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https://51zuoyejun.com/caseDetail/5459.html | 1,701,573,608,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100484.76/warc/CC-MAIN-20231203030948-20231203060948-00389.warc.gz | 109,881,857 | 7,020 | # 程序代写案例-MATH3871/MATH5960-Assignment 1
MATH3871/MATH5960
Assignment 1
Assignment 1
This assignment covers material in Lectures 1–3. Assignment is worth 15% of final course
of grade will be allocated for neat and concise presentation. Please refer to the
following instructions:
• Assignment to be submitted via Moodle by 7 October 11:55PM AEDT
• Include in your assignment, any relevant R code, R output, and mathematical derivations.
Embed the code and plots into your assignment (please don’t attach R markdown or other
R script files)
• The total number of submitted pages should not exceed 6 A4 pages. Any pages submitted
in excess of 6 pages will not be graded.
• Print, sign and attach this cover sheet with your assignment (not included in page count).
• Refer to course handout for grading of late submissions
Plagiarism Statement
I declare that this assessment item is my own work, except where acknowledged, and has not
been submitted for academic credit elsewhere. I acknowledge that the assessor of this item may,
for the purpose of assessing this item reproduce this assessment item and provide a copy to
another member of UNSW; and/or communicate a copy of this assessment item to a plagiarism
checking service (which may then retain a copy of the assessment item on its database for the
purpose of future plagiarism checking).
I certify that I have read and understood UNSW Rules in respect of Student Academic Mis-
conduct.
Name (print clearly):
Student Number:
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1
1. Inference: Let θ be the true proportion of people over the age of 40 in your community
with hypertension. Consider the following thought experiment:
(a) Though you may have little or no expertise in this area, give an initial point estimate
of θ.
(b) Now suppose a survey to estimate θ is established in your community, and of the first
5 randomly selected people, 4 are hypertensive. How does this information affect
(c) Finally, suppose that at the survey’s completion, 400 of 1000 people have emerged
as hypertensive. Now what is your estimate of θ?
2. Multivariate Priors: Let x1, . . . , xn ∈ Rd be n iid d-dimensional vectors. Suppose that
we wish to model xi ∼ Nd(µ,Σ) for i = 1, . . . , n where µ ∈ R is an unknown mean vector,
and Σ is a known positive semi-definite covariance matrix.
(a) Adopting the conjugate prior µ ∼ Nd(µ0,Σ0) show that the resulting posterior dis-
tribution for µ|x1, . . . , xn is Nd(µˆ, Σˆ) where
µˆ = (Σ−10 + nΣ
−1)−1(Σ−10 µ0 + nΣ
−1x¯)
and
Σˆ = (Σ−10 + nΣ
−1)−1.
(b) Derive Jeffreys’ prior piJ(µ) for µ.
places on the internet. One such place is https://en.wikipedia.org/wiki/Matrix calculus.
3. Importance Sampling: There are many ways to compute or estimate pi. A very sim-
ple estimation procedure is via importance sampling. Suppose that samples x1, . . . , xn
were obtained uniformly inside a square with side length 2r (see diagram), where each
xi = (x
(1)
i , x
(2)
i ) for i = 1, . . . , n.
r
2
Now define bi = 1 if xi is also inside the circle of radius r, and bi = 0 otherwise. Then
pˆ = 1
n
∑n
i=1 bi is an estimate of the ratio of the area of the circle to the area of the square.
Given that we know the true value of p for this setting, we can then obtain an estimate
of pi.
(a) Show that the estimate of pi is given by 4pˆ.
(b) Estimate pi using n = 1, 000 samples.
(c) Using the central limit theorem, determine the Monte Carlo sampling variability of
pˆi (i.e. derive the asymptotic distribution of pˆi as n gets large).
(d) Construct a histogram of 1, 000 estimates of pˆi, each based on n = 1, 000 samples.
Superimpose the Monte Carlo sampling variability distribution from part (c) under
the assumption that the true value for p=0.7854, and verify that it matches the
experimental result.
(e) Without using the true value of p, based on the Monte Carlo sampling variability,
determine what sample size, n, is needed if we require to estimate pi to within 0.01
with at least 95% probability.
(Hint: You will need to use a value for p in order to obtain this value. Choose the
value of p that gives the most conservative value of n, so that you can be sure that
you have estimated pi to the desired accuracy.)
3
Email:51zuoyejun
@gmail.com | 1,097 | 4,197 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2023-50 | latest | en | 0.926802 |
https://www.lmfdb.org/ModularForm/GL2/Q/holomorphic/273/2/bw/a/ | 1,624,397,293,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488519735.70/warc/CC-MAIN-20210622190124-20210622220124-00629.warc.gz | 736,534,764 | 55,679 | # Properties
Label 273.2.bw.a Level $273$ Weight $2$ Character orbit 273.bw Analytic conductor $2.180$ Analytic rank $0$ Dimension $4$ CM discriminant -3 Inner twists $4$
# Related objects
## Newspace parameters
Level: $$N$$ $$=$$ $$273 = 3 \cdot 7 \cdot 13$$ Weight: $$k$$ $$=$$ $$2$$ Character orbit: $$[\chi]$$ $$=$$ 273.bw (of order $$12$$, degree $$4$$, minimal)
## Newform invariants
Self dual: no Analytic conductor: $$2.17991597518$$ Analytic rank: $$0$$ Dimension: $$4$$ Coefficient field: $$\Q(\zeta_{12})$$ Defining polynomial: $$x^{4} - x^{2} + 1$$ Coefficient ring: $$\Z[a_1, \ldots, a_{7}]$$ Coefficient ring index: $$1$$ Twist minimal: yes Sato-Tate group: $\mathrm{U}(1)[D_{12}]$
## $q$-expansion
Coefficients of the $$q$$-expansion are expressed in terms of a primitive root of unity $$\zeta_{12}$$. We also show the integral $$q$$-expansion of the trace form.
$$f(q)$$ $$=$$ $$q + ( -2 \zeta_{12} + \zeta_{12}^{3} ) q^{3} + ( -2 \zeta_{12} + 2 \zeta_{12}^{3} ) q^{4} + ( \zeta_{12} - 3 \zeta_{12}^{3} ) q^{7} + 3 q^{9} +O(q^{10})$$ $$q + ( -2 \zeta_{12} + \zeta_{12}^{3} ) q^{3} + ( -2 \zeta_{12} + 2 \zeta_{12}^{3} ) q^{4} + ( \zeta_{12} - 3 \zeta_{12}^{3} ) q^{7} + 3 q^{9} + ( 4 - 2 \zeta_{12}^{2} ) q^{12} + ( 4 \zeta_{12} - 3 \zeta_{12}^{3} ) q^{13} + ( 4 - 4 \zeta_{12}^{2} ) q^{16} + ( 3 + 5 \zeta_{12} - 5 \zeta_{12}^{2} - 3 \zeta_{12}^{3} ) q^{19} + ( -4 + 5 \zeta_{12}^{2} ) q^{21} + 5 \zeta_{12} q^{25} + ( -6 \zeta_{12} + 3 \zeta_{12}^{3} ) q^{27} + ( -2 + 6 \zeta_{12}^{2} ) q^{28} + ( 1 - \zeta_{12} - 6 \zeta_{12}^{2} - 5 \zeta_{12}^{3} ) q^{31} + ( -6 \zeta_{12} + 6 \zeta_{12}^{3} ) q^{36} + ( -7 - 7 \zeta_{12} + 4 \zeta_{12}^{2} + 3 \zeta_{12}^{3} ) q^{37} + ( -7 + 2 \zeta_{12}^{2} ) q^{39} + ( 6 + 6 \zeta_{12}^{2} ) q^{43} + ( -4 \zeta_{12} + 8 \zeta_{12}^{3} ) q^{48} + ( -3 - 5 \zeta_{12}^{2} ) q^{49} + ( -8 + 6 \zeta_{12}^{2} ) q^{52} + ( -8 - \zeta_{12} + \zeta_{12}^{2} + 8 \zeta_{12}^{3} ) q^{57} + ( 18 \zeta_{12} - 9 \zeta_{12}^{3} ) q^{61} + ( 3 \zeta_{12} - 9 \zeta_{12}^{3} ) q^{63} + 8 \zeta_{12}^{3} q^{64} + ( -2 + 7 \zeta_{12} - 7 \zeta_{12}^{2} + 2 \zeta_{12}^{3} ) q^{67} + ( -1 - \zeta_{12} + 9 \zeta_{12}^{2} - 8 \zeta_{12}^{3} ) q^{73} + ( -5 - 5 \zeta_{12}^{2} ) q^{75} + ( -10 + 4 \zeta_{12} + 6 \zeta_{12}^{2} + 6 \zeta_{12}^{3} ) q^{76} + ( -7 \zeta_{12} - 7 \zeta_{12}^{3} ) q^{79} + 9 q^{81} + ( -2 \zeta_{12} - 8 \zeta_{12}^{3} ) q^{84} + ( 6 - 11 \zeta_{12}^{2} ) q^{91} + ( -4 + 4 \zeta_{12} + 11 \zeta_{12}^{2} + 7 \zeta_{12}^{3} ) q^{93} + ( 8 - 8 \zeta_{12} - 11 \zeta_{12}^{2} - 3 \zeta_{12}^{3} ) q^{97} +O(q^{100})$$ $$\operatorname{Tr}(f)(q)$$ $$=$$ $$4q + 12q^{9} + O(q^{10})$$ $$4q + 12q^{9} + 12q^{12} + 8q^{16} + 2q^{19} - 6q^{21} + 4q^{28} - 8q^{31} - 20q^{37} - 24q^{39} + 36q^{43} - 22q^{49} - 20q^{52} - 30q^{57} - 22q^{67} + 14q^{73} - 30q^{75} - 28q^{76} + 36q^{81} + 2q^{91} + 6q^{93} + 10q^{97} + O(q^{100})$$
## Character values
We give the values of $$\chi$$ on generators for $$\left(\mathbb{Z}/273\mathbb{Z}\right)^\times$$.
$$n$$ $$92$$ $$106$$ $$157$$ $$\chi(n)$$ $$-1$$ $$\zeta_{12}$$ $$-\zeta_{12}^{2}$$
## Embeddings
For each embedding $$\iota_m$$ of the coefficient field, the values $$\iota_m(a_n)$$ are shown below.
For more information on an embedded modular form you can click on its label.
Label $$\iota_m(\nu)$$ $$a_{2}$$ $$a_{3}$$ $$a_{4}$$ $$a_{5}$$ $$a_{6}$$ $$a_{7}$$ $$a_{8}$$ $$a_{9}$$ $$a_{10}$$
11.1
−0.866025 − 0.500000i −0.866025 + 0.500000i 0.866025 + 0.500000i 0.866025 − 0.500000i
0 1.73205 1.73205 1.00000i 0 0 −0.866025 + 2.50000i 0 3.00000 0
149.1 0 1.73205 1.73205 + 1.00000i 0 0 −0.866025 2.50000i 0 3.00000 0
158.1 0 −1.73205 −1.73205 + 1.00000i 0 0 0.866025 2.50000i 0 3.00000 0
254.1 0 −1.73205 −1.73205 1.00000i 0 0 0.866025 + 2.50000i 0 3.00000 0
$$n$$: e.g. 2-40 or 990-1000 Significant digits: Format: Complex embeddings Normalized embeddings Satake parameters Satake angles
## Inner twists
Char Parity Ord Mult Type
1.a even 1 1 trivial
3.b odd 2 1 CM by $$\Q(\sqrt{-3})$$
91.bd odd 12 1 inner
273.bw even 12 1 inner
## Twists
By twisting character orbit
Char Parity Ord Mult Type Twist Min Dim
1.a even 1 1 trivial 273.2.bw.a yes 4
3.b odd 2 1 CM 273.2.bw.a yes 4
7.c even 3 1 273.2.bv.a 4
13.f odd 12 1 273.2.bv.a 4
21.h odd 6 1 273.2.bv.a 4
39.k even 12 1 273.2.bv.a 4
91.bd odd 12 1 inner 273.2.bw.a yes 4
273.bw even 12 1 inner 273.2.bw.a yes 4
By twisted newform orbit
Twist Min Dim Char Parity Ord Mult Type
273.2.bv.a 4 7.c even 3 1
273.2.bv.a 4 13.f odd 12 1
273.2.bv.a 4 21.h odd 6 1
273.2.bv.a 4 39.k even 12 1
273.2.bw.a yes 4 1.a even 1 1 trivial
273.2.bw.a yes 4 3.b odd 2 1 CM
273.2.bw.a yes 4 91.bd odd 12 1 inner
273.2.bw.a yes 4 273.bw even 12 1 inner
## Hecke kernels
This newform subspace can be constructed as the kernel of the linear operator $$T_{2}$$ acting on $$S_{2}^{\mathrm{new}}(273, [\chi])$$.
## Hecke characteristic polynomials
$p$ $F_p(T)$
$2$ $$T^{4}$$
$3$ $$( -3 + T^{2} )^{2}$$
$5$ $$T^{4}$$
$7$ $$49 + 11 T^{2} + T^{4}$$
$11$ $$T^{4}$$
$13$ $$169 - 22 T^{2} + T^{4}$$
$17$ $$T^{4}$$
$19$ $$1369 + 74 T + 2 T^{2} - 2 T^{3} + T^{4}$$
$23$ $$T^{4}$$
$29$ $$T^{4}$$
$31$ $$169 + 286 T + 137 T^{2} + 8 T^{3} + T^{4}$$
$37$ $$2209 + 94 T + 101 T^{2} + 20 T^{3} + T^{4}$$
$41$ $$T^{4}$$
$43$ $$( 108 - 18 T + T^{2} )^{2}$$
$47$ $$T^{4}$$
$53$ $$T^{4}$$
$59$ $$T^{4}$$
$61$ $$( -243 + T^{2} )^{2}$$
$67$ $$169 - 286 T + 242 T^{2} + 22 T^{3} + T^{4}$$
$71$ $$T^{4}$$
$73$ $$2116 - 1564 T + 338 T^{2} - 14 T^{3} + T^{4}$$
$79$ $$21609 + 147 T^{2} + T^{4}$$
$83$ $$T^{4}$$
$89$ $$T^{4}$$
$97$ $$27889 - 4676 T + 221 T^{2} - 10 T^{3} + T^{4}$$ | 2,852 | 5,631 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2021-25 | latest | en | 0.400111 |
http://statsmodels.sourceforge.net/devel/examples/generated/example_gmle.html | 1,455,285,138,000,000,000 | text/html | crawl-data/CC-MAIN-2016-07/segments/1454701164268.69/warc/CC-MAIN-20160205193924-00234-ip-10-236-182-209.ec2.internal.warc.gz | 216,068,056 | 6,058 | # Generic Maximum Likelihood Models¶
This tutorial explains how to quickly implement new maximum likelihood models in statsmodels. The GenericLikelihoodModel class eases the process by providing tools such as automatic numeric differentiation and a unified interface to scipy optimization functions. Using statsmodels, users can fit new MLE models simply by “plugging-in” a log-likelihood function.
## Negative Binomial Regression for Count Data¶
Consider a negative binomial regression model for count data with log-likelihood (type NB-2) function expressed as:
with a matrix of regressors , a vector of coefficients , and the negative binomial heterogeneity parameter .
Using the nbinom distribution from scipy, we can write this likelihood simply as:
In [1]: import numpy as np
In [2]: from scipy.stats import nbinom
In [3]: def _ll_nb2(y, X, beta, alph):
...: mu = np.exp(np.dot(X, beta))
...: size = 1 / alph
...: prob = size / (size + mu)
...: ll = nbinom.logpmf(y, size, prob)
...: return ll
...:
## New Model Class¶
We create a new model class which inherits from GenericLikelihoodModel:
In [4]: from statsmodels.base.model import GenericLikelihoodModel
In [5]: class NBin(GenericLikelihoodModel):
...: def __init__(self, endog, exog, **kwds):
...: super(NBin, self).__init__(endog, exog, **kwds)
...: def nloglikeobs(self, params):
...: alph = params[-1]
...: beta = params[:-1]
...: ll = _ll_nb2(self.endog, self.exog, beta, alph)
...: return -ll
...: def fit(self, start_params=None, maxiter=10000, maxfun=5000, **kwds):
...: if start_params == None:
...: # Reasonable starting values
...: start_params = np.append(np.zeros(self.exog.shape[1]), .5)
...: start_params[0] = np.log(self.endog.mean())
...: return super(NBin, self).fit(start_params=start_params,
...: maxiter=maxiter, maxfun=maxfun,
...: **kwds)
...:
Two important things to notice:
• nloglikeobs: This function should return one evaluation of the negative log-likelihood function per observation in your dataset (i.e. rows of the endog/X matrix).
• start_params: A one-dimensional array of starting values needs to be provided. The size of this array determines the number of parameters that will be used in optimization.
That’s it! You’re done!
## Usage Example¶
The Medpar dataset is hosted in CSV format at the Rdatasets repository. We use the read_csv function from the Pandas library to load the data in memory. We then print the first few columns:
In [6]: import pandas as pd
In [7]: url = 'http://vincentarelbundock.github.com/Rdatasets/csv/COUNT/medpar.csv'
Out[9]:
Unnamed: 0 los hmo white died age80 type type1 type2 type3 provnum
0 1 4 0 1 0 0 1 1 0 0 30001
1 2 9 1 1 0 0 1 1 0 0 30001
2 3 3 1 1 1 1 1 1 0 0 30001
3 4 9 0 1 0 0 1 1 0 0 30001
4 5 1 0 1 1 1 1 1 0 0 30001
The model we are interested in has a vector of non-negative integers as dependent variable (los), and 5 regressors: Intercept, type2, type3, hmo, white.
For estimation, we need to create 2 numpy arrays (pandas DataFrame should also work): a 1d array of length N to hold los values, and a N by 5 array to hold our 5 regressors. These arrays can be constructed manually or using any number of helper functions; the details matter little for our current purposes. Here, we build the arrays we need using the Patsy package:
In [10]: import patsy
In [11]: y, X = patsy.dmatrices('los~type2+type3+hmo+white', medpar)
In [12]: print y[:5]
[[ 4.]
[ 9.]
[ 3.]
[ 9.]
[ 1.]]
In [13]: print X[:5]
[[ 1. 0. 0. 0. 1.]
[ 1. 0. 0. 1. 1.]
[ 1. 0. 0. 1. 1.]
[ 1. 0. 0. 0. 1.]
[ 1. 0. 0. 0. 1.]]
Then, we fit the model and extract some information:
In [14]: mod = NBin(y, X)
In [15]: res = mod.fit()
Optimization terminated successfully.
Current function value: 3.209014
Iterations: 805
Function evaluations: 1238
Extract parameter estimates, standard errors, p-values, AIC, etc.:
In [16]: res.params
Out[16]: array([ 2.3103, 0.2213, 0.7061, -0.068 , -0.129 , 0.4458])
In [17]: res.bse
Out[17]: array([ 0.0679, 0.0506, 0.0761, 0.0533, 0.0685, 0.0198])
In [18]: res.pvalues
Out[18]: array([ 0. , 0. , 0. , 0.2018, 0.0598, 0. ])
In [19]: res.aic
Out[19]: 9604.9532058301575
As usual, you can obtain a full list of available information by typing dir(res).
To ensure that the above results are sound, we compare them to results obtained using the MASS implementation for R:
url = 'http://vincentarelbundock.github.com/Rdatasets/csv/COUNT/medpar.csv'
f = los~factor(type)+hmo+white
library(MASS)
mod = glm.nb(f, medpar)
coef(summary(mod))
Estimate Std. Error z value Pr(>|z|)
(Intercept) 2.31027893 0.06744676 34.253370 3.885556e-257
factor(type)2 0.22124898 0.05045746 4.384861 1.160597e-05
factor(type)3 0.70615882 0.07599849 9.291748 1.517751e-20
hmo -0.06795522 0.05321375 -1.277024 2.015939e-01
white -0.12906544 0.06836272 -1.887951 5.903257e-02
### Numerical precision¶
The statsmodels and R parameter estimates agree up to the fourth decimal. The standard errors, however, agree only up to the second decimal. This discrepancy may be the result of imprecision in our Hessian numerical estimates. In the current context, the difference between MASS and statsmodels standard error estimates is substantively irrelevant, but it highlights the fact that users who need very precise estimates may not always want to rely on default settings when using numerical derivatives. In such cases, it may be better to use analytical derivatives with the LikelihoodModel class. | 1,830 | 6,019 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2016-07 | longest | en | 0.588377 |
https://stats.stackexchange.com/questions/146222/truncated-binomial-samples-with-glm | 1,719,094,826,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862420.91/warc/CC-MAIN-20240622210521-20240623000521-00322.warc.gz | 461,778,656 | 39,760 | # truncated binomial samples with GLM
We have a binomial process that yields samples of 60 trials. To save time, once 2 failures have been observed the process is reset.
So if a test series hits 2 failures early, the resultant sample ends up being truncated.
ex <- data.frame(FAIL = c(1,1,0,2,0,2), PASS = c(59,59,60,5,60,2))
Samples that have 2 failures could be anywhere between 2/58 to 2/2.
This causes a problem. Because I can't know if the 2/2 sample wouldn't have been maybe 5/55 if the series wasn't terminated early.
Population P for each sample is also different due to some changing IVs.
I'm having a hard time thinking about how to analyze these samples.
I know odds ratios would be valid but I have lots of zero count cells when stratifying 2x2 tables.
Is this the proper way to weight for changes in N/variance with GLM?
GLM(FAIL/PASS)~IV1+IV2+IV3, family=quasibinomial, ex)
I'm confused about when to use weights or offset or both.
e.g.
GLM(FAIL/PASS)~IV1+IV2+IV3+offset(FAIL+PASS), family=quasibinomial, ex)
or
GLM(FAIL/PASS)~IV1+IV2+IV3, weights=(FAIL+PASS), family=quasibinomial, ex)
• If a trial terminates early, what you actually observe is the number of trials to the second failure, rather than the number of failures in a fixed number of trials. If all finished early, or none did, this would be easy (in the first case you could fit say a negative binomial model, in the second a straight binomial GLM). However, you may be able to treat it as a censored binomial response, perhaps. Commented Apr 14, 2015 at 0:48
• One could treat it as a censored negative binomial, the response variable is the number of successes before the second failure, but if we do not reach 2 failures in 60 trials we only observe that the waiting time is $\ge 58$! (since the next trial could be the second fail. Commented Aug 1, 2017 at 15:46
Let $N_i$ be the number of successes observed before the second fail, which will have a negative binomial distribution. But if you reach 60 trials without seeing the second failure, the observation is censored, and you observe that $N_i \ge 59$ (if you have seen one failure so far) or $N_i \ge 60$ (otherwise). Then you can use likelihood methods with censored observations, there are many example on this site, see for example ML estimate of exponential distribution (with censored data) | 626 | 2,354 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2024-26 | latest | en | 0.925602 |
https://countrymusicstop.com/whats-25-of-660-update-new/ | 1,679,799,086,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945381.91/warc/CC-MAIN-20230326013652-20230326043652-00561.warc.gz | 227,101,560 | 21,940 | Home » Whats 25 Of 660? Update New
# Whats 25 Of 660? Update New
Let’s discuss the question: whats 25 of 660. We summarize all relevant answers in section Q&A of website Countrymusicstop.com in category: MMO. See more related questions in the comments below.
## What number is 25 percent of 680?
25 percent of 680 is 170.
## What number is 15% of 660?
15 percent of 660 is 99.
### Percentage Trick – Solve precentages mentally – percentages made easy with the cool math trick!
Percentage Trick – Solve precentages mentally – percentages made easy with the cool math trick!
Percentage Trick – Solve precentages mentally – percentages made easy with the cool math trick!
## How do you work out 75% of 440?
440 = 100% (1). x = 75% (2).
Frequently Asked Questions on What is 75 percent of 440?
1. How do I calculate percentage of a total?
2. What is 75 percent of 440? 75 percent of 440 is 330.
3. How to calculate 75 percent of 440? Multiply 75/100 with 440 = (75/100)*440 = (75*440)/100 = 330.
## How do you calculate 25 of a number?
To calculate 25 percent of a number, simply divide it by 4. For example, 25 percent of 12 is 12 divided by 4, or 3.
## What number is 25% of 600?
25 percent of 600 is 150.
## What number is 8 percent of 50?
Percentage Calculator: What is 8 percent of 50? = 4.
## What number is 50% of 96?
50 percent of 96 is 48.
## What number is 15% of 240?
15 percent of 240 is 36.
## What number is 15 percent of 60?
Answer: 15% of 60 is 9.
## What number is 20% of 120?
Answer: 20% of 120 is 24.
Let’s find 20% of 120.
## What number is 20% of 130?
20 percent of 130 is 26.
## What number is 65% of 200?
Percentage Calculator: What is 65 percent of 200? = 130.
### How to Calculate 25% Off
How to Calculate 25% Off
How to Calculate 25% Off
## How do I calculate % off?
How to calculate percent off?
1. Divide the number by 100 (move the decimal place two places to the left).
2. Multiply this new number by the percentage you want to take off.
3. Subtract the number from step 2 from the original number. This is your percent off number.
## How do you calculate a perimeter?
The perimeter is the length of the outline of a shape. To find the perimeter of a rectangle or square you have to add the lengths of all the four sides. x is in this case the length of the rectangle while y is the width of the rectangle.
## How do you find 25% of 40?
Frequently Asked Questions on What is 25 percent of 40?
1. How do I calculate percentage of a total?
2. What is 25 percent of 40? 25 percent of 40 is 10.
3. How to calculate 25 percent of 40? Multiply 25/100 with 40 = (25/100)*40 = (25*40)/100 = 10.
## What percentage is 120 of are90?
Steps to solve “what percent is 90 of 120?” If you are using a calculator, simply enter 90÷120×100 which will give you 75 as the answer.
## What number is 25% of 150?
25 percent of 150 is 37.5.
## What number is 25 percent of 100?
25 percent of 100 is 25. 3.
## What percentage is 88 out of 66?
Percentage Calculator: 88 is what percent of 66? = 133.33.
## How do you find 40% 15?
Frequently Asked Questions on What is 40 percent of 15?
1. How do I calculate percentage of a total?
2. What is 40 percent of 15? 40 percent of 15 is 6.
3. How to calculate 40 percent of 15? Multiply 40/100 with 15 = (40/100)*15 = (40*15)/100 = 6.
## What number is 25% of 120?
25 percent of 120 is 30.
### Finding a Fraction of a Number
Finding a Fraction of a Number
Finding a Fraction of a Number
## What number is 10 percent of 40?
10 percent of 40 is 4.
## What number is 5% of 80?
Percentage Calculator: What is 5. percent of 80? = 4.
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### Paired t-Test in 4 Steps in Excel 2010 and Excel 2013
This is one of the following seven articles on Paired (2-Sample Dependent) t-Tests in Excel
Paired t-Test in 4 Steps in Excel 2010 and Excel 2013
Excel Normality Testing of Paired t-Test Data
Paired t-Test Excel Calculations, Formulas, and Tools
Paired t-Test – Effect Size in Excel 2010, and Excel 2013
Paired t-Test – Test Power With G-Power Utility
Wilcoxon Signed-Rank Test in 8 Steps As a Paired t-Test Alternative
Sign Test in Excel As A Paired t-Test Alternative
# Paired t-Test in 4 Steps in Excel
This hypothesis test determines whether the mean of a sample of differences between pairs of data (x_bardiff) is equal to (two-tailed test) or else greater than or less than (one-tailed test) than a constant. Before-and-after fitness levels of individuals undergoing a training program would be an example of paired data. The sample evaluated would be the group of differences between the before-and-after scores of the individuals. This is called the difference sample.
x_bardiff = Observed Difference Sample Mean
df = n – 1
Null Hypothesis H0: x_bardiff = Constant
The Null Hypothesis is rejected if any of the following equivalent conditions are shown to exist:
1) The observed x_bardiff is beyond the Critical Value.
2) The t Value (the Test Statistic) is farther from zero than the Critical t Value.
3) The p value is smaller than α for a one-tailed test or α/2 for a two-tailed test.
## Example of Paired, 1-Tailed t-Test in Excel
This problem is very similar to the problem solved in the z-test section for a paired, one-tailed z-test. Similar problems were used in each of these sections to show the similarities and also contrast the differences between the paired z-Test and t-test as easily as possible.
A new clerical program was introduced to a large company with the hope that clerical errors would be reduced. 5,000 clerical workers in the company underwent the training program. 17 Clerical employees who underwent the training were randomly selected. The average number of clerical errors that each of these 17 employees made per month for six months prior to the training and also for six months following the training were recorded. Each of the 17 employees had a similar degree of clerical experience within the company and performed nearly the same volume and type of clerical work in the before and after months.
Based upon the results of the 17 sampled clerical employees, determine with 95 percent certainty whether the average number of monthly clerical mistakes was reduced for the entire 5,000 clerical employees who underwent the training.
It is the difference that we are concerned with. A hypothesis test will be performed on the sample of differences. The distributed variable will be designated as x_bardiff and will represent that average difference between After and Before samples.
x_bardiff was calculated by subtracting the Before measurement from the After measurement. This is the intuitive way to determine if a reduction in error occurred.
This problem illustrates why the t-test is nearly always used instead of a z-Test to perform a two-dependent-sample (paired) hypothesis test of mean. The z-Test requires the population standard deviation of the differences between the pairs be known. This is often not the case, but is required for a paired z-Test . The t-test requires only the sample standard deviation of the sample of paired differences be known.
Before and After Results and Their Differences Are As Follows:
(Click On Image To See a Larger Version)
Running the Excel data analysis tool Descriptive Statistics on the column of Difference data produces the following output:
Running the Excel data analysis tool Descriptive Statistics on the column of Difference data will provide the Sample Mean, the Sample Standard Deviation, the Standard Error, and the Sample Size. It will even provide half the width of a confidence interval about the mean based on this sample for any specified level of certainty if that option is specified. The output of this tool appears as follows:
(Click On Image To See a Larger Version)
It is the difference that we are concerned with. A hypothesis test will be performed on the sample of differences. The distributed variable will be designated as x_bardiff and will represent that average difference between After and Before samples.
x_bardiff was calculated by subtracting the Before measurement from the After measurement. This is the intuitive way to determine if a reduction in error occurred
### Summary of Problem Information
x_bardiff = sample mean =AVERAGE() = -3.35
sdiff = sample standard deviation = STDEV.S() = 6.4
n = sample size = number of pairs = COUNT() = 17
df = n – 1 = 16
SEdiff = Standard Error = sdiff / SQRT(n) = 6.4 / SQRT(16)
SEdiff = 1.55
Note that this calculation of the Standard Error using the sample standard deviation, sdiff, is an estimate of the true Standard Error which would be calculated using the population standard deviation, σdiff.
Level of Certainty = 0.95
Alpha = 1 - Level of Certainty = 1 – 0.95 = 0.05
As with all Hypothesis Tests of Mean, we must satisfactorily answer these two questions and then proceed to the four-step method of solving the hypothesis test that follows.
The Initial Two Questions That Must be Answered Satisfactorily
1) What Type of Test Should Be Done?
2)Have All of the Required Assumptions For This Test Been Met?
The Four-Step Method For Solving All Hypothesis Tests of Mean
Step 1) Create the Null Hypothesis and the Alternate Hypothesis
Step 2 – Map the Normal or t Distribution Curve Based on the Null Hypothesis
Step 3 – Map the Regions of Acceptance and Rejection
Step 4 – Perform the Critical Value Test, the p Value Test, or the Critical t Value Test
The Initial Two Questions That Need To Be Answered Before Performing the Four-Step Hypothesis Test of Mean are as follows:
### Question 1) Type of Test?
a) Hypothesis Test of Mean or Proportion?
This is a Hypothesis Test of Mean because each individual observation (each sampled difference) within the sample can have a wide range of values. Data points for Hypothesis Tests of Proportion are binary: they can take only one of two possible values.
b) One-Sample or Two-Sample Test?
This is a two-sample hypothesis test because the data exists in two groups of measurements. One sample group contains Before measurements and the other sample group contains After measurements.
c) Independent (Unpaired) Test or Dependent (Paired) Test?
This is a paired (dependent) hypothesis test because each Before observation has a related After observation made on the same person.
d) One-Tailed or Two-Tailed Test?
The problem asks to determine whether there has been a reduction in clerical mistake from Before to After. This is a directional inequality making this hypothesis test a one-tailed test. If the problem asked whether Before and After were simply different, the inequality would be non-directional and the resulting hypothesis test would be a two-tailed test. A two-tailed test is more stringent than a one-tailed test.
e) t-Test or z-Test?
Assuming that the difference population or difference sample can pass a normality test, a hypothesis test of mean must be performed as a t-Test when the difference sample size (n = number of difference pairs) is small (n < 30) or if the variance of differences is unknown.
In this case the difference sample size (the number of data pairs) is small as n = 17 data sample pairs. This Hypothesis Test of Mean must therefore be performed as a t-Test and not as a z-Test.
The t Distribution with degrees of freedom = df = n – 1 is defined as the distribution of random data sample of sample size n taken from a normal population.
The means of samples taken from a normal population are also distributed according to the t Distribution with degrees of freedom = df = n – 1.
The Test Statistic (the t Value, which is based upon the difference sample mean (x_bardiff) because it equals (x_bardiff – Constant)/(SEdiff) will therefore also be distributed according to the t Distribution. A t-Test will be performed if the Test Statistic is distributed according to the t Distribution.
The distribution of the Test Statistic for the difference sample taken from a normal population of differences is always described by the t Distribution. The shape of the t Distribution converges to (very closely resembles) the shape of the standard normal distribution when the difference sample size becomes large (n > 30).
The Test Statistic’s distribution can be approximated by the normal distribution only if the difference sample size is large (n > 30) and the population standard deviation, σ, is known. A z-Test can be used if the Test Statistic’s distribution can be approximated by the normal distribution. A t-Test must be used in all other cases.
It should be noted that a paired t-Test can always be used in place of a paired z-Test. All z-Tests can be replaced be their equivalent t-Tests. As a result, some major commercial statistical software packages including the well-known SPSS provide only t-Tests and no direct z-Tests.
This hypothesis test is a t-Test that is two-sample, paired (dependent), one-tailed hypothesis test of mean.
### Question 2) Test Requirements Met?
a) t-Distribution of Test Statistic
A t-Test can be performed if the distribution of the Test Statistic (the t value) can be approximated under the Null Hypothesis by the t Distribution. The Test Statistic is derived from the mean of the difference sample and therefore has the same distribution that the difference sample mean would have if multiple similar samples were taken from the same population of differences between data sample pairs.
The difference sample size indicates how to determine the distribution of the difference sample mean and therefore the distribution of the Test Statistic as follows:
When Difference Sample Size Is Large
When the difference sample size is large (n > 30 meaning that there are more than 30 pairs of data), the distribution of means of similar samples drawn from the same population of differences is described by the t Distribution. As per the Central Limit Theorem, as the difference sample size increases, the distribution of the difference sample means converges to the normal distribution as does the t Distribution. When the difference sample size approaches infinity, the t Distribution converges to the standard normal distribution.
When the difference sample size is large, the distribution of the distribution sample mean, and therefore the distribution of the Test Statistic, is always described by the t Distribution. A t-Test can therefore always be used when the difference sample size is large, regardless of the distribution of the population of differences or the difference sample.
When the Difference Sample Size is Small
The data in a difference sample taken from a normally-distributed population of paired differences will be distributed according to the t Distribution regardless of the difference sample size.
The means of similar difference samples randomly taken from a normally-distributed population of paired differences are also distributed according to the t Distribution regardless of the difference sample size.
The difference sample mean, and therefore the Test Statistic, are distributed according to the t Distribution if the population of paired differences is normally distributed.
The population of paired differences is considered to be normally distributed if any of the following are true:
1) Population of Paired Differences Is Normally Distributed
2) Difference Sample Is Normally Distributed
If the difference sample passes a test of normality then the population of paired difference from which the difference sample was taken can be assumed to be normally distributed.
The population of paired differences or the difference sample must pass a normality test before a t-Test can be performed. If the only data available are the data of the single difference sample taken, then difference sample must pass a normality test before a t-Test can be performed.
### Evaluating the Normality of the Sample Data
In the next blog article the following five normality tests will be performed on the difference sample data here:
An Excel histogram of the sample data will be created.
A normal probability plot of the sample data will be created in Excel.
The Kolmogorov-Smirnov test for normality of the sample data will be performed in Excel.
The Anderson-Darling test for normality of the sample data will be performed in Excel.
The Shapiro-Wilk test for normality of the sample data will be performed in Excel.
We now proceed to complete the four-step method for solving all Hypothesis Tests of Mean. These four steps are as follows:
Step 1) Create the Null Hypothesis and the Alternate Hypothesis
Step 2 – Map the Normal or t Distribution Curve Based on the Null Hypothesis
Step 3 – Map the Regions of Acceptance and Rejection
Step 4 – Determine Whether to Accept or Reject theNull Hypothesis By Performing the Critical Value Test, the p Value Test, or the Critical t Value Test
Proceeding through the four steps is done is follows:
### Step 1 – Create the Null and Alternate Hypotheses
The Null Hypothesis is always an equality and states that the items being compared are the same. In this case, the Null Hypothesis would state that the there is no difference between before and after data. We will use the variable x_bardiff to represent the mean between the before and after measurements. The Null Hypothesis is as follows:
H0: x_bardiff = Constant = 0
The Alternate Hypothesis is always in inequality and states that the two items being compared are different. This hypothesis test is trying to determine whether there has been a reduction in clerical errors, i.e., the After measurements are, on average, smaller than the Before measurements. The Alternate Hypothesis is as follows:
H1: x_bardiff < Constant , which is 0
H1: x_bardiff < 0
The Alternative Hypothesis is directional (“greater than” or “less than” instead of “not equal,” which is non-directional) and the hypothesis test is therefore a one-tailed test. The “less than” operator indicates that this is a one-tailed test with the Region of Rejection (the alpha region) entirely contained in the left tail. A “greater than” operator would indicate a one-tailed test focused on the right tail.
It should also be noted that a two-tailed test is more rigorous (requires a greater differences between the two entities being compared before the test shows that there is a difference) than a one-tailed test.
It is important to note that the Null and Alternative Hypotheses refer to the means of the population of paired differences from which the difference samples were taken. A population of paired differences would be the differences of data pairs in a population of data pairs.
A paired t-Test determines whether to reject or fail to reject the Null Hypothesis that states that that population of paired differences from which the difference sample was taken has a mean equal to the Constant. The Constant in this case is equal to 0. This means that the Null Hypothesis states that the average difference between data pairs of an entire population from which the sample of data pairs were drawn is zero.
Parameters necessary to map the distributed variable, x_bardiff, are the following:
x_bardiff = sample mean =AVERAGE() = -3.35
sdiff = sample standard deviation = STDEV.S() = 6.4
n = sample size = number of pairs = COUNT() = 17
df = n – 1 = 16
SEdiff = Standard Error = sdiff / SQRT(n) = 6.4 / SQRT(16)
These parameters are used to map the distributed variable, x_bardiff, to the t Distribution curve as follows:
### Step 2 – Map the Distributed Variable to t-Distribution
A t-Test can be performed if the difference sample mean, and the Test Statistic (the t Value) are distributed according to the t Distribution. If the difference sample has passed a normality test, then the difference sample mean and closely-related Test Statistic are distributed according to the t Distribution.
The t Distribution always has a mean of zero and a standard error equal to one. The t Distribution varies only in its shape. The shape of a specific t Distribution curve is determined by only one parameter: its degrees of freedom, which equals n – 1 if n = sample size.
The means of similar, random difference samples taken from a normal population of paired differences are distributed according to the t Distribution. This means that the distribution of a large number of means of difference samples of size n taken from a normal population will have the same shape as a t Distribution with its degrees of equal to n – 1.
The difference sample mean and the Test Statistic are both distributed according to the t Distribution with degrees of freedom equal to n – 1 if the sample or population is shown to be normally distributed. This step will map the sample mean to a t Distribution curve with a degrees of freedom equal to n – 1.
The t Distribution is usually presented in its finalized form with standardized values of a mean that equals zero and a standard error that equals one. The horizontal axis is given in units of Standard Errors and the distributed variable is the t Value (the Test Statistic) as follows in this Excel-generated graph:
(Click On Image To See a Larger Version)
A non-standardized t Distribution curve would simply have its horizontal axis given in units of the measure used to take the samples. The distributed variable would be the sample mean, x_bardiff.
The variable x_bardiff is distributed according to the t Distribution. Mapping this distributed variable to a t Distribution curve is shown as follows in this Excel-generated graph:
(Click On Image To See a Larger Version)
This non-standardized t Distribution curve has its mean set to equal the Constant taken from the Null Hypothesis, which is:
H0: x_bardiff = Constant = 0
This non-standardized t Distribution curve is constructed from the following parameters:
Mean = Constant = 0
Standard Errordiff = 1.55
Degrees of Freedom = 16
Distributed Variable = x_bardiff
### Step 3 – Map the Regions of Acceptance and Rejection
The goal of a hypothesis test is to determine whether to accept or reject the Null Hypothesis at a given level of certainty. If the two things being compared are far enough apart from each other, the Null Hypothesis (which states that the two things are not different) can be rejected. In this case we are trying to show graphically how different x_bardiff (-3.35) is from the hypothesized mean of 0.
The non-standardized t Distribution curve can be divided up into two types of regions: the Region of Acceptance and the Region of Rejection. A boundary between a Region of Acceptance and a Region of Rejection is called a Critical Value.
The above distribution curve that maps the distribution of variable x_bardiff can be divided up into two types of regions: the Region of Acceptance and the Region of Rejection.
If x_bardiff’s value of -3.35 falls in the Region of Acceptance, we must accept the Null Hypothesis. If x_bardiff’s value of -3.35 falls in the Region of Rejection, we can reject the Null Hypothesis.
The total size of the Region of Rejection is equal to Alpha. In this case Alpha, α, is equal to 0.05. This means that the Region of Rejection will take up 5 percent of the total area under this t distribution curve.
This 5 percent is entirely contained in the outer left tail. The outer left tail contains the 5 percent of the curve that is the Region of Rejection.
Calculate the Critical Value
The boundary between Region of Acceptance and Region of Rejection is called Critical Value. The location of this Critical Value need to be calculated as follows.
One-tailed, Left tail Critical Value = x_bardiff - (Number of Standard Errors from Mean to Region of Rejection) * SEdiff
One-tailed, Left tail Critical Value = x_bardiff + T.INV(α,df) * SEdiff
One-tailed, Left tail Critical Value = 0 + T.INV(0.05, 16) * 1.55
One-tailed, Left tail Critical Value = -2.711
The Region of Rejection is therefore everything that is to the left of -2.711.
The distribution curve with the blue 95-percent Region of Acceptance and the yellow 5-percent Region of Rejection entirely contained in the left tail is shown is as follows in this Excel-generated graph:
(Click On Image To See a Larger Version)
### Step 4 – Determine Whether to Reject Null Hypothesis
The object of a hypothesis test is to determine whether to accept of reject the Null Hypothesis. There are three equivalent-Tests that determine whether to accept or reject the Null Hypothesis. Only one of these tests needs to be performed because all three provide equivalent information. The three tests are as follows:
1) Compare x-bardiff With Critical Value
Reject the Null Hypothesis if the sample mean, x_bardiff = -3.35, falls into the Region of Rejection.
Equivalently, reject the Null Hypothesis if the sample mean, x_bardiff, is further the curve’s mean of 0 than the Critical Value.
The Critical Values have been calculated to be -2.71 on the left. x_bardiff (-3.35) is further from the curve mean (0) than left Critical Value (-2.71). The Null Hypothesis would therefore be rejected.
2) Compare t Value With Critical t Value
The t Value corresponds to the standardized value of the sample mean, x_bardiff = -3.35. The t Value is the number of Standard Errors that x_bardiff is from the curve’s mean of 0.
The Critical t Value is the number of Standard Errors that the Critical Value is from the curve’s mean.
Reject the Null Hypothesis if the t Value is farther from the standardized mean of zero than the Critical t Value.
Equivalently, reject the Null Hypothesis if the t Value is closer to the standardized mean of zero than the Critical t Value.
t Value (Test Statistic) = (x_bardiff) / SEdiff = (-3.35)/1.55
t Value (Test Statistic) = -2.159
This means that the sample mean, x_bardiff, is 2.159 standard errors to the left of the curve mean of 0.
One-tailed, left-tail Critical t Value = T.INV(α,df)
One-tailed, left-tail Critical t Value = T.INV(0.05, 16) = -1.76
This means that the boundary of the Region of Rejection are 1.76 standard errors to the left of the curve mean of 0 since this is a one-tailed test in the left tail.
The Null Hypothesis is rejected because the t Value is farther from curve mean the Critical t Values indicating that x_bardiff is in the Region of Rejection.
