Silly bugs everywhere :(
• » » 2 years ago, # ^ | ← Rev. 2 → +1 You don't need to do calculate any prime gaps, just work backwards from y. Because you know that there won't be a huge prime gap, you are sure to find something pretty quickly. http://codeforces.com/contest/937/submission/35691076 Just be sure to leave the loop once you find it and to only check up to sqrt(y) when searching for divisors
• » » » 2 years ago, # ^ | 0 How to calculate the time complexity of your code Intrincantation
• » » » » 2 years ago, # ^ | 0 In the worst case, outer loop runs until we find a prime number, so when calculating the complexity you need to use prime gaps, but you don't need to actually know the exact value to implement it, just know it is relatively small. Let P(n) be the largest prime gap. Complexity will be O(P(y) * sqrt(y)). Because you are checking up to sqrt(y) to find divisors, and you are doing this at most P(y) times.
» 2 years ago, # | +10 How to solve Div-1 C?
• » » 2 years ago, # ^ | ← Rev. 3 → +16 First turn the string into a permutation, that we want to transform into 1,2,...,n. Then, starting from n/2, start constructing the string. In a single step, do: 6789...5... --> ...6789...5 --> 5......6789 --> 98765...... This reverses the string we've built, and adds the number we wanted to add to the end of the reversed string. Use this to append small and large numbers alternatingly. It takes 3*n = 6000 operations.
• » » 2 years ago, # ^ | 0 I had similar idea to other one.But I wanted to share my way of thinking.I could easily come up with 4*n steps for building from one corner to another. But the 4*n was coming because each time the string was coming reversed so that wasted n steps to straighten in it. So doing from middle and adding a letter on each side we can get rid of straightening here.4*n approach: assume u have come to this form S.... S is suffix of required string . Now I will describe a method to convert it to Y.... where Y is bigger suffix by 1 than S. So now we find character before S in t and flip it there so now ....Y... we have.Now we have to bring Y to start for that we reverse whole string(...rev(Y)...) and bring rev(Y) to end (....rev(Y)) and now flip from just before Y giving Y..... . So main idea is getting rid of reverse whole string by doing operations from both sides.
• » » » 2 years ago, # ^ | +8 I extend the prefix like you do in your 4N approach, but each time I add a letter from a different end considering that s[0] and s[n-1] are adjacents. At the end the string may need to be reversed and/or rotated. Reversing it is simple: op(n). Then if it needs to be rotated X times, you can do that with 3 operations: op(x), op(n-x), op(n).
» 2 years ago, # | +17 Problem E is fucking awesome (no sarcasm)
» 2 years ago, # | 0
» 2 years ago, # | ← Rev. 2 → 0 Why is this solution is wrong?! Div2Dhttps://ideone.com/kZvFal Line: 50 :'(
» 2 years ago, # | 0 Again a good problemset... Thanks codeforces.
» 2 years ago, # | 0 What is hack test for Div.2 D???
• » » 2 years ago, # ^ | ← Rev. 3 → +17 Something like4 41 22 3 401 21You can't win, but you can draw by cycle 2->4->2
• » » » 2 years ago, # ^ | 0 My code gives the right answer, thanks :D I was fearing something harder
• » » » » 2 years ago, # ^ | 0 You can check also odd length cycle, then you must win.
• » » » » » 2 years ago, # ^ | +1 too many bugs in my solution, yet it passes pretests.seems like bad pretests
• » » 2 years ago, # ^ | +6 Suppose, it was against invalid back path construction. It is not enough to build back path until start vertex is found, the additional condition of stop must be the first player in start vertex. Otherwise for the path like (1)->(2)->(3)->(1)->(4)->(5) the first player can win in case of complete path walking, but back path construction, with mistake mentioned above, brings to only (1)->(4)->(5) which is not the win of the first player.
» 2 years ago, # | +10 What were the hacks for B?
• » » 2 years ago, # ^ | ← Rev. 2 → -7 What was the solution for DIV2.B?? was there primality check involved?
• » » » 2 years ago, # ^ | 0 You don't have to do a full-blown primality check. Just iterate in descending order through every x for which p < x <= y, and check if it has any divisor d for which 2 <= d <= min(p, sqrt(x)).
• » » 2 years ago, # ^ | -10
» 2 years ago, # | 0 May anyone tell me the core issue to raise a TLE in pretest 4 of Div1B/Div2D? ;)Here is my last solution, sorry for the code being so messy...
• » » 2 years ago, # ^ | 0 Maybe you didn't judge the condition when there is at least one circle in the graph,but there don't exist anyway to get to a point with a 0 outdegree,then in this case he could at least draw?I didn't pass all the pretests until the end of the contest(it may be something wrong with the output of the path),but I failed on pretest 4 previously,and after I found this condition and judged it I passed pretest 4...Hope it can help you,and sorry for my bad English
• » » » 2 years ago, # ^ | 0 Well, seemed like I got the issues. I did check for such, but I was totally careless when handling my condition-checker in return commaands.Not sure if it could save me from TLE-ing, but at least it was wrong.Thanks ;)
» 2 years ago, # | -51 MATH, MATH, everywhere MATH.
» 2 years ago, # | ← Rev. 2 → +15 Hack for D div 2: 5 5 1 2 2 3 5 1 4 1 2 0 1 Expected : Win 1 2 3 4 2 5
• » » 2 years ago, # ^ | ← Rev. 2 → +6 What is the answer? Edit: You already edited your comment. Thanks !
• » » » 2 years ago, # ^ | +6 Sorry I forget to write the answer ... I edited it
• » » 2 years ago, # ^ | 0 Thanks, i hope my solution is right.
• » » 2 years ago, # ^ | 0 how the answer can be win ? when petya moves from 1 to 2 vasya will move from 2 to 5, as both will play optimally. so how can petya win?
• » » » 2 years ago, # ^ | 0 They don't play optimally, Petya plays for both of them.
• » » » » 2 years ago, # ^ | ← Rev. 3 → +3 ohh ! I wrote the condition for optimally well. As i thought that petya checks whether he will win the game no matter what vasya will move after sleep, or make it draw. And how did it pass the pretests ? As the checking condition itself is entirely wrong
• » » » » » 2 years ago, # ^ | -8 And how did it pass the pretests ? pretests are weak in this one, probably intentionally. which is a good idea for div1-B but not good for div2-D.. but i guess thats a disadvantage of mixed round.
• » » » » » » 2 years ago, # ^ | +1 pretests are weak. But the checking condition itself is entirely wrong.
• » » » » » » » 2 years ago, # ^ | 0 coincidence
• » » » » » 2 years ago, # ^ | ← Rev. 2 → 0 Your solution fails for this:3 31 22 1 301Answer is Draw.Idk how you passed pretests, I guess they are weak
• » » » » » » 2 years ago, # ^ | ← Rev. 2 → 0 It fails for all cases as my checking condition itself is entirely wrong, as i misunderstood it.
» 2 years ago, # | +12 https://postimg.org/image/ts93ksg85/ still waiting :)
» 2 years ago, # | 0 Can someone please explain the logic for solving Div2.D/Div1.B ?
• » » 2 years ago, # ^ | ← Rev. 3 → 0 You need to reach a childless node in an odd number of steps to Win. You can track the parents of all nodes that can be visited starting from s in an odd and even number of steps and their in a 2*n array.If one childless node can be visited in an odd number of steps, you win. Else, you draw if there is a cycle attainable from s. Otherwise, you lose.Edit : my condition for cycle detection was wrong. You have to check explicitely for cycles starting from s to differentiate draw from loss
• » » » 2 years ago, # ^ | 0 Nice. Thanks !
• » » » 2 years ago, # ^ | 0 Not really. If a cycle has a way in and a way out, which lead to a childless node by an odd number of steps, you can still win.Also, if a cycle has an odd number of sides, any path from it to an arbitrary childless node can lead to victory as well.
• » » » » 2 years ago, # ^ | +8 how to check if there is an odd cycle?
• » » » » » 2 years ago, # ^ | +2 (Actually, I wasn't able to solve it probably — it got TLE verdicts).Since I can't handle cycles properly, I used Kosaraju's algorithm and divided vertices into SCCs.Then, the condition to call each DFS function is to check if there has been a path from s to that vertex with odd/even number of edges (based on the oddity of the currently working-on-vertex).
• » » » » » » 2 years ago, # ^ | 0 can u tell me why my code is wronghttp://codeforces.com/contest/937/submission/35735167
• » » » » » 2 years ago, # ^ | +5 can u tell me why my code is wrong http://codeforces.com/contest/937/submission/35735167
• » » » » » » 2 years ago, # ^ | 0 For the failed case, Someone else marked vis[node] = 2; for node = 45. So, when you tried to end game by visiting 45, it was already visited in some path so it was never even tested. Basically, you aren't trying certain paths once that node is visited in some other path.
• » » » » 2 years ago, # ^ | +11 If you (implicitly) create a bipartite graph G', with twice the number of the original graph, one vertex for each (vertex, parity) setting, and the same relative neighborhoods and perform a DFS on that graph, this case will be treated without having to explicitly detect odd cycles.
• » » » » » 2 years ago, # ^ | -8 Hi, Can you suggest some readings/problems based on the similar idea? A lot of programmers have used this technique, is it well known?
• » » » » 2 years ago, # ^ | +3 My algorithm will go at most twice through every edge : once when you reach him with an odd number of steps, and once when you reach him with an even number of steps.I believe this takes care of these two cases
• » » » 2 years ago, # ^ | 0 How do u calculate if a childless node can be reached in odd number of step(I mean there could be cycles)?Thanx in advance!!
• » » » » 2 years ago, # ^ | 0 Is this solution correct?We have to reach the start node from childless node in odd number of steps.So lets say we color the nodes starting from the childless node.lets say i color the childless node with black.So i'll color its adjacent white.So the question comes that whether the source vertex can coloured white.So just do a sort of bfs with two visited array ,vis1 to denote if it has been coloured black,vis2 to denote if it has been colored white.push all childless node and start the normal bfs,i mean if i am black and my adjacent are not white then push them,else dont.
• » » » » 2 years ago, # ^ | 0 each time you visit a node, mark all his children to be visited with the opposite parity, if they have not been visited with this parity. You'll find out in O(n+m) if each node can be reached in an odd or even number of steps, then you can check the childless nodes one by one
• » » » » » 2 years ago, # ^ | 0 will this handle the cases where there is an odd cycle in the path from s to any vertex?
» 2 years ago, # | +1 Could anyone give any hint to solve div2 D?? Thanx in advance!!
• » » 2 years ago, # ^ | +4 Just replace any vertex v to vertex (v, 0), (v, 1) and any edge u -> v to two edges: (u, 0) -> (v, 1) and (u, 1) -> (v, 0). Now you can do simple Dijkstra from (start, 0).
» 2 years ago, # | 0 If you pass pretests on a submission and then get compilation error on the next, does the accepted solution go through system testing or does cf take the last submission even if it gives WA/compilation error?
• » » 2 years ago, # ^ | 0 the last accepted submission
» 2 years ago, # | ← Rev. 4 → +5 In div-2 D, when I tried to hack and got unsuccessful attempt, I realize I misunderstood...I thought the real Vasya starts after some point, So for have a draw, in every possibility they cannot get out of the loop . 3 31 22 1 301my answer is Lose for it although the real answer is Draw :(
» 2 years ago, # | 0 Another hack for B. Although i din solve D, as i found the test case in which it will fail 35 min before the end and i could not find the solution, this is it. 8 8 1 2 1 3 2 4 6 1 5 1 1 1 7 1 8 0 4 ans is win and 4 5 1 2 3 6 7 8
» 2 years ago, # | ← Rev. 2 → 0 For div2d .. if: find any path to a leaf and it's vasya's turn --> winelse if: cycle --> drawelse: --> loseWhat's wrong ?Edit: never mind. My code contained a bug
• » » 2 years ago, # ^ | +3 else if: cycle --> draw You must also check if you can reach that cycle.
• » » » 2 years ago, # ^ | 0 When you start from S, any cycle you reach is certainly reachable
• » » 2 years ago, # ^ | 0 if: find any path to a leaf and it's vasya's turn --> win also, any path can contain itself contain a cycle (might not be required to traverse same cycle more than once though, not sure)
• » » 2 years ago, # ^ | 0 The fact is that while traversing if you en-counter a odd length cycle, it can be used to change the parity and reach the same state with the other parity of the solution.
» 2 years ago, # | ← Rev. 2 → 0 How people could solve problem C in 1 minute? Like LodThe.
• » » 2 years ago, # ^ | 0 idkits fking impossible
• » » » 2 years ago, # ^ | 0 Possible because questions were from some innopolis olympiad.
» 2 years ago, # | 0 in D div 2, assuming both petya and vasya play optimally well right?
• » » 2 years ago, # ^ | +21 Vasya doesn't even play how he can play optimally lol.
• » » 2 years ago, # ^ | +3 No we have to favour petya (check test case 1) vasya can move 1 -> 2 -> 5 but we favour petya thats why 1 -> 2 -> 4 -> 5
• » » » 2 years ago, # ^ | 0 since petya plays first, how will he wants to lose. And what i thought is petya is checking whether he will win, if tomorrow vasya gets up from sleep and play optimally. AND by the way how did it pass the pretests ?
• » » » » 2 years ago, # ^ | 0 you are checking petya can win or not irrespective of how vasya will play, if vasya will play optimally there is no way petya can win in first test case.
• » » » » » 2 years ago, # ^ | 0 He can. if petya goes to 3, instead of 2, He will definitely win, As petya moves first.
• » » » » » » 2 years ago, # ^ | ← Rev. 2 → +3 yes if petya will play 1->3 than vasya can't, but both the moves were made by patya, so petya is playing optimally, and vasya's moves always favouring petya
» 2 years ago, # | ← Rev. 2 → 0 In problem D : my codes second one is correct :( :( https://www.diffchecker.com/FXKB7hnMUPD : second one also wrong ! :))
» 2 years ago, # | 0 long queue but nice round, hope rated!
» 2 years ago, # | +6 4 4 2 2 3 1 4 1 4 0 1 hack for div2 D: ans is lose
• » » 2 years ago, # ^ | +3 Doing the hard work but getting WA due to wrongly implementing cycle detection as in a non directed graph = priceless
• » » » 2 years ago, # ^ | +16 pretests were too weak.. i was using 0 indexed vertices instead of 1 indexed vertices in dfs and still it passed the pretests
» 2 years ago, # | 0 Can anybody tell me what is the time complexity of this code for DIV2 B.
• » » 2 years ago, # ^ | 0 around 300 * sqrt(10^9)
» 2 years ago, # | +17 Late hack announcement! Little too late!!!I submitted Div1B in 01:22:38 and it was hacked in 01:31:59. RoomBut I didn't get any hack announcement notification during the contest time. I checked my submission couple of times in Room standings and Common standings but didn't see that it had been hacked. Even, I refreshed the main Problems page and standings page after the contest had ended, till then it was in "pretests passed" state instead of "hacked". I only got a hacked notification during the system testing. I didn't take any screenshots for proof as now I think I should have (sorry). Now I am thinking, how much a late hack announcement/update can affect a contestant? Also had anyone else experienced this?
• » » 2 years ago, # ^ | +5 My Div1B was also hacked without giving me a pop-up notification whereas it should be given immediately right after a submission is hacked in normal practice. Luckily, in my case the verdict showed correctly with "hacked" in "My submission" page. As a result, I was not affected by the missing of the pop-up notification after all.
» 2 years ago, # | +38 Poland stronk!
• » » 2 years ago, # ^ | +32 Tanks in advance!
» 2 years ago, # | 0 Ooops! Something has broken down in Codeforces :<
» 2 years ago, # | +8 In Div2 problem C: what is the correct output for this test case: 999999999999999999 999999999999999998 1000000000000000000
• » » 2 years ago, # ^ | 0 1000000000000000000.93750000000
• » » » 2 years ago, # ^ | 0 i got 1000000000000000000.93750000000 and 1000000000000000000.00000000000 and 1000000000000000001.00000000000 from different accepted codes
• » » » » 2 years ago, # ^ | 0 hmm.. I think its something about the precision. Mine returns the .9375 one
• » » » » 2 years ago, # ^ | +6 Thats because mod(x-y)/max(1,y) should be less than pow(10,-9) which is true for all accepted codex,y are expected and actual output
• » » 2 years ago, # ^ | 0 I am getting 999999999999999998
• » » 2 years ago, # ^ | ← Rev. 2 → 0 1000000000000000001 here (on AC code).
• » » 2 years ago, # ^ | +3 1000000000000000001 to be exact, but just 1.0E18 works
» 2 years ago, # | +70 How did LoDThe solved div2A at 00:02:12 and div2C at 00:03:38 in div2?No one in the top20 positions in div1 solved div2C in under 4 minutes, and he even solved A before!
• » » 2 years ago, # ^ | ← Rev. 6 → +26 He took part in offline part. Just cheats and no more. P.S. His name is Igor Balyuk
• » » » 2 years ago, # ^ | +24 Shouldn't this be taken care of? Because it'd affect the rating change. I mean if he took part in the actual contest before, then it's not fair.
• » » » » 2 years ago, # ^ | +4 hope :p
• » » » » 2 years ago, # ^ | +20 Of course it wasn't fair. I believe somebody should fix it.
• » » » » » 2 years ago, # ^ | 0 And the rating has been updated, they won't fix it :( And I'm only 2 short from being a Candidate Master :( It's not fair. I would have been Candidate Master if he didn't participate :( I'm so frustrated now...Can't anything be done?
• » » » » » 2 years ago, # ^ | +84 I write to niyaznigmatul and KAN about this problem and don't get any feedback. Lets think it's legal to cheat on rounds.
» 2 years ago, # | +3 D went from ~400 ACs to ~70. E F F C Y C L E S
» 2 years ago, # | ← Rev. 3 → +5 I got WA for problem DIV2 D on test 28 which was: Participant's output Win 233 971 277 477 871 502 673 37 219 Jury's answer Win 233 864 714 151 370 604 141 233 971 277 477 871 502 673 37 219what was wrong ?
• » » 2 years ago, # ^ | +3 Your answer is an odd length, the correct answer must be even length...
• » » 2 years ago, # ^ | 0 To win you have to have a path of even length
• » » » 2 years ago, # ^ | +5 I made a terrible mistake while printing path :(
• » » » » 2 years ago, # ^ | +1 If it makes you feel any better, I made the exact same mistake. :(
• » » 2 years ago, # ^ | 0 Well, to win you need go from 233(turn First Player) to 233(turn Second Player).
» 2 years ago, # | ← Rev. 2 → 0 Why am I getting the wrong answer for div2B. I used the segmented sieve and I am getting the correct answer when I run it on my pc.Here is the code..
• » » 2 years ago, # ^ | 0 You mean to say you're getting the correct answer for the case that is failing?
• » » » 2 years ago, # ^ | 0 Yes.
» 2 years ago, # | +23 Testcases for E are too many..
» 2 years ago, # | ← Rev. 2 → +75 Currently, Div.1 contest doesn't have a final result yet, because of this.If cki86201's solution for Div1E is correct, he/she will make it to the 3rd place. Cheers! ;)UPD: Accepted. Congrats!
• » » 2 years ago, # ^ | +134 Thanks!!
» 2 years ago, # | -10 i got wrong answer in C-Div2 because the output number was in double format like that 7.67538e+017 insted of 767537647587662141 , so is that my fault or the judge fault
• » » 2 years ago, # ^ | +18 If you are using cout use cout<
• » » » 2 years ago, # ^ | ← Rev. 2 → 0 How to solve problem C ?UPD: Thanks for not replying to this poor. I got it.
• » » 2 years ago, # ^ | 0 same thing happend with me also..
» 2 years ago, # | ← Rev. 2 → -11 Running the same code on the computer gives different results from that when it's run from the judge! Link
• » » 2 years ago, # ^ | +17 In this function: int divs(int x) { for(int i=2;i
• » » » 2 years ago, # ^ | +9 totally agree with u, in my computer I get -1; So I change this code a little bit, 35711545 , It passes test 1.Hope it is helpful @ LouaiZahran
• » » » » 2 years ago, # ^ | +11 Thanks a lot! Akikaze Wavator
» 2 years ago, # | +3 Rating changes take forever when you want them to be quick.
• » » 2 years ago, # ^ | +3 You may use this site to get an approximation https://cf-predictor-frontend.herokuapp.com According to it you'll become purple with an increase of 150 points Congratulations :)
» 2 years ago, # | 0 Yes.
• » » 2 years ago, # ^ | 0 Never purple again! Congrats!
» 2 years ago, # | +64 Hmm.. Feels like someone was writing div1 and div2 contest in the same time http://codeforces.com/contest/937/submission/35707300 http://codeforces.com/contest/936/submission/35706280
• » » 2 years ago, # ^ | +10 It doesn't make sense. Why did he do that? I could understand testing div1 solutions on div2 pretests, but he submited div1 earlier than div2...
» 2 years ago, # | ← Rev. 2 → 0 Can somebody please tell me why my code for D gets a TLE at pretest 4? Link- http://codeforces.com/contest/937/submission/35711753For some reason, the DFS function is running into an infinite loop- but according to me that shouldn't be, as I am using the recur[u] to act like visited[u] in a way. I used the fact that, the only way when recur[u] will be 0 again after becoming 1 is when the DFS of the subtree of u ends- else its a cycle which we detect and exit such visited nodes.Any test case or help is appreciated. Thank you :)(I know this can Time out for larger cases, like, vertex s leading to n/2 nodes, and all those n/2 nodes leading down to same chain of length n/2. Like a long-tailed kite. But why is it getting TLE at pretest 4 with n=50 is bugging me :( )
• » » 2 years ago, # ^ | +3 After done With current node you are allowing others to come back to the same node later on. Many node can come back which points to this node causing infinite’s loop.
» 2 years ago, # | ← Rev. 6 → +23 Hi guys, while you are waiting for editorial, I'll try to help you to solve problem D.So, our task is to find a uneven route to a leaf-vertex (a vertex that has no edges coming from it)The answer is Draw, if the graph doesn't contain leaves at all.If the graph has a leaf (leaves). But a route to each is only even, we need to check whether the graph has a cycle. If it does — the answer is Draw else the answer is Lose.If the graph has a leaf with an odd route to it the answer is Win.Now our task is to output the way to our leaf. We should be careful with it and not break a while(probably :P) cycle if our route at the moment is even! How can we code the written above? I think, that one of the easiest way is to write a bfs and check the following:"If we can reach a V-vertex with an uneven route then we can reach TO-vertex with an even route. And vice a verse"My solution is here. Feel free to ask me some questions. GLHF
• » » 2 years ago, # ^ | 0 dfs on 2-layers graph looks simpler)We can save sequence at once we found that answer is "Win".
» 2 years ago, # | 0 Too eager to know solution for Div2 E/ Div1 C.. help required
• » » 2 years ago, # ^ | +3 The cleanest solution I've seen is if you perform the operation on x = n, then on x = i, then on x = 1, you move the ith character to the front, e.g. if we want to move the c to the front: front|c|back (ignore the |'s)-> kacb|c|tnorf (x = n) -> frontkacb|c| (x = i) -> |c|backfront (x = 1)Using this operation, you can repeatedly build the string in reverse order, using at most 3*N < 6100 steps.
• » » » 2 years ago, # ^ | +3 Oops formatting:front|c|back-> kacb|c|tnorf (x = n)-> frontkacb|c| (x = i)-> |c|backfront (x = 1)
• » » » » 2 years ago, # ^ | ← Rev. 4 → 0 Thanks. Got it . xD
• » » » » » 2 years ago, # ^ | +8 I think it would be x=i since after the first operation the character is in the n-i'th position.
• » » » » » » 2 years ago, # ^ | ← Rev. 3 → +23 Got it man. Thanks. by the way your last ans is wrong. it should be |c|frontkacb
» 2 years ago, # | +6 I am seeing some pretty complicated submissions for problem B — Vile Grasshoppers. The main idea is that the distance between consecutive primes upto 10^9 is never more than 300 or so. This allows us to start from y and keep decrementing, till we get a number who's smallest factor is greater than P. Worst case, we will not have to do more than 300 root(n) factorisations. Explanation and code.
» 2 years ago, # | +1 Lost ~1300 points because I didn't use std::fixed in the output of Div2/C...
• » » 2 years ago, # ^ | 0 me too :( now my rating change is -35 instead of predicted +35.
» 2 years ago, # | 0 Can anyone tell me why my solution is failing at test 28? http://codeforces.com/contest/936/submission/35722758
• » » 2 years ago, # ^ | ← Rev. 2 → 0 Cause in your solution you do not consider paths like 1->2->3->1->4->5 when you can return to previously passed vertex and still win, I guess so.Check this out: 5 52 2 41 31 11 50 1If it returns "Draw", your solution is wrong.
» 2 years ago, # | +2 When can we expect editorials ?
» 2 years ago, # | 0 Hey in the C problem of div. 2 one of my friend's code got accepted and as I was going through the test cases I found this(refer link). How is this possible? https://drive.google.com/open?id=1scjEnYJuRpkfcu2gPGvJRpBm94NcH1sH
• » » 2 years ago, # ^ | ← Rev. 3 → 0 Everyone who solved C has this problem, but the answers are definetely right
» 2 years ago, # | 0 How to solve Div2 C?
• » » 2 years ago, # ^ | ← Rev. 5 → +14 Our goal is to find how long will be a chicken in a fast mode and in slow :D.So, in first k minutes chicken will be prepared for ((k/t)*100)%.fastMode=((k/t)*100)Now we need to find a moment of time when Julia will again switch chicken to fast mode. This moment will be f:f=ceil(k/d)*d => (f-k) minutes chicken will be in a slow mode and will be prepared for ((f-k)/(2*t)*100) %.slowMode=((f-k)/(2*t)*100);So total time of that cycle is k+(f-k). After each cycle our chicken will be ready for fastMode+slowMode%.Now let's find out how many full cycles we will be able to "put" in our 100%. That is floor(100/(fastMode+slowMode))At that moment our answer=floor(100/(fastMode+slowMode))*fFinally we need to check how many percent of chicken preparation left left=100-floor(100/(fastMode+slowMode))*(fastMode+slowMode). If it is equal less than (k/t)*100 then we just add to our answer (left/(fastMode))*k, else ans+=k, left-=fastMode and then ans+=(left/(slowMode))*(f-k)Yeah, I see it looks quite complicated , but you are free to ask questions. Maybe my solution could also help.** P.S. When I was using % (multiplying by 100) i received a WA on test case #32. On codeforces compiler answer was "nan" while on mine it was OK :( **UPD Corrected a formula
• » » » 2 years ago, # ^ | +3 If d > k then your algorithm will give f = 1 => f-k = 1-k(negative for k > 1) minutes chicken will be in slow mode. Isn't it undesirable? BTW thanx for great explanation.
• » » » » 2 years ago, # ^ | +5 Yeah, I did a mistake Correct is: f=ceil(k/d)*dThank you for your answer.
• » » » 2 years ago, # ^ | 0 This solution is correct but the formulas can be simpler:Indeed, the chicken is cooked in fast mode for k minutes then, for d — k mod d minutes. Indeed, after k + (d — k mod d) minutes, you will be on a multiple of d. So the lenght of a cycle is of k + d — k % d. The formula for the number of cycles : cycles = t / (k + (d — k%d)/2)
• » » » 2 years ago, # ^ | 0 When I ran your code for k = 2, d = 5 and t = 10 it outputs 14 whereas the answer should be 12, shouldn't it? or am I missing something?
• » » » » 2 years ago, # ^ | ← Rev. 2 → 0 The correct answer is 14.After first cycle: chicken is prepared for 4/20+3/20=7/20; TimeTaken=5;Second cycle: chicken is prepared for 14/20. TimeTaken=10;Third cycle onlyFastMode: chicken is prepared for 18/20. TimeTaken=12;We need to prepared chicken for 2/20, while using fullSlowMode we will prepare it for 21/20. That's why we need to use only 2/3 of our SlowMode. So the time of this part of SlowMode will be 2/3*3=2.TimeTaken=12+2=14
• » » » » » 2 years ago, # ^ | ← Rev. 4 → 0 As d = 5 when Julia goes to kitchen at t = 5 she will find the stove to be on. So she will not do anything and return. So The time period of one cycle comes out to be = 10. I guess you have taken time period of your cycle = d(= 5) which I think is be wrong in this case... How I visualized it is as follows When the clock is high it means gas is on and when it is low it means gas is off.
• » » » » » » 2 years ago, # ^ | ← Rev. 2 → 0 You misunderstood the problem. The stove turns off by itself, but only Julia can turn it on. When t=4, who turned it on on your graphic?
• » » » » » » » 2 years ago, # ^ | 0 Yeah my bad I misunderstood the problem. Anyway thanx for helping me.
» 2 years ago, # | +15 Editorial?
» 2 years ago, # | 0 can someone tell me why my code is wrong and how can i rectify it thanx in advance..http://codeforces.com/contest/937/submission/35735167
» 2 years ago, # | ← Rev. 2 → +6 17 people contributed to this contest and there is no editorial after one day!!UPD1 : a semi editorial is posted,hope to see the complete one!
• » » 2 years ago, # ^ | +26 Perhaps Brooks's law applies to writing editorials, too?
• » » » 2 years ago, # ^ | 0 although i had to use wikipedia to understand your comment's meaning, but i'm agree with you :D
• » » » 2 years ago, # ^ | +10 I think it's more of bystander effect.
• » » » » 2 years ago, # ^ | +15 I wouldn't call this round an accident. I quite liked problem C.
» 2 years ago, # | 0 Why 35710602 WA ? Div2D
• » » 2 years ago, # ^ | ← Rev. 2 → 0 Your problem is that you overlooked the case when you have a cycle with odd number of vertices where you should transverse the cycle to change the turn from Petya to Vasya and vice versa.Consider the following testcase: 5 4 2 2 4 1 3 1 1 1 5 0 1 This corresponds to a graph that has a cycle 1->2->3->1 but it still had an answer 1->2->3->1->4->5 Your Code gives a wrong answer in that case as it prints "Draw" instead, you just didn't handle the cycles with odd number of vertices case.
» 2 years ago, # | 0 When will we get tutorials for the problems ?
» 2 years ago, # | ← Rev. 2 → 0 i was waiting for editorial but after a day now i decide to ask my question here...how to solve Div2.D(Div1.B)?
• » » 2 years ago, # ^ | +34
• » » » 2 years ago, # ^ | 0 thanks... better than nothing!!
» 2 years ago, # | +1 where's the tutorial to the problems ?
» 2 years ago, # | ← Rev. 2 → +2 pls, upload the editorials...
» 11 months ago, # | 0 niyaznigmatul Div. 1 E doesn't appear to be viewable.
• » » 4 months ago, # ^ | 0 I'm having the same trouble.niyaznigmatul Could you fix it? | |
# Frontiers of Earth Science
ISSN 2095-0195 (Print)
ISSN 2095-0209 (Online)
CN 11-5982/P
Postal Subscription Code 80-963
Formerly Known as Frontiers of Earth Science in China
2018 Impact Factor: 1.205
#### , Volume 13 Issue 2 dateToAbbreviationMonthYear('2019-06-15 00:00:00.0','CurrentIssueDate');
RESEARCH ARTICLE
Select Chemical and minero-petrographical changes on granulite rocks affected by weathering processes Carmine APOLLARO, Francesco PERRI, Emilia LE PERA, Ilaria FUOCO, Teresa CRITELLI Front. Earth Sci.. 2019, 13 (2): 247-261. https://doi.org/10.1007/s11707-018-0745-5 Abstract HTML PDF (14030KB) The purpose of this work is to study the weathering processes of the granulite rocks of the Serre Massif (southern Calabria, Italy) using a multidisciplinary approach based on field studies, geochemical modeling, and minero-petrographical analyses. The granulite rocks are plagioclase-rich with minor amphibole, clinopyroxene, orthopyroxene, biotite, and garnet and their texture are coarse-grained. The reaction path modeling was performed to simulate the evolution of groundwaters upon interaction with local granulite by means of the software package EQ3/6, version 8.0a. Simulations were performed in kinetic (time) mode under a closed system at a constant temperature of 11.5°C, (which reproduces the average temperature of local area) and fixing the fugacity of CO2 at 10−2.34 bar (mean value). During the most advanced stage of weathering the main mineralogical changes are: partial destruction and transformation of biotite and plagioclase associated with neoformation of ferruginous products and secondary clay minerals producing a change in the origin rock fabric. The secondary solid phases observed during the geochemical modeling (kaolinite, vermiculite and ferrihydrite) are similar to those found in this natural system. Thus, the soil-like material mainly characterized by mostly sand to gravel grain-size fractions is the final result of the weathering processes.
Select Evolution model of a modern delta fed by a seasonal river in Daihai Lake, North China: determined from ground-penetrating radar and trenches Beibei LIU, Chengpeng TAN, Xinghe YU, Xin SHAN, Shunli LI Front. Earth Sci.. 2019, 13 (2): 262-276. https://doi.org/10.1007/s11707-018-0740-x Abstract HTML PDF (4952KB) While deltas fed by seasonal rivers are common in modern sedimentary environments, their characteristics remain unclear as compared to those fed by perennial rivers. This study identifies a small delta discharged by a seasonal stream flowing into Daihai Lake, in northern China, which is driven by ephemeral and high-energy flood events. Detailed 3D facies architecture was analyzed using ground-penetrating radar (GPR) and sedimentary logs from outcrop and trenches. Four types of radar surfaces, including truncations of underlying inclined strata, weak reflections, and depositional surface of downlap and onlap, were identified. Six radar facies (high-angle oblique-tangential, low-angle subparallel, gently plane parallel, plane-parallel, chaotic, and continuous strong reflection) were identified based on distinctive reflections, including amplitude, continuity, dip, and termination patterns. Five depositional units (Unit A to E) were documented from proximal to distal delta. Seasonal discharge signatures include significant grain-size decrease over short distance, abundant Froude supercritical flow sedimentary structures, poorly developed barforms, and small-scale scour and fill structures. Records of lake-level and sediment budget were evaluated over the past 60 years. In highstand stage (1960–1980), amalgamated channel (Units A and B), and delta front (Unit C) were deposited. In slope stage (1980–1996), the lower deposits (Units A, B, C) were eroded by Unit D with a distinct truncation surface. In lowstand stage, most eroded sediments bypassed the incised channel and accumulated in the distal part, in which a new depositional unit was formed (Unit E). The model demonstrates that deltas fed by seasonal rivers tend to accumulate large amounts of sediments carried by high magnitude floods within short periods.
Select Bias characterization of ATMS low-level channels under clear-sky and cloudy conditions Qi LI, Xiaolei ZOU Front. Earth Sci.. 2019, 13 (2): 277-289. https://doi.org/10.1007/s11707-019-0750-3 Abstract HTML PDF (6831KB) The Advanced Technology Microwave Sounder (ATMS) onboard the Suomi National Polar-Orbiting Partnership satellite is a cross-track scanning instrument containing 22 sounding channels in total. In this study, the bias characteristics of channels 1–6, which could have significant cloud contamination in heavy precipitation, are first analyzed based on the differences between ATMS observations (O) and model simulations (B) under clear-sky conditions over oceans. Latitudinal dependencies of the biases of window channels 1–3 are greater than those of channels 4–6. Biases of all nadir-only observations examined in different latitudinal bands [μ1(ϕ)] are positive and no more than 7.0 K. Biases at higher latitudes are larger. Channels 1–5 have a generally symmetric scan bias pattern [μ2(α)]. The global distributions of brightness temperature differences after subtracting the biases, i.e., O-B-m1(ϕ)-μ2(α), for channels 3–6 spatially match the liquid water path distributions. Excluding ice-affected observations, channel 3–6 O-B differences systematically increase as the liquid water path increases under cloudy conditions. Further investigation is needed to apply these findings for ATMS data assimilation under both clear-sky and cloudy conditions.
Select The relationships between urban-rural temperature difference and vegetation in eight cities of the Great Plains Yaoping CUI, Xiangming XIAO, Russell B DOUGHTY, Yaochen QIN, Sujie LIU, Nan LI, Guosong ZHAO, Jinwei DONG Front. Earth Sci.. 2019, 13 (2): 290-302. https://doi.org/10.1007/s11707-018-0729-5 Abstract HTML PDF (5673KB) Interpreting the relationship between urban heat island (UHI) and urban vegetation is a basis for understanding the impacts of underlying surfaces on UHI. The calculation of UHI intensity (UHII) requires observations from paired stations in both urban and rural areas. Due to the limited number of paired meteorological stations, many studies have used remotely sensed land surface temperature, but these time-series land surface temperature data are often heavily affected by cloud cover and other factors. These factors, together with the algorithm for inversion of land surface temperature, lead to accuracy problems in detecting the UHII, especially in cities with weak UHII. Based on meteorological observations from the Oklahoma Mesonet, a world-class network, we quantified the UHII and trends in eight cities of the Great Plains, USA, where data from at least one pair of urban and rural meteorological stations were available. We examined the changes and variability in urban temperature, UHII, vegetation condition (as measured by enhanced vegetation index, EVI), and evapotranspiration (ET). We found that both UHI and urban cold islands (UCI) occurred among the eight cities during 2000–2014 (as measured by impervious surface area). Unlike what is generally considered, UHII in only three cities significantly decreased as EVI and ET increased (p<0.1), indicating that the UHI or UCI cannot be completely explained simply from the perspective of the underlying surface. Increased vegetative cover (signaled by EVI) can increase ET, and thereby effectively mitigate the UHI. Each study station clearly showed that the underlying surface or vegetation affects urban-rural temperature, and that these factors should be considered during analysis of the UHI effect over time.
Select Effects of sea level rise on storm surge and waves within the Yangtze River Estuary Yongming SHEN, Gefei DENG, Zhihao XU, Jun TANG Front. Earth Sci.. 2019, 13 (2): 303-316. https://doi.org/10.1007/s11707-018-0746-4 Abstract HTML PDF (2678KB) Sea level rise (SLR) can cause water depth increase (WDI) and coastal inundation (CI). By applying the coupled FVCOM+SWAN model, this study investigates the potential impacts of WDI and CI, induced by a 1.0 m SLR, on storm surge and waves within the Yangtze River Estuary. A 1.0 m WDI decreases the maximum storm surge by 0.15 m and increases the maximum significant wave height by 0.35 m. The CI effect size is smaller when compared with WDI. CI decreases the maximum storm surge and significant wave height by 0.04 and 0.07 m, respectively. In the near-shore area, WDI significantly alters the local hydrodynamic environment, thereby stimulating changes in maximum storm surges and wave heights. Low-lying regions are negatively impacted by CI. Conversely, in deep-water areas, the relative change in water depth is minimal and the effect of CI is gradually enhanced. The combined effect of WDI and CI decreases the maximum surge by 0.31 m and increases the maximum significant wave height by 0.21 m. As a result, CI may be neglected when designing deep-water infrastructures. Nonetheless, the complex interactions between adoption and neglect of CI should be simulated to achieve the best seawall flood control standards and design parameters.
Select A fast and simple algorithm for calculating flow accumulation matrices from raster digital elevation Guiyun ZHOU, Hongqiang WEI, Suhua FU Front. Earth Sci.. 2019, 13 (2): 317-326. https://doi.org/10.1007/s11707-018-0725-9 Abstract HTML PDF (1355KB) Calculating the flow accumulation matrix is an essential step for many hydrological and topographical analyses. This study gives an overview of the existing algorithms for flow accumulation calculations for single-flow direction matrices. A fast and simple algorithm for calculating flow accumulation matrices is proposed in this study. The algorithm identifies three types of cells in a flow direction matrix: source cells, intersection cells, and interior cells. It traverses all source cells and traces the downstream interior cells of each source cell until an intersection cell is encountered. An intersection cell is treated as an interior cell when its last drainage path is traced and the tracing continues with its downstream cells. Experiments are conducted on thirty datasets with a resolution of 3 m. Compared with the existing algorithms for flow accumulation calculation, the proposed algorithm is easy to implement, runs much faster than existing algorithms, and generally requires less memory space.
Select Land use and land cover classification using Chinese GF-2 multispectral data in a region of the North China Plain Kun JIA, Jingcan LIU, Yixuan TU, Qiangzi LI, Zhiwei SUN, Xiangqin WEI, Yunjun YAO, Xiaotong ZHANG Front. Earth Sci.. 2019, 13 (2): 327-335. https://doi.org/10.1007/s11707-018-0734-8 Abstract HTML PDF (2472KB) The newly launched GF-2 satellite is now the most advanced civil satellite in China to collect high spatial resolution remote sensing data. This study investigated the capability and strategy of GF-2 multispectral data for land use and land cover (LULC) classification in a region of the North China Plain. The pixel-based and object-based classifications using maximum likelihood (MLC) and support vector machine (SVM) classifiers were evaluated to determine the classification strategy that was suitable for GF-2 multispectral data. The validation results indicated that GF-2 multispectral data achieved satisfactory LULC classification performance, and object-based classification using the SVM classifier achieved the best classification accuracy with an overall classification accuracy of 94.33% and kappa coefficient of 0.911. Therefore, considering the LULC classification performance and data characteristics, GF-2 satellite data could serve as a valuable and reliable high-resolution data source for land surface monitoring. Future works should focus on improving LULC classification accuracy by exploring more classification features and exploring the potential applications of GF-2 data in related applications.
Select Analysis of the relation between ocean internal wave parameters and ocean surface fluctuation Yufei ZHANG, Bing DENG, Ming ZHANG Front. Earth Sci.. 2019, 13 (2): 336-350. https://doi.org/10.1007/s11707-018-0735-7 Abstract HTML PDF (2364KB) The relation between ocean internal waves (IWs) and surface fluctuation is studied using a quasi-incompressible two-dimensional linear ocean wave model. The main conclusions are as follows: the IW parameters can be obtained by solving the boundary value problem of ordinary differential equations with the frequency, wave number, and amplitude of the surface fluctuation. When the ocean surface fluctuation state is given, the ocean IW presents a different structure, i.e., the uncertainty of the solution, which reflects the characteristics of the inverse problem. To obtain a definite solution, this study proposes constraint conditions for the inverse problem, namely, the relationship among background flow, buoyancy frequency, sea surface height, and geostrophic parameters. The necessary and sufficient conditions for the existence of IWs and external waves (surface wave) can be obtained according to the different constraint conditions. The amplitude of the surface fluctuation is positively correlated with IWs, and they share the same frequency and wave number. We also examined the relationship between the vertical structure, the maximum amplitude, and the constraint conditions. For a certain wave number, when the ocean environment is defined, the natural frequency (characteristic frequency) of IWs can be obtained. If the frequency of the surface fluctuation is similar or equal to the natural frequency, the resonance phenomenon will occur and can result in very strong IWs. The presented theory can serve as a basis for the analytical estimation of IWs.
Select Comparison of C- and L-band simulated compact polarized SAR in oil spill detection Xiaochen WANG, Yun SHAO, Fengli ZHANG, Wei TIAN Front. Earth Sci.. 2019, 13 (2): 351-360. https://doi.org/10.1007/s11707-018-0733-9 Abstract HTML PDF (1874KB) This paper presents the compact polarized (CP) pseudo quad-pol parameters for the detection of marine oil spills and segregation of lookalikes using simulated CP SAR data from full-polarized (FP) SAR imagery. According to the CP theory, 11 polarized parameters generally used for the detection of oil spills were derived from reconstructed pseudo quad-pol data for both C and L bands. In addition, the reconstruction performance between C and L bands was also compared by evaluating the reconstruction accuracy of retrieved polarized parameters. The results show that apart from $σHV$ and $RH$, other polarized parameters of $σHH$, $σVV$, $H$, $α$, $ϕH−V$, $r$, $ρH−V$, and $γ$ can be reconstructed with satisfactory accuracy for both C and L bands. Furthermore, C band has a higher reconstruction accuracy than L band, especially for $ϕH−V$. Moreover, the effect of reconstruction of polarized parameters on oil spill classification was also evaluated using the maximum likelihood classification (MLC) method. According to the evaluation of kappa coefficients and mapping accuracy, it is recommended to use $σHH$, $σVV$, $H$, $ρH−V$, and $γ$ of the C band CP SAR for marine oil spill classification.
Select Metal accumulation in Asiatic clam from the Lower Min River (China) and implications for human health Yue ZENG, Zhongtao LI, Qianfeng WANG, Changcheng XU, Yunqin LI, Jia TANG Front. Earth Sci.. 2019, 13 (2): 361-370. https://doi.org/10.1007/s11707-018-0724-x Abstract HTML PDF (1259KB) Considering growing concerns regarding polluted estuaries and their adverse effects on public health, this study aimed to identify concentrations of metal (Zn, Fe, Cr, Ni, Cd, Mn, As, Cu, and Pb) in Asiatic clams sampled along the Lower Min River, China. Multivariate methods were used to identify and apportion pollution sources. Noncarcinogenic and carcinogenic health risk assessments were performed to gauge adverse consumer health effects. Results showed that Cr, Pb, and Zn concentrations were higher than the limits prescribed in Chinese government guidelines. In comparison with concentrations of selected metals in other rivers, Cr, Pb, Zn, and As concentrations in clams were generally higher. Pollution assessment using the metal pollution index showed that sampling sites surrounding developing industrial and residential areas were the most polluted. Principal component analysis indicated significant anthropogenic metal contributions in clams. Health risk assessment indicated significant risk for clam consumers along the Lower Min River in terms of hazard quotient and carcinogenic risk and, thus, clam consumption from the study area should be avoided. The present findings would help in establishing environmental monitoring plans and contribute to preserving public health as well as the development of water conservation strategies to alleviate the metal pollution.