3) Compare p Value With Alpha
The p Value is the percent of the curve that is beyond x_bardiff (-3.35). If the p Value is smaller than Alpha, the Null Hypothesis is rejected.
p Value = T.DIST.RT(ABS(t Value), df)
p Value = T.DIST.RT(ABS(-2.159), 16)
p Value = 0.023
The p Value (0.023) is smaller than Alpha (0.05) Region of Rejection in the right tail and we therefore reject the Null Hypothesis. A graph below shows that the red p Value (the curve area beyond x_bar) is smaller than the yellow Alpha, which is the 5 percent Region of Rejection in the left tail. This is shown in the following Excel-generated graph of this non-standardized t Distribution curve:
(Click On Image To See a Larger Version)
Excel Master Series Blog Directory
Statistical Topics and Articles In Each Topic
• Histograms in Excel
• Bar Chart in Excel
• Combinations & Permutations in Excel
• Normal Distribution in Excel
• t-Distribution in Excel
• Binomial Distribution in Excel
• z-Tests in Excel
• t-Tests in Excel
• Hypothesis Tests of Proportion in Excel
• Chi-Square Independence Tests in Excel
• Chi-Square Goodness-Of-Fit Tests in Excel
• F Tests in Excel
• Correlation in Excel
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• Simple Linear Regression in Excel
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• Two-Factor ANOVA With Replication in Excel
• Two-Factor ANOVA Without Replication in Excel
• Randomized Block Design ANOVA in Excel
• Repeated-Measures ANOVA in Excel
• ANCOVA in Excel
• Normality Testing in Excel
• Nonparametric Testing in Excel
• Post Hoc Testing in Excel
• Creating Interactive Graphs of Statistical Distributions in Excel
• Solving Problems With Other Distributions in Excel
• Optimization With Excel Solver
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• Analyzing Data With Pivot Tables
• SEO Functions in Excel
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• VLOOKUP | 5,474 | 24,686 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2017-39 | longest | en | 0.91468 |
http://pmean.com/00/collinearity.html | 1,624,237,846,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488259200.84/warc/CC-MAIN-20210620235118-20210621025118-00217.warc.gz | 36,034,725 | 2,439 | StATS: What is collinearity? (created 2000-01-27)
Dear Professor Mean, Could you describe the term collinearity for me? I understand that it has to do with variables which are not totally independent, but that is all I know!
Collinearity is a situation where there is close to a near perfect linear relationship among some or all of the independent variables in a regression model. In practical terms, this means there is some degree of redundancy or overlap among your variables.
Some authors describe this as multicollinearity, near collinearity, or ill conditioning. Coming up with four different technical terms for the same condition is one way that we statisticians keep our discipline mysterious and awe inspiring.
Collinearity can appear as a very high correlation among two independent variables, but it doesn't have to work that way. Another type of collinearity is when several of the variables add up to something that it very close to a constant value.
Collinearity is not a fatal flaw, but it does cause a loss in power and it makes interpretation more difficult.
A simple example of collinearity is when you are using both gestational age and birth weight as independent variables. These two measures are highly correlated, of course, since low gestational ages tend to be associated with low birth weights.
Interpretation is difficult in this situation, because when both variables are in a regression model, the parameter for birth weight is measuring the effect of a change in one unit in birth weight on the dependent variable, assuming that all of the other variables are held constant. It's hard to envision what it means to change birth weight while gestational age is held constant. What you are looking at, in effect, is the size of a baby for a gestational age.
Collinearity also causes a loss in power. When you have overlap among some of the variables, it takes more data to disentangle the individual effects of these variables. Think of it as a table where you push two of the four legs away from the corners and close to the middle of the table. Such a table will be very unstable.
In the previous example, we have very few 1000 gram babies who are 40 weeks gestational age and very few 2500 gram babies who are 32 weeks gestational age. Without data at these two "corners" of the table, it's hard to get stable statistical estimates.
It should be noted, though, that you can make sense of your data, even when you have collinearity. It just takes more data and a bit of care in interpretation. Some health outcomes, it turns out, are related more closely to gestational age than to birth weight. It's not how small you are that is as important as how early you make your entry into the world. Keep in mind that I'm not a doctor (see my disclaimer), so check my limited knowledge of medicine out with the experts. Especially if you are a newborn baby.
This page was written by Steve Simon while working at Children's Mercy Hospital. Although I do not hold the copyright for this material, I am reproducing it here as a service, as it is no longer available on the Children's Mercy Hospital website. Need more information? I have a page with general help resources. You can also browse for pages similar to this one at Category: Modeling issues. | 688 | 3,285 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2021-25 | latest | en | 0.959861 |
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Unformatted text preview: pherical object with free charges Spherical object is polarized when a charged rod is brought near it Negative charges are depleted by grounding Spherical object is isolated from ground Spherical object is left positively charged January 22, 2014 Physics 104, Spring 2014 15 Force on charges: Coulomb’s Law • Coulomb’s law gives quantitative description of the electrical force between charged particles • Unit of charge is called a Coulomb • Gives the magnitude and the direction of the force between charges q1 and q2 separated by a distance r. • Magnitude: q1 q2 k = 9x109 Nm2/C2 F=k 2 r • direction: along the line joining the charges • Like charges repel, unlike charges attract January 22, 2014 Physics 104, Spring 2014 16 Example Force on electron in Hydrogen atom Qp=1.6x10-19 C + F Qe = -1.6x10-19 C - r = 1x10-10 m F = (9x109)(1.6x10-19)(1.6x10-19)/(10-10)2 N = 2.3x10-8 N (to the left) January 22, 2014 Physics 104, Spring 2014 17 Once you know the force…. • Laws of mechanics (recall 103) describe the subsequent motion of charged...
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## This note was uploaded on 02/14/2014 for the course PHYSICS 103 taught by Professor Citrinini during the Spring '09 term at University of Wisconsin.
Ask a homework question - tutors are online | 416 | 1,490 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2017-17 | longest | en | 0.899634 |
https://blog.flyingcoloursmaths.co.uk/multiplying-halves-halves-secrets-mathematical-ninja/ | 1,620,326,275,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988759.29/warc/CC-MAIN-20210506175146-20210506205146-00539.warc.gz | 161,131,896 | 3,778 | The student, by now, knew better than to pick up the calculator. “There’s got to be a way of doing that quickly.”
The Mathematical Ninja looked a little disappointed; he’d built an elaborate electromagnetic pulse generator for the express purpose of killing calculators in a mysterious flash. “You’re right,” he said, grudgingly.
“$7.5 \times 3.5$,” he muttered again, as if asking for a clue.
“Brackets”, sotto-voced the Mathematical Ninja, helpfully.
“Oh… so do it as $(7 + \frac12)(3 + \frac12)$? That… give me a second.” He reached for a pen and paper. “That’s 21, plus half of seven, plus half of 3, plus half of a half. 21 + 3.5 + 1.5 + … is it 0.25?”
“We call it a quarter where I come from,” said the Mathematical Ninja. “Yep, 26 and a quarter. There’s a slightly quicker way, though - rather than work out a half of seven and a half of three separately, you can glom them together and say ‘half of ten’.”
“Nice… but doesn’t that only come out nicely if they’re both odd or both even?”
The Mathematical Ninja was, quietly, impressed; this was a slightly intelligent question.
“So, if you had $7.5 \times 4.5$, you’d need to work out half of 11, which is a pain.”
“It’s not that bad,” said the Mathematical Ninja. “But, if you didn’t like it, you could do it as $(8 - \frac12)(4 + \frac12)$, so you’d have 32, plus half the difference, minus a quarter.”
“35.75?” said the student, reaching for his calculator to check. “No, wait! 33.75!” ((Thanks, Claudio, for the correction.))
The Mathematical Ninja let it pass, just this once. | 422 | 1,548 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2021-21 | longest | en | 0.936662 |
http://mathhelpforum.com/number-theory/132423-jacobi-symbol.html | 1,529,639,906,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864343.37/warc/CC-MAIN-20180622030142-20180622050142-00267.warc.gz | 199,347,435 | 12,107 | 1. ## Jacobi Symbol
Evaluate the Jacobi symbol ((n−1)(n+1)/n) for any odd natural number n.
Trying out some numbers, I THINK it alternates between 1 and -1, but how can we PROVE it formally?
Any help is appreciated!
[also under discussion in math link forum]
2. $\displaystyle \left(\frac{(n-1)(n+1)}{n}\right) = \left(\frac{n-1}{n}\right)\left(\frac{n+1}{n}\right) = \left(\frac{-1}{n}\right)\left(\frac{1}{n}\right) = \left(\frac{-1}{n}\right)$
For $\displaystyle n$ odd, $\displaystyle \left(\frac{-1}{n}\right) = (-1)^\frac{n-1}{2} = \begin{cases} \;\;\,1 & \text{if }n \equiv 1 \pmod 4\\ -1 &\text{if }n \equiv 3 \pmod 4\end{cases}$.
For $\displaystyle n$ even, take $\displaystyle n=2k$, then $\displaystyle \left(\frac{-1}{n}\right) = \left(\frac{-1}{2k}\right) = \left(\frac{-1}{2}\right)\left(\frac{-1}{k}\right) = \left(\frac{-1}{k}\right) = \begin{cases} \;\;\,1 & \text{if }k \equiv 1 \pmod 4\\ -1 &\text{if }k \equiv 3 \pmod 4\end{cases}$.
3. Originally Posted by chiph588@
$\displaystyle \left(\frac{(n-1)(n+1)}{n}\right) = \left(\frac{n-1}{n}\right)\left(\frac{n+1}{n}\right) = \left(\frac{-1}{n}\right)\left(\frac{1}{n}\right) = \left(\frac{-1}{n}\right)$
For $\displaystyle n$ odd, $\displaystyle \left(\frac{-1}{n}\right) = (-1)^\frac{n-1}{2} = \begin{cases} \;\;\,1 & \text{if }n \equiv 1 \pmod 4\\ -1 &\text{if }n \equiv 3 \pmod 4\end{cases}$.
For $\displaystyle n$ even, take $\displaystyle n=2k$, then $\displaystyle \left(\frac{-1}{n}\right) = \left(\frac{-1}{2k}\right) = \left(\frac{-1}{2}\right)\left(\frac{-1}{k}\right) = \left(\frac{-1}{k}\right) = \begin{cases} \;\;\,1 & \text{if }k \equiv 1 \pmod 4\\ -1 &\text{if }k \equiv 3 \pmod 4\end{cases}$.
Thank you!
But I thought the Jacobi symbol is defined only for ODD positive integers at the bottom. In your proof, why is there a case where "n" is even??? How is this possible??
4. Originally Posted by kingwinner
Thank you!
But I thought the Jacobi symbol is defined only for ODD positive integers at the bottom. In your proof, why is there a case where "n" is even??? How is this possible??
Oops! You're right!
5. But I'm concerned with one special case. For the case n=1, ((n−1)(n+1)/n)=(0/1)
Is (0/1)=1 or (0/1)=0 ?? Which one is the correct answer and why?
Thanks!
6. Originally Posted by kingwinner
But I'm concerned with one special case. For the case n=1, ((n−1)(n+1)/n)=(0/1)
Is (0/1)=1 or (0/1)=0 ?? Which one is the correct answer and why?
Thanks!
$\displaystyle \left(\frac{a}{n}\right) = \begin{cases} \;\;\,0\mbox{ if } \gcd(a,n) \ne 1 \\\pm1\mbox{ if } \gcd(a,n) = 1\end{cases}$
So sub in $\displaystyle 0$ and $\displaystyle 1$ and see what you get.
7. Originally Posted by chiph588@
$\displaystyle \left(\frac{a}{n}\right) = \begin{cases} \;\;\,0\mbox{ if } \gcd(a,n) \ne 1 \\\pm1\mbox{ if } \gcd(a,n) = 1\end{cases}$
So sub in $\displaystyle 0$ and $\displaystyle 1$ and see what you get.
gcd(0,1)=1, so that rule says that (0/1)=+1 OR -1, but how do we know whether it is +1 or -1?
8. Jacobi symbol
Apparently the answer is $\displaystyle 0$. I haven't had too much exposure to the Jacobi symbol so I can't really tell you why. Check out the site for yourself though.
9. $\displaystyle \left( \frac{0}{1} \right)=1$ because the set of prime factors of $\displaystyle 1$ is empty. if $\displaystyle n > 1$ is an odd integer, then $\displaystyle \left( \frac{0}{n} \right)=0.$
10. Originally Posted by NonCommAlg
$\displaystyle \left( \frac{0}{1} \right)=1$ because the set of prime factors of $\displaystyle 1$ is empty. if $\displaystyle n > 1$ is an odd integer, then $\displaystyle \left( \frac{0}{n} \right)=0.$
Is there any reason why (0/1)=1?? Is this simply becuase by convention, we define it to be that way?
1 has no prime factorization, so the product is empty. Is it conventional to define the "empty" product to be equal to +1??
Also, is it true that, by definition, (a/1)=1 for any integer a?
Can someone clarify this? Thank you!
11. Originally Posted by kingwinner
Is there any reason why (0/1)=1?? Is this simply becuase by convention, we define it to be that way?
1 has no prime factorization, so the product is empty. Is it conventional to define the "empty" product to be equal to +1??
Also, is it true that, by definition, (a/1)=1 for any integer a?
Can someone clarify this? Thank you!
12. Originally Posted by NonCommAlg
Is there any real reason why we define (a/1)=1 for any integer a?
Why not define it to be 0? why not -1?
13. Originally Posted by kingwinner
Is there any real reason why we define (a/1)=1 for any integer a?
Why not define it to be 0? why not -1? | 1,579 | 4,618 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2018-26 | latest | en | 0.429575 |
https://stackoverflow.com/questions/69094771/count-number-of-pairs-across-elements-in-a-list-in-r | 1,721,661,675,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517878.80/warc/CC-MAIN-20240722125447-20240722155447-00491.warc.gz | 468,596,085 | 46,667 | # Count number of pairs across elements in a list in R?
Similar questions have been asked about counting pairs, however none seem to be specifically useful for what I'm trying to do.
What I want is to count the number of pairs across multiple list elements and turn it into a matrix. For example, if I have a list like so:
``````myList <- list(
a = c(2,4,6),
b = c(1,2,3,4),
c = c(1,2,5,7),
d = c(1,2,4,5,8)
)
``````
We can see that the pair `1:2` appears 3 times (once each in `a`, `b`, and `c`). The pair `1:3` appears only once in `b`. The pair `1:4` appears 2 times (once each in `b` and `d`)... etc.
I would like to count the number of times a pair appears and then turn it into a symmetrical matrix. For example, my desired output would look something like the matrix I created manually (where each element of the matrix is the total count for that pair of values):
``````> myMatrix
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8]
[1,] 0 3 1 2 2 0 1 1
[2,] 3 0 1 3 2 1 1 1
[3,] 1 1 0 1 0 0 0 0
[4,] 2 3 1 0 0 0 0 1
[5,] 2 2 0 0 0 0 1 1
[6,] 0 1 0 0 0 0 0 0
[7,] 1 1 0 0 1 0 0 0
[8,] 1 1 0 1 1 0 0 0
``````
Any suggestions are greatly appreciated
• probably made a mistake .. `tcrossprod(sapply(myList, function(x) table(factor(x, 1:8))))` Commented Sep 7, 2021 at 23:20
• `microbenchmark` shows that @user20650 `trcrossprod` is 300 times faster than @akrun 's `Reduce` code is. Commented Sep 8, 2021 at 13:50
• I think the answers supplied only work if you guarantee there are no duplicate entries in each source vector. If there are, e.g. `myList\$a = c(2,4,2)` then you can't distinguish between the unwanted diagonals and the actual pair `2,2` found . Commented Sep 8, 2021 at 17:53
• Oh, and I think won't work if the input vectors are not ordered, either Commented Sep 8, 2021 at 17:59
Inspired by @akrun's answer, I think you can use a crossproduct to get this very quickly and simply:
``````out <- tcrossprod(table(stack(myList)))
diag(out) <- 0
# values
#values 1 2 3 4 5 6 7 8
# 1 0 3 1 2 2 0 1 1
# 2 3 0 1 3 2 1 1 1
# 3 1 1 0 1 0 0 0 0
# 4 2 3 1 0 1 1 0 1
# 5 2 2 0 1 0 0 1 1
# 6 0 1 0 1 0 0 0 0
# 7 1 1 0 0 1 0 0 0
# 8 1 1 0 1 1 0 0 0
``````
Use `combn` to get the combinations, as well as reversing each combination.
Then convert to a `data.frame` and `table` the results.
``````tab <- lapply(myList, \(x) combn(x, m=2, FUN=\(cm) rbind(cm, rev(cm)), simplify=FALSE))
tab <- data.frame(do.call(rbind, unlist(tab, rec=FALSE)))
table(tab)
# X2
#X1 1 2 3 4 5 6 7 8
# 1 0 3 1 2 2 0 1 1
# 2 3 0 1 3 2 1 1 1
# 3 1 1 0 1 0 0 0 0
# 4 2 3 1 0 1 1 0 1
# 5 2 2 0 1 0 0 1 1
# 6 0 1 0 1 0 0 0 0
# 7 1 1 0 0 1 0 0 0
# 8 1 1 0 1 1 0 0 0
``````
• After revisiting the code, it seems that under certain conditions (which I haven't figured out yet), akrun's code was producing small errors where it wouldn't sum the total number of pairs correctly. That coupled with the speed increase made me change this to the best answer. Thanks! Commented Oct 16, 2021 at 15:44
We could loop over the `list`, get the pairwise combinations with `combn`, `stack` it to a two column dataset, convert the 'values' column to `factor` with `levels` specified as 1 to 8, get the frequency count (`table`), do a cross product (`crossprod`), convert the output back to logical, and then `Reduce` the `list` elements by adding elementwise and finally assign the `diag`onal elements to 0. (If needed set the `names` attributes of `dimnames` to NULL
``````out <- Reduce(`+`, lapply(myList, function(x)
crossprod(table(transform(stack(setNames(
combn(x,
2, simplify = FALSE), combn(x, 2, paste, collapse="_"))),
values = factor(values, levels = 1:8))[2:1]))> 0))
diag(out) <- 0
names(dimnames(out)) <- NULL
``````
-output
``````> out
1 2 3 4 5 6 7 8
1 0 3 1 2 2 0 1 1
2 3 0 1 3 2 1 1 1
3 1 1 0 1 0 0 0 0
4 2 3 1 0 1 1 0 1
5 2 2 0 1 0 0 1 1
6 0 1 0 1 0 0 0 0
7 1 1 0 0 1 0 0 0
8 1 1 0 1 1 0 0 0
``````
• Master akrun, could you please check my answer. Am I completely wrong. I tried to count all the combinations. Thanks in advance! Commented Sep 7, 2021 at 22:14
• @TarJae Perhaps you may need to use `pivot_wider` at the end Commented Sep 7, 2021 at 22:15
• Why not use the crossproduct to do all the hard work? - ``diag<-`(tcrossprod(table(stack(myList))), 0)` . I totally overlooked that possibility, but I think it works. Commented Sep 7, 2021 at 23:48
• @thelatemail next step: do a time-test for `tcrossprod` vs your `lapply(combn)` code. I'm betting on crossproduct :-) Commented Sep 8, 2021 at 12:58
• see comment at question: that `tcrossprod` is 300 times faster than this. Commented Sep 8, 2021 at 13:53
I thought of a solution based on @TarJae answer, is not a elegant one, but it was a fun challenge!
## Libraries
``````library(tidyverse)
``````
## Code
``````map_df(myList,function(x) as_tibble(t(combn(x,2)))) %>%
count(V1,V2) %>%
{. -> temp_df} %>%
bind_rows(
temp_df %>%
rename(V2 = V1, V1 = V2)
) %>%
full_join(
expand_grid(V1 = 1:8,V2 = 1:8)
) %>%
replace_na(replace = list(n = 0)) %>%
arrange(V2,V1) %>%
pivot_wider(names_from = V1,values_from = n) %>%
as.matrix()
``````
## Output
`````` V2 1 2 3 4 5 6 7 8
[1,] 1 0 3 1 2 2 0 1 1
[2,] 2 3 0 1 3 2 1 1 1
[3,] 3 1 1 0 1 0 0 0 0
[4,] 4 2 3 1 0 1 1 0 1
[5,] 5 2 2 0 1 0 0 1 1
[6,] 6 0 1 0 1 0 0 0 0
[7,] 7 1 1 0 0 1 0 0 0
[8,] 8 1 1 0 1 1 0 0 0
``````
• Why bother with the rename? Just swap the input order of subsequent calls maybe? Commented Sep 8, 2021 at 12:56
First identify the possible combination of each vector from the list to a `tibble` then I bind them to one `tibble` and `count` the combinations.
``````library(tidyverse)
a <- as_tibble(t(combn(myList[[1]],2)))
b <- as_tibble(t(combn(myList[[2]],2)))
c <- as_tibble(t(combn(myList[[3]],2)))
d <- as_tibble(t(combn(myList[[4]],2)))
bind_rows(a,b,c,d) %>%
count(V1, V2)
``````
`````` V1 V2 n
<dbl> <dbl> <int>
1 1 2 3
2 1 3 1
3 1 4 2
4 1 5 2
5 1 7 1
6 1 8 1
7 2 3 1
8 2 4 3
9 2 5 2
10 2 6 1
11 2 7 1
12 2 8 1
13 3 4 1
14 4 5 1
15 4 6 1
16 4 8 1
17 5 7 1
18 5 8 1
`````` | 2,694 | 6,475 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2024-30 | latest | en | 0.907793 |
https://www.theflatearthsociety.org/forum/index.php?topic=23279.0;prev_next=next | 1,701,994,984,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100705.19/warc/CC-MAIN-20231207221604-20231208011604-00662.warc.gz | 1,131,867,747 | 8,480 | Orbits and mass calculations of celestial bodies.
• 9 Replies
• 2852 Views
Marcus Aurelius
• 4546
• My Alts: Tom Bishop, Gayer, theonlydann
Orbits and mass calculations of celestial bodies.
« on: September 19, 2008, 12:34:58 PM »
I read this wiki page on Astronomy and gravitation. I find it interesting that these methods were used to calculate the mass of the planets and sun, and their distance from the earth and the sun.
It appears to contradict FE beliefs about the size of the sun. The question is, how did FE'ers come about the conclusion that the Sun is 32 miles in diameter, and much closer to the earth than in the RE model?
http://en.wikipedia.org/wiki/Gravitation_(astronomy)
?
Rig Navigator
• 808
Re: Orbits and mass calculations of celestial bodies.
« Reply #1 on: September 19, 2008, 12:51:35 PM »
The question is, how did FE'ers come about the conclusion that the Sun is 32 miles in diameter, and much closer to the earth than in the RE model?
They did it as a simple exercise in planar trigonometry. Here is the diagram...
Once you have a distance to the Sun using that calculation, you can calculate how large the Sun should be given its angular size and that distance.
Marcus Aurelius
• 4546
• My Alts: Tom Bishop, Gayer, theonlydann
Re: Orbits and mass calculations of celestial bodies.
« Reply #2 on: September 19, 2008, 12:58:20 PM »
I can't see the picture.
Can this method that the use also calculate and predict the movement of the planets?
?
Rig Navigator
• 808
Re: Orbits and mass calculations of celestial bodies.
« Reply #3 on: September 19, 2008, 01:05:42 PM »
I can't see the picture.
Hmm, interesting. I can.
Quote
Can this method that the use also calculate and predict the movement of the planets?
No, but it could be used to calculate the height of the Moon, stars and planets. I will give you one guess as to what answer you will get.
?
Rig Navigator
• 808
Re: Orbits and mass calculations of celestial bodies.
« Reply #4 on: September 19, 2008, 01:08:20 PM »
There is a copy of a different diagram and the description of the process here...
http://www.sacred-texts.com/earth/za/za23.htm
sokarul
• 19301
• Extra Racist
Re: Orbits and mass calculations of celestial bodies.
« Reply #5 on: September 19, 2008, 01:12:30 PM »
The question is, how did FE'ers come about the conclusion that the Sun is 32 miles in diameter, and much closer to the earth than in the RE model?
They did it as a simple exercise in planar trigonometry. Here is the diagram...
Once you have a distance to the Sun using that calculation, you can calculate how large the Sun should be given its angular size and that distance.
What's funny is that the stupid bending light theory from robosteve would nullify the results.
ANNIHILATOR OF SHIFTER
It's no slur if it's fact.
?
Rig Navigator
• 808
Re: Orbits and mass calculations of celestial bodies.
« Reply #6 on: September 19, 2008, 01:13:59 PM »
What's funny is that the stupid bending light theory from robosteve would nullify the results.
Well, that has often been the problem with the "bendy light" theory of OBLSeve. It contradicts most FE canon.
sokarul
• 19301
• Extra Racist
Re: Orbits and mass calculations of celestial bodies.
« Reply #7 on: September 19, 2008, 01:33:05 PM »
What's funny is that the stupid bending light theory from robosteve would nullify the results.
Well, that has often been the problem with the "bendy light" theory of OBLSeve. It contradicts most FE canon.
Very true.
ANNIHILATOR OF SHIFTER
It's no slur if it's fact.
Marcus Aurelius
• 4546
• My Alts: Tom Bishop, Gayer, theonlydann
Re: Orbits and mass calculations of celestial bodies.
« Reply #8 on: September 19, 2008, 01:52:56 PM »
Wow, I disagree with most, if not all of those experiments. Plus, didn't his Bedford results get debunked later on? If the RE earth is as large as RE'er predict, it would not be possible to see the curvature at ground level.
?
Rig Navigator
• 808
Re: Orbits and mass calculations of celestial bodies.
« Reply #9 on: September 19, 2008, 03:15:29 PM »
Wow, I disagree with most, if not all of those experiments. Plus, didn't his Bedford results get debunked later on? If the RE earth is as large as RE'er predict, it would not be possible to see the curvature at ground level.
You can still observe the effects of that curvature from ground level though, and that is what he is trying to explain. The observable effect of ships disappearing beneath the horizon is what this was originally designed to explain. Unfortunately, it requires proof that light is bent in the manner which he postulates, and that proof just doesn't exist and contradicts most observations about the behavior of light. | 1,242 | 4,726 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2023-50 | longest | en | 0.897076 |
http://healthybusinesstips.com/bitcoin-mining-sites-bitcoin-history.html | 1,544,914,630,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376827137.61/warc/CC-MAIN-20181215222234-20181216004234-00363.warc.gz | 127,778,708 | 7,507 | ```Exchange hacks. As stated above, an exchange hack has nothing to do with the integrity of the Bitcoin system… but the market freaks out regardless. This trend seems to minimize as users see that cryptos recover from exchange hacks. As exchanges evolve and become more secure, this threat becomes less of an issue. Additionally, outside investments funneling into exchanges are providing the capital for them to grow stronger.
```
A variant race attack (which has been called a Finney attack by reference to Hal Finney) requires the participation of a miner. Instead of sending both payment requests (to pay Bob and Alice with the same coins) to the network, Eve issues only Alice's payment request to the network, while the accomplice tries to mine a block that includes the payment to Bob instead of Alice. There is a positive probability that the rogue miner will succeed before the network, in which case the payment to Alice will be rejected. As with the plain race attack, Alice can reduce the risk of a Finney attack by waiting for the payment to be included in the blockchain.[16]
Mining a block is difficult because the SHA-256 hash of a block's header must be lower than or equal to the target in order for the block to be accepted by the network. This problem can be simplified for explanation purposes: The hash of a block must start with a certain number of zeros. The probability of calculating a hash that starts with many zeros is very low, therefore many attempts must be made. In order to generate a new hash each round, a nonce is incremented. See Proof of work for more information.
In the process of mining, each Bitcoin miner is competing with all the other miners on the network to be the first one to correctly assemble the outstanding transactions into a block by solving those specialized math puzzles. In exchange for validating the transactions and solving these problems. Miners also hold the strength and security of the Bitcoin network. This is very important for security because in order to attack the network, an attacker would need to have over half of the total computational power of the network. This attack is referred to as the 51% attack. The more decentralized the miners mining Bitcoin, the more difficult and expensive it becomes to perform this attack.
But here, Carlson and his fellow would-be crypto tycoons confronted the bizarre, engineered obstinacy of bitcoin, which is designed to make life harder for miners as time goes by. For one, the currency’s mysterious creator (or creators), known as “Satoshi Nakamoto,” programmed the network to periodically—every 210,000 blocks, or once every four years or so—halve the number of bitcoins rewarded for each mined block. The first drop, from 50 coins to 25, came on November 28, 2012, which the faithful call “Halving Day.” (It has since halved again, to 12.5, and is expected to drop to 6.25 in June 2020.)
Exchange hacks. As stated above, an exchange hack has nothing to do with the integrity of the Bitcoin system… but the market freaks out regardless. This trend seems to minimize as users see that cryptos recover from exchange hacks. As exchanges evolve and become more secure, this threat becomes less of an issue. Additionally, outside investments funneling into exchanges are providing the capital for them to grow stronger.
To add a new block to the chain, a miner has to finish what’s called a cryptographic proof-of-work problem. Such problems are impossible to solve without applying a ton of brute computing force, so if you have a solution in hand, it’s proof that you’ve done a certain quantity of computational work. The computational problem is different for every block in the chain, and it involves a particular kind of algorithm called a hash function.
But, as always, the miners’ biggest challenge came from bitcoin itself. The mere presence of so much new mining in the Mid-Columbia Basin substantially expanded the network’s total mining power; for a time, Carlson’s mine alone accounted for a quarter of the global bitcoin mining capacity. But this rising calculating power also caused mining difficulty to skyrocket—from January 2013 to January 2014, it increased one thousandfold—which forced miners to expand even faster. And bitcoin’s rising price was now drawing in new miners, especially in China, where power is cheap. By the middle of 2014, Carlson says, he’d quadrupled the number of servers in his mine, yet had seen his once-massive share of the market fall below 1 percent.
The incremental complexity and technological know-how needed for this method are both downsides to the paper wallet approach. Cold storage solutions and hardware wallets are less nimble than other options, too; if the price of bitcoin were crashing, for example, you might find yourself slower to the draw than if you merely kept your BTC on a site like Coinbase.
Ledger’s main competitor in the market space is the original Trezor hardware wallet. One of the key advantages of the Ledger over the Trezor is the freedom to create your own unique passphrases. Both the Ledger and the Trezor require 20 passphrases for recovery and reset purposes; however, the Trezor package sends the user a random list. The Ledger gives the user the freedom to create their own. Additionally, if aesthetics matter to you, the Ledger sports an arguably sleeker design than the Trezor.
Of course, by the end of 2017, the players who were pouring into the basin weren’t interested in building 5-megawatt mines. According to Carlson, mining has now reached the stage where the minimum size for a new commercial mine, given the high levels of difficulty, will soon be 50 megawatts, enough for around 22,000 homes and bigger than one of Amazon Web Services’ immense data centers. Miehe, who has become a kind of broker for out-of-town miners and investors, was fielding calls and emails from much larger players. There were calls from China, where a recent government crackdown on cryptocurrency has miners trying to move operations as large as 200 megawatts to safer ground. And there was a flood of interest from players outside the sector, including big institutional investors from Wall Street, Miami, the Middle East, Europe and Japan, all eager to get in on a commodity that some believe could touch \$100,000 by the end of the year. And not all the interest has been so civil. Stories abound of bitcoin miners using hardball tactics to get their mines up and running. Carlson, for example, says some foreign miners tried to bribe building and safety inspectors to let them cut corners on construction. “They are bringing suitcases full of cash,” Carlson says, adding that such ploys invariably backfire. Adds Miehe, “I mean, you know how they talk about the animal spirits—greed and fear? Well, right now, everyone is in full-greed mode.”
```At this point, the actual mining begins. In essence, each miner now tries to demonstrate to the rest of the network that his or her block of verified payments is the one true block, which will serve as the permanent record of those 2,000 or so transactions. Miners do this by, essentially, trying to be the first to guess their block’s numerical password. It’s analogous to trying to randomly guess someone’s computer password, except on a vastly larger scale. Carlson’s first mining computer, or “rig,” which he ran out of his basement north of Seattle, could make 12 billion “guesses” every second; today’s servers are more than a thousand times faster.
```
No one knows. Not conclusively, at any rate. Satoshi Nakamoto is the name associated with the person or group of people who released the original Bitcoin white paper in 2008 and worked on the original Bitcoin software that was released in 2009. The Bitcoin protocol requires users to enter a birthday upon signup, and we know that an individual named Satoshi Nakamoto registered and put down April 5 as a birth date. And that's about it.
A mining pool sets a difficulty level between 1 and the currency’s difficulty. If a miner returns a block which scores a difficulty level between the pool’s difficulty level and the currency’s difficulty level, the block is recorded as a ‘share’. There is no use whatsoever for these share blocks, but they are recorded as proof of work to show that miners are trying to solve blocks. They also indicate how much processing power they are contributing to the pool the better the hardware, the more shares are generated.
# Just because miners want power doesn’t mean they get it. Some inquiries are withdrawn. And all three county public utilities have considerable discretion when it comes to granting power requests. But by law, they must consider any legitimate request for power, which has meant doing costly studies and holding hearings—sparking a prolonged, public debate over this new industry’s impact on the basin’s power economy. There are concerns about the huge costs of new substations, transmission wires and other infrastructure necessary to accommodate these massive loads. In Douglas County, where the bulk of the new mining projects are going in, a brand new 84-megawatt substation that should have been adequate for the next 30 to 50 years of normal population growth was fully subscribed in less than a year.
```On paper, the Mid-Columbia Basin really did look like El Dorado for Carlson and the other miners who began to trickle in during the first years of the boom. The region’s five huge hydroelectric dams, all owned by public utility districts, generate nearly six times as much power as the region’s residents and businesses can use. Most of the surplus is exported, at high prices, to markets like Seattle or Los Angeles, which allows the utilities to sell power locally at well below its cost of production. Power is so cheap here that people heat their homes with electricity, despite bitterly cold winters, and farmers have been able to irrigate the semi-arid region into one of the world’s most productive agricultural areas. (The local newspaper proudly claims to be published in “the Apple Capital of the World and the Buckle on the Power Belt of the Great Northwest.”) And, importantly, it had already attracted several power-hungry industries, notably aluminum smelting and, starting in the mid-2000s, data centers for tech giants like Microsoft and Intuit.
```
Bitcoin mining is the process of updating the ledger of Bitcoin transactions known as the blockchain. Mining is done by running extremely powerful computers (known as ASICs) that race against other miners in an attempt to guess a specific number. The first miner to guess the number gets to update the ledger of transactions and also receives a reward of newly minted Bitcoins (currently the reward is 12.5 Bitcoins).
Electrum is a software wallet that enables you to set up a strong level of security very quickly. During the simple installation process, you are given a twelve word phrase that will allow you to recover all of your bitcoins in the event that your computer fails. Your wallet is also encrypted by default which helps protect your coins against hackers. Electrum is available for Windows, OSX, and Linux and is our recommended software wallet for beginners. Click here to download the right version for your operating system.
Bitcoin mining is intentionally designed to be resource-intensive and difficult so that the number of blocks found each day by miners remains steady. Individual blocks must contain a proof of work to be considered valid. This proof of work is verified by other Bitcoin nodes each time they receive a block. Bitcoin uses the hashcash proof-of-work function.
# With the largest variety of markets and the biggest value - having reached a peak of 18 billion USD - Bitcoin is here to stay. As with any new invention, there can be improvements or flaws in the initial model however the community and a team of dedicated developers are pushing to overcome any obstacle they come across. It is also the most traded cryptocurrency and one of the main entry points for all the other cryptocurrencies. The price is as unstable as always and it can go up or down by 10%-20% in a single day.
A wallet stores the information necessary to transact bitcoins. While wallets are often described as a place to hold[87] or store bitcoins,[88] due to the nature of the system, bitcoins are inseparable from the blockchain transaction ledger. A better way to describe a wallet is something that "stores the digital credentials for your bitcoin holdings"[88] and allows one to access (and spend) them. Bitcoin uses public-key cryptography, in which two cryptographic keys, one public and one private, are generated.[89] At its most basic, a wallet is a collection of these keys. | 2,618 | 12,739 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2018-51 | latest | en | 0.955535 |
http://www.sliderules.info/a-to-z/scales.htm | 1,686,005,337,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224652184.68/warc/CC-MAIN-20230605221713-20230606011713-00310.warc.gz | 90,198,048 | 3,203 | Scales
A standard naming system has developed for scales over the years. For example most calculations are done using C and D scales, the A and B scales usually represent the values on the C and D scales squared, and so on. Not all makers, particularly earlier ones, respected this naming convention or even gave their scales any name at all. The following table shows the normal designation and alternative names commonly found. If you have a scale whose name is not on the list, look at the column "Numbering (for identification)" which should enable you to identify it.
You should be aware that since, for most of the scales, the position of the decimal point is not clear from the scale marking some scales may be marked as, for example, p to p. A scale like this would start at 3.14, have numbers increasing to 9, a 1 in the middle of the scale and then numbers increasing to 3.14.
The numbering is given from left to right so, for example, reciprocal scales are shown as running from 10 to 1.
This image shows a rule which follows the standard naming convention.
The next image shows a rule on which the scales are not named but, for example, the upper scale of the stock can be identified as a cube (K) scale as it has 1 to 1 repeated 3 times. (The cursor has been removed for clarity).
Normal Designation Alternative Designation Numbering (for identification) Uses Notes A x2 1 to 100, 1 to 10 repeated twice Squares and square roots (on stock) B x2 1 to 100, 1 to 10 repeated twice Squares and square roots (on slide) C x 1 to 10 Multiplication and division (on slide) C1,C2 Ö10 to Ö10 C scales shifted by Ö10 CI RC, 1/x, R 10 to 1 Reciprocals (on slide) CF px Pi to Pi Multiplication and division by pi CIF Pi to Pi Multiplication and division by pi D x 1 to 10 Multiplication and division (on stock) D1,D2 Ö10 to Ö10 D scales shifted by Ö10 DF px Pi to Pi Multiplication and division by pi DI 10 to 1 Reciprocals (on stock) Dynamo/Motor 20 to 100 to 20 Electrical efficiency Often in well of stock Fahr/Cent Various Celcius (centigrade) to Fahrenheit conversion, Change of resistance with temperature. Often on electrical rules. ISd 1.0 to 0.0 (short scale on slide) Inverse sines On PIC/Thornton rules ITd 0 to 1.76 (short scale on slide) Inverse tangents On PIC/Thornton rules K x3, d 3 1 to 1000, 1 to 1 repeated 3 times Cubes and cube roots L lg x,log 0.0 to 1.0 0.0 to 0.5 and 0.5 to 1 Logarithms (base 10) Two ranges are only for use with W scales. LC 25000 to 2.5 (2px)2 For electronic calculations (Cooke). LG H 2.5 to 2.5 1/(2px)2 For electronic calculations (Pickett). Ln 0 to 2.3 Logarithms (base e) LL00 e-x .999 to .99 Raising to a power (x<1) LL01 e-0.1x .99 to .91 Raising to a power (x<1) LL02 e-0.01x .91 to .37 Raising to a power (x<1) LL03 e-0.001x .37 to .0001 Raising to a power (x<1) LL0 e0.001x 1.001 to 1.01 Raising to a power (x>1) LL1 e0.01x 1.01 to 1.12 Raising to a power (x>1) LL2 E1,e0.1x ,LU 1.1 to 3.0 (slight variations in upper and lower bound) Raising to a power (x>1) LL3 E2,ex ,LL 2.6 to 105 (slight variations in upper and lower bound) Raising to a power (x>1) P f(x),Ö(1-s2),Ö(1-x2) .995 to 0 Right angled triangles S sin 5.5° to 90° Sine Sd sin 0° to 90° (short scale on slide) Sines On PIC/Thornton rules Sh SH, Sh1, Sh2 0.1 to 0.89 and 0.89 to 3.0 Hyperbolic sine Sq1,Sq2 Ö 1 to 3.16 and 3.16 to 10 Square and square roots Sq on K&E and Ö on Pickett. ST S&T, sin tan, arc, SRT, s-t .55° to 6°, 40' to 5° 45' (slight variations in upper and lower bound) Sin and tangent for small angles T tan, T1, T2,tg 5.5° to 45° sometimes also 45° to 84.3° Tangent Td tan 0° to 60° (short scale on slide) Tangents On PIC/Thornton rules Th TH 0.1 to 3.0 Hyperbolic tan Volt 05. to 10 Voltage loss Often in well of stock W1,W1',W2,W2' P1,P2,Q1,Q2 Ö10x and Öx 2.8 to 11 and 0.9 to 3.5 or1 to 3.16 and 3.16 to 10 20" scales on slider of 10" rule or 10" scales on 5" rule Extended range on Faber Castell 2/83N Y 10% to 0% (small scale on stock) Interest rate calculations On PIC/Thornton rules Z 10 to 0 (small scale on stock) Interest rate calculations On PIC/Thornton rules 2p fx 2p 1.6 to 1.6 1/(2px) For electronic calculations. square root 3Ö cube root 1 to 2.16, 2.16 to 4.64 and 4.64 to 10 Cube and cube roots Mainly on Pickett rules
An explanation of how to read the scales is given here. | 1,408 | 4,339 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2023-23 | latest | en | 0.933566 |
https://www.studypool.com/discuss/1175425/looking-for-angle-measures?free | 1,481,325,286,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542851.96/warc/CC-MAIN-20161202170902-00011-ip-10-31-129-80.ec2.internal.warc.gz | 1,008,278,806 | 14,531 | ##### looking for angle measures
Mathematics Tutor: None Selected Time limit: 1 Day
1- An angle measures . What is the measure of its supplement?
2-An angle measures . What is the measure of its complement?
Oct 3rd, 2015
What we are dealing with here is geometry. Now even though we do not have a lot of data we want to define what an angle measurement is as it relates to shapes and angle measures. Now let's suppose we had an angle which was a supplemental angle. For example angle AB is 100 degrees its supplement would be an angle measure of 80 degrees for angle BC so that it would add up to 180.
On the other hand you also have a complementary angle which has to add up to 90 degrees. For example angle CD could measure 45 degrees, what would be the complement. It would be another acute angle measuring 45 degrees so clearly supplemental and complimentary angles are fairly easy.
There is a website I would like for you to see in order to help you see more clearer how this all fits together I think you will find it very helpful. This website explains clearly by definition how to apply the steps for various functions in Math. There is also another website that you can take a look at on www.kutasofware.com. This is a math website as well that eplains angles very well.
https://www.mathsisfun.com/definitions/complementary-angle.html
Sep 18th, 2015
Hello,
Please tell me what you think of those websites when you get a chance to look at them. I think you will find them to be very helpful.
Mr. James Price
Study Pool Tutor
Sep 18th, 2015
...
Oct 3rd, 2015
...
Oct 3rd, 2015
Dec 9th, 2016
check_circle | 404 | 1,636 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2016-50 | longest | en | 0.966137 |
https://se.mathworks.com/matlabcentral/answers/821355-how-can-i-get-the-value-by-choosing-t-and-x-how-should-i-set-the-graph-range | 1,670,147,814,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710968.29/warc/CC-MAIN-20221204072040-20221204102040-00204.warc.gz | 532,520,082 | 26,889 | # How can I get the value by choosing 't' and 'x'? ,How should I set the graph range?
2 views (last 30 days)
Seung Ho Kim on 4 May 2021
Answered: VBBV on 26 Sep 2022
L = 30;
n = 10;
T0=500;
T1s=300;
T2s=400;
dx=L/n;
alpha=0.000085;
t_final=4800;
dt=0.1;
x = dx/2:dx:L-dx/2;
T = ones(n,1)*T0;
dTdt=zeros(n,1);
t = 0:dt:t_final;
for j = 1:length(t)
for i = 2:n-1
dTdt(i)=alpha*(-(T(i)-T(i-1))/dx^2+(T(i+1)-T(i))/dx^2);
end
dTdt(1)=alpha*(-(T(1)-T1s)/dx^2+(T(2)-T(1))/dx^2);
dTdt(n)=alpha*(-(T(n)-T(n-1))/dx^2+(T2s-T(n))/dx^2);
T=T+dTdt*dt;
figure(1)
plot(x,T)
xlabel('distance(n)')
ylabel('Temperature(\circC)')
pause(1)
end
J Chen on 4 May 2021
Move the plot command outside the loop. Store the calculation at each time step in a buffer. For example:
hist = nan(n,length(t));
for
.