Select Development of a groundwater flow and reactive solute transport model in the Yongding River alluvial fan, China Haizhu HU, Xiaomin MAO, Qing YANG Front. Earth Sci.. 2019, 13 (2): 371-384. https://doi.org/10.1007/s11707-018-0718-8 Abstract HTML PDF (3413KB) The Yongding River in the western suburbs of Beijing has been recharged with reclaimed water since 2010 for the purpose of ecological restoration. Where the reclaimed water is not well treated, it poses a danger to the aquifer underneath the river. To provide a reliable tool which could be used in future research to quantify the influence of reclaimed water in the Yongding River on the local groundwater environment, a transient groundwater flow and reactive solute transport model was developed using FEFLOWTM in the middle-upper part of the Yongding River Alluvium Fan. The numerical model was calibrated against the observed groundwater levels and the concentrations of typical solutes from June 2009 to May 2010 and validated from June 2010 to December 2010. The average RMSE and R2 of groundwater level at four observation wells are 0.48 m and 0.61, respectively. The reasonable agreement between observed and simulated results demonstrates that the developed model is reliable and capable of predicting the behavior of groundwater flow and typical contaminant transport with reactions. Water budget analysis indicates that the water storage in this aquifer had decreased by 43.76×106 m3 from June 2009 to December 2010. The concentration distributions of typical solutes suggest that the middle and southern parts of the unconfined aquifer have been polluted by previous discharge of industrial and domestic sewage. The results underscore the necessity of predicting the groundwater response to reclaimed water being discharged into the Yongding River. The study established a coupled groundwater flow and reactive solute transport model in the middle-upper part of the Yongding River Alluvium Fan, one of the drinking water supply sites in Beijing city. The model would be used for risk assessment when reclaimed water was recharged into Yongding River.
Select Ecological vulnerability analysis of Beidagang National Park, China Xue YU, Yue LI, Min XI, Fanlong KONG, Mingyue PANG, Zhengda YU Front. Earth Sci.. 2019, 13 (2): 385-397. https://doi.org/10.1007/s11707-018-0726-8 Abstract HTML PDF (1494KB) Ecological vulnerability analysis (EVA) is vital for ecological protection, restoration, and management of wetland-type national parks. In this study, we assessed the ecological vulnerability of Beidagang National Park based upon remote sensing (RS) and geographic information system (GIS) technologies. To quantify the ecological vulnerability, 10 indices were collected by the ‘exposure-sensitivity-adaptive capacity’ model and spatial principal component analysis (SPCA) was then applied to calculate the ecological vulnerability degree (EVD). Based on the numerical values, EVD of the study area was classified into five levels: moderate, light, medium, strong, and extreme. Results showed that the average EVD value was approximately 0.39, indicating overall good ecological vulnerability in Beidagang National Park. To be specific, 80.42% of the whole area was assigned to a moderate level of EVD with the highest being the tourism developed areas and the lowest being the reservoirs and offshore areas. Ecological vulnerability of the region was determined to be affected by the natural environment and anthropogenic disturbance jointly. The primary factors included tourism disturbance, traffic interference, exotic species invasion, land use/land cover, and soil salinization. We expected to provide some insights of the sustainable development of Beidagang National Park and would like to extend the results to other wetland-type national parks in the future.
Select Impact of seasonal water-level fluctuations on autumn vegetation in Poyang Lake wetland, China Xue DAI, Rongrong WAN, Guishan YANG, Xiaolong WANG, Ligang XU, Yanyan LI, Bing LI Front. Earth Sci.. 2019, 13 (2): 398-409. https://doi.org/10.1007/s11707-018-0731-y Abstract HTML PDF (3005KB) Water level fluctuations (WLF) are natural patterns that are necessary for the survival of various plants, and WLF guarantee both the productivity and the biodiversity of wetlands. However, the underlying mechanisms of how changes in vegetation are linked to seasonal WLF remain unclear. Using vegetation and hydrological data from 1989 to 2009, we identified the key seasonal fluctuations and their impacts on vegetation in the Poyang Lake wetland by utilizing a tree-based hierarchical model. According to our results: 1) WLF in summer had significant impacts on both sedges and reeds. The severe summer floods promoted the expansion of sedges, while they inhibited the expansion of reeds; 2) WLF in autumn also greatly impacted sedges, while reeds were severely affected in spring. Specifically, we found that low water levels in autumn led to the expansion of sedges, and low water levels in spring led to the expansion of reeds. The results were well corroborated through comparisons of the vegetation distribution patterns over the last two decades (i.e., the 1990s and 2000s), which may shed light on corresponding water resource and wetland management.
Select Terrain relief periods of loess landforms based on terrain profiles of the Loess Plateau in northern Shaanxi Province, China Jianjun CAO, Guoan TANG, Xuan FANG, Jilong LI, Yongjuan LIU, Yiting ZHANG, Ying ZHU, Fayuan LI Front. Earth Sci.. 2019, 13 (2): 410-421. https://doi.org/10.1007/s11707-018-0732-x Abstract HTML PDF (2640KB) The Loess Plateau is densely covered by numerous types of gullies which represent different soil erosion intensities. Therefore, research on topographic variation features of the loess gullies is of great significance to environmental protection and ecological management. Using a 5 m digital elevation model and data from a national geographic database, this paper studies different topographical areas of the Loess Plateau, including Shenmu, Suide, Yanchuan, Ganquan, Yijun, and Chunhua, to derive representative gully terrain profile data of the sampled areas. First, the profile data are standardized in MATLAB and then decomposed using the ensemble empirical mode decomposition method. Then, a significance test is performed on the results; the test confidence is 95% to 99%. The most reliable decomposition component is then used to calculate the relief period and size of the gullies. The results showed that relief periods of the Chunhua, Shenmu, Yijun, Yuanchuan, Ganquan, and Suide gullies are 1110.14 m, 1096.85 m, 1002.49 m, 523.48 m, 498.12 m, and 270.83 m, respectively. In terms of gully size, the loess landforms are sorted as loess fragmented tableland, aeolian and dune, loess tableland, loess ridge, loess hill and loess ridge, and loess hill, in descending order. Taken together, the gully terrain features of the sample areas and the results of the study are approximately consistent with the actual terrain profiles. Thus, we conclude that ensemble empirical mode decomposition is a reliable method for the study of the relief and topography of loess gullies.
Select The 2015/16 El Niño-related glacier changes in the tropical Andes Bijeesh Kozhikkodan VEETTIL, Jefferson Cardia SIMÕES Front. Earth Sci.. 2019, 13 (2): 422-429. https://doi.org/10.1007/s11707-018-0738-4 Abstract HTML PDF (673KB) Significant changes in the area and snowline altitude of two glacierized mountains – Nevado Champara (Cordillera Blanca, Peru) and Cerro Tilata (Cordillera Real, Bolivia) – in the tropical Andes, before and after the recent El Niño in 2015/16 period, have been analysed using Sentinel 2A and Landsat data. It is seen that the recent El Niño has been accompanied by higher fluctuation in glacier coverage on Nevado Champara and the loss of glacier coverage on Cerro Tilata was very high during the past 16 years. Rise in snowline altitude of selected glaciers was very high after the 2015/16 El Niño. Increase in the area covered by snow and ice during the La Niña periods were not enough to cover the ice loss occurred during the previous El Niño events and the strongest El Niño in 2015/16 was followed by a significant loss of ice-covered areas in the tropical Andes. Freshwater resources in this region will be affected in the near future if the current trends in glacier decline continue. Adaptation strategies needs to be implemented to reduce the impacts of the continuing loss of glacierized on regional communities in the tropical Andean region.
Select Vegetation and soil wind erosion dynamics of sandstorm control programs in the agro-pastoral transitional zone of northern China Zhitao WU, Mingyue WANG, Hong ZHANG, Ziqiang DU Front. Earth Sci.. 2019, 13 (2): 430-443. https://doi.org/10.1007/s11707-018-0715-y Abstract HTML PDF (4127KB) To combat soil erosion and desertification, large-scale sandstorm control programs have been put in place since 2000 in the agro-pastoral transitional zone of northern China. Vegetation dynamics as well as soil wind erosion control effects are very important for assessing the ecological success of sandstorm control programs in China. However, no comprehensive evaluation of vegetation dynamics and soil wind erosion control effects in this region has been achieved. In this study, we illustrate the vegetation and soil wind erosion dynamics of sandstorm control programs in the northern Shanxi Province using remote sensing data and soil wind erosion models. There was a significant increase in vegetation cover for 63.59% of the study area from 2001 to 2014 and a significant decrease for 2.00% of the study area. The normalized difference vegetation index (NDVI) showed that the largest increase occurred in autumn. Soil wind erosion mass decreased from 20.90 million tons in 2001 to 7.65 million tons in 2014. Compared with 2001, the soil wind erosion moduli were reduced by 43.05%, 36.16%, and 62.66% in 2005, 2010, and 2014, respectively. Spatially, soil wind erosion in most of the study area was alleviated between 2001 and 2014. The relationship between NDVI and soil wind erosion mass showed that the increased vegetation coverage reduced the soil wind erosion mass. In addition, wind was the main driving force behind the soil wind erosion dynamics. The results indicate that the vegetation coverage has increased and soil wind erosion mass has been reduced following the implementation of the sandstorm control programs. However, the ecological effects of the sandstorm control programs may vary over different periods. While the programs appear to be beneficial in the short term, there may be unintended consequences in the long term. Research on the sustainability of the ecological benefits of sandstorm control programs needs to be conducted in the future.
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# Limits of a simple damped system
Definition: Let $F_n(s) = \frac{1}{s^{n+1}(1+s)^n}$ be the Laplace transform of $f_n(t)$.
Required Result: To show $\lim_{n\rightarrow\infty}f_n(n+n/e) < o(n)$.
Ideas: Let $G_n(s)=\frac{1}{s^{n+1}}$. Using standard Laplace transform properties:
1. $g_n(t)= \frac{t^{n}}{n!}$. Using Stirling's approximation, $\lim_{n\rightarrow\infty}g(n/e) = 0$.
2. $h_n(t) = g_n(t-n)$. We have $H_n(s) = \frac{1}{s^{n+1}e^{sn}}$ and $\lim_{n\rightarrow\infty}h_n(n+n/e) = 0$.
Note that $F_n(s)$ and $H_n(s)$ 'look' similar. The $(1+s)$ term corresponds to damping whereas $e^s$ term corresponds to delay. Can we use this idea to prove the result?
Any ideas/references would be appreciated! | |
# Documentation
Lean.SubExpr
A position of a subexpression in an expression.
See docstring of SubExpr for more detail.
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The coordinate 3 = maxChildren - 1 is reserved to denote the type of the expression.
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The Pos representing the root subexpression.
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The coordinate deepest in the Pos.
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partial def Lean.SubExpr.Pos.foldl {α : Type} (f : αNatα) (a : α) (p : Lean.SubExpr.Pos) :
α
Fold over the position starting at the root and heading to the leaf
partial def Lean.SubExpr.Pos.foldr {α : Type} (f : Natαα) (p : Lean.SubExpr.Pos) (a : α) :
α
Fold over the position starting at the leaf and heading to the root
partial def Lean.SubExpr.Pos.foldlM {α : Type} [inst : ] {M : TypeType u_1} [inst : ] (f : αNatM α) (a : α) (p : Lean.SubExpr.Pos) :
M α
monad-fold over the position starting at the root and heading to the leaf
partial def Lean.SubExpr.Pos.foldrM {α : Type} {M : TypeType u_1} [inst : ] (f : NatαM α) (p : Lean.SubExpr.Pos) (a : α) :
M α
monad-fold over the position starting at the leaf and finishing at the root.
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Returns true if pred is true for each coordinate in p.
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Creates a subexpression Pos from an array of 'coordinates'. Each coordinate is a number {0,1,2} expressing which child subexpression should be explored. The first coordinate in the array corresponds to the root of the expression tree.
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Decodes a subexpression Pos as a sequence of coordinates cs : Array Nat. See Pos.fromArray for details. cs[0] is the coordinate for the root expression.
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structure Lean.SubExpr :
An expression and the position of a subexpression within this expression.
Subexpressions are encoded as the current subexpression e and a position p : Pos denoting e's position with respect to the root expression.
We use a simple encoding scheme for expression positions Pos: every Expr constructor has at most 3 direct expression children. Considering an expression's type to be one extra child as well, we can injectively map a path of childIdxs to a natural number by computing the value of the 4-ary representation 1 :: childIdxs, since n-ary representations without leading zeros are unique. Note that pos is initialized to 1 (case childIdxs == []).
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Returns true if the selected subexpression is the topmost one.
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@[inline]
abbrev Lean.SubExpr.PosMap (α : Type u) :
Map from subexpr positions to values.
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• One of the hypotheses.
hyp:
• A subexpression of the type of one of the hypotheses.
hypType:
• A subexpression of the value of one of the let-bound hypotheses.
hypValue:
• A subexpression of the goal type.
target:
A location within a goal.
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A location within a goal state. It identifies a specific goal together with a GoalLocation within it.
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def Lean.Expr.traverseAppWithPos {M : TypeType u_1} [inst : ] (visit : ) (p : Lean.SubExpr.Pos) (e : Lean.Expr) :
Same as Expr.traverseApp but also includes a SubExpr.Pos argument for tracking subexpression position.
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Computer Science
Material in Unit 6 is optional. You are not expected to know this for a test and there are no labs or projects using this material.
# Using pandas to plot climate change data
This guide shows how to use pandas to plot climate change data. If you are new to pandas, see introduction to pandas.
This guide is NOT designed to give you a thorough introduction to plotting with pandas. It uses some more advanced techniques. Rather, I just want to show you some real-world applications of pandas using a topic I am passionate about. That’s a great way to learn how to plot in Python — find something you’re interested in and then dive in.
This notebook is inspired by the R hockeystick library, which makes it really easy to plot some common climate change graphs.
You can view a Google Colab notebook of this guide page.
## BYU Forum from Dr. Katharine Hayhoe
To gain some context on the importance of climate change, watch this BYU forum from Dr. Katharine Hayhoe. Dr. Hayhoe is a renowned climate scientist and a Christian. She gave this talk in fall semester 2022 and received a standing ovation. Dr. Hayhoe explained how being a Christian means loving all of God’s creations. This led her to become a climate scientist and she believes addressing climate change is “a true expression of God’s love.”
## CO2 Emissions
We are first going to plot worldwide $CO_2$ emissions. This dataset includes years and the $CO_2$ emitted each year, for each country. We are going to look at just the worldwide data.
import pandas as pd
# CO2 data can be found at
# https://github.com/owid/co2-data
# An explanation of columns is at
# https://github.com/owid/co2-data/blob/master/owid-co2-codebook.csv
def plot_global_co2_per_year(csv_filename):
# convert the CSV file into a Pandas dataframe
# get the subset of the data where the country column is equal to 'World'
world = data[data["country"].eq("World")]
# create a new column where co2 is measured in gigatons instead of megatons
world["co2"].div(1000)
# create a line plot with x = year and y = GTco2
# set the figure size, title, xlabel, ylabel
plot = world.plot.line(
x="year",
y="co2",
figsize=(10, 4),
title="Atmospheric $CO_2$ Emissions, World ",
xlabel="year",
ylabel="Gt $CO_2$ per year",
)
# get a figure for the plot
fig = plot.get_figure()
# save the figure to a file
fig.savefig("global-co2-per-year.png")
if __name__ == "__main__":
plot_global_co2_per_year("owid-co2-data.csv")
This creates the following plot:
Some of the things you see in the code above include:
• data = pd.read_csv('https://raw.githubusercontent.com/owid/co2-data/master/owid-co2-data.csv') — We can load datasets over the Internet.
• world = data[data["country"].eq("World")] — This gets a subset of the data where the country column is equal to World. The dataset contains data for every country, but we just want to look at the worldwide data.
• world["co2"].div(1000) — This divides the entire co2 column by 1000, so we can convert megatons to gigatons. It modifies the column instead of returning a new dataframe.
• $CO_2$ — This uses LaTeX math notation, which is valid in CoLab notebooks and in strings when plotting.
## Global temperature anomalies
Next we are going to plot land-surface air and sea water temperature anomalies. This data shows how much the temperature of the earth (including both air and sea) has warmed relative to a mean of the data from 1951 to 1980.
We are going to plot both the raw data — the anomaly per year — and a curve that fits the data with a LOWESS smoothing, which is a form of regression analysis. This can show a smoothed trend over time.
import pandas as pd
# pip install statsmodels
from statsmodels.nonparametric.smoothers_lowess import lowess
# Global Land-Surface Air and Sea Water Temperature Anomalies
# (Land-Ocean Temperature Index, L-OTI)
# This has the global surface temperature relative to 1951-1980 mean
# https://data.giss.nasa.gov/gistemp/
# The specific file used here is
# https://data.giss.nasa.gov/gistemp/tabledata_v4/GLB.Ts+dSST.csv
# The text file contains some additional explanation
# https://data.giss.nasa.gov/gistemp/tabledata_v4/GLB.Ts+dSST.txt
def plot_global_loti_per_year(csv_filename):
# convert the CSV file into a Pandas dataframe
# we need to skip the first row!
# the column labeled J-D is the anomaly for that entire year
# the data description says to divide by 100 to convert to centigrade
# skip years where annual average not computed
data = data[data["J-D"] != "***"]
# create a new column where the anomaly is measured in degrees centigrade
# must convert column to a float first, since Pandas sees it as a string
# create a line plot with x = year and y = J-D-centigrade
# set the figure size, title, xlabel, ylabel
ax = data.plot.line(
x="Year",
figsize=(10, 4),
title="Global surface temperature relative to 1951 - 1980 mean ",
xlabel="year",
ylabel="temperature anomaly (°C)",
)
# get x and y values for LOWESS
x = data['Year'].values
# run LOWESS
y_hat = lowess(y, x, frac=1/5)
# add LOWESS to data frame
data['lowess'] = y_hat[:,1]
data.plot(ax=ax, x='Year', y='lowess')
# get a figure for the plot
fig = ax.get_figure()
# save the figure to a file
fig.savefig("global-loti-per-year.png")
if __name__ == "__main__":
plot_global_loti_per_year("GLB.TS+dSST.csv")
This creates the following plot:
Some of the things you see in the code above include:
• data = data[data["J-D"] != "***"] — We are using the J-D column because this has data averaged for the whole year, January through December. We are going to ignore any data that is incomplete, which is marked in this dataset with ***.
• data["J-D-centigrade"] = data["J-D"].astype(float).div(100) — This creates a new column where the anomaly is measured in degrees centigrade. We first use astype() to convert the column to a float, since Pandas sees it as a string. This is because values start with - for negative numbers.
• We want to compute the LOWESS smoothing using a numpy array. numpy is a library that provides fast scientific computation for Python.
• x = data['Year'] and y = data['J-D-centigrade'] — Gets separate columns for x and y values for the LOWESS smoothing.
• y_hat = lowess(y, x, frac=1/5) — Calculates the LOWESS smoothing, using the x and y values and a parameter frac equal to 1/5 — the lower the value, the greater the smoothing. The lowess function comes from a statsmodels library. It returns a two-dimensional array of (x, y) values.
• data['lowess'] = y_hat[:,1] — This adds a column with the y values of the LOWESS calculation to the dataframe. The notation y_hat[:,1] says to get all the rows, and the first column, from the y_hat array.
• data.plot(ax=ax, x='Year', y='lowess') — This adds an additional line plot onto our previous plot. The ax = ax sets the axis to the the same as the previous plot.
## Global Warming Stripes
We now create a plot of ‘warming stripes’, which were created by Professor Ed Hawkins, a climate scientist. You can create your own stripes at #ShowYourStripes.
import pandas as pd
# Be sure to run:
# pip install matplotlib
import matplotlib.pyplot as plt
from matplotlib.patches import Rectangle
from matplotlib.collections import PatchCollection
from matplotlib.colors import ListedColormap
# code from
# https://matplotlib.org/matplotblog/posts/warming-stripes/
# This data is published by the Met Office in the UK. It shows
# combined sea and land surface temperatures as +/- from the average for
#the years 1961 to 1990.
# An explanation of the columns is here:
# first and last years in the dataset
FIRST = 1850
LAST = 2021
# Reference period for the center of the color scale
FIRST_REFERENCE = 1971
LAST_REFERENCE = 2000
LIM = 0.7 # degrees
def global_warming_stripes(fwf_filename):
# read a fixed-width file (fwf) into a Pandas data frame
# a fixed-width file has file formatted into columns using spaces
# index_col -- which column to use for row labels
# usecols -- which columns to use from the file
# names -- how to name the columns
# header -- indicates there is no first row with column names
data = pd.read_fwf(fwf_filename, index_col=0, usecols=(0, 1),
# create a new table that has all the columns where the year is between
# FIRST and LAST, dropping any that don't have data
data = data.loc[FIRST:LAST, 'anomaly'].dropna()
# take the mean of the anomaly data for the years between FIRST_REFERENCE
# and LAST_REFERENCE
reference = data.loc[FIRST_REFERENCE:LAST_REFERENCE].mean()
# create a set of colors for the bars
# the colors in this colormap come from http://colorbrewer2.org
# the 8 more saturated colors from the 9 blues / 9 reds
cmap = ListedColormap([
'#08306b', '#08519c', '#2171b5', '#4292c6',
'#6baed6', '#9ecae1', '#c6dbef', '#deebf7',
'#fee0d2', '#fcbba1', '#fc9272', '#fb6a4a',
'#ef3b2c', '#cb181d', '#a50f15', '#67000d',
])
# create a figure , giving its size
fig = plt.figure(figsize=(10, 1))
# add axes -- the dimensions are [left, bottom, width, height]
ax = fig.add_axes([0, 0, 1, 1])
# turn off showing lines, ticks, labels, etc for the axes
ax.set_axis_off()
# create a collection of rectangles, one for each year
rectangles = []
for year in range(FIRST, LAST + 1):
# create a Rectangle with (x, y) starting point, width, height
rectangles.append(Rectangle((year, 0), 1, 1))
collection = PatchCollection(rectangles)
# set data, colormap and color limits for the collection
collection.set_array(data)
collection.set_cmap(cmap)
collection.set_clim(reference - LIM, reference + LIM)
# set the limits on the axes
ax.set_ylim(0, 1)
ax.set_xlim(FIRST, LAST + 1)
# save the figure
fig.savefig('global-warming-stripes.png')
if __name__ == '__main__':
This creates the following plot:
Some of the things you see in the code above include:
• import matplotlib.pyplot as plt — We import various items from the matplotlib library. This is a low level, and thus very powerful, library for plotting in Python.
• pd.read_fwf() — we use read_fwf to read the dataset because the file is NOT stored as a CSV, it is stored as a fixed-with file, meaning the columns are all a fixed width, using spaces.
• data = data.loc[FIRST:LAST, 'anomaly'].dropna() — This gets certain rows from the dataframe, dropping any that are missing data.
• reference = data.loc[FIRST_REFERENCE:LAST_REFERENCE].mean() — We are going to be plotting bars of various colors. So we use this to get the mean of the anomaly for the period 1971 — 2000. We will use this as the middle temperature and plot colder temperatures in shades of blue, with warmer temperatures in red.
• cmap = ListedColormap(...) — This creates a colormap, meaning a list of colors, using their hex values.
• fig = plt.figure(figsize=(10, 1)) — Creates a blank figure of a given size.
• ax = fig.add_axes([0, 0, 1, 1]) — This adds an axis object to the figure, which is a container for the X, Y axes, the plot, the legend, and so forth. A plot can contain multiple “axes”, meaning it can have multiple subfigures. See here for a good visualization.
• Rectangle((year, 0), 1, 1) — Each bar on the chart has a size, starting from the (year, 0) point in (x, y) space, and extending 1 unit high and 1 unit wide.
• PatchCollection(rectangles) — This is a collection of rectangles, so we can operate on them all at once.
• collection.set_array(data) — This maps each rectangle in the collection to one of the anomaly values from our dataset.
• collection.set_cmap(cmap) — This provides our colormap (a list of colors) to the collection of rectangles.
• collection.set_clim(reference - LIM, reference + LIM) — This maps each rectangle to a color. From above, each rectangle is associated with an anomaly value. This command indicates that low range for the anomaly (reference - LIM) is set to the first color in the map, and the high range for the anomaly (reference + LIM) is set to the last color in the map. The rest are colors in between.
• ax.add_collection(collection) — This adds all of the rectangles to the plot. | |
268 views
The questions are based on the following information.
What is the simple annual rate of increase in the number of Agricultural loans from $1970$ to $1983$?
1. $132$%
2. $81$%
3. $75$%
4. $1056$%
1
508 views
2
431 views | |
# Sum Of Roots
Level pending
[A Problem From London University]
Find the sum of the fifth powers of the roots of the equation
× | |
# American Institute of Mathematical Sciences
February 2016, 10(1): 1-25. doi: 10.3934/ipi.2016.10.1
## On the choice of the Tikhonov regularization parameter and the discretization level: A discrepancy-based strategy
1 Computational Science Center, University of Vienna, Oskar Morgenstern-Platz 1, 1090 Vienna, Austria 2 Institute of Mathematics, Statistics and Physics, Federal University of Rio Grande, Av. Italia km 8, 96201-900 Rio Grande, Brazil 3 Instituto Nacional de Matemática Pura e Aplicada, Rio do Janeiro, RJ 22460-320
Received October 2014 Revised September 2015 Published February 2016
We address the classical issue of appropriate choice of the regularization and discretization level for the Tikhonov regularization of an inverse problem with imperfectly measured data. We focus on the fact that the proper choice of the discretization level in the domain together with the regularization parameter is a key feature in adequate regularization. We propose a discrepancy-based choice for these quantities by applying a relaxed version of Morozov's discrepancy principle. Indeed, we prove the existence of the discretization level and the regularization parameter satisfying such discrepancy. We also prove associated regularizing properties concerning the Tikhonov minimizers. We conclude by presenting some numerical examples of interest.
Citation: Vinicius Albani, Adriano De Cezaro, Jorge P. Zubelli. On the choice of the Tikhonov regularization parameter and the discretization level: A discrepancy-based strategy. Inverse Problems & Imaging, 2016, 10 (1) : 1-25. doi: 10.3934/ipi.2016.10.1
##### References:
show all references
##### References:
[1] Vinicius Albani, Adriano De Cezaro. A connection between uniqueness of minimizers in Tikhonov-type regularization and Morozov-like discrepancy principles. Inverse Problems & Imaging, 2019, 13 (1) : 211-229. doi: 10.3934/ipi.2019012 [2] Stefan Kindermann, Andreas Neubauer. On the convergence of the quasioptimality criterion for (iterated) Tikhonov regularization. Inverse Problems & Imaging, 2008, 2 (2) : 291-299. doi: 10.3934/ipi.2008.2.291 [3] Stefan Kindermann, Antonio Leitão. Convergence rates for Kaczmarz-type regularization methods. Inverse Problems & Imaging, 2014, 8 (1) : 149-172. doi: 10.3934/ipi.2014.8.149 [4] Len Margolin, Catherine Plesko. Discrete regularization. Evolution Equations & Control Theory, 2019, 8 (1) : 117-137. doi: 10.3934/eect.2019007 [5] Guozhi Dong, Bert Jüttler, Otmar Scherzer, Thomas Takacs. Convergence of Tikhonov regularization for solving ill-posed operator equations with solutions defined on surfaces. Inverse Problems & Imaging, 2017, 11 (2) : 221-246. doi: 10.3934/ipi.2017011 [6] Markus Grasmair. Well-posedness and convergence rates for sparse regularization with sublinear $l^q$ penalty term. Inverse Problems & Imaging, 2009, 3 (3) : 383-387. doi: 10.3934/ipi.2009.3.383 [7] Jussi Korpela, Matti Lassas, Lauri Oksanen. Discrete regularization and convergence of the inverse problem for 1+1 dimensional wave equation. Inverse Problems & Imaging, 2019, 13 (3) : 575-596. doi: 10.3934/ipi.2019027 [8] Frank Pörner, Daniel Wachsmuth. Tikhonov regularization of optimal control problems governed by semi-linear partial differential equations. Mathematical Control & Related Fields, 2018, 8 (1) : 315-335. doi: 10.3934/mcrf.2018013 [9] Thorsten Hohage, Mihaela Pricop. Nonlinear Tikhonov regularization in Hilbert scales for inverse boundary value problems with random noise. Inverse Problems & Imaging, 2008, 2 (2) : 271-290. doi: 10.3934/ipi.2008.2.271 [10] Armin Lechleiter, Marcel Rennoch. Non-linear Tikhonov regularization in Banach spaces for inverse scattering from anisotropic penetrable media. Inverse Problems & Imaging, 2017, 11 (1) : 151-176. doi: 10.3934/ipi.2017008 [11] Daniela Calvetti, Erkki Somersalo. Microlocal sequential regularization in imaging. Inverse Problems & Imaging, 2007, 1 (1) : 1-11. doi: 10.3934/ipi.2007.1.1 [12] Tim Hoheisel, Christian Kanzow, Alexandra Schwartz. Improved convergence properties of the Lin-Fukushima-Regularization method for mathematical programs with complementarity constraints. Numerical Algebra, Control & Optimization, 2011, 1 (1) : 49-60. doi: 10.3934/naco.2011.1.49 [13] Ke Zhang, Maokun Li, Fan Yang, Shenheng Xu, Aria Abubakar. Electrical impedance tomography with multiplicative regularization. Inverse Problems & Imaging, 2019, 13 (6) : 1139-1159. doi: 10.3934/ipi.2019051 [14] James Broda, Alexander Grigo, Nikola P. Petrov. Convergence rates for semistochastic processes. Discrete & Continuous Dynamical Systems - B, 2019, 24 (1) : 109-125. doi: 10.3934/dcdsb.2019001 [15] Paul-Eric Chaudru De Raynal. Weak regularization by stochastic drift : Result and counter example. Discrete & Continuous Dynamical Systems - A, 2018, 38 (3) : 1269-1291. doi: 10.3934/dcds.2018052 [16] Nicolas Fournier. A new regularization possibility for the Boltzmann equation with soft potentials. Kinetic & Related Models, 2008, 1 (3) : 405-414. doi: 10.3934/krm.2008.1.405 [17] Gabriel Peyré, Sébastien Bougleux, Laurent Cohen. Non-local regularization of inverse problems. Inverse Problems & Imaging, 2011, 5 (2) : 511-530. doi: 10.3934/ipi.2011.5.511 [18] Xiangtuan Xiong, Jinmei Li, Jin Wen. Some novel linear regularization methods for a deblurring problem. Inverse Problems & Imaging, 2017, 11 (2) : 403-426. doi: 10.3934/ipi.2017019 [19] T. J. Sullivan, M. Koslowski, F. Theil, Michael Ortiz. Thermalization of rate-independent processes by entropic regularization. Discrete & Continuous Dynamical Systems - S, 2013, 6 (1) : 215-233. doi: 10.3934/dcdss.2013.6.215 [20] Stephen Schecter, Bradley J. Plohr, Dan Marchesin. Computation of Riemann solutions using the Dafermos regularization and continuation. Discrete & Continuous Dynamical Systems - A, 2004, 10 (4) : 965-986. doi: 10.3934/dcds.2004.10.965
2018 Impact Factor: 1.469 | |
# what is a clear liquid that attacks cellular respiration
I'm doing a biology assignment and need to find out a clear liquid that attacks cellular respiration. It needs to be toxic and either absorbed through the skin or inhaled.
• I'd try asking this on Biology.SE, not here. And nevertheless, you should give candidates that you've thought of already and discarded for one reason or the other. This isn't a "do my homework" site. – tschoppi Mar 10 '14 at 21:37
• As below, but potassium cyanide solution. Liquid HCN can explosively polymerize, doi:10.1021/je60007a031 – Uncle Al Mar 11 '14 at 19:21
• @tschoppi: Or a "plan my murder" site... – Aesin Mar 11 '14 at 20:46
If you do not want the more effective acetylcholinesterase inhibitors, metal carbonyls might be an option: Nickeltetracarbonyl, $\ce{Ni(CO)4}$, is a colourless liquid with a boiling point of 43 °C. | |
# Trend Base Analysis & Divergence
Education
ASX:FWD FLEETWOOD LIMITED
109 views
Today, I show this example for you to understand why bullish divergence is important for an entry order.
FWD .AX is selected because VectorVest indicator RT crossover occurred on 4 May which is the same day that Stochastic (8,3,3) crossover occurred
Value of RT changed from 0.97 (3 May) to 1.03 (4 May)
Value of RT yesterday was 1.53
If you hold this stock since the crossovers occurred on 4 May, you gain 30% return so far.
For beginners, you may also try the second entry on 29 July. Why? It is simply because of the bullish divergence . And also, the price actions struggled at the Fib Ret. 38.2% as well.
Your homework is to analyse JIN.AX, using the same strategies above and the markups on the graph. Show me why JIN.AX gained 10% in one day on 11 July. | |
# Why are powers of coprime ideals are coprime? [duplicate]
Possible Duplicate:
Comaximal ideals in a commutative ring
On this webpage http://www.imsc.res.in/~kapil/geometry/caag/finite.html it's a stated fact that if $I$ and $J$ are coprime ideals of a ring $A$, then $I^k$ and $J^\ell$ are also coprime for any integers $k,\ell$.
There's no proof given. Why is this fact true?
-
## marked as duplicate by Bill Dubuque, Willie WongMar 6 '12 at 8:20
Let $R$ be your ring. I am assuming my ring is commutative. Saying $I,J$ are coprime ideals is the same as saying that there is no prime $q \subset R$ such that $I + J \subset q$. Well suppose there exists a prime $p \subset R$ such that $I^l + J^k \subset p$. Then $I^l \subset p$ and $J^k \subset p$. But this means $I \subset p$ and $J \subset p$ (why?), so $I + J \subset p$, contradicting the fact that $I,J$ are coprime.
-
I hadn't thought of this. I like it! – Dylan Moreland Mar 6 '12 at 3:32
I sometimes get confused when raising binomial or multi-nomial expressions to powers, so I like this one. Thanks @DylanMoreland. – Rankeya Mar 6 '12 at 3:38
To fill in the gap you left me, if $a\in I$, then $a^l\in p$, but since $p$ is prime, necessarily $a\in p$. Same for $J$. I like this too! Thanks. – Cye Mar 6 '12 at 4:37
Yup. The more general statement is this: if $A$ is a ring and $\alpha_1,...,\alpha_n \subset A$ are ideals, $p \subset A$ a prime ideal and $\alpha_1...\alpha_n \subset p$, then $\exists i \in \{1,...,n\}$ such that $\alpha_i \in p$. This is a useful result to keep in mind. There is an analogous statement with products replaced by finite intersections. – Rankeya Mar 6 '12 at 4:46
You have $I + J = A$. Take each side to the $(k + \ell - 1)$-th power; then each term on the left is contained in either $I^k$ or $J^\ell$, and it follows that $I^k + J^\ell \supset A$. You could instead work with an expression $x + y = 1$, where $x \in I$ and $y \in J$, if you prefer.
-
If you are familar with ideal radicals then the proof is a one-liner:
$$\rm rad\ (I^m +\: \cdots\: + J^n) \ \supset\ I +\:\cdots\:+ J\: =\: 1\ \ \Rightarrow\ \ I^m +\:\cdots\: + J^n\: =\: 1$$
Alternatively, and much more generally, it may be viewed as an immediate consequence of the Freshman's Dream binomial theorem $\rm\ (A + B)^n = A^n + B^n\$ which is true more generally if $\rm\: A+B\:$ is invertible (e.g. $(1)$ or any principal ideal), e.g.
$\rm\qquad\qquad (A + B)^4 \ =\ A^4 + A^3 B + A^2 B^2 + AB^3 + B^4$
$\rm\qquad\qquad\phantom{(A + B)^4 }\ =\ A^2\ (A^2 + AB + B^2) + (A^2 + AB + B^2)\ B^2$
$\rm\qquad\qquad\phantom{(A + B)^4 }\ =\ (A^2 + B^2)\ \:(A + B)^2$
So $\rm\qquad\ \ \ {(A + B)^2 }\ =\ \ A^2 + B^2\$ if $\rm\ A+B\$ is cancellative, e.g. if $\rm\ A+B = 1$
The same proof works generally since, as above
$\rm\qquad\qquad (A + B)^{2n}\ =\ A^n\ (A^n + \:\cdots\: + B^n) + (A^n +\:\cdots\: + B^n)\ B^n$
$\rm\qquad\qquad\phantom{(A + B)^{2n}}\ =\ (A^n + B^n)\ (A + B)^n$
For more see this prior answer.
-
First prove that if $I$ is coprime to $J$ and $J'$ then $I$ is coprime to $JJ'$.
Then use induction to show $I$ and $J^\ell$ are coprime for any $\ell \ge 1$, and then use induction on $k$ in the same way.
Alternatively:
$$A=A^{k+\ell-1}=(I+J)^{k+\ell-1} \subseteq I^k+J^\ell$$
- | |
## dark plasma physics
According to plasma metaphysics, a significant proportion of dark matter is in the form of a plasma of super (high energy) particles. … aa/article/view/4550, Eugene Oks. Research in Astronomy and Astrophysics. Nobody knows exactly what dark matter is. In 2020, in a paper published in Research in Astronomy and Astrophysics, I showed that for hydrogen atoms, the singular solution of the Dirac equation outside the proton is legitimate not just for the ground state, but for all excited discrete and continuous states of zero angular momentum—that is, for the so-called S-states. The best definition of plasma might be: a system of particles with high level of average kinetic energy. It is often created by applying a voltage between two electrodes in a glass tube containing a low-pressure gas. We find that the observed excess can be explained if kinetic mixing is in the approximate range: $10^{-12} \lesssim \epsilon \lesssim 10^{-10}$. A potential of several hundred volts is applied between the two electrodes. These bodies include the "bioplasma" bodies and "astral" bodies found in the metaphysical literature. Plasma physics is a rich and diverse field of enquiry, with its own special twist. However, with today's instruments it can be declassified from dark matter. In that paper, I demonstrated that with the allowance for the finite size of the proton, the singular solution of the Dirac equation outside the proton becomes legitimate for states of zero angular momentum (in that paper I focused at the ground state). In that 2001 paper, I showed that the allowance for the second flavor of hydrogen atoms eliminates this huge discrepancy. Dark matter does not respond to the electromagnetic force, or does so only very weakly. Plasma is found almost everywhere on Earth and in space - only the invisible dark matter is more abundant. He founded/co … For presenting the key part of this theorem, I will review the following: In quantum mechanics, any physical quantity corresponds to an operator, which is a set of rules to transform the so-called wave function into another wave function. In general, most distros and desktop environments come with atrociously pale fonts, gray on gray. Research in the field of atomic, molecular, nuclear, particle and plasma physics from the Faculty of Science, University of Melbourne. The resulting charged ions an… Plasma is the fourth state of matter, consisting of electrically charge remnants of atoms in the form of electrons and ions. I did not change any physical laws. • The fluid govern equations of a positron beam plasma are reduced to the NLSE. Dark plasma has very low density but tends to form in clumps. The ‘antigravitational’ force due to its pressure gradient then represents dark energy, and its gravitational force due to the energy density represents dark matter. April 16, 2012. Plasma metaphysics is the application of plasma and dark matter physics to the study of our high energy subtle bodies and their corresponding environments. For the schedule of course offerings, please see the department website. Basic research is carried out and applied to all of these areas by CU scientists in the Physics Department and collaborators in Institutes such as LASP and departments such as APS. The universe has five times more dark matter than normal matter. Development is one of the oldest traits humanity possesses. So, as I write, an article has appeared in The Daily Galaxy dated 11 June, 2019 titled “The Magnetic Cosmos” “. He also developed a large number of advanced spectroscopic methods for diagnosing various laboratory and astrophysical plasmas – the methods that were then used and are used by many experimental groups around the world. Plasma can be artificially generated by heating a neutral gas or subjecting it to a strong electromagnetic field to the point where an ionized gaseous substance becomes increasingly electrically conductive. Dark matter is just as mysterious as the origin of the big bang. Detecting compact dark objects in the universe. While the plasma in our world is hot and short-lived, scientists speculate that since dark plasma has less density, it can remain in the plasma state even at room temperature for long periods of time. How to make glow in the dark slime. If X-ray galaxies were observed in the nineteenth century to affect nearby stars through gravity it would have been classified under matter that we would now classify as dark matter because it would not have emitted any electromagnetic waves that would have been detectable using the scientific instruments available in the nineteenth century. Bowman et al found that the amplitude of the absorption profile of the 21-cm line was more than a factor of two greater than the largest predictions. ... Fireworks in a Dark Universe. The publication first offers information on the introduction to plasma physics and basic properties of the equilibrium plasma. | |
# What name would you give a compound consisting of two bromine atoms and one calcium atom?
I would call $C a B {r}_{2}$ $\text{calcium bromide}$.
$\text{Calcium bromide}$ is formally an ionic compound that is composed of $C {a}^{2 +}$ and $2 \times B {r}^{-}$ ions. | |
# Chord Socket¶
This is an extension of the MeshSocket which syncronizes a common dict. It works by providing some extra handlers to store data. This exposes the entire dict API.
Note
This is a fairly inefficient architecture for read intensive applications. For large databases which are infrequently changed, this is ideal. For smaller networks where there is significant access required, you should use the sync socket.
## Basic Usage¶
There are three limitations compared to a normal dict.
1. Keys can only be bytes-like objects
2. Keys are automatically translated to bytes
3. Fetching values is significantly slower than for a dict
The only API differences between this and MeshSocket are for access to this dictionary. They are as follows.
### get() / __getitem__()¶
A value can be retrieved by using the get() method, or alternately with __getitem__(). These calls are about O(log(n)) hops, as they approximately halve their search area with each hop.
>>> foo = sock.get('test key', None) # Returns None if there is nothing at that key
>>> bar = sock[b'test key'] # Raises KeyError if there is nothing at that key
>>> assert bar == foo == sock[u'test key'] # Because of the translation mentioned below, these are the same key
It is important to note that keys are all translated to bytes before being used, so it is required that you use a bytes-like object. It is also safer to manually convert unicode keys to bytes, as there are sometimes inconsistencies betwen the Javascript and Python implementation. If you notice one of these, please file a bug report.
### set() / __setitem__()¶
A value can be stored by using the set() method, or alternately with __setitem__(). Like the above, these calls are about O(log(n)) hops, as they approximately halve their search area with each hop.
>>> sock.set('test key', 'value')
>>> sock[b'test key'] = b'value'
>>> sock[u'测试'] = 'test'
Like above, keys and values are all translated to bytes before being used, so it is required that you use a bytes-like object.
### __delitem__()¶
This deletes an association. Like the above, this call is about O(log(n)).
>>> del sock['test']
### update()¶
The update method is simply a wrapper which updates based on a fed dict. Essentially it runs the following:
>>> for key, value in update_dict.items():
... sock[key] = value
### keys() / values() / items()¶
These methods are analagous to the ones in Python’s dict. The main difference is that they emulate the Python 3 behavior. So if you call these from Python 2, they will still return an iterator, rather than a list.
In addition, you should always surround values() and items() in a try-catch for KeyError and socket.error. Because the data is almost always stored on other nodes, you cannot guaruntee that an item in keys() is retrievable.
### pop() / popitem()¶
These methods are also analagous to the ones in Python’s dict. The main difference is that, like the above, you should always surround these in a try-catch for KeyError and socket.error.
## Events¶
In addition to the above, and those of MeshSocket, the ChordSocket object has two Events.
First there’s Event 'add'(). This is called whenever an association is added. Because the value is not necessarily stored by you, it is not given as an argument to this event.
>>> @sock.on('add')
>>> def handle_new_key(conn, key):
... # conn is a reference to the socket
... print("A new key was added: {}".format(key))
...
This class has one other event: Event 'delete'(). This is called every time an association is removed.
>>> sock.on('delete')
>>> def handle_deleted_key(conn, key):
... # conn is a reference to the socket
... print("A key was deleted: {}".format(key))
... | |
# What's one thing you like about Socratic?
Apr 5, 2017
(I'm going to answer my own question:)
#### Explanation:
One thing I really love about Socratic is that it reminds me somewhat of the Avengers (from Marvel Studios). Everyone is answering a question and doing their part and at the end, it ends up helping a lot of people. :)
Apr 6, 2017
I like helping people by answering their questions! :)
Apr 6, 2017
I like Socratic very much it is Because it Feels Good:-
#### Explanation:
I love to clear the doubts of people from all around the World :)
Apr 6, 2017
It's free, the interface allows answers to be well written and helping others understand difficult concepts gives me a sense of satisfaction.
#### Explanation:
I love teaching students, non trad, adults, children and anyone wanting to learn. There is just some satisfaction I get when I teach someone a concept and that person goes ,"Ohh, now I understand."
Every teacher has his/her own teaching style. Because of this, students can either get it the first time or struggle with understanding it. By breaking down answers, it allows me to teach what I know in a way which makes sense to others. Most importantly it keeps me motivated to learn topics which I don't understand while still keeping the information I do know fresh in my mind.
It challenges me
#### Explanation:
There are many ways that I enjoy Socratic:
• It keeps my mind active
• I enjoy the sense of satisfaction of being able to touch so many lives and in a little way each time help to make someone else's life just a little bit better
• It helps to expand my knowledge and understanding. Every once in a while, there is a problem that gets posed that forces me to think and rethink about it - each time garnering a deeper understanding of maths and patterns. The one that is currently making my brain itch I've been working on for five days now and while I've made some progress, I've yet to definitively solve it. For anyone who wants to tackle it along with me, here's the question (make sure to read the supplemental information https://socratic.org/questions/you-have-n-pieces-of-pie-to-be-given-out-to-k-people-and-all-n-pieces-are-given-)
Apr 7, 2017
The thrill of being included and working with so many exceptionally educated and talented professors, teachers, students, and parents.
#### Explanation:
We will all be students for a lifetime.
Thank you to teachers who facilitate valuable learning.
Thank you to parents who will always help us and support us for as long as they can.
Apr 9, 2017
In eager to answer a question, I search about its solution. That increases my knowledge which I love.
Apr 11, 2017
• Socratic is ace app.
#### Explanation:
It induces interest in doing studies. It helps to tackle the problems.It increase our knowledge .
Apr 11, 2017
People can check my answers and correct them.
#### Explanation:
No offense. People are able to correct mistakes. "The err is human"...
May 18, 2017
Jun 18, 2017
Everything!!!
#### Explanation:
It is really interesting!!! It helps me to ask questions which I have doubt on. The members using socratic are really helpful. Their explanation is worth-reading. It increases the interest for learning.
And the best part of socratic is that it makes you more interested in learning than using social media. It is in such a format that appeals learners. It helps to explore our creativity and develop our perception about certain topics. It is making the world narrower and promoting cooperation among learners from all around the world. It a platform to test how much you are capable of competing with the world.
It is a one of the most fantastic learning material in this century where students are engaged in chatting in social media instead of learning something new.
It increases the willingness of a learner to help others with their problems and get helped.