.
hist(:,j) = T;
end
plot(:,400)
VBBV on 26 Sep 2022
L = 30;
n = 10;
T0=500;
T1s=300;
T2s=400;
dx=L/n;
alpha=0.000085;
t_final=4800;
dt=300; % use a coarse timestep for faster loop execuetion
x = dx/2:dx:L-dx/2;
T = ones(n,1)*T0;
dTdt=zeros(n,1);
t = 0:dt:t_final;
for j = 1:length(t)
hold on % use a hold on command
for i = 2:n-1
dTdt(i)=alpha*(-(T(i)-T(i-1))/dx^2+(T(i+1)-T(i))/dx^2);
end
dTdt(1)=alpha*(-(T(1)-T1s)/dx^2+(T(2)-T(1))/dx^2);
dTdt(n)=alpha*(-(T(n)-T(n-1))/dx^2+(T2s-T(n))/dx^2);
T=T+dTdt*dt;
%figure(1)
plot(x,T)
end
xlabel('distance(n)'); ylabel('Temperature(\circC)')
Use a hold on command at the beginning of first for loop. Assume a coarser time step for faster loop execution. | 634 | 1,481 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2022-49 | latest | en | 0.545741 |
https://www.coursehero.com/file/5914451/ECE474S09-Lecture22/ | 1,527,144,285,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794865928.45/warc/CC-MAIN-20180524053902-20180524073902-00613.warc.gz | 719,270,130 | 180,401 | {[ promptMessage ]}
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ECE474S09_Lecture22
# ECE474S09_Lecture22 - ECE 474 Principles of Electronic...
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ECE 474: Principles of Electronic Devices Prof. Virginia Ayres Electrical & Computer Engineering Michigan State University [email protected]
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Lecture 22: Chp. 03: Current State-of-Art Transistors Use n and p to get current I : I = q( n μ + p μ ) E , E = electric field (V/cm), μ = mobility (cm cm )
I = q(n 0 <velocity electron >+ p 0 <velocity hole >) Area I = q(n 0 μ n + p 0 μ p ) E Area J = I/Area = q(n 0 μ n + p 0 μ p ) E = σ E Lecture 21:
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Battery only: what happens? V = IR, R= ρ L /wt Hole motion, electron motion what you’d expect from the +- terminals of the battery I = q(n 0 μ n + p 0 μ p ) E Area + -
Battery only: what happens? V = IR, R= ρ L /wt Hole motion, electron motion what you’d expect from the +- terminals of the battery I = q(n 0 μ n + p 0 μ p ) E Area New: F = q E + -
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Battery only: what happens? V = IR, R= ρ L /wt Hole motion, electron motion what you’d expect from the +- terminals of the battery I = q(n 0 μ n + p 0 μ p ) E Area New: F = q E F x = q E x + - +z +y +x
Magnetic field B + battery: what happens?
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Jill Tulane University ‘16, Course Hero Intern | 692 | 2,692 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2018-22 | latest | en | 0.839607 |
http://blog.lib.umn.edu/sali0090/synbio/cat_general.html | 1,432,568,064,000,000,000 | text/html | crawl-data/CC-MAIN-2015-22/segments/1432207928520.68/warc/CC-MAIN-20150521113208-00205-ip-10-180-206-219.ec2.internal.warc.gz | 28,935,756 | 13,084 | ## November 18, 2005
### Why Math in Synthetic Bio?
In the past 3 years, I've developed some very advanced methods for computing the stochastic dynamical behavior of "small" physical or chemical systems, which especially includes biological systems. The methods are an improvement over existing ones, decreasing the amount of computational time by 1000x or more for some systems, but still retaining accuracy. Now, I'm working on more methods to better predict the long time behavior of biological systems.
But why is using math important in synthetic biology?
The idea is to develop an accurate model of a biological system (usually focusing on a small subsystem of a single cell) and then predict what the dynamics will be over time. If you can predict what the system will do before you build it, you save yourself both time and money. The model should be a "first principles" one based on the molecular interactions of each DNA site, protein, RNA, etc molecule in the system. That way, if you know the interactions of a DNA site in one model, then you should be able to put the same DNA site in a different model and still predict what will happen. (No lumped interactions!) Of course, we're still constrained by the limited amount of information we have on molecular "parts". That's ok for now, because (one day) we should have that information. Until then, we will need to be good engineers and make guesses (yes, guesses) on what those interactions might be and how they affect the system dynamics.
You would be surprised as to how much guessing goes into making 70 story buildings, cars that move at 120mph, and lots of other contraptions that will easily kill you if built incorrectly. The "engineer guess" is making sure that if you're 500% wrong that nothing bad will ever happen. The technical term is robustness. But, in practice, you assume the unknown quantity can take values within a very large range and then you make sure that nothing breaks for any value in that range. Of course, you have to pick which quantities to make your design robust to. That's where you get this tradeoff between "robustness" and "fragility".
But we should be rigorous about our "guessing". We should be able to identify _all_ possible behaviors that exist when varying the value of a specific quantity. What might the quantity be, as related to synthetic biology? It could be the binding affinity of a protein to a DNA site. It could be the enzymatic Kcat of a phosphorylation reaction. It could be the influx of a regulatory protein from the extracellular space. It is any parameter in our model that is not entirely known.
The math behind computing _all_ possible behaviors of a system while varying one or more parameters is called bifurcation analysis. The subject has always interested me and it's actually extremely useful in real life. Computing that your reactor has a subcritical Hopf bifurcation at a critical parameter value tells you that if your parameter is past this point, your reactor will suddenly blow up and kill lots of people. Whoever said math wasn't useful? In practice, they make sure the parameter never goes near that critical value..not even remotely near it. So reactors generally don't blow up. Whew, that's good to know.
How is bifurcation analysis related to synthetic biology? Say you wanted to create a gene therapy system that consisted of a biosensor + regulated production of a therapeutic protein. There might be 120 parameters in your system. You might have good information about 60 of those, so-so information about 30, and the rest...who knows. But you want the gene therapy to work no matter what. Even if you incorrectly measured the interaction between molecule A and B. Even if you get a mutation that changes an interaction between molecule C and D. It just has to work. If it doesn't, someone can die. So can you determine the behavior of the model over all unknown parameters and make sure that the system will never break? If the number of unknown parameters is 30, well ... that's a 30 dimensional space to worry about. Mathematically, you could do it...but it would take a while. Are there ways to speed up the process? Absolutely. (I won't go into details here.) But my main point is that bifurcation analysis is extremely important for designing synthetic biological systems.
But, wait, I did say "stochastic dynamics" and the bifurcation analysis of stochastic systems is ... not well developed yet. That's because when you're working with probability distributions, the idea of a "qualitative change" in the solution gets harder to define. And working with these types of systems is harder in general. So that is what I am currently doing. And it is going well. :)
The math can get very heady and I worry that people who lack the background will become turned off by it. But the final product is very useful: You can find all possible behaviors of a "small" physical or chemical system (such as biological system) using a combination of existing and new simulation techniques. The words "all possible" are extremely important. As it turns out, if your system has two stable states, a "good" one and a "bad" one, then, because of the random nature of the interactions, the dynamics might first go to the "good" state, but later go to the "bad" one. Not good. But you can minimize the "escape" from the good to bad state if you design the system well enough. This type of "escape" doesn't happen if you describe the system using deterministic dynamics, but it happens in real life. (One more reason why stochastic descriptions are important.)
If you've read this far, then my guess is that you're somewhat interested in mathematics. Study it! (Especially non-linear dynamics, bifurcation analysis, and stochastic processes.) They are useful for real life applications, especially in new fields such as synthetic biology. If you're looking for a book on non-linear dynamics I would suggest the one by Steven Strogatz. The material is at the intermediate-advanced undergraduate level with lots of pictures.
So to answer my initial question: "Why Math in Synthetic Biology?". Math is needed because there's no way to build large, complex dynamical systems without first understanding (and then predicting) what those dynamics will be. Math then gives us the tools to gaurantee certain types of behaviors even if we don't exactly know all of the parameters of a model of the system. Also, using math generates testable and precisely quantitative predictions about the biological system of interest.
(This post is very hodgepodge and not very technical. For technical details on the bifurcation analysis of stochastic systems, you'll just have to wait for the paper. In the mean time, there's two papers on the stochastic numerical methods that are 1000x+ faster than the original one. My group also published a paper on the design principles behind an oscillating gene network. If you search PubMed for 'Salis H', they should come up.)
Posted by at 11:59 AM
## August 6, 2005
### Microfluidics & Laser Manipulation of Cells: The Road to Automated Molecular Biology?
Is automated DNA synthesis, transformation, selection, PCR, ligations, and imaging that far away? How would molecular biology change if a single PhD student could design and synthesize an entire plasmid or section of chromosome in a week? Besides exponentially speeding up the productivity of researchers (and encouraging PhD advisors to get even more paper greedy ;) ), it would have tremendous effects on the whole idea of synthetic biology. Design it. Make it. It works. Time To Completion? A month? A week? If you ask the typical biologist about the prospects of automating the process of genetic engineering, they'll say it's crazy talk. But two recent presentations at two separate conferences outlined the first steps towards the beginning of a new era in biology.
The first conference was on a new cell-manipulation system that uses lasers to selectively toast, lyse, or heat individual cells. The conference is named Biochemical Engineering XIV and took place in Harrison Springs, British Columbia (near Vancouver). The presenter is a founder of a company named Cyntellect and he demonstrated how a high powered laser, driven by the precise movement of mirrors, can target a single yeast or mammalian cell for treatment. The laser has three modes of operation: it can completely break open and evaporate the fluid of a cell, killing it instantly; it can hit the cell with a burst of mechanical force that lyses the cell open, but leaves the intracellular space generally unharmed (including the DNA!); and it can gently heat up a single cell to a nice 42 deg Celsius. Using those three operational modes, the device can automate the transformation and selection process _on the single cell level_. Cool, huh? There's a built in optical microscope to image either staining or fluorescence and the laser system can target groups of cells that have a particular fluorescence (which looked similar to the gating function of a cytometer).
So you can heat shock, transform, image, select, lyse, and ... uh oh..that's all it can do for now. Either way, I was very excited by the time the guy finished his presentation (I mean, hey, he's using fricken laser beams). The real reason, though, is that the field of microfluidics has steadily advanced to the point where you can start pumping in and out reagents of all types into a chamber and control the reactions. Cyntellect only supports 96-well plates for now, but there's no reason why someone couldn't put a nice microfluidic device in there with a good sized viewing window for the laser. The laser system's chamber is temperature controlled (with a good response time!) so we're talking about _everything_ one needs to continue the process of genetic engineering: heat shock, transform, image, select, lyse, ... PCR, purify, and repeat!
The second conference was on "Molecular Recognition & Biosensors" in Santa Barbara, CA. The conference was sponsored by the Army's AHPCRC. In my mind, I knew microfluidics have been making steady strides, but seeing the recent accomplishments really surprised me. A professor at Cal Tech named Yo-Chong Tai presented his work on _integrated_ microfluidic devices at a recent conference The components of the integrated microfluidic device, such as the pumps, valves, pipes, and other chambers, are molded together in a single chip using micrometer-resolution plastic deposition. To strut his stuff, he showed off his mini-HPLC (High Performance/Pressure Liquid Chromotagraphy). The mini-HPLC is 3 cm in length, uses ~100 nl of sample, and the results match the big commercial ones. Crazy, huh? How far away is he from making a tiny electrophoresis chamber? Probably not too far.
Using an integrated microfluidic device combined with laser manipulation of individual cells, one can perform all of the necessary tasks that are needed to genetically engineer cells. How long will it be until someone decides to put these two innovations together and optimize the automation of the entire process? I give it 4-6 years, tops. Genetic engineering (cloning) is such a tedious, mind numbing, and repetitive task that any efficient (and cheap) automation would have a huge, earth-shattering effect on the entire field.
Honestly, I can't wait. I don't do half the cloning that some of my other friends do, but I still think it's the most repetitive and least satisfying part of molecular biology. If it works, you're happy...but only because you don't have to repeat it. The real satisfying accomplishment is doing the experiment and gaining knowledge (and something to write a paper about, heh). And if it doesn't work...ugh, wasted time!
-Howard
Posted by at 12:40 AM
## July 22, 2005
### A Quick Update
Hello all,
I apologize for the huge delays in posting, but I promise that it hasn't been in vain.
For the past five months, I've been going back and forth between a wet and dry lab, continuing to develop new stochastic numerical methods while helping to build a gene network in E. coli. I started my graduate studies in the computational area of systems / synthetic biology and moving into the experimental side has certainly been a change of pace. Some of the other bio people are pretty shocked to find me pipetting. When they ask why, I usually say something glib like "What use is designing something if you never build it?" Which is more or less true. However, in between experiments, I've been designing new networks and developing new stochastic methods. The work is going well. I (and my advisor, of course) have two papers in the review process and two more being written. I'll wait until they're published online before blogging about them (probably safer that way, heh).
I still believe that synthetic biology will never mature into an engineering science unless we use computational tools to quantitatively predict the behavior of a biological system. We need to create the same sort of tools, like CAD, that the auto industry uses to design cars. They spend 95% of their time on the computer, testing new hypothetical designs, and they only build prototypes when they see something work well. The same can be said for other industries, including building chemical or pharma plants and computer chips. The same process must develop for synthetic bio. We need more experiments to determine how simple biological systems work, but then we must incorporate that information into realistic models, powered by sophisticated mathematics. Without the combination of both, we'll be flailing in the dark, using the trial and error system that is rarely productive, often boring, and never generalizable.
I also have some thoughts on the differences between wet and dry lab experiments. I'll save that for a future post, I think.
I recently received a bunch of questions about stochastic methods from Dr. Herbert Sauro, who (like me) is a member of the sbml-discuss mailing list. They were very good questions and I think I did a decent job of answering them so I think I'll link them here
Ok, that's all for now (I so suck at blogging).
Posted by at 8:55 PM
## February 3, 2005
### And the Fun Begins
I apologize for the long delay in posting. Right now, I'm finishing up a manuscript on the design of biological AND gates that activate gene expression if and only if two or more transcription factors are present. I won't mention any details until the paper is accepted, but these facts should pique your interest:
-- They require no special cooperativity or allosteric changes (ie. no intensive engineering)
-- They can be made for two or three inputs (and probably more, but I didn't want to push it)
After the paper's out the door, it's time for the real fun: experiments!
So far, most of the work has been computational: developing new algorithms and software to quickly design and analyze synthetic biological systems. The methods have been shown to be accurate, but there's always a lingering doubt that the in silico results won't match the in vivo results. Now, it's time to go back to the (wet) lab and really see if it all works. The Fun Begins.
To test out our methodoloy, we thawed out some strains of E. coli producing GFP under control of the lac operon and sent it through a flow cytometer (courtesy of Dr. Friedrich Srienc, thank you very much). The pictures it produces are simply fantastic.
It's amazing how stochastic simulations and flow cytometry complement one another. A stochastic simulation will produce the probability distribution of the solution at whatever time points you want, but requires numerous independent trials (usually 10,000, each representing a single cell). A flow cytometer, on the other hand, can count hundreds of thousands of cells and measure the fluorescence in each individual cell, producing a very smooth distribution, but only for a relatively few number of time points. Using the simulations, you can gain a high-resolution (in time) picture of the dynamics. Using the flow cytometer, you can validate the results at specific time points and easily produce very smooth distributions. Very nice.
If you're interested in computational design of biological systems, check out http://hysss.sourceforge.net. HySSS, or Hybrid Stochastic Simulation for Supercomputers, is a software package that creates, simulates, and analyzes biological or chemical systems. It uses the hybrid homogeneous stochastic simulation previously mentioned. That paper actually was published on February 1st (not Jan 15th, woops) in J Chem Phys.
Posted by at 9:58 PM
## November 20, 2004
### Good News!
Paper #2 titled 'Accurate hybrid stochastic simulation of a coupled system chemical or biochemical reactions' is slated to be in the January 15th issue of the Journal of Chemical Physics.
In short, it's a novel method improving upon the stochastic simulation algorithm of Gillespie as well as previous hybrid methods. It approximates fast reactions as a continuous Markov process (governed by a system of stochastic differential equations) while still representing the slow reactions as a jump Markov process. Partitioning of the system into fast & slow reactions is dynamic and it introduces a new way to quickly monitor when slow reactions occur while still retaining all of their time-dependence on the fast reactions.
On a related note, I saw Dr. Linda Petzold give a presentation today on a partial stochastic equilibrium and its usage to speed up the stochastic simulation algorithm. She only got through maybe half of it because the moderator kept asking boring questions in the middle of her talk (she obviously thought her questions were very insightful and fruitful, but they weren't). Of course, then the moderator proclaims the five minute warning with only 2/3 of the presentation left. Dr. Petzold was much too nice with her. I would have asked to have the conversation after the 15 minute time limit.
For those of you who aren't modelers, the stochastic simulation algorithm is a computational method that simulates the dynamics of a system of bio/chemical reactions. It's especially useful when the numbers of participating molecules are few because alternative simulation methods (like reaction rate equations / ODEs) fail in the 'small' regime. One can use these types of methods to simulate gene expression or signal transduction or the cell cycle. If you can break it down into a system of reactions, then the SSA can simulate it very nicely. The only problem is that it can be very slow in certain circumstances. The challenge now is to identify the circumstances which cause it to slow down and speed it up somehow. Hybrid methods or the partial stochastic equilibrium assumption are two approximations that seek to make the simulation go faster without losing too much accuracy.
Posted by at 11:56 PM
## October 22, 2004
I've updated the Links and Books sections to be more relevant.
Be sure to check out MIT's Registry of Biological Parts. It's a great idea.
I've also linked the research site of Yiannis Kaznessis, my advisor. In addition to synthetic biology, my group also performs docking calculations for protein-protein pairs and protein-DNA pairs as well as bioinformatics techniques to analyze microarray data. One obstacle to designing synthetic bio systems is the lack of the necessary 'parts'. Part of our integrative approach is to use computational design of proteins/DNA sequences to predict which modifications should be made to provide the needed parts for a particular design.
My (soon to be) published papers are in the list of publications. The first one is available in PDF form and should be published soon (we've corrected the galleys already, but they've been dragging their feet on this 'special' issue of theirs. It's taken a year after acceptance for us to get back the galleys!).
Posted by at 8:03 PM
## October 19, 2004
### Before there were Answers, there were Questions
Before I begin reviewing the field of Synthetic Biology and discussing topics of interest, I just wanted to start by listing some of the questions that drive the research. Before you can find useful answers, you must always ask good questions.
What is Synthetic Biology and what do we want to gain by it? How is it useful? To what extent is it 'molecular biology' under a different name and why is it called 'synthetic'? What are we creating and how do we do it? What sort of methods are used?
I'll start answering those questions, but since the field is so new, feel free to pipe up and throw in your own two cents. I won't go into too many specifics or else I'll be writing all night. ;)
Synthetic biology is the study or design of biological molecules whose function did not previously (knowingly) exist in nature. This includes proteins that have catalytic activity (enzymes) or new structural binding properties, such as DNA binding regulators. This also includes engineered mRNA or DNA that has a specific, designed function. The most common example of an engineering biological system is a 'gene network' or 'gene circuit', which is a system of one or more genes whose function has been engineered to perform a specific task.
Why is synthetic biology useful? First, if we can first design a biological system and then build it, then we know how it all (mostly) works. By first predicting what will happen before you build it and then building it, you not only state the hypothesis that a) if built according to the design, it will work, but also that b) the biological system exists as you represented it in the design. So if the system doesn't behave as one might think, then something unknown must exist. Like all good scientific efforts, we have a hypothesis. But as engineers, if something doesn't work, we can investigate the problem and determine the solution.
Secondly, biology naturally interfaces with other biology. The most effective treatments of disease will naturally be biological molecules whose purpose has either evolved to treat the disease or which has been designed (by us) to do so. Not only can we engineer molecules to activate/inhibit/bind/etc, treating a disease, but we may also engineer the production, degradation, and localization of that molecule to control its effects and prevent the cure from becoming worse than the disease. We may also construct biological devices that detect the presence of other biological or chemical molecules (a biosensor), which would have tremendous use in medical diagnostics or defense.
So we can use synthetic biology to study biology while we build new and useful biological devices.
Which brings me to another aspect of synthetic biology: There's really a lot of engineers doing it. I'm also an engineer so I'm happy about that, but I've become accustomed to entering a seminar and being the only engineer there. Why does synthetic biology attract engineers, then? Well, I've been using the word 'design' over and over so that should be a clue.
How is synthetic biology different from molecular biology? Well, ... synthetic biology IS molecular biology, except more quantitative and precise. If you're reading a journal catering to molecular biologists, you'll typically see a model as a slightly cartoonish diagram depicting the interactions between a collection of proteins/etc and arrows showing the order of events. Basic questions are left unanswered by such models: How strong are the interactions? For every protein/etc in the diagram, are there additional interactions that will affect the model? Even though the interactions are listed, what are the dynamics that result? These answers may be counter-intuitive. One part of synthetic biology is to create a more defined, quantitative (and predictive) model of biological systems. Depending upon the level of detail, one could include the kinetic constants of all interactions (reaction/binding events), all unique chemical species, diffusion of all species, and membraned compartments. One outstanding question is what amount of detail is necessary to get predictive results. Conversely, what approximations may we make without sacrificing accuracy?
Finally, what are some of the methods that we use? To build these designs, common genetic engineering techniques are employed to cut and paste DNA into vectors, transform vectors (plasmids) into an organism, and (possibly) integrate the vector into the genome of the organism. These techniques have been used for the past 50 years and, while there are difficulties in extreme cases, it is relatively easy to construct something interesting. (By easy, I mean, it won't take one person their _entire_ PhD program...maybe only a year or two. ;) )
The real obstacle is not the experimental construction, but the design of the DNA sequences. In order to quickly design the system and avoid excessive experimental trial and error, mathematical tools must be used to analyze a design and ascertain whether it will function as expected. Relying on experimental construction alone would result in years of wasted effort. When using mathematical tools, there are two main questions: What quantitative, mechanistic model best predicts the dynamic behavior of the particular biological system of interest? What mathematical representation (and simulation) best reflects the process that occurs? These are two separation questions because one may take a good model and generate faulty equations with it, where the assumptions used in forming those equations are wrong. Solving those equations perfectly would reflect an incorrect answer, even though the model is perfectly valid. Conversely, forming and solving the most accurate and complete equations with the least number of assumptions would be futile if the model itself was not accurate and predictive.
This is where my research starts. As you could tell, I haven't gone into details. Over the next few weeks, I plan to review specific topics within the synthetic biology area, including experimental construction of different gene networks, the mathematical theory behind the most advanced simulators, and the guiding principles behind the optimal design of gene networks.
The format will be informal. I'll reference where it's necessary and include pictures when I can find them. For the math, there's no LaTex and so I will probably just upload PDFs.
This blog is not a substitute for my published papers. I spend a lot more time on them than I do on this (wisely, as you may agree). Feel free to post comments below.
Posted by at 8:39 PM
## September 25, 2004
### Welcome!
Somehow you have stumbled onto this blog. That's good, I guess. First off, let me introduce myself: My name is Howard Salis and I'm a graduate student within the Department of Chemical Engineering and Materials Science here at the U of Minnesota.
My research is on the study and design of systems of genes, also called 'gene networks' or 'regulatory networks' or just plain gene expression. Some people have recently branded this area of reseach as "Synthetic Biology" because a lot of knowledge may be gained by building systems from scratch. Well, no one has built a biological system from scratch yet..so it mostly means that we're building systems that have never existed before in nature, but using parts that nature has graciously given us. I can just about hear the red whirling lights of fear dancing in your head. Every time I mention my research to *cough* lay people I elicit two widely different responses: It's either 'Wow, that's so cool!' or 'MY GOD, you'll kill us all!'. I'm sure people working on the first microwave got the same response. I plan to use this blog to both document my own thoughts on the science behind synthetic biology and, if I get any readers, to discuss the ethical consequences of the now having the capability to design biological systems from a rational perspective.
But, and you should understand this, there's a lot of mathematics behind the 'rational design' part of synthetic biology. It's not easy to predict the de novo structure of a protein, the kinetic constants of molecular interactions, or the dynamics of gene expression. There's some crazy math involved and I may delve into the vortex of technicality and jargon. O well, you're reading this far so you can't be completely bored. I may also use this blog to document my thoughts on papers I've recently read. I have about 300 papers in my desk cabinet and I've honestly come to the point where it's very difficult to pick up one that I've read and say "I remember reading this." That's bad. Notes are good. Blogs are good. Why not do both at the same time?
Sounds good.
Posted by at 1:39 AM | 5,912 | 28,736 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2015-22 | latest | en | 0.951238 |
https://oeis.org/A286313/internal | 1,660,828,774,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882573197.34/warc/CC-MAIN-20220818124424-20220818154424-00783.warc.gz | 392,947,481 | 3,075 | The OEIS is supported by the many generous donors to the OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A286313 Union of A078561 and A078562. 1
%I
%S 19,31,43,61,73,79,127,157,163,229,271,349,373,379,433,439,499,607,
%T 643,673,733,751,937,967,1009,1093,1213,1279,1291,1429,1489,1543,1549,
%U 1597,1609,1657,1777,1861,1987,2131,2203,2287,2341,2347,2371,2383,2389
%N Union of A078561 and A078562.
%C Number of terms among first 10^k primes, k=1..8:
%C 0, 1, 17, 105, 646, 4385, 31721, 240346, 1884832.
%C E.g., k=1, first 10 primes are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, and only 19 is a term of the sequence. - _Zak Seidov_, May 08 2017
%C Primes p such that prime(p+2) = p + 10. - _Harvey P. Dale_, Jan 13 2022
%H Harvey P. Dale, <a href="/A286313/b286313.txt">Table of n, a(n) for n = 1..2000</a>
%H R. J. Mathar, <a href="/A022004/a022004_1.pdf">Table of Prime Gap Constellations</a>
%p select(p -> isprime(p) and isprime(p+10) and (isprime(p+4) xor isprime(p+6)), [seq(i,i=5..10000,2)]); # _Robert Israel_, May 08 2017
%t Select[Prime[Range[1000]], NextPrime[#, 2] == # + 10 &]
%t Select[Partition[Prime[Range[400]],3,1],#[[1]]+10==#[[3]]&][[All,1]] (* _Harvey P. Dale_, Jan 13 2022 *)
%Y Cf. A078561 and A078562.
%K nonn
%O 1,1
%A _Zak Seidov_, May 06 2017
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Last modified August 18 09:18 EDT 2022. Contains 356204 sequences. (Running on oeis4.) | 665 | 1,656 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2022-33 | latest | en | 0.605951 |
https://www.homydezign.com/math/925933-what-percent-of-300-is-18/ | 1,675,842,635,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500719.31/warc/CC-MAIN-20230208060523-20230208090523-00504.warc.gz | 805,527,009 | 14,209 | # what percent of 300 is 18?
Here are the answers to the questions:
what percent of 300 is 18?
Solution for 18 is what percent of 300
100%/x%=300/18
(100/x)*x=(300/18)*x - we multiply both sides of the equation by x
100=16.6666666667*x - we divide both sides of the equation by (16.6666666667) to get x
100/16.6666666667=x
6=x
x=6
now we have:
18 is 6% of 300 | 134 | 376 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2023-06 | latest | en | 0.698694 |
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18 Jun 2008, 15:28
I would strongly suggest you do a search for ChineseBurned's template for AWA. I studied that for about an hour along with the template from the Princeton Review book. That was enough!! And I honestly believe it will be enough for anyone. As long as you have a good structure to your essay, write a lot and use some of the key words, you will definitely get above a 5. You can almost write anything (god knows I wrote every stupid idea that came to my mind to fill in the space) Got a 6 on my AWA.
Once again, IMHO do not waste time on the AWA. Especially if you are a native speaker or can write decently, it really should not be a focus in your preparation.
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19 Jun 2008, 11:48
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19 Jun 2008, 12:57
I scored a 6. It's is easy. Just use the template.
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19 Jun 2008, 13:09
I agree with the above. If you are a native english speaker there should be no problem getting a 6. Use the template, write a lot of words, use keywords (Moreover, Alternatively, Furthermore) , write 4 paragraphs, use examples for issue essay, even if they seem to be abstract.
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http://centralokgo.org/m0z3npe/ia6qh.php?page=7%2F8-%2B-7%2F12-as-a-mixed-number-b642af | 1,632,620,261,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057787.63/warc/CC-MAIN-20210925232725-20210926022725-00216.warc.gz | 11,937,813 | 6,749 | Talk about it with a partner. Solving a Single Variable Equation : 4.2 Solve : 3x-7 = 0 Add 7 to both sides of the equation : 3x = 7 This calculator will automatically update the answer or solution when any of the inputs change. 7; 8; Page 7 of 8; Adding and subtracting fractions. The mixed number subtraction work with steps shows the complete step-by-step calculation for finding the mixed numbers difference of two mixed numbers $5\frac{3}{7}$ and $6\frac{4}{5}$ using the mixed numbers subtraction rule. Example of numbers and how can they be expressed as fractions: For instance when you are referring to a decimal number such as 0.05, this is as fractional number even though you do not see its numerator and denominator as it can easily be written as 5/100. Enter each fraction or mixed number in your list, each separated by a comma. Convert to a Mixed Number 12/7. Also a table with the result fraction converted in to decimals an percent is shown. If entering a mixed number, enter the whole number, followed by a space, followed by the fraction (for two and one half, enter 2 1/2). Place this digit in the quotient on top of the division symbol. Set up the division problem in long division format. Whose method do you prefer? The inputs include the whole number, numerator or … Keeping 8 the same for the denominator, introduce the 71. Divide by . Subtract from . The result of division of is with a remainder of . 71/8 We are given the mixed number which is 87/8. ... {12} + \frac{4}{12} = \frac{7}{12}\] Mixed numbers. 3x-7 ———— • 12 = 0 • 12 12 Now, on the left hand side, the 12 cancels out the denominator, while, on the right hand side, zero times anything is still zero. This would make the fraction In that case, you could convert it into a whole number or mixed number fraction. The mixed number's fraction is the remainder over the original denominator. For example, looking at the number 0.1234, the number 4 is in the fourth decimal place which constitutes 10 4, or 10,000. 2 Complete the calculations. a) 1 2 5 + 2 3 10 = b) 2 2 5 + 2 3 10 = c) 1 3 4 +3 3 20 = e) 4 1 4 2 11 16 d) 1 3 16 + 4 3 4 = f) 1 4 7/12 = Proper Fraction The result is a (mixed) fraction reduced to it’s simplest form. A fraction is the result of a division of two whole numbers. 7/12 Therefore, this equation is true: 7/12 = 7/12 If the numerator is greater than or equal to the denominator of a fraction, then it is called an improper fraction. Simply determine what power of 10 the decimal extends to, use that power of 10 as the denominator, enter each number to the right of the decimal point as the numerator, and simplify. About fractions. If one number can be written in this form then it is called a rational number. Fraction calculator that shows work to find the sum of two fractions, difference between two fractions, product of two fractions and quotient when fraction divided by a fraction by arithmetic operations like addition, subtraction, multiplication and division. The mixed number's whole part is self explanatory. Multiply the newest quotient digit by the divisor. The equation now takes the shape : 3x-7 = 0. Enter fractions without spaces before or after the division sign (/). 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https://study.com/academy/practice/quiz-worksheet-hinge-theorem.html | 1,585,840,469,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370506988.10/warc/CC-MAIN-20200402143006-20200402173006-00472.warc.gz | 713,334,200 | 24,187 | # Comparing Triangles with the Hinge Theorem
Instructions:
question 1 of 3
### Which of the following statements is true?
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### 2. ΔABC and ΔDEF are shown in the image. AB is congruent to DE and BC is congruent to EF. ∠ABC has measure 54 degrees and ∠DEF has measure 78 degrees. Based on all of this information, which of the following statements is true?
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The hinge theorem and its use in triangle comparison is the subject of this quiz and worksheet. Questions focus on the converse side of the theorem, along with practice questions involving given triangles and the application of the theorem.
## Quiz and Worksheet Goals
You can check out the subjects listed below in the quiz questions:
• A characteristic of the hinge theorem
• The converse of the hinge theorem
• Triangle congruence and the hinge theorem
• Using the hinge theorem to compare triangles
## Skills Practiced
• Problem solving - use the hinge theorem to solve practice problems
• Information recall - remember what you have learned about triangle congruence
• Defining key concepts - ensure that you can name a characteristic about the hinge theorem | 332 | 1,559 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2020-16 | longest | en | 0.923809 |
https://www.schoolfamily.com/blog/entry/2014/04/10/tens-and-ones-an-easy-way-to-remember-place-value | 1,527,280,744,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867217.1/warc/CC-MAIN-20180525200131-20180525220131-00603.warc.gz | 815,738,668 | 26,270 | 2 minutes reading time (325 words)
# Tens and Ones: An Easy Way To Remember Place Value
Young students should understand that, when looking at a two-digit number, the left number represents “10’s”, and the right number represents “1’s.” This is a critical math skill needed for subsequent math advancement.
Here is a simple way to help your child practice this concept. First practice counting by tens, to 100, until your child can easily and fluently do it herself (10, 20, 30, etc.).
• Take an 8 ½ x 11 inch paper and fold it, vertically, in the middle. You should have two equal columns. Trace the fold line, top to bottom, so the columns can clearly be seen.
• On the top of the left column, print the word “tens” using all lowercase letters.
• On top of the right column, print the word “ones.”
• Say the number 24 (as an example) to your child.
• With a pencil, make two thin vertical rectangles, about an inch long, to represent two tens, in the left-hand tens column.
• Make four small dots in the right-hand ones column, directly across from the two rectangles in the tens column.
• Count the vertical rectangles by tens and count the dots by one. Help your child count the number using the rectangles and dots. Start with the tens column and move across to the ones column. “Ten, twenty, then twenty-one, twenty-two, twenty-three, twenty-four.”
• After practicing a few different two-digit numbers together, say a two-digit number, and see if she can draw the rectangles and dots in the correct columns. Practice until she can easily show you the tens and ones in a two-digit number.
Knowing that two-digit numbers can be broken apart into tens and ones, then put back together, gives your child a deeper concept of how math operations work. Understanding place value goes beyond memorization and teaches the “why” of addition and subtraction.
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Calculate arithmetic sequence from the series of numbers easily with this online calculator.. Code to add this calci to your website Just copy and paste the below code to your webpage where you want to display this calculator. Sequence calculator online - get the n-th term of an arithmetic, geometric, or fibonacci sequence, as well as the sum of all terms between the starting number and the nth term. Easy to use sequence calculator. Fourier Series Calculator is an online application on the Fourier series to calculate the Fourier coefficients of one real variable functions. Also can be done the graphical representation of the function and its Fourier series with the number of coefficients desired Inputs
Interactive, free online graphing calculator from GeoGebra: graph functions, plot data, drag sliders, and much more! My calculator shows the result a number that is around 40 digits. Even with calculator, most regular calculators wouldn't be able to display all of its digits. Comment. A geometric sequence is a number sequence in which each successive number after the first number is the multiplication of the previous number with a fixed, non-zero number (common ratio). The general form of a geometric sequence can be written as: Sequence solver by AlteredQualia. Find the next number in the sequence using difference table. Please enter integer sequence (separated by spaces or commas). Free series convergence calculator - test infinite series for convergence step-by-step. This website uses cookies to ensure you get the best experience. By using this website, you agree to our Cookie Policy. Learn more Accept. Number Line. Graph.
## The list of online calculators for sequences and series.
Number Series Calculator is designed to assist in preparation for IQ Tests and Number Series Aptitude Tests. This application works in offline and online modes . Sequences have many applications in various mathematical disciplines due to their properties of convergence. A series is convergent if the sequence converges Series Calculator computes sum of a series over the given interval. It is capable of computing sums over finite, infinite (inf) and parametrized sequencies (n). Find the next number in the sequence (using difference table). Please enter integer sequence (separated by spaces or commas): Example ok sequences: 1, 2, 3 In online mode you can search our number series database using a question mark, for example, 1,?,?,4 or ?,?,3,4. To solve problems with letter sequences, you Enter numbers separated by comma, space or line break: About Sum Calculator. The Sum Calculator is used to calculate the total sum of any set of numbers.
### Just enter the nth term and get the fibonacci series on screen less than 1 second of time. Get more mathematical & scientific online calculators & converters.
In mathematics, a series is, roughly speaking, a description of the operation of adding infinitely To emphasize that there are an infinite number of terms, a series may be called an infinite series. Such a series is Paul's Online Math Notes. 15 May 2019 I create online courses to help you rock your math class. Read more. Once we confirm that our alternating series meets these Your calculator manual is the best reference. Select an item: press the number next to the item, or move the cursor there and press Sequences and Series. Recursion Calculator. A recursion is a special class of object that can be defined by two properties: Base case. Special rule to determine all other cases. As you can see from the following chart, beyond 7 characters, the possible combinations become too large to be practical for an online calculator. Characters Find the Taylor series expansion of any function around a point using this online calculator.
### This is a free online math calculator together with a variety of other free math calculators that compute standard deviation, percentage, fractions, and time, along with hundreds of other calculators addressing finance, fitness, health, and more.
where n is the number of terms, a1 is the first term and an is the last term. Example 1: Find the sum of the first 20 terms of the arithmetic series if
## Use this step-by-step Geometric Series Calculator, to compute the sum of an we are summing an infinite number of terms, we use notation, as always in Math.
Get the free "Pattern Finder" widget for your website, blog, Wordpress, Blogger, or iGoogle. Find more Mathematics widgets in Wolfram|Alpha. The main purpose of this calculator is to find expression for the n th term of a given sequence. Also, it can identify if the sequence is arithmetic or geometric. The calculator will generate all the work with detailed explanation. Instructions: This algebra calculator will allow you to compute elements of an arithmetic sequence. You need to provide the first term of the sequence (\(a_1\)), the difference between two consecutive values of the sequence (\(d\)), and the number of steps (\(n\)). The list of online calculators for sequences and series. Calculate arithmetic sequence from the series of numbers easily with this online calculator.. Code to add this calci to your website Just copy and paste the below code to your webpage where you want to display this calculator.
Series Calculator computes sum of a series over the given interval. It is capable of computing sums over finite, infinite (inf) and parametrized sequencies (n). Find the next number in the sequence (using difference table). Please enter integer sequence (separated by spaces or commas): Example ok sequences: 1, 2, 3 In online mode you can search our number series database using a question mark, for example, 1,?,?,4 or ?,?,3,4. To solve problems with letter sequences, you Enter numbers separated by comma, space or line break: About Sum Calculator. The Sum Calculator is used to calculate the total sum of any set of numbers. Below is the simple calculator which can help you to find missing numbers in an integer sequence. Let's suppose you have text file of consecutive numbers, like n =0 ∞ Go. Related » Graph » Number Line » Examples ». G o t a d i f f e r e n t a n s w e r ? C h e c k i f i t ′ s c o r r e c t. Correct Answer :) Let's Try Again :(. | 1,291 | 6,263 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2024-33 | latest | en | 0.894657 |
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Initial Position: ( in, in) Initial Angle: deg Current Position: (18 in, 0 in) Current Angle: 90 deg Wheel Radius: 1.75 in1.625 in2.0 in Track Width: in
Arithmetic Sequence Challenge
```/* Code generated by RoboBlockly v2.0 */
/* Problem Statement:
Run the program and find the pattern for the sequence of
numbers. Use the pattern to find the fourth number in the
sequence.*/
#include <chplot.h>
double n;
CPlot plot;
int count;
int count2;
// This is a sample solution
printf("Please complete the sequence: 20, 5, -10, ___ . The 4th number is -25.");
printf("The numbers of stars will appear in order of the sequence. If the number is negative, it will be hexagons.");
for(i = 0; i <= 3; i++) {
//plot.backgroundImage("stars.png");
n = 20 + -15 * i;
if (n >= 0) {
count = 0;
while(count < n) {
plot.strokeColor(randcolor());
plot.star(randdouble(-12, 24), randdouble(-12, 24), 1.5, 0);
count = count + 1;
}
}
else {
double repeat_end = abs(n);
count2 = 0;
while(count2 < repeat_end) {
plot.strokeColor(randcolor());
plot.regularPolygon(randdouble(-12, 24), randdouble(-12, 24), 5, 1, 30);
count2 = count2 + 1;
}
}
delaySeconds(1);
}
plot.strokeColor("yellow");
plot.text("You are correct! Well done!", PLOT_TEXT_LEFT, -6, 8);
plot.label(PLOT_AXIS_XY, "");
plot.grid(PLOT_OFF);
plot.tics(PLOT_AXIS_XY, PLOT_OFF);
plot.axis(PLOT_AXIS_XY, PLOT_OFF);
plot.axisRange(PLOT_AXIS_XY, -12, 24);
plot.ticsRange(PLOT_AXIS_XY, 6);
plot.sizeRatio(1);
plot.plotting();```
Load Blocks Symbol Symbol+Word Word Hardware LArduino Workspace Show Ch Save File
Problem Statement:
Run the program and find the pattern for the sequence of numbers. Use the pattern to find the fourth number in the sequence.
Time | 823 | 2,636 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2020-10 | latest | en | 0.581697 |
https://mathoverflow.net/questions/330132/difference-between-local-cohomology-and-cohomology-with-support-in-a-family | 1,721,781,147,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518130.6/warc/CC-MAIN-20240723224601-20240724014601-00613.warc.gz | 326,555,321 | 25,233 | # Difference between local cohomology and cohomology with support in a family
Let $$X$$ be a topological space. A collection of closed subsets of $$X$$ is called a family of supports (in the sense of Cartan) if: (1) the union of any two elements of $$\Phi$$ is an element of $$\Phi$$, and (2) any closed subset of $$X$$ contained in an element of $$\Phi$$ is itself an element of $$\Phi$$.
Given a family of supports $$\Phi$$, one can define a functor $$\Gamma_{\Phi}$$ from the category of sheaves of abelian groups on $$X$$ to the category of abelian groups as follows: $$\Gamma_{\Phi}(\mathcal{F})=\{\sigma\in\Gamma(X,\mathcal{F})\mid\operatorname{Supp}(\sigma)\in\Phi\}.$$ This functor is easily seen to be left exact and one can derive it using resolutions by flasque sheaves. The derived functors of $$\Gamma_{\Phi}$$ applied to a sheaf of abelian groups $$\mathcal{F}$$ are denoted by $$H^p_{\Phi}(X,\mathcal{F})$$ and called the cohomology groups of $$X$$ with coefficients in $$\mathcal{F}$$ and support in $$\Phi$$.
Let $$Z$$ be a subset of $$X$$ and let $$\Phi_Z$$ be the set of all closed subsets of $$X$$ that are contained in $$Z$$. Then $$\Phi_Z$$ is a family of supports and one can consider the cohomology groups $$H^p_{\Phi_Z}(X,\mathcal{F})$$, for any sheaf of abelian groups $$\mathcal{F}$$ on $$X$$. When $$Z$$ is closed in $$X$$, Grothendieck defines the functor $$\Gamma_Z$$ in SGA2 as follows, and notes that the functor $$\Gamma_{{\Phi}_Z}$$ and $$\Gamma_Z$$ are the same: $$\Gamma_{Z}(\mathcal{F})=\{\sigma\in\Gamma(X,\mathcal{F})\mid\operatorname{Supp}(\sigma)\subset Z\}.$$ Then he says: "we want to generalize this definition to the case where $$Z$$ is locally closed in $$X$$."
Question 1. Which definition exactly does he want to generalize? If he wants tp generalize the definition of $$\Gamma_Z$$ to the case where $$Z$$ is locally closed, why doesn't he just consider the functor $$\Gamma_{{\Phi}_Z}$$ in this case, as well?
In any case, when $$Z$$ is locally closed in $$X$$, Grothendieck doesn't define the functor $$\Gamma_Z$$ as $$\Gamma_{{\Phi}_Z}$$. Instead, he chooses an open set $$V$$ that contains $$Z$$ as a closed subset and defines $$\Gamma_Z(\mathcal{F})=\Gamma_Z(\mathcal{F}|_V).$$ Then he shows this definition is independent of the choice of $$V$$.
When $$Z$$ is closed in $$X$$, if $$V$$ is any open subset of $$X$$ that contains $$Z$$ as a closed subset, and $$\sigma\in\Gamma(V,\mathcal{F})$$ is any section with support contained in $$Z$$, then it is easy to see that $$\sigma$$ is the restriction of a section in $$\Gamma(X,\mathcal{F})$$. This is no longer true if $$Z$$ is locally closed in $$X$$, showing that $$\Gamma_{\Phi_Z}(\mathcal{F})$$ and $$\Gamma_Z(\mathcal{F})$$ are not the same in this case.
Question 2. What is the reason that Grothendieck didn't use the functor $$\Gamma_{\Phi_Z}$$ in the case when $$Z$$ is locally closed? In other words, what properties of local cohomology did he have in mind, which he couldn't obtain by using the functor $$\Gamma_{\Phi_Z}$$? Where can we see the difference between local cohomology groups with support in a locally closed subset $$Z$$, and the cohomology groups $$H^p_{\Phi_Z}(X,\mathcal{F})$$?
I think here's a simple example that might illustrate the difference. Let's take $$X$$ to be the real line and $$Z$$ to be the open interval $$(0,1)$$. Let $$\Phi$$ be the closed subsets of $$X$$. Then $$\Phi_Z$$ consists of closed subsets of $$\mathbb{R}$$ contained in $$(0,1)$$. Since these are bounded, $$\Phi_Z$$ winds up being the compact subsets of $$(0,1)$$ and so $$H_{\Phi_Z}^i(Z;\mathcal{F})=H^i_c(Z;\mathcal{F}|_Z)$$, i.e. you get cohomology with compact supports. By contrast, in this case we have $$V=Z$$ and so $$\Gamma_Z(\mathcal{F}|_V)$$ is just the sections of $$\mathcal{F}|_Z$$. In other words, the resulting cohomology is the ordinary cohomology $$H^i(Z;\mathcal{F}|_Z)$$.