I am really grateful to the makers of SOCRATIC and hope that SOCRATIC will develop more and add other subjects to it to make it more interesting!! ^__^ :)
Jun 20, 2017
A combination of things:
1. It's free!
3. The span of answer difficulties accommodates for a large audience.
1)
Unlike websites like Chegg, Quora, etc. that have you subscribe for a (sometimes paid) service to get answers, Socratic just puts the answers up for free, with no account required (just encouraged).
Furthermore, the answers are under a creative commons license, so you can use the material in your work as long as you cite us (including link back to http://socratic.org/), make clear what you modified, and you don't sell it for money!
2)
Most of us are doing this because we want to, not because we have to.
We enjoy helping other people (for the most part), so if you have any pressing questions, bring them our way, and we'll give it our best shot!
3)
Part of our answering strategy is to reach as large of an audience as possible in terms of intelligence.
StackExchange is an example of a site that sometimes gets too technical. We wouldn't like it if someone reads something that appears too technical on our site and decides to go somewhere that explains it in an easier, and possibly less effective way.
We would rather make sense to as many grade levels as possible.
4)
We encourage explaining something to the best of your ability.
That means not just simple "here's the answer", but "here's the answer, and why it's (probably) right".
This is an issue in other Q&A websites that aren't fortunate enough to have, say, math formatting, which makes their math answers hard to read, or diligent enough people to share their thought process.
Jun 20, 2017
I don't think there's just one thing I like:)
#### Explanation:
Ever since I found Socratic, I've developed a strong attachment to it. The community of teachers and students is unlike any you'll find in school, and everyone here is so eager to help and learn!
Another important detail is that it's free! No other web site on the planet provides such detailed explanations to as many answers as Socratic, and to have it not require monthly payments, or any payments, sends it past the competition!
As my bio states, knowledge is a blessing from God and a gift that is worth sharing:)
I've never met such a community of people that are so helpful and encouraging toward others, and my only wish is that it grows even larger!
Jun 20, 2017
I like Socratic for more reasons than one!
For one thing, it gives me ample chances to practice my math skills prior to exams. I know firsthand as a contributor here that we all go over each other's answers, so it's nice to know what you're doing is correct.
Secondly, I love to help in any way possible, and this is such a friendly and accommodating community to help out in.
Socratic's math formatting system sure is a pleasure to work with. It's so easy in comparison to certain other formatting programs that I've used elsewhere.
I also like the thank-you note system, because it really gives me a sense that I'm making impact on the ground-level.
To conclude, Socratic is a wonderful website and I hope to keep contributing for a long time still to come:)
Jun 20, 2017
Hmmm...
#### Explanation:
I've been on Socratic for less than half a year but already I feel that it is the best site for seeking/giving help with homework. I remember discovering Socratic by accident because I was looking for an answer to a question. At first, I saw Socratic as a resource for my own benefit but then I came across a question and I felt the will to help answer it. Long story short, I soon found myself helping others with their homework.
• With Socratic, I very much enjoy the fact that I can help people around the world because it makes me happy that I can contribute my knowledge and make a small difference in the world by helping someone learn.
• It makes me really think about the content. When I attempt explain something, I try to be as clear and concise as possible. I have to really think about the content in a way that I otherwise wouldn't. I think, in terms of how this benefits me, is that writing answers helps be become more analytical and detail-orientated. Also, I understand the content more once I answered the question precisely because I thought deep about the content.
• As some have mentioned, Socratic is FREE!!! It's great that Socratic is free because it means everyone can contribute and learn without any restrictions. I firmly believe that knowledge should be free and not limited to those who can afford it.
• Lastly, I admire the community. There are many people, such as myself, who are eager to learn and teach. Socratic is filled with a wide range of contributors from experienced experts to active owls (as I like to call them) to occasional contributors. Each have their own ways to teach which allows for multiple perspectives and approaches towards given problems. They seem to really enjoy what they do and I think that's what makes Socratic the best thing out there today.
Jun 22, 2017
It makes me happy to see lots of different countries pay attention to answers and contribute questions.
#### Explanation:
Norway, for instance, is such a great country that whenever I put an answer, I check it how many times my answers were visited in Norway. I have almost 690 answers and my answers have been visited in Norway (one check per answer) frequently.
In my country, however, my answers have been visited only ten-twenty maybe thirty times (total).
When I provide an answer, someone can warn me if I made a mistake. I try to fix my mistakes and learn something.
That is great.
Jul 17, 2017
I like the way that it actually allows you to give a detailed answer!
#### Explanation:
There is a current fashion for supposing that everything is extremely simple, and can be boiled down to simple one-sentence answers.
This completely removes the ability to build in context, and also to include information that allows people to know the background, as well as ways to avoid being misled or confused by information which, if we were only working off single sentence answers, would be a major risk. This was a big disadvantage with Yahoo Answers.
Another MAJOR issue with Yahoo Answers is that there is (was) a growing tendency there for people to "thumbs down" answers that were scientifically correct. This led to people reading misleading information, which is no help to them at all in their learning. That is the main reason that I deleted my Yahoo account.
Socratic appears to be a good platform for sharing knowledge, and it is enjoyable to use.
Jul 25, 2017
Not one there are many things, which make me like Socratic.
#### Explanation:
1. I loved teaching even when I was a student. And though I had to chose a profession, which could pay me nicely, Socratic helps me return to my interests after my retirement.
2. In India educatiion is highly commercialised. The focus is more on qualifying or preparing for an examination or qualification but not on grasping concepts and their application. This is something I do not like, but in Socratic I can go my way.
3. Socratic also keeps my mind active in variety of ways. In fact many of the concepts which I had not used for last many decades have become sharpened and answers posted by other specialists and those already in the field of education have helped me in enancing my knowledge further.
4. Even after answering a question, queries by some as well answers by others on the same question helps me in looking at answers afresh from a different angle / perspective.
5. Last but not the least, it helps me reach out to my audience across te world irrespective of language barriers. (I have tried to answer a few questions asked in different language by using Google Translate and answering it in English.) I enjoy helping peopla across the world and I love my popularity even in non-English speaking places like Heifi and Bucharest.
Jul 26, 2017
Hmmm... there's a lot of things that I like about Socratic!
#### Explanation:
The 7 things that I LOVE about Socratic!
1. Whenever I answer a question, it helps me remember the things that I learned already and it refreshes my knowledge!
2. It's totally free! Isn't that awesome!
3. Has this magic which makes me "addicted" to it!
4. Everyone is very friendly... :D
5. The achievement badges and stuff motivate people a lot to answer questions and stuff. I like the idea of featured answers!
6. The atmosphere. It is very "learny" but I also feel that Socratic is fun in a way!
7. The questions are not like essays or something. Even if your answer has a few errors, people almost always fix it. :D
SOCRATIC IS A-W-E-S-O-M-E, AWESOME!!!
Mar 21, 2018
I like helping others and receiving assistance in return.
#### Explanation:
I like the feeling of being productive and helpful at the same time in a kind, accepting community. It feels nice to be able to do something to help others (kind of like online volunteering). | |
# Transformations of conics?
• May 15th 2011, 08:55 PM
homeylova223
Transformations of conics?
Write an equation of the translated or rotated graph in general form.
y=3x^2-2x+5 T(2,-3)
I know the graph is a parabola. But I am not sure where I would "plug in" h and k into this equation. To get the translated equation.
• May 16th 2011, 01:47 AM
earboth
Quote:
Originally Posted by homeylova223
Write an equation of the translated or rotated graph in general form.
y=3x^2-2x+5 T(2,-3)
I know the graph is a parabola. But I am not sure where I would "plug in" h and k into this equation. To get the translated equation.
1. You know
$\left|\begin{array}{l}\bar{x} = x+2 \\ \bar{y} =y-3\end{array}\right. ~\implies~ \left|\begin{array}{l}\bar{x}-2 = x \\ \bar{y}+3 =y\end{array}\right.$
2. Plug in the terms for x and y and change the variable:
$y+3 = 3(x-2)^2-2(x-2)+5$
Expand and collect like terms.
• May 16th 2011, 04:33 PM
homeylova223
I am having difficulty with one more question
y^2+8x=0 rotated on pi/6
I did this
x= cos(pi/6)x+sin(pi/6)y
square root 3 / 2 x+1/2y
For y I got -1/2X+ square root 3/2y
Then I plug it in
(-1/2x+ square root 3/2y)^2+8( square root 3 /2 x +1/2y)
But I am having trouble figuring out the equation.
• May 16th 2011, 11:13 PM
earboth
Quote:
Originally Posted by homeylova223
I am having difficulty with one more question
y^2+8x=0 rotated on pi/6
I did this
x= cos(pi/6)x+sin(pi/6)y
square root 3 / 2 x+1/2y
For y I got -1/2X+ square root 3/2y
Then I plug it in
(-1/2x+ square root 3/2y)^2+8( square root 3 /2 x +1/2y)
But I am having trouble figuring out the equation.
1. Do yourself (and us) a favour and start a new thread if you have a new question.
2. I assume that the center of rotation is the origin(?). If so you know:
$\left|\begin{array}{l}\bar{x} = x \cdot \cos\left(\frac{\pi}6\right) - y \cdot \sin\left(\frac{\pi}6\right) \\ \bar{y} = x \cdot \sin\left(\frac{\pi}6\right) + y \cdot \cos\left(\frac{\pi}6\right) \end{array}\right.$
With $\cos\left(\frac{\pi}6\right)=\frac12 \sqrt{3}$ and $\sin\left(\frac{\pi}6\right) = \frac12$
you'll get:
$\left|\begin{array}{l}x = \frac12 ( \bar{x} \sqrt{3} + \bar{y}) \\ y= \frac12 (\bar{y} \sqrt{3} - \bar{x} )\end{array}\right.$
3. Plug in these terms into the original equation which yields:
$\frac14 (\bar{y} \sqrt{3} - \bar{x})^2+ 4( \bar{x} \sqrt{3} + \bar{y}) = 0$
4. Expand the brackets and collect like terms:
$x^2 - 2·√3·x·y + 16·√3·x + 3·y^2 + 16·y = 0$
• May 17th 2011, 11:29 AM
homeylova223
I have a question when you wrote the equation why is it square root of 3 instead of square root of 3 over 2? I mean it makes sense in the final answer.
Also would instead of having 1/2 in the equation would you not have 2 because if you plug it back into the equation with two it makes more sense to me
As I checked and found out the answer to this problem is
x^2-2square root 3 xy +3(y)^2+ 16square root 3 x+16y
And if you use 2 instead of 1/2 in the equation I would get that answer.
• May 17th 2011, 11:48 PM
earboth
Quote:
Originally Posted by earboth
...
4. Expand the brackets and collect like terms:
$x^2 - 2·√3·x·y + 16·√3·x + 3·y^2 + 16·y = 0$
Quote:
Originally Posted by homeylova223
...
As I checked and found out the answer to this problem is
x^2-2square root 3 xy +3(y)^2+ 16square root 3 x+16y
And if you use 2 instead of 1/2 in the equation I would get that answer.
You probably have noticed that I made a mistake typing the equation (I used √3 instead of \sqrt{3}) So Here is the correct equation:
$x^2 - 2\cdot \sqrt{3} \cdot x·y + 16\cdot \sqrt{3} \cdot x + 3·y^2 + 16·y = 0$
Sorry for the confusion. | |
# Method of Undetermined Coefficients help
1. Jul 21, 2011
Ok, please view the attached image for the Question, and for the given solution.
I need some help understanding the solution.
I can get the Complimentary equation with no problems, I understand how to do that.
However, some questions
1) Why do we first ignore the sin(3x) in our particular solution? so that we only have y'' + 9y = cos(2x) ?
I first attempted to look for a particular solution of the form
yparticular = Acos(2x) + Bsin(2x) + Ccos(3x) + Dsin(3x)
And I managed to solve A = 1/5, but sin(3x) was left over and all the terms making it up had been cancelled out. I assume this is what it means by "satisfies the homogenous equation" but I fail to see how the dude who wrote the answer knew that from the beginning.
But my main question really is why are we allowed to look for a particular solution that ignores the sin(3x) ?
2) My second question, is why do attempt to find a particular solution of the form Ax.sin(3x) + Bx.cos(3x) ?
Ie. Why do we suddenly include an 'x' in there? what was our thought process that led us to try this?
Thanks to anyone who helps explain this. Cheers!
#### Attached Files:
• ###### aaaaaa.png
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2. Jul 21, 2011
### Kreizhn
The differential equation is linear right? Let's break this into two differential equations. Consider
$$y'' + 9y = \cos(2x)$$
and let the solution to this be $y_{p,1}$. Similarly, let the solution to
$$y'' + 9y = 2\sin(3x)$$
be given by $y_{p,2}$. Now notice that
\begin{align*} (y_{p,1} + y_{p_2})'' + 93(y_{p,1} + y_{p_2}) &= [y_{p,1}'' + 9y_{p,1}] + [y_{p,2}'' + 9y_{p,2}] \\ &= \cos(2x) + 2\sin(3x) \end{align*}
Thus the particular solution to each part can be solved separately, then added to get the total particular solution. Your method would also work normally, but here we need to be careful.
Indeed, the homogeneous equation is when you "keep" all the parts that involve y, and set the rest to zero. Namely, the homogeneous equation for this is
$$y'' + 9y = 0$$
Now there are many ways to solve this, and indeed one solution is $\sin(3x)$. The problem now is that when we try to find a particular solution $y_p = A \sin(3x) + B\cos(3x)$ we don't get anything useful out since $y_p'' + 9y_p = 0$. Thus we need to look at something more complicated. In particular, what can we differentiate in order to get a sin(3x) term that isn't sin(3x) itself? Well, how about xsin(3x)?
Hopefully I made this clear from before, but the point is that you can't look for solutions just involving sin(3x) or cos(3x) because they satisfy y'' + 9y = 0, so if you sub them into the DE, they disappear. Thus we want to try terms that when differentiated, give a sin(3x) & cos(3x) term, but are not sin(3x) or cos(3x) themselves. So try xsin(3x). Well
$$\frac{d}{dx} x\sin(3x) = \sin(3x) + 3x\cos(3x)$$ | |
Write 35 percent as a fraction in the lowest term
35% expressed as a fraction is 35/100
To put it in lowest terms you need to find the largest common factor of 35 and 100
The only factors of 35 are 1, 35, 5 and 7
The highest factor of 35 that will divide evenly into 100 is 5
35/5
100/5
= 7/20
.
To solve this, we simply have to put the percentage over 100, since there are only 2 digits to the left of the decimal point to make this a whole number.
35% = $$\frac{35}{100}$$
We can see by looking at this that both the numerator and the denominator are divisible by 5, so we can reduce.
$$\frac{35 / 5}{100 / 5} = \frac{7}{20}$$
35% as a fraction in the lowest term would be $$\frac{7}{20}$$. | |
# Debian on Android and my quest for a full-fledged terminal Python IDE¶
So when I recently discovered that my Asus Transformer Prime has Python on it, but that it only has version 2.6.2 and is missing some needed libraries, I could tell that setting up a sweet Python programming environment was going to be possible, but a little longer road than I thought it was going to be. Here’s how I successfully overcame multiple obstacles to get things working.
A few things indicated to me that what I wanted to accomplish would still be possible:
• I found this blog post that talks about all sorts of add-ons to Vim that make it a really powerful IDE in a terminal environment. I like the idea of doing things this way so I’m dependent on what is available in the Android app store or what I could figure out how to create myself. Plus I would never have time to write my own Android app that filled that purpose to my satisfaction.
The first major hurdle with this idea is that Vim has to be compiled with Python support. I knew (and later verified) that the Vim app wasn’t going to provide that, and I also knew that trying to cross-compile Vim on another machine and figure out how to get it on my tablet and working as expected wasn’t going to be fun at all.
The second major hurdle with this was figuring out how to get the add-ons installed such that Vim knew about them. With the differences in structure of the Android OS, I simply did not want to try to figure out what was going to work.
• The answer to many of the above listed problems seemed to be in an app I found called Linux Installer. The idea is that with root access on an Android device, you can create a loop device where you can install a full-fledged distribution of Linux , then mount the loop device and then utilize chroot to use that file system when working in something like Terminal Emulator. With this, I can install specific packages from a mainstream distribution that already satisfy the requirements of my project.
The first obstacle I faced was rooting my tablet. As of this writing, there isn’t a direct method of rooting the current version of the Prime’s firmware, 9.4.2.28, or the previous version, 9.4.2.21. Searching around on XDA Developers turned up some instructions on downgrading the firmware of the Prime to a “root-vulnerable” version. Then I rooted it, installed Root Keeper (which protects your root privileges between system upgrades), and did a system update to get the latest firmware again.
At this point I was ready to install Debian using Linux Installer. I had read some of the reviews of Linux Installer so I would know a little better what to expect when running it, and some people reported having problems with the program stalling and not successfully completing the installation. Looking through the sparse documentation, however, shows the following instruction just before the “Quick Tutorial” section:
If some step fails, retry it twice. If a step fails three consecutive times, quit the application by pressing the back button several times, and restart it.
While this isn’t the most eloquent of instructions, I took it to heart and got started. Sure enough, the installer had problems getting through various steps of the process, stalling for literally days but still having the appearance of making progress. I could never tell how long each step should take under ideal conditions or if allowing it to run and run was actually doing any good.
I found the solution to be some mixture of the following steps, all the while taking measures to keep the tablet powered and connected to the Internet:
• Let the installer go for a while and give it some time to try and resolve the step it’s working on.
• After a while, stop the application from either the menu, “pressing the back button several times,” or pressing the Home button and terminating the application using some other method.
• Restart the application. If it is unable to start running normally again, try exiting the program again and starting it up. If this still doesn’t do the trick, reboot the tablet. There were several times that I had to do this more than once to get Linux Installer working normally again.
• Once it’s running normally, remount the loop device and start the installation over. Chances are about 50% that it will get further than the previous installation attempt.
To give you an idea of where you are in the overall installation process, you can take a look at the console log. In it at the top you should see several lines in blue text. These are the messages that appear on the main window of the application in the top left corner that indicate what the installer is currently working on. One thing to take note of is that they are logged in reverse chronological order, which is contrary to typical log files. The installer will go through the following steps (assuming you did the default Debian installation like I did) in this order meaning the log will show the following in reverse order:
1. Retrieving Release
2. Validating Packages
3. Resolving dependencies of required packages…
4. Resolving dependencies of base packages…
5. Found additional required dependencies: insserv libbz2-1.0 libdb4.8 libslang2
I still haven’t undertaken including all the add-ons to Vim so that it acts as the IDE I’m hoping to have for Python development, nor have I figured out how to access /sdcard from Debian. I’m sure I’ll be posting something in the near future with any details of what I had to do to get things working. Until then, I hope the above helps someone to not get discouraged with Linux Installer, because it really is cool once you have everything working. Happy hacking! | |
# 62 Non-right Triangles: Law of Sines
### Learning Objectives
In this section, you will:
• Use the Law of Sines to solve oblique triangles.
• Find the area of an oblique triangle using the sine function.
• Solve applied problems using the Law of Sines.
Suppose two radar stations located 20 miles apart each detect an aircraft between them. The angle of elevation measured by the first station is 35 degrees, whereas the angle of elevation measured by the second station is 15 degrees. How can we determine the altitude of the aircraft? We see in (Figure) that the triangle formed by the aircraft and the two stations is not a right triangle, so we cannot use what we know about right triangles. In this section, we will find out how to solve problems involving non-right triangles.
### Using the Law of Sines to Solve Oblique Triangles
In any triangle, we can draw an altitude, a perpendicular line from one vertex to the opposite side, forming two right triangles. It would be preferable, however, to have methods that we can apply directly to non-right triangles without first having to create right triangles.
Any triangle that is not a right triangle is an oblique triangle. Solving an oblique triangle means finding the measurements of all three angles and all three sides. To do so, we need to start with at least three of these values, including at least one of the sides. We will investigate three possible oblique triangle problem situations:
1. ASA (angle-side-angle) We know the measurements of two angles and the included side. See (Figure).
2. AAS (angle-angle-side) We know the measurements of two angles and a side that is not between the known angles. See (Figure).
3. SSA (side-side-angle) We know the measurements of two sides and an angle that is not between the known sides. See (Figure).
Knowing how to approach each of these situations enables us to solve oblique triangles without having to drop a perpendicular to form two right triangles. Instead, we can use the fact that the ratio of the measurement of one of the angles to the length of its opposite side will be equal to the other two ratios of angle measure to opposite side. Let’s see how this statement is derived by considering the triangle shown in (Figure).
Using the right triangle relationships, we know that$\,\mathrm{sin}\,\alpha =\frac{h}{b}\,$and$\,\mathrm{sin}\,\beta =\frac{h}{a}.\,\,$Solving both equations for$\,h\,$gives two different expressions for$\,h.$
$h=b\mathrm{sin}\,\alpha \text{ and }h=a\mathrm{sin}\,\beta$
We then set the expressions equal to each other.
$\begin{array}{ll}\text{ }b\mathrm{sin}\,\alpha =a\mathrm{sin}\,\beta \hfill & \hfill \\ \text{ }\left(\frac{1}{ab}\right)\left(b\mathrm{sin}\,\alpha \right)=\left(a\mathrm{sin}\,\beta \right)\left(\frac{1}{ab}\right)\begin{array}{cccc}& & & \end{array}\hfill & \text{Multiply both sides by}\,\frac{1}{ab}. \hfill \\ \text{ }\,\frac{\mathrm{sin}\,\alpha }{a}=\frac{\mathrm{sin}\,\beta }{b}\hfill & \hfill \end{array}$
Similarly, we can compare the other ratios.
$\frac{\mathrm{sin}\,\alpha }{a}=\frac{\mathrm{sin}\,\gamma }{c}\text{ and }\frac{\mathrm{sin}\,\beta }{b}=\frac{\mathrm{sin}\,\gamma }{c}$
Collectively, these relationships are called the Law of Sines.
$\frac{\mathrm{sin}\,\alpha }{a}=\frac{\mathrm{sin}\,\beta }{b}=\frac{\mathrm{sin}\,\lambda }{c}$
Note the standard way of labeling triangles: angle$\,\alpha \,$(alpha) is opposite side$\,a;\,$angle$\,\beta \,$(beta) is opposite side$\,b;\,$and angle$\,\gamma \,$(gamma) is opposite side$\,c.\,$See (Figure).
While calculating angles and sides, be sure to carry the exact values through to the final answer. Generally, final answers are rounded to the nearest tenth, unless otherwise specified.
### Law of Sines
Given a triangle with angles and opposite sides labeled as in (Figure), the ratio of the measurement of an angle to the length of its opposite side will be equal to the other two ratios of angle measure to opposite side. All proportions will be equal. The Law of Sines is based on proportions and is presented symbolically two ways.
$\frac{\mathrm{sin}\,\alpha }{a}=\frac{\mathrm{sin}\,\beta }{b}=\frac{\mathrm{sin}\,\gamma }{c}$
$\frac{a}{\mathrm{sin}\,\alpha }=\frac{b}{\mathrm{sin}\,\beta }=\frac{c}{\mathrm{sin}\,\gamma }$
To solve an oblique triangle, use any pair of applicable ratios.
### Solving for Two Unknown Sides and Angle of an AAS Triangle
Solve the triangle shown in (Figure) to the nearest tenth.
The three angles must add up to 180 degrees. From this, we can determine that
$\begin{array}{l}\begin{array}{l}\hfill \\ \beta =180°-50°-30°\hfill \end{array}\hfill \\ \,\,\,\,=100°\hfill \end{array}$
To find an unknown side, we need to know the corresponding angle and a known ratio. We know that angle $\alpha =50°$and its corresponding side$a=10.\,$We can use the following proportion from the Law of Sines to find the length of$\,c.\,$
$\begin{array}{llllll}\,\,\frac{\mathrm{sin}\left(50°\right)}{10}=\frac{\mathrm{sin}\left(30°\right)}{c}\hfill & \hfill & \hfill & \hfill & \hfill & \hfill \\ c\frac{\mathrm{sin}\left(50°\right)}{10}=\mathrm{sin}\left(30°\right)\hfill & \hfill & \hfill & \hfill & \hfill & \text{Multiply both sides by }c.\hfill \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,c=\mathrm{sin}\left(30°\right)\frac{10}{\mathrm{sin}\left(50°\right)}\hfill & \hfill & \hfill & \hfill & \hfill & \text{Multiply by the reciprocal to isolate }c.\hfill \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,c\approx 6.5\hfill & \hfill & \hfill & \hfill & \hfill & \hfill \end{array}$
Similarly, to solve for$\,b,\,$we set up another proportion.
$\begin{array}{ll}\begin{array}{l}\hfill \\ \,\text{ }\frac{\mathrm{sin}\left(50°\right)}{10}=\frac{\mathrm{sin}\left(100°\right)}{b}\hfill \end{array}\hfill & \hfill \\ \text{ }b\mathrm{sin}\left(50°\right)=10\mathrm{sin}\left(100°\right)\hfill & \text{Multiply both sides by }b.\hfill \\ \text{ }b=\frac{10\mathrm{sin}\left(100°\right)}{\mathrm{sin}\left(50°\right)}\begin{array}{cccc}& & & \end{array}\hfill & \text{Multiply by the reciprocal to isolate }b.\hfill \\ \text{ }b\approx 12.9\hfill & \hfill \end{array}$
Therefore, the complete set of angles and sides is
$\begin{array}{l}\begin{array}{l}\hfill \\ \alpha =50°\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,a=10\hfill \end{array}\hfill \\ \beta =100°\,\,\,\,\,\,\,\,\,\,\,\,b\approx 12.9\hfill \\ \gamma =30°\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,c\approx 6.5\hfill \end{array}$[/hidden-answer]
### Try It
Solve the triangle shown in (Figure) to the nearest tenth.
$\begin{array}{l}\alpha ={98}^{\circ }\,\,\,\,\,\,\,\,\,\,\,\,a=34.6\\ \beta ={39}^{\circ }\,\,\,\,\,\,\,\,\,\,\,\,b=22\\ \gamma ={43}^{\circ }\,\,\,\,\,\,\,\,\,\,\,\,\,\,c=23.8\end{array}$
### Using The Law of Sines to Solve SSA Triangles
We can use the Law of Sines to solve any oblique triangle, but some solutions may not be straightforward. In some cases, more than one triangle may satisfy the given criteria, which we describe as an ambiguous case. Triangles classified as SSA, those in which we know the lengths of two sides and the measurement of the angle opposite one of the given sides, may result in one or two solutions, or even no solution.
### Possible Outcomes for SSA Triangles
Oblique triangles in the category SSA may have four different outcomes. (Figure) illustrates the solutions with the known sides$\,a\,$and$\,b\,$and known angle$\,\alpha .$
### Solving an Oblique SSA Triangle
Solve the triangle in (Figure) for the missing side and find the missing angle measures to the nearest tenth.
Use the Law of Sines to find angle$\,\beta \,$and angle$\,\gamma ,\,$and then side$\,c.\,$Solving for$\,\beta ,\,$we have the proportion
$\begin{array}{r}\hfill \frac{\mathrm{sin}\,\alpha }{a}=\frac{\mathrm{sin}\,\beta }{b}\\ \hfill \frac{\mathrm{sin}\left(35°\right)}{6}=\frac{\mathrm{sin}\,\beta }{8}\\ \hfill \frac{8\mathrm{sin}\left(35°\right)}{6}=\mathrm{sin}\,\beta \,\\ \hfill 0.7648\approx \mathrm{sin}\,\beta \,\\ \hfill {\mathrm{sin}}^{-1}\left(0.7648\right)\approx 49.9°\\ \hfill \beta \approx 49.9°\end{array}$
However, in the diagram, angle$\,\beta \,$appears to be an obtuse angle and may be greater than 90°. How did we get an acute angle, and how do we find the measurement of$\,\beta ?\,$Let’s investigate further. Dropping a perpendicular from$\,\gamma \,$and viewing the triangle from a right angle perspective, we have (Figure). It appears that there may be a second triangle that will fit the given criteria.
The angle supplementary to$\,\beta \,$is approximately equal to 49.9°, which means that$\,\beta =180°-49.9°=130.1°.\,$(Remember that the sine function is positive in both the first and second quadrants.) Solving for$\,\gamma ,$ we have
$\gamma =180°-35°-130.1°\approx 14.9°$
We can then use these measurements to solve the other triangle. Since$\,{\gamma }^{\prime }\,$is supplementary to the sum of$\,{\alpha }^{\prime }\,$and$\,{\beta }^{\prime },$ we have
${\gamma }^{\prime }=180°-35°-49.9°\approx 95.1°$
Now we need to find$\,c\,$and$\,{c}^{\prime }.$
We have
$\begin{array}{l}\frac{c}{\mathrm{sin}\left(14.9°\right)}=\frac{6}{\mathrm{sin}\left(35°\right)}\hfill \\ \text{ }c=\frac{6\mathrm{sin}\left(14.9°\right)}{\mathrm{sin}\left(35°\right)}\approx 2.7\hfill \end{array}$
Finally,
$\begin{array}{l}\frac{{c}^{\prime }}{\mathrm{sin}\left(95.1°\right)}=\frac{6}{\mathrm{sin}\left(35°\right)}\hfill \\ \text{ }{c}^{\prime }=\frac{6\mathrm{sin}\left(95.1°\right)}{\mathrm{sin}\left(35°\right)}\approx 10.4\hfill \end{array}$
To summarize, there are two triangles with an angle of 35°, an adjacent side of 8, and an opposite side of 6, as shown in (Figure).
However, we were looking for the values for the triangle with an obtuse angle$\,\beta .\,$We can see them in the first triangle (a) in (Figure).[/hidden-answer]
### Try It
Given$\,\alpha =80°,a=120,\,$and$\,b=121,\,$find the missing side and angles. If there is more than one possible solution, show both.
Solution 1
$\begin{array}{ll}\alpha =80°\hfill & a=120\hfill \\ \beta \approx 83.2°\hfill & b=121\hfill \\ \gamma \approx 16.8°\hfill & c\approx 35.2\hfill \end{array}$
Solution 2
$\begin{array}{l}{\alpha }^{\prime }=80°\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{a}^{\prime }=120\hfill \\ {\beta }^{\prime }\approx 96.8°\,\,\,\,\,\,\,\,\,\,\,\,\,{b}^{\prime }=121\hfill \\ {\gamma }^{\prime }\approx 3.2°\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{c}^{\prime }\approx 6.8\hfill \end{array}$[/hidden-answer]
### Solving for the Unknown Sides and Angles of a SSA Triangle
In the triangle shown in (Figure), solve for the unknown side and angles. Round your answers to the nearest tenth.
In choosing the pair of ratios from the Law of Sines to use, look at the information given. In this case, we know the angle$\,\gamma =85°,\,$and its corresponding side$\,c=12,\,$and we know side$\,b=9.\,$We will use this proportion to solve for$\,\beta .$
$\begin{array}{ll}\,\,\,\frac{\mathrm{sin}\left(85°\right)}{12}=\frac{\mathrm{sin}\,\beta }{9}\begin{array}{cccc}& & & \end{array}\hfill & \text{Isolate the unknown}.\hfill \\ \,\frac{9\mathrm{sin}\left(85°\right)}{12}=\mathrm{sin}\,\beta \hfill & \hfill \end{array}$
To find$\,\beta ,\,$apply the inverse sine function. The inverse sine will produce a single result, but keep in mind that there may be two values for$\,\beta .\,$It is important to verify the result, as there may be two viable solutions, only one solution (the usual case), or no solutions.
$\begin{array}{l}\beta ={\mathrm{sin}}^{-1}\left(\frac{9\mathrm{sin}\left(85°\right)}{12}\right)\hfill \\ \beta \approx {\mathrm{sin}}^{-1}\left(0.7471\right)\hfill \\ \beta \approx 48.3°\hfill \end{array}$
In this case, if we subtract$\,\beta \,$from 180°, we find that there may be a second possible solution. Thus,$\,\beta =180°-48.3°\approx 131.7°.\,$To check the solution, subtract both angles, 131.7° and 85°, from 180°. This gives
$\alpha =180°-85°-131.7°\approx -36.7°,$
which is impossible, and so$\,\beta \approx 48.3°.$
To find the remaining missing values, we calculate$\,\alpha =180°-85°-48.3°\approx 46.7°.\,$Now, only side$\,a\,$is needed. Use the Law of Sines to solve for$\,a\,$by one of the proportions.
$\begin{array}{l}\begin{array}{l}\hfill \\ \begin{array}{l}\hfill \\ \frac{\mathrm{sin}\left(85°\right)}{12}=\frac{\mathrm{sin}\left(46.7°\right)}{a}\hfill \end{array}\hfill \end{array}\hfill \\ \,a\frac{\mathrm{sin}\left(85°\right)}{12}=\mathrm{sin}\left(46.7°\right)\hfill \\ \text{ }\,\,\,\,\,\,a=\frac{12\mathrm{sin}\left(46.7°\right)}{\mathrm{sin}\left(85°\right)}\approx 8.8\hfill \end{array}$
The complete set of solutions for the given triangle is
$\begin{array}{l}\begin{array}{l}\hfill \\ \alpha \approx 46.7°\text{ }a\approx 8.8\hfill \end{array}\hfill \\ \beta \approx 48.3°\text{ }b=9\hfill \\ \gamma =85°\text{ }c=12\hfill \end{array}$[/hidden-answer]
### Try It
Given$\,\alpha =80°,a=100,\,\,b=10,\,$find the missing side and angles. If there is more than one possible solution, show both. Round your answers to the nearest tenth.
$\beta \approx 5.7°,\gamma \approx 94.3°,c\approx 101.3$
### Finding the Triangles That Meet the Given Criteria
Find all possible triangles if one side has length 4 opposite an angle of 50°, and a second side has length 10.
Using the given information, we can solve for the angle opposite the side of length 10. See (Figure).
$\begin{array}{l}\,\frac{\mathrm{sin}\,\alpha }{10}=\frac{\mathrm{sin}\left(50°\right)}{4}\hfill \\ \,\,\mathrm{sin}\,\alpha =\frac{10\mathrm{sin}\left(50°\right)}{4}\hfill \\ \,\,\mathrm{sin}\,\alpha \approx 1.915\hfill \end{array}$
We can stop here without finding the value of$\,\alpha .\,$Because the range of the sine function is$\,\left[-1,1\right],\,$it is impossible for the sine value to be 1.915. In fact, inputting$\,{\mathrm{sin}}^{-1}\left(1.915\right)\,$in a graphing calculator generates an ERROR DOMAIN. Therefore, no triangles can be drawn with the provided dimensions.[/hidden-answer]
### Try It
Determine the number of triangles possible given$\,a=31,\,\,b=26,\,\,\beta =48°.\,\,$
two
### Finding the Area of an Oblique Triangle Using the Sine Function
Now that we can solve a triangle for missing values, we can use some of those values and the sine function to find the area of an oblique triangle. Recall that the area formula for a triangle is given as$\,\text{Area}=\frac{1}{2}bh,\,$where$\,b\,$is base and$\,h\,$is height. For oblique triangles, we must find$\,h\,$before we can use the area formula. Observing the two triangles in (Figure), one acute and one obtuse, we can drop a perpendicular to represent the height and then apply the trigonometric property$\,\mathrm{sin}\,\alpha =\frac{\text{opposite}}{\text{hypotenuse}}\,$to write an equation for area in oblique triangles. In the acute triangle, we have$\,\mathrm{sin}\,\alpha =\frac{h}{c}\,$or$c\mathrm{sin}\,\alpha =h.\,$However, in the obtuse triangle, we drop the perpendicular outside the triangle and extend the base$\,b\,$to form a right triangle. The angle used in calculation is$\,{\alpha }^{\prime },\,$or$\,180-\alpha .$
Thus,
$\text{Area}=\frac{1}{2}\left(\text{base}\right)\left(\text{height}\right)=\frac{1}{2}b\left(c\mathrm{sin}\,\alpha \right)$
Similarly,
$\text{Area}=\frac{1}{2}a\left(b\mathrm{sin}\,\gamma \right)=\frac{1}{2}a\left(c\mathrm{sin}\,\beta \right)$
### Area of an Oblique Triangle
The formula for the area of an oblique triangle is given by
$\begin{array}{l}\text{Area}=\frac{1}{2}bc\mathrm{sin}\,\alpha \hfill \\ \,\,\,\,\,\,\,\,\,\,\,\,=\frac{1}{2}ac\mathrm{sin}\,\beta \hfill \\ \,\,\,\,\,\,\,\,\,\,\,\,=\frac{1}{2}ab\mathrm{sin}\,\gamma \hfill \end{array}$
This is equivalent to one-half of the product of two sides and the sine of their included angle.
### Finding the Area of an Oblique Triangle
Find the area of a triangle with sides$\,a=90,b=52,\,$and angle$\,\gamma =102°.\,$Round the area to the nearest integer.
Using the formula, we have
$\begin{array}{l}\text{Area}=\frac{1}{2}ab\mathrm{sin}\,\gamma \hfill \\ \text{Area}=\frac{1}{2}\left(90\right)\left(52\right)\mathrm{sin}\left(102°\right)\hfill \\ \text{Area}\approx 2289\,\,\text{square}\,\,\text{units}\hfill \end{array}$[/hidden-answer]
### Try It
Find the area of the triangle given$\,\beta =42°,\,\,a=7.2\,\text{ft},\,\,c=3.4\,\text{ft}.\,$Round the area to the nearest tenth.
about$\,8.2\,\,\text{square}\,\text{feet}$
### Solving Applied Problems Using the Law of Sines
The more we study trigonometric applications, the more we discover that the applications are countless. Some are flat, diagram-type situations, but many applications in calculus, engineering, and physics involve three dimensions and motion.
### Finding an Altitude
Find the altitude of the aircraft in the problem introduced at the beginning of this section, shown in (Figure). Round the altitude to the nearest tenth of a mile.
To find the elevation of the aircraft, we first find the distance from one station to the aircraft, such as the side$\,a,$ and then use right triangle relationships to find the height of the aircraft,$\,h.$
Because the angles in the triangle add up to 180 degrees, the unknown angle must be 180°−15°−35°=130°. This angle is opposite the side of length 20, allowing us to set up a Law of Sines relationship.
$\begin{array}{l}\begin{array}{l}\begin{array}{l}\hfill \\ \hfill \end{array}\hfill \\ \text{ }\frac{\mathrm{sin}\left(130°\right)}{20}=\frac{\mathrm{sin}\left(35°\right)}{a}\hfill \end{array}\hfill \\ a\mathrm{sin}\left(130°\right)=20\mathrm{sin}\left(35°\right)\hfill \\ \text{ }a=\frac{20\mathrm{sin}\left(35°\right)}{\mathrm{sin}\left(130°\right)}\hfill \\ \text{ }a\approx 14.98\hfill \end{array}$
The distance from one station to the aircraft is about 14.98 miles.
Now that we know$\,a,\,$we can use right triangle relationships to solve for$\,h.$
$\begin{array}{l}\mathrm{sin}\left(15°\right)=\frac{\text{opposite}}{\text{hypotenuse}}\hfill \\ \mathrm{sin}\left(15°\right)=\frac{h}{a}\hfill \\ \mathrm{sin}\left(15°\right)=\frac{h}{14.98}\hfill \\ \text{ }\,\text{ }h=14.98\mathrm{sin}\left(15°\right)\hfill \\ \text{ }\,h\approx 3.88\hfill \end{array}$
The aircraft is at an altitude of approximately 3.9 miles.
The diagram shown in (Figure) represents the height of a blimp flying over a football stadium. Find the height of the blimp if the angle of elevation at the southern end zone, point A, is 70°, the angle of elevation from the northern end zone, point$\,B,\,$is 62°, and the distance between the viewing points of the two end zones is 145 yards.
161.9 yd.
Access these online resources for additional instruction and practice with trigonometric applications.
### Key Equations
Law of Sines $\begin{array}{l}\frac{\mathrm{sin}\,\alpha }{a}=\frac{\mathrm{sin}\,\beta }{b}=\frac{\mathrm{sin}\,\gamma }{c}\,\hfill \\ \frac{a}{\mathrm{sin}\,\alpha }=\frac{b}{\mathrm{sin}\,\beta }=\frac{c}{\mathrm{sin}\,\gamma }\hfill \end{array}$ Area for oblique triangles $\begin{array}{r}\hfill \text{Area}=\frac{1}{2}bc\mathrm{sin}\,\alpha \\ \hfill \text{ }=\frac{1}{2}ac\mathrm{sin}\,\beta \\ \hfill \text{ }=\frac{1}{2}ab\mathrm{sin}\,\gamma \end{array}$
### Key Concepts
• The Law of Sines can be used to solve oblique triangles, which are non-right triangles.
• According to the Law of Sines, the ratio of the measurement of one of the angles to the length of its opposite side equals the other two ratios of angle measure to opposite side.
• There are three possible cases: ASA, AAS, SSA. Depending on the information given, we can choose the appropriate equation to find the requested solution. See (Figure).
• The ambiguous case arises when an oblique triangle can have different outcomes.
• There are three possible cases that arise from SSA arrangement—a single solution, two possible solutions, and no solution. See (Figure) and (Figure).
• The Law of Sines can be used to solve triangles with given criteria. See (Figure).
• The general area formula for triangles translates to oblique triangles by first finding the appropriate height value. See (Figure).
• There are many trigonometric applications. They can often be solved by first drawing a diagram of the given information and then using the appropriate equation. See (Figure).
### Section Exercises
#### Verbal
Describe the altitude of a triangle.
The altitude extends from any vertex to the opposite side or to the line containing the opposite side at a 90° angle.
Compare right triangles and oblique triangles.
When can you use the Law of Sines to find a missing angle?
When the known values are the side opposite the missing angle and another side and its opposite angle.
In the Law of Sines, what is the relationship between the angle in the numerator and the side in the denominator?
What type of triangle results in an ambiguous case?
A triangle with two given sides and a non-included angle.
#### Algebraic
For the following exercises, assume$\,\alpha \,$is opposite side$\,a,\beta \,$is opposite side$\,b,\,$and$\,\gamma \,$is opposite side$\,c.\,$Solve each triangle, if possible. Round each answer to the nearest tenth.
$\alpha =43°,\gamma =69°,a=20$
$\alpha =35°,\gamma =73°,c=20$
$\beta =72°,a\approx 12.0,b\approx 19.9$
$\alpha =60°,\,\,\beta =60°,\,\gamma =60°$
$a=4,\,\,\alpha =\,60°,\,\beta =100°$
$\gamma =20°,b\approx 4.5,c\approx 1.6$
$b=10,\,\beta =95°,\gamma =\,30°$
For the following exercises, use the Law of Sines to solve for the missing side for each oblique triangle. Round each answer to the nearest hundredth. Assume that angle$\,A\,$is opposite side$\,a,\,$angle$\,B\,$is opposite side$\,b,\,$and angle$\,C\,$is opposite side$\,c.$
Find side$\,b\,$when$\,A=37°,\,\,B=49°,\,c=5.$
$b\approx 3.78$
Find side$\,a$ when$\,A=132°,C=23°,b=10.$
Find side$\,c\,$when$\,B=37°,C=21°,\,b=23.$
$c\approx 13.70$
For the following exercises, assume$\,\alpha \,$is opposite side$\,a,\beta \,$is opposite side$\,b,\,$and$\,\gamma \,$is opposite side$\,c.\,$Determine whether there is no triangle, one triangle, or two triangles. Then solve each triangle, if possible. Round each answer to the nearest tenth.
$\alpha =119°,a=14,b=26$
$\gamma =113°,b=10,c=32$
one triangle,$\,\alpha \approx 50.3°,\beta \approx 16.7°,a\approx 26.7$
$b=3.5,\,\,c=5.3,\,\,\gamma =\,80°$
$a=12,\,\,c=17,\,\,\alpha =\,35°$
two triangles,$\,\gamma \approx 54.3°,\beta \approx 90.7°,b\approx 20.9$or${\gamma }^{\prime }\approx 125.7°,{\beta }^{\prime }\approx 19.3°,{b}^{\prime }\approx 6.9$
$a=20.5,\,\,b=35.0,\,\,\beta =25°$
$a=7,\,c=9,\,\,\alpha =\,43°$
two triangles,$\beta \approx 75.7°, \gamma \approx 61.3°,b\approx 9.9$or${\beta }^{\prime }\approx 18.3°,{\gamma }^{\prime }\approx 118.7°,{b}^{\prime }\approx 3.2$
$a=7,b=3,\beta =24°$
$b=13,c=5,\gamma =\,10°$
two triangles,$\,\alpha \approx 143.2°,\beta \approx 26.8°,a\approx 17.3\,$or$\,{\alpha }^{\prime }\approx 16.8°,{\beta }^{\prime }\approx 153.2°,{a}^{\prime }\approx 8.3$
$a=2.3,c=1.8,\gamma =28°$
$\beta =119°,b=8.2,a=11.3$
no triangle possible
For the following exercises, use the Law of Sines to solve, if possible, the missing side or angle for each triangle or triangles in the ambiguous case. Round each answer to the nearest tenth.
Find angle$A$when$\,a=24,b=5,B=22°.$
Find angle$A$when$\,a=13,b=6,B=20°.$
$A\approx 47.8°\,$or$\,{A}^{\prime }\approx 132.2°$
Find angle$\,B\,$when$\,A=12°,a=2,b=9.$
For the following exercises, find the area of the triangle with the given measurements. Round each answer to the nearest tenth.
$a=5,c=6,\beta =\,35°$
$8.6$
$b=11,c=8,\alpha =28°$
$a=32,b=24,\gamma =75°$
$370.9$
$a=7.2,b=4.5,\gamma =43°$
#### Graphical
For the following exercises, find the length of side$\,x.\,$Round to the nearest tenth.
$12.3$
$12.2$
$16.0$
For the following exercises, find the measure of angle$\,x,\,$if possible. Round to the nearest tenth.
$29.7°$
$x=76.9°\text{or }x=103.1°$
Notice that$\,x\,$is an obtuse angle.
$110.6°$
For the following exercises, find the area of each triangle. Round each answer to the nearest tenth.
$A\approx 39.4,\text{ }C\approx 47.6,\text{ }BC\approx 20.7$
$57.1$
$42.0$
$430.2$
#### Extensions
Find the radius of the circle in (Figure). Round to the nearest tenth.
Find the diameter of the circle in (Figure). Round to the nearest tenth.
$10.1$
Find$\,m\angle ADC\,$in (Figure). Round to the nearest tenth.
Find$\,AD\,$in (Figure). Round to the nearest tenth.
$AD\approx \text{ }13.8$
Solve both triangles in (Figure). Round each answer to the nearest tenth.
Find$\,AB\,$in the parallelogram shown in (Figure).
$AB\approx 2.8$
Solve the triangle in (Figure). (Hint: Draw a perpendicular from$\,H\,$to$\,JK).\,$Round each answer to the nearest tenth.
Solve the triangle in (Figure). (Hint: Draw a perpendicular from$\,N\,$to$\,LM).\,$Round each answer to the nearest tenth.
$L\approx 49.7,\text{ }N\approx 56.3,\text{ }LN\approx 5.8$
In (Figure),$\,ABCD\,$is not a parallelogram.$\,\angle m\,$is obtuse. Solve both triangles. Round each answer to the nearest tenth.
#### Real-World Applications
A pole leans away from the sun at an angle of$\,7°\,$to the vertical, as shown in (Figure). When the elevation of the sun is$\,55°,\,$the pole casts a shadow 42 feet long on the level ground. How long is the pole? Round the answer to the nearest tenth.
51.4 feet
To determine how far a boat is from shore, two radar stations 500 feet apart find the angles out to the boat, as shown in (Figure). Determine the distance of the boat from station$\,A\,$and the distance of the boat from shore. Round your answers to the nearest whole foot.
(Figure) shows a satellite orbiting Earth. The satellite passes directly over two tracking stations$\,A\,$and$\,B,\,$which are 69 miles apart. When the satellite is on one side of the two stations, the angles of elevation at$\,A\,$and$\,B\,$are measured to be$\,86.2°\,$and$\,83.9°,\,$respectively. How far is the satellite from station$\,A\,$and how high is the satellite above the ground? Round answers to the nearest whole mile.
The distance from the satellite to station$\,A\,$is approximately 1716 miles. The satellite is approximately 1706 miles above the ground.
A communications tower is located at the top of a steep hill, as shown in (Figure). The angle of inclination of the hill is$\,67°.\,$A guy wire is to be attached to the top of the tower and to the ground, 165 meters downhill from the base of the tower. The angle formed by the guy wire and the hill is$\,16°.\,$Find the length of the cable required for the guy wire to the nearest whole meter.
The roof of a house is at a$\,20°\,$angle. An 8-foot solar panel is to be mounted on the roof and should be angled$\,38°\,$relative to the horizontal for optimal results. (See (Figure)). How long does the vertical support holding up the back of the panel need to be? Round to the nearest tenth.
2.6 ft
Similar to an angle of elevation, an angle of depression is the acute angle formed by a horizontal line and an observer’s line of sight to an object below the horizontal. A pilot is flying over a straight highway. He determines the angles of depression to two mileposts, 6.6 km apart, to be$\,37°$and$\,44°,$as shown in (Figure). Find the distance of the plane from point$\,A\,$to the nearest tenth of a kilometer.
A pilot is flying over a straight highway. He determines the angles of depression to two mileposts, 4.3 km apart, to be 32° and 56°, as shown in (Figure). Find the distance of the plane from point$\,A\,$to the nearest tenth of a kilometer.
5.6 km
In order to estimate the height of a building, two students stand at a certain distance from the building at street level. From this point, they find the angle of elevation from the street to the top of the building to be 39°. They then move 300 feet closer to the building and find the angle of elevation to be 50°. Assuming that the street is level, estimate the height of the building to the nearest foot.
In order to estimate the height of a building, two students stand at a certain distance from the building at street level. From this point, they find the angle of elevation from the street to the top of the building to be 35°. They then move 250 feet closer to the building and find the angle of elevation to be 53°. Assuming that the street is level, estimate the height of the building to the nearest foot.
371 ft
Points$\,A\,$and$\,B\,$are on opposite sides of a lake. Point$\,C\,$is 97 meters from$\,A.\,$The measure of angle$\,BAC\,$is determined to be 101°, and the measure of angle$\,ACB\,$is determined to be 53°. What is the distance from$\,A\,$to$\,B,\,$rounded to the nearest whole meter?
A man and a woman standing$\,3\frac{1}{2}\,$miles apart spot a hot air balloon at the same time. If the angle of elevation from the man to the balloon is 27°, and the angle of elevation from the woman to the balloon is 41°, find the altitude of the balloon to the nearest foot.
5936 ft
Two search teams spot a stranded climber on a mountain. The first search team is 0.5 miles from the second search team, and both teams are at an altitude of 1 mile. The angle of elevation from the first search team to the stranded climber is 15°. The angle of elevation from the second search team to the climber is 22°. What is the altitude of the climber? Round to the nearest tenth of a mile.
A street light is mounted on a pole. A 6-foot-tall man is standing on the street a short distance from the pole, casting a shadow. The angle of elevation from the tip of the man’s shadow to the top of his head of 28°. A 6-foot-tall woman is standing on the same street on the opposite side of the pole from the man. The angle of elevation from the tip of her shadow to the top of her head is 28°. If the man and woman are 20 feet apart, how far is the street light from the tip of the shadow of each person? Round the distance to the nearest tenth of a foot.
24.1 ft
Three cities,$\,A,B,$and$\,C,$are located so that city$\,A\,$is due east of city$\,B.\,$If city$\,C\,$is located 35° west of north from city$\,B\,$and is 100 miles from city$\,A\,$and 70 miles from city$\,B,$how far is city$\,A\,$from city$\,B?\,$Round the distance to the nearest tenth of a mile.
Two streets meet at an 80° angle. At the corner, a park is being built in the shape of a triangle. Find the area of the park if, along one road, the park measures 180 feet, and along the other road, the park measures 215 feet.
19,056 ft2
Brian’s house is on a corner lot. Find the area of the front yard if the edges measure 40 and 56 feet, as shown in (Figure).
The Bermuda triangle is a region of the Atlantic Ocean that connects Bermuda, Florida, and Puerto Rico. Find the area of the Bermuda triangle if the distance from Florida to Bermuda is 1030 miles, the distance from Puerto Rico to Bermuda is 980 miles, and the angle created by the two distances is 62°.
445,624 square miles
A yield sign measures 30 inches on all three sides. What is the area of the sign?
Naomi bought a modern dining table whose top is in the shape of a triangle. Find the area of the table top if two of the sides measure 4 feet and 4.5 feet, and the smaller angles measure 32° and 42°, as shown in (Figure).
8.65 ft2
### Glossary
altitude
a perpendicular line from one vertex of a triangle to the opposite side, or in the case of an obtuse triangle, to the line containing the opposite side, forming two right triangles
ambiguous case
a scenario in which more than one triangle is a valid solution for a given oblique SSA triangle
Law of Sines
states that the ratio of the measurement of one angle of a triangle to the length of its opposite side is equal to the remaining two ratios of angle measure to opposite side; any pair of proportions may be used to solve for a missing angle or side
oblique triangle
any triangle that is not a right triangle | |
# Figure 4
\textit{} Number of standard deviations $n_{\sigma}$ of the modeled correlation function for a given set of scattering parameters (effective range $d_0$and scattering length $f_0$)with respect to the data, together with various model calculations~ and measurements~. The gray shaded area corresponds to the region where the Lednick\'{y} model predicts a negative correlation function for \pp collisions at $\sqrt{s} = 7\,$TeV. | |
# Derriving time-dilation formula for velocity
I took a stab at deriving the time dilation formula for velocity. This is entirely my own approach though I suspect it is nothing new/special in terms of how it is derived. Just wanted to have some fun and see if i could figure it out for myself. So I knew it was all based on the idea that the speed of light is a constant in any frame of reference. So I drew this diagram showing two frames of reference.
So what I did was I tried to calculate the amount of time it would take for light to travel between the two people in the spaceship (A and B). The distance between them is $D_x$. That's the distance light travels in the rest frame. Then I calculated the distance it would travel as an outside observer would see it (accounting for the velocity of the space ship). For that the distance was longer, but the speed of light the same. The ratio between the difference between the apparent time from the two observers would be the time dilation. Here is my math.
$$T_0 = \frac{D_x}{C}$$
$$D_y = V_y \cdot T_0$$
$$D_y = V_y \cdot \frac{D_x}{C}$$
$$D_1 = \sqrt{{D_{x}}^2 + {D_{y}}^2}$$
$$D_1 = \sqrt{{D_{x}}^2 + {(V_y \cdot \frac{D_x}{C})}^2}$$
$$T_1 = \frac{D_1}{C}$$
$$T_1 = \frac{\sqrt{{D_{x}}^2 + {(V_y \cdot \frac{D_x}{C})}^2}}{C}$$
$$T_1 = \frac{\sqrt{{D_{x}}^2 + {V_y}^2 \cdot \frac{{D_x}^2}{C^2} }}{C}$$
$$T_1 = \frac{\sqrt{{D_{x}}^2 + {V_y}^2 \cdot \frac{1}{C^2} \cdot {D_x}^2 }}{C}$$
$$T_1 = \frac{D_x \cdot \sqrt{1 + {V_y}^2 \cdot \frac{1}{C^2} }}{C}$$
$$T_1 = \frac{D_x \cdot \sqrt{1 + \frac{{V_y}^2}{C^2} }}{C}$$
$$\Delta T = \frac{T_0}{T_1}$$
$$\Delta T = \frac{\frac{D_x}{C}}{\frac{D_x \cdot \sqrt{1 + \frac{{V_y}^2}{C^2} }}{C}}$$
$$\Delta T = \frac{D_x}{C} \cdot \frac{C}{D_x \cdot \sqrt{1 + \frac{{V_y}^2}{C^2} }}$$
$$\Delta T = \frac{C D_x}{C D_x \cdot \sqrt{1 + \frac{{V_y}^2}{C^2} }}$$
$$\Delta T = \frac{1}{\sqrt{1 + \frac{{V_y}^2}{C^2} }}$$
So just some notes, subscript 0 here, such as $T_0$ refers to the time in the rest frame, and subscript 1 here, such as $T_1$ represents the frame of reference where the ship is moving.
As you can see the final equation I get is almost exactly right but i seem to have resulted in a plus sign where there should be a minus sign. After going through my work I can't find the source of my error. What did I do wrong?
This approach is actually well-thought out and it's pretty much on point. The problem, though, occurs very early.
There are two inertial frames that we're considering. I'll refer to one as observer frame and the other one as the ship frame.
$$T_{0} = \frac{D_{x}}{C}$$
So you define $T_{0}$ to be the time for light to go between the two people with regards to the ship frame.
$$D_{y} = V_{y} \cdot T_{0}$$
If I am not mistaken, I would guess that the reasoning for this was as follows: given the ship frame, in the time it took for light to go between the two people, they flew a distance $D_{y}$.
Unfortunately, $D_{y}$ is defined as the vertical distance with respect to the observer frame. This does not hold for the ship frame.
To explain this a little further, in relativity we let go of certain assumptions. One of the assumptions is that time is universal in all frames. The things is, though, that once you let go of that assumption, there are other consequences.
One of those consequences is that distance is not the same in all frames, just like time. More specifically, if a ship is moving in the $y$-axis, length parallel to the $y$-axis is not the same in all frames (although length parallel to the $x$-axis is okay).
Going back to the derivation, the approach is the right idea; you just have to avoid the mistake above. The distance $D_{x}$ applies to both frames, but $D_{y}$ applies to the observer frame only.
• That makes perfect sense. The distance the ship travels along Y is the time it takes the light to travel but not in the time in the ship frame but the time in the observer frame. So my assumption there is wrong. Though it does surprise me that despite this mistake the answer I got was still so close to the correct answer. I guess im left wondering how I can actually fix this "proof" to be correct now without rewriting the whole thing (if possible). Regardless this answer looks valid to me. May 11 '17 at 1:15
I think your mistake may be in the first few lines. You set your distance in the Y direction equal to your velocity in the Y direction times the time interval as experienced by the people in the rest frame; shouldn't it be equal to the Y velocity in the moving frame times the time experienced in the moving frame? I may be entirely off-base here though. I did my own similar derivation and ended up with the right answer (the time dilation factor is the same as yours except there's a negative sign instead of a positive sign). My phone's just charging up now so I'll post an image of the derivation shortly (my phone's the only thing I can take a picture with, and I did the derivation on a piece of graphing paper). I'm really sorry if my writing is terrible, or if it's not entirely clear. My writings always been abysmal, in high school I just did everything on the computer since nobody could read my writing. Note; as a previous comment stated, the distance in the X direction is the same in both reference frames since there the velocity is purely in the Y-direction in the second frame. That's an assumption when I replace D squared by C squared times Time in Frame 1 (T1) squared.
• Oh thanks, I'll look this over tomorrow, but it does indeed look very similar to what I was doing. May 11 '17 at 3:54
• It is pretty similar. I'll try to make one with neater writing tomorrow and post it as well. May 11 '17 at 3:56 | |
# Method used by FindFit
Using FindFit, Mathematica selects an optimal algorithm when Automatic is selected. I'd like to know the name of the method applied to the FindFit computation. Is there any option to explain that?
-
I don't think you can hijack FindFit[] to say exactly what method it uses. However, the docs say that by default it uses SVD for linear least squares, and Levenberg-Marquardt for nonlinear least squares problems. – J. M. is back. Apr 17 '13 at 2:50
I think we should explain the name of algorithm in academic papers. How can researchers clear this issue when they use FindFit? – S.Orii Apr 17 '13 at 3:50
You can use an explicit setting of Method (e.g. Method -> "LevenbergMarquardt") when invoking FindFit[]. – J. M. is back. Apr 17 '13 at 4:28
A merit to use FindFit is of automatic selection of algorithm when it executes under a condition like "a>0". We may find the algorithm selected by an explicit setting mentioned above. Is it the only way to find the selected algorithm? – S.Orii Apr 17 '13 at 7:30
Well, since FindFit[] is effectively a black box, there's not much hope in figuring out what's going on under the hood. If you're going to be writing about this in some paper, just say you used the Automatic setting, and mention the claim in the docs about the methods internally used if need be. – J. M. is back. Apr 17 '13 at 8:09
As noted by @Ｊ.Ｍ., the docs say that the default for FindFit[] is SVD for linear least squares and Levenberg-Marquardt for nonlinear least squares. If it is important for the end-user to know explicitly the method used, then the user should explicitly define the method. | |
grit - GBA Raster Image Transmogrifier
Grit and its GUI version Wingrit are my image converters for the GBA (and NDS I guess). They can do most of the simple things like reading an image (pretty much any type of bitmap thanks to FreeImage) and converting it to binary data of various bitdepths which can be directly put into VRAM, but also more complicated matters such as tiling and metatiling (for 1D object mapping for example), making a tilemap along with a reduced tileset (or using an external tileset), popular map layouts, and compression compatible with the GBA's BIOS routines. The capability for an NDS alphabit has been added recently as well. Output can be C/asm arrays, raw binary, GBFS, and a RIFF-based format called GRF.
Not good enough? Well, the source code is available too, so you're free to modify it. The code should be platform independent right now, or at least very nearly so; the catch being that you might have add or remove some type definitions and maybe create your own makefile for compilation.
Documentation
Current version: 0.8.6
Previous versions
27 thoughts on “grit - GBA Raster Image Transmogrifier”
1. To make Grit compile with GCC 4.4, I had to make the following changes:
srcgrit/grit_main.cpp, line 793, add a cast to first argument:
args.insert((strvec::iterator)&args[ii], files.begin(), files.end());
2. Because I'm getting this error when I try to compile on linux:
srcgrit/grit_main.cpp: 793: error: no matching function for call to 'std::vector<char*,std::allocator >::insert(char**, __gnu_cxx::__normal_iterator<char**, std::vector<char*, std::allocator > >, __gnu_cxx::__normal_iterator<char**, std::vector<char*, std::allocator > > >)
/usr/include/c++/4.3/bits/vector.tcc:94: note: candidates are: __gnu_cxx::__normal_iterator<typename std::_Vector_base::__normal_iterator<typename std::_Vector_base > std::vector::insert(__gnu_cxx::normal_iterator<typename std::_Vector_base::_Tp_alloc_type::pointer, std::vector >, const_Tp&) [with _Tp = char*, _Alloc = std::allocator]
C++ libs related?
3. I'm using gcc 4.3 but carl's fix mentioned above works.
Should've tried that, sorry!
4. I cannot compile grit on linux:
either it stops with
g++ -s -static -o grit build/grit_main.o build/cli.o build/fi.o -L. -lgrit -lcldib -lfreeimage
/usr/lib/gcc/x86_64-pc-linux-gnu/4.3.4/../../../../x86_64-pc-linux-gnu/bin/ld: cannot find -lfreeimage
, indicating that libfreeimage is only bundled as Windows-dll.
having installed freeimage-3.13.1 via http://bugs.gentoo.org/show_bug.cgi?id=307487, compilations stops with
g++ -s -static -o grit build/grit_main.o build/cli.o build/fi.o -L. -lgrit -lcldib -lfreeimage
./libgrit.a(grit_xp.o): In function grit_xp_h(GritRec*)':
grit_xp.cpp:(.text+0x1a6b): warning: the use of tmpnam'
is dangerous, better use mkstemp'
/usr/lib/gcc/x86_64-pc-linux-gnu/4.3.4/../../../../lib64/libfreeimage.a(PluginEXR.o): In function C_OStream::write(char const*, int)'
:
(.text+0x26a): undefined reference to Iex::throwErrnoExc(std::basic_string<char, std::char_traits<char>, std::allocator<char> > const&)'
and lots of other undefined references in libfreeimage.a (complete buildlog at
The makefile looks like you link statically on linux anyway, so how about bundling libfreeimage.a in a usable version as well ?
5. Unfortunately, I'm limited to Windows, so I can't really help much with linux. That's more wintermute's department. Unfortunately, mkstemp() isn't standard, which is why I'm using tmpnam. IIRC, the warning isn't that serious.
The slew of errors in FreeImage looks like an improper build of the library, particularly of some of the sub-libraries it uses (libpng and libjpeg, for example). But again I can't built on linux, so there's little I can do here. This is also the reason I can't bundle a linux-usable libfreeimage.a – well, that and I don't know if it'll work for the various linuxes out there. I imagine if there were a one-size-fits-all that it'd have been included in the FreeImage distribution.
You may have better luck on the devkitPro forum.
6. 'Cause I keep getting an error message saying it was configured incorrectly and should reinstall...
7. You did something kind of silly when you where naming your classes in cldib... In cldib_wu.c and cldib_quant.h you use the WuQuantizer class which is the same name as freeimage which is WuQuantizer which ended up as a what I like to call a naming collision.
The fix is simple just change the name of WuQuantizer to some like this CldibWuQuantizer. This should get rid of the compiler linker error and it will compile to linux just fine. Here a list of files that need to be edited.
cldib_wu.cpp
cldib_quant.h
cldi_conv.cpp line 344
8. Is there a way of using transparency info in the png, to set the transparency of the loaded texture?
9. Alas, no.
It's something I've been meaning to do, but this is not a minor operation, and time has been kind of lacking :\
10. To get this working in CentOS 6, had to install freeimage-dev from EPEL, and make the following chagnes to the makefile:
LIBDIRS := .
change to
LIBDIRS := . /usr/lib/
$(CXX)$(LDFLAGS) -o $@$(GRIT_OBJ) $(LIBPATHS) -lgrit -lcldib -lfreeimage change to$(CXX) $(LDFLAGS) -o$@ $(GRIT_OBJ)$(LIBPATHS) -lgrit -lcldib -shared -lfreeimage
-T
11. Note on compiling for OSX 10.6 and newer:
sudo port install freeimage
In Makefile change OSX section to:
ifneq (,$(findstring Darwin,$(UNAME)))
SDK := /Developer/SDKs/MacOSX10.6.sdk
OSXCFLAGS := -mmacosx-version-min=10.6 -isysroot $(SDK) -arch x86_64 OSXCXXFLAGS :=$(OSXCFLAGS)
CXXFLAGS += -fvisibility=hidden
LDFLAGS += -mmacosx-version-min=10.6 -Wl,-syslibroot,$(SDK) -arch x86_64 endif In Makefile, change LIBDIRS to: LIBDIRS := . /opt/local/lib Then make clean, make, and good to go! 12. Just compiled on OS X 10.9 with instructions similar to the above comment. The SDK directory is now: SDK := /Applications/Xcode.app/Contents/Developer/Platforms/MacOSX.platform/Developer/SDKs/MacOSX10.9.sdk and also I used Homebrew instead of MacPorts to install libfreeimage, which places it in /usr/local/lib instead of /opt/local/lib, so I set LIBDIRS accordingly. 13. I have a problem with make to Linux: g++ -E -MMD -MF build/grit_main.d -Icldib -Ilibgrit -Iextlib -DGRIT_VERSION=\"0.8.6\" -DGRIT_BUILD=\"20100317\" srcgrit/grit_main.cpp > /dev/null g++ -Icldib -Ilibgrit -Iextlib -DGRIT_VERSION=\"0.8.6\" -DGRIT_BUILD=\"20100317\" -o build/grit_main.o -c srcgrit/grit_main.cpp g++ -s -static -o grit build/grit_main.o build/cli.o build/fi.o -L. -L/usr/lib -lgrit -lcldib -shared -lfreeimage ./libgrit.a(grit_xp.o) : Dans la fonction « grit_xp_c(GritRec*) » : grit_xp.cpp:(.text+0x6f2): avertissement : the use of tmpnam' is dangerous, better use mkstemp' /usr/bin/ld: /usr/lib/gcc/x86_64-linux-gnu/7/crtbeginT.o: relocation R_X86_64_32 against symbole caché __TMC_END__' can not be used when making a shared object /usr/bin/ld : échec de l'édition de liens finale : Section non-représentable sur la sortie collect2: error: ld returned 1 exit status Makefile:127: recipe for target 'grit' failed make: *** [grit] Error 1 14. I boߋught the package of frozen Hakеe at Costco. 15. I'm impressed, I have to admit. Rarely do I come across a blog that's both equally educative and entertaining, and without a doubt, you've hit the nail on the head. The problem is something which too few people are speaking intelligently about. I'm very happy that I stumbled across this in my hunt for something concerning this. 16. I'm amazed, I must say. Seldom do I come across a blog that's both equally educative and entertaining, and without a doubt, you've hit the nail on the head. The problem is something that not enough people are speaking intelligently about. Now i'm very happy I stumbled across this in my search for something concerning this. 17. I've recently built on MacOS 10.14 following these changes: install freeimage using brew: brew install freeimage find the xcode sdk path by running: xcrun --show-sdk-path the path I got was: "/Applications/Xcode.app/Contents/Developer/Platforms/MacOSX.platform/Developer/SDKs/MacOSX10.14.sdk" Use that path to make the following changes in the makefile: Line 4 ----------------------------------------------------------- From: CXXFLAGS := -Icldib -Ilibgrit -Iextlib To : CXXFLAGS := -Icldib -Ilibgrit -Iextlib -stdlib=libc++ Lines 32-36 ----------------------------------------------------------- From: SDK := /Developer/SDKs/MacOSX10.4u.sdk OSXCFLAGS := -mmacosx-version-min=10.4 -isysroot$(SDK) -arch i386 -arch ppc
OSXCXXFLAGS := $(OSXCFLAGS) CXXFLAGS += -fvisibility=hidden LDFLAGS += -mmacosx-version-min=10.4 -Wl,-syslibroot,$(SDK) -arch i386 -arch ppc
To:
SDK := /Applications/Xcode.app/Contents/Developer/Platforms/MacOSX.platform/Developer/SDKs/MacOSX10.14.sdk
OSXCFLAGS := -mmacosx-version-min=10.9 -isysroot $(SDK) -arch x86_64 OSXCXXFLAGS :=$(OSXCFLAGS)
CXXFLAGS += -fvisibility=hidden
LDFLAGS += -mmacosx-version-min=10.9 -Wl,-syslibroot,$(SDK) -arch x86_64 Note: use the path you got from running "xcrun --show-sdk-path" on the SDK := line Line 102 ----------------------------------------------------------- From: LIBDIRS := . To: LIBDIRS := . /usr/local/lib Line 108 ----------------------------------------------------------- From: CPPFLAGS +=$(INCLUDE)
To:
CPPFLAGS += \$(INCLUDE) -stdlib=libc++
Hope this helps others looking to build in MacOS.
18. Hi!
So I'm coding some virtual boy. I'm having trouble understanding how I can get the color value of each pixel from the data of: "grit my_image.png -fh! -ftc -gB2 -p!"
I tried doing like gameboy does like here https://www.huderlem.com/demos/gameboy2bpp.html, but it does not correctly show the image I made via GRIT.
"title_screen_384x224Tiles" is the array generated by the above grit on an image which is 384x224 pixels.
Here's my function so far:
https://pastebin.com/sAyCD6NE
Can you help me write a function to get the color value for each pixel so I can print it into the framebuffer like I'm trying in that pastebin. | |
R igraph manual pages
Use this if you are using igraph from R
Description
Classify dyads in a directed graphs. The relationship between each pair of vertices is measured. It can be in three states: mutual, asymmetric or non-existent.
Usage
dyad_census(graph)
Arguments
graph The input graph. A warning is given if it is not directed.
Value
A named numeric vector with three elements:
mut The number of pairs with mutual connections. asym The number of pairs with non-mutual connections. null The number of pairs with no connection between them.
Author(s)
Gabor Csardi csardi.gabor@gmail.com
References
Holland, P.W. and Leinhardt, S. A Method for Detecting Structure in Sociometric Data. American Journal of Sociology, 76, 492–513. 1970.
Wasserman, S., and Faust, K. Social Network Analysis: Methods and Applications. Cambridge: Cambridge University Press. 1994.
triad_census for the same classification, but with triples.
Examples
g <- sample_pa(100) | |
# Math Help - Fibers
1. ## Fibers
7. $f: A \rightarrow B$ is a surjective map of sets. We want to prove that $a \sim b$ if and only if $f(a) = f(b)$ is an equivalence relation whose equivalence classes are fibers of $f$.
2. Here it is: $
\begin{array}{l}
\left( {\forall x \in A} \right)\left[ {f(x) = f(x)} \right] \\ | |
New Titles | FAQ | Keep Informed | Review Cart | Contact Us Quick Search (Advanced Search ) Browse by Subject General Interest Logic & Foundations Number Theory Algebra & Algebraic Geometry Discrete Math & Combinatorics Analysis Differential Equations Geometry & Topology Probability & Statistics Applications Mathematical Physics Math Education
An Introduction to the Theory of Infinite Series
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AMS Chelsea Publishing
1926; 535 pp; hardcover
Volume: 335
ISBN-10: 0-8218-3976-4
ISBN-13: 978-0-8218-3976-8
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Order Code: CHEL/335.H
This edition consists largely of a reproduction of the first edition (which was based on lectures on Elementary Analysis given at Queen's College, Galway, from 1902-1907), with additional theorems and examples. Additional material includes a discussion of the solution of linear differential equations of the second order; a discussion of elliptic function formulae; expanded treatment of asymptomatic series; a discussion of trigonometrical series, including Stokes's transformation and Gibbs's phenomenon; and an expanded Appendix II that includes an account of Napier's invention of logarithms.
Reviews
"The book is especially good at counterexamples, and includes many of these to warn against pitfalls in reasoning and to show that all the hypotheses of the theorems are really needed. One especially nice feature is the use of Tannery's theorem, on interchanging limit and summation, throughout the book."
-- MAA Reviews
• Sequences and limits
• Series of positive terms
• Series in general
• Absolute convergence
• Double series
• Infinite products
• Series of variable terms
• Power series
• Special power series
• Trigonometrical formulae
• Complex series and products
• Special complex series and functions
• Non-convergent series
• Asymptotic series
• Trigonometrical series
• Appendix I. Arithmetic theory of irrational numbers and limits
• Appendix II. Definitions of the logarithmic and exponential functions
• Appendix III. Some theorems on infinite integrals and gamma-functions
• Miscellaneous examples
• Index of special integrals, products, and series
• General index | |
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(Universidade Federal de São Carlos, UFSCar, Programa de Pós-Graduação em Matemática - PPGM, Câmpus São Carlos, 12/07/2019)
This work is dedicated to the study of dynamical properties of some partial differential equations (PDE, for short) of parabolic type, local or nonlocal ones. We prove existence, uniqueness and establish the asymptotic ...
• #### Uma estimativa do tipo L1 para potencial de Riesz
(Universidade Federal de São Carlos, UFSCar, Programa de Pós-Graduação em Matemática - PPGM, Câmpus São Carlos, 13/08/2019)
In this work we will present a L1-type estimates for the Riesz potentials envolving the Riesz transform. This estimate is an improvement of result due to Stein and Weiss the result of Stein and Weiss that provides Riesz ...
• #### Resolubilidade contínua e singularidades removíveis para a equação divergente
(Universidade Federal de São Carlos, UFSCar, Programa de Pós-Graduação em Matemática - PPGM, Câmpus São Carlos, 12/08/2019)
In this text we study some results obtained by Nguyen Cong Phuc and Monica Torres in the paper "Characterizations of the Existence and Removable Singularities of Divergence-measure Vector Fields" regarding the characterisation ... | |
# End vertex
In graph theory , a vertex with degree $1$ is called an end vertex (plural end vertices ). In the figure below, vertices $b,c,d,e$ and $f$ are end vertices. | |
# [Fontinst] glyph name aliasing
Alexej Kryukov akrioukov at newmail.ru
Tue May 3 20:00:04 CEST 2005
Dear Lars,
aliasing, and only now found it in the mailing list archive.
> what will happen when some virtual font calls upon the glyph in
> slot 17 (which fontinst was told by fasr8r.mtx is dotlessi)? The
> PS interpreter will get element 17 of the TeXBase1Encoding array:
> this is the name /dotlessi. Then it tries to look up the /dotlessi
> entry in the font's CharStrings dictionary---but there isn't one,
> since noone renamed the glyphs in the actual font! Instead the
> substitution glyph /.notdef is rendered.
I don't see any problem here. Of course I understand, that, if I
have processed a font with non-standard glyph names, I should
also create a special *.enc file for it instead of relying on
standard *.enc files (like 8r.enc) or on fontinst \etxtoenc macro.
Since the enc file format is quite clear (at least much more simple
than the etx format), this can be easily done by hand, and, once
created, such *.enc file may be used in all subsequent versions
of the font package, no matter, how many times the fonts itself
are regenerated.
> A solution which enables add-on packages such as fnstcorr.tex
> to do this cleanly, namely the \aliased command, has been in place in
> the core since v1.915 and is described in the current manual
> (Subsection 6.4), but fnstcorr.tex still hasn't been updated to take
>
> It should be sufficient to change
>
> \def\charnamealias#1{%
> \@ifundefined{GN@\charnameprefix#1}%
> {\@ifundefined{GN@#1}{#1}{\csname GN@#1\endcsname}}% was just {#1}
> {\csname GN@\charnameprefix#1\endcsname}}
>
> to
>
> \def\charnamealias#1{
> \string\aliased{#1}{
> \@ifundefined{GN@\charnameprefix#1}%
> {\@ifundefined{GN@#1}{#1}{\csname GN@#1\endcsname}}% was just {#1}
> {\csname GN@\charnameprefix#1\endcsname}
> }
> }
First, the definition you propose should at least be modified in order
to prevent writing useless \aliasing commands even for those glyphs
which already have "standard" (form fontinst's point of view) names.
Second, using \aliasing means that I can get with single mtx for
a specific set of glyphs, but still need separate etx files
for almost any specific font I am installing. Unfortunately, this
solves only a half of the problem. It is very inconvenient to
write a new etx file just because of a buggy font with incorrect
glyph names: much less convenient than creating only an *.enc file
for dvips.
And, of course, there is a problem of distribution. If I have used
a specific *.etx file for installing a publicly available font package,
I probably should make this file publicly available too. However, it
is inconvenient to include it into the font package itself, since this
file is not needed for its proper functioning (unlike the corresponding
enc file, which *is* needed anyway). So, how should I distribute it?
This is one more reason to have as few etx files as possible and as
many enc files as needed.
--
Regards,
Alexej Kryukov <akrioukov at newmail dot ru>
Moscow State University
Historical Faculty | |
# Density operator
1. Sep 21, 2007
### PhysiSmo
We are given that 2 systems can only be found in the states $$|00\rangle, |01\rangle, |10\rangle, |11\rangle$$. We are also given that the density operator is
$$\rho=\frac{1}{2}\left(|00\rangle \langle 00|+|11\rangle \langle 00|+|00\rangle \langle 11|+|11\rangle \langle 11|\right)$$.
a)Write the matrix form of the density operator. Prove that it describes a pure state. Which one?
b)By taking the partial trace of one system, show that the yielding state is a mixed state. Which is the matrix for this new state?
Solution.
a)We express the matrix form by taking inner products, so that the $$\rho_{nm}$$ element of the matrix is $$\rho_{nm}=\langle n|\rho |m \rangle$$. We find then
$$\rho= \begin{bmatrix} 1/2 & 0 \\ 0 & 1/2 \end{bmatrix}$$
which clearly describes a pure state, since $$Tr(\rho^2)=1$$.
How can one find then the state vectors? Since $$\rho=|\psi \rangle \langle \psi |$$ in general, is it true to say that $$\rho |\psi \rangle = |\psi \rangle$$?
b)Unfortunately, I don't have a clue for this one. What do we mean by taking the partial trace of one system? Any help please?
2. Sep 21, 2007
### genneth
a) Indeed it is. The pure state is the eigenvector of $$\rho$$ which has eigenvalue of 1 (think about what eigenvalues of the density matrix mean, what the eigenstates mean, and it should be obvious why this must be so, beyond just the form of the matrix representation).
b) http://beige.ucs.indiana.edu/M743/node80.html
3. Sep 22, 2007
### PhysiSmo
Thank you very much! | |
## Abstract
We consider the abelian group $PT$ generated by quasi-equivalence classes of pretriangulated DG categories with relations coming from semiorthogonal decompositions of corresponding triangulated categories. We introduce an operation of “multiplication” • on the collection of DG categories, which makes this abelian group into a commutative ring. A few applications are considered: representability of “standard” functors between derived categories of coherent sheaves on smooth projective varieties and a construction of an interesting motivic measure. | |
# Massless integrals in dim-reg
Consider the massless divergent integral $$\int dk^4 \frac{1}{k^2},$$ which occurs in QFT. We can't regularize this integral with dim-reg; the continuation from the massive to the massless case is ill-defined. It can be shown, however, that no "inconsistencies" occur if $$\int dk^4 \frac{1}{k^2} "=" 0.$$ I think this is the now proven 't Hooft-Veltman conjecture. I don't understand this "equation". The integral is certainly not zero (although inconsistencies might occur in dim-reg if it weren't treated as zero in some contexts.)
Suppose I think about $\lambda\phi^4$ theory, with no bare mass for the field. Is it reasonable to claim that I will chose dim-reg and calculate the one-loop correction to the mass as $$\delta m^2 \propto \int dk^4 \frac{1}{k^2} = 0 \qquad ?$$ Such that the particle stays massless, with no fine-tuning. I think this is simply incorrect - an unreasonable application of the 't Hooft-Veltman conjecture (I've no reason to worry about the consistency of dim-reg because I'm not regulating any integrals with it). Surely the 't Hooft-Veltman conjecture can only be applied in particular contexts? I can't start making any calculation in QFT, see an integral such as $$\int dk^n \frac{1}{k^a} = 0 \text{ for n>a},$$ and set it zero, citing dim-reg and 't Hooft-Veltman?
P.S. This is not a straw man. I read people saying such things in the context of classical scale invariance.
The problem with such an integral is that it is both UV and IR divergent. We therefore need to introduce two regulators (to regulate both divergences). To regulate the UV divergence, we use dimreg. To regulate the IR divergence, we give the particle a mass and then take the massless limit. Doing both of them, the integral becomes $$I = \lim_{m^2 \to 0} \int d^d k \frac{1}{k^2 - m^2 + i \varepsilon}$$ We can now Wick rotate and we get (I won't keep track of overall constants) $$I \sim \int d^d k_E \frac{1}{k_E^2 + m^2} = \Omega_{d-2} \int_{0}^{\infty} \frac{k_E^{d-1} d k_E}{k_E^2 + m^2} \sim m^2 \left[\frac{2}{\epsilon} + \log \frac{m^2}{\mu^2} + {\cal O}(\epsilon) \right]$$ Now, we take the massless limit to make the integral well-defined and we get $$I = 0$$
More generally, it is easy to show (just using dimensional analysis) that if we introduce the mass to regulate the IR divergence $$\int d^n k \frac{1}{k^a} = \lim_{m^2 \to 0} \int d^n k \frac{1}{\left( k^2 - m^2 \right)^{a/2}} \propto \lim_{m^2 \to 0} m^{n-a}$$ Thus, as long as $n > a$, this limit gives us 0.
• Thanks for answering. But the addition of a finite mass in the denominator is unsatisfactory. The limits $m\to0$ and $\epsilon\to0$ don't commute, so the result is ambiguous at best. – innisfree Sep 20 '14 at 15:51
• You can't do the massless integral with dim-reg. We just have the 't Hooft-Veltman conjecture. – innisfree Sep 20 '14 at 15:52
• What if I use the "prescription" to regulate the UV first and then regulate the IR (...one problem at a time?) – Siva Sep 20 '14 at 15:57
• @Siva I think you mean IR then UV, i.e. $m\to0$ first? Well, I suppose it works, but it's not much different form the conjecture no inconsistencies arise if set massless integrals to zero. – innisfree Sep 20 '14 at 16:12
• Btw, the integrals I am most interested in are the corrections to the mass, which don't have IR divgerences. – innisfree Sep 20 '14 at 16:28
Like OP says, using dimensional regularization to wave away the naturalness problem would be quite reckless. I will elaborate below:
Case 1: If you honestly believe that there is no new degree of freedom beyond the standard model (at least till the Planck scale, where all hell might break loose, for all you care) AND you believe in the 't Hooft-Veltman conjecture (as stated in the question) then you might claim that there is no hierarchy problem since there cannot be any divergent contributions.
Case 2: However, note that dimensional regularization gets rid of infinite pieces, not large constant pieces. So, if you had any new degree of freedom in your theory, entering at some mass scale $\Lambda$, then you might see the integral get cut-off at that scale in the UV -- then no conjecture will save you and you have a large finite contribution of order $\lambda \Lambda^2$, which is the essence of the naturalness problem.
Personally, Case-1 seems extremely unlikely to me, but this hasn't been proven wrong. If you pragmatically worry about having more/new degrees of freedom at higher energies that will talk to your scalar field (Even if they're not directly coupled, they might talk through gravity, for all I care :-?) then you have to come up with some mechanism (other than Case-1) to solve the naturalness problem.
• Thanks, but to be honest, I'm looking for a more specific, detailed critique of the application of dim-reg in classically massless $\phi^4$ theory. – innisfree Sep 20 '14 at 17:25
• Ah, I see. I don't have a general picture of the kind of applications you're talking about. Some links might help. – Siva Sep 20 '14 at 18:25
• Observation: When the higgs mass is set to zero, the SM is classically scale invariant. So, the question is, can that symmetry be broken quantum mechanically, and if so, will it be suppressed in any manner. – Siva Sep 21 '14 at 4:16 | |
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# [REL][BETA 2]GTASA SaveGame ENEX Updater 0.1 beta2
## Recommended Posts
Guys I released the latest version of G^seu. It got fixed most of the bugs contained in the beta1. Also it is powered with some new features like. Editing existing ENEXs. An added feature enables you to Lock/Unlcok any ENEXs. whether new or existing.
Also its now kind of complete tool with proper documentation.
or check out my blogg
My Blogg
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Initial Report on Beta2
The program stripped all my comments from gta.dat. This is not acceptable. The tool did not make a backup of gta.dat. When I selected Cancel after Updating the tool made the changes anyway. Nothing was altered so I'm not sure why the tool messed up my file.
I tried adding a new enex to an existing IPL file and altering an existing line but no modifications were made to the IPL file. I fear was that the tool will strip the comments from my IPL files. The commented lines must remain in place and in order. Consider them as disabled lines that may be switched for another line. Same goes for gta.dat.
My enex lines are heavily annotated with "end-line comments." This text is ending up mixed with the Hour Off field and all spaces are being stripped. Unfortunately when I commented my files I used // to denote a comment instead of using the proper format of # to denote comments. Xmen has adjusted his tool to handle only the // comment marker, and at this point I don't feel like updating all my IPL files again. Here's an example from LAs.ipl:
enex# 1833.54, -1843.38, 12.5595, 0, 1, 1, 8, 1831.54, -1843.38, 12.5595, 90, 0, 4, "X7_11B", 0, 2, 0, 24 // Unity 24-7#1833.54, -1843.38, 12.5595, 0, 1, 1, 8, 1831.54, -1843.38, 12.5595, 90, 0, 98, "DS2LS2", 0, 2, 0, 24 // ^warp to Unmarked El Que Barber (allow cars and bikes)1833.54, -1843.38, 12.5595, 0, 1, 1, 8, 1831.54, -1843.38, 12.5595, 90, 0, 354, "DS2LS2", 0, 2, 0, 24 // ^also accept NPC group1684.74, -2099, 12.8507, 0, 2, 2, 8, 1684.76, -2101.99, 12.8507, -180, 0, 4, "SVLAMD", 0, 2, 0, 24 // El Corona saveend
At the end of the process the program doesn't appear have a method to reload and start again. The user must quit the application and start again in order to make additional changes.
I haven't really got into what the tool is doing or how it is working. I've got too many other conflicts to make much progress. I don't want to be too critical of the tool, especially considering the unique posibility of reworking enex connections on existing saves, but my first impression is that this is an extremely dangerous tool. I've got a pretty good handle on how to alter my enexes in the IPL files and in the save file, but I'm really nervous about using this tool properly.
The save I modified with the tool was that save where I had the nested hideout inside of Caligula's Penthouse Suites. I'm not sure if the change I made worked or not as CJ was not able to leave the hideout. The destination for the LA Big House had been changed to the Queens Wardrobe and Queen's main door doesn't have a connection on this map so CJ was trapped. This is an unusual situation, but it brings up a potential problem.
For starters, the Queens Wardrobe link would have been the last matching name, but the LA Big House should have already had it's link set to the first enex with a match, not the last. This shouldn't matter on new saves, but it does break with the standard format for matching enex connections
The other problem is that the tool is reassigning the exit destination for the interior in which CJ has saved. On a game save where CJ has saved at Mulholland he'll probably end up in SF after using the tool, and a save at El Corona is likely to end up with an exit location in LV. This could be a bit dangerous if those areas are not unlocked. Setting the link to the first match should avoid spawning CJ in the wrong city, but it still seems weird to change the location of the save house used. Perhaps the nested enex chain can be used to repair the save location after all the links are redrawn.
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The other problem is that the tool is reassigning the exit destination for the interior in which CJ has saved. On a game save where CJ has saved at Mulholland he'll probably end up in SF after using the tool, and a save at El Corona is likely to end up with an exit location in LV. This could be a bit dangerous if those areas are not unlocked. Setting the link to the first match should avoid spawning CJ in the wrong city, but it still seems weird to change the location of the save house used. Perhaps the nested enex chain can be used to repair the save location after all the links are redrawn.
Thankz a lot. I have almost fixed the above problem (atleast for single savehouse). I am wondering why i didn't noticed these problems. I stayed upto 3.45 AM that night/day checking for buggs before i uploaded the file.
Other problems are also under concideration.
I found something about extra burglary. May be you already knew it. In an original save after 1st mision, some of the burglary enex have destination as FFFF. But In savefiles containing extra burglary ENEXs, some of these FFFF are replaced with those extra ones. This assignment is random as far i know. It is different in savefiles. And the extra enexs are connected back to these existing saves.
But in some cases, the extra burglary enexs are connected to already paired ones like TATOO. But the destination of TATOO is same as before, not altered.
And one more question, the number of saved extra Burglary enexs is always 10 ???
Edited by ATHMystikal
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Something about the internal working of tool.
How it finds out matched pairs
First it scan all ipls and make a list of ENEX numbers who have the same GXT name.
Then each item in list will be another list which may look like this(for SVLAMD)
0A 00 04 00 SVLAMD LAs.IPL El Corona Safe House24 00 04 00 SVLAMD LAw.ipl Verona Beach Safe House2E 00 04 00 SVLAMD LAw2.ipl Santa Maria Safe House97 00 04 00 SVLAMD vegasE.ipl Rockshore West saveAB 00 04 00 SVLAMD vegasW.ipl Redsands West saveAD 00 04 00 SVLAMD vegasW.ipl Whitewood Estates save72 01 00 40 SVLAMD savehous.ipl Verdant Bluffs Int
Then it find out the interior ENEX and connect all others to it.
0A 00 04 00 72 01 SVLAMD LAs.IPL El Corona Safe House24 00 04 00 72 01 SVLAMD LAw.ipl Verona Beach Safe House2E 00 04 00 72 01 SVLAMD LAw2.ipl Santa Maria Safe House97 00 04 00 72 01 SVLAMD vegasE.ipl Rockshore West saveAB 00 04 00 72 01 SVLAMD vegasW.ipl Redsands West saveAD 00 04 00 72 01 SVLAMD vegasW.ipl Whitewood Estates save72 01 00 40 SVLAMD savehous.ipl Verdant Bluffs Int
The next step caused all the trouble.
It take the nearest ENEX of the interior ENEX and connect the interior ENEX with it
0A 00 04 00 72 01 SVLAMD LAs.IPL El Corona Safe House24 00 04 00 72 01 SVLAMD LAw.ipl Verona Beach Safe House2E 00 04 00 72 01 SVLAMD LAw2.ipl Santa Maria Safe House97 00 04 00 72 01 SVLAMD vegasE.ipl Rockshore West saveAB 00 04 00 72 01 SVLAMD vegasW.ipl Redsands West saveAD 00 04 00 72 01 SVLAMD vegasW.ipl Whitewood Estates save72 01 00 40 AD 00 SVLAMD savehous.ipl Verdant Bluffs Int
This caused all trubles. Now I fixed it by reading the savehouse chain and reassigning the connection.
But i kept the same method for connecting other pairs. I think the change is only needed for current svehouse ENEX chain only.
my god the guys at R* are great, how did they do it perfectly!!!!!