Either of these is reasonable to study, but I think this suggests that the latter construction is more in line with what one usually considers to be the cohomology of a subspace. I'd suggest having a look at something like Section II.10 of Bredon's Sheaf Theory, which contains a lot of related material.
• Thank you for your response. One of the goals of my question was to try to understand what should be "called" local cohomology. For example, suppose we want to generalize the definition of local cohomology to the case where the support is not locally closed. What should we be looking for? In other words, what characterizes local cohomology? Do you have any comments on this question? Commented May 26, 2019 at 16:48
• Again, I'd suggest looking at Bredon's book. It also has a lot of material about sheaf cohomology on subspaces. Commented May 27, 2019 at 4:36 | 1,394 | 4,780 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 78, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2024-30 | latest | en | 0.904683 |
https://math.stackexchange.com/questions/2746932/subgroups-and-invariants-in-a-unitary-group-u3 | 1,618,778,178,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038860318.63/warc/CC-MAIN-20210418194009-20210418224009-00485.warc.gz | 465,031,923 | 39,333 | # Subgroups and invariants in a unitary group U(3)
This is related to the post, but an enriched version of the problem. Now we require the richer form of $P_1,P_2,P_3,P_4,P_5,P_6$.
Let $$G=U(3),$$ be the unitary group. Here we consider $G$ in terms of the fundamental representation of U(3). Namely, all of $g \in G$ can be written as a rank-3 (3 by 3) matrices.
Can we find some subgroup of Lie group, $$k \in K \subset G= U(3)$$ such that
$$k^T \{P_1, P_2, P_3,P_4,P_5,P_6, -P_1, - P_2, - P_3,-P_4,-P_5,-P_6 \} k =\{P_1, P_2, P_3,P_4,P_5,P_6, -P_1, - P_2, - P_3,-P_4,-P_5,-P_6\}.$$ This means that set $\{P_1, P_2, P_3,P_4,P_5,P_6, -P_1, - P_2, - P_3,-P_4,-P_5,-P_6\}$ is invariant under the transformation by $k$. Here $k^T$ is the transpose of $k$. What is the full subset (or subgroup) of $K$?
Here we define: $$P_1 = \left( \begin{array}{ccc} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \\ \end{array} \right),\;\;\;\; P_2 = \left( \begin{array}{ccc} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 1 & 0 & 0 \\ \end{array} \right),\;\;\;\; P_3 = \left( \begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \\ \end{array} \right).$$ $$P_4 =\sqrt{2} \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{array} \right),\;\;\;\; P_5 = \sqrt{2}\left( \begin{array}{ccc} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \\ \end{array} \right),\;\;\;\; P_6 = \sqrt{2}\left( \begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \\ \end{array} \right).$$
This means that $k^T P_a k= \pm P_b$ which may transform $a$ to a different value $b$, where $a,b \in \{1,2,3,4,5,6 \}$. But overall the full set $\{P_1, P_2, P_3,P_4,P_5,P_6, -P_1, - P_2, - P_3,-P_4,-P_5,-P_6\}$ is invariant under the transformation by $k$.
There must be a trivial element $k=$ the rank-3 identity matrix. But what else can it allow?
How could we determine the complete $K$?
The answer (and the method) is the same as the previous question.
The subgroup $K$ of $U(3)$ containing invariant matrices are isomorphic to the finite group $$\mathbb{Z}_4\times S_4 \cong\langle i\rangle\times D(2,3,4)$$ where $\langle i\rangle=\{\pm I,\pm iI\}\cong\mathbb{Z}_4$ and $D(2,3,4)$ is the von Dyck group which is isomorphic to $S_4$.
More specifically, $D(2,3,4)=\langle a,b,c \mid a^2=b^3=c^4=abc=I\rangle$ is represented in $U(3)$ as follows: $$a = \begin{pmatrix} -1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix}, \quad b = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{pmatrix}, \quad c = \begin{pmatrix} 0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix}$$
1. It is a fact from elementary linear algebra that $P_a$ and $k^TP_ak$ have the same rank because $k\in U(3)$ is non-singular. Notice that $P_1,P_2,P_3$ are of rank $2$, but $P_4,P_5,P_6$ are of rank $1$. Thus, if $k^TP_ak=\pm P_b$, then either $a,b\in\{1,2,3\}$ or $a,b\in\{4,5,6\}$.
2. All the matrices preserving $\{P_i\mid 1\leq i\leq 6\}$ also preserve $\{P_i\mid 1\leq i\leq 3\}$.
3. Because we already know all the matrices preserving $\{P_i\mid 1\leq i\leq 3\}$ in the previous question, it suffices to check whether those matrices preserve $\{P_i\mid 4\leq i\leq 6\}$ or not.
4. Note that the three generators $a,b,c$ preserve $\{P_i\mid 4\leq i\leq 6\}$ as follows $$\begin{gather*} a^TP_4a=P_4, \quad b^TP_4b=P_5, \quad c^TP_4c=P_5 \\ a^TP_5a=P_6, \quad b^TP_5b=P_6, \quad c^TP_5c=P_4 \\ a^TP_6a=P_5, \quad b^TP_6b=P_4, \quad c^TP_6c=P_6 \end{gather*}$$
5. It is trivial that $\langle i\rangle=\{\pm I,\pm iI\}$ preserve $\{P_i\mid 4\leq i\leq 6\}$.
6. Therefore we have the same solution as the previous question.
• @ ChoF +1, thanks -- I agree that S4 and Z4 are part of answers. How do we know that the $P_1,P_4,P_5$ cannot transform to each other? – annie marie heart Apr 22 '18 at 15:54
• The S4 and Z4 cannot transform P1,P4,P5 to each other. But P1,P4,P5 are part of triplet $3$ under the SU(2) bifundamentals, $2 \otimes 2 =1 \oplus 3$, where 3 is the triplet under adjoint representation of SU(2), thus vector representation of SO(3). – annie marie heart Apr 22 '18 at 15:57
• Similarly, I am conjecturing that P3, P5, P6 can transform to each other under a U(3) subgroup, and P2, P4, P6 can transform to each other under a U(3) subgroup. Or it could be that there is a subtle way to define the SU(2)'s quotient group SO(3) and its finite subgroup within the U(3). See also /math.stackexchange.com/questions/2746912/ – annie marie heart Apr 22 '18 at 16:09
• @annieheart $P_1,P_2,P_3$ have rank $2$ and $P_4,P_5,P_6$ have rank $1$. Notice that $k^TP_ak$ preserves the rank information. So $P_1,P_2,P_3$ cannot transform to $P_4,P_5,P_6$. – ChoF Apr 22 '18 at 23:15
• @annieheart This question is finding the group of invariant matrices which preserve the set $\{\pm P_i\mid 1\leq i\leq 6\}$. I don't know why you say "The S4 and Z4 cannot transform P1,P4,P5 to each other." Actually $Z_4$ which is generated by the diagonal matrix $i\times I$ transforms $\{\pm P_1,\pm P_4,\pm P_5\}$ to itself. Moreover, the generators $a,b,c$ of $D(2,3,4)\simeq S^4$ transform the set $\{\pm P_1,\pm P_2,\pm P_3\}$ to itself and $\{\pm P_4,\pm P_5,\pm P_6\}$ to itself. – ChoF Apr 22 '18 at 23:27 | 2,119 | 5,133 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2021-17 | latest | en | 0.640615 |
https://www.itbaoku.cn/post/978363.html | 1,695,995,705,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510516.56/warc/CC-MAIN-20230929122500-20230929152500-00078.warc.gz | 893,351,964 | 14,678 | # 合并list1中的两个数组列表，同时保持排序。[英] merge two arraylist lists in list1 while it remain sorted
### 问题描述
``` public static void merge (ArrayList<Integer> list1, ArrayList<Integer> list2)
{
int i;
int n=list1.size();
int pos , j=0;
for (pos =0 ;pos<n ; pos++)
{
for ( i=0 ; i<n ; i++)
if (list1.get(j)>list2.get(pos))
else
j++;
}
}
```
## 推荐答案
```public static void merge(List<Integer> l1, List<Integer> l2) {
for (int index1 = 0, index2 = 0; index2 < l2.size(); index1++) {
if (index1 == l1.size() || l1.get(index1) > l2.get(index2)) {
}
}
}
```
```public static void merge(List<Integer> l1, List<Integer> l2) {
for (int index2 = 0; index2 < l2.size(); index2++) {
for (int index1 = 0; ; index1++) {
if (index1 == l1.size() || l1.get(index1) > l2.get(index2)) {
break;
}
}
}
}
```
## 其他推荐答案
```list1.addAll(list2);
Collections.sort(list1);
```
## 其他推荐答案
```public static void merge (ArrayList<Integer> list1, ArrayList<Integer> list2)
{
Collections.sort(list1);
}
```
### 问题描述
In my assignment the third step is to Call the method merge to merge the two lists in list1 so that the list1 remains sorted.
I write my code but it doesn't work well , the output show wrong because it important to be sorted
``` public static void merge (ArrayList<Integer> list1, ArrayList<Integer> list2)
{
int i;
int n=list1.size();
int pos , j=0;
for (pos =0 ;pos<n ; pos++)
{
for ( i=0 ; i<n ; i++)
if (list1.get(j)>list2.get(pos))
else
j++;
}
}
```
## 推荐答案
You only need one for loop assuming both lists are sorted:
```public static void merge(List<Integer> l1, List<Integer> l2) {
for (int index1 = 0, index2 = 0; index2 < l2.size(); index1++) {
if (index1 == l1.size() || l1.get(index1) > l2.get(index2)) {
}
}
}
```
If l2 isn't sorted, you need two loops:
```public static void merge(List<Integer> l1, List<Integer> l2) {
for (int index2 = 0; index2 < l2.size(); index2++) {
for (int index1 = 0; ; index1++) {
if (index1 == l1.size() || l1.get(index1) > l2.get(index2)) {
break;
}
}
}
}
```
## 其他推荐答案
Easy fix: sort afterwards.
```list1.addAll(list2);
Collections.sort(list1);
```
Use sets to avoid duplicates.
## 其他推荐答案
```public static void merge (ArrayList<Integer> list1, ArrayList<Integer> list2)
{ | 700 | 2,206 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2023-40 | latest | en | 0.243222 |
https://rdrr.io/cran/SyScSelection/f/vignettes/Usage.Rmd | 1,632,852,335,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780060877.21/warc/CC-MAIN-20210928153533-20210928183533-00094.warc.gz | 524,563,695 | 6,870 | Usage In SyScSelection: Systematic Scenario Selection for Stress Testing
knitr::opts_chunk\$set(
collapse = TRUE,
comment = "#>"
)
library(SyScSelection)
Example ellipsodial mesh for a normal distribution:
• Estimate the mean and covariance matrix from the data:
mu <- colMeans(data)
sig <- cov(data)
• The number of dimensions, d, is taken directly from the data:
d <- length(data[1,])
• Get the size parameter for a normal dist’n at a 95% threshold:
calpha <- sizeparam_normal_distn(.95, d)
• Create a hyperellipsoid object. Note that the constructor takes the inverse of the disperion matrix:
hellip <- hyperellipsoid(mu, solve(sig), calpha)
• Scenarios are calculated as a mesh of fineness 3. The number of scenarios is a function of the dimensionality of the hyperellipsoid and the fineness of the mesh:
scenarios <- hypercube_mesh(3, hellip)
Example ellipsodial mesh for a t distribution:
• Estimate the mean, covariance, and degrees of freedom from the data:
mu <- colMeans(data)
sig <- cov(data)
nu <- dim(data)[1] - 1
• The number of dimensions, d, is taken directly from the data:
d <- length(data[1,])
• Get the size parameter for a normal dist’n at a 95% threshold:
calpha <- sizeparam_t_distn(.95, d, nu)
• Create a hyperellipsoid object. Note that the constructor takes the inverse of the disperion matrix:
hellip <- hyperellipsoid(mu, solve(sig), calpha)
• Scenarios are calculated as a mesh of fineness 3. The number of scenarios is a function of the dimensionality of the hyperellipsoid and the fineness of the mesh:
scenarios <- hypercube_mesh(3, hellip)
Try the SyScSelection package in your browser
Any scripts or data that you put into this service are public.
SyScSelection documentation built on Oct. 26, 2020, 5:08 p.m. | 442 | 1,759 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2021-39 | latest | en | 0.644837 |
https://www.gov.uk/hmrc-internal-manuals/tonnage-tax-manual/ttm13020 | 1,544,939,905,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376827281.64/warc/CC-MAIN-20181216051636-20181216073636-00080.warc.gz | 908,537,293 | 9,276 | This part of GOV.UK is being rebuilt – find out what beta means
HMRC internal manual
Partnerships: Outline
Computation of profit
Tonnage tax partner
For the purpose of calculating the profits of a tonnage tax company, the profits of a partnership are calculated as if the partnership were a tonnage tax company.
The profits calculated in this way are then apportioned to the company (and to any other tonnage tax partner) in the normal way, so that:
• its share of any partnership tonnage tax profits is treated as (part of) its own tonnage tax profits,
• an appropriate proportion of the partnership’s ‘relevant shipping profits’ (see TTM06010) will be franked by those tonnage tax profits, and
• its share of any other profits or losses is treated as its own profits or losses arising outside the ring-fence
If there is more than one tonnage tax company in the partnership then each such company calculates its partnership profits on the basis of a single computation worked out using the tonnage tax rules.
If other corporate partners become tonnage tax companies at a later date, or if new tonnage tax companies join the partnership, those companies will simply change from computing their share of the profits computed on normal lines to computing their share of the profits computed on tonnage tax lines.
Non-tonnage tax partners
If the partnership includes partners other than tonnage tax companies, then a separate computation of the whole partnership profit is required, computed on normal lines. The non-tonnage tax partners’ shares of profit are a share of that separate computation.
Multiple computations
In theory, there could be several types of partner in one corporate partnership. A separate computation will be required for each type of partner, for example
• Tonnage tax company – computes its share as if the partnership were a tonnage tax company
• Non-tonnage tax company – computes its share using normal rules for computation of total partnership profits (under CTA09/S1259)
• Individual – computes his share using normal rules for computation of total partnership profits (under CTA09/S1258)
Non-resident partners
A separate computation will also be required to calculate the profits of a non- resident partner.
The normal rule is that a non-resident should compute his share of the partnership profits using the normal rules for computation of total partnership profits, but only to the extent that those profits are derived from the carrying on of a trade or business in the United Kingdom (see DT1750).
However, the normal rule may be amended by the terms of the Double Taxation Agreement (if any) with the non-resident’s country of residence. In many cases, the specific DTA may direct that profits derived from the operation of ships in international traffic will fall to be excluded from the computation of the non- resident’s share of the partnership profits.
References
FA00/SCH22/PARA131 (calculation of partnership profit) TTM17731 SI00/2303/REG10 (rules for calculating tonnage tax profits) TTM18010 | 630 | 3,056 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2018-51 | latest | en | 0.955618 |
https://rpg.stackexchange.com/questions/70291/what-circumstances-make-heighten-turning-worthwhile | 1,701,618,621,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100508.23/warc/CC-MAIN-20231203125921-20231203155921-00265.warc.gz | 551,308,090 | 40,077 | # What circumstances make Heighten Turning worthwhile?
The feat Heighten Turning has the following as part of its benefit:
When you turn or rebuke undead, you may choose a number no higher than your cleric level. Add that number to your turning check, while subtracting it from your turning damage roll. (Libris Mortis 27)
So we're all on the same page, here's a summary of the special attack turn undead:
• When a cleric makes a turn undead attempt, the cleric first determines the maximum Hit Dice of the largest single undead that can be affected by making a turning check (1d20 + Charisma modifier), the result found on a chart that yields a Hit Dice outcome of between the cleric's level −4 and the cleric's level +4.
• Then the cleric determines how many total Hit Dice of undead are actually affected by making a turning damage roll (2d6 + cleric level + Charisma modifier). There is no upper limit to this result.
The feat Heighten Turning lets the cleric increase the turning check by +1 but decrease the turning damage roll by −1. The cleric can do this to a value up to the cleric's level. Reversing the polarity would've made the feat's use obvious in the face of a horde of low-powered undead creatures, but the way it's written instead makes the feat just weird: The cleric tries to turn a lone powerful undead creature while simultaneously limiting the size of the creature and his ability to affect it.
What are the optimal circumstances in which to use the feat Heighten Turning? That is, what numbers have to be present on both the cleric's side and the enemies' side for the feat Heighten Turning to be a good idea? Is the feat one that looked good on paper but is useless in actual play? Is there ever a certain point at which, or a certain build wherein, it's the ideal feat?
• I've included the tag statistics because I think this might ultimately be a gaming math question (with a lot of variables). However, answers avoiding statistics are fine. Oct 25, 2015 at 12:10
• I just noted that there is somehow no mention in the feat description of when you decide to use it or how much to add. It could thus be useful in ensuring that you reach high enough. Otherwise it might also be useful for a cleric with low charisma targeting a single undead. Oct 25, 2015 at 13:17
• @MatthieuM. I don't think that rules quirk does the cleric much good as, because turning takes place in steps, the bonus is always applied before the penalty. So the cleric could, I guess, impose a penalty on his turning damage roll without gaining the bonus on the turning check, but I don't know of a situation in which that's useful. Of course, I don't know of a situation in which the feat is useful, so there's that. (An interesting idea, though.) Oct 25, 2015 at 13:21
• @HeyICanChan It is useful to purposefully ruin your turning damage in some situations, e.g. to precisely turn all the enemy zombies near you but not the allied zombies near the enemy. Such situations, where excess damage is bad, are very rare, however. Oct 26, 2015 at 3:52
– Ruut
Oct 26, 2015 at 6:12
What are the optimal circumstances in which to use the feat Heighten Turning?
The optimal circumstances would be obviously a situation, where you reap only the benefits. But is this possible? It is.
The turning check as D20 has a 15% probability to end up with a value of three or less, the damage roll is guaranteed to be 3 though, assuming you are always at least level 1.
Level 2 Cleric versus 4 enemies with 1 hit die
A level 2 cleric is guaranteed to turn 4 hit dice, but has a chance of 30% to fail completely, by rolling 1-6 on the D20. This chance to fail is constant.
Chance to fail completely: 30%
Chance to turn all enemies: 70%
(100% - 30% chance to fail completely)
The group has a 30% chance to end up with 4 enemies. The group either ends up with no enemies or all of them.
Let's now use the feature to add the +1/-1 modifier to the rolls:
Chance to fail completely: 25%
Chance to turn all enemies: 73%
Okay, we improved the maximum hit dice odds by the 5%, we only need to not roll 1-5 now. But how can we get a better result by adding a negative modifier? Well, you don't, you still decrease your odds, but in sum and comparison to before, it doesn't matter.
There is a chance of 1/36 to roll 1 + 1 as damage roll, which is 3% rounded up. Only this way you end up with less than 4 total hit dice. So, there is a 97% chance to roll any other combination that saves your day. There is also a 75% chance to not fail the maximum hit dice check. The probability of not failing both checks (97% * 75%) is 73%.
Whenever the enemies total hit dice are near your guaranteed damage roll, you can reap most benefits.
Otherwise, the feature gives you options. 4 enemies together are stronger than the 4 individuals. It's usually more beneficial for the group to have a guaranteed reduction of enemies instead of sometimes many and sometimes few enemies.
Level 2 Cleric versus 1 enemy with 6 hit dice
This is a simple one. Your probability to make the maximum hit dice check is zero. Which is a really bad probability. But you only need to roll a 4 or more on 2d6 for the damage roll, which is quite a good chance - 91%! But it's a waste. What's the point of a nearly guaranteed damage roll if there is a guaranteed turning check failure?
A +2/-2 modifier will give you at least a chance of 5% to make the maximum hit dice check. And the chance to at least roll 6 as damage roll is 72%, which isn't that bad. Your overall chance (5% * 72%) is still crappy 3%. But 3% is more than 0%.
• Interesting. So, in a sense, like the similar feat Power Attack, the feat's best used when you already know you're going to win or when you've no chance of winning without it. Oct 25, 2015 at 16:24
Speaking simply, on the high end or the low end.
When you desperately want to turn something with more HD than you have cleric levels, (aka turning a single high-CR undead), you might not need that much 'turning damage' aka 'amount of hd turned' but want a higher chance to actually affect it at all. The only real way to calculate this is through metagaming, or a GM who treats knowledge checks as 'you know the monster's stats'.
On the low end, when you just don't want to roll like, a 2, and not be able to turn the low-HD ghouls despite being a higher cleric level than they have HD, you can just take away a few 'turning damage' points to push that from 'needing to roll a 4 or better' to 'needing to roll a 1'. Aka, making the initial turning check a shoe-in.
Is the feat any good?
Nope. Get the ones which give +effective turning levels. Get the one that activates the relic item that does the same. Counting as a higher level cleric makes the turning check relatively pointless, with the amount of etls you can get for the cost of a feat. And if you're caring about turning, be a Sun Cleric and just straight up burn undead into actual cinders with Greater Turning.
Is there a certain build where it's good
Sorta. The maxed out charisma turning cleric build. Where you already have all the stuff to make your turning as murderous as possible, in an undead-heavy campaign. The high charisma means you can spend points of 'turning damage' without making it useless, so the 'high end' undead can be affected by your 1d20+charisma+heighten turning bonus'd check. Since 2d6 is unlikely to roll less than four, you're effectively going to be making the turning check 1d20+charismax2, with when you're really going for it, 1d20+charismax2+3 (average roll on 2d6 being 7), and with a charisma of 5 and average of 10.5, pushing the turning check into 'very unlikely to get below level+3' territory. | 1,860 | 7,688 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2023-50 | longest | en | 0.934052 |
https://en.wikipedia.org/wiki/Omega-categorical_theory | 1,493,230,134,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917121528.59/warc/CC-MAIN-20170423031201-00527-ip-10-145-167-34.ec2.internal.warc.gz | 809,616,526 | 11,062 | # Omega-categorical theory
In mathematical logic, an omega-categorical theory is a theory that has exactly one countably infinite model up to isomorphism. Omega-categoricity is the special case κ = ${\displaystyle \aleph _{0}}$ = ω of κ-categoricity, and omega-categorical theories are also referred to as ω-categorical. The notion is most important for countable first-order theories.
## Equivalent conditions for omega-categoricity
Many conditions on a theory are equivalent to the property of omega-categoricity. In 1959 Erwin Engeler, Czesław Ryll-Nardzewski and Lars Svenonius, proved several independently.[1] Despite this, the literature still widely refers to the Ryll-Nardzewski theorem as a name for these conditions. The conditions included with the theorem vary between authors.[2][3]
Given a countable complete first-order theory T with infinite models, the following are equivalent:
• The theory T is omega-categorical.
• Every countable model of T has an oligomorphic automorphism group.
• Some countable model of T has an oligomorphic automorphism group.[4]
• The theory T has a model which, for every natural number n, realizes only finitely many n-types, that is, the Stone space Sn(T) is finite.
• For every natural number n, T has only finitely many n-types.
• For every natural number n, every n-type is isolated.
• For every natural number n, up to equivalence modulo T there are only finitely many formulas with n free variables, in other words, for every n, the nth Lindenbaum-Tarski algebra of T is finite.
• Every model of T is atomic.
• Every countable model of T is atomic.
• The theory T has a countable atomic and saturated model.
• The theory T has a saturated prime model.
## Notes
1. ^ Rami Grossberg, José Iovino and Olivier Lessmann, A primer of simple theories
2. ^ Hodges, Model Theory, p. 341.
3. ^ Rothmaler, p. 200.
4. ^ Cameron (1990) p.30 | 461 | 1,887 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 1, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2017-17 | latest | en | 0.891475 |
http://stackoverflow.com/questions/12845141/how-to-complete-a-countdown-challenge-in-c-programming?answertab=active | 1,395,058,825,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1394678705235/warc/CC-MAIN-20140313024505-00055-ip-10-183-142-35.ec2.internal.warc.gz | 130,744,601 | 16,717 | how to complete a countdown challenge in c programming
Could somebody help me with solving this. I'm almost there I just need a little push. I'm supposed to have the output count down from 99 and count down by 3(I solved that correctly), but I also need the output to print "found one" if it comes across a number divisible by 5! I need help with that. heres what I have so far. Note that the code doesn't print "found one" even when i have the right function in that solves dividends.
``````#include <objc/objc.h>
#include <stdio.h>
int main(int argc, const char * argv[])
{
int i = 99;
while (i > -1) {
printf("%d\n", i);
i = i-3;
}
{
if (i%5==0) printf("found one\n");
}
return 0;
}
``````
-
You should probably learn the basic syntax of the language first, and then ask questions later, instead of typing in jibberish and then posting it to SO. – Paul Tomblin Oct 11 '12 at 18:30
look, i know the basic syntax of the language...btw im sorry for shouting(i didn't realize my CAPS key was turned on – user1706978 Oct 11 '12 at 20:06
@PaulTomblin: That's hardly jibberish, but rather a valid novice question. Unfortunately (imho) the homework tag is gone. When we had that tag, one could just not read questions with that tag if one did not want to spend time helping novices. – Eric J. Oct 11 '12 at 21:09
@EricJ, the question as originally posted contained something that was so far distant from any valid syntax for any known computer language as to make you question whether the person asking the question had even read chapter one of his text book. Thus, "jibber ish". – Paul Tomblin Oct 11 '12 at 21:28
well i guess you dont know c then – user1706978 Oct 11 '12 at 23:17
show 1 more comment
3 Answers
Use `if(i%5 == 0)` instead. This checks to see if it is divided by 5 – that there are no remainders.
-
thank you so much – user1706978 Oct 11 '12 at 20:02
add comment
I'll give you a hint. The modulus (%) operator returns the remainder after integer division. So 11 % 10 = 1, and 15 % 5 = 0. Given your code, you should be able to go from there!
-
thank you so much – user1706978 Oct 11 '12 at 20:02
add comment
You are looking for the modulo operator
``````if (i % 5 == 0) // The number is divisible by 5
``````
The modulo operator calculates the remainder after division. If the remainder is zero, then the integer number was exactly divisible by the modulo operand, in this case 5.
-
thank you so much – user1706978 Oct 11 '12 at 20:04
thanks for reminding me...since you seem like a person who doesn't taunt those who are relatively new to the language, ill ask you one more thing. I inserted that into my code, but now the code doesn't say "found one" when it has found a dividend of 5. heres my code now. – user1706978 Oct 11 '12 at 20:14
Your new code will only ever test the final value if i because you don't have `if (i%5==0) printf("found one\n");` inside your while loop. – Eric J. Oct 11 '12 at 21:07
Also... don't forget to upvote helpful answers and accept the most helpful/most correct one. – Eric J. Oct 11 '12 at 21:09
u are a lifesaver!!! thanks – user1706978 Oct 11 '12 at 21:27
add comment | 889 | 3,142 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2014-10 | latest | en | 0.930923 |
https://ewtaf.com/rules-and-guidelines/impulse/ | 1,660,274,825,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571538.36/warc/CC-MAIN-20220812014923-20220812044923-00597.warc.gz | 246,960,789 | 21,492 | # Impulse
Rules
• An impulse always subdivides into five waves.
• Wave 1 always subdivides into an impulse or (rarely) a diagonal triangle.
• Wave 3 always subdivides into an impulse.
• Wave 5 always subdivides into an impulse or a diagonal triangle.
• Wave 2 always subdivides into a zigzag, flat or combination.
• Wave 4 always subdivides into a zigzag, flat, triangle or combination.
• Wave 2 never moves beyond the start of wave 1.
• Wave 3 always moves beyond the end of wave 1.
• Wave 3 is never the shortest wave.
• Wave 4 never moves beyond the end of wave 1.
• Never are waves 1, 3 and 5 all extended.
Guidelines
• Wave 4 will almost always be a different corrective pattern than wave 2.
• Wave 2 is usually a zigzag or zigzag combination.
• Wave 4 is usually a flat, triangle or flat combination.
• Sometimes wave 5 does not move beyond the end of wave 3 (in which case it is called a truncation).
• Wave 5 often ends when meeting or slightly exceeding a line drawn from the end of wave 3 that is parallel to the line connecting the ends of waves 2 and 4, on either arithmetic or semilog scale.
• The center of wave 3 almost always has the steepest slope of any equal period within the parent impulse except that sometimes an early portion of wave 1 (the “kickoff”) will be steeper.
• Wave 1, 3 or 5 is usually extended. (An extension appears “stretched” because its corrective waves are small compared to its impulse waves. It is substantially longer, and contains larger subdivisions, than the non-extended waves).
• Often, the extended subwave is the same number (1, 3 or 5) as the parent wave.
• Rarely do two subwaves extend, although it is typical for waves 3 and 5 both to extend when they are of Cycle or Supercycle degree and within a fifth wave of one degree higher.
• Wave 1 is the least commonly extended wave.
• When wave 3 is extended, waves 1 and 5 tend to have gains related by equality or the Fibonacci ratio.
• When wave 5 is extended, it is often in Fibonacci proportion to the net travel of waves 1 through 3.
• When wave 1 is extended, it is often in Fibonacci proportion to the net travel of waves 3 thorough 5.
• Wave 4 typically ends when it is within the price range of subwave four of 3.
• Wave 4 often subdivides the entire impulse into Fibonacci proportion in time and/or price.
Russian
• ## Chat
Only for members | 585 | 2,356 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2022-33 | longest | en | 0.896922 |
https://www.scribd.com/document/348470113/9709-w16-ms-33 | 1,552,927,248,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912201455.20/warc/CC-MAIN-20190318152343-20190318174343-00378.warc.gz | 893,327,585 | 63,377 | You are on page 1of 6
# Cambridge International Examinations
## Cambridge International Advanced Level
MATHEMATICS 9709/33
Paper 3 October/November 2016
MARK SCHEME
Maximum Mark: 75
Published
This mark scheme is published as an aid to teachers and candidates, to indicate the requirements of the
examination. It shows the basis on which Examiners were instructed to award marks. It does not indicate the
details of the discussions that took place at an Examiners meeting before marking began, which would have
considered the acceptability of alternative answers.
Mark schemes should be read in conjunction with the question paper and the Principal Examiner Report for
Teachers.
Cambridge will not enter into discussions about these mark schemes.
Cambridge is publishing the mark schemes for the October/November 2016 series for most
Cambridge IGCSE, Cambridge International A and AS Level components and some Cambridge O Level
components.
## UCLES 2016 [Turn over
Page 2 Mark Scheme Syllabus Paper
Cambridge International A Level October/November 2016 9709 33
## Marks are of the following three types:
M Method mark, awarded for a valid method applied to the problem. Method marks are not lost
for numerical errors, algebraic slips or errors in units. However, it is not usually sufficient for a
candidate just to indicate an intention of using some method or just to quote a formula; the
formula or idea must be applied to the specific problem in hand, e.g. by substituting the
relevant quantities into the formula. Correct application of a formula without the formula being
quoted obviously earns the M mark and in some cases an M mark can be implied from a
A Accuracy mark, awarded for a correct answer or intermediate step correctly obtained.
Accuracy marks cannot be given unless the associated method mark is earned (or implied).
## B Mark for a correct result or statement independent of method marks.
When a part of a question has two or more method steps, the M marks are generally
independent unless the scheme specifically says otherwise; and similarly when there are several
B marks allocated. The notation DM or DB (or dep*) is used to indicate that a particular M or B
mark is dependent on an earlier M or B (asterisked) mark in the scheme. When two or more
steps are run together by the candidate, the earlier marks are implied and full credit is given.
The symbol implies that the A or B mark indicated is allowed for work correctly following on
from previously incorrect results. Otherwise, A or B marks are given for correct work only. A and
B marks are not given for fortuitously correct answers or results obtained from incorrect
working.
## Note: B2 or A2 means that the candidate can earn 2 or 0.
B2/1/0 means that the candidate can earn anything from 0 to 2.
The marks indicated in the scheme may not be subdivided. If there is genuine doubt whether a
candidate has earned a mark, allow the candidate the benefit of the doubt. Unless otherwise
indicated, marks once gained cannot subsequently be lost, e.g. wrong working following a correct
form of answer is ignored.
Wrong or missing units in an answer should not lead to the loss of a mark unless the scheme
specifically indicates otherwise.
For a numerical answer, allow the A or B mark if a value is obtained which is correct to 3 s.f., or
which would be correct to 3 s.f. if rounded (1 d.p. in the case of an angle). As stated above, an A
or B mark is not given if a correct numerical answer arises fortuitously from incorrect working. For
Mechanics questions, allow A or B marks for correct answers which arise from taking g equal to
9.8 or 9.81 instead of 10.
UCLES 2016
Page 3 Mark Scheme Syllabus Paper
Cambridge International A Level October/November 2016 9709 33
The following abbreviations may be used in a mark scheme or used on the scripts:
## AEF/OE Any Equivalent Form (of answer is equally acceptable) / Or Equivalent
AG Answer Given on the question paper (so extra checking is needed to ensure that the
detailed working leading to the result is valid)
CAO Correct Answer Only (emphasising that no follow through from a previous error is
allowed)
## SOI Seen or implied
SR Special Ruling (detailing the mark to be given for a specific wrong solution, or a case
where some standard marking practice is to be varied in the light of a particular
circumstance)
Penalties
## MR 1 A penalty of MR 1 is deducted from A or B marks when the data of a question or part
question are genuinely misread and the object and difficulty of the question remain
unaltered. In this case all A and B marks then become follow through marks. MR is
not applied when the candidate misreads his own figures this is regarded as an error in
accuracy. An MR 2 penalty may be applied in particular cases if agreed at the
coordination meeting.
## PA 1 This is deducted from A or B marks in the case of premature approximation. The PA 1
penalty is usually discussed at the meeting.
UCLES 2016
Page 4 Mark Scheme Syllabus Paper
Cambridge International A Level October/November 2016 9709 33
## 1 Use law of the logarithm of a quotient M1
y+2
Remove logarithms and obtain a correct equation, e.g. e z = A1
y +1
2 ez
Obtain answer y = , or equivalent A1 [3]
ez 1
## 2 Use correct quotient or product rule M1
Obtain correct derivative in any form A1
1
Use Pythagoras to simplify the derivative to , or equivalent A1
1 + cos x
Justify the given statement, 1 < cosx < 1 statement, or equivalent A1 [4]
## 3 Use the tan 2A formula to obtain an equation in tan only M1
Obtain a correct horizontal equation A1
Rearrange equation as a quadratic in tan , e.g. 3tan 2 + 2tan 1 = 0 A1
Solve for (usual requirements for solution of quadratic) M1
Obtain answer, e.g. 18.4 A1
Obtain second answer, e.g. 135 , and no others in the given interval A1 [6]
## 4 (i) Commence division by x 2 x + 2 and reach a partial quotient 4x2 + kx M1
Obtain quotient 4 x + 4 x + a 4 or 4 x + 4 x + b / 2
2 2
A1
Equate x or constant term to zero and solve for a or b M1
Obtain a = 1 A1
Obtain b = 6 A1 [5]
## (ii) Show that x 2 x + 2 = 0 has no real roots B1
Obtain roots 12 and 32 from 4 x 2 + 4 x 3 = 0 B1 [2]
dy 1
5 (i) State equation = xy B1 [1]
dx 2
(ii) Separate variables correctly and attempts to integrate one side of equation M1
Obtain terms of the form a ln y and bx 2 A1
Use x = 0 and y = 2 to evaluate a constant, or as limits, in expression containing
a ln y or bx2 M1
Obtain correct solution in any form, e.g. ln y = 14 x 2 + ln 2 A1
1 x2
Obtain correct expression for y, e.g. y = 2e 4 A1 [5]
(iii) Show correct sketch for x 0. Needs through (0, 2) and rapidly increasing
positive gradient. B1 [1]
UCLES 2016
Page 5 Mark Scheme Syllabus Paper
Cambridge International A Level October/November 2016 9709 33
1
6 (i) State or imply du = dx B1
2 x
Substitute for x and dx throughout M1
Justify the change in limits and obtain the given answer A1 [3]
B
(ii) Convert integrand into the form A + M1*
u +1
Obtain integrand A =1, B = 2 A1
Integrate and obtain u 2ln(u + 1) A1 + A1
Substitute limits correctly in an integral containing terms au and bln(u + 1),
where ab 0 DM1
Obtain the given answer following full and correct working A1 [6]
[The f.t. is on A and B.]
## 7 (i) State modulus 2 2 , or equivalent B1
State argument 13 (or 60) B1 [2]
## (b) EITHER: Substitute for z and multiply numerator and denominator by
conjugate of iz M1
Simplify the numerator to 4 3 + 4 i or the denominator to 8 A1
Obtain final answer 12 3 + 12 i A1
OR: Substitute for z, obtain two equations in x and y and solve for x or
for y M1
Obtain x = 12 3 or y = 12 A1
Obtain final answer 1
2
3+ 1
2
i A1 [4]
## (iii) Show points A and B in relatively correct positions B1
Carry out a complete method for finding angle AOB, e.g. calculate the
z
argument of M1
iz
Obtain the given answer A1 [3]
A Bx + C
8 (i) State or imply the form + 2 B1
x+2 x +4
Use a correct method to determine a constant M1
Obtain one of A = 2, B = 1, C = 1 A1
Obtain a second value A1
Obtain a third value A1 [5]
UCLES 2016
Page 6 Mark Scheme Syllabus Paper
Cambridge International A Level October/November 2016 9709 33
(ii) Use correct method to find the first two terms of the expansion of ( x + 2) 1 ,
(1 + 12 x ) 1 , (4 + x 2 ) 1 or (1 + 14 x 2 ) 1 M1
Obtain correct unsimplified expansions up to the term in x 2 of each partial
fraction A1 + A1
Multiply out fully by Bx + C, where BC 0 M1
Obtain final answer 34 14 x + 165 x 2 , or equivalent A1 [5]
1
[Symbolic binomial coefficients, e.g. are not sufficient for the M1. The f.t.
1
is on A, B, C.]
[In the case of an attempt to expand (3 x 2 + x + 6)( x + 2) 1 ( x 2 + 4) 1 , give
M1A1A1 for the expansions, M1 for multiplying out fully, and A1 for the final
## 9 (i) Differentiate both equations and equate derivatives M1*
k
Obtain equation cos a a sin a = 2 A1 + A1
a
k
State a cos a = and eliminate k DM1
a
Obtain the given answer showing sufficient working A1 [5]
(ii) Show clearly correct use of the iterative formula at least once M1
Obtain answer 1.077 A1
Show sufficient iterations to 5 d.p. to justify 1.077 to 3 d.p., or show there is a
sign change in the interval (1.0765, 1.0775) A1 [3]
## (iii) Use a correct method to determine k M1
Obtain answer k = 0.55 A1 [2]
## 10 (i) Express general point of l in component form e.g. (1 + 2 , 2 , 1 + ) B1
Using the correct process for the modulus form an equation in M1*
Reduce the equation to a quadratic, e.g. 6 2 + 2 4 = 0 A1
Solve for (usual requirements for solution of a quadratic) DM1
Obtain final answers i + 3j and 73 i + 43 j + 53 k A1 [5]
(ii) Using the correct process, find the scalar product of a direction vector for l and a
normal for p M1
Using the correct process for the moduli, divide the scalar product by the product
of the moduli and equate the result to 23 M1
2a 1 + 1 2
State a correct equation in any form, e.g. = A1
(a 2 + 1 + 1). (22 + (1) 2 + 1) 3
Solve for a2 M1
Obtain answer a = 2 A1 [5]
UCLES 2016 | 2,846 | 9,988 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2019-13 | latest | en | 0.914327 |
https://link.springer.com/article/10.1007/s10951-021-00719-1 | 1,696,039,232,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510575.93/warc/CC-MAIN-20230930014147-20230930044147-00456.warc.gz | 395,573,540 | 92,109 | ## 1 Introduction
Creating and optimizing timetables is substantial for planning and operating public transportation networks. Both in local traffic and in long-distance train networks, timetables are often periodic, i.e., the schedule of trips repeats after a certain time period T, e.g., 60 min.
Mathematically, periodic timetabling is captured by the periodic event scheduling problem (PESP, Serafini & Ukovich, 1989). The idea behind PESP is to model arrival and departure events of trips in a public transportation network as vertices (events) of a directed graph. Dependencies between events, such as driving of a vehicle or changing at a station, are modeled as arcs (activities) connecting pairs of events. These activities come with restrictions on their duration, e.g., driving from one station to the next might take at least 7 min. Then, a solution to PESP is an assignment of times in [0, T) to each event (a periodic timetable) such that the activity duration restrictions are respected. We refer to Sect. 2 for rigorous formulations.
Deciding whether a periodic timetable exists is an NP-complete problem, even if $$T \ge 3$$ is not considered as part of the input. This result can be proved by a polynomial-time reduction in T -Vertex Coloring, where a T-coloring of a graph corresponds to event times in $$\{0, 1, \dots , T-1\}$$ (Odijk, 1994). This suggests a close relationship between PESP and coloring problems. In fact, both PESP and Vertex Coloring are solvable in linear time on trees, regardless of T. Furthermore, for any T, deciding if a graph admits a T-coloring is fixed-parameter tractable when parameterized by treewidth. More precisely, there is a function f such that for a given graph G on n vertices with a nice tree decomposition of treewidth $$\le k$$ and a natural number T, there is an $$O(f(k, T) \cdot n)$$ algorithm deciding the T-Vertex Coloring problem on G (Arnborg & Proskurowski, 1989).
This motivates the investigation of parameterized complexity of PESP. In fact, the treewidth turns out to be rather small on the instances of the benchmarking library PESPlib (Goerigk, 2012). This library is a collection of 20 periodic timetabling problems, none of which could be solved to optimality in the past years. In Sect. 3, we recall some standard notions from parameterized complexity, list values or give bounds for several typical graph parameters for the smallest PESPlib instance R1L1, and give an overview on the complexity landscape for these parameters. The corresponding hardness results, algorithms, and the computation of the parameters make up the content of the following three sections of the paper.
We show in Sect. 4 that, in contrast to Vertex Coloring, PESP is not fixed-parameter tractable when parameterized by treewidth unless P $$=$$ NP. Even if the event-activity network has treewidth 2 or branchwidth 2, i.e., every connected component is series–parallel, it is an NP-complete problem to decide whether a feasible periodic timetable exists. When considering reduced PESP instances, where the activities carry only lower bounds, but no (non-trivial) upper bounds, we prove that it is NP-complete to decide if there exists a periodic timetable whose weighted periodic slack is below a given threshold. Both proofs work by a reduction in the Subset Sum problem. As a by-product, we also obtain an NP-hardness result on networks of carvingwidth 3. As a consequence, if P $$\ne$$ NP, then there are only pseudo-polynomial-time algorithms available for both the feasibility and the reduced optimality problem. Moreover, the existence of fixed-parameter tractable algorithms allowing a unary encoding of the period time T is shown to be unlikely, as we show that deciding the feasibility of a PESP instance is W[1]-hard w.r.t. the vertex cover number and hence for the treewidth.
We construct a dynamic program that solves PESP based on branch decompositions in Sect. 5. This provides a pseudo-polynomial algorithm for networks of bounded branchwidth or treewidth. A straightforward tree decomposition analogue is presented in the Appendix. We also prove that the feasibility version of PESP is fixed-parameter tractable when parameterized by the cyclomatic number, i.e., the dimension of the cycle space of the event-activity network. For computing periodic timetables with minimum weighted periodic slack, we give a polynomial-time algorithm when the cyclomatic number is bounded.
As PESP instances arising from public transportation networks typically have a special structure, we show in Sect. 6 that on this type of event-activity networks, the branchwidth can be related to invariants of an underlying line network: Roughly speaking, the number of lines at a station of a public transport network is a (sharp) lower bound on the branchwidth of the event-activity network. The carvingwidth of the often planar line network provides an upper bound, which is also sharp. Finally, we exploit the relation to line networks in order to compute bounds on the parameters discussed in this paper for the smallest PESPlib instance R1L1.
The presentation finishes with a few concluding remarks in Sect. 7.
## 2 The periodic event scheduling problem
The periodic event scheduling problem (PESP) was introduced in Serafini and Ukovich (1989). PESP instances comprise the following ingredients:
• a directed graph G (often called event-activity network) with vertex set V(G) (events) and arc set A(G) (activities),
• a period time $$T \in {\mathbb {N}}$$,
• lower bounds $$\ell \in {\mathbb {Z}}_{\ge 0}^{A(G)}$$ with $$\ell _a < T$$ for all $$a \in A(G)$$,
• upper bounds $$u \in {\mathbb {Z}}_{\ge 0}^{A(G)}$$ with $$u_a \ge \ell _a$$ for all $$a \in A(G)$$,
• weights $$w \in {\mathbb {Q}}_{\ge 0}^{A(G)}$$.
In the application of periodic timetabling in public transport, the events are typically arrivals or departures of a vehicle at a station, and activities model driving between stations, dwelling at a station, passenger transfers, or safety constraint such as minimum distances between vehicles (Liebchen & Möhring, 2007). The weights often reflect the number of passengers using an activity.