Edited by ATHMystikal
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Guys I released the latest version of G^seu. It got fixed most of the bugs contained in the beta1. Also it is powered with some new features like. Editing existing ENEXs. An added feature enables you to Lock/Unlcok any ENEXs. whether new or existing.
Also its now kind of complete tool with proper documentation.
or check out my blogg
My Blogg
To update savefiles to work with modified enex is very useful.
But it needs more description to know how to make enex settings and which IPL.
You should but you didnt, so I will clarify something.
1. XYZ2 are not the locations of the target. Its for coming back.
XYZ2 should be near XYZ1
2. The entry of GXT name is not exactly the gxt name, its the ENEX-NAME
Name - interior name, used to find the counterpart and to identify via mission script
Its the only information of the target in the enex string
3. The number entry of the Interior must be the interior number of the starting place
Thats my knowledge when I have a look into existing IPLs
So I made an enex in vegasW.ipl
892.02, 2032.08, 10.82, 0, 1.5, 1.5, 8, 892.02, 2027.08, 10.82, 90, 0, 4, "MDDOGS", 0, 2, 0, 24
Works fine.
To get in and out needs allways 2 enex (one on map and one in interior)
I tried to get back from "MDDOGS"
Enex marker appears but dont work
int_LA.ipl
1297.16, -790.57, 1084.01, 0, 1.0, 1.0, 8, 1299.16, -790.57, 1084.01, 90, 5, 4, "MDDOGS", 1, 2, 0, 24
other trial
1297.16, -790.57, 1084.01, 0, 1.0, 1.0, 8, 1299.16, -790.57, 1084.01, 90, 5, 4, "MDDOGS", 0, 0, 0, 24
No success, any idea ?
886.08, 2031.6, 10.82, 0, 1.5, 1.5, 8, 886.08, 2027.08, 10.82, 90, 0, 4, "CARLS", 0, 2, 0, 24879.43, 2031.6, 10.82, 0, 1.5, 1.5, 8, 879.43, 2027.08, 10.82, 90, 0, 4, "CASINO2", 0, 2, 0, 24
Works fine.
Arrived back to the modded start location in vegasW by using original enex for exit the interior.
Thats right. Also after make save in "CARLS" and load this save, arrived at the modded start location in vegasW.
Request:
-The edit button dont work. If I made a failure i had to delete the string and making a new. Its a terrible kind of editing.
-The chance for adding a complete string would be very useful
-Enex works only with existing Enex Names. Its not possible to get access to the girlfriend rooms
(spaceeinstein wrote: |GF1|Denise's| but it dont works) will there be a solution ?
Basicly your tool works fine. With the modded enex above i testet ca. 30 other entrances, in and out and didnt got any enex bug.
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Ive been working my self on ENEXfor my mission Your tool is very good i must say,is very usefull for every modder.
-1945.9468, -2457.6553, 30.7781,0, 1, 1, 8, -1953.1443, -2464.3528, 30.625,314.1245 ,0, 4, "MADDOGS" 0, 2, 0, 24
-1726.637, -2217.3491, 54.0557,0, 1, 1, 8, -1723.1042, -2217.2815, 52.529,100.3394 ,0, 4, "AMMUN1" 0, 2, 0, 24
-212.244, -1511.8516, 8.0453,0, 1, 1, 8, -1710.6908, -2225.739, 46.9365,278.5248 ,0, 4, "AMMUN1" 0, 2, 0, 24
1. XYZ2 are not the locations of the target. Its for coming back.
XYZ2 should be near XYZ1
format:for enexInX, InY, InZ, InR, XRad, YRad, C8, OutX, OutY, OutZ, Outangle, Int, Mark, Name, Sky, IntF, Ton, Toff
Any idea how to link CJ to real world to real world without link to another enex ?
Tryed diffent enex link big ear,tall building.
1571.24, -1336.67, 15.7288, 0, 3, 3, 8, 1577.24, -1330.67, 15.7288, -46, 0, 70, "SKYLAN2", 0, 2, 0, 24
-The edit button dont work. If I made a failure i had to delete the string and making a new. Its a terrible kind of editing.
Yes could you fix that ? The tool is very good & usefull no bugs other than that at all for me.
Edited by gtasbigfoot
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To ZAZ:
Thankz for the feedback
Well I designed this tool only for updating the savefiles. Adding enexs within the tool was not in my mind but I added it if somebody need it. But as I can see that you are interested in adding enex using tool, i will add it somemore functionality.
And about giving enex in single string. There is a better way.:
Create a complete IPL using IPL Helper , then add that file through the tool. It will automatically add all ENEXs in them. That IPL is added to gta.dat also, so you need only to specify the ipl everything else is automatic.
As you said there is a small problem with safehouse shuffling. But it is only at start . When you load savegame and get out of safehouse, there is a bug which takes you to exit of someother safehouse. But it will not affect other ENEXs.
But I can fix it soon.
GXT Name : Yes you are correct. But there is a GXT entry of samename which is used to display the text when you enter an enex. I will change it to 'ENEX Name'
in next version.
So I made an enex in vegasW.iplCODE
892.02, 2032.08, 10.82, 0, 1.5, 1.5, 8, 892.02, 2027.08, 10.82, 90, 0, 4, "MDDOGS", 0, 2, 0, 24
Works fine.
To get in and out needs allways 2 enex (one on map and one in interior)
I tried to get back from "MDDOGS"
Enex marker appears but dont work
int_LA.ipl
1297.16, -790.57, 1084.01, 0, 1.0, 1.0, 8, 1299.16, -790.57, 1084.01, 90, 5, 4, "MDDOGS", 1, 2, 0, 24
other trial
1297.16, -790.57, 1084.01, 0, 1.0, 1.0, 8, 1299.16, -790.57, 1084.01, 90, 5, 4, "MDDOGS", 0, 0, 0, 24
No success, any idea ?
I think the MDDOGS interior must be locked by the game.
Oh wait: You said you saw the marker? Then it is unlocked.
But why did you made the last two ENEX entries. You should not make interior ENEXs. Becaue those already exists in that name. Make only the exterior one.
If you want your own int-ext pair, Use different name from what game uses. Like 'MADDG1' or something like that.
If you want to make an external marker to take cj to the crib, Only the exterior marker is needed, because interior is already defined in the IPL 'DATA\MAPS\interior\int_LA.ipl'
If you want to make your own pair use this
Exterior:892.02, 2032.08, 10.82, 0, 1.5, 1.5, 8, 892.02, 2027.08, 10.82, 90,0 ,4,"MADDG1", 0, 2, 0, 24Interior:1298.08, -794.827, 1083.03, 0, 3, 1, 8, 1299.08, -795.227, 1083.03, -8, 5, 0, "MADDG1", 1, 0, 0, 24
-The edit button dont work. If I made a failure i had to delete the string and making a new. Its a terrible kind of editing.
-Enex works only with existing Enex Names. Its not possible to get access to the girlfriend rooms
(spaceeinstein wrote: |GF1|Denise's| but it dont works) will there be a solution ?
I didn't get it correctly. Will you please explain it a little more. Ok I will sure try to do that.
Basicly your tool works fine. With the modded enex above i testet ca. 30 other entrances, in and out and didnt got any enex bug.
Thankz very happy to hear that.
Edited by ATHMystikal
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To gtabigfoot:
Any idea how to link CJ to real world to real world without link to another enex ?
Yes I think I can help you, but I didn't tested it much.
*First you must create two ENEXs in an IPL
*They must have same ENEX names.
X1, Y1, Z1, 0, 3, 3, 8, X2, Y2, Z2, -46, 0, 70, "URNAME", 0, 2, 0, 24X3, Y3, Y3, 0, 3, 3, 8, X4, Y4, Z4, 90, 0, 66, "URNAME", 0, 2, 0, 24
X1,Y1,Z1 and X3,Y3,Z3 are marker location of first and second ENEXs respectively.
X2,Y2,Z2 and X4,Y4,Z4 are exit positon of First and second ENEXs respectively.
By saying 'exit' I mean it is the position you will be taken to by other ENEX.
For example. In this case, You enter first enex located at X1,Y1,Z1 and it will take you to X4,Y4,Z4 and
When you comeback, through ENEX at X3,Y3,Z3, you will be taken to X2,Y2.Z2.
Try to use ENEX Names less than or equal to 7 letters
Look carefully, you can see that the Marker type is 70 and 66 bigger number than normal ones. These can be other nos also
Mark explanation by OrionSR and pdescobar
Brief tutorial by pdescobar (17th post down)
Happy to hear that the tool was useful to you.
BTW Edit button dooesn't work???. I enabled it in beta2 which is published. Did you try that?
IMPORTANT: As OrionSR said, I too got a feeling that the tool is a bit dangerous. So take backup of data folder and savefiles before you tru to update.
Edited by ATHMystikal
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Thx alot works nice, been doing enex for a long time but i never have been able to link them correct to another in real world for some reason i keeped on spawning at a long tower in the desert near area 69.
These are the ones i did in linking real world-real world with your info.
-1638.4142, -2227.0591, 30.2185, 0, 3, 3, 8, -120.4738, -2664.1653, 71.9278 -46, 0, 70, "enex2", 0, 2, 0, 24-1301.7812, -1632.5659, 53.5525 0, 3, 3, 8, -227.1601, -2075.6287, 31.2067, 90, 0, 66, "enex2", 0, 2, 0, 24
BTW Edit button dooesn't work???. I enabled it in beta2 which is published. Did you try that?
Yes,just now works great!
IMPORTANT: As OrionSR said, I too got a feeling that the tool is a bit dangerous. So take backup of data folder and savefiles before you tru to update.
I have a two R* games folders i did not try the tool on the real one yet i make new ipls with notepad,20 mins later:but just tryed it on the real a min just now did not see anything worng but I will keep track of what you said:) thx for info!
Edited by gtasbigfoot
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Thx alot works nice, been doing enex for a long time but i never have been able to link them correct to another in real world for some reason i keeped on spawning at a long tower in the desert near area 69.
These are the ones i did in linking real world-real world with your info.
-1638.4142, -2227.0591, 30.2185, 0, 3, 3, 8, -120.4738, -2664.1653, 71.9278 -46, 0, 70, "enex2", 0, 2, 0, 24-1301.7812, -1632.5659, 53.5525 0, 3, 3, 8, -227.1601, -2075.6287, 31.2067, 90, 0, 66, "enex2", 0, 2, 0, 24
Those entries should work. Are they still spawning you at area 69 tower?
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-Enex works only with existing Enex Names. Its not possible to get access to the girlfriend rooms
(spaceeinstein wrote: |GF1|Denise's| but it dont works) will there be a solution ?
Links to the GF Interior should work just fine. The doors, both inside and out, will get locked by the game script unless they are renamed to something else.
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-Enex works only with existing Enex Names. Its not possible to get access to the girlfriend rooms
(spaceeinstein wrote: |GF1|Denise's| but it dont works) will there be a solution ?
Links to the GF Interior should work just fine. The doors, both inside and out, will get locked by the game script unless they are renamed to something else.
Ok guys, i understand.
I created a new ipl
# My_EnterINT.iplinstendcullendpathendgrgeendenex879.43, 2031.6, 10.82, 0, 1.5, 1.5, 8, 879.43, 2027.08, 10.82, 90, 0, 4, "GF1", 0, 2, 0, 24892.02, 2032.08, 10.82, 0, 1.5, 1.5, 8, 892.02, 2027.08, 10.82, 90, 0, 4, "MDDOG1", 0, 2, 0, 241297.16, -790.57, 1083.03, 0, 1.0, 1.0, 8, 1299.16, -795.227, 1083.03, 90, 5, 0, "MDDOG1", 1, 0, 0, 24endpickendcarsendjumpendtcycendauzoendmultend
Used a stripped main and started new game. The enex worked fine.
"GF1" works if it is enabled (with stripped main all enabled by default)
The new created "MDDOG1" works too, Interior and Exterior.
Removed stripped main and inserted original main
Used now Beta2 of SaveGame ENEX Updater, loaded gta.dat, savefile and added Existing IPL: My_EnterINT.ipl
Started game, loaded save but no enex appear.
Wrote a Cleo script to enable "GF1" and "MDDOG1"
Only "GF1" appears
"MDDOG1" not. If I go to XYZ1, CJ will be teleported to XYZ2 but not into interior.
Started new game with original main, now both enex works.
Disabled a lot of other enex with Cleo script and loaded the modded savegame.
No success. Only "GF1" appears.
New created enex pairs wont work with save file updating
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TO ZAZ: I will check that problem very soon. Thanks for reporting.
Did u try to add real world enex pairs. Or only int-ext pair has problem.
EDIT: I need a clarification. You said u added the enex and tried it on new game. Then u must have modified the gta.dat. Then u run the tool and loaded the modified gta.dat. So enex must have added 2 times. This may have caused the trouble
Edited by ATHMystikal
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TO ZAZ: I will check that problem very soon. Thanks for reporting.
Did u try to add real world enex pairs. Or only int-ext pair has problem.
Tried only int-ext pair
EDIT: I need a clarification. You said u added the enex and tried it on new game. Then u must have modified the gta.dat. Then u run the tool and loaded the modified gta.dat. So enex must have added 2 times. This may have caused the trouble
No. I first had to remove the modded line in gta.dat. Your tool wont load a savegame when it loaded first a modded gta.dat
Edited by ZAZ
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Thx alot works nice, been doing enex for a long time but i never have been able to link them correct to another in real world for some reason i keeped on spawning at a long tower in the desert near area 69.
These are the ones i did in linking real world-real world with your info.
-1638.4142, -2227.0591, 30.2185, 0, 3, 3, 8, -120.4738, -2664.1653, 71.9278 -46, 0, 70, "enex2", 0, 2, 0, 24-1301.7812, -1632.5659, 53.5525 0, 3, 3, 8, -227.1601, -2075.6287, 31.2067, 90, 0, 66, "enex2", 0, 2, 0, 24
Those entries should work. Are they still spawning you at area 69 tower?
Nope,now it works great now,I think what i did before was mix the enex up with were the it takes the player,I did not link them correct.
Used a stripped main and started new game. The enex worked fine.
"GF1" works if it is enabled (with stripped main all enabled by default)
The new created "MDDOG1" works too, Interior and Exterior.
Removed stripped main and inserted original main
Used now Beta2 of SaveGame ENEX Updater, loaded gta.dat, savefile and added Existing IPL: My_EnterINT.ipl
Started game, loaded save but no enex appear.
Wrote a Cleo script to enable "GF1" and "MDDOG1"
Only "GF1" appears
"MDDOG1" not. If I go to XYZ1, CJ will be teleported to XYZ2 but not into interior.
Started new game with original main, now both enex works.
Disabled a lot of other enex with Cleo script and loaded the modded savegame.
No success. Only "GF1" appears.
New created enex pairs wont work with save file updating
I know what you mean when ever I try to spawn enex to go to GF1 it dosen't apper for a test i even changed one of my other to go to GF1 but it didn't spawn changed it back to nomal and it spawned.
disabled 5 others with coding.
Edited by gtasbigfoot
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Guys,
This version of tool cannot create connection to special interiors like GF1. But i am trying to add that option.
The other thing, the int-ext bug, is definitely my fault. I made a small mistake in coding. But i will fix it and upload it within 1 day
Thanks for pointing the errors. As I am adding new features, bugs are increasing by about two times!! Quite funny. Thats why I as you to take backups.
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Guys,
This version of tool cannot create connection to special interiors like GF1. But i am trying to add that option.
Shure it can. As I said:
Used now Beta2 of SaveGame ENEX Updater, loaded gta.dat, savefile and added Existing IPL: My_EnterINT.ipl
Started game, loaded save but no enex appear.
Wrote a Cleo script to enable "GF1" and "MDDOG1"
Only "GF1" appears
"MDDOG1" not. If I go to XYZ1, CJ will be teleported to XYZ2 but not into interior.
It means: "GF1" works with savegame update, because it uses originaly enex name.
(in with modded enex, ex with original)
But the enex must be enabled. In stripped main its enable by default. In original main it disabled.
To enable in original savegame a cleo script is useful:
07FB: toggle_interior 'GF1' access 1
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Hey ZAZ, I fixed the problem with int-ext ENEX pair. As I told it was due to my mistake (I forgot to comment a line in my program which caused the trouble.)
But about GF1, I tested it with your IPL. So I am giving a description f what I saw
1. two new markers at Willofield Estates.
First one expected to take you to MadDogs
Second one to GF1
2. One new marker in MadDogs interior which expected to take you back to Willofoeld
the original one is present also.
The above description is according to bug fixed version.
But I tried GF1, it took me to GF1 interior. But there is no coming back. This is expected, because I have disabled the ability to connect special interiors. thats why I said tool is not able to make such connections.( i mean beta2 but beta1 may do it im not sure)
Why I disabled Special interiors:
Those interiors have no match in noraml IPL. You have no yellow marker which connect you to them. Those markers are generated temperorily during dating. So they have FFFF as their destination ID. Such ENEXs assume the original Destination ID through game play.
So I am able to make my tool connect those with your exterior marker. But I fear that it will interrupt the gameflow.(because after that the game may not be able to connect them during dating. )
So if you really want to do it, I will add an option which enables to connect/disconnect such special interior.
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I'm confused as to why the GF "special" interiors are treated differently. The GF Interiors, like the other unused interiors, have a destination index of -1 (FFFF). These enex connections function just like one-way links. Enter XYZ1 and exit at XYZ2 (see Big Ear Tower, SF Tower, DamIn/DamOut, Pier 69). If CJ uses an enex with a destination set to a one-way or unused link then the destination address of the one-way/unused enex will be set to the return address of the enex used.
That sounds confusing, try again: The destination addresses are always redirected to whichever enex linked to that door. This is how the fastfood, wardrobe, and common safehouses remember where to return CJ. As far as I can tell, this is the main method for redirecting enex connections during game play. There doesn't seem to be any scripting options available, but the game engine seems to be able manipulate things. BTW, I suspect a clever cleo script could rewrite the destination address using a hex patch.
*Sorry for mixing and matching my terms. The destination ID should be termed as an index (structure address + index*record length = record address). But the changes occur in memory which uses the memory address not the index; the index is calculated and written to the save file. Next time I update the wiki I'll be changing the term Destination ID to Destination Index.
It might be worth checking into the Millie Keycard scripts to see how her enex is handled. I haven't experimented with the HC scripts, but I suspect access to the interiors is controlled completely by the mission script. CJ is teleported to the interior for the mission and teleported back out after the mission. No enex links are used. Anyway, I anticipate no problems to linking to these interiors.
BTW, an alternative to using a CLEO script or stripped main to keep interiors open is to change the enex name to something else. This might result in confusing zone names when using the link, but you can always rename the links back to the original name after the startup scripts have completed. Also, I'm not sure if the character scripts such as store clerks will appear if the interior names are altered. I know the exterior and interior names must match, but I'm not sure if original names are required. This doesn't work for property exteriors; they use coordinates to lock and unlock the door.
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Thankz OrionSR I've reenabled the tools ability to connect to unused interiors like GF1. But since such interiors are initially locked, Even though u can go into it, the return ENEX will not appear. IOW you are trapped. So yopu must unlock such interior ENEX if you want to comeback. I hope my tool can unlock them.
I wiil upload the tool with those fixes within hours. After that I'm going to rewrite the entire code, because those hot fixes actually ruined the readability and performance.
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G^seu- Hot Fix released. (G^seu 0.1c)
I have uploaded the G^seu hot fix on gtagarage.com.
This will fix some of the bugs found in G^seu01.b.
The bug that spawned CJ at wrong safehouse is fixed.
The bug that crea problems with creating new Interior-Exterior pair is also fixed.
And I've reenabled ability to connect special Interiors.
How to connect Special Interiors:
-----------------------------------------
Add an exterior ENEX as normal with special Interior name.
Then find out the special interior ENEX, (it will be defined in one of the interior IPLs]. Select it. and choose 'Edit'. If it is locked, then unlock it and press 'OK'.
Then update the savefile.
Important
-------------
I found that all ENEXs that are added through 'Add Existing IPl file' are automatically locked by the tool. So you have to manually unlock them using 'Edit feature'. Sorry for the inconvenience.
I am going to rewrite the whole code for better performance and more reliability. So It take some time. So there won't be a new version soon before the final version. But the final version will have many exciting new features and better UI.
I will publish intermediate betas for testing. and I will give the link in this topic. So those who are willing to help can use them to test.
Edited by ATHMystikal
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## Section8.1Graphical and Numerical Solutions to Differential Equations
In Section 5.1, we were introduced to the idea of a differential equation. Given a function $$y = f(x)\text{,}$$ we defined a differential equation as an equation involving $$y, x\text{,}$$ and derivatives of $$y\text{.}$$ We explored the simple differential equation $$\yp = 2x\text{,}$$ and saw that a solution to a differential equation is simply a function that satisfies the differential equation.
### Subsection8.1.1Introduction and Terminology
#### Definition8.1.2.Differential Equation.
Given a function $$y=f(x)\text{,}$$ a differential equation is an equation relating $$x, y\text{,}$$ and derivatives of $$y\text{.}$$
• The variable $$x$$ is called the independent variable.
• The variable $$y$$ is called the dependent variable.
• The order of the differential equation is the order of the highest derivative of $$y$$ that appears in the equation.
Let us return to the simple differential equation
\begin{equation*} \yp = 2x\text{.} \end{equation*}
To find a solution, we must find a function whose derivative is $$2x\text{.}$$ In other words, we seek an antiderivative of $$2x\text{.}$$ The function
\begin{equation*} y = x^2 \end{equation*}
is an antiderivative of $$2x\text{,}$$ and solves the differential equation. So do the functions
\begin{equation*} y = x^2 + 1 \end{equation*}
and
\begin{equation*} y = x^2 - 2346\text{.} \end{equation*}
We call the function
\begin{equation*} y = x^2 + C\text{,} \end{equation*}
with $$C$$ an arbitrary constant of integration, the general solution to the differential equation.
In order to specify the value of the integration constant $$C\text{,}$$ we require additional information. For example, if we know that $$y(1) = 3\text{,}$$ it follows that $$C=2\text{.}$$ This additional information is called an initial condition.
#### Definition8.1.3.Initial Value Problem.
A differential equation paired with an initial condition (or initial conditions) is called an initial value problem.
The solution to an initial value problem is called a particular solution. A particular solution does not include arbitrary constants.
The family of solutions to a differential equation that encompasses all possible solutions is called the general solution to the differential equation.
#### Example8.1.4.A simple first-order differential equation.
Solve the differential equation $$\yp = 2y\text{.}$$
Solution.
The solution is a function $$y$$ such that differentiation yields twice the original function. Unlike our starting example, finding the solution here does not involve computing an antiderivative. Notice that “integrating both sides” would yield the result $$y = \int 2y\,dx\text{,}$$ which is not useful. Without knowledge of the function $$y\text{,}$$ we can't compute the indefinite integral. Later sections will explore systematic ways to find analytic solutions to simple differential equations. For now, a bit of thought might let us guess the solution
\begin{equation*} y = e^{2x}\text{.} \end{equation*}
Notice that application of the chain rule yields $$\yp = 2e^{2x} = 2y\text{.}$$ Another solution is given by
\begin{equation*} y = -3e^{2x}\text{.} \end{equation*}
In fact,
\begin{equation*} y = Ce^{2x}\text{,} \end{equation*}
where $$C$$ is any constant, is the general solution to the differential equation because $$\yp = 2Ce^{2x} = 2y\text{.}$$
If we are provided with a single initial condition, say $$y(0) = 3/2\text{,}$$ we can identify $$C=3/2$$ so that
\begin{equation*} y = \frac{3}{2}e^{2x} \end{equation*}
is the particular solution to the initial value problem
\begin{equation*} \yp = 2y, \text{ with } y(0) = \frac{3}{2}\text{.} \end{equation*}
Figure 8.1.5 shows various members of the general solution to the differential equation $$\displaystyle \yp = 2y\text{.}$$ Each $$C$$ value yields a different member of the family, and a different function. We emphasize the particular solution corresponding to the initial condition $$y(0)=3/2\text{.}$$
#### Example8.1.6.A second-order differential equation.
Solve the differential equation $$\yp' + 9y = 0\text{.}$$
Solution.
We seek a function whose second derivative is negative 9 multiplied by the original function. Both $$\sin(3x)$$ and $$\cos(3x)$$ have this feature. The general solution to the differential equation is given by
\begin{equation*} y = C_1\sin(3x) + C_2\cos(3x)\text{,} \end{equation*}
where $$C_1$$ and $$C_2$$ are arbitrary constants. To fully specify a particular solution, we require two additional conditions. For example, the initial conditions $$y(0)=1$$ and $$\yp(0)=3$$ yield $$C_1 = C_2 = 1\text{.}$$
The differential equation in Example 8.1.6 is second order, because the equation involves a second derivative. In general, the number of initial conditions required to specify a particular solution depends on the order of the differential equation. For the remainder of the chapter, we restrict our attention to first order differential equations and first order initial value problems.
#### Example8.1.7.Verifying a solution to the differential equation.
Which of the following is a solution to the differential equation
\begin{equation*} \yp + \frac{y}{x} - \sqrt{y} = 0\text{?} \end{equation*}
1. $$\displaystyle y = C \left ( 1 + \ln(x) \right )^2$$
2. $$\displaystyle y = \left ( \frac{1}{3}x + \frac{C}{\sqrt{x}} \right )^2$$
3. $$\displaystyle y = C e^{-3x} + \sqrt{\sin(x)}$$
Solution.
Verifying a solution to a differential equation is simply an exercise in differentiation and simplification. We substitute each potential solution into the differential equation to see if it satisfies the equation.
1. Testing the potential solution $$y = C \left ( 1 + \ln(x) \right )^2\text{:}$$
Differentiating, we have $$\displaystyle \yp = \frac{2C(1 + \ln(x))}{x}\text{.}$$ Substituting into the differential equation,
\begin{align*} \amp \frac{2C(1+\ln(x))}{x} + \frac{C(1+\ln(x))^2}{x} -\sqrt{C}(1+\ln(x))\\ \amp = (1+\ln(x))\left( \frac{2C}{x} + \frac{C(1+\ln(x))}{x} - \sqrt{C}\right)\\ \amp \neq 0\text{.} \end{align*}
Since it doesn't satisfy the differential equation, $$y = C(1 + \ln(x))^2$$ is not a solution.
2. Testing the potential solution $$\displaystyle y = \left ( \frac{1}{3}x + \frac{C}{\sqrt{x}} \right )^2\text{:}$$
Differentiating, we have $$\displaystyle \yp = 2\left ( \frac{1}{3}x + \frac{C}{\sqrt{x}} \right )\left ( \frac{1}{3} - \frac{C}{2x^{3/2}}\right )\text{.}$$ Substituting into the differential equation,
\begin{align*} \amp 2\left ( \frac{1}{3}x + \frac{C}{\sqrt{x}} \right )\left ( \frac{1}{3} - \frac{C}{2x^{3/2}}\right ) + \frac{1}{x}\left ( \frac{1}{3}x + \frac{C}{\sqrt{x}} \right )^2 - \left (\frac{1}{3}x + \frac{C}{\sqrt{x}}\right)\\ \amp = \left ( \frac{1}{3}x + \frac{C}{\sqrt{x}} \right ) \left ( \frac{2}{3} - \frac{C}{x^{3/2}} + \frac{1}{3} + \frac{C}{x^{3/2}} - 1 \right )\\ \amp = 0. \text{ (Note how the second parenthetical grouping above reduces to 0.) } \end{align*}
Thus $$\displaystyle y = \left ( \frac{1}{3}x + \frac{C}{\sqrt{x}} \right )^2$$ is a solution to the differential equation.
3. Testing the potential solution $$\displaystyle y = C e^{-3x} + \sqrt{\sin(x)}\text{:}$$
Differentiating, $$\displaystyle \yp = -3Ce^{-3x} + \frac{\cos(x)}{2\sqrt{\sin(x)}}\text{.}$$ Substituting into the differential equation,
\begin{equation*} -3Ce^{-3x} + \frac{\cos(x)}{2\sqrt{\sin(x)}} + \frac{C e^{-3x} + \sqrt{\sin(x)}}{x} - \sqrt{C e^{-3x} + \sqrt{\sin(x)}} \neq 0\text{.} \end{equation*}
The function $$\displaystyle y = C e^{-3x} + \sqrt{\sin(x)}$$ is not a solution to the differential equation.
#### Example8.1.8.Verifying a solution to a differential equation.
Verify that $$x^2+y^2 = Cy$$ is a solution to $$\displaystyle \yp = \frac{2xy}{x^2-y^2}\text{.}$$
Solution.
The solution in this example is called an implicit solution. That means the dependent variable $$y$$ is a function of $$x\text{,}$$ but has not been explicitly solved for. Verifying the solution still involves differentiation, but we must take the derivatives implicitly. Differentiating, we have
\begin{equation*} 2x + 2y\yp = C\yp\text{.} \end{equation*}
Solving for $$\yp\text{,}$$ we have
\begin{equation*} \yp = \frac{2x}{C-2y}\text{.} \end{equation*}
From the solution, we know that $$\displaystyle C = \frac{x^2+y^2}{y}\text{.}$$ Then
\begin{align*} \yp \amp = \frac{2x}{\displaystyle \frac{x^2+y^2}{y} - 2y}\\ \amp =\frac{2xy}{x^2+y^2-2y^2}\\ \amp = \frac{2xy}{x^2-y^2}\text{.} \end{align*}
We have verified that $$x^2+y^2 = Cy$$ is a solution to $$\displaystyle \yp = \frac{2xy}{x^2-y^2}\text{.}$$
### Subsection8.1.2Graphical Solutions to Differential Equations
In the examples we have explored so far, we have found exact forms for the functions that solve the differential equations. Solutions of this type are called analytic solutions. Many times a differential equation has a solution, but it is difficult or impossible to find the solution analytically. This is analogous to algebraic equations. The algebraic equation $$x^2 + 3x - 1 = 0$$ has two real solutions that can be found analytically by using the quadratic formula. The equation $$\cos(x) = x$$ has one real solution, but we can't find it analytically. As shown in Figure 8.1.9, we can find an approximate solution graphically by plotting $$\cos(x)$$ and $$x$$ and observing the $$x$$-value of the intersection. We can similarly use graphical tools to understand the qualitative behavior of solutions to a first order-differential equation.
Consider the first-order differential equation
\begin{equation*} \yp = f(x,y)\text{.} \end{equation*}
The function $$f$$ could be any function of the two variables $$x$$ and $$y\text{.}$$ Written in this way, we can think of the function $$f$$ as providing a formula to find the slope of a solution at a given point in the $$xy$$-plane. In other words, suppose a solution to the differential equation passes through the point $$(x_0,y_0)\text{.}$$ At the point $$(x_0,y_0)\text{,}$$ the slope of the solution curve will be $$f(x_0,y_0)\text{.}$$ Since this calculation of the slope is possible at any point $$(x,y)$$ where the function $$f(x,y)$$ is defined, we can produce a plot called a slope field (or direction field) that shows the slope of a solution at any point in the $$xy$$-plane where the solution is defined. Further, this process can be done purely by working with the differential equation itself. In other words, we can draw a slope field and use it to determine the qualitative behavior of solutions to a differential equation without having to solve the differential equation.
#### Definition8.1.10.Slope Field.
A slope field for a first-order differential equation $$\yp = f(x,y)$$ is a plot in the $$xy$$-plane made up of short line segments or arrows. At each point $$(x_0,y_0)$$ where $$f(x,y)$$ is defined, the slope of the line segment is given by $$f(x_0,y_0)\text{.}$$ Plots of solutions to a differential equation are tangent to the line segments in the slope field.
#### Example8.1.11.Sketching a slope field.
Find a slope field for the differential equation $$\displaystyle \yp = x+y\text{.}$$
Solution.
Because the function $$f(x,y) = x+y$$ is defined for all points $$(x,y)\text{,}$$ every point in the $$xy$$-plane has an associated line segment. It is not practical to draw an entire slope field by hand, but many tools exist for drawing slope fields on a computer. Here, we explicitly calculate a few of the line segments in the slope field.
• The slope of the line segment at $$(0,0)$$ is $$f(0,0) = 0 + 0 = 0\text{.}$$
• The slope of the line segment at $$(1,1)$$ is $$f(1,1) = 1 + 1 = 2\text{.}$$
• The slope of the line segment at $$(1,-1)$$ is $$f(1,-1) = 1 - 1 = 0\text{.}$$
• The slope of the line segment at $$(-2,-1)$$ is $$f(-2,-1) = -2 - 1 = -3\text{.}$$
Though it is possible to continue this process to sketch a slope field, we usually use a computer to make the drawing. Most popular computer algebra systems can draw slope fields. There are also various online tools that can make the drawings. The slope field for $$\yp = x+y$$ is shown in Figure 8.1.12.
#### Example8.1.13.A graphical solution to an initial value problem.
Approximate, with a sketch, the solution to the initial value problem $$\displaystyle \yp = x+y\text{,}$$ with $$y(1)=-1\text{.}$$
Solution.
The solution to the initial value problem should be a continuous smooth curve. Using the slope field, we can draw of a sketch of the solution using the following two criteria:
1. The solution must pass through the point $$(1,-1)\text{.}$$
2. When the solution passes through a point $$(x_0,y_0)$$ it must be tangent to the line segment at $$(x_0,y_0)\text{.}$$
Essentially, we sketch a solution to the initial value problem by starting at the point $$(1,-1)$$ and “following the lines” in either direction. A sketch of the solution is shown in Figure 8.1.14.
#### Example8.1.15.Using a slope field to predict long term behavior.
Use the slope field for the differential equation $$\yp = y(1-y)\text{,}$$ shown in Figure 8.1.16, to predict long term behavior of solutions to the equation.
Solution.
This differential equation, called the logistic differential equation, often appears in population biology to describe the size of a population. For that reason, we use $$t$$ (time) as the independent variable instead of $$x\text{.}$$ We also often restrict attention to non-negative $$y$$-values because negative values correspond to a negative population.
Looking at the slope field in Figure 8.1.16, we can predict long term behavior for a given initial condition.
• If the initial $$y$$-value is negative ($$y(0)\lt 0$$), the solution curve must pass though the point $$(0,y(0))$$ and follow the slope field. We expect the solution $$y$$ to become more and more negative as time increases. Note that this result is not physically relevant when considering a population.
• If the initial $$y$$-value is greater than 0 but less than 1, we expect the solution $$y$$ to increase and level off at $$y=1\text{.}$$
• If the initial $$y$$-value is greater than 1, we expect the solution $$y$$ to decrease and level off at $$y=1\text{.}$$
The slope field for the logistic differential equation, along with representative solution curves, is shown in Figure 8.1.17. Notice that any solution curve with positive initial value will tend towards the value $$y=1\text{.}$$ We call this the carrying capacity.
### Subsection8.1.3Numerical Solutions to Differential Equations: Euler's Method
While the slope field is an effective way to understand the qualitative behavior of solutions to a differential equation, it is difficult to use a slope field to make quantitative predictions. For example, if we have the slope field for the differential equation $$\yp = x+y$$ from Example 8.1.11 along with the initial condition $$y(0)=1\text{,}$$ we can understand the qualitative behavior of the solution to the initial value problem, but will struggle to predict a specific value, $$y(2)$$ for example, with any degree of confidence. The most straightforward way to predict $$y(2)$$ is to find the analytic solution to the the initial value problem and evaluate it at $$x=2\text{.}$$ Unfortunately, we have already mentioned that it is impossible to find analytic solutions to many differential equations. In the absence of an analytic solution, a numerical solution can serve as an effective tool to make quantitative predictions about the solution to an initial value problem.
There are many techniques for computing numerical solutions to initial value problems. A course in numerical analysis will discuss various techniques along with their strengths and weaknesses. The simplest technique is called Euler's Method.
Consider the first-order initial value problem
\begin{equation*} \yp = f(x,y), \text{ with } y(x_0) = y_0\text{.} \end{equation*}
Using the definition of the derivative,
\begin{equation*} \yp(x) = \lim_{h \to 0} \frac{y(x+h) - y(x)}{h}\text{.} \end{equation*}
This notation can be confusing at first, but “$$y(x)$$” simply means “the $$y$$-value of the solution when the $$x$$-value is $$x$$”, and “$$y(x+h)$$” means “the $$y$$-value of the solution when the $$x$$-value is $$x+h$$”.
If we remove the limit but restrict $$h$$ to be “small,” we have
\begin{equation*} \yp(x) \approx \frac{y(x+h) - y(x)}{h}\text{,} \end{equation*}
so that
\begin{equation*} f(x,y) \approx \frac{y(x+h)-y(x)}{h}\text{,} \end{equation*}
because $$\yp = f(x,y)$$ according to the differential equation. Rearranging terms,
\begin{equation*} y(x + h) \approx y(x) + h\,f(x,y)\text{.} \end{equation*}
This statement says that if we know the solution ($$y$$-value) to the initial value problem for some given $$x$$-value, we can find an approximation for the solution at the value $$x+h$$ by taking our $$y$$-value and adding $$h$$ times the function $$f$$ evaluated at the $$x$$ and $$y$$ values. Euler's method uses the initial condition of an initial value problem as the starting point, and then uses the above idea to find approximate values for the solution $$y$$ at later $$x$$-values. The algorithm is summarized in Key Idea 8.1.18.
#### Key Idea8.1.18.Euler's Method.
Consider the initial value problem
\begin{equation*} \yp = f(x,y) \text{ with } y(x_0)=y_0\text{.} \end{equation*}
Let $$h$$ be a small positive number and $$N$$ be an integer.
1. For $$i = 0, 1, 2, \ldots, N\text{,}$$ define
\begin{equation*} x_i = x_0 + ih\text{.} \end{equation*}
2. The value $$y_0$$ is given by the initial condition. For $$i = 0, 1, 2, \ldots, N-1\text{,}$$ define
\begin{equation*} y_{i+1} = y_i + hf(x_i,y_i)\text{.} \end{equation*}
This process yields a sequence of $$N+1$$ points $$(x_i,y_i)$$ for $$i= 0,1,2,\ldots,N\text{,}$$ where $$(x_i, y_i)$$ is an approximation for $$(x_i,y(x_i))\text{.}$$
Let's practice Euler's Method using a few concrete examples.
#### Example8.1.19.Using Euler's Method 1.
Find an approximation at $$x=2$$ for the solution to $$\yp = x + y$$ with $$y(1)=-1$$ using Euler's Method with $$h=0.5\text{.}$$
Solution.
Our initial condition yields the starting values $$x_0 = 1$$ and $$y_0 = -1\text{.}$$ With $$h = 0.5\text{,}$$ it takes $$N=2$$ steps to get to $$x=2\text{.}$$ Using steps 1 and 2 from the Euler's Method algorithm,
\begin{align*} x_0 \amp = 1 \amp y_0 \amp = -1\\ x_1 \amp = x_0 + h \amp y_1 \amp = y_0 + hf(x_0,y_0)\\ \amp = 1 + 0.5 \amp \amp = -1 + 0.5(1 - 1)\\ \amp = 1.5 \amp \amp = -1\\ x_2 \amp = x_0 + 2h \amp y_2 \amp = y_1 + hf(x_1,y_1)\\ \amp = 1 + 2(0.5) \amp \amp = -1 + 0.5(1.5 -1)\\ \amp = 2 \amp \amp = -0.75 \text{.} \end{align*}
Using Euler's method, we find the approximate $$y(2) \approx -0.75\text{.}$$
To help visualize the Euler's method approximation, these three points (connected by line segments) are plotted along with the analytical solution to the initial value problem in Figure 8.1.20.
This approximation doesn't appear terrific, though it is better than merely guessing. Let's repeat the previous example using a smaller $$h$$-value.
#### Example8.1.21.Using Euler's Method 2.
Find an approximation on the interval $$[1,2]$$ for the solution to $$\yp = x + y$$ with $$y(1)=-1$$ using Euler's Method with $$h=0.25\text{.}$$
Solution.
Our initial condition yields the starting values $$x_0 = 1$$ and $$y_0 = -1\text{.}$$ With $$h = 0.25\text{,}$$ we need $$N=4$$ steps on the interval $$[1,2]$$ Using steps 1 and 2 from the Euler's Method algorithm (and rounding to 4 decimal points), we have
\begin{align*} x_0 \amp = 1 \amp y_0 \amp = -1 \\ x_1 \amp = 1.25 \amp y_1 \amp = -1 + 0.25(1 - 1) \\ \amp \amp \amp = -1 \\ x_2 \amp = 1.5 \amp y_2 \amp = -1 + 0.25(1.25-1) \\ \amp \amp \amp = -0.9375 \\ x_3 \amp = 1.75 \amp y_3 \amp = -0.9375 + 0.25(1.5-0.9375) \\ \amp \amp \amp = -0.7969 \\ x_4 \amp = 2 \amp y_4 \amp = -0.7969 + 0.25(1.75 - 0.7969) \\ \amp \amp \amp = -0.5586 \text{.} \end{align*}
Using Euler's method, we find $$y(2) \approx -0.5586\text{.}$$
These five points, along with the points from Example 8.1.19 and the analytic solution, are plotted in Figure 8.1.22.
Using the results from Examples 8.1.19 and Example 8.1.21, we can make a few observations about Euler's method. First, the Euler approximation generally gets worse as we get farther from the initial condition. This is because Euler's method involves two sources of error. The first comes from the fact that we're using a positive $$h$$-value in the derivative approximation instead of using a limit as $$h$$ approaches zero. Essentially, we're using a linear approximation to the solution $$y$$ (similar to the process described in Section 4.4 on Differentials.) This error is often called the local truncation error. The second source of error comes from the fact that every step in Euler's method uses the result of the previous step. That means we're using an approximate $$y$$-value to approximate the next $$y$$-value. Doing this repeatedly causes the errors to build on each other. This second type of error is often called the propagated or accumulated error.
A second observation is that the Euler approximation is more accurate for smaller $$h$$-values. This accuracy comes at a cost, though. Example 8.1.21 is more accurate than Example 8.1.19, but takes twice as many computations. In general, numerical algorithms (even when performed by a computer program) require striking a balance between a desired level of accuracy and the amount of computational effort we are willing to undertake.
Let's do one final example of Euler's Method.
#### Example8.1.23.Using Euler's Method 3.
Find an approximation for the solution to the logistic differential equation
$$\yp = y(1-y)$$ with $$y(0) = 0.25\text{,}$$ for $$0 \leq y \leq 4\text{.}$$ Use $$N = 10$$ steps.
Solution.
The logistic differential equation is what is called an autonomous equation. An autonomous differential equation has no explicit dependence on the independent variable ($$t$$ in this case). This has no real effect on the application of Euler's method other than the fact that the function $$f(t,y)$$ is really just a function of $$y\text{.}$$ To take steps in the $$y$$ variable, we use
\begin{equation*} y_{i+1} = y_i + hf(t_i,y_i) = y_i + hy_i(1-y_i)\text{.} \end{equation*}
Using $$N=10$$ steps requires $$\displaystyle h = \frac{4-0}{10} = 0.4\text{.}$$ Implementing Euler's Method, we have
\begin{align*} x_0 \amp = 0 \amp y_0 \amp = 0.25 \\ x_1 \amp = 0.4 \amp y_1 \amp = 0.25 + 0.4(0.25)(1-0.25) \\ \amp \amp \amp = 0.325 \\ x_2 \amp = 0.8 \amp y_2 \amp = 0.325 + 0.4(0.325)(1-0.325) \\ \amp \amp \amp = 0.41275 \\ x_3 \amp = 1.2 \amp y_3 \amp = 0.41275 + 0.4(0.41275)(1-0.41275)\\ \amp \amp \amp = 0.50970 \\ x_4 \amp = 1.6 \amp y_4 \amp = 0.50970 + 0.4(0.50970)(1 - 0.50970) \\ \amp \amp \amp = 0.60966 \\ x_5 \amp = 2.0 \amp y_5 \amp = 0.60966 + 0.4(0.60966)(1-0.60966)\\ \amp \amp \amp = 0.70485 \\ x_6 \amp = 2.4 \amp y_6 \amp = 0.70485 + 0.4(0.70485)(1 - 0.70485) \\ \amp \amp \amp = 0.78806\\ x_7 \amp = 2.8 \amp y_7 \amp = 0.78806 + 0.4(0.78806)(1-0.78806) \\ \amp \amp \amp = 0.85487 \\ x_8 \amp = 3.2 \amp y_8 \amp = 0.85487 + 0.4(0.85487)(1-0.85487)\\ \amp \amp \amp = 0.90450\\ x_9 \amp = 3.6 \amp y_9 \amp = 0.90450 + 0.4(0.90450)(1 - 0.90450)\\ \amp \amp \amp = 0.93905 \\ x_{10}\amp = 4.0 \amp y_{10}\amp = 0.93905 + 0.4(0.93905)(1 - 0.93905) \\ \amp \amp \amp = 0.96194 \text{.} \end{align*}
These 11 points, along with the the analytic solution, are plotted in Figure 8.1.24. Notice how well they seem to match the true solution.
The study of differential equations is a natural extension of the study of derivatives and integrals. The equations themselves involve derivatives, and methods to find analytic solutions often involve finding antiderivatives. In this section, we focus on graphical and numerical techniques to understand solutions to differential equations. We restrict our examples to relatively simple initial value problems that permit analytic solutions to the equations, but we should remember that this is only for comparison purposes. In reality, many differential equations, even some that appear straightforward, do not have solutions we can find analytically. Even so, we can use the techniques presented in this section to understand the behavior of solutions. In the next two sections, we explore two techniques to find analytic solutions to two different classes of differential equations.
### Exercises8.1.4Exercises
#### Terms and Concepts
##### 1.
In your own words, what is an initial value problem, and how is it different than a differential equation?
##### 2.
In your own words, describe what it means for a function to be a solution to a differential equation.
##### 3.
How can we verify that a function is a solution to a differential equation?
##### 4.
Describe the difference between a particular solution and a general solution.
##### 5.
Why might we use a graphical or numerical technique to study solutions to a differential equation instead of simply solving the differential equation to find an analytic solution?
##### 6.
Describe the considerations that should be made when choosing an $$h$$ value to use in a numerical method like Euler's Method.
#### Problems
##### Exercise Group.
In the following exercises, verify that the given function is a solution to the differential equation or initial value problem.
###### 7.
$$y = Ce^{-6x^2}\text{;}$$ $$\yp = -12xy\text{.}$$
###### 8.
$$y = x\sin(x)\text{;}$$ $$\yp - x\cos(x)= (x^2+1)\sin(x) - xy\text{,}$$ with $$y(\pi) = 0\text{.}$$
###### 9.
$$2x^2-y^2=C\text{;}$$ $$y\yp-2x = 0$$
###### 10.
$$y=xe^x\text{;}$$ $$\yp' - 2\yp + y = 0$$
##### Exercise Group.
In the following exercises, verify that the given function is a solution to the differential equation and find the $$C$$ value required to make the function satisfy the initial condition.
###### 11.
$$y = 4e^{3x}\sin(x) + Ce^{3x}\text{;}$$ $$\yp - 3y = 4e^{3x}\cos(x)\text{,}$$ with $$y(0)=2$$
###### 12.
$$y(x^2+y) = C\text{;}$$ $$2xy + (x^2+2y)\yp= 0\text{,}$$ with $$y(1)=2$$
##### Exercise Group.
In the following exercises, sketch a slope field for the given differential equation. Let $$x$$ and $$y$$ range between $$-2$$ and $$2\text{.}$$
###### 13.
$$\yp = y-x$$
###### 14.
$$\yp = \displaystyle \frac{x}{2y}$$
###### 15.
$$\yp = \sin(\pi y)$$
###### 16.
$$\yp = \frac{y}{4}$$
##### Exercise Group.
Match each slope field below with the appropriate differential equation.
(a)
(b)
(c)
(d)
###### 17.
$$\yp = xy$$
###### 18.
$$\yp = -y$$
###### 19.
$$\yp = -x$$
###### 20.
$$\yp = x(1-x)$$
##### Exercise Group.
In the following exercises, sketch the slope field for the differential equation, and use it to draw a sketch of the solution to the initial value problem.
###### 21.
$$\yp = \displaystyle \frac{y}{x} - y\text{,}$$ with $$y(0.5)=1\text{.}$$
###### 22.
$$\yp = y\sin(x)\text{,}$$ with $$y(0)=1\text{.}$$
###### 23.
$$\yp = y^2-3y+2\text{,}$$ with $$y(0)=2\text{.}$$
###### 24.
$$\displaystyle \yp = -\frac{xy}{1+x^2}\text{,}$$ with $$y(0)=1\text{.}$$
##### Exercise Group.
In the following exercises, use Euler's Method to make a table of values that approximates the solution to the initial value problem on the given interval. Use the specified $$h$$ or $$N$$ value.
###### 25.
$$\yp = x+2y$$
$$y(0)=1$$
interval: $$[0,1]$$
$$h=0.25$$
###### 26.
$$\yp = xe^{-y}$$
$$y(0)=1$$
interval: $$[0,0.5]$$
$$N=5$$
###### 27.
$$\yp = y + \sin(x)$$
$$y(0)=2$$
interval: $$[0,1]$$
$$h = 0.2$$
###### 28.
$$\yp = e^{x-y}$$
$$y(0)=0$$
interval: $$[0,2]$$
$$h = 0.5$$
##### Exercise Group.
In the following exercises, use the provided solution $$y(x)$$ and Euler's Method with the $$h=0.2$$ and $$h=0.1$$ to complete the following table.
$$x$$ $$0.0$$ $$0.2$$ $$0.4$$ $$0.6$$ $$0.8$$ $$1.0$$ $$y(x)$$ $$h = 0.2$$ $$h = 0.1$$
###### 29.
$$\yp = xy^2$$
$$y(0)=1$$
Solution: $$\displaystyle y(x) = \frac{2}{1-x^2}$$
###### 30.
$$\displaystyle \yp = xe^{x^2}+\frac{1}{2}xy$$
$$\displaystyle y(0)=\frac{1}{2}$$
Solution: $$\displaystyle y(x) = \frac{1}{2}(x^2+1)e^{x^2}$$ | |
Many conferences and journal, at least in my field, impose a restriction on the number of pages of the manuscript. I personally don’t agree with the current practice of limiting the page number of manuscript. Although this is somehow understandable for printed journals, this is a totally unnecessary restriction for papers that are indexed online. If I want to read a paper nowadays, I would download an electronic copy from the internet. No one is going to read the actual paper proceedings anymore. That is why the practice of limiting the page number is obsolete.
Anyway when writing a paper, those extra white spaces are like the real state. Here are the five commandments to save more space in latex:
Thou shall play with the margins
Reducing the margins is the easiest and most effective way to reduce the total page number. Many reviewers won’t notice 0.5 inch reduction in margins. Even when they do, they would simply ignore it. To do this, you will need simplemargins.sty package.
\usepackage{simplemargins} \setleftmargin{0.7in} \setrightmargin{0.6in} \settopmargin{.7in} \setbottommargin{0.5in}
The document does not have to be symmetrical. We can have different margins at top and left vs bottom and right side of the paper. English readers usually put more weight on top left side of the document where as Arabic/Farsi readers put more visual weight on the top right side. My general recommendation is to put more margin on the top left side and reduce the margin at bottom and right side of the document.
Thou shall reduce the linespace.
The easiest way to reduce the linespacing is through setspace package.
\usepackage{setspace} \linespread{0.88} %1.3 is one and a half line spacing, 1.6 is double line spacing, etc.
Thou shall reduce the parskip
Latex automatically adds some spaces between paragraphs and lines. It also allows for some space relaxation to make the document look nicer. The default values for parskip settings comes from the latex class. We can save some space by disabling this feature.
\setlength{\parskip}{0cm plus0mm minus0mm}
Thou shall reduce the references font size
Latex supports 10 different font size, ranging from \tiny, \scriptsize, \footnotesize, \small to \huge and \Huge. To save some space, set the font size of algorithms, bibliographies, etc to scriptsize or footnotesize.
\tiny \bibliographystyle{abbrv} \bibliography{references}
You can also remove the spacing between references by adding the following code to your document’s header:
\let\OLDthebibliography\thebibliography \renewcommand\thebibliography[1]{ \OLDthebibliography{#1} \setlength{\parskip}{0pt} \setlength{\itemsep}{0pt plus 0.3ex} }
Thou shall remember the vspace
\vspace gives a very fine grained control over vertical spacing in latex. You can basically put \vspace{-1cm} anywhere that there is some extra white space and it will instantly saves those precocious millimeters! (Think after paragraphs, figures, equations, etc). This is specially useful for blind submission copy, where you can reduce the spacing between title and the abstract using
\title{Some title \vspace{-1ex}} | |
Friday, January 9, 2015
Convert Runtime Database to File Geodatabase
New at 10.3, you can convert a runtime database to a file geodatabase. The tool copies the contents of a runtime geodatabase into a new file geodatabase including the table names, etc... It's very simple to use:
import arcpy
arcpy.CopyRuntimeGdbToFileGdb_conversion(r"D:\data\RuntimeGDBs\delta.geodatabase",
'D:\data\copiedGDBs\deltaFGDB.gdb')
Here we copied a .geodatabase file (runtime database) to a file geodatabase. Now the data captured in the runtime db can be used in desktop or anything else that can't directly access runtime libraries.
Enjoy | |
# Find the kth prime and all the primes below it
It works by finding 2n+1 with n from 1 to k, but not where 2n+1 would be divisible by one of the already known primes. I'd like to know if the program can find every prime, and if so, how it compares to the efficiency of other algorithms.
n=1
k=input()
primes=[2]
def f(n):
b, a = 1, 1
for i in primes:
a=((2*n+1)%i)
b=b*a
return b
while k > n:
if f(n) >= 1:
primes.append((2*n)+1)
n=n+1
else:
n=n+1
k=k+1
print primes
Before we talk about efficiency, let's talk about something very important: readability. It took me an unreasonably long time to understand what it was your code was doing, due to lots of single-letter variable names, confusing flow control, and odd choices for functions.
Checking for primality
Let's start with f. f is apparently determining the primality of n, so let's call it is_prime. Furthermore, it's not actually determining the primality of n, it's determining the primality of 2n+1. That's too much for one function - let's change it's responsibility to take the number it tests. The way you're doing it (taking the % against every prime) is a good algorithm, but then you're multiplying the mods together. That doesn't make sense as an operation. You don't care about the product, you care if any of them are zero! And as soon as one of them is zero, you can return False.
Here's a clearer function:
def is_prime(n):
for p in primes:
if n%p == 0:
return False
return True
Or if you're generator inclined:
def is_prime(n):
return not any(n%p == 0 for p in primes)
Main Loop
We need to find k primes. You are keeping track of this by comparing k to n and occasionally incrementing k. That is confusing. Since we're already keeping track of all the primes, the termination condition should be:
len(primes) >= k
And then we can handle iteration by the odd numbers with itertools.count():
num_primes = input()
primes = [2]
def is_prime(n): ...
for n in itertools.count(start=3, step=2):
if is_prime(n):
primes.append(n)
if len(primes) >= num_primes:
break
This program is far easier to understand.
Efficiency
We could do better in terms of efficiency. At a first approach, we're checking all the odd numbers but we can drop the multiples of 3 earlier by simply writing our own generator that spits out 3, 5, and then alternates adding 2 and 4. That's a quick optimization that let's us keep the same algorithm.
Better would be to use an entirely different, and very well-known algorithm: the Sieve of Eratosthenes. It is a much better algorithm for this problem. | |
# find a basis for U span?
Find a basis for $U= span\{(1,3,2) ,(2,4 ,5) , (5,11,12) , (1,1,3)\}$?
i was trying this question many times, but i could not able to solve this question , i have no any idea or hint to solve this question. I know that all these vector are linearly independent, but i know that all linear independent vector are not basis. I was taking $(1,0,0) , (0,1,0)$ and $(0,0,1)$
If anbody help me i would be very thankful to him thanks in advance
• $(1,3,2) ,(2,4 ,5) , (5,11,12) , (1,1,3)$ are not linearly independent ($4$ vectors, dimension $3$). Try Gaussian elimination to find a basis. – mfl Aug 15 '17 at 8:49
Write the matrix with rows the coordinates of the vectors and perform row reduction. The given vectors corresponding to the non-zero rows make up a basis of the span. Alternatively the non-zero rows are the coordinates of the vectors of a basis (these vectors are linear combinations of the given vectors): $$\begin{bmatrix}1&3&2\\2&4&5\\5&11&12\\1&1&3\end{bmatrix}\rightsquigarrow\begin{bmatrix}1&3&2\\0&-2&1\\0&-4&2\\0&-2&1\end{bmatrix}\rightsquigarrow\begin{bmatrix}1&3&2\\0&-2&1\\0&0&0\\0&0&0\end{bmatrix}$$ Thus, the span of these vectors has dimension $2$, and we've found two bases:
• $\;\bigl\{(1,3,2), \,(2,4,5)\bigr\}$,
• $\;\bigl\{(1,3,2), \,(0,-2,1)\bigr\}$. | |
## 187 – Selected solutions to problems from Chapter 1
September 19, 2011
Professor Warren Esty, one of the authors of our main textbook, has made available a list of solutions to some of the problems from Chapter 1. They are most of the odd numbered problems. Please let him (or me) know if you find errors or typos, or if you have suggestions for alternative solutions or different approaches.
Solutions (Chapter 1)
## 187 – The resolution method
September 19, 2011
This note is based on lecture notes for the Caltech course Math 6c, prepared with A. Kechris and M. Shulman.
We would like to have a mechanical procedure (algorithm) for checking whether a given set of formulas logically implies another, that is, given ${A_1, \dots, A_n, A}$, whether
$\displaystyle (A_1\land \dots \land A_n)\Rightarrow A$
is a tautology (i.e., it is true under all truth-value assignments.)
This happens iff
$\displaystyle A_1\wedge\dots\wedge A_n\wedge\neg A\text{ is unsatisfiable.}$
So it suffices to have an algorithm to check the (un)satisfiability of a single propositional formula. The method of truth tables gives one such algorithm. We will now develop another method which is often (with various improvements) more efficient in practice.
It will be also an example of a formal calculus. By that we mean a set of rules for generating a sequence of strings in a language. Formal calculi usually start with a certain string or strings as given, and then allow the application of one or more “rules of production” to generate other strings.
A formula ${A}$ is in conjunctive normal form iff it has the form
$\displaystyle A_1\land A_2\land\dots\land A_n$
where each ${A_i}$ has the form
$\displaystyle B_1\lor B_2\lor\dots\lor B_k$
and each ${B_i}$ is either a propositional variable, or its negation. So ${A}$ is in conjunctive normal form iff it is a conjunction of disjunctions of variables and negated variables. The common terminology is to call a propositional variable or its negation a literal.
Suppose ${A}$ is a propositional statement which we want to test for satisfiability. First we note (without proof) that although there is no known efficient algorithm for finding ${A'}$ in cnf (conjunctive normal form) equivalent to ${A}$, it is not hard to show that there is an efficient algorithm for finding ${A^*}$ in cnf such that:
${A}$ is satisfiable iff ${A^*}$ is satisfiable.
(But, in general, ${A^*}$ has more variables than ${A}$.)
So from now on we will only consider ${A}$s in cnf, and the Resolution Method applies to such formulas only. Say
${A=(\ell_{1,1}\vee\dots\vee\ell_{1,n_1})\wedge\dots \wedge (\ell_{k,1}\vee\dots\vee\ell_{k,n_k})}$
with ${\ell_{i,j}}$ literals. Since order and repetition in each conjunct (1):
$\displaystyle \ell_{i,1}\vee\dots\vee\ell_{i,n_i}$
are irrelevant, we can replace (1) by the set of literals
$\displaystyle c_i=\{\ell_{i,1},\ell_{i,2},\dots ,\ell_{i,n_i}\}.$
Such a set of literals is called a clause. It corresponds to the formula (1). So the formula ${A}$ above can be simply written as a set of clauses (again since the order of the conjunctions is irrelevant): ${C=\{c_1,\dots ,c_k\}}$
$=\{\{\ell_{i,1},\dots \ell_{i,n_1}\},\dots ,\{\ell_{k,1},\dots , \ell_{k,n_k}\}\}$
Satisfiability of ${A}$ means then simultaneous satisfiability of all of its clauses ${c_1,\dots ,c_k}$, i.e., finding a valuation ${\nu}$ which makes ${c_i}$ true for each ${i}$, i.e., which for each ${i}$ makes some ${\ell_{i,j}}$ true.
Example 1 $A=(p_1\vee\neg p_2)\wedge (p_3\vee p_3)$
$c_1=\{p_1,\neg p_2\}$
$c_2=\{p_3\}$
$C=\{\{p_1,\neg p_2\},\{p_3\}\}.$
From now on we will deal only with a set of clauses ${C=\{c_1, c_2,\dots,c_n \}}$. It is possible to consider also infinite sets ${C}$, but we will not do that here.
Satisfying ${C}$ means (again) that there is a valuation which satisfies all ${c_1,c_2,\dots}$, i.e. if ${c_i=\ell_{i,1}\vee\dots \vee\ell_{i,n_i}}$, then for all ${i}$ there is ${j}$ so that it makes ${\ell_{i,j}}$ true.
Notice that if the set of clauses ${C_A}$ is associated as above to ${A}$ (in cnf) and ${C_B}$ to ${B}$, then
$\displaystyle A\wedge B\text{ is satisfiable iff }C_A\cup C_B\text{ is satisfiable.}$
By convention we also have the empty clause ${\Box}$, which contains no literals. The empty clause is (by definition) unsatisfiable, since for a clause to be satisfied by a valuation, there has to be some literal in the clause which it makes true, but this is impossible for the empty clause, which has no literals. For a literal ${u}$, let ${\bar u}$ denote its “conjugate”, i.e. $\bar u=\neg p,$ if $u=p,$ and $\bar u=p$ if $u=\neg p.$
Definition 1 Suppose now ${c_1,c_2,c}$ are three clauses. We say that ${c}$ is a resolvent of ${c_1,c_2}$ if there is a ${u}$ such that ${u\in c_1}$, ${\bar u\in c_2}$ and
$\displaystyle c=(c_1\setminus\{u\})\cup (c_2\setminus\{\bar u\}).$
We allow here the case ${c=\Box}$, i.e. ${c_1=\{u\},\ c_2= \{\bar u\}}$.
## 187 – The P vs NP problem
September 19, 2011
This note is based on lecture notes for the Caltech course Math 6c, prepared with A. Kechris and M. Shulman.
1. Decision problems
Consider a finite alphabet ${A}$, and “words” on that alphabet (the “alphabet” may consist of digits, of abstract symbols, of actual letters, etc).
We use the notation ${A^*}$ to indicate the set of all “words” from the alphabet ${A}$. Here, a word is simply a finite sequence of symbols from ${A}$. For example, if ${A}$ is the usual alphabet, then
$\displaystyle awwweeeedddfDDkH$
would be a word.
We are also given a set ${V}$ of words, and we say that the words in ${V}$ are valid. (${V}$ may be infinite.)
In the decision problem associated to ${V}$, we are given as input a word in this alphabet. As output we say yes or no, depending on whether the word is in ${V}$ or not (i.e., whether it is “valid”).
We are interested in whether there is an algorithm that allows us to decide the right answer.
## 414/514 – Metric spaces
September 19, 2011
This is homework 2, due Monday September 26 at the beginning of lecture.
Let $(X,d)$ be a metric space.
• Show that if $d_1:X\times X\to{\mathbb R}$ is defined by $\displaystyle d_1(x,y)=\frac{d(x,y)}{1+d(x,y)}$, then $d_1$ is also a metric on $X$.
• Show that if $U$ is open in $(X,d)$ then it is open in $(X,d_1)$, and viceversa.
Recall that $U$ is open iff it is a union of open balls. Use this to explain why it suffices to show that if $U$ is open in $(X,d)$ then for any $x\in U$ there is an $\epsilon>0$ such that
$B_\epsilon^{d_1}(x):=\{y\mid d_1(x,y)<\epsilon\}\subseteq U$,
and similarly, if $V$ is open in $(X,d_1)$ then for any $z\in V$ there is a $\delta>0$ such that
$B_\delta^d(z):=\{w\mid d(z,w)<\delta\}\subseteq V$.
In turn, explain why to show this it suffices to prove that for any $x\in X$ and any $\eta>0$ there is a $\rho>0$ such that
$B^{d_1}_\eta(x)\supseteq B^d_\rho(x)$
and, similarly, for any $\tau>0$ there is a $\mu>0$ such that
$B_\tau^d(x)\supseteq B^{d_1}_\mu(x)$.
Finally, prove this by showing that we can take $\rho=\epsilon$ (no matter what $x$ is) and similarly, find an appropriate $\mu$ that works for $\tau$ (again, independently of $x$).
• Illustrate the above in ${\mathbb R}^2$ as accurately as possible.
• Suppose that a sequence $(x_n)_{n\in{\mathbb N}}$ converges to $x$ in $(X,d)$ and to $x'$ in $(X,d_1)$. Show that $x=x'$.
• Is it true that a sequence $(x_n)_{n\in{\mathbb N}}$ is Cauchy in $(X,d)$ iff it is Cauchy in $(X,d_1)$? (Give a proof or else exhibit a counterexample, with a proof that it is indeed a counterexample.)
• $(*)$ Show that any dense $G_\delta$ subset of ${\mathbb R}$ has the same size as ${\mathbb R}$.
## Alan Turing year
September 5, 2011
Alan Turing was born in London on June 23, 1912. The Alan Turing Year 2012 celebrates his life and scientific achievements. Here is a link to the official page.
Widely recognize for his contributions during the Second War to the solving of German ciphers, Turing is also responsible for the development of computer science, having isolated the key notion of Turing machine, the mathematical formalization of algorithm. He is also known for the notion of Turing test, aimed at “recognizing” artificial intelligence.
Unfortunately, Turing’s story is marred by his end. Rather than being held as a war hero (or a renown academic), being a homosexual, in 1952 Turing was charged with gross indecency under Section 11 of the Criminal Law Amendment Act of 1885 and required to undergo hormonal treatment, including chemical castration. Turing committed suicide on June 7, 1954. On September 10, 2009, the British government finally issued an apology.
Here is a link to the excellent Wikipedia entry on Turing. And here is a link to the Alan Turing page, maintained by Andrew Hodges, author of Alan Turing: The Enigma.
## 414/514 – Cauchy sequences
September 2, 2011
This is homework 1, due Friday September 9 at the beginning of lecture.
We define absolute value as usual: Given a real $x$, we say that ${}|x|$ is $x$ if $x\ge0$ and is $-x$ otherwise.
Absolute values have useful properties: ${}|x|=|-x|\ge0$ for any $x$. Also, ${}|x|=0$ iff $x=0$. The key property is the triangle inequality: ${}|a+b|\le|a|+|b|$.
Formally, a sequence is a function $s:{\mathbb N}\to{\mathbb R}$. As usual, we write the sequence as $s_1,s_2,\dots$ rather than $s(0),s(1),\dots$
A sequence $s$ is a Cauchy sequence iff for all $\epsilon>0$ there is an $N\in{\mathbb N}$ such that whenever $n,m\in{\mathbb N}$ and $n,m>N$, we have $|s_n-s_m|<\epsilon.$
A sequence $s$ converges iff there is a real $r$ such that for all $\epsilon>0$ there is an $N\in{\mathbb N}$ such that whenever $n\in{\mathbb N}$ and $n>N$, we have $|r-s_n|<\epsilon$.
Note that these concepts also make sense in ${\mathbb Q}$. Now we require all the $s_n$ to be rational, and we require $\epsilon$ and $r$ to be rational as well.
• Show that if a sequence converges, then it is Cauchy.
• Give an example of a Cauchy sequence in ${\mathbb Q}$ that does not converge.
• Show that any Cauchy sequence in ${\mathbb R}$ converges.
Cauchy’s way of defining the reals was to use Cauchy sequences as the basic building blocks rather than cuts. Again, the idea is that we want to have all the limits, and in ${\mathbb Q}$ some of these limits are missing. In the case of cuts, the way of solving the presence of gaps in ${\mathbb Q}$ was by giving names to all the gaps (the cuts), and adding the names. The easiest repair to the lack of limits here will be the same: We give a name to the limits (the sequences themselves) and the reals will be just the sequences.
There is a problem here that does not occur with the construction using cuts, namely different sequences may have the same limit. We should identify all of them.
Recall that an equivalence relation on a set $X$ is a binary relation $\sim$ that is:
1. Reflexive: For any $x\in X$, $x\sim x$.
2. Symmetric: For any $x,y\in X$, if $x\sim y$ then also $y\sim x$.
3. Transitive: For any $x,y,z\in X$, if $x\sim y$ and $y\sim z$, then $x\sim z$.
If $\sim$ is an equivalence relation, the equivalence classes determined by $\sim$ are the sets ${}[x]=\{y\in X\mid x\sim y\}$. An intuitive way of thinking about this is that we are looking at $X$ from a distance, and so we cannot distinguish points that are close to one another, we just see them smashed together as a single point. Here, two points $x,y$ are close iff $x\sim y$.
Let $s$ and $t$ be two Cauchy sequences of rationals. Say that $s\sim t$ iff $s-t$ converges to $0$. Here, of course, $s-t$ is the sequence $r_1,r_2,\dots$ with $r_n=s_n-t_n$.
• Show that $\sim$ is an equivalence relation. Check that any Cauchy sequence $s$ is equivalent to infinitely many other sequences.
• Define ${\mathbb R}$ as the set of equivalence classes of the relation $\sim$. The elements of ${\mathbb R}$ are then Cauchy sequences or, more precisely, collections of Cauchy sequences. A typical element of ${\mathbb R}$ is a class ${}[s]$, and we think of ${}[s]$ as the limit of $s$. Of course, we have a copy of ${\mathbb Q}$ inside ${\mathbb R}$: We can identify the rational $q$ with the class $q^*=[s]$ of all sequences $s$ that converge to $q$.
• Define $+,\times,<$ in ${\mathbb R}$ and verify that with these definitions we have an ordered field.
• Verify that ${\mathbb R}$ is complete, meaning that the least upper bound property holds.
This gives a second sense in which ${\mathbb R}$ is complete: It contains the limits of all Cauchy sequences. A small but important point not mentioned above is the following: Given a sequence $s$ of rationals, let $s'$ be its “copy” inside ${\mathbb R}$, i.e., $(s')_n=s_n^*$. Then $s'$ is Cauchy iff $s$ is Cauchy, and $s$ converges to a rational $q$ iff $s'$ converges to $q^*$.
## Sets and games
September 1, 2011
I just finished a talk at the Department Colloquium on Sets and Games. I have posted the slides in my talks page. About a year ago I gave a talk in the Graduate Student Seminar on Determinacy (also available in my talks page). Though that talk was significantly less technical, it covers a nice bit of history that I had to skip in this case, and I think the two complement each other well.
The talk today covered some of my recent results with Richard Ketchersid on the structure of natural models of determinacy. (I have discussed technical details of the proofs in other talks, also available at the page linked to above.) At the end I touched on some recent results with Boban Velickovic on failures of square principles (inspired by similar recent results of Dilip Raghavan), and on the results of the SQuaRE group I am a part of: | |
# Generating Reflectance Curves from sRGB Triplets (Page 2)
#### sRGB Triplet from a Reflectance Curve
The reverse process of computing an sRGB triplet from a reflectance curve is straightforward. It requires two main components: (1) a mathematical model of the “standard observer,” which is an empirical mathematical relationship between color stimuli and color sensation (tristimulus values), and (2) a definition of the sRGB color space that specifies the reference illuminant and the mathematical transformation between tristimulus values and sRGB values.
The linear transformation relating a color stimulus, , to its corresponding tristimulus values, , is
The column vector has three elements, , , and . The matrix has three rows (called the three “color matching functions”) and columns, where is the number of discretized wavelength bands. In this study, all computations are performed with 36 wavelength bands of width 10 nm, running over the range 380 nm to 730 nm. The stimulus vector also has components, representing the total power of all wavelengths within each band. The specific color matching functions I use in this work are the CIE 1931 color matching functions. (Note that I’m using the symbol here; the standard matrix is x 3, and so indicates that it has been transposed.)
The stimulus vector can be constructed as the product of an diagonal illuminant matrix, , and an reflectance vector, . The computation of is usually normalized so the tristimulus value is equal to 1 when is a perfect reflector (contains all 1s). The normalizing factor, , is thus the inner product of the second row of and the illuminant vector, , yielding the alternate form of the tristimulus value equation
The transformation from tristimulus values to sRGB is a two-step process. First, a 33 linear transformation, , is applied to convert to , which is a triplet of “linear RGB” values:
The second step is to apply a “gamma correction” to the linear values, also known as “companding” or applying the “color component transfer function.” This is how it is done: for each , , and component of , let’s generically call it , if , use , otherwise use . This gives sRGB values in the range of 0 to 1. To convert them to the alternate integer range of 0 to 255 (used in most 24-bit color devices), we multiply each by 255 and round to the nearest integer.
sRGB to rgb
The inverse operation of converting sRGB to , expressed in code, is:
sRGB=sRGB/255; % convert 0-255 range to 0-1 range for i=1:3 if sRGB(i)<0.04045 rgb(i)=sRGB(i)/12.92; else rgb(i)=((sRGB(i)+0.055)/1.055)^2.4; end end
The expression relating and above can be simplified by combining the three matrices and the normalizing factor into a single matrix, so that
The formal definition of the sRGB color space uses an illuminant similar to daylight, called D65, as its “reference” illuminant. Here are the specific values for the matrix, the matrix, the D65 vector, and the matrix. The normalizing factor, , has a value of 10.5677. Most of the RGB-related theory I present here I learned from Bruce Lindbloom’s highly informative website.
Now that we have a simple expression for computing sRGB from a reflectance curve, we can use that as the basis of doing the opposite, computing a reflectance curve from an sRGB triplet. In the sections that follow, I will present five different algorithms for doing this. Each has its strengths and weaknesses. Once they are presented, I will then compare them to each other and to reflectance curves found in nature. | |
# Wavelengths of the visible spectrum
Why is data about wavelengths of different colors and the visible spectrum in general so different in different sources? On Wikipedia, the numbers differ by up to $\pm\mathrm{30~nm}$.
• It may all be because there's no distinguishable "line" between for example blue and violet. I imagine this is like saying all compounds with species with electronegativity difference of more than 1.9 are ionic; which is very crude, at times leads to wrong results and confusion. – M.A.R. Jun 18 '15 at 16:24
• Many different light wavelengths can produce the same perception of color. Besides, the color perception is not purely physical process, but rather psychophysical and physiological part can differ from one person to another. – Wildcat Jun 18 '15 at 16:29
• @Brian, I knew someone gonna say something like this. But using that logic I can copy Biology.SE site almost entirely and paste its content here, since ultimately all biological processes are rooted in chemical (and physical) processes. But we don't do so, we distinguish the scientific disciplines by focusing on the most important aspects and abstracting out non-essentials. And in this question the actual chemical way receptors work is non-essential. And besides, electronic excitation is a physical process. – Wildcat Jun 18 '15 at 16:46
• I'm voting to close this question as off-topic because I don't feel that it is about chemistry. There is much-much more physics & biology out there. – Wildcat Jun 18 '15 at 16:53
• There is a lot of distracting comment here. Distracting because there is a simple and not-entirely-unrelated-to-chemistry answer. Physiology and philosophy are irrelevant. The reason there are different definitions is because the colour scale is continuous and any division into simple colours is arbitrary and different people's definitions will differ. The specific wavelength of the bright lines in the emission spectrum of sodium, for example, is unambiguous, but whether we classify them as orange or yellow is arbitrary. – matt_black Jun 18 '15 at 23:15
The comments on the question discuss this may not be about chemistry.
One could argue that colors are chemistry related because they are associated with chemical substances: like minerals, chlorophyl, rust, dyes etc.
So colors seem to be properties of substances.
And that's fundamentally wrong.
Color is a function of the visual system of the viewer!
If that sounds surprising, think of color blind people, which see different colors.
Also, there are not just us - there are other animals, with different visual spectrum, partly including what we call UV.
## Colors are not well defined
Actually, there are more than enough ways, even standards, to define which part of the rainbow is green, exactly - people working with visual media like print design can tell details.
And there are way more details than anyone could expect.
So, without a common definition of the colors, it's not the wavelengths that vary in different sources - it's the colors themselves.
To illustrate that color is more complex when looking closely, below you see a diagram defining color names. It is part of a formal standard which is in heavy use: | |
# 400V digital potentiometer
Status
Not open for further replies.
#### lordsandman
##### Newbie level 3
hi everyone,
i want to design a digital power supply with 400V, max 2mA output and 0.1V resolution(12-bit). i know its hard to control 400V with MCU but output current so tiny. is it possible to divide 400V by digital pots? any specific high voltage IC or a method for transforming high voltages appropriate to digital ICs?
just an idea, it may be absurd
#### IanP
How about constructing a ladder consisting of 12 resistors connected in parallel with 12 reed relays that can be driven directly by a microcontroller?
:wink:
IanP
lordsandman
### lordsandman
Points: 2
#### lordsandman
##### Newbie level 3
hey Ian,
thats a solution but i need 10 independent outputs. it means 120 relays, and will a huge circuit, not practical in my case.
can i amplify a 10V DAC output to 400V(with 40 gain and regulated)?
#### FvM
##### Super Moderator
Staff member
Of course, you can use HV OPs, e.g. from APEX Apex Precision Power
Or design your own amplifier using discrete devices for the HV part.
lordsandman
### lordsandman
Points: 2
#### lordsandman
##### Newbie level 3
from the apex link, PA15 is what i need but 200$unit price. 10x200=2000$
#### FvM
##### Super Moderator
Staff member
I would think of an complementary class AB or B output stage with HV transistors as used in an audio amplifier with an OP gain stage. You need at least two output transistors, a sriver transistor and a current source transistor, two npn and two pnp HV transistors.
lordsandman
Points: 2 | |
How do you find the derivative of (2e^(3x) + 2e^(-2x))^4?
Aug 14, 2017
$4 {\left(2 {e}^{3 x} + 2 {e}^{- 2 x}\right)}^{3} \left(6 {e}^{3 x} - 4 {e}^{- 2 x}\right)$
Explanation:
$\text{differentiate using the "color(blue)"chain rule}$
$\text{given "y=f(g(x))" then}$
$\frac{\mathrm{dy}}{\mathrm{dx}} = f ' \left(g \left(x\right)\right) \times g ' \left(x\right) \leftarrow \text{ chain rule}$
$f \left(g \left(x\right)\right) = {\left(2 {e}^{3 x} + 2 {e}^{- 2 x}\right)}^{4}$
$\Rightarrow f ' \left(g \left(x\right)\right) = 4 {\left(2 {e}^{3 x} + 2 {e}^{- 2 x}\right)}^{3}$
$g \left(x\right) = 2 {e}^{3 x} + 2 {e}^{- 2 x}$
$\Rightarrow g ' \left(x\right) = 2 {e}^{3 x} . \frac{d}{\mathrm{dx}} \left(3 x\right) + 2 {e}^{- 2 x} . \frac{d}{\mathrm{dx}} \left(- 2 x\right)$
$\textcolor{w h i t e}{\Rightarrow g ' \left(x\right)} = 6 {e}^{3 x} - 4 {e}^{- 2 x}$
$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}} = 4 {\left(2 {e}^{3 x} + 2 {e}^{- 2 x}\right)}^{3} \left(6 {e}^{3 x} - 4 {e}^{- 2 x}\right)$ | |
Hide
# Problem ILinear Recurrences
## Input
The first line of input contains an integer $1 \le N \le 40$, the degree of the recurrence. The next line of input contains $N+1$ integers $a_0, a_1, \ldots , a_ N$ indicating that the linear recurrence is $x_{t} = a_0 + \sum _{i=1}^ N a_ i x_{t-i}$. The next line contains $N$ integers $x_0, \ldots , x_{N-1}$ giving the initial values for the recursion. All the coefficients $a_0, \ldots , a_ N$ and initial values $x_0, \ldots , x_{N-1}$ are integers between $-10^9$ and $10^9$ (inclusive).
The next line contains an integer $1 \le Q \le 10$, the number of queries. Then follow $Q$ lines of queries. Each query consists of two integers $T$, $M$ where $0 \le T \le 10^{18}$ gives the index and $1 \le M \le 10^{9}$ is a moduli.
## Output
For each query $T$, $M$, output a line containing $x_ T \bmod M$.
Sample Input 1 Sample Output 1
2
0 1 1
0 1
6
1 100000
2 100000
3 100000
4 100000
5 100000
6 100000
1
1
2
3
5
8
Sample Input 2 Sample Output 2
2
5 7 9
36713 5637282
4
1 10000
1375 1
3781 23
34683447233 1571385
7282
0
16
299255
Sample Input 3 Sample Output 3
3
1 2 3 4
0 0 0
1
42424242424242 1000000
552200
CPU Time limit 4 seconds
Memory limit 1024 MB
Statistics Show | |
# Determine if the following change is an oxidation, a reduction, or neither
The following reaction is not a complete reaction. Determine if the change is an oxidation, a reduction, or neither: $$\ce{CrO4^{-2} -> Cr2O7^{-2}}$$
I know that the answer to this problem is neither, but I don't have any idea as to how that answer is reached. Could anyone give me a hint? I am familiar with oxidation, reduction, and oxidation numbers.
• If you are familiar with oxidation numbers, what are the oxidation numbers of the chromium atoms in each compound? – jerepierre Feb 26 '15 at 1:20
• @jerepierre in the first compound +6, in the second compound +6 as well. – McB Feb 26 '15 at 1:22
• @jerepierre Does this mean that since there was no change in oxidation numbers, it is neither? – McB Feb 26 '15 at 1:27
As pointed out in the comments and the other answer, it is neither an oxidation, nor a reduction.
Q: So, what is it?
A: It is a condensation: two molecules of an ortho acid combine under loss of one molecule of water to yield a diacid (pyro acid).
This is frequently found in oxoacids, see
• $\ce{H3PO4}$ (phosphoric acid) and $\ce{H4P2O7}$ (diphosphoric acid)
• $\ce{H3AsO4}$ (arsenic acid) and $\ce{H4AsO7}$ (diarsenic acid)
• $\ce{H2SO4}$ (sulfuric acid) and $\ce{H2S2O7}$ (disulfuric acid)
Find the oxidation numbers of the substances in each compound.
You should find the oxidation number of Cr in each compound to be +6, and the oxidation number of O to be -2. Since there is no change in the oxidation numbers, this reaction is neither an oxidation or reduction. | |
# 21.3 Radioactive decay (Page 5/21)
Page 5 / 21
Because each nuclide has a specific number of nucleons, a particular balance of repulsion and attraction, and its own degree of stability, the half-lives of radioactive nuclides vary widely. For example: the half-life of ${}_{\phantom{\rule{0.5em}{0ex}}83}^{209}\text{Bi}$ is 1.9 $×$ 10 19 years; ${}_{\phantom{\rule{0.5em}{0ex}}94}^{239}\text{Ra}$ is 24,000 years; ${}_{\phantom{\rule{0.5em}{0ex}}86}^{222}\text{Rn}$ is 3.82 days; and element-111 (Rg for roentgenium) is 1.5 $×$ 10 –3 seconds. The half-lives of a number of radioactive isotopes important to medicine are shown in [link] , and others are listed in Appendix M .
Half-lives of Radioactive Isotopes Important to Medicine
Type The “m” in Tc-99m stands for “metastable,” indicating that this is an unstable, high-energy state of Tc-99. Metastable isotopes emit γ radiation to rid themselves of excess energy and become (more) stable. Decay Mode Half-Life Uses
F-18 β + decay 110. minutes PET scans
Co-60 β decay, γ decay 5.27 years cancer treatment
Tc-99m γ decay 8.01 hours scans of brain, lung, heart, bone
I-131 β decay 8.02 days thyroid scans and treatment
Tl-201 electron capture 73 hours heart and arteries scans; cardiac stress tests
Several radioisotopes have half-lives and other properties that make them useful for purposes of “dating” the origin of objects such as archaeological artifacts, formerly living organisms, or geological formations. This process is radiometric dating and has been responsible for many breakthrough scientific discoveries about the geological history of the earth, the evolution of life, and the history of human civilization. We will explore some of the most common types of radioactive dating and how the particular isotopes work for each type.
The radioactivity of carbon-14 provides a method for dating objects that were a part of a living organism. This method of radiometric dating, which is also called radiocarbon dating or carbon-14 dating, is accurate for dating carbon-containing substances that are up to about 30,000 years old, and can provide reasonably accurate dates up to a maximum of about 50,000 years old.
Naturally occurring carbon consists of three isotopes: ${}_{\phantom{\rule{0.5em}{0ex}}6}^{12}\text{C}\text{,}$ which constitutes about 99% of the carbon on earth; ${}_{\phantom{\rule{0.5em}{0ex}}6}^{13}\text{C}\text{,}$ about 1% of the total; and trace amounts of ${}_{\phantom{\rule{0.5em}{0ex}}6}^{14}\text{C}.$ Carbon-14 forms in the upper atmosphere by the reaction of nitrogen atoms with neutrons from cosmic rays in space:
${}_{\phantom{\rule{0.5em}{0ex}}7}^{14}\text{N}+{}_{0}^{1}\text{n}\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{}_{\phantom{\rule{0.5em}{0ex}}6}^{14}\text{C}+{}_{1}^{1}\text{H}$
All isotopes of carbon react with oxygen to produce CO 2 molecules. The ratio of ${}_{\phantom{\rule{0.5em}{0ex}}6}^{14}\text{C}{\text{O}}_{2}$ to ${}_{\phantom{\rule{0.5em}{0ex}}6}^{12}\text{C}{\text{O}}_{2}$ depends on the ratio of ${}_{\phantom{\rule{0.5em}{0ex}}6}^{14}\text{C}\text{O}$ to ${}_{\phantom{\rule{0.5em}{0ex}}6}^{12}\text{C}\text{O}$ in the atmosphere. The natural abundance of ${}_{\phantom{\rule{0.5em}{0ex}}6}^{14}\text{C}\text{O}$ in the atmosphere is approximately 1 part per trillion; until recently, this has generally been constant over time, as seen is gas samples found trapped in ice. The incorporation of ${}_{\phantom{\rule{0.5em}{0ex}}6}^{14}\text{C}{}_{\phantom{\rule{0.5em}{0ex}}6}^{14}\text{C}{\text{O}}_{2}$ and ${}_{\phantom{\rule{0.5em}{0ex}}6}^{12}\text{C}{\text{O}}_{2}$ into plants is a regular part of the photosynthesis process, which means that the ${}_{\phantom{\rule{0.5em}{0ex}}6}^{14}\text{C}\text{:}\phantom{\rule{0.2em}{0ex}}{}_{\phantom{\rule{0.5em}{0ex}}6}^{12}\text{C}$ ratio found in a living plant is the same as the ${}_{\phantom{\rule{0.5em}{0ex}}6}^{14}\text{C}\text{:}\phantom{\rule{0.2em}{0ex}}{}_{\phantom{\rule{0.5em}{0ex}}6}^{12}\text{C}$ ratio in the atmosphere. But when the plant dies, it no longer traps carbon through photosynthesis. Because ${}_{\phantom{\rule{0.5em}{0ex}}6}^{12}\text{C}$ is a stable isotope and does not undergo radioactive decay, its concentration in the plant does not change. However, carbon-14 decays by β emission with a half-life of 5730 years:
what is zero gravity
every object is that zero gravity
Rabiu
Probably when an object is in space and there are no nearby masses that pull her, and exert gravity
Abdelkarim
Alright. .good job
Rabiu
And all majesty to God, (وَهُوَ ٱلَّذِی خَلَقَ ٱلَّیۡلَ وَٱلنَّهَارَ وَٱلشَّمۡسَ وَٱلۡقَمَرَۖ كُلࣱّ فِی فَلَكࣲ یَسۡبَحُونَ) [سورة الأنبياء 33 And it is He who created the night and the day and the sun and the moon; all [heavenly bodies] in an orbit are swimming. General theory of relativity in Qur
Abdelkarim
what is lattice energy
why is CO a neutral oxide and CO2 an acidic oxide
Because when CO2 dissolves in water forming a weak acid. CO does not dissolve in water as it has strong triple bond.
Abdelkarim
What is acid
which donate H+ or accept lone pair of electron
Kajal
kinetic theory of matter and gas law
hi
Victoria
pls explain
Victoria
what is clay
material containing clay minerals. Clays develop plasticity when wet, due to a molecular film of water surrounding the clay particles, but become hard, brittle and non–plastic upon drying or firing. Most pure clay minerals are white or light-coloured, but natural clays show a variety of colours
Abdelkarim
due iron oxide. The four types of clay are Earthenware clay, Stoneware clay, Ball clay, and Porcelain. All of them can be used to make pottery, but the end result would differ a lot thanks to their different textures, colors, and flexibilities.
Abdelkarim
And do you know that god has created human from clay (وَلَقَدۡ خَلَقۡنَا ٱلۡإِنسَـٰنَ مِن صَلۡصَـٰلࣲ مِّنۡ حَمَإࣲ مَّسۡنُونࣲ) [سورة الحجر 26] And We did certainly create man out of clay from an altered black mud. You can install Quran from paly store for free with translations.
Abdelkarim
darw a periodic table
draw a periodic table
Hazard
You will arrange the elements into row and coloumns according to increasing proton number. You may want to use symbols or their names. Hydrogen, Helium, etc. God has created all these elements from nothing, in Islam we know God is the creator.
Abdelkarim
why are you drawing a periodic table? why not just print one from the internet and use as a reference
Jakhari
Great thought
Bright
how are you?
Marina
am fine
Agbo
Marina
my name is amel
Farid
l use the email of my husband
Farid
Define organic chemistry
It is the chemistry concerning molecules that have Carbon skeletons and hydrogen atoms. We find organic molecules like in plants, living derivatives, etc.
Abdelkarim
what's matter
Anything that can be to cutting from all dimensions to halve. So you end up with 4 cubes of 5 cm side. Repeat with one of the cubes. 10, 5, 2.5, .., 0 1st 2nd 3rd Nth Un= a(r) ^ n-1
Abdelkarim
Anything that has mass and can reflect or absorb waves. GOD created everything from nothing only he can destroy it as prooved.
Abdelkarim
Suppose you have a cube of side 10 cm. Then you start cutting from all dimensions to halve. So you end up with 4 cubes of 5 cm side. Repeat with one of the cubes. 10, 5, 2.5, .., 0 1st 2nd 3rd Nth Un= a(r) ^ n-1 0= 10 (1/2)^n-1 0= (1/2) ^ n-1 Log0= (n-1) Log(1/2) - infinity =( n-1)
Abdelkarim
matter is anything that has mass,volume and can occupy space
Getrude
what is electrolysis
good equation
Aliyu
differenciate between fat and oil
what is the meaning of coordinate bond
It is the alternative for dative which is a covalent bond but both electrons of the pair are from shared from the same (one) atom.
Abdelkarim
can someone please tell me what does an Entropy means
what is chemistry?
what is chemistry
Afiwape
chemistry is a brach of science which deal with the study of the nature, composition structure and with the force that hold the structure together and the change matter will undergo undedifferent conditions
Afiwape
And god has created everything from nothing
Abdelkarim | |
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# Linear system
Article Id: WHEBN0000722503
Reproduction Date:
Title: Linear system Author: World Heritage Encyclopedia Language: English Subject: Collection: Systems Theory Publisher: World Heritage Encyclopedia Publication Date:
### Linear system
A linear system is a mathematical model of a system based on the use of a linear operator. Linear systems typically exhibit features and properties that are much simpler than the general, nonlinear case. As a mathematical abstraction or idealization, linear systems find important applications in automatic control theory, signal processing, and telecommunications. For example, the propagation medium for wireless communication systems can often be modeled by linear systems.
## Contents
• Definition 1
• Time-varying impulse response 2
• Time-varying convolution integral 3
• Continuous time 3.1
• Discrete time 3.2
• Causality 4
## Definition
A general deterministic system can be described by an operator, H, that maps an input, x(t), as a function of t to an output, y(t), a type of black box description. Linear systems satisfy the property of superposition. Given two valid inputs
x_1(t) \,
x_2(t) \,
as well as their respective outputs
y_1(t) = H \left \{ x_1(t) \right \}
y_2(t) = H \left \{ x_2(t) \right \}
then a linear system must satisfy
\alpha y_1(t) + \beta y_2(t) = H \left \{ \alpha x_1(t) + \beta x_2(t) \right \}
for any scalar values \alpha \, and \beta \,.