### Definition 2.1
Given $$(G, T, \ell , u)$$ as above, a periodic timetable is a vector $$\pi \in [0, T)^{V(G)}$$ such that there exists a periodic tension $$x \in {\mathbb {R}}_{\ge 0}^{A(G)}$$ satisfying
\begin{aligned} \forall ij \in A(G): \ell _{ij} \le x_{ij} \le u_{ij} \text { and } \pi _j - \pi _i \equiv x_{ij} \bmod T.\nonumber \\ \end{aligned}
(1)
Intuitively, $$\pi$$ gives the cyclic order of the events, and x corresponds to the duration of the activities. If there exists a periodic timetable $$\pi$$, then a periodic tension can be computed by
\begin{aligned} x_{ij} := [\pi _j - \pi _i - \ell _{ij}]_T + \ell _{ij}, ij \in A(G), \end{aligned}
(2)
where $$[\cdot ]_T$$ denotes the modulo-T-operator taking values in [0, T). Conversely, from a vector $$x \in {\mathbb {R}}_{\ge 0}^{A(G)}$$ with $$\ell \le x \le u$$, one can construct a periodic timetable with tension x by a graph traversal; see also Lemma 2.7.
In terms of the incidence matrix B of G and its transpose $$B^t$$, condition (1) can be rewritten as
\begin{aligned} \ell \le x \le u \quad \text { and } \quad B^t \pi \equiv x \mod T. \end{aligned}
Since B and hence are $$B^t$$ are totally unimodular (Schrijver, 1986, Example 19.2) and the bounds $$\ell , u$$ are integer, it follows that if a periodic timetable exists, then there is also an integer timetable with an integer periodic tension. However, in general, it is not at all clear that a periodic timetable exists:
### Definition 2.2
($$T$$ -PESP-Feasibility) Given a tuple $$(G, T, \ell , u)$$ as above, decide if there exists a periodic timetable $$\pi$$.
### Theorem 2.3
(Odijk, 1994) $$T$$ -PESP-Feasibility is NP-complete for fixed $$T \ge 3$$.
### Proof
It is clear that $$T$$ -PESP-Feasibility is in NP, as a feasible timetable $$\pi$$ with periodic tension x serves as certificate. We recall the proof in order to emphasize the natural relationship between PESP and Vertex Coloring. Recall that the T-Vertex Coloring problem is to decide for an undirected graph H whether it admits a T-coloring, i.e., a map $$f: V(H) \rightarrow \{0, \dots , T-1\}$$ such that $$f(i) \ne f(j)$$ for every pair (ij) of adjacent vertices in H.
Fix $$T \ge 3$$. Given an undirected graph H, construct a $$T$$ -PESP-Feasibility instance $$(G, T, \ell , u)$$ as follows: G is obtained from H by arbitrarily directing the edges, and set $$\ell := 1$$, $$u := T-1$$. Then, H admits a T-coloring if and only if $$(G, T, \ell , u)$$ has a periodic timetable $$\pi$$. Namely, if $$f: V(H) \rightarrow \{0, \dots , T-1\}$$ is a T-coloring, then setting $$\pi _i := f(i)$$ for all $$i \in V(G)$$ is a periodic timetable, as for every arc $$ij \in A(G)$$ then holds $$x_{ij} = [\pi _j - \pi _i]_T = [f(j) - f(i)]_T \in [1, T-1]$$. Vice versa, if the PESP instance is feasible, then there is an integer timetable, giving rise to a T-coloring. $$\square$$
So far, we have neglected the weight vector $$w \in {\mathbb {Q}}_{\ge 0}^{A(G)}$$. The weights come into play in the optimization variant of PESP, which we state as its corresponding decision version. If x is a periodic tension, we call $$y := x - \ell \ge 0$$ the periodic slack.
### Definition 2.4
($$T$$ -PESP-Optimality) Given a tuple $$(G, T, \ell , u, w)$$ as above and a number M, find a periodic timetable $$\pi$$ with periodic slack y such that $$w^t y \le M$$.
Minimizing the weighted periodic slack, or equivalently, the weighted periodic tension, can be interpreted as minimizing the total travel time of all passengers in a public transportation network. Clearly, $$T$$ -PESP-Optimality is an NP-hard optimization problem. However, in many non-railway public transport networks, minimum distances are neglected for planning, and the driving and dwelling times of vehicles have a rather small span, so that they can be assumed as fixed. Contracting the corresponding activities yields a graph where only transfer activities remain, and these have typically no restrictions on their durations in the sense that lower and upper bounds differ by at least $$T-1$$. This motivates the following specialization of PESP, called reduced PESP in Pätzold and Schöbel (2016):
### Definition 2.5
($$T$$ -RPESP-Optimality) Given $$(G, T, \ell , u, w)$$ as above with $$u \ge \ell + T - 1$$ and a number M, find a periodic timetable $$\pi$$ with periodic slack y such that $$w^t y \le M$$.
Note that the feasibility problem is trivial to solve: In fact, any integral vector $$\pi \in [0, T)^{V(G)}$$ is a periodic timetable, because for any activity $$ij \in A(G)$$,
\begin{aligned} \ell _{ij} \le x_{ij} = [\pi _j - \pi _i - \ell _{ij}]_T + \ell _{ij} \le T - 1 + \ell _{ij} \le u_{ij}. \end{aligned}
### Theorem 2.6
(Nachtigall, 1993) For any fixed $$T \ge 3$$, $$T$$ -RPESP-Optimality is NP-hard.
### Proof
We adapt the proof of Nachtigall to our notions and notations. Fix some period time $$T \ge 3$$. We reduce $$T$$ -PESP-Feasibility to $$T$$ -RPESP-Optimality.
Let $$(G, T, \ell , u)$$ be a $$T$$ -PESP-Feasibility instance. Without loss of generality, assume that $$u - \ell < T$$, because if the instance is feasible, then there is a periodic tension x satisfying $$x < \ell + T$$ by (2). Add to each arc $$a \in A(G)$$ from i to j a reverse copy $${\overline{a}}$$ with $$\ell _{{\overline{a}}} := [-u_a]_T$$. Set all weights w to 1. For this $$T$$ -RPESP-Optimality instance, let $$\pi$$ be a periodic timetable with tension x as defined in (2). Then, for any original arc a holds $$\ell _{a} \le x_{a} < \ell _a + T$$ and $$x_{a} \equiv -x_{{{\overline{a}}}} \mod T$$. For the slacks $$y_a$$ and $$y_{{\overline{a}}}$$, we obtain
\begin{aligned} y_a + y_{{{\overline{a}}}} = [x_{a} - \ell _{a}]_T + [u_{a} - x_{a}]_T. \end{aligned}
Since $$0 \le x_a - \ell _a \le u_a - \ell _a < T$$ and $$-T< \ell _a - x_a \le u_a - x_a \le u_a - \ell _a < T$$,
\begin{aligned} y_a + y_{{{\overline{a}}}} = {\left\{ \begin{array}{ll} u_a - \ell _a &{} \quad \text { if } u_a - x_a \ge 0,\\ u_a - \ell _a + T &{} \quad \text { if } u_a - x_a < 0. \end{array}\right. } \end{aligned}
In particular, for the weighted slack of the $$T$$ -RPESP-Optimality instance holds
\begin{aligned} \sum _{a \in A(G)} (y_a + y_{{\overline{a}}}) \ge \sum _{a \in A(G)} (u_a - \ell _a), \end{aligned}
and equality holds if and only if $$x_a \le u_a$$ for all $$a \in A(G)$$. This means that the $$T$$ -PESP-Feasibility instance is feasible if and only if the described $$T$$ -RPESP-Optimality instance has weighted periodic slack at most $$M := \sum _{a \in A(G)} (u_a - \ell _a)$$. $$\square$$
We turn now to simple algorithms for $$T$$ -PESP-Optimality. Consider at first instances where undirecting the event-activity network G results in a tree (shortly, G is a tree):
### Lemma 2.7
Suppose that G is a tree on n vertices. Then, $$T$$ -PESP-Optimality on $$(G, T, \ell , u, w)$$ can be solved in O(n) time. Moreover, $$\ell$$ is an optimal periodic tension, and the minimum weighted periodic slack is 0.
### Proof
If undirecting G results in a tree on n vertices, then the transpose $$B^t$$ of the incidence matrix B of G is a $$(n-1) \times n$$ matrix of full rank $$n-1$$. In particular, $$B^t \pi = \ell$$ has a solution over $${\mathbb {Z}}$$, and reducing modulo T gives a feasible periodic timetable $$\pi ^*$$ with periodic tension $$\ell$$ and hence weighted periodic slack 0.
Avoiding linear algebra, $$\pi ^*$$ can as well be obtained by traversing the tree, starting with $$\pi ^*_v = 0$$ at an initial vertex v and setting $$\pi ^*_j := [\pi ^*_i + \ell _{ij}]_T$$ when traversing $$ij \in A(G)$$, and $$\pi ^*_j := [\pi ^*_i - \ell _{ij}]_T$$ if $$ji \in A(G)$$ is traversed. A depth-first traversal takes $$O(|V(G)| + |A(G)|) = O(n)$$ time. $$\square$$
On general networks, there are several ways to give naive exponential-time algorithms for $$T$$ -PESP-Optimality:
### Lemma 2.8
On instances $$(G, T, \ell , u, w)$$ with n events and m activities, $$T$$ -PESP-Optimality can be solved in
1. 1.
$$O^*(T^{n-1})$$, or
2. 2.
$$O^*(2^{n-1} n^{n-2})$$, or
3. 3.
$$O^*(3^m)$$ time,
where $$O^*(\cdot )$$ means $$O(\cdot )$$ ignoring polynomial factors.
### Proof
1. 1.
Enumerate all $$T^n$$ integral vectors in $$[0, T)^{V(G)}$$,
compute x by (2), and check the bounds. This is an $$O(mT^n)$$ algorithm. If $$\pi$$ is a periodic timetable, then for any $$d \in {\mathbb {R}}$$, $$\pi '$$ defined by $$\pi '_i := [\pi _i + d]_T$$ for all $$i \in V(G)$$ is a periodic timetable with the same periodic tension. In particular, only $$T^{n-1}$$ vectors have to be enumerated.
2. 2.
By Nachtigall (1998), if G is weakly connected and the instance is feasible, there is an optimal periodic tension $$x^*$$ and a spanning tree F of G such that $$x^*_a \in \{\ell _a, u_a\}$$ for all $$a \in A(F)$$. Enumerate all $$O(n^{n-2})$$ spanning trees F of G. For each such F, enumerate all $$2^{n-1}$$ vectors $$x \in \prod _{a \in A(F)} \{\ell _a, u_a\}$$. Interpreting x as a periodic tension defines a periodic timetable $$\pi \in [0, T)^{V(G)}$$ which can be computed by an $$O(n+m)$$ depth-first traversal as in the proof of Lemma 2.7 (replacing $$\ell$$ by x). Use (2) to compute the periodic tension $$x_a$$ of all O(m) remaining co-tree arcs $$a \notin A(F)$$ and check if the bounds are satisfied. If G is not connected, $$T$$ -PESP-Optimality can be solved on each component individually.
3. 3.
Let B denote the incidence matrix of G. Then, the modulo constraint $$B^t \pi \equiv x \bmod T$$ is satisfied if and only if there is a vector $$p \in {\mathbb {Z}}^{A(G)}$$ such that $$B^t \pi = x - Tp$$. Since any entry of $$B^t \pi$$ lies in the interval $$(-T, T)$$, and any tension computed by (2) satisfies $$x \in [0, 2T)$$, it suffices to consider $$p \in \{0, 1, 2\}^{A(G)}$$, cf. Liebchen (2006, Lemma 9.2). The algorithm is now to solve the problem
\begin{aligned}&\text {Minimize}&w^t y \\&\text {s.t.}&B^t \pi&= x - Tp,\\&\ell \le x&\le u \end{aligned}
for each fixed $$p \in \{0,1,2\}^{A(G)}$$. This is a series of $$3^m$$ linear programs. $$\square$$
Somewhat unsurprisingly, Lemma 2.8 implies that all three presented PESP variants are solvable in polynomial time when the number of events or the number of activities is fixed. In the remainder of the paper, we investigate several graph parameters and their effects on the parameterized complexity of PESP.
## 3 Parameters
### 3.1 Parameterized Complexity
Parameterized complexity is an active topic in the context of scheduling; see, for example, Mnich & van Bevern, 2018. Before introducing the parameters discussed in this paper, we recall several standard notions from the field of parameterized complexity and refer to the books (Downey & Fellows, 1999) and (Cygan et al., 2015) for further details.
A parameterized problem over an alphabet $$\varSigma$$ is a language $$L \subseteq \varSigma ^* \times {\mathbb {N}}$$. A parameterized problem L is called fixed-parameter tractable (FPT) if there are an algorithm A, a computable function $$f: {\mathbb {N}} \rightarrow {\mathbb {N}}$$, and a constant $$c \in {\mathbb {N}}$$ such that for each $$(x, k) \in \varSigma ^* \times {\mathbb {N}}$$, A decides whether $$(x, k) \in L$$ within $$f(k) \cdot {\text {size}}(x, k)^c$$ steps. In this case, we will call A an FPT algorithm. The class of FPT problems forms a complexity class, which will also be denoted by FPT.
A parameterized problem L belongs to the complexity class XP if there are an algorithm A and functions $$f, g: {\mathbb {N}} \rightarrow {\mathbb {N}}$$ such that A decides whether $$(x, k)\in L$$ within $$f(k) \cdot {\text {size}}(x,k)^{g(k)}$$ steps. Clearly, FPT $$\subseteq$$ XP.
Generalizing FPT algorithms to non-deterministic algorithms, we obtain the complexity class para-NP. We have FPT $$=$$ para-NP if and only if P $$=$$ NP, and no para-NP-hard problem can be in XP.
For parameterized problems $$L\subseteq \varSigma ^*_L \times {\mathbb {N}}$$ and $$M\subseteq \varSigma ^*_M \times {\mathbb {N}}$$, a parameterized reduction of L to M is an algorithm A that transforms an instance $$(x, k)\in \varSigma ^*_L \times {\mathbb {N}}$$ to an instance $$(x', k')\in \varSigma ^*_M \times {\mathbb {N}}$$ such that
1. 1.
$$(x, k) \in L$$ if and only if $$(x', k') \in M$$,
2. 2.
$$k' \le f(k)$$,
3. 3.
A requires $$f(k) \cdot {\text {size}}(x)^c$$ time,
where $$f: {\mathbb {N}} \rightarrow {\mathbb {N}}$$ is computable and $$c \in {\mathbb {N}}$$ is a constant, and both f and c are independent of (xk).
Finally, we define the complexity class W[1] as the class of all parameterized problems that admit a parameterized reduction to the k -Clique problem. It is known that FPT $$\subseteq$$ W[1] $$\subseteq$$ XP $$\cap$$ para-NP, and it is believed that FPT $$\ne$$ W[1].
### 3.2 Parameter search
We now return to PESP. The benchmarking library PESPlib (Goerigk, 2012) contains 20 difficult $$T$$ -PESP-Optimality instances, created with the LinTim toolbox with data of the German railway network (Goerigk et al., 2013). Despite several efforts, none of the instances is currently solved to proven optimality.
For the event-activity network of the smallest PESPlib instance R1L1, the values of several typical graph parameters in parameterized complexity are summarized in Table 1. Most of the parameters are easy to obtain; we will elaborate in Sect. 6 on the computation of the others. The cyclomatic number, also known as feedback edge set number, is a common measure for the difficulty of PESP instances, as it counts the number of integral variables used in a cycle-based mixed integer programming formulation (Borndörfer et al., 2019). However, parameters such as treewidth and branchwidth are much smaller. As $$T$$ -PESP-Optimality is solvable in linear time on trees by Lemma 2.7, it seems therefore reasonable to investigate parameterized algorithms in terms of treewidth. In particular, we could ask whether there is a polynomial-time algorithm for PESP on event-activity networks with bounded treewidth, i.e., $$T$$ -PESP-Optimality belongs to XP. Even better, we wonder if there is an FPT algorithm. Unfortunately, the answer is negative unless P = NP.
The complexity landscape of the parameters discussed in this paper is manifold. An overview of the results and the relations between the parameters is depicted in Fig. 1. We will introduce the parameters in the next two sections, presenting hardness results in Sect. 4 and algorithms in Sect. 5.
## 4 Hardness results
### Definition 4.1
(Garey & Johnson, 1979, SP13) The Subset Sum problem is the following: Given $$r \in {\mathbb {N}}$$, $$c \in {\mathbb {Z}}_{\ge 0}^r$$ and $$C \in {\mathbb {Z}}_{\ge 0}$$ with $$C \le \sum _{i=1}^r c_i$$, is there a $$z \in \{0, 1\}^r$$ such that $$c^t z = C$$?
The Subset Sum problem is weakly NP-complete (Karp, 1972). We will at first construct a polynomial-time reduction in Subset Sum to $$T$$ -PESP-Feasibility.
### Definition 4.2
For a Subset Sum instance (rcC) as above, define I(rcC) as the instance $$(G, T, \ell , u)$$ for $$T$$ -PESP-Feasibility as depicted in Fig. 2 with $$T := \sum _{i=1}^r c_i + 1$$.
### Lemma 4.3
The Subset Sum instance (rcC) has a solution if and only if $$T$$ -PESP-Feasibility has a solution on the instance I(rcC).
### Proof
Let $$\pi$$ be a periodic timetable for I(rcC). Then, $$[\pi _{i} - \pi _{i-1}]_T \in \{0, c_i\}$$ holds for all $$i \in \{1, \dots , r\}$$. Set
\begin{aligned} z_i := {\left\{ \begin{array}{ll} 1 &{} \quad \text { if } [\pi _{i} - \pi _{i-1}]_T = c_i ,\\ 0 &{} \quad \text { otherwise,} \end{array}\right. } \quad i = 1, \dots , r. \end{aligned}
For the arc (0, r), we then obtain
\begin{aligned} C&= [C]_T = [\pi _r - \pi _0]_T \\&= \left[ \sum _{i=1}^{r} (\pi _{i} - \pi _{i-1}) \right] _T \\&= \left[ \sum _{i=1}^{r} z_i c_i \right] _T = \sum _{i=1}^r z_i c_i, \end{aligned}
and found a positive answer to the Subset Sum problem on (rcC). Conversely, any vector $$z \in \{0,1\}^r$$ such that $$c^t z = C$$ yields a feasible periodic timetable $$\pi$$ by setting
\begin{aligned} \pi _0&:= 0 \quad \text { and } \\ \pi _{i}&:= \pi _{i-1} + z_i c_i, \quad i = 1, \dots , r.&\\[-25pt] \end{aligned}
$$\square$$
### Definition 4.4
(e.g., Robertson & Seymour, 1984) Given a graph G, a tree decomposition of G is a pair $$({\mathcal {T}}, {\mathcal {X}})$$ consisting of a tree $${\mathcal {T}}$$ and a family of bags $${\mathcal {X}} = (X_t)_{t \in V({\mathcal {T}})}$$ with $$X_t \subseteq V(G)$$ for each $$t \in V({\mathcal {T}})$$ such that
1. 1.
$$\bigcup _{t \in V({\mathcal {T}})} X_t = V(G)$$,
2. 2.
for each $$a \in A(G)$$, there is a bag $$X_t$$ containing both endpoints of a,
3. 3.
for each $$v \in V(G)$$, the subforest of $${\mathcal {T}}$$ induced by $$\{t \in V({\mathcal {T}}) \mid v \in X_t\}$$ is connected.
The width of a tree decomposition is $$\max _{t \in V}({\mathcal {T}}) |X_t| - 1$$, and the treewidth of G is defined as the minimum possible width of a tree decomposition, i.e.,
\begin{aligned} {\text {tw}}(G) := \min \left\{ \max _{t \in V({\mathcal {T}})} |X_t| \,\big |\, \begin{array}{cc} ({\mathcal {T}}, {\mathcal {X}}) \text { is a tree} \\ \text {decomposition of } G \end{array} \right\} -1. \end{aligned}
This definition applies to both undirected and directed graphs, and as well to multigraphs. According to Definition 4.4, the treewidth of a graph with multiple edges equals the treewidth of the graph where all multiple edges between two vertices are replaced by a single edge. The simple connected graphs of treewidth 1 are precisely the trees.
### Lemma 4.5
For any Subset Sum instance (rcC) with $$r \ge 2$$, the event-activity network G of the instance I(rcC) has treewidth 2.
### Proof
The path of length r depicted in Fig. 3, where each node is labeled with its bag, is a tree decomposition of G. Checking the properties of a tree decomposition is straightforward. The maximum bag size is 3, and hence $${\text {tw}}(G) \le 2$$. Removing one of the two arcs from i to $$i+1$$, $$i = 0, \dots , r-1$$, does not change the treewidth in the sense of Definition 4.4. As $$r \ge 2$$, this results in a simple graph containing a cycle, so that $${\text {tw}}(G) \ge 2$$. $$\square$$
An alternative way to see that the network G of I(rcC) has treewidth at most 2 is to observe that G is series–parallel (see, for example, Bodlaender & van Antwerpen-de Fluiter, 2001, Lemma 3.4). With Lemmas 4.3 and 4.5, we obtain:
### Theorem 4.6
$$T$$ -PESP-Feasibility is NP-complete on networks of treewidth 2.
Since the transformation in the proof of Theorem 2.6 reduces $$T$$ -PESP-Feasibility to the $$T$$ -RPESP-Optimality problem by adding antiparallel arcs, which does not alter the treewidth, we have moreover:
### Theorem 4.7
$$T$$ -RPESP-Optimality is NP-complete on networks of treewidth 2.
### Remark 4.8
If simple graphs are desired, one can dispose of the parallel arcs in I(rcC) by subdividing any arc with bounds $$[0, c_i]$$ into two arcs with bounds $$[0, c_i]$$ and [0, 0] without affecting the feasibility of the PESP instance. Moreover, one checks that this does not increase the treewidth of I(rcC).
### Remark 4.9
In the proof of Lemma 4.3, the period time T is chosen very large. The above NP-completeness theorems hence do not hold when T is fixed. We will give a pseudo-polynomial algorithm for $$T$$ -PESP-Optimality with bounded branchwidth in Sect. 5, showing that fixing both T and the treewidth results in a polynomial-time algorithm.
### Remark 4.10
We want to remark that there has already been a paper (van Heuven van Staereling, 2018) titled Tree Decomposition Methods for the periodic event scheduling problem. However, the title is misleading, as the algorithm there is about finding trees in the network rather than considering tree decompositions in the usual sense.
### Definition 4.11
(Robertson & Seymour, 1991) Given a graph G, a branch decomposition of G is a pair $$({\mathcal {B}}, \varphi )$$, where $${\mathcal {B}}$$ is a tree such that every non-leaf node has degree 3, and $$\varphi$$ is a bijection from the leaves of $${\mathcal {B}}$$ to A(G). Deleting an edge e of $${\mathcal {B}}$$ disconnects $${\mathcal {B}}$$ into two subtrees and hence partitions the leaves of $${\mathcal {B}}$$ into two sets. Applying $$\varphi$$, this yields a partition $$A = A_e^1 \overset{.}{\cup } A_e^2$$. This defines in turn a vertex separator $$S_e \subseteq V$$ as the set of vertices that are incident both to an edge in $$A_e^1$$ and to an edge in $$A_e^2$$.
The width of a branch decomposition $$({\mathcal {B}}, \varphi )$$ is defined as $$\max _{e \in E({\mathcal {B}})} |S_e|$$. The branchwidth of G is then the minimum possible width of a branch decomposition, i.e.,
\begin{aligned} {\text {bw}}(G) := \min \left\{ \max _{e \in E({\mathcal {B}})} |S_e| \,\big |\, \begin{array}{c} ({\mathcal {B}}, \varphi ) \text { is a branch} \\ \text {decomposition of } G \end{array} \right\} . \end{aligned}
Treewidth and branchwidth are related as follows:
### Theorem 4.12
(Robertson & Seymour, 1991, 5.1) If $${\text {bw}}(G) \ge 2$$, then
\begin{aligned} {\text {bw}}(G) \le {\text {tw}}(G) + 1 \le \left\lfloor \frac{3}{2} {\text {bw}}(G) \right\rfloor . \end{aligned}
It follows immediately that both $$T$$ -PESP-Feasibility and $$T$$ -RPESP-Optimality are NP-complete on networks with branchwidth 3. We prove below that the NP-completeness is already given for branchwidth 2.
### Lemma 4.13
For any Subset Sum instance (rcC), the event-activity network G of the instance I(rcC) has branchwidth 2.
### Proof
Figure 4 shows a branch decomposition of G, where the leaves are labeled with the corresponding arc and the edges are labeled with the cardinality of the corresponding vertex separator. This is clearly a branch decomposition. Checking the cardinalities of the vertex separators is again straightforward, and hence $${\text {bw}}(G) \le 2$$. The network G cannot have branchwidth 1, as in any branch decomposition, the edge incident to the leaf representing (0, r) always induces the vertex separator $$\{0, r\}$$ of size 2. $$\square$$
### Theorem 4.14
The problems $$T$$ -PESP-Feasibility and $$T$$ -RPESP-Optimality are NP-complete on networks with branchwidth at most 2.
### Proof
For $$T$$ -PESP-Feasibility, this follows by combining Lemma 4.3 with Lemma 4.13. Introducing anti-parallel arcs in the network of I(rcC) does not increase the branchwidth of 2: In the branch decomposition of the proof of Lemma 4.13, replace a leaf with a vertex of degree 3 adjacent to two new leaves corresponding to the two anti-parallel arcs. The new edges have vertex separators of size 2. Therefore, the transformation in the proof of Theorem 2.6 does not alter the branchwidth, and $$T$$ -RPESP-Optimality is NP-complete on networks with branchwidth at most 2. $$\square$$
Another approach to prove Lemma 4.13 and Theorem 4.14 is to exploit that a graph of treewidth at most 2 has branchwidth at most 2 (combine, for example, Bodlaender & van Antwerpen-de Fluiter, 2001, Lemma 3.5 with Robertson & Seymour, 1991, 4.2).
#### 4.1.4 Carvingwidth
Carvingwidth is defined analogously to branchwidth, by labeling the leaves of an unrooted binary tree with the vertices of the original graph instead of the edges.
### Definition 4.15
(Seymour & Thomas, 1994)
Given a graph G, a carving decomposition of G is a pair $$({\mathcal {C}}, \psi )$$, where $${\mathcal {C}}$$ is a tree such that every non-leaf node has degree 3, and $$\psi$$ is a bijection from the leaves of $${\mathcal {C}}$$ to V(G). Removing an edge e of $${\mathcal {C}}$$ induces a partition of the leaves of $${\mathcal {C}}$$, and hence via $$\psi$$ also a partition $$V(G) = V_e^1 \overset{.}{\cup } V_e^2$$. Let $$\delta (V_e^1)$$ ($$=\delta (V_e^2)$$) denote the set of cut edges.
The maximum cardinality of $$\delta (V_e^1)$$ taken over all $$e \in E({\mathcal {C}})$$ is the width of $$({\mathcal {C}}, \psi )$$. The carvingwidth of G is defined as
\begin{aligned} {\text {cw}}(G) := \min \left\{ \max _{e \in E({\mathcal {C}})} |\delta (V_e^1)| \,\big |\, \begin{array}{c} ({\mathcal {C}}, \psi ) \text { is a carving} \\ \text {decomposition of } G \end{array} \right\} . \end{aligned}
The definition of carvingwidth applies as well to multigraphs, but in contrast to branch- and treewidth, it is sensitive to multiple edges: The carvingwidth is at least the maximum vertex degree $$\varDelta (G)$$, as for each $$v \in V(G)$$, $${\text {cw}}(G) \ge \deg (v) = |\delta (v)|$$. Carvingwidth is related to branchwidth as follows:
### Theorem 4.16
(Nestoridis & Thilikos, 2014; Eppstein, 2018) Let G be a graph with maximum vertex degree $$\varDelta (G)$$. Then,
\begin{aligned} \max \left( \varDelta (G), \left\lceil \frac{1}{2}{\text {bw}}(G) \right\rceil \right) \le {\text {cw}}(G) \le \varDelta (G) \cdot {\text {bw}}(G). \end{aligned}
### Theorem 4.17
$$T$$ -PESP-Feasibility is NP-complete on networks of carvingwidth at most 3, and $$T$$ -RPESP-Optimality is NP-complete on networks of carvingwidth at most 6.
### Proof
Let $$r \ge 2$$ and consider again an instance of the form I(rcC) with event-activity network G. Then, $$\varDelta (G) = 4$$, so that $${\text {cw}}(G) \ge 4$$. For all $$i \in \{1, \dots , r-1\}$$, split vertex i into two new vertices $$i^+$$ and $$i^-$$, connected by a single directed arc $$(i^+,i^-)$$ with bounds $$\ell _{i^+,i^-} = u_{i^+, i^-} = 0$$. The splitting is done in such a way that the arcs entering i are now entering $$i^+$$, and the arcs leaving i are leaving $$i^-$$. Any periodic timetable has the same value at $$i^+$$ and $$i^-$$, so that the proof of Lemma 4.3 carries over. However, the modified graph has maximum degree 3. The carving decomposition in Fig. 5, where the edges labeled with the number of cut edges, has width 3. Hence, $$T$$ -PESP-Feasibility is NP-complete on networks of carvingwidth 3. As a consequence, keeping in mind the arc duplication occurring in the proof of Theorem 2.6, $$T$$ -RPESP-Optimality is NP-complete for carvingwidth 6. $$\square$$
We also obtain from the proof of Theorem 4.17 that $$T$$ -PESP-Feasibility is in fact NP-complete on networks with maximum degree 3.
### Remark 4.18
$$T$$ -PESP-Optimality is trivial to solve on graphs G with $${\text {cw}}(G) = 1$$, as then $$\varDelta (G) \le 1$$ by Theorem 4.16. If $${\text {cw}}(G) = 2$$, then $$\varDelta (G) \le 2$$, so that any weakly connected component of G, seen as an undirected graph, is either a path or a cycle. $$T$$ -PESP-Optimality is solvable in linear time on paths (Lemma 2.7). We will see later in Theorem 5.12 that $$T$$ -PESP-Optimality admits a polynomial-time algorithm when the cyclomatic number is bounded by a constant, and in particular on a single cycle.
#### 4.1.5 Diameter
The diameter of a graph is the maximum length of an undirected shortest path between two vertices.
### Lemma 4.19
$$T$$ -PESP-Feasibility is NP-hard on networks with diameter 1.
### Proof
For the first statement, we turn the instances I(rcC) used in Lemma 4.3 into complete graphs by adding activities a with $$\ell _a = 0$$ and $$u_a = T-1$$ and observe that this does not alter the feasibility. $$\square$$
In Sect. 5, we will give an FPT algorithm for $$T$$ -PESP-Optimality when both diameter and cyclomatic number are bounded.
### 4.2 W[1]-Hardness
List Coloring dates back to Vizing (1976) and Erdős et al. (1980). It is known to be W[1]-hard when parameterized by treewidth (Fellows et al., 2011), or by the vertex cover number (Fiala et al., 2011, Theorem 1). We will now describe a parameterized reduction in the List Coloring problem to $$T$$ -PESP-Feasibility, which will be useful for two purposes.
At first, we can deduce the W[1]-hardness of $$T$$ -PESP-Feasibility w.r.t. the vertex cover number.
Secondly, note that reducing the Subset Sum problem does still allow for FPT algorithms for $$T$$ -PESP-Feasibility w.r.t. treewidth or branchwidth when the period time T is encoded in unary. We show that in this situation, $$T$$ -PESP-Feasibility is W[1]-hard, i.e., there is most likely no such “pseudo”-FPT algorithm. In particular, we can only hope for “pseudo”-XP algorithms, where the exponent of T is a non-constant function of the parameter. We will indeed present such an algorithm in Sect. 5.
### Definition 4.20
The List Coloring problem is the following: Given a graph H together with a finite list $$L(v) \subseteq {\mathbb {Z}}_{\ge 0}$$ for each $$v \in V(H)$$, decide whether there is a vertex coloring of H such that each vertex v is colored with a color in L(v).
Given a List Coloring instance (HL), we construct a $$T$$ -PESP-Feasibility instance $$(G, T, \ell , u)$$ as follows:
Define $$T := \max _{v \in V(H)} L(v) + 1$$. Let G be any orientation of H. For each edge of H, we define for the corresponding activity a in A(G) the bounds $$\ell _a := 1$$ and $$u_a := T-1$$. Further add a new vertex $$v_0$$ to G. Let $$v \in V(H)$$ with $$L(v) = \{c_1, \dots , c_r\}$$, $$c_1< \dots < c_r$$. Add r parallel activities $$a_1, \dots , a_r$$ from $$v_0$$ to v, with bounds
\begin{aligned} \ell _{a_i} := c_i, \quad u_{a_i} := c_{i-1} + T, \quad i = 1,\dots ,r, \end{aligned}
where we set $$c_0 := c_r - T$$. This way, we model the disjunctive constraints of choosing a color in L(v) as a PESP instance (Liebchen & Möhring, 2007, §3.3).
We claim that (HL) has a feasible list coloring if and only if $$(G, T, \ell , u)$$ admits a feasible periodic timetable. Thus, let $$\pi \in [0, T)^{V(H)}$$ be a list coloring for (HL). If $$ij \in A(H)$$, then $$\pi _j \ne \pi _j$$, so that $$[\pi _j - \pi _i - 1]_T \le T-2$$ is a feasible periodic slack. Extend $$\pi$$ to a timetable on G by setting $$\pi _{v_0} := 0$$. Let $$v \in V(H)$$, and assume that v is colored with the jth color from its list, i.e., $$\pi _v = c_j \in L(v)$$. For $$i \in \{1, \dots , r\}$$, for the ith activity $$a_i$$ from $$v_0$$ to v, the periodic tension would be
\begin{aligned}{}[\pi _v - \pi _{v_0} - \ell _{a_i}]_T + \ell _{a_i}&= [c_j - c_i]_T + c_i \\&= {\left\{ \begin{array}{ll} c_j &{} \quad \text { if } i \le j,\\ c_j + T &{} \quad \text { if } i > j. \end{array}\right. } \end{aligned}
In the first case $$c_j \le T \le c_{i-1} + T$$, and in the second $$c_i > c_j$$ implies $$c_j + T \le c_{i-1} + T$$. We conclude that $$\pi$$ is a feasible periodic timetable.
Conversely, consider a feasible periodic timetable $$\pi \in [0, T)^{V(G)}$$. By a shift replacing $$\pi$$ by $$[\pi - \pi _{v_0}]_T$$, we can assume that $$\pi _{v_0} = 0$$. By restriction, using that $$\pi _j \ne \pi _i$$ for all $$ij \in A(H)$$, $$\pi$$ yields a coloring of H. It remains to check that this is a feasible list coloring. The periodic tension on an activity $$a_i$$ from $$v_0$$ to v must satisfy
\begin{aligned}{}[\pi _v - \pi _{v_0} - \ell _{a_i}]_T + \ell _{a_i} = [\pi _v - c_i]_T + c_i \le c_{i-1} + T \end{aligned}
for all $$i \in \{1, \dots , r\}$$. Suppose that there is an index j such that $$c_j \le \pi _v < c_{j+1}$$. Then, the above inequality for $$i = j+1$$ means
\begin{aligned} \pi _v + T = [\pi _v - c_{j+1}]_T + c_{j+1} \le c_j + T, \end{aligned}
hence $$\pi _v = c_j$$. If $$\pi _v \ge c_r$$, then $$\pi _v \le c_{0} + T = c_r$$. Finally, if $$\pi _v < c_1$$, then $$\pi _v \le c_r - T < 0$$, contradicting $$\pi _v \ge 0$$. This shows that $$\pi _v \in \{c_1, \dots , c_r\}$$, so that $$\pi$$ is indeed a feasible list coloring.
We have hence proved the following:
### Lemma 4.21
A List Coloring instance (HL) has a solution if and only if $$(G, T, \ell , u)$$ is feasible.
### Definition 4.22
For a graph G, its vertex cover number $${\text {vc}}(G)$$ is defined as the minimum cardinality of a vertex cover of G.
### Theorem 4.23
$$T$$ -PESP-Feasibility is W[1]-hard when parameterized by the vertex cover number.
### Proof
List Coloring is W[1]-hard when parameterized by the vertex cover number (Fiala et al., 2011, Theorem 1). Construct a $$T$$ -PESP-Feasibility instance $$(G, T, \ell , u)$$ from a List Coloring instance (HL) as in Lemma 4.21. Since all arcs in G not present in H are connected to $$v_0$$, we obtain $${\text {vc}}(G) \le {\text {vc}}(H) + 1$$. $$\square$$
It remains unclear whether $$T$$ -PESP-Feasibility or $$T$$ -PESP-Optimality admit polynomial-time algorithms for fixed vertex cover number, i.e., if these problems belong to XP. In Sect. 5.3, we show that fixing both vertex cover number and cyclomatic number enables an FPT algorithm for $$T$$ -PESP-Optimality.
#### 4.2.3 Treewidth and diameter revisited
The treewidth of a graph is known to be bounded by its vertex cover number:
### Lemma 4.24
(Fiala et al., 2011, §2) Let G be a graph. Then, $${\text {tw}}(G) \le {\text {vc}}(G)$$.
Moreover, the vertex cover number bounds the diameter, as any path of length d contains at least d/2 vertices of any vertex cover. As a direct consequence of Theorem 4.23, we hence obtain:
### Corollary 4.25
$$T$$ -PESP-Feasibility is W[1]-hard when parameterized by treewidth, branchwidth, or diameter.
## 5 Parameterized algorithms
### 5.1 A dynamic program for bounded branchwidth
#### 5.1.1 PESP and vertex separators
The Subset Sum problem is weakly NP-complete and can be solved by pseudo-polynomial-time algorithms. In the following, we present a dynamic program for $$T$$ -PESP-Optimality running in pseudo-polynomial time for event-activity networks of bounded branchwidth. Since $$T$$ -PESP-Optimality comprises both $$T$$ -PESP-Feasibility and $$T$$ -RPESP-Optimality, this implicitly gives pseudo-polynomial time algorithms for these problems as well.
The key insight for our dynamic programming approach is the following decomposition property: Let
$$I = (G, T, \ell , u, w)$$ be a $$T$$ -PESP-Optimality instance. For any partition $$A(G) = A^1 \overset{.}{\cup } A^2$$, we can partition I into two subinstances $$I^1$$ resp. $$I^2$$ restricted to the activities in $$A^1$$ resp. $$A^2$$. If y is a feasible periodic slack on I, then the restrictions $$y^1$$ resp. $$y^2$$ to $$I^1$$ resp. $$I^2$$ yield feasible periodic slacks with the property
\begin{aligned} \sum _{a \in A(G)} w_a y_a = \sum _{a \in A^1} w_a y^1_a + \sum _{a \in A^2} w_a y^2_a. \end{aligned}
(3)
On the level of timetables, we obtain from a timetable $$\pi$$ on I two timetables $$\pi ^1$$ resp. $$\pi ^2$$ on $$I^1$$ resp. $$I^2$$ such that $$\pi$$, $$\pi ^1$$, and $$\pi ^2$$ all coincide when restricted to the events of the vertex separator S associated with the partition of A(G) as in Definition 4.11.
Conversely, we can glue two periodic timetables $$\pi ^1$$ and $$\pi ^2$$ together to a timetable $$\pi$$ on I if the restrictions to a vertex separator S satisfy $$\pi ^1|_S = \pi ^2|_S$$. The corresponding periodic slacks then hold Eq. (3).
### Definition 5.1
Let $$I = (G, T, \ell , u, w)$$ be a $$T$$ -PESP-Optimality instance, and let $$S \subseteq V(G)$$. For a vector $$\rho \in [0, T)^S$$, define $${\text {OPT}}(I, S, \rho )$$ as the minimum weighted slack of a periodic timetable $$\pi$$ on I when additionally $$\pi |_S = \rho$$ is required.
For minimum weighted slacks, the above discussion shows the following:
### Lemma 5.2
Let $$I = (G, T, \ell , u, w)$$ be a feasible $$T$$ -PESP-Optimality instance. Let $$A(G) = A^1 \overset{.}{\cup } A^2$$ be a partition with vertex separator S giving subinstances $$I^1$$ and $$I^2$$ of I. Then,
\begin{aligned} {\text {OPT}}(I) = \min \{&{\text {OPT}}(I^1, S, \pi |_S) + {\text {OPT}}(I^2, S, \pi |_S) \mid \\&\pi \text { is a feasible per. timetable on } I\}. \end{aligned}
More generally, if additionally the timetable is fixed on $$W \subseteq V(G)$$ to $$\rho \in [0, T)^W$$, then
\begin{aligned} {\text {OPT}}(I, W, \rho ) = \min \{&{\text {OPT}}(I^1, S \cup W^1, \pi |_{S \cup W^1}) \\&+ {\text {OPT}}(I^2, S \cup W^2, \pi |_{S \cup W^2}) \mid \\&\pi \text { is a feasible periodic timetable}\\&\text {on } I \text { with } \pi |_W = \rho \}, \end{aligned}
where $$W^i = W \cap V^i$$ denotes the intersection of W with the set of events $$V^i$$ of $$I^i$$, $$i = 1,2$$.
#### 5.1.2 A branch decomposition approach
Since branch decompositions naturally encode vertex separators, we describe at first a branch-decomposition-based dynamic program for $$T$$ -PESP-Optimality. Let $$I = (G, T, \ell , u, w)$$ be a $$T$$ -PESP-Optimality instance. We assume that G is 2-edge-connected when seen as undirected graph. Let $$({\mathcal {B}}, \varphi )$$ be a branch decomposition of G with node set $$V({\mathcal {B}})$$ and edge set $$E({\mathcal {B}})$$. Subdivide an arbitrary edge of $$E({\mathcal {B}})$$ and call the new node $$\tau$$ the root. Recall from Definition 4.11 that every edge $$e \in E({\mathcal {B}})$$ corresponds to a partition $$A = A^1_e \overset{.}{\cup } A^2_e$$ with vertex separator $$S_e$$. We assume that $$A^1_e$$ is the subset of activities coming from the component of $${\mathcal {B}} \setminus \{e\}$$ not containing the root $$\tau$$.
### Algorithm 5.3
For each edge $$e \in E({\mathcal {B}})$$, we compute an $$|S_e|$$-dimensional table $$F_e$$ having an entry for each $$\pi \in \{0, \dots , T-1\}^{S_e}$$. The table $$F_e$$ is filled by a dynamic program starting from the edges $$e \in E({\mathcal {B}})$$ incident to leaves and with decreasing distance to the root:
1. 1.
If $$e \in E({\mathcal {B}})$$ is incident to a leaf corresponding via $$\varphi$$ to an activity $$ij \in A(G)$$, then set
\begin{aligned} F_e(\pi ) := {\left\{ \begin{array}{ll} w_{ij} y_{ij} &{} \quad \text { if } y_{ij} \le u_{ij} - \ell _{ij},\\ \infty &{} \text { otherwise,} \end{array}\right. } \end{aligned}
where $$y_{ij} := [\pi _j-\pi _i-\ell _{ij}]_T$$.
2. 2.
If e is incident to two edges $$e_1, e_2$$ with larger distance from the root, then set
\begin{aligned} F_e(\pi ) :=&\min \{ F_{e_1}(\pi '|_{S_{e_1}}) + F_{e_2}(\pi '|_{S_{e_2}}) \mid \\&\quad \; \pi ' \in \{0, \dots , T-1\}^{S_{e_1} \cup S_{e_2}}, \pi '|_{S_e} = \pi \}. \end{aligned}
3. 3.
If the tables of the two edges $$e_1, e_2$$ incident to $$\tau$$ have been computed, return
\begin{aligned} \min \{ F_{e_1}(\pi ) + F_{e_2}(\pi ) \mid \pi \in \{0, \dots , T-1\}^{S_{e_1}} \}. \end{aligned}
### Lemma 5.4
Let $$e \in E({\mathcal {B}})$$, $$\pi \in \{0, \dots , T-1\}^{S_e}$$. Denote by $$I_e$$ the subinstance of I containing precisely the activities in $$A^1_e$$.
1. 1.
If $$F_e(\pi ) < \infty$$, then $$F_e(\pi ) = {\text {OPT}}(I_e, S_e, \pi )$$.
2. 2.
If $$F_e(\pi ) = \infty$$, then $$I_e$$ is infeasible.
### Proof
Recall from Sect. 2 that it suffices to consider timetables with values in the discrete set $$\{0, \dots , T-1\}$$, as $$\ell$$ and u are integer.
If e is incident to a leaf associated with an activity $$ij \in A(G)$$, then $$A^1_e = \{ij\}$$ and $$S_e = \{i, j\}$$, as G is 2-edge-connected. Hence, $${\text {OPT}}(I_e, S_e, \pi )$$ is the minimum weighted slack of the activity ij when the timetable at i resp. j is fixed to $$\pi _i$$ resp. $$\pi _j$$. Therefore, we set $$F_e(\pi )$$ to $$w_{ij} [\pi _j - \pi _i - \ell _{ij}]_T$$ if the slack $$[\pi _j - \pi _i - \ell _{ij}]_T$$ is feasible and otherwise to $$\infty$$.
Otherwise, let e be adjacent to $$e_1, e_2 \in E$$, with $$e_1, e_2$$ having larger distance from the root than e. Then, $$A^1_e = A^1_{e_1} \overset{.}{\cup } A^1_{e_2}$$ and hence $$S_e \subseteq S_{e_1} \cup S_{e_2}$$. Moreover, $$(S_{e_1} \cup S_{e_2}) \setminus S_e$$ is the vertex separator of the partition of $$A^1_e = A^1_{e_1} \overset{.}{\cup } A^1_{e_2}$$. Applying Lemma 5.2 for $$W = S_e$$ and $$S = (S_{e_1} \cup S_{e_2}) \setminus S_e$$ yields the formula in Algorithm 5.3. $$\square$$
### Lemma 5.5
If $$k = \max _{e \in E({\mathcal {B}})} |S_e|$$ and $$m = |A(G)|$$, then Algorithm 5.3 computes $${\text {OPT}}(I)$$ or decides that I is infeasible in $$O(mT^{\lfloor 3k/2 \rfloor })$$ time.
### Proof
We first consider correctness. At the root $$\tau$$ with incident edges $$e_1, e_2$$, Algorithm 5.3 computes the tables $$F_{e_1}, F_{e_2}$$ such that $$F_{e_i}(\pi ) = {\text {OPT}}(I_{e_i}, S_{e_i}, \pi )$$ for all $$\pi$$ and $$i = 1,2$$ by Lemma 5.4. Observe that $$A^1_{e_1} = A^2_{e_2}$$, $$A^2_{e_1} = A^1_{e_2}$$, and $$S_{e_1} = S_{e_2}$$, so that by the first equation in Lemma 5.2,
\begin{aligned} {\text {OPT}}(I) =&\min \{ {\text {OPT}}(I_{e_1}, S_{e_1}, \pi ) + {\text {OPT}}(I_{e_2}, S_{e_2}, \pi ) \mid \\&\quad \; \pi \in \{0, \dots , T-1\}^{V} \text { feasible timetable} \}. \end{aligned}
This is precisely reflected in the third step of Algorithm 5.3, treating infeasible subinstances with infinite objective value.
Concerning running time, Step 1 can be done in $$O(T^2)$$ time and is called m times. Step 3 takes $$O(T^k)$$ time since $$|S_{e_1}| \le k$$. As any node in the rooted branch decomposition has either 0 or 2 children and there are m leaves, there are $$m-1$$ edges for which Step 2 is called. Each of the $$|S_e|$$ table entries requires to take a minimum over $$|(S_{e_1} \cup S_{e_2}) \setminus S_e|$$ previously computed table entries.
We claim that
\begin{aligned} 2 |S_{e_1} \cup S_{e_2}| \le |S_{e_1}| + |S_{e_2}| + |S_e|. \end{aligned}
If $$i \in S_{e_1} \setminus S_{e_2}$$, then i is adjacent to an activity $$a \in A^1_{e_1}$$ and $$a' \notin A^1_{e_1}$$. Since $$A^1_{e_1}$$ and $$A^1_{e_2}$$ are disjoint, $$a \notin A^1_{e_2}$$. As $$i \notin S_{e_2}$$, then also $$a' \notin A^1_{e_2}$$, and consequently $$a' \notin A^1_e = A^1_{e_1} \cup A^1_{e_2}$$. It follows that $$i \in S_e$$, and any such i appears hence twice in the right-hand side of the above inequality. By symmetry, the same holds for all $$i \in S_{e_2} \setminus S_{e_1}$$. Clearly, any $$i \in S_{e_1} \cap S_{e_2}$$ is counted both in $$|S_{e_1}|$$ and in $$|S_{e_2}|$$. This proves the claim. The relation between the separators is also depicted in Fig. 6.
Since the size of any of the three vertex separators is bounded by k, we obtain
\begin{aligned} |S_e| + |(S_{e_1} \cup S_{e_2}) \setminus S_e| = |S_{e_1} \cup S_{e_2}| \le \left\lfloor \frac{3k}{2} \right\rfloor . \end{aligned}
Thus, Step 2 accounts in total for a running time of $$O(mT^{\lfloor 3k/2 \rfloor })$$, and this dominates the other steps, as $$k \ge 2$$ since $$|S_e| = 2$$ for every edge e incident to a leaf of $${\mathcal {B}}$$. $$\square$$
Having presented the core dynamic program, we now turn to the surrounding problems:
Finding an optimal branch decomposition If $${\text {bw}}(G) \le k$$, then there is a linear-time algorithm computing a branch decomposition of width $$\le k$$ (Bodlaender & Thilikos, 1997).
Computing an optimal timetable By additional bookkeeping, we can not only compute the minimum weighted slack, but also a periodic timetable realizing this slack.
2-edge-connectedness It is clear from the description of $$T$$ -PESP-Optimality that the problem can be solved on each weakly connected component of G individually. Moreover, if one of these components is not 2-edge-connected, then the optimal periodic slack of any bridge will be zero (Borndörfer et al., 2019, §3.2). Hence, one can safely assume w.l.o.g. that G is 2-edge-connected. Note that this assumption implies $${\text {bw}}(G) \ge 2$$.
Fixing If $$\pi$$ is a feasible periodic timetable and $$d \in {\mathbb {R}}$$, then the timetable $$\pi '$$ defined by $$\pi '_i := [\pi _i + d]_T$$ for all $$i \in V(G)$$ is feasible as well and produces the same periodic slack. In particular, in Algorithm 5.3, for all $$e \in E$$, one can choose an event $$i \in S_e$$ and then fix $$\pi _i = 0$$. Thus, Algorithm 5.3 can be adapted to run in $$O(mT^{\lfloor 3k/2 \rfloor -1})$$ time.
As a consequence, in conjunction with Lemma 2.7 and Theorem 4.14, we obtain
### Theorem 5.6
For $$k \in {\mathbb {N}}$$, there is an $$O(mT^{\lfloor 3k/2 \rfloor -1})$$ algorithm solving $$T$$ -PESP-Optimality on networks G with m activities and $${\text {bw}}(G) \le k$$. In particular, if $$k \ge 2$$ is fixed, then the problems $$T$$ -PESP-Optimality, $$T$$ -PESP-Feasibility and $$T$$ -RPESP-Optimality are all weakly NP-complete.
#### 5.1.3 Bounding treewidth and carvingwidth
It is also possible to construct a tree-decomposition-based dynamic program for bounded treewidth. This tree decomposition version is expected to be asymptotically faster, but requires potentially more space. We refer to the Appendix for details.
We omit a carving-decomposition-based algorithm. Bounding the carvingwidth by k means that the branchwidth is bounded by 2k according to Theorem 4.16, and we can invoke Algorithm 5.3.
### 5.2 Cyclomatic number
We have already seen in Lemma 2.8 that fixing the number of events or the number of activities leads to FPT algorithms for $$T$$ -PESP-Optimality. In this subsection, we discuss fixed-parameter tractability by cyclomatic number. The cyclomatic number is a common measure for the difficulty of PESP instances, as it counts the number of integral variables used in a cycle-based mixed integer programming formulation (Borndörfer et al., 2019).
### Definition 5.7
Let G be a graph on n vertices, m edges and c weakly connected components. The cyclomatic number of G is defined as $$\mu (G) := m - n + c$$.
The following seems to be folklore:
### Lemma 5.8
Let G be a graph. Then, $${\text {tw}}(G) \le \mu (G) + 1$$.
### Proof
We give a proof by induction on $$\mu (G)$$. If $$\mu (G) = 0$$, then G is a forest and therefore $${\text {tw}}(G) = 1$$.
Now, let G be a graph with $$\mu (G) > 0$$. Then, G contains a cycle and hence an edge e such that $$G' := G \setminus \{e\}$$ has the same number of connected components as G, and $$\mu (G') = \mu (G) - 1$$. By induction hypothesis, $${\text {tw}}(G') \le \mu (G') + 1 = \mu (G)$$, so that we find a tree decomposition $$({\mathcal {T}}, {\mathcal {X}})$$ of $$G'$$ with maximum bag size at most $$\mu (G) + 1$$. If there is a bag containing both endpoints of e, then $$({\mathcal {T}}, {\mathcal {X}})$$ is also a valid tree decomposition for G, and so $${\text {tw}}(G) \le \mu (G)$$. Otherwise, let i be an endpoint of e and add i to each bag of $$({\mathcal {T}}, {\mathcal {X}})$$. This is a tree decomposition for G of width $$\mu (G) + 1$$, so that $${\text {tw}}(G) \le \mu (G) + 1$$. $$\square$$
While it is true that treewidth and branchwidth can be bounded in terms of each other (see Theorem 4.12), this does not hold for treewidth and cyclomatic number:
### Lemma 5.9
For $$k \ge 2$$, there is a class $${\mathcal {C}}_k$$ of simple connected graphs such that $${\text {tw}}(G) \le k$$ holds for all $$G \in {\mathcal {C}}_k$$, but for any $$N \in {\mathbb {N}}$$, there is a graph $$G \in {\mathcal {C}}$$ with $$\mu (G) \ge N$$.
### Proof
Let $${\mathcal {C}}_{k}$$ be the class of graphs G built from a finite disjoint union of cliques of size k with vertex sets $$V_1, \dots , V_r$$ together with one additional vertex v joined to each vertex from each clique. Let $${\mathcal {T}}$$ be a path on the vertices $$\{1, \dots , r\}$$, and set $$X_i := V_i \cup \{v\}$$, $$i = 1, \dots , r$$. Then, $$({\mathcal {T}}, {\mathcal {X}})$$ is a tree decomposition of G and, as $$|X_i| = k+1$$, we have $${\text {tw}}(G) \le k$$. The cyclomatic number of G is given by
\begin{aligned} \mu (G) = \left( r \cdot \frac{k(k-1)}{2} + rk\right) - (rk + 1) + 1 = r \cdot \frac{k(k-1)}{2}, \end{aligned}
and for $$k \ge 2$$, this goes to infinity as $$r \rightarrow \infty$$. $$\square$$
### Lemma 5.10
On networks where no vertex has degree 2, $$T$$ -PESP-Optimality is FPT when parameterized by the cyclomatic number.
### Proof
Let $$(G, T, \ell , u, w)$$ be a $$T$$ -PESP-Optimality instance. We can safely remove all $$i \in V(G)$$ with $$\deg (i) = 1$$, as in any optimal solution, the incident activity must have periodic slack 0. Hence, we can assume that G has minimum degree 3. By the Handshaking lemma,
\begin{aligned} 2m = \sum _{i \in V(G)} \deg (i) \ge 3n, \end{aligned}
and hence
\begin{aligned} \mu = m - n + c \ge \frac{n}{2} + 1, \end{aligned}
so that $$n \le 2\mu - 2$$. This means that fixing $$\mu$$ provides a fixed bound on the number n of events, and we conclude by Lemma 2.8. $$\square$$
Contracting vertices of degree 2 is a preprocessing technique that is applied to large PESP instances, for example, in Goerigk and Liebchen (2017) and Borndörfer et al. (2019).
### Theorem 5.11
$$T$$ -PESP-Feasibility is FPT when parameterized by the cyclomatic number.
### Proof
Let $$(G, T, \ell , u)$$ be a $$T$$ -PESP-Feasibility instance. Remove all events of degree 1 from G, as this does neither affect feasibility nor alter the cyclomatic number. Now, all degree 2 vertices of G are arranged on (undirected) paths between two vertices of degree $$\ge 3$$. Consider such a path from s to t with $$\deg (s), \deg (t) \ge 3$$, forward activities $$a_1, \dots , a_r$$ and backward activities $$b_1, \dots , b_s$$. Delete all intermediate vertices between s and t and insert a single activity a from s to t with
\begin{aligned} \ell _a := \sum _{i=1}^r \ell _{a_i} - \sum _{j=1}^s u_{b_j} \quad \text { and } \quad u_a := \sum _{i=1}^r u_{a_i} - \sum _{j=1}^s \ell _{b_j}. \end{aligned}
Clearly, if x is a feasible tension with $$\ell _{a_i} \le x \le u_{a_i}$$ and $$\ell _{b_j} \le x_{b_j} \le u_{b_j}$$ for all i and j, then also $$\ell _a \le x_a \le u_a$$ with $$x_a := \sum _{i=1}^r x_{a_i} - \sum _{j=1}^s x_{b_j}$$. Conversely, any $$x_a$$ with $$\ell _a \le x_a \le u_a$$ can be split into feasible tensions on all $$a_i$$ and $$b_j$$. Thus, this transformation preserves feasibility. Moreover, contracting vertices of degree 2 does not change the cyclomatic number, so that we can assume that G has minimum degree 3. Invoke Lemma 5.10. $$\square$$
### Theorem 5.12
$$T$$ -PESP-Optimality is in XP when parameterized by the cyclomatic number.
### Proof
Let F be a spanning forest of G and let $$\gamma _1, \dots , \gamma _\mu$$ be its fundamental cycles, seen as incidence vectors $$\{-1, 0, 1\}^{A(G)}$$. Then (e.g., Nachtigall, 1998), $$x \in {\mathbb {R}}^{A(G)}$$ is a feasible periodic tension if and only if
\begin{aligned} \ell \le x \le u \quad \text { and } \quad \forall i \in \{1, \dots , \mu \}: \gamma _i^t x \equiv 0 \mod T. \end{aligned}
Decomposing $$\gamma _i = \gamma _{i,+} - \gamma _{i, -}$$ into positive resp. negative part $$\gamma _{i,+}, \gamma _{i, -} \in \{0, 1\}^{A(G)}$$, the modulo constraints are equivalent to
\begin{aligned} \forall i \in&\{1, \dots , \mu \}:\\&\gamma _i^t x = Tz_i,\\&\left\lceil \frac{\gamma _{i,+}^t \ell - \gamma _{i, -}^t u }{T} \right\rceil \le z_i \le \left\lfloor \frac{\gamma _{i,+}^t u - \gamma _{i, -}^t \ell }{T} \right\rfloor , \\&z_i \in {\mathbb {Z}}. \end{aligned}
These are the so-called cycle inequalities (Odijk, 1994). In particular, for each i, one has to check at most
\begin{aligned}&\left\lfloor \frac{\gamma _{i,+}^t u - \gamma _{i, -}^t \ell }{T} \right\rfloor - \left\lceil \frac{\gamma _{i,+}^t \ell - \gamma _{i, -}^t u }{T} \right\rceil + 1 \\&\quad \le \frac{(\gamma _{i,+} + \gamma _{i, -})^t (u - \ell )}{T} + 1 \end{aligned}
values for $$z_i$$. Since, as in the proof of Theorem 2.6, we can assume w.l.o.g. that $$u - \ell < T$$, we have the estimate
\begin{aligned} z_i \le |\{a \in A(G): \gamma _{i,a} \ne 0\}| + 1 \le n + 1, \end{aligned}
as the $$\gamma _i$$ are simple cycles and hence contain at most n vertices.
The description of the polynomial-time algorithm is as follows: Enumerate all $$O((n+1)^\mu )$$ integral vectors
$$(z_1, \dots , z_\mu )$$ satisfying the cycle inequalities and solve the problem
\begin{aligned}&\text {Minimize}\quad&w^t x \\&\text {subject to} \quad&\ell \le x&\le u, \\&\gamma _i^t x&= Tz_i, \quad&\text { for all } i \in \{1, \dots , \mu \}. \end{aligned}
This is a minimum cost network tension problem and can be solved in polynomial time by network flow approaches (Hadjiat & Maurras, 1997; Nachtigall & Opitz, 2008). Alternatively, the above minimization problem can be solved by linear programming. $$\square$$
It remains open whether $$T$$ -PESP-Optimality can be solved with an FPT w.r.t. the cyclomatic number.
### 5.3 Cyclomatic number and diameter
The main obstacle for a FPT for $$T$$ -PESP-Optimality is that the cyclomatic number does not bound the number of vertices. However, if, for example, one additionally fixes the diameter, then also $$T$$ -PESP-Optimality becomes fixed-parameter tractable:
### Corollary 5.13
$$T$$ -PESP-Optimality is FPT when parameterized by cyclomatic number and diameter.
### Proof
We adapt the proof of Theorem 5.12. In our final estimate of $$z_i$$, the number of activities contained in $$\gamma _i$$ can be bounded from above by 2d if d denotes the diameter of the graph. $$\square$$
A by-product of Corollary 5.13 is that $$T$$ -PESP-Optimality is also fixed-parameter tractable when parameterized by cyclomatic number and vertex cover number.
Finally, we want to remark that fixing both the cyclomatic number and the diameter does not bound the number of vertices:
### Lemma 5.14
For any $$k \in {\mathbb {N}}$$, there is an infinite class of simple connected graphs of diameter at most 2 and cyclomatic number at most k.
### Proof
For $$r \ge k$$, let G be a star graph on r leaves. Connect k distinct pairs of leaves by an edge. Then, $$\mu (G) = (k + r) - (r + 1) + 1 = k$$ and G has diameter 2. $$\square$$
## 6 Structure of realistic event-activity networks
In this section, we discuss the size of the so far discussed graph parameters on realistic periodic timetabling instances. We consider networks with a special structure based on line networks. This structure is the direct outcome of a typical modeling process (Nachtigall, 1998; Liebchen & Möhring, 2007; Schöbel, 2017; Pätzold et al., 2017). For example, the railway networks in the benchmarking library PESPlib (Goerigk, 2012) are found as subgraphs of networks with this structure. For this type of networks, we give lower and upper bounds on the branchwidth in terms of the underlying line network. We use this theoretical result to compute bounds on the branchwidth of the smallest PESPlib instance R1L1.
### 6.1 Line-based event-activity networks
Public transportation systems of cities, but also railway services, are typically organized in lines.
### Definition 6.1
A line network $$(N, {\mathcal {L}})$$ is a directed multigraph N, together with a set $${\mathcal {L}}$$ of directed walks on G such that the arc set A(N) is the disjoint union of $$A(\ell )$$ over all lines $$\ell \in {\mathcal {L}}$$.
Line networks reflect the maps which public transport companies offer for passenger information, displaying stations and lines. Depending on the precise application, lines may also constitute non-simple paths or contain cycles (e.g., London’s Circle Line or Berlin’s Ringbahn). In the context of line planning, we interpret a line network as a frequency-expanded line plan, i.e., some lines might have the same vertex sequence. Given a line network $$(N, {\mathcal {L}})$$, construct an event-activity network G as follows:
1. 1.
For each line $$\ell \in {\mathcal {L}}$$ and each arc $$ij \in A(\ell )$$, create a departure event $$(i,\ell ,\text {dep})$$ and an arrival event $$(j,\ell ,\text {arr})$$, and connect these by a driving activity $$((i,\ell ,\text {dep}), (j,\ell ,\text {arr}))$$.
2. 2.
For each vertex $$i \in V(N)$$ and each line $$\ell \in \mathcal L$$, add a dwelling activity $$((i,\ell ,\text {arr}), (i,\ell ,\text {dep}))$$ if both events exist.
3. 3.
For each vertex $$i \in V(N)$$ and each pair $$(\ell _1, \ell _2)$$ of distinct lines, add a transfer activity
$$((i,\ell _1,\text {arr}), (i,\ell _2,\text {dep}))$$ if both events exist.
### Definition 6.2
An event-activity network G is line-based if it arises from a line network $$(N, {\mathcal {L}})$$ by the above construction. Shortly, G is based on $$(N, {\mathcal {L}})$$.
Denote by $$\deg ^+(i)$$ resp. $$\deg ^-(i)$$ the number of outgoing resp. ingoing arcs at i. We summarize some straightforward structural properties of line-based networks in the following lemma:
### Lemma 6.3
Let G be based on $$(N, {\mathcal {L}})$$.
1. 1.
G is bipartite, the parts being the departure and arrival events, respectively.
2. 2.
Every departure event has a unique outgoing activity, and every arrival event has a unique ingoing activity. In both cases, these are driving activities.
3. 3.
The driving activities in G form a perfect matching in G.
4. 4.
Deleting the driving activities from G and undirecting the arcs results in the disjoint union of complete bipartite graphs $$K_{\deg ^+(i), \deg ^-(i)}$$ over $$i \in V(N)$$.
### Remark 6.4
Our definition of line-based networks disregards headway activities, which are typically employed to model different line frequencies or safety distances. However, the railway instances found in the PESPlib do not feature this particular kind of activities. (See §6.3 for the analysis of R1L1.)
### 6.2 Branchwidth of line-based networks
To give bounds on the branchwidth of line-based event-activity networks, we start with a well-known result on the branchwidth of minors:
### Theorem 6.5
(Robertson & Seymour, 1991, 4.1) If G is a graph and H is a minor of G, then $${\text {bw}}(H) \le {\text {bw}}(G)$$.
By Lemma 6.3, this implies that if G is based on $$(N, {\mathcal {L}})$$, then
\begin{aligned} {\text {bw}}(G) \ge \max _{i \in V(N)} {\text {bw}}(K_{\deg ^+(i), \deg ^-(i)}). \end{aligned}
As we did not manage to find a reference in the literature for the branchwidth of complete bipartite graphs, we give a proof here:
### Lemma 6.6
The complete bipartite graph $$K_{p,q}$$ has branchwidth $$\min (p,q)$$.
### Proof
Assume $$p \le q$$. Let P and Q denote the two parts, $$|P| = p$$, $$|Q| = q$$. The vertex separator associated with any neighborhood $$\delta (w)$$ for $$w \in Q$$ is given by P. Moreover, if $$E \subsetneq \delta (w)$$ is a proper subset, then the cardinality of the corresponding vertex separator is $$|E| + 1 \le p$$. Take any ternary tree with q leaves labeled with the vertices in Q. Then, replace each leaf w by any ternary tree with p leaves, labeled by the edges in $$\delta (w)$$. The result is a branch decomposition of $$K_{p,q}$$ of width p. This shows $${\text {bw}}(K_{p,q}) \le p$$.
To show that $${\text {bw}}(K_{p,q}) \ge p$$, we make use of tangles. That is, if we can find a collection $${\mathcal {T}}$$ of subsets of $$E(K_{p,q})$$ such that
1. 1.
for each $$A \in {\mathcal {T}}$$, the size of the corresponding vertex separator is at most $$p-1$$,
2. 2.
for each $$A \subseteq E(K_{p,q})$$ inducing a separator of size $$\le p-1$$, either A or its complement is in $${\mathcal {T}}$$,
3. 3.
for any three sets $$A_1, A_2, A_3 \in {\mathcal {T}}$$, their union is not $$E(K_{p,q})$$,
4. 4.
for each $$A \in {\mathcal {T}}$$, the subgraph induced by A does not contain all vertices of $$K_{p,q}$$,
then $${\text {bw}}(K_{p,q}) \ge p$$ or $$p \le 2$$ holds (Robertson & Seymour, 1991, 4.3). Clearly, $${\text {bw}}(K_{1,q}) = 1$$, as these are star graphs, and $${\text {bw}}(K_{2,q}) = 2$$, as these are series–parallel and contain cycles. Hence, suppose $$p \ge 3$$ and define
\begin{aligned} {\mathcal {T}} := \{A \subseteq E(K_{p,q}) \mid \,&\text {sep}(A) \le p-1,\\&|V(K_{p,q}[A]) \cap P| \le p-1,\\&|V(K_{p,q}[A]) \cap Q| \le p-1 \}, \end{aligned}
where $$\text {sep}(A)$$ denotes the cardinality of the vertex separator associated with A, and $$K_{p,q}[A]$$ is the subgraph induced of $$K_{p,q}$$ by A. We check the above properties:
1. 1.
This is clear.
2. 2.
Let $$A \subseteq E(K_{p,q})$$ with $$\text {sep}(A) \le p-1$$ and suppose that $$K_{p,q}[A]$$ contains all vertices of P. Then, there must be a vertex $$v \in P$$ incident to some edge of A, but not contained in the separator. Hence, A must contain $$\delta (v)$$, and every vertex $$w \in Q$$ is incident to the edge $$vw \in A$$. Similarly, if A contains at least p vertices from Q, then it contains $$\delta (w)$$ for at least $$|V(K_{p,q}[A]) \cap Q| - (p-1)$$ vertices $$w \in Q$$. This means if $$\text {sep}(A) \le p-1$$ and $$A \notin {\mathcal {T}}$$, then A contains $$\delta (v)$$ for some $$v \in P$$ and $$\delta (w)$$ for at least $$q - (p - 1)$$ vertices $$w \in Q$$. The subgraph induced by the complement of A hence does not contain v, and it also does not contain at least $$q - (p-1)$$ vertices of Q. Therefore, the complement of A is in $${\mathcal {T}}$$.
3. 3.
It follows by the argument in 2. that for each $$A \in {\mathcal {T}}$$, the vertex separator is given by $$V(K_{p,q}[A])$$. Now $$K_{p,q}[A]$$ is a simple bipartite graph on at most $$p-1$$ vertices. It follows that $$|A| \le (p-1)^2/4$$. Now, if $$A_1, A_2, A_3 \in {\mathcal {T}}$$, the cardinality of their union is at most $$3(p-1)^2/4 < p^2 \le pq = |E(K_{p,q})|$$.
4. 4.
This follows as $$|V(K_{p,q}[A])| \le 2p-2 < 2p \le p+q = |V(K_{p,q})|$$.
Hence, we conclude $${\text {bw}}(K_{p,q}) = p$$. $$\square$$
### Theorem 6.7
Let G be a line-based event activity network, based on the line network $$(N, {\mathcal {L}})$$. Then,
\begin{aligned} \max _{i \in V(N)} \min (\deg ^{+}(i), \deg ^-(i)) \le {\text {bw}}(G) \le {\text {cw}}(N), \end{aligned}
and both bounds are sharp.
Recall from Sect. 4.1.4 that $${\text {cw}}(N)$$ denotes the carvingwidth of N, and that
\begin{aligned} {\text {cw}}(N) \ge \max _{i \in V(N)} (\deg ^+(i) + \deg ^-(i) ). \end{aligned}
Speaking more intuitively, the carvingwidth is hence bounded by the maximum number of lines departing and arriving at a stop of the line network. In practice, line networks are often planar, and the carvingwidth of planar graphs can be computed in polynomial time by the ratcatcher algorithm (Seymour & Thomas, 1994).
### Proof of Theorem 6.7
The lower bound follows from Lemma 6.3, Theorem 6.5, and Lemma 6.6. For $$r \in {\mathbb {N}}$$, let $$N_r$$ be a graph on the vertex set $$\{0, 1, \dots , 2r\}$$, and arcs (i, 0) for $$i \in \{1, \dots , r\}$$ and (0, i) for $$i \in \{r+1, \dots , 2r\}$$. The lines are given by $${\mathcal {L}} := \{(i,0,i+r) \mid i \in \{1, \dots , r\}\}$$. The resulting line-based event-activity network $$G_r$$ has the structure of a complete bipartite graph $$K_{r,r}$$ plus 2r activities connecting the $$K_{r,r}$$ with events of degree 1 each. It follows that
\begin{aligned} {\text {bw}}(G_r) = r = \max _{i \in V(N_r)} \min (\deg ^+(i), \deg ^-(i)). \end{aligned}
As $$N_r$$ is a star graph on 2r rays, its carvingwidth is easily determined to be 2r. Hence, we found a family of graphs for which the lower bound is sharp, and the upper bound is larger than the lower bound.
Concerning the upper bound, we first partition the activities of G: For each $$i \in V(N)$$, let $$A_i$$ denote the set of all activities incident to some departure event at i. Then, $$\{A_i \mid i \in V(N)\}$$ partitions A(G) because of the bipartite structure of G. For $$i \in V(N)$$, let $$({\mathcal {B}}_i, \varphi _i)$$ be a branch decomposition of the subgraph of G induced by $$A_i$$. Add a root $$b_i$$ to each $${\mathcal {B}}_i$$ by subdividing an arbitrary edge. Let $$({\mathcal {C}}, \psi )$$ be an optimal carving decomposition of N. Attach to every leaf v of $${\mathcal {C}}$$ the tree $$B_{\psi (v)}$$, identifying v with $$b_{\psi (v)}$$. This results in a branch decomposition $$({\mathcal {B}}, \varphi )$$ of G.
Now, let $$e \in E({\mathcal {B}})$$ and let $$A(G) = A_e^1 \overset{.}{\cup } A_e^2$$ be the induced partition. If $$e \in E({\mathcal {B}}_i)$$, then one of $$A_e^1$$, $$A_e^2$$ is contained in $$A_i$$, so that the vertex separator $$S_e$$ has size at most $$\deg ^+(i) + \deg ^-(i)$$: $$S_e$$ contains at most all $$\deg ^-(i)$$ arrival events at i, and every other vertex in $$S_e$$ must be either a departure event or the unique arrival event following a departure event. Observe that $$\deg ^+(i) + \deg ^-(i) \le {\text {cw}}(N)$$ due to Theorem 4.16. In the other case that $$e \in E({\mathcal {C}})$$, there is a subset $$W_e \subseteq V(G)$$ such that $$A_e^1 = \bigcup _{i \in W_e} A_i$$. Each vertex of the vertex separator $$S_e$$ is hence an arrival event at some $$i \in W_e$$. The set $$S_e$$ is in bijection to $$\delta (W_e)$$ by mapping $$(j, \ell , \text {arr})$$ to $$ij \in A(\ell ) \subseteq A(N)$$, where $$(i, \ell , \text {dep})$$ is the unique driving activity entering $$(j, \ell , \text {arr})$$. Hence, as $${\mathcal {C}}$$ was chosen to be optimal, $$|S_e| = |\delta (W_e)| \le cw(N)$$.
Finally, we show that the upper bound is sharp: For $$r \in {\mathbb {N}}$$, let $$N_r'$$ be a directed simple cycle on r vertices. Then, $${\text {cw}}(N_r') = 2$$. The event-activity network $$G_r'$$ based on $$N_r'$$ is then a directed simple cycle on 2r vertices and has branchwidth 2. On the other hand, the lower is not 2, as $$\max _{i \in V(N_r')} \min (\deg ^+(i), \deg ^-(i)) = 1$$. $$\square$$
### 6.3 Parameters of R1L1
#### 6.3.1 An upper bound on branchwidth
The network G in its original shape does not satisfy the properties of Lemma 6.3. However, with small modifications, we can find a line network N such that G is a subgraph of an event-activity network based on N.
Transfer activities $$a \in A(G)$$ are in practice typically recognized by a large span $$u_a - \ell _a$$. As the period time is $$T = 60$$, we let $$A_t := \{ij \in A(G) \mid u_a - \ell _a \ge 59\}$$. We call any vertex i with $$ij \in A_t$$ for some j an arrival event, and analogously any vertex j with $$ij \in A_t$$ for some i is called a departure event. This is well defined and extends to a bipartition of G into arrival resp. departure events.
Now, we interpret any activity from a departure to an arrival is called a driving activity. The set of driving activities is not quite a perfect matching in G: There are two arrival events (84 and 256) with two ingoing driving activities, and two arrival events (53 and 177) with no ingoing driving activity. This is due to four mysterious activities with lower and upper bound 0 breaking the structure at this particular spot; see Fig. 7. However, the vertices 53 to 84 and 177 to 256 can be removed from G by sequentially deleting vertices of degree 1. Since the remaining network has branchwidth at least two, removing vertices of degree 1 has no effect on branchwidth:
### Lemma 6.8
Let G be a connected graph, and let $$v \in V(G)$$ be a vertex of degree 1. If $${\text {bw}}(G \setminus \{v\}) \ge 2$$, then $${\text {bw}}(G) = {\text {bw}}(G\setminus \{v\})$$.
### Proof
By Theorem 6.5, $${\text {bw}}(G \setminus \{v\}) \le {\text {bw}}(G)$$. For the reverse inequality, consider a branch decomposition
$$({\mathcal {B}}, \varphi )$$ of $$G \setminus \{v\}$$. Since G is connected, v is adjacent to some vertex w of degree $$\ge 2$$. Choose an edge $$e \ne \{v, w\}$$ incident with w, and replace the leaf of $${\mathcal {B}}$$ corresponding to e by a node with the two children e and $$\{v, w\}$$. This is a branch decomposition of G. The size of the vertex separator corresponding to $$\{v, w\}$$ is 1, the size of the separator w.r.t. e is at most 2, and all other vertex separators remain unchanged. This shows $${\text {bw}}(G) \le {\text {bw}}(G \setminus \{v\})$$. $$\square$$
With this adjustment, G has 3552 vertices and 6273 arcs. G satisfies now properties 1–3 of Lemma 6.3. Deleting the driving activities yields a disjoint union of bipartite graphs $$G_i$$, but they are not all complete. We define N now as the network obtained from G by contracting all these bipartite graphs $$G_i$$ to a single vertex i. Then, G is a subgraph of an event-activity network based on N, choosing, for example, $${\mathcal {L}}$$ as the set of all single-arc walks. By Theorems 6.5 and 6.7, $${\text {bw}}(G) \le {\text {cw}}(N)$$.
The network N obtained in this way is unfortunately not planar. The maximum degree in N is 62, so that $${\text {cw}}(N) \ge 62$$. To compute an upper bound, we first preprocess N by removing vertices of degree 2, as this does not alter carvingwidth (Belmonte et al., 2013, Lemma 6). We use then the Kuratowski subgraph detection algorithm implemented in the Python package networkx (Hagberg et al., 2008) to recursively remove (multi-)arcs from N until the graph becomes planar. We prefer arcs of low multiplicity, and a sequence of 20 removals of simple arcs finally yields a planar graph $$N'$$. We implemented the ratcatcher method of Robertson and Seymour to compute $${\text {cw}}(N') = 62$$ and an optimal carving decomposition. As $$V(N') = V(N)$$, this is also a carving decomposition of N, but the width increases to 70. We hence conclude $${\text {cw}}(N) \in [62, 70]$$ and $${\text {bw}}(G) \le 70$$.
#### 6.3.2 A lower bound on branchwidth
Since G is only realized as a subgraph of an event-activity network based on N, we cannot invoke Theorem 6.7 directly. Of course, it remains true that $${\text {bw}}(G)$$ is at least the branchwidth of the (disconnected) subgraph $$G'$$ obtained by deleting the driving activities. $$G'$$ is reasonably small, but there seems to be no freely available software for exact branchwidth computations. However, there are treewidth codes, and we use the algorithm by Tamaki (2019), which has been implemented for the PACE 2017 challenge on exact treewidth computations (Dell et al., 2018). It turns out that
$${\text {tw}}(G') = 20$$, hence $${\text {bw}}(G) \ge 14$$.
The largest of the components of $$G'$$ is a bipartite graph with maximum part size 31. If this component were complete, then Theorem 6.7 would have predicted $${\text {bw}}(G) \ge 31$$.
To obtain a better bound on $${\text {bw}}(G)$$, we use balanced vertex separators:
### Lemma 6.9
(Robertson & Seymour, 1995, 3.1) Let G be a graph. Then, there is a vertex separator S with $$|S| \le {\text {bw}}(G)$$ and
\begin{aligned} \max (|V^1|, |V^2|) \le \frac{2}{3} |V(G)| - \frac{1}{2} |S|, \end{aligned}
where $$V(G) = V^1 \overset{.}{\cup } V^2 \overset{.}{\cup } S$$, and no vertex in $$V^1$$ is adjacent to a vertex in $$V^2$$ and vice versa.
We now compute a minimum cardinality vertex separator subject to the balance constraint of Lemma 6.9 by plugging in a straightforward integer program into the CPLEXFootnote 1 12.10 solver. We do not use the full network G as input, but take a smaller network that is obtained after standard preprocessing for $$T$$ -PESP-Optimality instances (Borndörfer et al., 2019, §3.2). This network is a minor of G, so that we obtain a valid bound on the branchwidth. CPLEX finds a vertex separator of cardinality 58 and is able to solve the instance to optimality. We conclude $${\text {bw}}(G) \ge 58$$.
#### 6.3.3 Treewidth and carvingwidth
Since $${\text {bw}}(G) \in [58, 70]$$, we obtain by Theorem 4.12 that $${\text {tw}}(G) \in [58, 104]$$. As the maximum degree in R1L1 is 26, $${\text {cw}}(G) \in [29, 1820]$$ by Theorem 4.16. Determining the exact treewidth of G turns out to be computationally infeasible. We instead use TCS-Meiji (Tamaki, 2019) and FlowCutter (Hamann & Strasser, 2018), the best two submissions of the PACE 2017 challenge on heuristic treewidth computations (Dell et al., 2018), with different random seeds to obtain a better upper bound on $${\text {tw}}(G)$$. The best bound we could find was $${\text {tw}}(G) \le 97$$.
#### 6.3.4 Practical implications
The instance R1L1 has a period time of $$T = 60$$. It becomes clear from Table 1 that none of the algorithms presented in Sect. 5 can be applied for solving R1L1 in practice. For example, storing $$T^{{\text {bw}}(G) - 1} \ge 60^{57}$$ table entries for the branch-decomposition-based algorithm 5.3 as 32-bit integers would require roughly $$9 \cdot 10^{101}$$ bytes of space.
## 7 Conclusion
The results of this paper underline that PESP is a notoriously hard problem. Although there are several primal heuristics available, the promising global approaches fail to compute provably optimal solutions: For example, mixed-integer programming formulations suffer from weak linear programming relaxations and transformations to Boolean satisfiability problems scale badly. It fits into this picture that exploiting structural parameters such as treewidth does not lead to a polynomial-time algorithm unless P $$\ne$$ NP, and to no pseudo-fixed-parameter algorithm unless FPT $$\ne$$ W[1]. Moreover, the dynamic programs of Sect. 5 are only of theoretical interest. It is even unclear for tentatively large parameters as cyclomatic number and vertex cover number if $$T$$ -PESP-Optimality becomes fixed-parameter tractable.
On the positive side, it has been demonstrated in Lindner and Liebchen (2019) that balanced edge separators lead to benefits when computing lower bounds of $$T$$ -PESP-Optimality instances. We think that this should also carry over to vertex separators, and that good heuristic tree or branch decompositions may be useful as a source for separators in order to tackle PESP by a divide-and-conquer approach. | 24,951 | 79,345 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2023-40 | latest | en | 0.917571 |
https://www.calculator.net/cd-calculator.html?cstartingprinciple=15%2C000&cinterestrate=4&ccompound=annually&cyears=5&cmonths=0&ctaxtrate=0&printit=0&x=Calculate | 1,720,832,248,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514459.28/warc/CC-MAIN-20240712224556-20240713014556-00279.warc.gz | 593,489,409 | 11,417 | # CD Calculator
Initial deposit Interest rate Compound annually (APY) semiannually quarterly monthly (APR) continuously Deposit length years months Marginal tax rate
## Results
End balance \$18,249.79 Total interest \$3,249.79
## Accumulation Schedule
YearDepositInterestEnding balance
1\$15,000.00\$600.00\$15,600.00
2\$0.00\$624.00\$16,224.00
3\$0.00\$648.96\$16,872.96
4\$0.00\$674.92\$17,547.88
5\$0.00\$701.92\$18,249.79
RelatedInvestment Calculator | Interest Calculator
### What is a Certificate of Deposit?
A certificate of deposit is an agreement to deposit money for a fixed period that will pay interest. Common term lengths range from three months to five years. The lengthier the term, the higher the exposure to interest rate risk. Generally, the larger the initial deposit, or the longer the investment period, the higher the interest rate. As a type of investment, CDs fall on the low-risk, low-return end of the spectrum. Historically, interest rates of CDs tend to be higher than rates of savings accounts and money markets, but much lower than the historical average return rate of the equity market. There are also different types of CDs with varying rates of interest or rates linked to indexes of various kinds, but the calculator can only do calculations based on fixed-rate CDs.
The gains from CDs are taxable as income in the U.S. unless they are in accounts that are tax-deferred or tax-free, such as an IRA or Roth IRA. For more information about or to do calculations involving a traditional IRA or Roth IRA, please visit the IRA Calculator or Roth IRA Calculator.
CDs are called "certificates of deposit" because before electronic transfers were invented, buyers of CDs were issued certificates in exchange for their deposits as a way for financial institutions to keep track of buyers of their CDs. Receiving actual certificates for making deposits is no longer practiced today, as transactions are done electronically.
### FDIC-Backed
One of the defining characteristics of CDs in the U.S. is that they are protected by the Federal Deposit Insurance Corporation (FDIC). CDs that originate from FDIC-insured banks are insured for up to \$250,000, meaning that if banks fail, up to \$250,000 of each depositor's funds is guaranteed to be safe. Anyone who wishes to deposit more than the \$250,000 limit and wants all of it to be FDIC-insured can simply buy CDs from other FDIC-insured banks. Due to this insurance, there are few lower-risk investments. Similarly, credit unions are covered by insurance from the National Credit Union Administration (NCUA insurance), which provides essentially the same insurance coverage on deposits as the FDIC.
### Where and How to Purchase CDs
CDs are typically offered by many financial institutions (including the largest banks) as fixed-income investments. Different banks offer different interest rates on CDs, so it is important to first shop around and compare maturity periods of CDs, especially their annual percentage yields (APY). This ultimately determines how much interest is received. The process of buying CDs is straightforward; an initial deposit will be required, along with the desired term. CDs tend to have various minimum deposit requirements. Brokers can also charge fees for CDs purchased through them.
"Buying" a CD is effectively lending money to the seller of the CD. Financial institutions use the funds from sold CDs to re-lend (and profit from the difference), hold in their reserves, spend for their operations, or take care of other miscellaneous expenses. Along with the federal funds rate, all of these factors play a part in determining the interest rates that each financial institution will pay on their CDs.
### History of CDs
Although they weren't called CDs then, a financial concept similar to that of a modern CD was first used by European banks in the 1600s. These banks gave a receipt to account holders for the funds they deposited, which they lent to merchants. However, to ensure that account holders did not withdraw their funds while they were lent out, the banks began to pay interest for the use of their money for a designated period of time. This sort of financial transaction is essentially how a modern CD operates.
A major turning point for CDs happened in the early twentieth century after the stock market crash of 1929, which was partly due to unregulated banks that didn't have reserve requirements. In response, the FDIC was established to regulate banks and give investors (such as CD holders) assurance that the government would protect their assets up to a limit.
Historically, rates of CD yields have varied greatly. During the high-inflation years of the late 1970s and 1980s, CDs had return rates of almost 20%. After that the CD rates declined steadily. In late 2007, just before the economy spiraled downward, they were at around 4%. In comparison, the average one-year CD yield is below 1% in 2021. It gradually increased in 2022, reaching more than 5% in 2023 and 2024 due to rising inflation. In the U.S., the Federal Reserve, which controls federal funds rates, calibrates them accordingly based on the economic climate.
### How to Use CDs
CDs are effective financial instruments when it comes to protecting savings, building short-term wealth, and ensuring returns without risk. With these key benefits in mind, it is possible to capitalize on CDs by using them to:
• supplement diversified portfolios to reduce total risk exposure. This can come in handy as retirees get closer to their retirement date and require a more guaranteed return to ensure they have savings in retirement to live off of.
• act as a short-term (5 years or less) place to put extra money that isn't needed or isn't required until a set future date. This can come in handy when saving for a down payment for a home or car several years in the future.
• estimate future returns accurately because most CDs have fixed rates. The result of this is a useful investment for people who prefer predictability.
As the maturity date for a CD approaches, CD owners have options of what to do next. In most cases, if nothing is done after the maturity date, the funds will likely be reinvested into another similar CD. If not, it is possible for buyers to notify the sellers to transfer the funds into a checking or savings account, or reinvest into a different CD.
### Withdrawing from a CD
Funds that are invested in CDs are meant to be tied up for the life of the certificate, and any early withdrawals are normally subject to a penalty (except liquid CDs). The severity of the penalty depends on the length of the CD and the issuing institution. As an aside, in certain rising interest rate environments, it can be financially beneficial to pay the early withdrawal penalty in order to reinvest the proceeds into new higher-yielding CDs or other investments.
While longer-term CDs offer higher returns, an obvious drawback to them is that the funds are locked up for longer. A CD ladder is a common strategy employed by investors that attempts to circumvent this drawback by using multiple CDs. Instead of renewing just one CD with a specific amount, the CD is split up into multiple amounts for multiple CDs in a setup that allows them to mature at staggered intervals. For example, instead of investing all funds into a 3-year CD, the funds are used to invest in 3 different CDs at the same time with terms of 1, 2, and 3 years. As one matures, making principal and earnings available, proceeds can be optionally reinvested into a new CD or withdrawal. CD laddering can be beneficial when more flexibility is required, by giving a person access to previously invested funds at more frequent intervals, or the ability to purchase new CDs at higher rates if interest rates go up.
### APY vs. APR
It is important to make the distinction between annual percentage yield (APY) and annual percentage rate (APR). Banks tend to use APR for debt-related accounts such as mortgages, credit cards, and car loans, whereas APY is often related to interest-accruing accounts such as CDs and money market investments. APY denotes the amount of interest earned with compound interest accounted for in an entire year, while APR is the annualized representation of the monthly interest rate. APY is typically the more accurate representation of effective net gains or losses, and CDs are often advertised in APY rates.