The system is then defined by the equation H(x(t)) = y(t), where y(t) is some arbitrary function of time, and x(t) is the system state. Given y(t) and H, x(t) can be solved for. For example, a simple harmonic oscillator obeys the differential equation:
m \frac{d^2(x)}{dt^2} = -kx.
If
H(x(t)) = m \frac{d^2(x(t))}{dt^2} + kx(t),
then H is a linear operator. Letting y(t) = 0, we can rewrite the differential equation as H(x(t)) = y(t), which shows that a simple harmonic oscillator is a linear system.
The behavior of the resulting system subjected to a complex input can be described as a sum of responses to simpler inputs. In nonlinear systems, there is no such relation. This mathematical property makes the solution of modelling equations simpler than many nonlinear systems. For time-invariant systems this is the basis of the impulse response or the frequency response methods (see LTI system theory), which describe a general input function x(t) in terms of unit impulses or frequency components.
Typical differential equations of linear time-invariant systems are well adapted to analysis using the Laplace transform in the continuous case, and the Z-transform in the discrete case (especially in computer implementations).
Another perspective is that solutions to linear systems comprise a system of functions which act like vectors in the geometric sense.
A common use of linear models is to describe a nonlinear system by linearization. This is usually done for mathematical convenience.
## Time-varying impulse response
The time-varying impulse response h(t2,t1) of a linear system is defined as the response of the system at time t = t2 to a single impulse applied at time t = t1. In other words, if the input x(t) to a linear system is
x(t) = \delta(t-t_1) \,
where δ(t) represents the Dirac delta function, and the corresponding response y(t) of the system is
y(t) |_{t=t_2} = h(t_2,t_1) \,
then the function h(t2,t1) is the time-varying impulse response of the system.
## Time-varying convolution integral
### Continuous time
The output of any continuous time linear system is related to the input by the time-varying convolution integral:
y(t) = \int_{-\infty}^{\infty} h(t,s) x(s) ds
or, equivalently,
y(t) = \int_{-\infty}^{\infty} h(t,t-\tau) x(t-\tau) d \tau
where
s = t-\tau
### Discrete time
The output of any discrete time linear system is related to the input by the time-varying convolution sum:
y[n] = \sum_{k=-\infty}^{\infty} { h[n,k] x[k] }
or equivalently,
y[n] = \sum_{m=-\infty}^{\infty} { h[n,n-m] x[n-m] }
where
k = n-m \,
represents the lag time between the stimulus at time m and the response at time n.
## Causality
A linear system is causal if and only if the system's time varying impulse response is identically zero whenever the time t of the response is earlier than the time s of the stimulus. In other words, for a causal system, the following condition must hold:
h(t,s) = 0\text{ for }t < s \, | |
# triangle help
• Jun 5th 2012, 12:52 AM
Domicro
triangle help
Hi. I've just got 20 math tasks that I need to solve until saturday. I don't know how to solve 4. Pls help. P. S. My english is not very good, sorry about that..
1. Man and woman are standing on the street. A man is tall 188 cm and his shadow is tall 1 m (100cm).. How tall is the woman if her shadow is 80 cm tall.
2. Area (surface) of similar triangels is 8cm2 and 18cm2. If extent (perimeter) of bigget triangle is 20cm, what's the extent (perimeter) of smaller triangle?
3. a= 28 cm, b = 52/3 cm, c = 80/3 cm. What's the radius of circular of his similar triangle if his extent (perimeter) is 126 cm2
4. In right-angeled triangle p = 3cm v = 4 cm. calculate a, b, c. (p=extent (perimeter) and v=height of the triangle)
• Jun 5th 2012, 12:58 AM
Prove It
Re: triangle help
Quote:
Originally Posted by Domicro
4. In right-angeled triangle p = 3cm v = 4 cm. calculate a, b, c. (p=extent (perimeter) and v=height of the triangle)[/FONT][/COLOR]
How could the perimeter of the triangle be less than one of the lengths?
I expect you meant that the base length was 3cm and the height was 4cm. Since it's a right angle triangle, you can use Pythagoras' Theorem to find the third length.
• Jun 6th 2012, 01:06 PM
earboth
Re: triangle help
Quote:
Originally Posted by Domicro
Hi. I've just got 20 math tasks that I need to solve until saturday. I don't know how to solve 4. Pls help. P. S. My english is not very good, sorry about that..
1. Man and woman are standing on the street. A man is tall 188 cm[COLOR=#333333][FONT=arial] and his shadow is tall 1 m (100cm).. How tall is the woman if her shadow is 80 cm tall.
...
Draw a sketch. Use proportions.
$\frac{length\ of\ woman}{length\ of\ man}=\frac{length\ of\ woman's\ shadow}{length\ of\ man's\ shadow}$
$\frac w{188}=\frac{80}{100}$
Solve for w. | |
#### MAT-06.EE.02.c
6th Grade MAT Targeted StandardsDomain (EE) Expressions and EquationsCluster: Apply and extend previous understandings of arithmetic to algebraic expressions Write, read, and evaluate expressions in which letters stand for numbers. MAT-06.EE.02.c Evaluate expressions at specific values of their variables. Include expressions that arise from formulas used in real-world problems. Perform arithmetic operations, including those involving whole-number exponents, in the conventional order when there are no parentheses to specify a particular order (Order of Operations). For example, use the formulas V = s³ and A = 6 s² to find the volume and surface area of a cube with sides of length s =1/2.
## Student Learning Targets
### Reasoning Targets
• I can use Order of Operations to simplify expressions that may include exponents.
### Skills Domain (Performance) Targets
• I can simplify expressions when a value is given for a variable.
• I can solve multistep expressions involving properties of operations. | |
/ deeplearning.ai深度学习笔记整理
# deeplearning.ai深度学习笔记（Course2 Week1）：Practical aspects of deep learning
1. 训练集、开发集、测试集的概念
2. Bias/Variance问题
3. 如何通过泛化（regularization）算法，解决High variance问题，以及常用的集中regularization算法。
4. 最小化J中的一些加快手段
• 标准化输入（Normalizing ）
• 验证梯度计算是否正确。
# 1- Setting up your Machine Learning Application
## 1.1- Train / Dev / Test sets
1. Applied ML is a highly iterative process
之所以说是highly interative,是因为要设置不少超参，不断试验。但超参也不是说靠猜。而是通过一些手段，使得这个过程更有效。
2. 数据集划分：
• Train set：训练集
• dev set：即cross validation set，测试不同的算法
• test set：测试集
1. 关于数据集的一些趋势：
• 传统来说（数据集较小），这三种测试数据的比例为：70%，20%和20%；但在目前大数据集的情况下，dev set和test set的比例并不需要太高，只要足够测试即可，很可能比例是98%，1%和1%，甚至更高。
• 实际生产环境，可能Training set和Dev/test sets来源不同，前者来自开发者，后者来自用户，导致数据集分布不同
1. 经验法则：尽量让Training set和Dev/test sets具有相同的分布
2. 某些情况下没有test set也是没问题的。这个时候只有training set和dev set，或者不严谨的，这个时候dev set被称作为test set（但和tain/dev/test中的test作用并不一样，实际作用还是dev）
## 1.2- Bias / Variance
Bias：偏差，High bias: underfitting；说明算法比数据简单，不足以描述数据。
Variance：方差，High variance: overfitting；说明算法超过了数据的实际复杂性，甚至将一些随机因素过度解释为了数据规律。
Train set error Dev set error type
low (1%) high (11%) high variance
high (15%) high, but near train set error (16%) high bias
high (15%) high, but much higher than Train set error (30%) high bias & high variance
low (0.5%) low (1%) low bias & low variance
1. Dev set error理论上通常是大于Train set error
## 1.3- Basic Recipe for Machine Learning
1. 针对bias和variance要选择对应的解决方法。
2. 在早期的ML，强调bias variance trade-off；但在现代deep learning，可以通过加大Neural Network或增加更多数据，在分别解决High bias和High variance的时候，并不会影响彼此。 这也是deep learning在supervised learning如此成功的重要原因。
Training a bigger network almost never hurts. And the main cost of training a neural network that's too big is just computational time, so long as you're regularizing.
# 2- Regularization your neural network
1. Regularzization可以有效解决overfitting问题。
2. 常用的regularization方法：
• L2 regularization
• dropout reuglarization
3. 我的理解：之所以会有过拟合问题，本质上是数据存在一定的随机性干扰（即在主要的规律外，还有一定的随机因素干扰了数据，而这些随机因素被算法当成规律学习了），而中和这种随机性的办法就是在算法中也增加一些“干扰”，这个“干扰”就是Regularization。
## 2.1- L2 Regularization
1. 举例：
• logistic regression做regularization
在cost function增加：
$$\frac{\lambda}{2m}||w||^2_2$$
• Neural Network，增加Frobenius norm
$$\frac{\lambda}{2m} \sum\limits_{l = 1}^{m}||W^{[l]}||^2_F$$
其中
$$||W^{[l]}||^2_F = \sum\limits_{i = 1}^{n^{[l]}} \sum\limits_{j = 1}^{n^{[l-1]}} (W^{[l]}_{ij})^2$$
1. Why L2 regularization reduces overfitting?
Regularization其实是让函数变得简化
L2-regularization relies on the assumption that a model with small weights is simpler than a model with large weights. Thus, by penalizing the square values of the weights in the cost function you drive all the weights to smaller values. It becomes too costly for the cost to have large weights! This leads to a smoother model in which the output changes more slowly as the input changes.
L2 regularization的不足：要通过不断的选用不同的λ进行测试，计算量加大了。
## 2.2- Dropout Regularization
1. Dropout Regularization：在每轮迭代计算时，随机的将Network中一些neuron剔除，效果就好像用了一个更小的Network。（我有个疑问：为什么不直接较小的Network？）
Drop-out on the second hidden layer：
At each iteration, you shut down (= set to zero) each neuron of a layer with probability $$1 - keep_prob$$ or keep it with probability $$keep_prob$$ (50% here). The dropped neurons don't contribute to the training in both the forward and backward propagations of the iteration.
Drop-out on the first and third hidden layers:
$$1^{st}$$ layer: we shut down on average 40% of the neurons. $$3^{rd}$$ layer: we shut down on average 20% of the neurons.
1. implementing dropout("Inverted dropout")
illustrate with 3rd layer, 以0.8的概率保留neuron（keep_prob=0.8）
2. 在test time，不需要做regularization
3. dropout intuition
Can’t rely on any one feature, so have to spread out weights. And by spreading all the weights, this will tend to have an effect of shrinking the squared norm of the weights. And so, similar to what we saw with L2 regularization.
4. 不同的layer，可以选用不同的keep-pro；比如neuron较多的层，可以将keep-pro设置的小一些，而本身就没有几个neuron的（比如最后一层），则将keep-pro倾向于设置为1。另外，input layter也尽量设置为keep-pro==1。
5. dropout在computer vision用的非常多；但其他领域不轻易使用。
6. dropout的缺点是：导致cost function J不稳定，在估计J是否收敛时，可能会有不稳定。一般会在检查收敛的时候，把dropout关掉，确认收敛后在打开。
## 2.3- Other regularization methods
1. Data augmentation
通过对已有数据的人工加工，形成更多的训练数据，变相实现了增加数据量。比如对已有图片的翻转、裁剪形成新的数据。
2. Early stopping
在观察training set error的同时，将dev set error也输出观察。通常dev set error会先下降再上升。因此在这个点提前停止会得到一个相对好的结果。
通常，随着迭代的次数增加，W通常会越来越大；因此同L2 regularization类似（让W变小了），early stop也变相让W还没变大的时候提前停止。
early stop的缺点是：这个过程是和最小化J这个任务相悖的。但与L2 regularization相比，也有它的优势，即不需要通过迭代选择最好的λ，一次迭代就能确定early stop的点。
# 3- setting up your optimization problem
## 3.1- Normalizing inputs
1. normalizing = 中心化（均值为0）+归一化（方差为1）
注意：train set和dev/test set应该用一样的normalizing方法。
2. why normalize inputs?
## 3.2- Vanishing / Exploding gradients
Vanishing / Exploding gradients：在层数很深的neural network，可能因为input数据>1或<1的区别，深层的activation function会指数级的变的很大或很小，进而activation function的梯度要么很大，要么很小，使得梯度下降算法性能下降。
## 3.3- Weight Initialization for Deep Networks
$$W^{[l]} = np.random.randn(shape) *\sqrt{\frac{1}{n^{[l-1]}}}$$
$$W^{[l]} = np.random.randn(shape) *\sqrt{\frac{2}{n^{[l-1]}}}$$
• Different initializations lead to different results
• Random initialization is used to break symmetry and make sure different hidden units can learn different things
• Don't intialize to values that are too large
• He initialization works well for networks with ReLU activations.
$$\frac{\partial J}{\partial \theta} = \lim_{\varepsilon \to 0} \frac{J(\theta + \varepsilon) - J(\theta - \varepsilon)}{2 \varepsilon}$$
对每个参数进行近似导数求导，然后比较和实际计算的梯度相比：
$$\frac{||d_{approx} - d||_2} {||d_{approx}||_2 + ||d||_2}$$
上面的式子的结果应该和$$\varepsilon$$大概一个数量级
2. 注意：
• 不要漏掉 regularization 的部分
• dropout regularization不适用，因为J是不稳定的。
# 4- Heros of Deep Learning: Yoshua Bengio interview
1. Yoshua Bengio 是 Ian Goodfellow 的导师，也是 Deep learning 这本书的的第二作者。
2. When I started my graduate studies in 1985, I started reading neural net papers, and that's where I got all excited, and it became really a passion.
3. We start with experiments, with intuitions, and theory sort of comes later. We now understand a lot better, for example, why Backdrop is working so well, why depth is so important.
4. what were the biggest surprises of what turned out to be wrong, compared to what we knew 30 years ago?
The ReLU was working a lot better than the sigmoids and tanh.
5. 最骄傲的研究成果：
• Long-term dependencies
• curse of dimensionality
• joint distributions with neural nets
• word embeddings for joint distributions for words
• the vanishing gradient in deep nets.
• the denoising auto-encoders
• the GANs
6. Neural nets we used to think as machines that can map a vector to a vector. But really with attention mechanisms, you can now handle any kind of data structure. And this is really opening up a lot of interesting avenues.
7. 虽然，现在业界在supervised learning上很成功，但unsupervised learning很重要，It's about actually building a mental construction that explains how the world works by observation.
8. what in deep learning today excites you the most?
I feel like the current state of the science of deep learning is far from where I'd like to see it. And I have the impression that our systems right now make the kind of mistakes that suggest they have a very superficial understanding of the world.
they would become much easier to tackle if we had systems that had a better understanding of how the world works.
9. Advice for people entering AI
• First of all, there are different motivations and different things you can do. What you need to become a deep learning researcher may not be the same as if you want to be an engineer who's going to use deep learning to build products. T
• You have to practice programming the things yourself. So don't just use one of the programming frameworks where you can do everything in a few lines of code, but you don't really know what just happened. Trying to derive the thing yourself from first principles, if you can. That really helps. But yeah, the usual things you have to do like reading, looking at other people's code, writing your own code, doing lots of experiment, making sure you understand everything you do. So especially for the science part of it, trying to ask why am I doing this, why are people doing this? Maybe the answer is somewhere in the book and you have to read more. （再次提到实践的重要性）
• deep learning 这本书的看法： I feel like there is more people reading this book than people who can read it right now.
• Proceedings of the ICLR (International Conference on Learning Representations) conference is probably the best concentrated place of good papers. （ICLR 国际会议很重要，需要关注）
• Don't be afraid by the math. Just develop the intuitions, and then the math become really easier to understand once you get the hang of what's going on at the intuitive level. （Intuition很重要）
• You don't need five years of PhD to become proficient at deep learning.
• 需要的数学基础: probability, algebra, optimization and calculus. | |
# 7.1 Electric potential energy (Page 4/5)
Page 4 / 5
${W}_{2}=k\frac{{q}_{1}{q}_{2}}{{r}_{12}}=\left(9.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{9}\frac{\text{N}·{\text{m}}^{2}}{{\text{C}}^{2}}\right)\frac{\left(2.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-6}\phantom{\rule{0.2em}{0ex}}\text{C}\right)\left(3.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-6}\phantom{\rule{0.2em}{0ex}}\text{C}\right)}{1.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-2}\phantom{\rule{0.2em}{0ex}}\text{m}}=5.4\phantom{\rule{0.2em}{0ex}}\text{J}\text{.}$
Step 3. While keeping the charges of $+2.0\phantom{\rule{0.2em}{0ex}}\mu \text{C}$ and $+3.0\phantom{\rule{0.2em}{0ex}}\mu \text{C}$ fixed in their places, bring in the $+4.0\text{-}\mu \text{C}$ charge to $\left(x,y,z\right)=\left(1.0\phantom{\rule{0.2em}{0ex}}\text{cm},1.0\phantom{\rule{0.2em}{0ex}}\text{cm},0\right)$ ( [link] ). The work done in this step is
$\begin{array}{cc}{W}_{3}\hfill & =k\frac{{q}_{1}{q}_{3}}{{r}_{13}}+k\frac{{q}_{2}{q}_{3}}{{r}_{23}}\hfill \\ & =\left(9.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{9}\frac{\text{N}·{\text{m}}^{2}}{{\text{C}}^{2}}\right)\left[\frac{\left(2.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-6}\phantom{\rule{0.2em}{0ex}}\text{C}\right)\left(4.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-6}\phantom{\rule{0.2em}{0ex}}\text{C}\right)}{\sqrt{2}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-2}\phantom{\rule{0.2em}{0ex}}\text{m}}+\frac{\left(3.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-6}\phantom{\rule{0.2em}{0ex}}\text{C}\right)\left(4.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-6}\phantom{\rule{0.2em}{0ex}}\text{C}\right)}{1.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-2}\phantom{\rule{0.2em}{0ex}}\text{m}}\right]=15.9\phantom{\rule{0.2em}{0ex}}\text{J}\text{.}\hfill \end{array}$
Step 4. Finally, while keeping the first three charges in their places, bring the $+5.0\text{-}\mu \text{C}$ charge to $\left(x,y,z\right)=\left(0,\phantom{\rule{0.2em}{0ex}}1.0\phantom{\rule{0.2em}{0ex}}\text{cm},\phantom{\rule{0.2em}{0ex}}0\right)$ ( [link] ). The work done here is
$\begin{array}{cc}{W}_{4}\hfill & =k{q}_{4}\left[\frac{{q}_{1}}{{r}_{14}}+\frac{{q}_{2}}{{r}_{24}}+\frac{{q}_{3}}{{r}_{34}}\right],\hfill \\ & =\left(9.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{9}\frac{\text{N}·{\text{m}}^{2}}{{\text{C}}^{2}}\right)\left(5.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-6}\phantom{\rule{0.2em}{0ex}}\text{C}\right)\left[\frac{\left(2.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-6}\phantom{\rule{0.2em}{0ex}}\text{C}\right)}{1.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-2}\phantom{\rule{0.2em}{0ex}}\text{m}}+\frac{\left(3.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-6}\phantom{\rule{0.2em}{0ex}}\text{C}\right)}{\sqrt{2}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-2}\phantom{\rule{0.2em}{0ex}}\text{m}}+\frac{\left(4.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-6}\phantom{\rule{0.2em}{0ex}}\text{C}\right)}{1.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-2}\phantom{\rule{0.2em}{0ex}}\text{m}}\right]=36.5\phantom{\rule{0.2em}{0ex}}\text{J}\text{.}\hfill \end{array}$
Hence, the total work done by the applied force in assembling the four charges is equal to the sum of the work in bringing each charge from infinity to its final position:
${W}_{\text{T}}={W}_{1}+{W}_{2}+{W}_{3}+{W}_{4}=0+5.4\phantom{\rule{0.2em}{0ex}}\text{J}+15.9\phantom{\rule{0.2em}{0ex}}\text{J}+36.5\phantom{\rule{0.2em}{0ex}}\text{J}=57.8\phantom{\rule{0.2em}{0ex}}\text{J}\text{.}$
## Significance
The work on each charge depends only on its pairwise interactions with the other charges. No more complicated interactions need to be considered; the work on the third charge only depends on its interaction with the first and second charges, the interaction between the first and second charge does not affect the third.
Check Your Understanding Is the electrical potential energy of two point charges positive or negative if the charges are of the same sign? Opposite signs? How does this relate to the work necessary to bring the charges into proximity from infinity?
positive, negative, and these quantities are the same as the work you would need to do to bring the charges in from infinity
Note that the electrical potential energy is positive if the two charges are of the same type, either positive or negative, and negative if the two charges are of opposite types. This makes sense if you think of the change in the potential energy $\text{Δ}U$ as you bring the two charges closer or move them farther apart. Depending on the relative types of charges, you may have to work on the system or the system would do work on you, that is, your work is either positive or negative. If you have to do positive work on the system (actually push the charges closer), then the energy of the system should increase. If you bring two positive charges or two negative charges closer, you have to do positive work on the system, which raises their potential energy. Since potential energy is proportional to 1/ r , the potential energy goes up when r goes down between two positive or two negative charges.
On the other hand, if you bring a positive and a negative charge nearer, you have to do negative work on the system (the charges are pulling you), which means that you take energy away from the system. This reduces the potential energy. Since potential energy is negative in the case of a positive and a negative charge pair, the increase in 1/ r makes the potential energy more negative, which is the same as a reduction in potential energy.
The result from [link] may be extended to systems with any arbitrary number of charges. In this case, it is most convenient to write the formula as
${W}_{12\cdots N}=\frac{k}{2}\sum _{i}^{N}\sum _{j}^{N}\frac{{q}_{i}{q}_{j}}{{r}_{ij}}\phantom{\rule{0.2em}{0ex}}\text{for}\phantom{\rule{0.2em}{0ex}}i\ne j.$
The factor of 1/2 accounts for adding each pair of charges twice.
## Summary
• The work done to move a charge from point A to B in an electric field is path independent, and the work around a closed path is zero. Therefore, the electric field and electric force are conservative.
• We can define an electric potential energy, which between point charges is $U\left(r\right)=k\frac{qQ}{r}$ , with the zero reference taken to be at infinity.
• The superposition principle holds for electric potential energy; the potential energy of a system of multiple charges is the sum of the potential energies of the individual pairs.
## Conceptual questions
Would electric potential energy be meaningful if the electric field were not conservative?
No. We can only define potential energies for conservative fields.
Why do we need to be careful about work done on the system versus work done by the system in calculations?
Does the order in which we assemble a system of point charges affect the total work done?
No, though certain orderings may be simpler to compute.
## Problems
Consider a charge ${Q}_{1}\left(+5.0\phantom{\rule{0.2em}{0ex}}\mu \text{C}\right)$ fixed at a site with another charge ${Q}_{2}$ (charge $+3.0\phantom{\rule{0.2em}{0ex}}\mu \text{C}$ , mass $6.0\phantom{\rule{0.2em}{0ex}}\mu \text{g}\right)$ moving in the neighboring space. (a) Evaluate the potential energy of ${Q}_{2}$ when it is 4.0 cm from ${Q}_{1}.$ (b) If ${Q}_{2}$ starts from rest from a point 4.0 cm from ${Q}_{1},$ what will be its speed when it is 8.0 cm from ${Q}_{1}$ ? ( Note: ${Q}_{1}$ is held fixed in its place.)
a. $U=3.4\phantom{\rule{0.2em}{0ex}}\text{J;}$
b. $\frac{1}{2}m{v}^{2}=k{Q}_{1}{Q}_{2}\left(\frac{1}{{r}_{i}}-\frac{1}{{r}_{f}}\right)\to v=750\phantom{\rule{0.2em}{0ex}}\text{m/s}$
Two charges ${Q}_{1}\left(+2.00\phantom{\rule{0.2em}{0ex}}\mu \text{C}\right)$ and ${Q}_{2}\left(+2.00\phantom{\rule{0.2em}{0ex}}\mu \text{C}\right)$ are placed symmetrically along the x -axis at $x=±3.00\phantom{\rule{0.2em}{0ex}}\text{cm}$ . Consider a charge ${Q}_{3}$ of charge $+4.00\phantom{\rule{0.2em}{0ex}}\mu \text{C}$ and mass 10.0 mg moving along the y -axis. If ${Q}_{3}$ starts from rest at $y=2.00\phantom{\rule{0.2em}{0ex}}\text{cm,}$ what is its speed when it reaches $y=4.00\phantom{\rule{0.2em}{0ex}}\text{cm?}$
To form a hydrogen atom, a proton is fixed at a point and an electron is brought from far away to a distance of $0.529\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-10}\phantom{\rule{0.2em}{0ex}}\text{m,}$ the average distance between proton and electron in a hydrogen atom. How much work is done?
$U=4.36\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-18}\phantom{\rule{0.2em}{0ex}}\text{J}\phantom{\rule{0.2em}{0ex}}$
(a) What is the average power output of a heart defibrillator that dissipates 400 J of energy in 10.0 ms? (b) Considering the high-power output, why doesn’t the defibrillator produce serious burns?
it is the work done in moving a charge to a point from infinity against electric field
what is the weight of the earth in space
As w=mg where m is mass and g is gravitational force... Now if we consider the earth is in gravitational pull of sun we have to use the value of "g" of sun, so we can find the weight of eaeth in sun with reference to sun...
Prince
g is not gravitacional forcé, is acceleration of gravity of earth and is assumed constante. the "sun g" can not be constant and you should use Newton gravity forcé. by the way its not the "weight" the physical quantity that matters, is the mass
Jorge
Yeah got it... Earth and moon have specific value of g... But in case of sun ☀ it is just a huge sphere of gas...
Prince
Thats why it can't have a constant value of g ....
Prince
not true. you must know Newton gravity Law . even a cloud of gas it has mass thats al matters. and the distsnce from the center of mass of the cloud and the center of the mass of the earth
Jorge
please why is the first law of thermodynamics greater than the second
define electric image.obtain expression for electric intensity at any point on earthed conducting infinite plane due to a point charge Q placed at a distance D from it.
explain the lack of symmetry in the field of the parallel capacitor
pls. explain the lack of symmetry in the field of the parallel capacitor
Phoebe
does your app come with video lessons?
What is vector
Vector is a quantity having a direction as well as magnitude
Damilare
tell me about charging and discharging of capacitors
a big and a small metal spheres are connected by a wire, which of this has the maximum electric potential on the surface.
3 capacitors 2nf,3nf,4nf are connected in parallel... what is the equivalent capacitance...and what is the potential difference across each capacitor if the EMF is 500v
equivalent capacitance is 9nf nd pd across each capacitor is 500v
santanu
four effect of heat on substances
why we can find a electric mirror image only in a infinite conducting....why not in finite conducting plate..?
because you can't fit the boundary conditions.
Jorge
what is the dimensions for VISCOUNSITY (U)
Branda
what is thermodynamics
the study of heat an other form of energy.
John
heat is internal kinetic energy of a body but it doesnt mean heat is energy contained in a body because heat means transfer of energy due to difference in temperature...and in thermo-dynamics we study cause, effect, application, laws, hypothesis and so on about above mentioned phenomenon in detail.
ing
It is abranch of physical chemistry which deals with the interconversion of all form of energy
Vishal
what is colamb,s law.?
it is a low studied the force between 2 charges F=q.q`\r.r
Mostafa
what is the formula of del in cylindrical, polar media | |
# Fractions and roots
I have this problem:
$$\frac{\sqrt{18}+\sqrt{98}+\sqrt{50}+4}{2\sqrt{2}}$$
I'm able to get to this part by myself:
$$\frac{15\sqrt{2}+2}{2\sqrt{2}}$$
But that's when I get stuck. The book says that the next step is:
$$\frac{15\sqrt{2}}{2\sqrt{2}}+ \frac{2}{2\sqrt{2}}$$
But I don't understand why you can take the 2 out of the original fraction, make it the numerator of its own fraction and having root of 2 as the denominator of said fraction.
• $\dfrac{a+b}{c}=\dfrac{a}{c}+\dfrac{b}{c}$ – egreg Oct 4 '13 at 23:38
• Such an "obvious" solution! Thanks, egreg! – Jose Oct 4 '13 at 23:48
• Important detail: a "problem" really needs an English-language direction or question of what to do, and you didn't include that. You cannot look at a mathematical expressions and infer "what to do" from it, so pay attention to the actual directions. A short example I run in class is to ask: "What's the degree of $3x^2+5x^2$?", and a lot of students answer, $8x^2$, which isn't a degree at all. – Daniel R. Collins Sep 21 '15 at 6:33
$$\frac{\sqrt{18}+\sqrt{98}+\sqrt{50}+4}{2\sqrt{2}}= \frac{3\sqrt{2}+7\sqrt{2}+5\sqrt{2}+4}{2\sqrt{2}}= \frac{15\sqrt{2}+4}{2\sqrt{2}}$$ Now the standard procedure is to remove the radical in the denominator: $$\frac{15\sqrt{2}+4}{2\sqrt{2}}= \frac{15\sqrt{2}+4}{2\sqrt{2}}\frac{\sqrt{2}}{\sqrt{2}}= \frac{15\sqrt{2}\cdot\sqrt{2}+4\sqrt{2}}{2\sqrt{2}\cdot\sqrt{2}}= \frac{30+4\sqrt{2}}{4}=\frac{15+2\sqrt{2}}{2}$$ One can do it differently: set $a=\sqrt{2}$, so you can write
$$\frac{15\sqrt{2}+4}{2\sqrt{2}}= \frac{15a+a^4}{a^3}= \frac{15+a^3}{a^2}= \frac{15+2\sqrt{2}}{2}$$
The final result can also be written by using $\frac{a+b}{c}=\frac{a}{c}+\frac{b}{c}$ so $$\frac{15+2\sqrt{2}}{2}=\frac{15}{2}+\sqrt{2}$$ Whether you want to do this last transformation depends on what you have to do with this number.
Why shouldn't it be $+4$ at the end of the second expression? That should carry down to the third. From the second to the third you are just splitting one fraction into two.
To me the next obvious step is to remove the radical in the denominator so
$$\frac{15\sqrt{2}+2}{2\sqrt{2}} = \frac{15+\sqrt{2}}{2}.$$
Personally I would stop there, but you could separate this into $\dfrac{15}{2}+\dfrac{\sqrt{2}}{2}$ or into $\dfrac{15}{2}+\dfrac{1}{\sqrt{2}}$ if you really wanted to. | |
# Ross Rheingans-Yoo
197Joined Dec 2020
# Posts 2
Sorted by New
I'd argue that you need to use a point estimate to decide what bets to make, and that you should make that point estimate by (1) geomean-pooling raw estimates of parameters, (2) reasoning over distributions of all parameters, then (3) taking arithmean of the resulting distribution-over-probabilities and (4) acting according to that mean probability.
I think "act according to that mean probability" is wrong for many important decisions you might want to take - analogous to buying a lot of trousers with 1.97 legs in my example in the essay. No additional comment if that is what you meant though and were just using shorthand for that position.
Clarifying, I do agree that there are some situations where you need something other than a subjective p(risk) to compare EV(value|action A) with EV(value|action B). I don't actually know how to construct a clear analogy from the 1.97-legged trousers example if the variable we're meaning is probabilities (though I agree that there are non-analogous examples; VOI for example).
I'll go further, though, and claim that what really matters is what worlds the risk is distributed over, and that expanding the point-estimate probability to a distribution of probabilities, by itself, doesn't add any real value. If it is to be a valuable exercise, you have to be careful what you're expanding and what you're refusing to expand.
More concretely, you want to be expanding over things your intervention won't control, and then asking about your intervention's effect at each point in things-you-won't-control-space, then integrating back together. If you expand over any axis of uncertainty, then not only is there a multiplicity of valid expansions, but the natural interpretation will be misleading.
For example, say we have a 10% chance of drawing a dangerous ball from a series of urns, and 90% chance of drawing a safe one. If we describe it as (1) "50% chance of 9.9% risk, 50% chance of 10.1% risk" or (2) "50% chance of 19% risk, 50% chance of 1% risk" or (3) "10% chance of 99.1% risk, 90% chance of 0.1% risk", what does it change our opinion of <intervention A>? (You can, of course, construct a two-step ball-drawing procedure that produces any of these distributions-over-probabilities.)
I think the natural intuition is that interventions are best in (2), because most probabilities of risk are middle-ish, and worst in (3), because probability of risk is near-determined. And this, I think, is analogous to the argument of the post that anti-AI-risk interventions are less valuable than the point-estimate probability would indicate.
But that argument assumes (and requires) that our interventions can only chance the second ball-drawing step, and not the first. So using that argument requires that, in the first place, we sliced the distribution up over things we couldn't control. (If that is the thing we can control with our intervention, then interventions are best in the world of (3).)
Back to the argument of the original post: You're deriving a distribution over several p(X|Y) parameters from expert surveys, and so the bottom-line distribution over total probabilities reflects the uncertainty in experts' opinions on those conditional probabilities. Is it right to model our potential interventions as influencing the resolution of particular p(X|Y) rolls, or as influencing the distribution of p(X|Y) at a particular stage?
I claim it's possible to argue either side.
Maybe a question like "p(much harder to build aligned than misaligned AGI | strong incentives to build AGI systems)" (the second survey question) is split between a quarter of the experts saying ~0% and three-quarters of the experts saying ~100%. (This extremizes the example, to sharpen the hypothetical analysis.) We interpret this as saying there's a one-quarter chance we're ~perfectly safe and a three-quarters chance that it's hopeless to develop and aligned AGI instead of a misaligned one.
If we interpret that as if God will roll a die and put us in the "much harder" world with three-quarters probability and the "not much harder" world with one-quarters probability, then maybe our work to increase the we get an aligned AGI is low-value, because it's unlikely to move either the ~0% or ~100% much lower (and we can't change the die). If this was the only stage, then maybe all of working on AGI risk is worthless.
But "three-quarter chance it's hopeless" is also consistent with a scenario where there's a three-quarters chance that AGI development will be available to anyone, and many low-resourced actors will not have alignment teams and find it ~impossible to develop with alignment, but a one-quarter chance that AGI development will be available only to well-resourced actors, who will find it trivial to add on an alignment team and develop alignment. But then working on AGI risk might not be worthless, since we can work on increasing the chance that AGI development is only available to actors with alignment teams.
I claim that it isn't clear, from the survey results, whether the distribution of experts' probabilities for each step reflect something more like the God-rolls-a-die model, or different opinions about the default path of a thing we can intervene on. And if that's not clear, then it's not clear what to do with the distribution-over-probabilities from the main results. Probably they're a step forward in our collective understanding, but I don't think you can conclude from the high chances of low risk that there's a low value to working on risk mitigation.
I agree that geomean-of-odds performs better than geomean-of-probs!
I still think it has issues for converting your beliefs to actions, but I collected that discussion under a cousin comment here: https://forum.effectivealtruism.org/posts/Z7r83zrSXcis6ymKo/dissolving-ai-risk-parameter-uncertainty-in-ai-future?commentId=9LxG3WDa4QkLhT36r
An explicit case where I think it's important to arithmean over your subjective distribution of beliefs:
• coin A is fair
• coin B is either 2% heads or 98% heads, you don't know
• you lose if either comes up tails.
So your p(win) is "either 1% or 49%".
I claim the FF should push the button that pays us $80 if win, -$20 if lose, and in general make action decisions consistent with a point estimate of 25%. (I'm ignoring here the opportunity to seek value of information, which could be significant!).
It's important not to use geomean-of-odds to produce your actions in this scenario; that gives you ~9.85%, and would imply you should avoid the +$80;-$20 button, which I claim is the wrong choice.
Thanks for clarifying "geomean of probabilities" versus "geomean of odds elsethread. I agree that that resolves some (but not all) of my concerns with geomeaning.
I think the way in which I actually disagree with the Future Fund is more radical than simple means vs geometric mean of odds - I think they ought to stop putting so much emphasis on summary statistics altogether.
I agree with your pro-distribution position here, but I think you will be pleasantly surprised by how much reasoning over distributions goes into cost-benefit estimates at the Future Fund. This claim is based on nonpublic information, though, as those estimates have not yet been put up for public discussion. I will suggest, though, that it's not an accident that Leopold Aschenbrenner is talking with QURI about improvements to Squiggle: https://github.com/quantified-uncertainty/squiggle/discussions
So my subjective take is that if the true issue is "you should reason over distributions of core parameters", then in fact there's little disagreement between you and the FF judges (which is good!), but it all adds up to normality (which is bad for the claim "moving to reasoning over distributions should move your subjective probabilities").
If we're focusing on the Worldview Prize question as posed ("should these probability estimates change?"), then I think the geo-vs-arith difference is totally cruxy -- note that the arithmetic summary of your results (9.65%) is in line with the product of the baseline subjective probabilities for the prize (something like a 3% for loss-of-control x-risk before 2043; something like 9% before 2100).
I do think it's reasonable to critique the fact that those point probabilities are presented without any indication that the path of reasoning goes through reasoning over distributions, though. So I personally am happy with this post calling attention to distributional reasoning, since it's unclear in this case whether that is an update. I just don't expect it to win the prizes for changing estimates.
Because I do think distributional reasoning is important, though, I do want to zoom in on the arith-vs-geo question (which I think, on reflection, is subtler than the position I took in my top-level comment). Rather than being a minor detail, I think this is important because it influences whether greater uncertainty tends to raise or lower our "fair betting odds" (which, at the end of the day, are the numbers that matter for how the FF decides to spend money).
I agree with Jamie and you and Linch that when pooling forecasts, it's reasonable (maybe optimal? maybe not?) to use geomeans. So if you're pooling expert forecasts of {1:1000, 1:100, 1:10}, you might have a subjective belief of something like "1:100, but with a 'standard deviation' of 6.5x to either side". This is lower than the arithmean-pooled summary stats, and I think that's directionally right.
I think this is an importantly different question from "how should you act when your subjective belief is a distribution like that. I think that if you have a subjective belief like "1%, but with a 'standard deviation' of 6.5x to either side", you should push a button that gives you $98.8 if you're right and loses$1.2 if you're wrong. In particular, I think you should take the arithmean over your subjective distribution of beliefs (here, ~1.4%) and take bets that are good relative to that number. This will lead to decision-relevant effective probabilities that are higher than geomean-pooled point estimates (for small probabilities).
If you're combining multiple case parameters multiplicatively, then the arith>geo effect compounds as you introduce uncertainty in more places -- if the quantity of interest is x*y, where x and y each had expert estimates of {1:1000, 1:100, 1:10} that we assume independent, then arithmean(x*y) is about twice geomean(x*y). Here's a quick Squiggle showing what I mean: https://www.squiggle-language.com/playground/#code=eNqrVirOyC8PLs3NTSyqVLIqKSpN1QELuaZkluQXwUQy8zJLMhNzggtLM9PTc1KDS4oy89KVrJQqFGwVcvLT8%2FKLchNzNIAsDQM9A0NNHQ0jfWPNOAM9U82YvJi8SqJUVQFVVShoKVQCsaGBQUyeUi0A3tIyEg%3D%3D
For this use-case (eg, "what bets should we make with our money"), I'd argue that you need to use a point estimate to decide what bets to make, and that you should make that point estimate by (1) geomean-pooling raw estimates of parameters, (2) reasoning over distributions of all parameters, then (3) taking arithmean of the resulting distribution-over-probabilities and (4) acting according to that mean probability.
In the case of the Worldview Prize, my interpretation is that the prize is described and judged in terms of (3), because that is the most directly valuable thing in terms of producing better (4)s.
It sounds like the headline claim is that (A) we are 33.2% to live in a world where the risk of loss-of-control catastrophe is <1%, and 7.6% to live in a world where the risk is >35%, and a whole distribution of values between, and (B) that it follows from A that the correct subjective probability of loss-of-control catastrophe is given by the geometric mean of the risk, over possible worlds.
The ‘headline’ result from this analysis is that the geometric mean of all synthetic forecasts of the future is that the Community’s current best guess for the risk of AI catastrophe due to an out-of-control AGI is around 1.6%. You could argue the toss about whether this means that the most reliable ‘fair betting odds’ are 1.6% or not (Future Fund are slightly unclear about whether they’d bet on simple mean, median etc and both of these figures are higher than the geometric mean[9]).
I want to argue that the geometric mean is not an appropriate way of aggregating probabilities across different "worlds we might live in" into a subjective probability (as requested by the prize). This argument doesn't touch on the essay's main argument in favor of considering distributions, but may move the headline subjective probability that it suggests to 9.65%, effectively outside the range of opinion-change prizes, so I thought it worth clarifying in case I misunderstand.
Consider an experiment where you flip a fair coin A. If A is heads you flip a 99%heads coin B; if A is tails you flip a 1%heads coin B. We're interested in forming a subjective probability that B is heads.
The answer I find intuitive for p(B=heads) is 50%, which is achieved by taking the arithmetic average over worlds. The geometric average over worlds gives 9.9% instead, which doesn't seem like "fair betting odds" for B being heads under any natural interpretation of those words. What's worse, the geometric-mean methodology suggests a 9.9% subjective probability of tails, and then p(H)+p(T) does not add to 1.
(If you're willing to accept probabilities that are 0 and 1, then an even starker experiment is given by a 1% chance to end up in a world with 0% risk and a 99% chance to end up in a world with 100% risk -- the geometric mean is 0.)
Footnote 9 of the post suggests that the operative meaning of "fair betting odds" is sufficiently undefined by the prize announcement that perhaps it refers to a Brier-score bet, but I believe that it is clear from the prize announcement that a X bet is the kind under consideration. The prize announcement's footnote 1 says "We will pose many of these beliefs in terms of <u>subjective probabilities, which represent betting odds</u> that we consider fair in the sense that we’d be roughly indifferent between betting in favor of the relevant propositions <u>at those odds</u> or betting against them."
I don't know of a natural meaning of "bet in favor of P at 97:3 odds" other than "bet to win $97N if P and lose$3N if not P", which the bettor should be indifferent about if . Is there some other bet that you believe "bet in favor of P at odds of X:Y" could mean? In particular, is there a meaning which would support forming odds (and subjective probability) according to a geometric mean over worlds?
(I work at the FTX Foundation, but have no connection to the prizes or their judging, and my question-asking here is as a EA Forum user, not in any capacity connected to the prizes.)
To qualify, please please publish your work (or publish a post linking to it) on the Effective Altruism Forum, the AI Alignment Forum, or LessWrong with a "Future Fund worldview prize" tag. You can also participate in the contest by publishing your submission somewhere else (e.g. arXiv or your blog) and filling out this submission form. We will then linkpost/crosspost to your submission on the EA Forum.
I think it would be nicer if you say your P(Doom|AGI in 2070) instead of P(Doom|AGI by 2070), because the second one implicitly takes into account your timelines.
I disagree. (At least, if defining "nicer" as "more useful to the stated goals for the prizes".)
As an interested observer, I think it's an advantage to take timelines into account. Specifically, I think the most compelling way to argue for a particular P(Catastrophe|AGI by 20__) to the FF prize evaluators will be:
• states and argues for a timelines distribution in terms of P(AGI in 20__) for a continuous range of 20__s
• states and argues for a conditional-catastrophe function in terms of P(Catastrophe|AGI in 20__) over the range
• integrates the product over the range to get a P(Catastrophe|AGI by 20__)
• argues that the final number isn't excessively sensitive to small shifts in the timelines distribution or the catastrophe-conditional-on-year function.
An argument which does all of this successfully is significantly more useful to informing the FF's actions than an argument which only defends a single P(Catastrophe|20__).
I do agree that it would be nice to have the years line up, but as above I do expect a winning argument for P(Catastrophe|AGI by 2070) to more-or-less explicitly inform a P(Catastrophe|AGI by 2043), so I don't expect a huge loss.
(Not speaking for the prizes organizers/evaluators, just for myself.)
are there candidate interventions that only require mobile equipment and not (semi-)permanent changes to buildings?
Fortunately, yes. Within-room UVC (upper-room 254nm and lower-room 222nm) can be provided by mobile lights on tripod stands.
This is what the JHU Center for Health Security did for their IAQ conference last month. (Pictures at https://twitter.com/DrNikkiTeran/status/1567864920087138304 )
(Speaking for myself and not my employer.)
US tax law requires that US citizens pay income tax and capital gains tax, regardless of their physical/legal residency. Some limited deductions apply, but don't change the basic story.
Are you proposing to bite the bullet on the $100/hr card charge scenario by the$50/hr staffer (paid "$47.5/hr plus perks" at the EA org)? "Market rate" of$50/hr for labor netting \$500/hr of value seems well within the distribution I'd expect (not to mention that EA orgs might value that work even more than any org in industry ever will, perhaps because we're counting the consumer surplus and un-capturable externalities and the industry employer won't). | |
# Software to Find Kernels/Co-Kernels of Boolean Expressions
Is there any (free) software available that calculates all the possible kernel/co-kernel pairs of a boolean expression?
-
A free tool is misII - Multiple-level Combinational Logic Optimization Program. It is contained in the Windows freeware package Logic Friday 1 or in the Berkeley Octtools.
The terms Kernel and Co-Kernel are explained here.
Edited:
The misII documentation is available online.