### Compounding Frequency
The calculator contains options for different compounding frequencies. As a rule of thumb, the more frequently compounding occurs, the greater the return. To understand the differences between compounding frequencies or to do calculations involving them, please use our Compound Interest Calculator.
### Types of CDs
• Traditional CD—Investors receive fixed interest rates over a specified period of time. Money can only be withdrawn without penalty after maturity, and there are also options to roll earnings over for more terms. Traditional CDs that require initial deposits of \$100,000 or more are often referred to as "jumbo" CDs, and usually have higher interest rates.
• Bump-Up CD—Investors are allowed to "bump up" preexisting interest rates on CDs to match higher current market rates. Bump-up CDs offer the best returns for investors who hold them while interest rates increase. Compared to traditional CDs, these generally receive lower rates.
• Liquid CD—Investors can withdraw from liquid CDs without penalties, but they require maintaining a minimum balance. Interest rates are relatively lower than other types of CDs, but for the most part, still higher than savings accounts or money market investments.
• Zero-Coupon CD—Similar to zero-coupon bonds, these CDs contain no interest payments. Rather, they are reinvested in order to earn more interest. Zero-coupon CDs are bought at fractions of their par values (face value, or amount received at maturity), and generally have longer terms compared to traditional CDs, which can expose investors to considerable risk.
• Callable CD—Issuers that sell callable CDs can possibly recall them from their investors after call-protection periods expire and before they mature, resulting in the return of the initial deposit and any subsequent interest earnings. To make up for this, sellers offer higher rates for these CDs than other types.
• Brokered CD—These are different in that they are sold in brokerage accounts and not through financial institutions such as banks or credit unions. An advantage to brokered CDs is that there is exposure to a wide variety of CDs instead of just the CDs offered by individual banks.
### Alternatives to CDs
• Paying off Debt—Especially for high-interest debt, paying off existing debt is a great alternative to CDs because it is essentially a guaranteed rate of return, compared to any further investment. Comparatively, even the interest rate of a low rate loan, such as a home mortgage, is normally higher than CDs, making it financially rewarding to pay off a loan than to collect interest from CD.
• Money Market Accounts—Investors who like the security of a CD and are okay with slightly lower returns can consider money market accounts, which are certain types of FDIC-insured savings accounts that have restrictions such as limits on how funds can be withdrawn. They are generally offered by banks.
• Bonds—Similar to CDs, bonds are relatively low-risk financial instruments. Bonds are sold by the government (municipal, state, or federal) or corporate entities.
• Peer-to-Peer Lending—Peer-to-peer (P2P) lending is a fairly new form of lending that arose from advances in internet technology that enables lenders and borrowers to link up on an online platform. Peer borrowers request loans through the platform, and lenders can fund the loans they find desirable. Each P2P lending service will come with rules in order to regulate cases of default.
• Bundled Mortgages—Commonly available through mutual funds, bundled mortgages are securities that are traded in a similar manner as bonds but generally yield more than Treasury securities. Although they received a lot of negative publicity for the role they played in the 2008 financial crisis, mortgage securities have bounced back through more stringent regulations. Bundled mortgages are backed by the Government National Mortgage Association (Ginnie Mae).
Listed above are just some of the low-risk alternatives to CDs. There are much more investment options for those that can tolerate higher risk. | 2,566 | 12,536 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2024-30 | latest | en | 0.940004 |
https://what.wqidian.com/category/percentages/ | 1,708,516,759,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473472.21/warc/CC-MAIN-20240221102433-20240221132433-00422.warc.gz | 636,374,552 | 51,415 | ## What is 20 of 12000
Have you ever been asked the simple question, “What is 20 of 12,000?” It may seem like a straightforward math problem, but when put on the spot it can make even the most math-savvy individuals second guess themselves. So, what’s the answer? It’s 2,400. But don’t worry if you didn’t know right off the bat, we’ve all been there.
## What is 65 of 65
What is 65 of 65? It’s a simple question with a simple answer: 1. But it also represents completion, the end of a journey. It’s a milestone worth celebrating and reflecting on. What did you learn? How did you grow? Take a moment to appreciate your accomplishment and look forward to the next chapter.
## What is 10 of 1200.00
Attempting to solve mathematical equations can be daunting for some, but fear not! The task at hand is a simple one: what is 10 of 1200.00? It’s a relatively minuscule calculation, but a fundamental one nonetheless. So, brace yourself and let’s dive straight into it.
## What is 20 of \$80
Ah, the age-old question: what exactly is 20% of 80 dollars? It may seem like a simple math problem, but the answer is often shrouded in confusion. Luckily, the solution is quite simple: it’s sixteen dollars. But don’t let the simplicity fool you – understanding percentages can open up a whole new world of financial literacy.
## What Percent of 16 is 4
If you’re trying to figure out what percent of 16 is 4, the answer is 25%. To understand this, think of a whole pizza representing 100%. If you eat 4 slices out of 16, you have consumed 25% of the pizza. It’s important to understand basic math concepts like percentages as they come up in everyday life, from calculating tips to assessing discounts.
## What is 85 of 80
What is 85 of 80? This may seem like a mathematically impossible question. After all, how can something be more than 100%? However, in this case, it’s all about context. When we use the phrase “85 of 80,” we’re not talking about a percentage, but rather a ratio. So, what’s the answer? It’s simply 1.0625.
## What is 25 of 80
25 of 80, a simple mathematical query that can send our brains in a tizzy. It’s okay to admit that our basic arithmetic skills may have gotten a little rusty over the years. So whether you’re attempting to solve a complex problem or just trying to split the bill with friends, let’s break down this age-old question once and for all.
## What Percent of 25 is 5
If you asked someone what percent of 25 is 5, they might think it’s a tricky math problem. But in reality, it’s quite simple. 5 represents 20% of 25. This means that if you divide 5 by 25, the answer is 0.2 (or 20% as a decimal). Knowing how to calculate percentages is a valuable skill, as it comes up often in daily life, from calculating tips to determining discounts on purchases.
## What Percent of 160 is 20
If you’re wondering what percent of 160 is 20, it’s simple math. All you have to do is divide 20 by 160 and then multiply the answer by 100. The result is 12.5%, meaning 20 is 12.5% of 160. Keep this formula in mind next time you’re faced with this problem!
## What is 25 off of 60
Imagine walking into your favourite store and spotting an eye-catching item – originally priced at \$60. Now, imagine discovering that the store is offering a 25% discount on that same item. The math may seem daunting at first, but fear not! The answer to the question: ‘What is 25% off of \$60?’ is a simple \$15. So go ahead, treat yourself to that discounted item – you’ve earned it! | 884 | 3,501 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.71875 | 5 | CC-MAIN-2024-10 | latest | en | 0.931059 |
http://oxfordindex.oup.com/view/10.1093/oi/authority.20110803115334982 | 1,550,717,577,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247497858.46/warc/CC-MAIN-20190221010932-20190221032932-00057.warc.gz | 194,731,114 | 12,685 | Overview
# vector projection
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## Quick Reference
(of a vector on a vector)
Given non-zero vectors a and b, let and be directed line-segments representing a and b, and let θ be the angle between them (θ in radians, with 0≤θπ). Let C be the projection of B on the line OA. The vector projection of b on a is the vector represented by Since |OC|=|OB|cosθ, this vector projection is equal to |b|cosθ times the unit vector a/|a|. Thus the vector projection of b on a equalsThe scalar projection of b on a is equal to (a. b)/|a|, which equals |b|cosθ. It is positive when the vector projection of b on a is in the same direction as a, and negative when the vector projection is in the opposite direction to a; its absolute value gives the length of the vector projection of b on a.
Subjects: Mathematics.
##### Reference entries
Users without a subscription are not able to see the full content. Please, subscribe or login to access all content. | 247 | 1,047 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2019-09 | latest | en | 0.898381 |
https://www.azcalculator.com/calc/parts-per-trillion.php | 1,709,184,654,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474784.33/warc/CC-MAIN-20240229035411-20240229065411-00776.warc.gz | 654,282,174 | 4,065 | # Parts Per Trillion (PPT) Calculator
Posted by Dinesh on
ppt is a value that represents the concentration of one part per trillion parts (1/1,000,000,000,000) of something in water or soil. One ppt is equivalent to 1 milligram of something per liter of water or 1 milligram of something per kilogram of soil.
### Formula:
PPT = (Defective Parts / Total Parts) × 1,000,000,000,000
Defective Rate = (Defective Parts / Total Parts)
Defective Rate (%) = (Defective Parts / Total Parts) × 100 | 135 | 492 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2024-10 | latest | en | 0.833653 |
https://statsinthewild.com/2021/04/15/lets-clear-up-what-efficacy-means-when-we-talk-about-vaccine-efficacy/ | 1,685,671,745,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224648245.63/warc/CC-MAIN-20230602003804-20230602033804-00527.warc.gz | 591,683,070 | 31,434 | # Let’s clear up what efficacy means when we talk about vaccine efficacy
## What does 95% efficacy even mean?
But what does this MEAN?
In my casual observation, it seems to me that there are a lot of people who see these numbers and think, quite reasonably, that 95% effective means that 5% of the people who get the vaccine will get Covid-19. Or, if you were to get the Johnson and Johnson vaccine, there is still a 33.7% chance that you’ll get Covid-19. So, they then make the argument that if there is still about a 1 in 3 chance that you’ll get Covid even AFTER the vaccine, why even bother getting the vaccine?
Well, that’s not a correct interpretation of efficacy rate.
I will illustrate this with some simple examples.
## Example 1
Let’s say that we find 10,000 people and we inject them with a placebo. And we find another 10,000 people and we inject them with a vaccine. We follow all 20,000 for 90 days to see if they develop the disease of interest (in this case Covid-19).
Let’s say that 5,000 people who received the placebo get the disease while only 250 of the vaccinated group get the disease. In this case we have the following quantities:
Incidence rate UNvaccinated: 5,000 / 10,000 = 0.5 (or 50%)
Incidence rate vaccinated: 250 / 10,000 = 0.025 (or 2.5%)
(Note: Incidence rates are also known as “attack rates”. I didn’t know that until this morning. I’ve always just called these incidence rates).
Now using these incidence rates, we can calculate something called relative risk (RR):
RR = Incidence rate vaccinated / Incidence rate UNvaccinated = 0.025 / 0.5 = 0.05
The efficacy is then defined as 1 – RR = 1 – 0.05 = 0.95 (or 95%).
So in this scenario the vaccine was “95% effective” while 2.5% of the vaccinated group developed the disease.
(Note: You can also calculate efficacy this way and get the exact same answer:
Efficacy = (Incidence rate UNvaccinated – Incidence rate vaccinated) / Incidence rate UNvaccinated = (0.5 – 0.025) / (0.5) = 0.95
It’s exactly the same result.)
## Example 2
let’s look at a second example with the same initial set up: we find 10,000 people and we inject them with a placebo. And we find another 10,000 people and we inject them with a vaccine. We follow all 20,000 for 90 days to see if they develop the disease of interest (in this case Covid-19).
Let’s say that 100 people who received the placebo get the disease while only 5 of the vaccinated group get the disease. In this case we have the following quantities:
Incidence rate UNvaccinated: 100 / 10,000 = 0.01 (or 1%)
Incidence rate vaccinated: 5 / 10,000 = 0.0005 (or 0.05%)
Now using these incidence rates, we can calculate something called relative risk (RR):
RR = Incidence rate vaccinated / Incidence rate UNvaccinated = 0.0005 / 0.01 = 0.05
The efficacy is then defined as 1 – RR = 1 – 0.05 = 0.95 (or 95%).
So in this scenario the vaccine was ALSO “95% effective” while only 0.05% of the vaccinated group developed the disease.
## Takeaways
• In the first example given here, 2.5% of the vaccinated group developed the disease, and in the second example, 0.05% of the vaccinated group developed the disease, but in BOTH EXAMPLES the efficacy was 95%.
• Vaccine efficacy is a RELATIVE reduction in risk when compared to a placebo group.
• There are many different incidence rates that will result in a 95% efficacy.
• This is why a vaccine that has efficacy of 50% is really an incredible vaccine. It doesn’t mean that 50% of the people who get the vaccine will get the disease; it means that the relative risk has been reduced by 50%! Which is a ton!
• Someone should get on national television and explain this to the American people.
Cheers. | 976 | 3,708 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2023-23 | latest | en | 0.9448 |
http://www.boxingscene.com/weight-loss/7083.php | 1,412,147,249,000,000,000 | text/html | crawl-data/CC-MAIN-2014-41/segments/1412037663365.9/warc/CC-MAIN-20140930004103-00473-ip-10-234-18-248.ec2.internal.warc.gz | 457,772,026 | 16,387 | Bookmark Website | Free Registration | The Team The Lounge | Champions | The Wire | Schedule | Audio | Arcade | The Top Ten | Historical | Email | Video
# Why You Must Eat Before Going to Bed?
You have probably heard the old myth, that you should not eat before bed, because all the food you eat will be stored as fat. That is simply not true. People burn quite a number of calories while sleeping or resting in bed. If you know how many calories you are going to burn while sleeping, you can eat the same number of calories without getting fatter.
For bodybuilders and athletes, it is important to eat before bed. Doing so will bathe the muscles with complete amino acids for the whole night. I am going to develop a mathematical formula, so that you will know how many calories you can eat before bed, without getting fatter.
First, we need to calculate your Resting Metabolic Rate(RMR). There are many formulas, and for the sake of simplicity, let's use Owen's equation.
Owen's equation(kilograms):
RMR (male) = 879 + (10.2 * (body weight in kilograms))
RMR (female) = 795 + (7.18 * (body weight in kilograms))
Owen's equation(pounds):
RMR (male) = 879 + (4.64 * (body weight in pounds))
RMR (female) = 795 + (3.26 * (body weight in pounds))
One day has 1440 minutes. If we divide RMR by 1440, we will calculate how many calories per minute does a person burn at rest. Now, that we know this number, it is time to calculate how many calories a person will burn while sleeping. Sleeping has a MET value of 0.9, which means a calorie burn rate of 0.9 * (RMR/1440) calories per minute.
Finally we have to multiply the calorie burn rate per minute by the number of minutes that a person sleeps. We arrive at the formula: 0.9 * (RMR/1440) * number of minutes sleeping
We can further refine this formula because we haven't taken into consideration the Thermic Effect of Food. For most mixed meals, the body burns around 10% of the eaten calories to digest the food. To calculate the precise number of calories we need to eat to maintain zero calorie balance while sleeping, we will solve this equation:
Calories Eaten(CE) - TEF = 0.9 * (RMR/1440) * number of minutes sleeping (noms)
TEF = Calories Eaten/10
CE - CE/10 = 0.9 * (RMR/1440) * noms
9*CE/10 = 0.9 * (RMR/1440) * noms
0.9 * CE = 0.9 * (RMR/1440) * noms
CE = (RMR/1440) * noms
The formula turned out to be simpler than we expected. Another simplification is to base the formula on hours, not minutes. In this case:
Calories Eaten Before Bed = (RMR/24) * Hours Slept
It is time for a real world example. Consider a 200 pound male bodybuilder, that sleeps 7 hours a night.
Resting Metabolic Rate = 879 + (4.64 * 200) = 1807
Calories Eaten Before Bed = (RMR/24) * 7 = 527 calories
In this specific case, our case study bodybuilder can safely eat 527 calories before going to bed, and we are sure that he will maintain zero calorie balance.
What type of meal is best to eat before bed?
I suggest a protein and fat meal, providing complete slowly digested proteins. The best foods before bed are: all meats, eggs and low-carb cheeses.
Hristo Hristov owns X3MSoftware, a company specializing in developing diet and fitness tracking software. Hristo has a degree in Computer Science and passion for strength training. Hristo has designed and written Fitness Assistant, X3MSoftware's leading software product. Download your demo at Download Diet Software and Fitness Software by X3MSoftware
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https://techcommunity.microsoft.com/t5/excel/my-formula-is-broken-not-sure-house-to-fix/m-p/4093256/highlight/true | 1,717,079,608,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971667627.93/warc/CC-MAIN-20240530114606-20240530144606-00779.warc.gz | 493,737,531 | 58,443 | # My formula is broken, not sure house to fix
Copper Contributor
# My formula is broken, not sure house to fix
Here is my formula
=IFS(D22-D18>D18,{" "},D22-D18<D18,{"0"},D22-D18=D17,{"5000000"})
It is broken at the first part, I need it to come to a number that is between \$3M and \$5M, how do I get a number? The following section is the one that works D22-D18<d18,{"0"}. Can anyone help with this formula, not sure what I am doing wrong.
Thank you for your assistance, much appreciated.
7 Replies
# Re: My formula is broken, not sure house to fix
I don't know what you are trying to do / why it is 'broken'. The formula itself 'works' but not sure why you did a few things:
a) why is each answer formatted as a set of 1 (i.e. {" "} ) you don't need the set around those.
b) why is the 3rd condition =D17 instead of D18? you realize that if D22-D18 = D18 and D18 <> D17 then there is no correct response excel will return #N/A
c) are you sure you want those numbers 0 and 5000000 formatted as TEXT?
so maybe you want:
=IFS(D22-D18>D18,"", D22-D18<D18,0, D22-D18=D18, 5000000)
# Re: My formula is broken, not sure house to fix
The only way to get "5000000" is if D18=D17 and D22=2*D17.
I doubt this is what you intend.
Other observations might be that treating scalar values as arrays {"5000000"}. Also, does the result need to be text?
# Re: My formula is broken, not sure house to fix
this is what I am attempting to do is take the d22 less d18 if larger then the difference between d22, d18 and d17 if smaller then 0, or if it higher the d17 make it the amount in d17.
using=IFS(D22-D18>D18,"",D22-D18<D18,0,D22-D18=D17,500000)
I get the 500,000 but I can't seem to find the formula for the difference between d22-d18>d17,'"
can you help?
# Re: My formula is broken, not sure house to fix
the correct formula may be =IFS (AND(D22-D18>3000000, D22-D18>5000000)
# Re: My formula is broken, not sure house to fix
The now information is useful, but I still do not really understand exactly what it is you hope to see. My formula returned 500,000 but I wouldn't know whether that is good or bad.
``````= IFS(
paidLoss-retention > limit, limit,
paidLoss-retention < retention, 0,
TRUE, 500000
)``````
# Re: My formula is broken, not sure house to fix
I am attempting to make a figure out how to write a formal to determine if the paid loss is below retention, for a zero to enter, if not what is the difference between paid loss, retention and limit, and if the loss is higher then the limit to have the limit entered.
Does that makes sense?
# Re: My formula is broken, not sure house to fix
``````= IFS(
paidLoss-retention < 0, 0,
paidLoss-retention > limit, limit,
TRUE, paidLoss-retention
)``````
If that is correct, an alternative formula might be
``= MEDIAN(0, limit, paidLoss- retention)``
[p.s. I hope that the use of names makes the formula more meaningful written here, evenif you doen't use them in your workbooks] | 848 | 2,962 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2024-22 | latest | en | 0.944666 |
https://www.unitconverters.net/radiation-absorbed-dose/milligray-to-rad.htm | 1,725,949,868,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651224.59/warc/CC-MAIN-20240910061537-20240910091537-00262.warc.gz | 974,536,509 | 2,879 | Please provide values below to convert milligray [mGy] to rad [rd], or vice versa.
0.01 mGy0.001 rd
0.1 mGy0.01 rd
1 mGy0.1 rd
2 mGy0.2 rd
3 mGy0.3 rd
5 mGy0.5 rd
10 mGy1 rd
20 mGy2 rd
50 mGy5 rd
100 mGy10 rd
1000 mGy100 rd
How to Convert Milligray to Rad
1 mGy = 0.1 rd
1 rd = 10 mGy
Example: convert 15 mGy to rd:
15 mGy = 15 × 0.1 rd = 1.5 rd | 177 | 349 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2024-38 | latest | en | 0.271748 |
https://www.redhotpawn.com/forum/science/entropy-=%3E-energy.109856 | 1,540,177,824,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583514497.14/warc/CC-MAIN-20181022025852-20181022051352-00328.warc.gz | 1,040,316,016 | 8,812 | # Entropy => Energy?
AThousandYoung
Science 15 Mar '09 20:30
1. AThousandYoung
West Coast Rioter
15 Mar '09 20:30
Prove or disprove:
All events that result in an increase of entropy can theoretically be harnessed to produce usable energy.
2. 15 Mar '09 20:48
Define "usable".
3. 15 Mar '09 21:12
A possible answer to your question: if you consider some system in thermal equilibrium with its surroundings, the helmholz free energy is given by F = U - TS, an increase in entropy will reduce the helmholz free energy and therefore reduce the amount of work that can be done.
4. AThousandYoung
West Coast Rioter
16 Mar '09 02:11
Originally posted by KazetNagorra
Define "usable".
Gibbs free energy might fit the bill.
5. 16 Mar '09 09:44
Originally posted by AThousandYoung
Gibbs free energy might fit the bill.
Pretty much the same applies as for the post above, though it's a different equilibrium. The factor -TS is also in the Gibbs free energy, so it's reduced by an increase in entropy.
6. 20 Mar '09 19:21
If you look at entropy at a molecular level, it is basically the breaking of bonds between atoms to create simpler molecules. Every time you break a molecular bond, energy is liberated because there is a transfer of electrons.
I think this can prove that THEORETICALLY, every increase in entropy can be harnessed into useful energy.
7. 20 Mar '09 20:04
Originally posted by dannyUchiha
If you look at entropy at a molecular level, it is basically the breaking of bonds between atoms to create simpler molecules. Every time you break a molecular bond, energy is liberated because there is a transfer of electrons.
I think this can prove that THEORETICALLY, every increase in entropy can be harnessed into useful energy.
Actually entropy is simply proportional to the log of the number of states available to the system, which may or may not correspond to the number of bonds. A process in which entropy increases can be used for useful energy in some cases but not always, how are you going to get useful energy out of the melting of ice cubes in a glass of water?
8. AThousandYoung
West Coast Rioter
21 Mar '09 06:11
Originally posted by KazetNagorra
Actually entropy is simply proportional to the log of the number of states available to the system, which may or may not correspond to the number of bonds. A process in which entropy increases can be used for useful energy in some cases but not always, how are you going to get useful energy out of the melting of ice cubes in a glass of water?
That's right. Some increases in entropy require energy input. I guess you've disproven my hypothesis.
9. 21 Mar '09 16:31
Originally posted by KazetNagorra
Actually entropy is simply proportional to the log of the number of states available to the system, which may or may not correspond to the number of bonds. A process in which entropy increases can be used for useful energy in some cases but not always, how are you going to get useful energy out of the melting of ice cubes in a glass of water?
You're right. I was thinking of some processes that were useful, but clearly not all are. Thanks for clearing that up.
10. 23 Mar '09 08:59
Originally posted by dannyUchiha
If you look at entropy at a molecular level, it is basically the breaking of bonds between atoms to create simpler molecules. Every time you break a molecular bond, energy is liberated because there is a transfer of electrons.
That is not true at all. Entropy has no direct relationship to the number of chemical bonds or the complexity of molecules. If you were right then many chemical reactions would violate the second law.
11. 23 Mar '09 12:21
That is not true at all. Entropy has no direct relationship to the number of chemical bonds or the complexity of molecules. If you were right then many chemical reactions would violate the second law.
Of course it doesn't have a direct relationship with the number of bonds or with the complexity of the molecule. That's not what I said.
What I said was that it involved breaking of bonds. As you may know, entropy is the increase in disorder of a system. This may or may not liberate energy, but I do believe it involves some sort of interaction between atoms (breaking and/or formation of bonds)
Can you give examples when this does not apply?
12. 23 Mar '09 12:23
That is not true at all. Entropy has no direct relationship to the number of chemical bonds or the complexity of molecules. If you were right then many chemical reactions would violate the second law.
Also, you might want to reread the second law of thermodynamics.
http://en.wikipedia.org/wiki/Second_law_of_thermodynamics
13. DeepThought
23 Mar '09 15:37
Originally posted by dannyUchiha
What I said was that it involved breaking of bonds. As you may know, entropy is the increase in disorder of a system. This may or may not liberate energy, but I do believe it involves some sort of interaction between atoms (breaking and/or formation of bonds)
Can you give examples when this does not apply?
Entropy, as KazetNagorra stated above, is the log of the number of available states. It is fairly easy to show that it is a non-decreasing function of time. The breaking of molecular bonds is not necessary for an increase in entropy.
Here is an example: Imagine helium in a thermally isolated cylinder. First you compress the gas very slowly the volume drops and the pressure and temperature increase, you can then slowly reverse the process and the gas will end up in the state it started in. Now instead of compressing it slowly compress it quickly. This will cause an irreversible change. When you try to restore the gas to its earlier state you will find that the pressure and temperature are different to the initial state. Helium is a noble gas and will not form molecules.
14. 23 Mar '09 17:051 edit
Originally posted by dannyUchiha
Of course it doesn't have a direct relationship with the number of bonds or with the complexity of the molecule. That's not what I said.
What I said was that it involved breaking of bonds. As you may know, entropy is the increase in disorder of a system. This may or may not liberate energy, but I do believe it involves some sort of interaction between ...[text shortened]... atoms (breaking and/or formation of bonds)
Can you give examples when this does not apply?
It's a bit inaccurate to view entopy as a measure of disorder. Although it usually comes down to this, you must remember that disorder on a microscopic scale can result in ordering on a macroscopic scale. For example, the demixing of a fatty liquid in water results in an increase in entropy (also note that no chemical bonds are involved in this process).
15. 24 Mar '09 13:06
Originally posted by dannyUchiha
Also, you might want to reread the second law of thermodynamics.
http://en.wikipedia.org/wiki/Second_law_of_thermodynamics
I have gone and read it and I see nothing that contradicts what I said.
Put Hydrogen and Oxygen in a box in an isolated part of space and light it with a spark. Over time it will burn to form water. This either contradicts what you were saying or contradicts the second law. | 1,636 | 7,122 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2018-43 | latest | en | 0.901677 |
https://www.convertunits.com/from/ton/square+foot+%5Bshort%5D/to/attobar | 1,652,869,193,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662521883.7/warc/CC-MAIN-20220518083841-20220518113841-00365.warc.gz | 827,058,499 | 17,517 | ## ››Convert ton/square foot [short] to attobar
ton/square foot [short] attobar
Did you mean to convert ton/square foot [long] ton/square foot [short] to attobar
How many ton/square foot [short] in 1 attobar? The answer is 1.044271713652E-18.
We assume you are converting between ton/square foot [short] and attobar.
You can view more details on each measurement unit:
ton/square foot [short] or attobar
The SI derived unit for pressure is the pascal.
1 pascal is equal to 1.044271713652E-5 ton/square foot [short], or 10000000000000 attobar.
Note that rounding errors may occur, so always check the results.
Use this page to learn how to convert between tons/square foot and attobars.
Type in your own numbers in the form to convert the units!
## ››Quick conversion chart of ton/square foot [short] to attobar
1 ton/square foot [short] to attobar = 9.5760517777778E+17 attobar
2 ton/square foot [short] to attobar = 1.9152103555556E+18 attobar
3 ton/square foot [short] to attobar = 2.8728155333333E+18 attobar
4 ton/square foot [short] to attobar = 3.8304207111111E+18 attobar
5 ton/square foot [short] to attobar = 4.7880258888889E+18 attobar
6 ton/square foot [short] to attobar = 5.7456310666667E+18 attobar
7 ton/square foot [short] to attobar = 6.7032362444444E+18 attobar
8 ton/square foot [short] to attobar = 7.6608414222222E+18 attobar
9 ton/square foot [short] to attobar = 8.6184466E+18 attobar
10 ton/square foot [short] to attobar = 9.5760517777778E+18 attobar
## ››Want other units?
You can do the reverse unit conversion from attobar to ton/square foot [short], or enter any two units below:
## Enter two units to convert
From: To:
## ››Definition: Attobar
The SI prefix "atto" represents a factor of 10-18, or in exponential notation, 1E-18.
So 1 attobar = 10-18 bars.
The definition of a bar is as follows:
The bar is a measurement unit of pressure, equal to 1,000,000 dynes per square centimetre (baryes), or 100,000 newtons per square metre (pascals). The word bar is of Greek origin, báros meaning weight. Its official symbol is "bar"; the earlier "b" is now deprecated, but still often seen especially as "mb" rather than the proper "mbar" for millibars.
## ››Metric conversions and more
ConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 100 kg, US fluid ounce, 6'3", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more! | 789 | 2,688 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2022-21 | latest | en | 0.726596 |
https://www.mersenneforum.org/showthread.php?s=cea5da68e1179bfad7f9155c227b70e1&t=3861&page=2 | 1,675,743,054,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500384.17/warc/CC-MAIN-20230207035749-20230207065749-00590.warc.gz | 855,333,942 | 11,872 | mersenneforum.org > Math List of primes
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2005-03-16, 20:44 #12 grandpascorpion Jan 2005 Transdniestr 7678 Posts Doh, I found much better (and pretty obvious) way to store them: http://www.rsok.com/~jrm/printprimes.html
2005-03-17, 23:29 #13
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Quote:
Originally Posted by ewmayer Regarding the storage issue, there's no need to store the primes explicitly - you simply store a table of differences between consecutive primes. ... ...nearly all the prime gaps will be smallish (say less than 512), so a byte suffices to store any gap that is < 512 in gap/2 form. What do we do if a gap is larger? Well, since we know a gap must be > 0, we can reserve a zero-valued byte as a representing 512-plus-whatever-the-bext-byte-times-2-is. We can have as many zeros in a row as needed, i.e. can cover gaps of arbitrary size this way. Since gaps this large are very rare, that only trivially increases our storage.
The one subtlety that arises here is: what to do with a gap of precisely length 512? (Length 2^(n+1) for the general n-bit-field scheme.) Without modification, that would lead to an infinite string of zero bytes. To handle that, we only allow each nonzero byte to represent gaps from length 2 (byte = 00000001 in binary) to 510 (11111110), and use 11111111 to represent a *zero* terminating a string of zero-valued bytes (each of which denotes a length-512 interval), i.e. the byte pair 00000000 11111111 represents a 512-gap, the byte triplet 00000000 00000000 11111111 represents a 1024-gap, and so forth. It's still very simple to implement in code, and has one additional nice property which I neglected to mention in my initial post: in the lower limit of a 1-bit field length (n = 1), the resulting bit table is exactly the same one that results from simply labeling all the odd integers in the interval in question with a 0 (not prime) or 1 (prime).
2005-03-18, 10:17 #14
Unregistered
32×5×167 Posts
Quote:
Originally Posted by grandpascorpion Since 6 is still the "jumping champion" in this range, I think you could save even more space by using a 4-bit solution. ...I could be wrong here. I haven't crunched any numbers but on the surface, it looks like most of the time you would use half the space.
If your aim is to store primes as compactly as possible, the 4-bit solution wins for this size range. I've looked at the range (2*10^13, 2*10^13+10^6) which yields 32,764 primes, with a maximal gap of 376. About 65% of the gaps are 2-30, which would each require half a byte.
I've assumed a simple scheme where the next half byte codes gaps 32-60, the third half byte 62-90 etc. This scheme needs about 25200 bytes to store the range or about 77% of the full-byte scheme.
2005-03-18, 14:04 #15
Uncwilly
6809 > 6502
"""""""""""""""""""
Aug 2003
101×103 Posts
5×2,179 Posts
Quote:
Originally Posted by Unregistered I've assumed a simple scheme where the next half byte codes gaps 32-60, the third half byte 62-90 etc. This scheme needs about 25200 bytes to store the range or about 77% of the full-byte scheme.
Is it possible to post some of the code that you used?
Last fiddled with by Uncwilly on 2005-03-18 at 14:04
2005-03-18, 17:42 #16
Unregistered
100010111111002 Posts
Quote:
Originally Posted by Uncwilly Is it possible to post some of the code that you used?
I didn't use any code: I generated the primes using dsouza123's program, copied them into excel and counted the frequencies.
They were
2 to 30 21202
32 to 60 7604
62 to 90 2635
92 to 120 907
122 to 150 273
152 to 180 95
152+ 48
total: 32764
so 21202 gaps will code in 4 bits, 7604 in 8 and so on
2005-03-18, 19:55 #17 grandpascorpion Jan 2005 Transdniestr 1111101112 Posts Unregistered, I think there's an even better way to do this, albeit a tad harder to implement. Start with a set of numbers relatively prime to some product of the first n primes, like 30030 or 510510. That set will always be the primes to that point minus the prime factors plus 1. The sieved set size for 30030 is 3243. So, 1 in sieve gap terms translates roughly on average to 9. Since you dealing with a sieved range, it stands to reason that these sort of gaps would be much smaller. I would imagine a 3 bit would be implementable for gaps 1-7 3 bits - sieve gaps 1-7 8 bits - sieve gaps 8-23 12 bits - sieve gaps 24-128 16 bits - sieve 129-1024 (rarely necessary) I did a check: the minimum gap check for 23 for a sieved set of 30030 is 96. So, all gaps under <=96 could be handled in 8 bits. Similarly, for 3 bits, all gaps under 7 (a real gap of 26) could be handled. Note: These are min. values, the max values for the 30030-sieved gaps of 7 and 23 are 130 and 318 respectively. I'll try to implement this this weekend and see how compact it would be.
2005-03-19, 00:18 #18 Richard Cameron Mar 2005 2×5×17 Posts Grandpascorpion I think your approach could reduce the storage requirement quite a bit. But it all depends on the distribution of gaps, not just the average sizes. So I'm interested to see what results you get. Richard Cameron -thought it would be polite to register.
2005-03-20, 00:50 #19 grandpascorpion Jan 2005 Transdniestr 1111101112 Posts Hi Richard, I got off on the wrong foot with the sieved range. I was working with the primes themselves not the set of relatively prime numbers. I worked with 30030 sieve (2*3*5*7*11*13). I ran a very similar range (a little bit different due a quick-and-dirty sieve implementation) There were 33424 primes from 19999999999980 to 20000001020999 Max sieve gap is: 72 and max real gap is 376 444 Bit Total: 143212, 353 Bit total: 146644, Avg. is 4 Sieveless Bit Count: 203204 The sieveless count is 4 bits for 2-30, 8 for 32-62, 12 for 64-94 (so this would be take a bit less space than your scheme) As you can see, it takes lroughly 16kb to store the primes within roughly a 1Mb range. Not using a sieve, takes more than 25k. It may be overkill but I'll try the next higher range 510510 (30030*17) and see how much that would improve the storage. BTW, for the 30030, sieve range the 353 scheme (3 bits < 8, 8 bits < 24, 11 bits < 64) does better through about 1.1-1.2*10^12.
Similar Threads Thread Thread Starter Forum Replies Last Post MattcAnderson MattcAnderson 0 2017-05-27 14:00 Lennart Twin Prime Search 58 2017-05-05 14:15 gd_barnes Sierpinski/Riesel Base 5 2 2008-07-01 04:09 rogue Sierpinski/Riesel Base 5 1 2007-02-23 00:35 hyh1048576 Software 5 2003-08-20 13:38
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𝛢𝛼 𝛣𝛽 𝛤𝛾 𝛥𝛿 𝛦𝜀𝜖 𝛧𝜁 𝛨𝜂 𝛩𝜃𝜗 𝛪𝜄 𝛫𝜅 𝛬𝜆 𝛭𝜇 𝛮𝜈 𝛯𝜉 𝛰𝜊 𝛱𝜋 𝛲𝜌 𝛴𝜎𝜍 𝛵𝜏 𝛶𝜐 𝛷𝜙𝜑 𝛸𝜒 𝛹𝜓 𝛺𝜔 | 2,469 | 7,413 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2023-06 | latest | en | 0.865952 |
https://www.aqua-calc.com/one-to-one/surface-density/troy-ounce-per-square-centimeter/microgram-per-square-angstrom/1 | 1,638,310,356,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964359073.63/warc/CC-MAIN-20211130201935-20211130231935-00052.warc.gz | 717,033,508 | 8,026 | # 1 troy ounce per square centimeter in micrograms per square angstrom
## troy ounces/centimeter² to microgram/angstrom² unit converter
1 troy ounce per square centimeter [oz t/cm²] = 3 × 10−9 microgram per square angstrom [µg/Ų]
### troy ounces per square centimeter to micrograms per square angstrom surface density conversion cards
• 1
through
25
troy ounces per square centimeter
• 1 oz t/cm² to µg/Ų
= 3 × 10-9 µg/Ų
• 2 oz t/cm² to µg/Ų
= 6 × 10-9 µg/Ų
• 3 oz t/cm² to µg/Ų
= 9 × 10-9 µg/Ų
• 4 oz t/cm² to µg/Ų
= 1.2 × 10-8 µg/Ų
• 5 oz t/cm² to µg/Ų
= 1.6 × 10-8 µg/Ų
• 6 oz t/cm² to µg/Ų
= 1.9 × 10-8 µg/Ų
• 7 oz t/cm² to µg/Ų
= 2.2 × 10-8 µg/Ų
• 8 oz t/cm² to µg/Ų
= 2.5 × 10-8 µg/Ų
• 9 oz t/cm² to µg/Ų
= 2.8 × 10-8 µg/Ų
• 10 oz t/cm² to µg/Ų
= 3.1 × 10-8 µg/Ų
• 11 oz t/cm² to µg/Ų
= 3.4 × 10-8 µg/Ų
• 12 oz t/cm² to µg/Ų
= 3.7 × 10-8 µg/Ų
• 13 oz t/cm² to µg/Ų
= 4 × 10-8 µg/Ų
• 14 oz t/cm² to µg/Ų
= 4.4 × 10-8 µg/Ų
• 15 oz t/cm² to µg/Ų
= 4.7 × 10-8 µg/Ų
• 16 oz t/cm² to µg/Ų
= 5 × 10-8 µg/Ų
• 17 oz t/cm² to µg/Ų
= 5.3 × 10-8 µg/Ų
• 18 oz t/cm² to µg/Ų
= 5.6 × 10-8 µg/Ų
• 19 oz t/cm² to µg/Ų
= 5.9 × 10-8 µg/Ų
• 20 oz t/cm² to µg/Ų
= 6.2 × 10-8 µg/Ų
• 21 oz t/cm² to µg/Ų
= 6.5 × 10-8 µg/Ų
• 22 oz t/cm² to µg/Ų
= 6.8 × 10-8 µg/Ų
• 23 oz t/cm² to µg/Ų
= 7.2 × 10-8 µg/Ų
• 24 oz t/cm² to µg/Ų
= 7.5 × 10-8 µg/Ų
• 25 oz t/cm² to µg/Ų
= 7.8 × 10-8 µg/Ų
• 26
through
50
troy ounces per square centimeter
• 26 oz t/cm² to µg/Ų
= 8.1 × 10-8 µg/Ų
• 27 oz t/cm² to µg/Ų
= 8.4 × 10-8 µg/Ų
• 28 oz t/cm² to µg/Ų
= 8.7 × 10-8 µg/Ų
• 29 oz t/cm² to µg/Ų
= 9 × 10-8 µg/Ų
• 30 oz t/cm² to µg/Ų
= 9.3 × 10-8 µg/Ų
• 31 oz t/cm² to µg/Ų
= 9.6 × 10-8 µg/Ų
• 32 oz t/cm² to µg/Ų
= 1 × 10-7 µg/Ų
• 33 oz t/cm² to µg/Ų
= 1.03 × 10-7 µg/Ų
• 34 oz t/cm² to µg/Ų
= 1.06 × 10-7 µg/Ų
• 35 oz t/cm² to µg/Ų
= 1.09 × 10-7 µg/Ų
• 36 oz t/cm² to µg/Ų
= 1.12 × 10-7 µg/Ų
• 37 oz t/cm² to µg/Ų
= 1.15 × 10-7 µg/Ų
• 38 oz t/cm² to µg/Ų
= 1.18 × 10-7 µg/Ų
• 39 oz t/cm² to µg/Ų
= 1.21 × 10-7 µg/Ų
• 40 oz t/cm² to µg/Ų
= 1.24 × 10-7 µg/Ų
• 41 oz t/cm² to µg/Ų
= 1.28 × 10-7 µg/Ų
• 42 oz t/cm² to µg/Ų
= 1.31 × 10-7 µg/Ų
• 43 oz t/cm² to µg/Ų
= 1.34 × 10-7 µg/Ų
• 44 oz t/cm² to µg/Ų
= 1.37 × 10-7 µg/Ų
• 45 oz t/cm² to µg/Ų
= 1.4 × 10-7 µg/Ų
• 46 oz t/cm² to µg/Ų
= 1.43 × 10-7 µg/Ų
• 47 oz t/cm² to µg/Ų
= 1.46 × 10-7 µg/Ų
• 48 oz t/cm² to µg/Ų
= 1.49 × 10-7 µg/Ų
• 49 oz t/cm² to µg/Ų
= 1.52 × 10-7 µg/Ų
• 50 oz t/cm² to µg/Ų
= 1.56 × 10-7 µg/Ų
• 51
through
75
troy ounces per square centimeter
• 51 oz t/cm² to µg/Ų
= 1.59 × 10-7 µg/Ų
• 52 oz t/cm² to µg/Ų
= 1.62 × 10-7 µg/Ų
• 53 oz t/cm² to µg/Ų
= 1.65 × 10-7 µg/Ų
• 54 oz t/cm² to µg/Ų
= 1.68 × 10-7 µg/Ų
• 55 oz t/cm² to µg/Ų
= 1.71 × 10-7 µg/Ų
• 56 oz t/cm² to µg/Ų
= 1.74 × 10-7 µg/Ų
• 57 oz t/cm² to µg/Ų
= 1.77 × 10-7 µg/Ų
• 58 oz t/cm² to µg/Ų
= 1.8 × 10-7 µg/Ų
• 59 oz t/cm² to µg/Ų
= 1.84 × 10-7 µg/Ų
• 60 oz t/cm² to µg/Ų
= 1.87 × 10-7 µg/Ų
• 61 oz t/cm² to µg/Ų
= 1.9 × 10-7 µg/Ų
• 62 oz t/cm² to µg/Ų
= 1.93 × 10-7 µg/Ų
• 63 oz t/cm² to µg/Ų
= 1.96 × 10-7 µg/Ų
• 64 oz t/cm² to µg/Ų
= 1.99 × 10-7 µg/Ų
• 65 oz t/cm² to µg/Ų
= 2.02 × 10-7 µg/Ų
• 66 oz t/cm² to µg/Ų
= 2.05 × 10-7 µg/Ų
• 67 oz t/cm² to µg/Ų
= 2.08 × 10-7 µg/Ų
• 68 oz t/cm² to µg/Ų
= 2.12 × 10-7 µg/Ų
• 69 oz t/cm² to µg/Ų
= 2.15 × 10-7 µg/Ų
• 70 oz t/cm² to µg/Ų
= 2.18 × 10-7 µg/Ų
• 71 oz t/cm² to µg/Ų
= 2.21 × 10-7 µg/Ų
• 72 oz t/cm² to µg/Ų
= 2.24 × 10-7 µg/Ų
• 73 oz t/cm² to µg/Ų
= 2.27 × 10-7 µg/Ų
• 74 oz t/cm² to µg/Ų
= 2.3 × 10-7 µg/Ų
• 75 oz t/cm² to µg/Ų
= 2.33 × 10-7 µg/Ų
• 76
through
100
troy ounces per square centimeter
• 76 oz t/cm² to µg/Ų
= 2.36 × 10-7 µg/Ų
• 77 oz t/cm² to µg/Ų
= 2.39 × 10-7 µg/Ų
• 78 oz t/cm² to µg/Ų
= 2.43 × 10-7 µg/Ų
• 79 oz t/cm² to µg/Ų
= 2.46 × 10-7 µg/Ų
• 80 oz t/cm² to µg/Ų
= 2.49 × 10-7 µg/Ų
• 81 oz t/cm² to µg/Ų
= 2.52 × 10-7 µg/Ų
• 82 oz t/cm² to µg/Ų
= 2.55 × 10-7 µg/Ų
• 83 oz t/cm² to µg/Ų
= 2.58 × 10-7 µg/Ų
• 84 oz t/cm² to µg/Ų
= 2.61 × 10-7 µg/Ų
• 85 oz t/cm² to µg/Ų
= 2.64 × 10-7 µg/Ų
• 86 oz t/cm² to µg/Ų
= 2.67 × 10-7 µg/Ų
• 87 oz t/cm² to µg/Ų
= 2.71 × 10-7 µg/Ų
• 88 oz t/cm² to µg/Ų
= 2.74 × 10-7 µg/Ų
• 89 oz t/cm² to µg/Ų
= 2.77 × 10-7 µg/Ų
• 90 oz t/cm² to µg/Ų
= 2.8 × 10-7 µg/Ų
• 91 oz t/cm² to µg/Ų
= 2.83 × 10-7 µg/Ų
• 92 oz t/cm² to µg/Ų
= 2.86 × 10-7 µg/Ų
• 93 oz t/cm² to µg/Ų
= 2.89 × 10-7 µg/Ų
• 94 oz t/cm² to µg/Ų
= 2.92 × 10-7 µg/Ų
• 95 oz t/cm² to µg/Ų
= 2.95 × 10-7 µg/Ų
• 96 oz t/cm² to µg/Ų
= 2.99 × 10-7 µg/Ų
• 97 oz t/cm² to µg/Ų
= 3.02 × 10-7 µg/Ų
• 98 oz t/cm² to µg/Ų
= 3.05 × 10-7 µg/Ų
• 99 oz t/cm² to µg/Ų
= 3.08 × 10-7 µg/Ų
• 100 oz t/cm² to µg/Ų
= 3.11 × 10-7 µg/Ų
#### Foods, Nutrients and Calories
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15996 foods that contain Vitamin B-12. List of these foods starting with the highest contents of Vitamin B-12 and the lowest contents of Vitamin B-12, and Recommended Dietary Allowances (RDAs) for Vitamin B12
#### Gravels, Substances and Oils
CaribSea, Freshwater, Eco-Complete Cichlid, Ivory Coast weighs 1 281.48 kg/m³ (80.00018 lb/ft³) with specific gravity of 1.28148 relative to pure water. Calculate how much of this gravel is required to attain a specific depth in a cylindricalquarter cylindrical or in a rectangular shaped aquarium or pond [ weight to volume | volume to weight | price ]
Turf weighs 400 kg/m³ (24.97118 lb/ft³) [ weight to volume | volume to weight | price | density ]
Volume to weightweight to volume and cost conversions for Refrigerant R-13, liquid (R13) with temperature in the range of -95.56°C (-140.008°F) to 4.45°C (40.01°F)
#### Weights and Measurements
The stone per square angstrom surface density measurement unit is used to measure area in square angstroms in order to estimate weight or mass in stones
The kinematic viscosity (ν) is the dynamic viscosity (μ) divided by the density of the fluid (ρ)
st/l to oz/cm³ conversion table, st/l to oz/cm³ unit converter or convert between all units of density measurement.