Interactive sample session with misII:
misII> read_eqn
z = (a+b+c)(d+e)f+g;
^Z
misII> print_kernel -as z
Subkernels of {z}
(a f + b f + c f) * (d + e)
(d f + e f) * (a + b + c)
misII> print_kernel -a z
Kernels of {z}
(a f) * (d + e)
(b f) * (d + e)
(c f) * (d + e)
(d f) * (a + b + c)
(e f) * (a + b + c)
(f) * (a d + a e + b d + b e + c d + c e)
(-1-) * (a d f + a e f + b d f + b e f + c d f + c e f + g)
-
I have Logic Friday - I don't believe it does kernel and co-kernel calculations. I haven't used Octtools, but I don't see any evidence on their website that it does this either. I'm not looking for logic optimization, but for one specific step of it. – John Roberts Mar 16 '13 at 3:24
Below the 'Logic Friday 1' installation directory there is a misII directory. Look into the misII help file to find out how the 'print kernels' operation works. – Axel Kemper Mar 16 '13 at 9:21
I take it back. Good work man. – John Roberts Mar 16 '13 at 13:53
Would you happen to know how to use the gkx function as well? I can't seem to get it to work. – John Roberts Mar 16 '13 at 15:27
No, I don't have more information than source code and manual (page 13): www2.engr.arizona.edu/~veda/cadtools/sis/tutorials/sis.pdf 'gkx -?' gives a short help – Axel Kemper Mar 16 '13 at 16:05 | |
# Did I properly implement double integral equation in dblquad Python? Helicoidal Surface Theory [closed]
I show the equations (simplified Helicoidal Surface Theory for implementation purposes) that I want to calculate numerically using Python.
and the code:
from scipy.integrate import quad
import numpy as np
from scipy import interpolate
import time
start_time = time.time()
input="-0.5 0.0 \
-0.3 0.9 \
0.0 0.8 \
0.3 0.4 \
0.5 0.02"
input_coordinates = np.genfromtxt(input.splitlines()).reshape(-1,2) # shape to 2 columns, any number of rows
x_coordinates = input_coordinates[:,0]
H_values = input_coordinates[:,1]
H_interpolation = interpolate.InterpolatedUnivariateSpline(x_coordinates, H_values)
def complex_dblquad(func, a, b, g, h, **kwargs):
def real_func(z, x):
return np.real(func(z, x))
def imag_func(z, x):
return np.imag(func(z, x))
real_integral = dblquad(real_func, a, b, g, h, **kwargs)
imag_integral = dblquad(imag_func, a, b, g, h, **kwargs)
return (real_integral[0] + 1j*imag_integral[0], real_integral[1:], imag_integral[1:])
complex_integral = complex_dblquad(lambda z,x: np.sqrt(1+z*z)**2*(2/np.sqrt(1+z*z))**2*H_interpolation(x)*np.exp(1j*2/np.sqrt(1+z*z)*x), 0, 1, -0.5, 0.5)
print("--- %s seconds ---" % (time.time() - start_time))
The question is - is it implemented correctly? If not - what is wrong?
• I don't think you should be dumping your code here to get it checked for correctness. If you think there is a mistake then say why you think there is a mistake, or focus on one specific, short section of code. Jun 9 at 14:09
It looks correct, but it is difficult to interpret it is not a good practice to expand all the variables in one single expression, you can use the use definitions that match your equations more easily.
def integrand_function(z, x):
Mr = np.sqrt(1 + z**2)
kx = 2/Mr;
# merging the integrand of Psi in order to use the dblquad function.
return 2 * Mr**2 * kx**2 H_interpolation(x)*np.exp(1j * kx * x)
Then your complex integral is evaluated with
complex_integral = complex_dblquad(integrand_function, 0, 1, -0.5, 0.5)
And hopefully you can prove more easily the correctness this way.
• Thank you very much! It helps a lot! Jun 8 at 11:34 | |
# Keep Image Size the same even after zooming in and out
I have a mat-sidenav component filled with mat-card components. Outside the sidenav, I have the rest of my page. What I would like is for my sidenav card components to keep the exact same size as someone is zooming in and out the page. I have illustrated what I mean:
The image on the left shows my page before zooming, the right side is after the zoom. The content outside the sidenav (the boxes) zoom and scale accordingly, my sidenav itself has a max width set that it does not exceed even after zooming (I have achieved this). My problem lies in the mat card components (represented as circles). I would like my mat cards to also remain the same size after zoom as seen in the image, however, they grow like my components outside the sidenav. My mat cards are also wrapped individually in an <a> tag. I need it as if zooming in and out does not effect the sidenav nor its contents at all, how can I achieve this?
Source: New feed
Source Url Keep Image Size the same even after zooming in and out | |
## Shell Canada Outlines Further Oil Sands Expansion
##### 25 January 2007
Overview of oil sands production (mining) process. Click to enlarge. Source: Shell Canada.
Shell Canada is planning to expand its minable oil sands production to approximately 770,000 barrels a day—more than four times the current output—and also increasing upgrading capacity to approximately 700,000 barrels a day.
In April 2005, Shell Canada announced an incremental expansion strategy designed to increase minable bitumen production and upgrading capacity through a series of expansion projects. Shell Canada’s first 100,000-barrel-a-day expansion, AOSP (Athabasca Oil Sands Project) Expansion 1, was approved in November 2006 with full start-up expected in 2010.
OSP Expansion 1 will increase production to 255,000 bpd, up from 155,000 bpd at a cost of between C$10-C$12.8 billion (US$8.8-US$11.3 billion)—an increase ranging from 42% to 83% over the company’s 2005 estimate last year of C$7 billion. (Earlier post.) Beyond AOSP Expansion 1, Shell Canada plans additional oil sands expansions that could potentially increase minable bitumen production to approximately 770,000 barrels a day. This is about 200,000 barrels per day more than previously projected as the top end. In addition to existing regulatory approvals and expansion plans, Shell Canada’s growth strategy includes Jackpine Mine Expansion and an additional mine, called Pierre River Mine, on the west side of the Athabasca River. Bitumen upgrading capacity could also potentially be increased to about 700,000 barrels a day through a series of Shell Canada-owned upgrader projects in the Fort Saskatchewan area, east of Edmonton. New upgrading facilities would process Shell Canada’s share of future Athabasca minable bitumen production as well as bitumen from the company’s in situ oil sands developments. As previously announced, heavy oil upgrading and refining options in Ontario are also being considered. We are issuing these Public Disclosures so that we can start the process of our next phase of oil sands development. Shell Canada has major land holdings in the Athabasca region and we estimate that our minable bitumen resource base could ultimately support 770,000 barrels a day of production. This will provide a secure source of energy and economic benefit for Canadians for decades to come. —Clive Mather, Shell Canada President and CEO The regulatory review process for the expansion begins with the issuance of the Public Disclosure document. Shell Canada is working to prepare regulatory submissions for this project, and public consultations will be ongoing through 2007 and 2008. With bitumen production and upgrader production increasing, Shell will face a significant increase in greenhouse gas emissions. Shell Canada plans to design its upgrader expansions to be carbon capture ready. Implementation and use of CO2 capture technologies depends on the establishment of appropriate government policy and supporting framework, as well as project economics. —Upgrader Public Disclosure Document Shell Canada has set two targets for greenhouse gas emissions: one for its total business, one for the oil sands component. In 2005, Shell Canada’s emissions from its base businesses were 7.6 million tonnes, 217 thousand tonnes less than in 2004 and 5.6% below its 1990 level. The company’s target for 2008 is 7.567 million tonnes. Oil sands emissions in 2005 were, however, 3.68 million tonnes, 5% more than estimated emissions at start-up. Shell has set a 2010 target for oil sands emissions of 1.750 million tonnes. Shell Canada holds a 60% interest in the Athabasca Oil Sands Project (AOSP), a joint venture with Chevron Canada Limited (20%) and Western Oil Sands L.P. (20%). The AOSP consists of the Muskeg River Mine located north of Fort McMurray in northern Alberta, and the Scotford Upgrader near Edmonton, Alberta. Shell Canada also has in situ bitumen operations near Peace River and Cold Lake in Alberta. Chevron Canada Limited and Western Oil Sands L.P. have the option to participate with Shell Canada in the development of the Jackpine Mine Expansion and the Pierre River Mine. The day before the announcement, the board of Shell Canada had agreed to a C$8.7-billion buyout offer from Royal Dutch Shell (RDS). RDS already owns 78% of Shell Canada, and wants to buy the rest. On Monday, RDS upped its offers from C$40 per share to C$45. That deal now goes to the shareholders.
Resources:
Considering the near future massive arrival of Hybrids, PHEVs and BEVs + the current unacceptable high GHG level in Canada (specially in Alberta), this may not be a very good decision.
Canada DOES NOT NEED more oil from the tar sands. The current oil surplus does not have to be increased for export to (..... + .... + ...., specially from high pollution sources such as tar sands.
In the long run, Canada has all the land area (and more as climate warms up) required to produce enough biofuels such as cellulosic ethanol and butanol to meet the national requirements and export large qunatities to our neighbours.
If Shell were to spend $8.8 billion (US) in pilot BTL plant(s), they could have comparable, or better ROI. They could get 100-200k bpd crude equivalent from such an investment vs 100k bpd from bitumen. This entails setting up a collection system, and using dry forestry and agri waste as feedstock. Unacceptable to who? If it was so unacceptable to the Canadians, they would put a stop to it. As far as them not needing the extra oil, maybe you are right, but so what? Perhaps they would like to have the money they could earn by selling the surplus - most people would rather have Canadian oil than Middle Eastern oil. And near massive arrival of Hybrids and BEVs? Where are they? Most of what I have seen on this site are concept vehicles. I'll believe you when I can walk into any auto makers dealer and purchase something like that, not just a handful of them, as large as the manufacturer may be. We are always "near" something...... Allen, why didn't Shell think of that? You should call their Ivy League trained Accounting and Finance departments and explain it to them!!! Comparable or better ROI........give me a break. It is a no brainer for anyone. Of course it is economical and makes sense as most other oil sources are either at their peak production or already declining. The total supply cannot increase and we will be lucky to simply maintain the current total oil supply. Meanwhile you have ever increasing car production (China) and no effort to downsize (SUVs in NA). Regarding the new car technology, it is still progressing slowly and not being moved to production phase. There are still few options and/or very weak applications (so called soft hybrids). Their prices are still too high and only affordable to NA. And so on. Eventually cars will switch to the new technology away from gasoline/diesel, but not until price shocks wake up people and cause a couple of depressions for good measure. Meanwhile all oil suppliers will be bathing in money and this project will do very well indeed. For the tar sands production, count with me ... Current production is 155,000 bpd times 365 days divided by roughly 7 barrels of crude (specific weight of crude slightly less that 1): so about 8 million metric tonnes per year in oil produced. To produce/upgrade this, they now declare emissions of 3.68 million tonnes of greenhouse gas (GHG). Suppose this 3.68 million tonnes is the result of molecules consisting mainly of H2C subsets (approx atomic weight of 14) being burnt into CO2 (approx atomic weight of 46) + H2O (not a GHG). This gives an ratio of 14 units of oil to to produce 46 units of GHG. So, to account for the 3.68 million tonnes of GHG, you need to burn approx 1.1 million tonnes of oil, producing/upgrading 8 million tonnes of oil in the process. Even though the up-front number of 3.68 million tonnes of GHG seems huge, think what is produced when the resulting 8 million tonnes of product is burnt. In that light, the 8-for-1 cost even seems to become acceptable. Joff, you are onto something with your comments. The only PHEV available now is the Tesla Roadster, which there are a total of a few hundred of$100,000. The earliest possible mass production PHEVs will be in 2 years. Even then, how long until they are ubiquitous enough to actually decrease the TOTAL OIL used by Americans, much less the amount of oil used by fast progressing economies like China and India. And in case anyone has forgotten, there are over 2 billion Indians and Chinese!!! If by 2020 there is 20% car ownership in those countries, that's like adding more cars to the world than there are in 2 United States. Hmmm.
Tech of today will ease transitions for tomorrow, but total world oil use isn't going to decrease until I'm old and gray. BTW I'm only 28 right now.
DB -
I don't mean to nitpick, but the Tesla Roadster is not yet out and is a BEV design. There is no ICE at all in that vehicle.
As for the rapid motorization of China and India, I'd argue that it is precisely why it's risky to continue powering all vehicles everywhere using refined fossil hydrocarbons. Long before you are old and gray, there will be tensions, possibly military conflict between the major fossil hydrocarbon consumer countries if someone doesn't come up with alternative technology pretty darn quick. The urgency is the result of the long lifespans of vehicles and the caution with which auto makers adopt innovations.
The US and China already fought a proxy war over oil in Sudan, which has led directly to the genocide in Darfur. Iraq/Iran is next, the President issued orders today to kill or capture Iranian operatives in Iraq. China is Iran's biggest customer. In addition, India and Pakistan want to build a pipeline for Iranian natural gas. The EU wants a natural gas pipeline into Northern Iraq and ultimately into Iran as well. Et cetera ad nauseam.
If Allen's comment that BTL is just as profitable as tapping into new fossil hydrocarbon sources sounds wildly optimistic, perhaps that's because we pay the price for access to this energy wealth in terms of our environment/climate, political dependency on autocratic regimes and/or military expenditure. None of these directly affect the bottom line of any oil company, nor prices at the pump. But then, you do also pay taxes, don't you?
Money money MONEY!!!!!its money thats talking and ALOT of it.
And this has nothing to do with suvs or the us. Its simply that now people can make quite alot from tar and so tar is being mined faster nad faster. Even if the us was running on 100% fat white guy sweat the oil wouls still be made for shipment to china.
lambreja:
The current oil extraction from the Alberta tar sands is a bit over 1 million barrel/day going up to something beteewn 3 to 5 million barrels/day by 2020.
At the current rate, it is already a very messy business. The size and number of new chemical ponds-lakes is unbelievable. By 2020 an area almost the size of Florida will be a dirty mess for many generations.
If the nearby Athabascan River (collecting the polluted water) would flow South instead of North, a lot more people would take notice.
Almost 50% of all GHG produced in Canada come from Alberta with only 10% of the population. That is not an acceptable situation. It will get even a lot worse when tar sands oil production is increased fourfolds. At that rate, Alberta may produce about 75% of Canadian GHG by 2020-2030.
Assuming that it can be done, who will pay to clean up this huge mess? The end users? (95% USA) The oil companies? The Canadian taxpayers? (the remaining 5%) The American taxpayers? All of us?
The royalties (about 5 cents/gallon) paid to the Alberta Heritage Fund will never be enough to cover the downstream clean up cost. The latest estimates are 20 to 30 times more. This indicates that a special clean up fund of up to $1/gal should be created. Unfortunately, oil would have to be over$100/barrel to generate enough profits to pay for the downstream cleanup cost.
A massive progressive switch to PHEVs + BEVs is a much better.
From what I have read the end game cleanup should be simpler then many they deal with. The test will come when the first site runs dry and they reclaim it.
Harvey:
Annual Canadian GHG emissions from combustion of fossil fuels is close to 760 million tons of CO2 equivalent. Alberta oil sand exploration emits annually about 29 million tons of CO2 for production of about 1 million barrel/d. | |
## Accardi complementarity for -1/2 < mu < 0 and related results
##### Authors
Cepeda, Lenin A. Echavarria
Ruiz, Claudio Pita
Sontz, Stephen B.
##### Description
We show (an earlier conjecture of the last two authors) that the momentum and position operators of mu-deformed quantum mechanics for -1/2 < mu < 0 are not Accardi complementary. We also prove some related formulas that were conjectured by the same authors.
Comment: 13 pages
##### Keywords
Mathematical Physics, 81Sxx | |
• # question_answer If $S=\sum\limits_{n=0}^{\infty }{\frac{{{(\log x)}^{2n}}}{(2n)\,!},}$ then $S$ = A) $x+{{x}^{-1}}$ B) $x-{{x}^{-1}}$ C) $\frac{1}{2}(x+{{x}^{-1}})$ D) None of these
Solution :
We have $S=\sum\limits_{n=0}^{\infty }{\frac{{{(\log x)}^{2n}}}{(2n)!}=\left( \frac{{{e}^{\log x}}+{{e}^{-\log x}}}{2} \right)}=\frac{x+{{x}^{-1}}}{2}$.
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MathSciNet bibliographic data MR1317645 58F05 (58F25) Ginzburg, Viktor L. An embedding $S^{2n-1}\to {\bf R}^{2n}$$S^{2n-1}\to {\bf R}^{2n}$, $2n-1\geq 7$$2n-1\geq 7$, whose Hamiltonian flow has no periodic trajectories. Internat. Math. Res. Notices 1995, no. 2, 83–97. Article
For users without a MathSciNet license , Relay Station allows linking from MR numbers in online mathematical literature directly to electronic journals and original articles. Subscribers receive the added value of full MathSciNet reviews.
American Mathematical Society 201 Charles Street Providence, RI 02904-6248 USA | |
# Tag Archives: collatz
## Some Pictures of the 3n+1 Problem
The 3n+1 conjecture (a.k.a. the Collatz conjecture) is easy to state, unsolved, probably difficult to prove if true, and can provide us with some pretty pictures. I'm only going to state it and show some pretty pictures. First, what's this conjecture? It's about a simple algorithm: choose a positive integer $n$. If $n=1$, […] | |
Lemma 42.63.2. Let $(S, \delta )$ be as above. Let $X$ be a scheme locally of finite type over $S$. Then we have a canonical identification
$A^ p(X \to S) = \mathop{\mathrm{CH}}\nolimits _{1 - p}(X)$
for all $p \in \mathbf{Z}$.
Proof. Consider the element $[S]_1 \in \mathop{\mathrm{CH}}\nolimits _1(S)$. We get a map $A^ p(X \to S) \to \mathop{\mathrm{CH}}\nolimits _{1 - p}(X)$ by sending $c$ to $c \cap [S]_1$.
Conversely, suppose we have $\alpha \in \mathop{\mathrm{CH}}\nolimits _{1 - p}(X)$. Then we can define $c_\alpha \in A^ p(X \to S)$ as follows: given $X' \to S$ and $\alpha ' \in \mathop{\mathrm{CH}}\nolimits _ n(X')$ we let
$c_\alpha \cap \alpha ' = \alpha \times \alpha '$
in $\mathop{\mathrm{CH}}\nolimits _{n - p}(X \times _ S X')$. To show that this is a bivariant class we write $\alpha = \sum _{i \in I} n_ i[X_ i]$ as in Definition 42.8.1. In particular the morphism
$g : \coprod \nolimits _{i \in I} X_ i \longrightarrow X$
is proper. Pick $i \in I$. If $X_ i$ dominates an irreducible component of $S$, then the structure morphism $p_ i : X_ i \to S$ is flat and we have $\xi _ i = p_ i^* \in A^ p(X_ i \to S)$. On the other hand, if $p_ i$ factors as $p'_ i : X_ i \to s_ i$ followed by the inclusion $s_ i \to S$ of a closed point, then we have $\xi _ i = (p'_ i)^* \circ c_ i \in A^ p(X_ i \to S)$ where $c_ i \in A^1(s_ i \to S)$ is the gysin homomorphism and $(p'_ i)^*$ is flat pullback. Observe that
$A^ p(\coprod \nolimits _{i \in I} X_ i \to S) = \prod \nolimits _{i \in I} A^ p(X_ i \to S)$
Thus we have
$\xi = \sum n_ i \xi _ i \in A^ p(\coprod \nolimits _{i \in I} X_ i \to S)$
Finally, since $g$ is proper we have a bivariant class
$g_* \circ \xi \in A^ p(X \to S)$
by Lemma 42.33.4. The reader easily verifies that $c_\alpha$ is equal to this class (please compare with the proof of Lemma 42.63.1) and hence is itself a bivariant class.
To finish the proof we have to show that the two constructions are mutually inverse. Since $c_\alpha \cap [S]_1 = \alpha$ this is clear for one of the two directions. For the other, let $c \in A^ p(X \to S)$ and set $\alpha = c \cap [S]_1$. It suffices to prove that
$c \cap [X'] = c_\alpha \cap [X']$
when $X'$ is an integral scheme locally of finite type over $S$, see Lemma 42.35.3. However, either $p' : X' \to S$ is flat of relative dimension $\dim _\delta (X') - 1$ and hence $[X'] = (p')^*[S]_1$ or $X' \to S$ factors as $X' \to s \to S$ and hence $[X'] = (p')^*(s \to S)^*[S]_1$. Thus the fact that the bivariant classes $c$ and $c_\alpha$ agree on $[S]_1$ implies they agree when capped against $[X']$ (since bivariant classes commute with flat pullback and gysin maps) and the proof is complete. $\square$
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# Laplace and Poisson's equation.
1. Dec 10, 2013
### yungman
1. The problem statement, all variables and given/known data
I want to verify for $v=\ln( r)$, that
a)$\nabla^2 v=0$ for $v$ on a disk center at $(x_0,y_0)$. Therefore $v$ is Harmonic.
b)$\nabla^2 v=\frac{1}{r^2}$ for $v$ on a sphere center at $(x_0,y_0,z_0)$. Therefore $v$ is not Harmonic.
3. The attempt at a solution
a) For circular disk, $v=\frac{1}{2}\ln[(x-x_0)^2+(y-y_0)^2]$
$$\nabla v=\frac{\hat x (x-x_0)+\hat y(y-y_0)}{[(x-x_0)^2+(y-y_0)^2]^2}$$
$$\nabla^2 v=\nabla\cdot\nabla v=\frac{[(y-y_0)^2-(x-x_0)^2]+[(x-x_0)^2-(y-y_0)^2]}{[(x-x_0)^2+(y-y_0)^2]^2}$$=0
b)For a sphere, $v=\frac{1}{2}\ln[(x-x_0)^2+(y-y_0)^2+(z-z_0)^2]$
$$\nabla v=\frac{\hat x (x-x_0)+\hat y(y-y_0)+\hat z(z-z_0)^2}{[(x-x_0)^2+(y-y_0)^2+(z-z_0)^2]^2}$$
$$\nabla^2 v=\nabla\cdot\nabla v=\frac{[(y-y_0)^2+(z-z_0)^2-(x-x_0)^2]+[(x-x_0)^2+(z-z_0)^2-(y-y_0)^2]+[(x-x_0)^2+(y-y_0)^2-(z-z_0)^2]}{[(x-x_0)^2+(y-y_0)^2+(z-z_0)^2]^2]}$$
$$\Rightarrow\;\nabla^2 v=\frac{(x-x_0)^2+(y-y_0)^2+(z-z_0)^2}{[(x-x_0)^2+(y-y_0)^2+(z-z_0)^2]^2]}=\frac{1}{r^2}$$
So $v$ is Harmonic only for a two dimension disk, not in three dimension sphere. | |
# Free Alabama CDL General Knowledge Practice Test 2023
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List of questions
1.
A vehicle is loaded with very little weight on the drive axle. What may happen?
2.
Which of these statements about managing space to the sides is true?
3.
"According to the driver’s manual, why should you limit the use of your horn?"
4.
When the roads are slippery, you should:
5.
Special permits will be required for loads that are:
6.
A hazard:
7.
"The driver’s manual suggests several things to do when you pass a vehicle. Which of these is NOT one of them?"
8.
The aggregate WLL for logs loaded lengthwise must be at least:
9.
You are driving a vehicle that could be driven safely at 55 mph on an open road. But traffic is heavy and other vehicles drive at the speed of 35 mph, though the speed limit is 55 mph. The safest speed for your vehicle is more likely to be:
10.
When securing a load of lengthwise longwood logs weighing 42,500 lbs, what is the minimum aggregate WLL required for tiedowns? (rounded up to nearest lb)
11.
What is the aggregate WLL of one 7/15 inch Grade 43 chain, and three 5/16 inch Grade 70 chains?
12.
In mountain driving you will have to use a lower gear to drive safely on a grade. Which of these does NOT affect your choice of gear?
13.
If you jam on the brakes and the wheels lock up and skid:
14.
Service brake stopping action should be tested at:
15.
When you sight-side back your vehicle into an alley stopping as close as possible to the rear of the alley it is called:
16.
When entering or crossing traffic, observe all of the following EXCEPT _________.
17.
What factor has the greatest affect on how much weight is moved per hole in the tandem slider rail?
18.
What is the most important reason to inspect your truck or bus?
19.
Load securement for cargo weighing 29,650 lbs must be able to withstand upward force of how many lbs?
20.
As part of the log packing requirements:
21.
Which of these statements about double clutching and shifting is true?
22.
There are two types of jackknife, they are:
23.
Rims with welding repairs or missing lugs:
24.
Which of these is a good thing to remember about using mirrors?
25.
Drivers who have trouble sleeping at night, fall asleep at odd times, or snore loudly, gasp, and choke in their sleep should:
26.
When securing a non-cubic boulder with an unstable base, the four surrounding chains must have a WLL of at least:
27.
The object of the pre-trip inspection is:
28.
A tiedown with a marked WLL of 8,000 lbs directly attached to an article has an actual WLL of:
29.
If you need to leave the road in a traffic emergency, you should:
30.
Some large trucks have curved mirrors. These mirrors:
31.
Should brake adjustment be checked often?
32.
The maximum allowable overall length for semitrailers is:
33.
A tractor trailer traveling 55 mph will travel how far between when the brain perceives a hazard and stopping?
34.
Specific examples of vehicles which drivers should pay close attention to as possible hazards include:
35.
Under the adverse driving conditions exception, how many additional hours can you drive?
36.
Cargo that isn't prevented from forward movement, and is 4 feet long and weighs 1,500 lbs requires a minimum of how many tiedowns?
37.
When securing metal coils in a sided vehicle without anchor points:
38.
When parked on the side of a two-lane road carrying traffic in both directions, place warning devices _________.
39.
When securing metal coils, it is acceptable to use nailed blocking or cleats as the sole means to secure:
40.
When backing, all of the following are important safety rules:
41.
Backing your vehicle in a straight line without crossing over or touching the exercise boundaries is called:
42.
To avoid a crash, you had to drive onto the right shoulder. You are now driving at 40 mph on the shoulder. How should you move back into the pavement?
43.
When securing flattened or crushed cars on a vehicle with containment on 2 sides, how many tiedowns, at minimum, are required?
44.
HOS regulations were designed to do all of the following, except:
45.
What violation occurs in the example below?
Day 1
46.
All of the following are concerns for driving on offramps or onramps EXCEPT:
47.
You should avoid driving through deep puddles or flowing water. But if you must, which of these steps can help to keep your brakes working?
48.
There will be times when you must take a minimum of a 30 minute break off duty before performing which of the following tasks?
49.
You have steer tires rated at 6,150 pounds each, and you're in the state of West Virginia which says the legal limit for the steer axle is 20,000 pounds. What's the maximum legal weight you can carry on your steer axle in West Virginia?
50.
The maximum number of allowable looks and/or pull-ups on any particular exam exercise is: | |
Technical Article
# Turn Your PWM into a DAC
April 06, 2016 by Robert Keim
## If your microcontroller doesn’t have a digital-to-analog converter, you can make a passable replacement with nothing more than a pulse-width-modulated signal and a low-pass filter.
If your microcontroller doesn’t have a digital-to-analog converter, you can make a passable replacement with nothing more than a pulse-width-modulated signal and a low-pass filter.
### When You Lack a DAC
Even in this age of highly integrated mixed-signal integrated circuits, it is not uncommon to come across a microcontroller that does not include a digital-to-analog converter. Programmable logic is even more problematic in this respect; I have never heard of an FPGA or CPLD that has a DAC module. And even when a microcontroller does have a DAC, there may be only one or two channels—in contrast to integrated ADC peripherals, which usually incorporate a multiplexer that allows one ADC module to convert analog signals connected to several, or maybe even dozens, of port pins. So what are you supposed to do when you find a microcontroller that is perfect for your application in every way except that it doesn’t have an integrated DAC? Well, the most obvious option is to use an external DAC. A quick Digi-Key search indicates that you have at least a thousand to choose from, some of which cost less than a dollar and come in tiny SC70, MSOP, SOT, or DFN packages. But there are times when you really don’t want to add another chip to the design. Maybe your microcontroller doesn’t have the three unused pins you will need for SPI communication; maybe you are in a rush and don’t want to pay for overnight shipping; maybe you need six separate DAC outputs but don’t have enough board space for a six-channel device. In any event, if an external DAC is simply out of the question, you have an alternative.
### PWM, Resistor, Capacitor
The bare minimum here is a resistor, a capacitor, and some sort of pulse-width-modulation functionality. You will certainly have no shortage of PWM capabilities if you’re using an FPGA or CPLD. For processors, I think that every microcontroller I’ve ever worked with included PWM hardware, but I suppose there must be some parts that don’t. So that’s the first thing to check—if your microcontroller doesn’t have PWM, you’re out of luck (unless you want to rig up some sort of bit-banging PWM routine, but seriously, if you’re in that boat just use an external DAC). Next, you need a way to low-pass filter the PWM signal. A basic single-pole RC filter could be fine if you don’t mind some ripple on the output, so if all you can fit into your board or budget is a resistor and capacitor, the PWM DAC is still a viable option. However, a better filter means a better DAC, and it might be worth your while to bring in an inductor or an op-amp so you can have two poles instead of one.
### PWM–The Basics
You probably already know what pulse-width modulation is; nevertheless, we’ll briefly review the essential concepts to make sure that we have a solid foundation when we look at how exactly a low-pass filter turns a digital signal into a programmable analog voltage.
A typical digital clock signal is a sequence of periods in which the duration of the logic-high voltage is equal to the duration of the logic-low voltage. In contrast, a PWM signal is a sequence of periods in which the duration of the logic-high (or logic-low) voltage varies according to external conditions, and these variations can be used to transmit information. If you are familiar with radio circuitry, you know that information is transmitted by means of sinusoidal signals to which some type of modulation is applied. This situation is analogous to PWM functionality—instead of amplitude or frequency modulation we have pulse-width modulation. You may find it helpful to think in terms of this conceptual similarity: We all know that an analog audio signal can be transmitted from an antenna to a car radio by first modulating a carrier wave and then processing the received signal in a way that removes the carrier and recovers the original audio information. Likewise, we can generate a programmable analog voltage by pulse-width modulating a digital carrier wave then “transmitting” this modulated signal to a low-pass filter.
In the above diagram, logic high is identified as the “ON” or active state, and logic low is the “OFF” or inactive state. In the first period, the duration of the active state is equal to the duration of the inactive state. Then, for the next two periods, the active-state duration increases by one grid width; this means that the inactive-state duration must decrease by one grid width, because the PWM carrier frequency (and thus the PWM period) is constant. In the context of our PWM DAC, we don’t really need to know the absolute active and inactive durations; what matters is the ratio between the durations, which we discuss in terms of the PWM duty cycle:
$duty\ cycle=\frac{T_{ON}}{T_{ON}+T_{OFF}}$
### From Duty Cycle to Analog Voltage
The nominal DAC voltage observed at the output of the low-pass filter is determined by just two parameters, namely, the duty cycle and the PWM signal’s logic-high voltage; in the diagram, this logic-high voltage is denoted by A for “amplitude.” The relationship between duty cycle, amplitude, and nominal DAC voltage is fairly intuitive: In the frequency domain, a low-pass filter suppresses higher-frequency components of an input signal. The time-domain equivalent of this effect is smoothing, or averaging—thus, by low-pass filtering a PWM signal we are extracting its average value. Let’s assume the duty cycle is 50% (i.e., active duration equals inactive duration) and we are working with 3.3 V logic. You can probably guess what the nominal DAC voltage will be: 1.65 V, because the signal spends half of its time at 3.3 V and half at 0 V, and thus the smoothed-out version will end up right in the middle. We can generalize this as follows:
$desired\ DAC\ voltage=A\times duty\ cycle$
One of the first specs you look at when choosing a DAC is the “resolution,” which is a somewhat vague term expressed in the somewhat vague unit of “bits.” What we really mean by “resolution” is “how many distinct output voltages (or currents) can the DAC generate?” The number of “bits” refers to the data register that controls the digital-to-analog circuitry, such that a 10-bit DAC can generate 210 = 1024 distinct output voltages. If you understand this much, you will see that we can readily identify the equivalent resolution of a PWM DAC.
Let’s assume that the PWM signal shown in the diagram is restricted to pulse widths that are a multiple of one grid. This means that the duty cycle can assume 8 distinct values: 0%, ~14%, ~29%, ~43%, ~57%, ~71%, ~86%, and 100%. Each duty cycle corresponds to a particular output voltage, so what we have here is a 3-bit DAC, because 23 = 8.
To determine the resolution of your real-life PWM DAC, just apply this same analysis: how many distinct duty cycles can you generate? It is usually not too difficult to answer this question, because the central element in a standard PWM hardware block is an N-bit counter that controls the width of the pulse, meaning that the equivalent DAC resolution is 2N. For example, the datasheet for Atmel’s SAM4S microcontroller series includes the following characteristics for its PWM controller:
That 16-bit counter means 16 bits of resolution, or 216 = 65,536 distinct voltages ranging from 0 V to VDDIO (which can be anywhere from 1.62 V to 3.6 V). As another example, Silicon Labs’ EFM8UB1 microcontrollers feature variable PWM resolution, and consequently the reference manual is explicit regarding the resolution specs:
### Conclusion
At this point, it may seem like there is hardly any use for a regular DAC when you can get 16 bits of resolution from a PWM-plus-RC-filter implementation. But of course, this is not the whole story—there is more to a DAC than resolution. In the next article, we will use simulations to more thoroughly explore concepts, circuits, and performance limitations related to PWM digital-to-analog conversion.
Next Article in Series: Low-Pass Filter a PWM Signal into an Analog Voltage
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### Continuity and Differentiability
goals
Rewrite the following equations in the logarithm from :
$5^0 \, = \, 1$
If $\log^2 \left(\dfrac{a}{b}\right) + \log^2 \left(\dfrac{b}{c}\right) + \log^2 \left(\dfrac{c}{a}\right) + \log^2 a + log^2 b + \log^2 c - log \, abc + k \ge 0$ where a, b, c, k are real position numbers then value of 'k' can be
Find derivative of following function with respect to x
Find $\dfrac{dy}{dx}$ of the functions given
Differentiate the following function w.r.t.x.
$\dfrac{1}{(x^{2}+3^{2})}$ | |
# ComePLAY Titanium Handlebars vs Aluminium vs 500T Press
## Introduction
A viewer of the Hambini Youtube channel was on the lookout for some new titanium handlebars for his bicycle and found a Chinese company called Comeplay which specializes in titanium production. The handlebars retail at 180 and are available in a range of sizes and angles to accommodate most user requirements There are a lot of choices when it comes to bike handlebars. You can go with aluminium, carbon fibre, or even titanium. So why choose Titanium? Titanium is strong and lightweight, making it the perfect choice for handlebars. Comeplay Titanium Handlebars are some of the strongest on the market (see below) and cheap too! ## Handlebar Materials Aluminium is by far the most popular choice for handlebar materials. Caron fibre is lightweight and comes with a significant cost increase. By comparison, Titanium has been prohibitively expensive. ### Titanium Titanium is a strong, lightweight metal that has many desirable properties. It is used in a variety of applications, from motorsport to aerospace. Titanium is known for its high strength-to-weight ratio, making it a popular choice for many products. It is also corrosion-resistant and non-toxic, making it a safe choice for many products. Titanium is an excellent choice for products that need to be strong and lightweight. ### Aluminium 6061 Aluminium is strong for its weight. It has a yield strength of about 240MPa 38,000 pounds per square inch (psi). This means that it can be bent or stretched until it reaches 240MPa without breaking. Aluminium also has a tensile strength of about 290MPa 45,000 psi, meaning it can be pulled until it reaches 45,000 psi without breaking. These strengths make aluminium a good choice for parts that need to be strong yet lightweight. ### Carbon Fibre Carbon fibre composites are widely used in the aerospace, automotive, and sporting goods industries because of their high strength-to-weight ratios. The carbon fibres are extremely strong and stiff, while the resin matrix is relatively lightweight. This combination results in composites that are up to five times stronger than steel and twice as stiff. However, the resin matrix can also be brittle, so care must be taken during fabrication and handling to avoid damage. ## Handlebar Geometry Handlebars are semi-made to order and standard sizes can be found off the shelf, whilst any amount of customization is available. In this case, the review sample had 42cm C-C with a central 31.8mm clamp diameter with a 1″ reduction in flaring towards drops which were formed by composite radial bends having differing radii. The boundary geometry was generally very good, the ultimate centre of the handlebar drop end had a misalignment of approximately 2mm in the lateral direction. Following external consultation, this tolerance is considered to be excellent. ## Handlebar Testing Without a dedicated jig to test handlebars, it is quite difficult to ascertain the strength and stiffness between suppliers. Hence it was decided to subject a set of Aluminium handlebars and Titanium handlebars to some basic stiffness testing with a large hydraulic press. The test plotted the system pressure against the ram position. This test did not look at the vibration damping (compliance) or comfort aspects of the handlebars as these elements are very much user preference. Some photos during testing against an aluminium handlebar are shown below. Note the extensive flexion in the aluminium handlebar vs the ComePLAY Ti Handlebar Photos of the creased structure are shown below ## Analysis of Results The Titanium handlebar is considerably stronger and stiffer than the aluminium handlebar. The Ti handlebar is over twice as stiff and when loaded beyond it’s elastic limit, it deforms locally. The aluminium handlebar has a much lower ultimate tensile stress and flexes quite quickly before sustaining a general buckling failure accompanied by a fracture of the bar. In contrast, the titanium handlebar suffered buckling of the skin. It’s likely a rider would feel the reduced wind up (slack to take up the load) with the Ti Handlebar over the aluminium handlebar. This “feeling” is often perceived by cyclists to be the inherent stiffness. ## Overall There is a general thought amongst many people of poor quality emanating from Chinese manufacturers. There is little doubt that ComePLAY’s marketing would be considered to be borderline comical by western standards but their handlebars were found to be well-made, very stiff and extremely strong. The cost of180 is very reasonable given the raw material cost. It is not difficult to recommend these handlebars.
The likelihood is most people will not notice the difference between titanium handlebars and aluminium handlebars from a structural perspective unless they are heavy or engaging in hard efforts like sprinting. The low wind up defelction is a nice benefit.
The handlebars do have a slight weight penalty although the unit which was tested is quite wide. | |
# Evaluation of a hypergeometric function
I am working with functions like
f[z_] = Hypergeometric2F1[4, 4, 8, z]
Here is a plot of this function over the interval $$z \in [0,1]$$:
Plot[f[z], {z, 0, 1}]
As you can see, Mathematica has difficulties evaluating it in the region $$z \approx 0$$. This is surprising, because the hypergeometric function admits by definition a simple series expansion around $$z = 0$$, $$f(z) = \sum_{n = 0}^\infty \frac{7!}{(3!)^2} \frac{(n+1)(n+2)(n+3)}{(n+4)(n+5)(n+6)(n+7)} z^n = 1 + 2 z + \frac{25}{9} z^2 + \ldots$$
The problem is that Mathematica does not use this defining property of the hypergeometric function, but instead it "simplifies" it to
f[z] = (140*(-60*z + 60*z^2 - 11*z^3 - 60*Log[1 - z] + 90*z*Log[1 - z] - 36*z^2*Log[1 - z] + 3*z^3*Log[1 - z]))/(3*z^7)
and it turns out that cancellations between large numbers occur in this expression when $$z \approx 0$$.
How can I instruct mathematica to not perform this "simplification" in general? Is there a way I can use the command Hold or something similar?
What I want to do eventually is evaluate numerically some integrals in which $$f(z)$$ appears in the integrand, so I need a robust way of evaluating the function over the interval $$z \in [0,1]$$.
• The symbolic calculation can be stopped by f[z_] := Hypergeometric2F1[4, 4, 8, z] or more strictly Clear[f]; f[z_?NumericQ] := Hypergeometric2F1[4, 4, 8, z], but these solutions may not be suitable for the real problem. I think it's better to make the question more specific e.g. show an integral you want to calculate. (The integral should be simplified as much as possible of course. ) BTW somewhat related: mathematica.stackexchange.com/q/117888/1871 – xzczd Mar 12 at 15:57
• Your function can be expressed in terms of the Legendre function of the second kind. You could try reformulating in terms of LegendreQ[]. – J. M. will be back soon Mar 12 at 15:59
• @J.M.isslightlypensive, the Legendre functions get auto-expanded in the same way: the relevant one here is LegendreQ[3, 1 - 2/z], which automatically becomes $-\frac{10}{x^2}+\frac{\left(1-\frac{2}{x}\right) \left(x^2-10 x+10\right) \left(\frac{1}{2} \log \left(2-\frac{2}{x}\right)-\frac{1}{2} \log \left(\frac{2}{x}\right)\right)}{x^2}+\frac{10}{x}-\frac{11}{6}$ upon evaluation and then has the same numerical issues. – Roman Mar 13 at 10:50
The commands
Plot[Hypergeometric2F1[4, 4, 8, z], {z, 0, 1}, WorkingPrecision -> 15]
or/and
f[z_] = Hypergeometric2F1[4, 4, 8, z]; Plot[f[z], {z, 0, 1}, WorkingPrecision -> 25]
do the job.
Integrate[Hypergeometric2F1[4, 4, 8, z], {z, 0, 1}]
$$\frac{133}{9}$$
NIntegrate[Hypergeometric2F1[4, 4, 8, z], {z, 0, 1}, WorkingPrecision -> 15]
$$14.7777777777777$$
I'm not sure if this will help, but it might be worth checking out.
f[z_] = Hypergeometric2F1[4, 4, 8, z] // FunctionExpand
(*(140 (-(11 z^3) + 3 z^3 Log[1 - z] + 60 z^2 - 36 z^2 Log[1 - z] -
60 z + 90 z Log[1 - z] - 60 Log[1 - z]))/(3 z^7)*)
<< PrecisionPlot
PrecisionPlot[f[z], {z, 0, 1}]
You can get PrecisionPlot from http://library.wolfram.com/infocenter/MathSource/715/
The problem seems to be precision rather than the function Mathematica uses. Compare:
f[1/1000] // N
(*2.47395*10^8*)
with
f[1/1000] // N[#, 50] &
(*1.0020027811148271960352800702800453256995763414200*)
and while f[0] doesn't work, the limit does.
Limit[f[z], z -> 0]
(*1*)
Although PrecisionPlot is smoother, it doesn't really do a great job of matching the very small values of z.
• PrecisionPlot[] was useful back in the day when Plot[] did not have the WorkingPrecision` option; nowadays, the built-in function is adequate. – J. M. will be back soon Mar 13 at 14:03 | |
# Estate
Semicircle estate must be fence. The straight section has 26 meters long fence. How many meters of fence should buy?
Result
x = 40.8 m
#### Solution:
$x = \pi r = \pi \dfrac{ 26}{2} = 40.8 \ \text{m}$
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# Ecosystem#
Fugue can be used in conjuction with a lot of other Python libraries. Some of these integrations are native where Fugue can be used as a backend. For the others, there is no native integration but they can be used together with minimal lines of code, normally through the transform() function.
Have questions? Chat with us on Github or Slack:
## Data Validation#
Pandera
Pandera is a lightweight data validation framework originally designed to provide a minimal interface in validating Pandas DataFrames. Pandera has seen expanded to Spark and Dask libraries through Koalas and Modin, but it can also be used pretty seamlessly with Fugue. Fugue also supports validation by partition.
## Machine Learning#
PyCaret
PyCaret is an open-source low-code machine learning library that allows users to train dozens of models in a few lines of code. With a native integration, Fugue users and distribute the machine learning training over Spark, Dask or Ray.
Nixtla
Nixtla is a project focused on state-of-the-art time series modelling. The current Fugue integration is around their statistical forecasting packages named statsforecast. Fugue lets users apply AutoARIMA and ETS models to forecast millions of independent timeseries on top of distributed compute.
## Orchestration#
Prefect
[Prefect] is an open-source workflow orchestration framework used for scheduling and monitoring tasks. The prefect-fugue collection allows users to iterate locally, and then bring the code to Databricks or Coiled for execution when production ready.
Ploomber (Coming Soon) | |
# Chapter 11 - Section 11.3 - Series - Exercise Set: 5
$20$
#### Work Step by Step
We sum from i = low bound to i = high bound. $(2^2-3) + (3^2-3) + (4^2-3) = (4-3) + (9-3) + (16-3) = 1 + 6 + 13 = 20$
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback. | |
# All Questions
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# Group Theory
1. Mar 3, 2005
### PhysKid24
Hi. Can anyone help me figure out how to find the conjugacy classes for a certain group and the elements in each class. I'm looking at the dihedral group of degree 5 (D5). I found the 10 elements in the group, but I dont know how to get the conjugacy classes and the elements in them? Can anyone help? Thanks. In the group, a^5=e; b^2=e; ab=ba^-1
2. Mar 4, 2005
### matt grime
Do it. Take an element x, work out all its conjugates. Use the relations to help:
aba^{-1} = ba^{-2}
so b and ba^{-2} are conjugate. Rinse and repeat, There are two cases depending on n.
3. Mar 11, 2005
### ComputerGeek
you are assuming the group is Abelian. Do you know for a fact that D5 in this situation is Abelian?
4. Mar 11, 2005
### Dr Transport
point groups are Abelian if memory serves me correctly.
5. Mar 11, 2005
### matt grime
I am most definitely NOT assuming the group is abelian.
HINT: G is abelian IFF aba^{-1}=b for all a and b. I think you'll find I wrote
aba^{-1} = ba^{-2}
6. Mar 11, 2005
### ComputerGeek
wow... some how I thought you wrote:
b = ba^{-2}
Don't know where that came from
Onto another question though. How do you know that ba^{-2} is the conjugate of b from:
aba^{-1} = ba^{-2}
I am not seeing the steps between.
7. Mar 11, 2005
### matt grime
It is a dihedral group with generators a and b satisfying a^n=e=b^2 ab=ba^{-1], (n=5 for this particular example).
If you don't see why b and ba^{-2} are conjugate then this implies in my mind that you do not know what conjugate means.
b=ba^{-2} simply implies that a^2=e, that is all, by the way, nothing to do with abelian or otherwise.
Last edited: Mar 11, 2005
8. Mar 22, 2005
### Oxymoron
Matt, can I ask a question while you are here?
On the subject of Dihedral groups, I am considering $$\mathcal{D}_n$$, the dihedral group of order $$2n$$.
How would I go about finding the normal subgroups of $$\mathcal{D}_n$$. Do I consider the two distinct cases separately? That is, first let $$n=even$$ then work out the rotation maps $$a$$ and relfection maps $$b$$, and then let $$n=odd$$ and do the same thing?
What kind of things should I recognize (if any)? And will I be surprised?
I am just starting to think about these things, so I have no idea what to expect.
Thanks for any insight.
9. Mar 22, 2005
### Oxymoron
Continuing on from what I said.
if I consider the case where $$n$$ is even. Then obviously
$$a^n = e$$
$$b^2 = e$$
$$bab^{-1} = a^{-1}$$
and for $$n$$ odd, is it different? I dont even know if I'm on the right track.
10. Mar 23, 2005
### matt grime
A simple test for normality of a subgroup:
N is normal in G if and only if N is a union of conjugacy classes (this is one proof that A_5 is simple)
Your second post doesn't seem related to the first. The relations defining the dihedral group say nothing about whether n is even or odd.
11. Mar 23, 2005
### mathwonk
you might also look at the geomketric picture of this group, i.e. its action on a polygon of n sides. |
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