#### Calculators
Calculate the area of a square, its perimeter and diagonal | 3,621 | 6,271 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2021-49 | latest | en | 0.304155 |
https://rdrr.io/cran/coxme/src/tests/test2.R | 1,548,165,698,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583850393.61/warc/CC-MAIN-20190122120040-20190122142040-00042.warc.gz | 619,235,845 | 14,050 | tests/test2.R In coxme: Mixed Effects Cox Models
```library(coxme)
options(na.action='na.exclude')
aeq <- function(x,y) all.equal(as.vector(x), as.vector(y))
#
# Similar to test1.s, but with the rats data. This has enough groups to
# force sparse matrix computations.
#
femrat <- rats[rats\$sex=='f',]
contr.none <- function(n,contrasts=T) {
if(is.numeric(n) && length(n) == 1.)
levs <- 1.:n
else {
levs <- n
n <- length(n)
}
contr <- array(0., c(n, n), list(levs, levs))
contr[seq(1., n^2., n + 1.)] <- 1.
contr
}
options(contrasts=c('contr.none', 'contr.poly'))
theta <- pi/2
# First with no sparse
fit0 <- coxme(Surv(time, status) ~ rx + (1|litter), data=femrat,
vfixed=theta, iter=0, sparse=c(100, .001))
tfit <- coxph(Surv(time, status) ~ factor(litter) + rx,
data=femrat, x=T, iter=0)
dt0 <- coxph.detail(tfit)
aeq(apply(dt0\$score,2,sum), fit0\$u)
h0 <- apply(dt0\$imat,1:2,sum) + diag(c(rep(1/theta, 50),0))
aeq(as.matrix(gchol(h0)), as.matrix(fit0\$hmat))
aeq(diag(gchol(h0)), diag(fit0\$hmat))
aeq(diag(fit0\$var), diag(solve(h0)))
# then sparse
fit0 <- coxme(Surv(time, status) ~ rx + (1|litter), data=femrat,
vfixed=theta, iter=0, sparse=c(20, .1))
h0[1:50,1:50] <- diag(diag(h0)[1:50])
aeq(as.matrix(gchol(h0)), as.matrix(fit0\$hmat))
aeq(diag(gchol(h0)), diag(fit0\$hmat))
aeq(diag(fit0\$var), diag(solve(h0)))
# Now iteration 1
fit1 <- coxme(Surv(time, status) ~ rx + (1|litter), data=femrat,
vfixed=theta, iter=1, sparse=c(10, .1))
update0 <- solve(fit0\$hmat, fit0\$u)
update0[1:50] <- update0[1:50] - mean(update0[1:50])
aeq(update0, c(unlist(ranef(fit1)), fixef(fit1)))
tfit <- coxph(Surv(time, status) ~ factor(litter) + rx,
data=femrat, x=T, iter=0,
init=c(unlist(ranef(fit1)), fixef(fit1)))
dt1 <- coxph.detail(tfit)
aeq(apply(dt1\$score,2,sum)- c(unlist(ranef(fit1)), 0)/theta, fit1\$u)
h1 <- apply(dt1\$imat,1:2,sum) + diag(c(rep(1/theta, 50),0))
h1[1:50,1:50] <- diag(diag(h1)[1:50])
aeq(as.matrix(gchol(h1)), as.matrix(fit1\$hmat))
aeq(diag(gchol(h1)), diag(fit1\$hmat))
aeq(diag(fit1\$var), diag(solve(h1)))
# And iteration 2
fit2 <- coxme(Surv(time, status) ~ rx + (1|litter), data=femrat,
vfixed=theta, iter=2)
update1 <- solve(fit1\$hmat, fit1\$u)
update1[1:50] <- update1[1:50] - mean(update1[1:50])
aeq(update1, c(unlist(ranef(fit2)), fixef(fit2)) -
c(unlist(ranef(fit1)), fixef(fit1)))
#
# Same computation, using a specified matrix
#
fit2b <- coxme(Surv(time, status) ~ rx + (1|litter), data=femrat,
vfixed=theta, iter=2, varlist=bdsI(seq(1, 99, 2)))
all.equal(fit2b\$u, fit2\$u)
all.equal(fit2b\$variance, fit2\$variance)
all.equal(fit2b\$loglik, fit2\$loglik)
```
Try the coxme package in your browser
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coxme documentation built on May 13, 2018, 5:03 p.m. | 1,077 | 2,770 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2019-04 | longest | en | 0.216118 |
https://www.physicsforums.com/threads/how-dark-energy-make-universe-to-inflate.428092/ | 1,508,533,766,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187824325.29/warc/CC-MAIN-20171020192317-20171020212317-00135.warc.gz | 970,239,438 | 20,597 | # How dark energy make universe to inflate?
1. Sep 11, 2010
### sweet springs
Hi,
How dark energy make universe inflate? As it is positive energy, it seems to make universe contract by its gravity.
Regards.
2. Sep 11, 2010
### friend
If dark energy is made of particles AND is responsible for space itself expanding (or contracting), then it seems there must be some interaction between space and particle, that particles can convert to space and visa versa.
3. Sep 12, 2010
### Chalnoth
This stems from the Friedmann equations. The first of which is this:
$$H^2 = \rho$$
...where I have ignored constants for simplicity. $H$ is the Hubble parameter:
$$H = {1 \over a} {da \over dt}$$
...while $\rho$ is the energy density of the universe.
To give you an idea of how this works, let's take a universe where there is nothing but normal matter (or dark matter, take your pick). In such a situation $$\rho \propto 1/a^3$$. That is, the density of the matter decreases as as the volume of the universe increases. So with a matter-dominated universe we have:
$$H(a)^2 = {H_0^2 \over a^3}$$
Here I have introduced the Hubble constant, which is the value of the Hubble parameter now, when $a = 1$. If we substitute the definition of the Hubble parameter, some quick calculation leads to:
$${da \over dt} = H_0 a^{-1 \over 2}$$
With a little bit of calculus, we can then integrate this equation to get how the scale factor $a$ depends upon time:
$$a(t) \propto t^{2 \over 3}$$
Note that this power of time is smaller than one: this means that over time, the scale factor increases less and less. The universe decelerates. This is exactly what we expect from gravity, and is, in fact, the same whether we calculate based on General Relativity of Newtonian gravity. This should, I hope, make some degree of sense.
By contrast, we can take the simpler case where the energy density is a constant:
$$H(a)^2 = H_0^2$$
$$H(a) = H_0$$
$${1 \over a} {da \over dt} = H_0$$
$${da \over dt} = H_0 a$$
This is a classic differential equation: the rate of change in the scale factor is proportional to the scale factor. This is exponential growth. So as the universe expands, the scale factor increases, and the rate of change of the scale factor also increases! Thus, from the exact same law of gravity that predicts that a universe that dilutes will slow down, we end up with a universe that expands faster and faster if we just don't let the energy density dilute as it expands.
4. Sep 13, 2010
### sweet springs
Hi.
Thank you so much, Chalnoth.
As for dark energy, density is conserved and total energy is not conserved, I see.
Some explanations refer to pressure of dark energy. Can someone tell me how pressure work in the theory ?
Regards.
I
5. Sep 13, 2010
### Chalnoth
Yes. This article on energy conservation in General Relativity may interest you:
http://www.phys.ncku.edu.tw/mirrors/physicsfaq/Relativity/GR/energy_gr.html
Well, one way to look at it is that the pressure is the reason why energy isn't conserved. Imagine, if you will, that we place some dark energy in a closed box. This dark energy, due to its nature, will place a negative pressure on the sides of the box. That is, it pulls the sides of the box inward in proportion to its energy density.
Let's say that this box, to begin with, has a volume of $a^3$: it's a cube with sides length $a$. Now, what happens if we increase the size of the box? Let's take the simplest case, where we only increase the size of the box in one direction, so that the new volume is $a^2(a+b)$, where here we've lengthened the box in one direction by $b$. To do this, however, we had to apply a force to the side of the box, a force equal to the pressure times the area. Applying this force to cause the box to expand means that we did work on the box, increasing its energy. The amount that the energy increased is the force times the distance, which is equal to the pressure times the area of the side times the distance it increased, or:
$$\Delta E = \rho a^2b$$
...where I have substituted the energy density for the pressure, since they have the same magnitude. It should be no surprise, then, that this is exactly the energy increase in the dark energy inside the box from the increase in size: since the energy density of the stuff in the box stays the same, and the volume increases by $$a^2 b$$.
So we can understand the energy increase in the dark energy as coming from the force which is required to expand it. In an expanding universe, this force is gravity, and we can understand the added energy for dark energy as coming from gravitational potential energy.
6. Sep 14, 2010
### sweet springs
Hi Chalnoth.
Thanks again.
As for dark energy dE= roh dV=-pdV p<0, I see.
Let me show further question.
1. One should interprete that the inflation in one of two ways ( roh or p ) or that generated dark energy has pressure p (roh and p) ?
2. What direction does pressure p have ? Does it lie in the universe or perdinicular to it ?
Regards.
7. Sep 14, 2010
### Chalnoth
For inflation/dark energy, $\rho = -p$ (or close to it).
Pressure is omni-directional. There is no notion of a direction "outside" the universe in General Relativity.
8. Sep 15, 2010
### sweet springs
Hi.
Thank you so much. I summarize my understanding. Dark energy has uniform energy density roh and pressure p=-roh direction of which is omni-directional and inward to any point considered.
I've read dark energy cause effective repulsive force. I imagine that any distance between a pair of gravitationally interacting bodies increases by the inflation. Prolonged distance r weakens gravitational force of -r^-2 effectively. Does it sound good ?
Regards.
9. Sep 15, 2010
### Chalnoth
I don't think so. I seem to recall that in Newtonian gravity, it has the effect of adding a repulsive force proportional to the distance between two objects. That is:
$$F = (G m_1 m_2)\left({-1 \over r^2} + {r \Lambda}\right)\hat{r}$$
I'm having a hard time looking it up online, though, so I could be wrong here. But I think it's roughly correct.
10. Sep 15, 2010
### George Jones
Staff Emeritus
11. Sep 15, 2010
### Chalnoth
Thanks! I see that I at least got the form correct, provided that the spherical case translates to the two particle case in the obvious way.
12. Sep 15, 2010
### Ich
IMHO, the first Friedmann equation is not a good start for the problem. Better use the second, which explicitly states that deceleration = density plus three times pressure. It's the negative pressure that drives the acceleration, not density being constant. The latter is just a consequence of the continuity equation for p=-rho.
Last edited by a moderator: Apr 25, 2017
13. Sep 15, 2010
### sweet springs
Hi.
Thanks George for showing a "springy" repulsive force by cosmic constant.
For tasting it better, let me ask how cosmic constant relates to dark energy density and its pressure ?
Regards.
Last edited: Sep 15, 2010
14. Sep 17, 2010
### sweet springs
Hi.
Einstein equation is
Gμν = 8πG/c^4 (Tμν+ ρgμν) where rho is energy density of dark energy, or
Gμν + Λgμν = 8πG/c^4 Tμν
so Λ = - 8πG/c^4 ρ
If Λ is negative, it causes attractive force.
Where did I make a mistake ?
Regards.
15. Sep 17, 2010
### Chalnoth
Er, that first equation is just wrong, namely:
$$G_{\mu\nu} = {8\pi G \over c^4} \left(T_{\mu\nu} + \rho g_{\mu\nu}\right)$$
The second equation is the correct form of Einstein's equations.
The energy density $\rho$ is part of the stress energy tensor $T_{\mu\nu}$, not separate from it.
16. Sep 18, 2010
### sweet springs
Hi, Chalnoth. Thanks for your patience.
So in the correct formula Gμν + Λgμν = 8πG/c^4 Tμν, cosmological constant is in LHS and dark energy is included in Tμν in RHS. They are two independent quantity. George taught us that positive Λ brings spring like repulsive force but none about effect of dark energy. Am I right ?
Regards.
Last edited: Sep 18, 2010
17. Sep 18, 2010
### Chalnoth
You specifically meant $\rho_{\Lambda}$, then? In that case, you have to be a bit more careful and look at the specific way in which you would construct a stress-energy tensor that has a constant energy density in time and space.
You're right that you pick up one minus sign by moving the $\Lambda$ term to the other side of the equation, but you can't simply equate that to energy density.
18. Sep 18, 2010
### George Jones
Staff Emeritus
Dark energy that satisfies the condition that Chalnoth gave is indistinguishable from a positive cosmological constant. To see why, take the cosmological constant term to the other side of Einstein's equation.
I think that it will take a quantum theory of gravity to disentangle things.
Didn't see Chalnoth's previous post.[/edit]
19. Sep 18, 2010
### sweet springs
Hi.
I read you mean Gμν = 8πG/c^4 Tμν - Λgμν = 8πG/c^4 (traditional Tμν + Tμν of dark energy )
where Tμν of dark energy = (8πG/c^4)^-1 ( - Λgμν )
Here we see T00 of dark energy, dark energy density, is negative. I expect it is ρ as we named in this thread and positive.
What's wrong ?
Regards.
20. Sep 18, 2010
### Chalnoth
That depends upon your convention for the signature of the metric. Make sure you've properly kept track of all of the indices. | 2,456 | 9,269 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2017-43 | longest | en | 0.903612 |
https://electronics.stackexchange.com/questions/646991/what-is-the-correct-relationship-between-active-reactive-and-apparent-power-in | 1,721,869,038,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518532.61/warc/CC-MAIN-20240724232540-20240725022540-00561.warc.gz | 194,613,599 | 45,118 | # What is the correct relationship between active, reactive and apparent power in non-linear AC systems?
I am somewhat troubled by the 'basic' formula that describes the relationship between apparent power (S), active power (P) and reactive power (Q) in AC systems. This formula is S^2 = P^2 + Q^2 and makes sense to me in linear systems with or without phase shifts. But quite some modern loads contain large amounts of harmonic currents that in my view can not create any work (active power) if the voltage is purely sinusoidal. But if apparent power is defined as S = V_rms * I_rms, the harmonic content adds up in I_rms, thus making S > P. A simple way to think about this, is that the difference can be called reactive power Q by adhering to S^2 = P^2 + Q^2. But I wonder whether that would be correct. Since I always believed Q is defined as reactive power with the vector component of the current perpendicular to the voltage by the relationships P = V_rms * I_rms * cos(phi) and Q = V_rms * I_rms * sin(phi). I subsequently would believe higher harmonics in currents combined with the same harmonics in the voltage could have these relationships as well, such that higher harmonics could create both active and reactive power. But if the voltage is pure sinusoidal, it has no higher harmonics and the current harmonics will not create reactive power (nor active power) in my view. It would be great if this was generally seen as 'distortion power' D that adds to P and Q in a quadratic way (analogous to displacement and distortion power factors), but I have not seen this.
My question is whether there is a definition of power generated by harmonics that will fix the difference between active and apparent power. And I am wrong assuming that if that difference (in a quadratic way) is called Q, that the definition of reactive power should be different?
I have made a non-linear system that has a sinusoidal voltage and a square current that are in-phase and I did some calculations to get an idea of the 'missing' power. The active power is in-line with what I expect and I can calculate it in two ways. Firstly by integrating the voltage and current over a multiple of their period and dividing it by the elapsed time (simple average of the product in this case actually). Secondly by multiplying the RMS voltage by the RMS value of the fundamental harmonic of the current, knowing I do not have to multiply that by cos(phi) since these are in-phase. These values match and I have showed them in the output of my copied script below. There is no active power by higher harmonics, since the voltage does not have higher harmonics. And the same is true for reactive power of both fundamental and higher harmonics: this is zero for the fundamental harmonic since the voltage and current are in-line. And is zero for higher harmonics given that the harmonic values of the voltage are zero and thus the product will be zero.
import numpy as np
import scipy as sp
import matplotlib.pyplot as plt
def rms(x):
return np.sqrt(np.mean(x**2))
def fundamental(x):
N = len(x)//periods
X = abs(2/N*sp.fft.fft(x, N))
return X[1]*np.sin(w*t)
freq = 60
w = 2*np.pi*freq
periods = 4
t = np.linspace(0, periods/freq, 10000)
v = np.sin(w*t)
i = sp.signal.square(w*t)
plt.plot(t, v, label='v(t)')
plt.plot(t, i, label='i(t)')
plt.legend()
plt.show()
print('RMS Voltage =', rms(v),
'\nRMS Current =', rms(i))
print('Active Power:',
'\n integral of v*i of one period divided by that period = ', np.mean(v*i))
print('Active Power by the fundamental current (phase angle is zero):',
'\n V_1_rms * I_1_rms = ', rms(v) * rms(fundamental(i)))
print('Active Power by higher harmonic currents:',
'\n V_n_rms * I_n_rms for n>=2 = 0 * I_n_rms = 0')
print('Apparent Power:',
'\n V_rms * I_rms =', rms(v) * rms(i))
Based on this example I would come to the following result for 'missing' power:
Is there a definition for this value that I calculated? It is quite substantial at almost 50% of the active power.
• "But if the voltage is pure sinusoidal, it has no higher harmonics and the current harmonics will not create reactive power in my view." Did you mean active? You can indeed treat each harmonic independently: apply Parseval's theorem. Related, though not an exact answer: electronics.stackexchange.com/a/633117/311631 Commented Dec 20, 2022 at 6:19
• S, P, and Q are only defined for fundamental... All other are used for "distortion". Commented Dec 20, 2022 at 9:40
• True power is what it is (harmonics or not). Apparent power is what it is (harmonics or not) and, if you want to call what makes up the difference reactive power, then so be it. Are you trying to fix something that isn't broken I wonder? Is there any use for reactive power that shouts at needing a strict definition? Commented Dec 20, 2022 at 10:24
• @JeromeBu1982 S^2=P^2+Q^2+D^2. I think this is correct (Parseval theorem or identity?). It is a long time ago, I used some power analyzer. Don't remember how all parameters were measured ... Will search. Commented Dec 20, 2022 at 12:27
• Here are, for reference, the measured parameters with Fluke 430 Power Analyser i.sstatic.net/ecbnI.png Commented Dec 20, 2022 at 12:45
The missing power can simply be called 'Distortion Power' $$\D\$$, unit VA. Per the following equation: $$V_{rms}^2I_{rms}^2 = S^2 = D^2 +\sum_h P_h^2+\sum_h Q_h^2$$ , where h is the number of the harmonic (source: https://electronics.stackexchange.com/a/489895/199042)
In the somewhat extreme example in the question, with a sinusoidal voltage and an in-phase square wave current, the RMS values, active power (P) and the 'missing' power (D) were calculated by a 10,000 sample simulation. These values are: (rounded to 3 decimals): $$\begin{matrix} V_{rms} & = & 0.707\\ I_{rms} & = & 1\\ S & = & 0.707\\ P & = & 0.637\\ Q & = & 0\\ D & = & 0.308 \end{matrix}$$
These values are correct and can also be derived mathematically: $$\begin{matrix} \displaystyle V_{rms} &=& \frac{1}{2}\sqrt{2} \mbox{ (known quantity for sinusoid with amplitude 1)}\\ I_{rms} &=& 1 \mbox{ (sqrt of area under the squared values divided by period)}\\ S &=& \displaystyle V_{rms}*I_{rms}\\ &=&\displaystyle \frac{1}{2}\sqrt{2} \\ P^2 &=& \displaystyle\sum_hP_h^2 \\ &=& \displaystyle \sum_h V_h^2 I_h^2\cos(\theta_h) \\ &=& \displaystyle V_1^2I_1^2\\\ & & \mbox{since } V_h=0 \mbox{ for } h>=2 \mbox{ and }\theta_1=0^\circ \end{matrix}$$
That leaves the calculation of the fundamental current $$\I_1\$$. This can be done by the fourier series for a square wave f(t). By expressing a square wave f(t) as an infinite sum of sineways with harmonic frequencies per: $$f(t)=\frac{4}{\pi}\sum_{h=1,3,5,...}^\infty\frac{1}{h}\sin{\frac{h\pi t}{T}}$$ Where $$\I_1\$$ is the RMS value of the fundamental current, we can call $$\i_1(t)\$$ the fundamental current in the time domain. This leads to:
$$\begin{matrix} \displaystyle i_1(t) = \frac{4}{\pi}\frac{1}{h}\sin{\frac{h\pi t}{T}}\ \mbox{ with } h=1 \\ \end{matrix}$$ $$\I_1\$$ is then simply $$\\frac{1}{2}\sqrt{2}\$$ times the amplitude of $$\i_1(t)\$$: $$I_1 = \frac{1}{2}\sqrt{2}\cdot \frac{4}{\pi}$$ Thus: $$\begin{matrix} P &=& V_1\cdot I_1\\ & =& \frac{1}{2}\sqrt{2}\cdot \frac{1}{2}\sqrt{2}\cdot \frac{4}{\pi}\\ &=&\frac{2}{\pi}\\ &\approx & 0.637 \end{matrix}$$ $$\D\$$ is then simply: $$\begin{matrix} D^2&=&S^2-P^2-Q^2\\ &=&{(\frac{1}{2}\sqrt{2})}^2 - (\frac{2}{\pi})^2\\ D&\approx &0.308 \end{matrix}$$ And for reactive power: $$\begin{matrix} Q^2 &=& \displaystyle\sum_hQ_h^2 \\ &=& \displaystyle \sum_h V_h^2 I_h^2\sin(\theta_h) \\ &=& \displaystyle 0\\ & & \mbox{since } V_h=0 \mbox{ for } h>=2 \mbox{ and }\theta_1=0^\circ \end{matrix}$$
This confirms the notion that the equation $$\S^2=P^2+Q^2\$$ simply does not hold in non-linear AC systems. It is true that quite some power can be in the form of non-reactive, higher harmonic power while not creating any work. Still many will call the quadratic difference between $$\S\$$ and $$\P\$$ simply $$\Q\$$. This is incorrect for many modern non-linear AC systems. $$\Q\$$ is the reactive component, orthogonal to the active power.
• P, Q, and D are all "orthogonal" in geometric representations (3 axes) ... Commented Dec 22, 2022 at 14:12
• Thank you for pointing me to orthogonality. I used it implicitly. Now with Parceval’s theorem it is explicit. It makes the formulas work. And yet, I think the classical thinking of Q is just related the displacement of the fundamental. With the risk of putting the reactive harmonics in the “D bucket”. That would break orthogonality again. I am interested still how modern text books separate the classical thinking of displacement by machinery with the (mostly?) non-displaced harmonics by modern switched power supplies Commented Dec 22, 2022 at 14:49
Ah, I happened across this reference which may be handy:
"Power measurement techniques for non-sinusoidal conditions", S. Svensson, doctoral thesis, Chalmers University of Technology, Electric Power Engineering, Göteborg, Sweden, 1999 | 2,538 | 9,027 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 22, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2024-30 | latest | en | 0.960295 |
https://stuffsure.com/what-is-0-125-as-a-fraction/ | 1,712,937,856,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816024.45/warc/CC-MAIN-20240412132154-20240412162154-00053.warc.gz | 487,128,141 | 16,410 | # What is 0.125 as a Fraction?
0.125 as a fraction is 1/8. To convert 0.125 to a fraction, divide the decimal by the number of decimal places, which is 8 in this case.
Checkout this video:
## What is 0.125 as a Fraction?
0.125 as a fraction is 1/8. To convert 0.125 to a fraction, divide the decimal by the number following it, in this case 10, and then reduce the fraction.
## How to Convert 0.125 to a Fraction?
To convert 0.125 to a fraction, follow these steps:
-Write down the number as a decimal: 0.125
-Multiply both sides by 10: 10 × 0.125 = 1.25
-Remove the decimal point: 10 × 1.25 = 12.5
-Simplify by cancelling out any common factors: 10 × (1/8) = (5/4)
## How to Express 0.125 as a Fraction in Lowest Terms?
To change 0.125 to a fraction, we count the number of decimal places. There are three decimal places, so we count up three places from the right side of the decimal point and put a caret (^) above the 3:
0.125 = 1/8
To express 0.125 as a fraction in lowest terms, divide both the numerator and denominator by their greatest common factor:
1/8 = 1 ÷ 1 = 1/1
## What is 0.125 as a Percent?
To convert a decimal to a percent, multiply by 100 and add the “%” symbol. In this case, 0.125 * 100 = 12.5%.
## How to Convert 0.125 to a Percent?
To convert 0.125 to a percent, multiply by 100. So, 0.125 × 100 = 12.5%.
## What is 0.125 as a Mixed Number?
To change a decimal to a mixed number, follow these steps:
First, write down the decimal divided by 1, and then find how many times 1 goes into the number to the left of the decimal point. Write this result as the first number of the mixed number. Then, take the answer from step one and subtract it from the original decimal. This will give you the decimal portion of the mixed number. Lastly, write down this new decimal as the second number of the mixed number, making sure to place it to the right of the whole number with a space in between.
In other words, 0.125 as a mixed number would be written as 1 4/10 (one and four tenths).
## How to Convert 0.125 to a Mixed Number?
To convert a decimal to a mixed number, follow these steps:
1. Write down the decimal divided by 1, and find how many times 1 goes into the number above the decimal point.
2. Multiply this number by the decimal, and write it below the decimal point.
3. Subtract this new number from the original decimal, and write the answer above the line with the denominator being 1.
4. Find how many times the denominator of the original fraction goes into this new numerator. Write this number over to the right of the line with an equal sign (=), then write down this same number as a fraction with a denominator that is equal to what was originally under the line for the fraction part of your mixed number equation.. | 737 | 2,776 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.46875 | 4 | CC-MAIN-2024-18 | longest | en | 0.908057 |
https://sepwww.stanford.edu/sep/prof/pvi/rand/paper_html/node19.html | 1,721,210,023,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514759.39/warc/CC-MAIN-20240717090242-20240717120242-00424.warc.gz | 441,823,684 | 3,226 | Next: Spectral estimation Up: SPECTRAL FLUCTUATIONS Previous: Paradox: large n vs.
## An example of the bandwidth/reliability tradeoff
Letting n go to infinity does not take us to the expectation .The problem is, as we increase n, we increase the frequency resolution but not the statistical resolution (i.e., the fluctuation around ). To increase the statistical resolution, we need to simulate ensemble averaging. There are two ways to do this:
1.
Take the sample of n points and break it into k equal-length segments of n/k points each. Compute an for each segment and then average all k of the together. The variance of the average spectrum is equal to the variance of each spectrum () divided by the number of segments, and so the fluctuation is substantially reduced.
2.
Form from the n-point sample. Replace each of the n/2 independent amplitudes by an average over its k nearest neighbors. This could also be done by tapering the autocorrelation.
taper
Figure 10
Spectral smoothing by tapering the autocorrelation. is constant and specified on the top row. Successive rows show increasing while decreases. The width of a superimposed box roughly gives , and its height roughly gives .
The second method is illustrated in Figure 10. This figure shows a noise burst of 240 points. Since the signal is even, the burst is effectively 480 points wide, so the autocorrelation is 480 points from center to end: the number of samples will be the same for all cases. The spectrum is very rough. Multiplying the autocorrelation by a triangle function effectively smooths the spectrum by a sinc-squared function, thus reducing the spectral resolution (). Notice that is equal here to the width of the sinc-squared function, which is inversely proportional to the length of the triangle ().
However, the first taper takes the autocorrelation width from 480 lags to 120 lags. Thus the spectral fluctuations should drop by a factor of 2, since the count of terms sk in is reduced to 120 lags. The width of the next weighted autocorrelation width is dropped from 480 to 30 lags. Spectral roughness should consequently drop by another factor of 2. In all cases, the average spectrum is unchanged, since the first lag of the autocorrelations is unchanged. This implies a reduction in the relative spectral fluctuation proportional to the square root of the length of the triangle ().
Our conclusion follows:
The trade-off among resolutions of time, frequency, and spectral amplitude is
(47)
Next: Spectral estimation Up: SPECTRAL FLUCTUATIONS Previous: Paradox: large n vs.
Stanford Exploration Project
10/21/1998 | 570 | 2,614 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2024-30 | latest | en | 0.918634 |
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Replies: 78 Last Post: Mar 25, 2013 1:38 AM
Messages: [ Previous | Next ]
Robert Hansen Posts: 10,611 From: Florida Registered: 6/22/09
Posted: Mar 13, 2013 5:07 PM
Someone said that it could be done in 5 races, just pick the top 3 times. Unfortunately, this problem isn't stated as a mathematical problem.
Assuming that we do not know the time and assuming that "top 3" means that they have beaten everyone else directly or indirectly...
Step 1 - 5 races. These are the groups and they are in order 1st, 2nd, 3rd, ...
Step 2 - 1 race of 1st place winners of Step 1. Winner is #1. The rider that was 2nd in his group now replaces him.
Step 3 - 1 race of remaining 1st place and the replacement. Winner is #2. The rider that was behind him in the original group replaces him.
Step 4 - Same as step 3, Winner is #3.
That would be 8 races.
In each race the original racers are advancing up one as a member in their original group wins. Basically a merge sort with 5 groups.
Bob Hansen
On Mar 13, 2013, at 11:02 AM, Joe Niederberger <niederberger@comcast.net> wrote:
> Getting back to the original problem...
>
> Johnykeets says:
>> and finally Race 7:2nd and 3rd cyclists from the category of winner of Race 6, 1st and 2nd cyclists from the category of runners-up of Race 6, 3rd of Race 6,
> Other wise, we may also try this If we have two best players in one team we get win...ners in 7 races. After sixth race we get topper who initially belonged to say race 1, and then rest 4 competes with the second position holder of the race 1. so we get top three. Please let me know about the more easy methods in solving this problem.
>
> I too think you can do it with 7 races also but I cannot follow your description of the setup for race 7.
>
> Anyone care to prove that it cannot be done in 6?
>
> Cheers
> Joe N
Date Subject Author
2/24/13 hydraflap
2/24/13 Richard Strausz
2/25/13 James Elander
2/27/13 Wayne Bishop
3/1/13 GS Chandy
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3/8/13 GS Chandy
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3/8/13 Robert Hansen
3/8/13 Joe Niederberger
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3/12/13 Joe Niederberger
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3/14/13 Robert Hansen
3/14/13 Robert Hansen
3/15/13 GS Chandy
3/14/13 Joe Niederberger
3/14/13 Robert Hansen
3/14/13 Joe Niederberger
3/15/13 GS Chandy
3/25/13 VALERY | 1,248 | 3,614 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2017-39 | longest | en | 0.96792 |
https://www.think-accounting.com/formulas/excel-tips-and-tricks-demystifying-the-averageif-formula/ | 1,708,514,379,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473472.21/warc/CC-MAIN-20240221102433-20240221132433-00100.warc.gz | 1,056,053,454 | 24,779 | # Excel Tips and Tricks: Demystifying the AVERAGEIF Formula
Table of Content
In the vast realm of Excel, there lies a formula that has confounded many users, leaving them scratching their heads in perplexity. Yes, I'm talking about the enigmatic AVERAGEIF formula. Fear not, for today we shall embark on a journey to unravel the secrets of this elusive function, equipping you with the knowledge and confidence to wield it like a pro. So, fasten your seat belts, grab your calculators, and let's demystify the AVERAGEIF Formula!
## Mastering AVERAGEIF Function
First things first, let's begin our quest by understanding the syntax of the AVERAGEIF formula. In its simplest form, AVERAGEIF allows you to calculate the average of a range of values that meet certain criteria. The syntax goes something like this: `=AVERAGEIF(range, criteria, [average_range])`. The "range" refers to the cells that you want to evaluate, "criteria" is the condition that must be met, and "average_range" (optional) is the range to be averaged. Piece of cake, right?
But wait, there's more! AVERAGEIF has a few tricks up its sleeve that can make your life a whole lot easier. Let's dive into some practical examples to see it in action.
Imagine you have a spreadsheet with sales data for different products. You want to calculate the average sales for a specific product. This is where AVERAGEIF comes to the rescue. By using the criteria parameter, you can specify the product you're interested in and get the average sales for that particular product.
Let's say you have a range of cells (A2:A10) containing the product names and a corresponding range of cells (B2:B10) with the sales data. To calculate the average sales for a specific product, you can use the following formula: `=AVERAGEIF(A2:A10, "Product A", B2:B10)`. This will give you the average sales for "Product A" only.
But what if you want to calculate the average sales for multiple products? No worries, AVERAGEIF can handle that too. You can use wildcards in the criteria parameter to match multiple products. For example, if you want to calculate the average sales for all products starting with the letter "P", you can use the formula: `=AVERAGEIF(A2:A10, "P*", B2:B10)`. This will give you the average sales for all products starting with "P".
Another handy feature of AVERAGEIF is the ability to use logical operators in the criteria parameter. This allows you to define more complex conditions. For instance, if you want to calculate the average sales for products with sales greater than 1000, you can use the formula: `=AVERAGEIF(B2:B10, ">1000")`. This will give you the average sales for products that meet the specified condition.
Now that you know the ins and outs of the AVERAGEIF function, you can unleash its power to analyze your data and make informed decisions. Whether you're calculating average sales, average ratings, or any other average based on specific criteria, AVERAGEIF will be your go-to tool.
## Understanding the Syntax of AVERAGEIF
Imagine you have a spreadsheet listing the scores of students from different classes. You want to know the average score of all the students who scored above 80. This is where AVERAGEIF comes to the rescue! Use the formula `=AVERAGEIF(A2:A10, ">80")`, and voilà! The average score of the overachievers is magically calculated.
But wait, there's more! What if you want to only average the scores of students from a specific class? Easy peasy lemon squeezy! Just add another condition to the mix. Use the formula `=AVERAGEIF(A2:A10, ">80", B2:B10)` to calculate the average score for the top performers in a particular class. Now you're cooking with gas!
### Practical Examples of AVERAGEIF in Action
Let's dive a little deeper into the realm of AVERAGEIF and explore its versatility. Imagine you have a spreadsheet showing the sales figures of a company over a period of time. You want to calculate the average sales for each month. This is where AVERAGEIF shines! Use the formula `=AVERAGEIF(A2:A31, "January", B2:B31)` to calculate the average sales for the month of January. Repeat the same formula for the other months, and you'll have a monthly sales report at your fingertips!
But wait, there's more! What if you want to calculate the average sales for multiple months? Fear not, for AVERAGEIF can handle that too! Use the formula `=AVERAGEIF(A2:A31, "January", B2:B31)+AVERAGEIF(A2:A31, "February", B2:B31)` (and so on) to calculate the average sales for each individual month. Pretty nifty, huh?
### Tips & Tricks for Using AVERAGEIF Effectively
Now that you've become well-versed in the ways of AVERAGEIF, here are a few tips and tricks to help you make the most of this powerful function.
1. Remember to enclose text criteria in double quotation marks, such as "`January`". Excel needs those quotes to understand that you're looking for specific text.
2. For numeric criteria, write the condition without quotation marks. For example, `">80"`.
3. If you're dealing with a large dataset, consider using cell references instead of typing the criteria directly into the formula. This way, you can easily update the criteria without having to rewrite the entire formula.
4. Combine the AVERAGEIF formula with other functions, such as MAX and MIN, to gain deeper insights into your data. It's like unraveling the secret codes of the Excel universe!
5. Use the wildcard characters, such as "*" and "?", in your criteria to match a pattern of text or unknown characters. This can save you oodles of time and effort.
### Avoiding Common Mistakes with AVERAGEIF
While AVERAGEIF is a mighty tool in your Excel arsenal, it's not without its potential pitfalls. Here are a few common mistakes to watch out for:
• Double-check your ranges to ensure that you're referencing the correct cells. A simple typo can lead to erroneous results.
• Be mindful of the syntax and ensure that you're using the correct operators. Mixing up greater than (>) and less than (<) can create some unintended chaos.
• When using wildcard characters, be careful with the placement. For example, "`*apple*`" will match any text that contains the word "apple", while "`apple*`" will only match text that starts with "apple". It's like the difference between searching for a needle in a haystack and searching for a needle in a stack of hay!
### Troubleshooting AVERAGEIF: Why Isn't It Working?
Now, dear reader, I must address an important question: what do you do when AVERAGEIF refuses to cooperate? Fear not, as I shall bestow upon you some troubleshooting wisdom:
1. Check your criteria. It's possible that you've made a typo or overlooked a small detail. Even the tiniest mistake can cause the formula to misbehave, so give it a thorough once-over.
2. Make sure your ranges are properly formatted. If Excel doesn't recognize your data as numbers or text, it may not evaluate the criteria correctly.
3. If you're using cell references, verify that the cell references are correct and haven't been accidentally altered. Remember, Excel is unforgiving when it comes to mismatched ranges!
4. If all else fails, don't hesitate to seek help from the vast online Excel community. They have the wisdom and experience to guide you through even the trickiest Excel conundrums.
## Exploring AVERAGEIF and Its Related Formulae
Now that you're well-acquainted with the AVERAGEIF formula, why stop there? Let's take our Excel skills to the next level by exploring its related formulae and their synergistic powers.
### How to Combine AVERAGEIF with Other Functions
To truly unlock the full potential of AVERAGEIF, you need to team it up with its fellow Excel superstars. One such dynamic duo is AVERAGEIFS and SUMIF. These functions allow you to perform even more complex calculations based on multiple criteria.
For example, let's say you have a dataset containing sales figures for different regions and product categories. You want to calculate the average sales for a specific region and product category combination. This is where AVERAGEIFS comes to the rescue! The syntax is similar to AVERAGEIF, but with an added twist: `=AVERAGEIFS(average_range, criteria_range1, criteria1, criteria_range2, criteria2)`. Simply specify the various criteria ranges and their corresponding values, and AVERAGEIFS will work its magic!
But wait, there's more! Let's not forget about our old friend SUMIF. Imagine you have a spreadsheet with a list of products and their prices. Your task is to calculate the total price of all the products that meet a certain criteria, such as being on sale. This is where SUMIF comes in handy! Use the formula `=SUMIF(range, criteria, [sum_range])` to calculate the desired sum. Summing up sales has never been easier!
### Advanced Applications of AVERAGEIF in Data Analysis
Now that you're an AVERAGEIF aficionado, let's delve into some advanced applications in the realm of data analysis. Brace yourself for the mind-boggling possibilities!
Imagine you have a dataset containing customer feedback ratings on various aspects of your product or service. You want to analyze the average rating for each aspect. Strap on your seat belts, my friend, because AVERAGEIF is about to blow your mind!
Use the formula `=AVERAGEIF(aspect_range, aspect, rating_range)` to calculate the average rating for each aspect. With just a flick of your wrist, you'll have the power to identify the rockstar aspects that deserve a standing ovation, as well as the areas that need a little extra love and attention.
### Comparing AVERAGEIF with Similar Functions
Now, let's gather our wits and embark on a brief exploration of how AVERAGEIF compares to its fellow Excel functions.
One commonly used cousin of AVERAGEIF is AVERAGEIFS (with an 'S'), which allows you to calculate the average of a range based on multiple criteria. With AVERAGEIFS, you can specify multiple criteria ranges and their corresponding values, unlocking a whole new world of analysis possibilities.
Another sibling in the formula family is COUNTIF. While AVERAGEIF calculates the average of a range based on a specified condition, COUNTIF counts the number of cells that meet a certain criteria. Think of it as AVERAGEIF's diligent sibling, focused on keeping track of numbers rather than calculating their average.
And there you have it, fellow Excel enthusiasts! Armed with a newfound understanding of the AVERAGEIF formula and its related functions, you are ready to conquer the realm of data analysis! So go forth, crunch those numbers, and may your spreadsheets be forever error-free!
###### Simon Taylor
Hi there!
I'm Simon, your not-so-typical finance guy with a knack for numbers and a love for a good spreadsheet. Being in the finance world for over two decades, I've seen it all - from the highs of bull markets to the 'oh no!' moments of financial crashes. But here's the twist: I believe finance should be fun (yes, you read that right, fun!).
As a dad, I've mastered the art of explaining complex things, like why the sky is blue or why budgeting is cool, in ways that even a five-year-old would get (or at least pretend to). I bring this same approach to THINK, where I break down financial jargon into something you can actually enjoy reading - and maybe even laugh at!
So, whether you're trying to navigate the world of investments or just figure out how to make an Excel budget that doesn’t make you snooze, I’m here to guide you with practical advice, sprinkled with dad jokes and a healthy dose of real-world experience. Let's make finance fun together!
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Your navigator through the financial jungle. Discover helpful tips, insightful analyses, and practical tools for taxes, accounting, and more. Empowering you to make informed financial decisions every step of the way.
Categories | 2,592 | 11,819 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2024-10 | longest | en | 0.90636 |
http://plus.maths.org/content/lights-identity-crisis | 1,418,981,292,000,000,000 | text/html | crawl-data/CC-MAIN-2014-52/segments/1418802768378.98/warc/CC-MAIN-20141217075248-00009-ip-10-231-17-201.ec2.internal.warc.gz | 227,135,425 | 9,663 | # Light's identity crisis
Issue 5
May 1998
A moving charged particle, such as an electron, experiences electric and magnetic forces. In the middle of the 19th century, the Scottish physicist James Clerk Maxwell wrote down a set of equations which unified these two forces into a single theory. This led him to understand light waves (and radio waves) as electromagnetic oscillations propagating through a vacuum.
### Light as electromagnetic waves
Waves on the surface of a pond are a sequence of peaks and troughs that move along in the shape of a sine wave. Light waves are mathematically similar, except that they are electromagnetic: the sinusoidal oscillation is of the amplitude of the electromagnetic field.
Figure 1: Light waves. In each box the sum of the first and second waves is the third wave.
When two waves of the same wavelength meet, their combined electromagnetic field depends on their relative phase. If the peaks of the two waves coincide, we say that the waves are in phase, and then the two waves reinforce each other, as in the left-hand box in the figure. But if the peaks of one wave coincide with the troughs of the other the waves are said to be out of phase, and they cancel each other, as in the right-hand box.
Try it : with a graphing calculator compare the graphs of sin(x) and sin(x)+sin(x). Then compare the graphs of sin(x), sin(x+pi) and sin(x)+sin(x+pi). What do you see?
Figure 2: The grooves on a CD are only 0.5 micro-metres wide and make a good diffraction grating, resulting in coloured interference patterns.
Because of the wave-like character of light, it is diffracted when it is shone through a pair of closely separated slits: if a screen is placed some way behind the slits, a pattern of light and dark fringes appears on it. The spacing of these fringes is calculated from the wavelength lambda of the light. Dark fringes appear at points on the screen where the light received from the two slits is exactly out of phase. These points have distances from the two slits that differ by:
where n is an integer.
Figure 3: The double-slit experiment
Figure 4: Diffraction patterns are often used to analyse the structure of materials. Subtle differences in structure can change the properties of a material dramatically. Researchers at Heriot-Watt University are using the technique to help Cadbury make sure their chocolate always solidifies in the tastiest way.
### Light as particles
So, by the beginning of the 19th century, it was a well-established notion that light is wave-like. But in the early years of the 20th century it became apparent that it is also particle-like. A key experiment was the photoelectric effect, in which electrons escape from the surface of a metal when it is bombarded with light.
A metal consists of a huge number of atoms effectively anchored to fixed sites by the electric forces caused by all the other atoms. The outermost orbital electrons of the atoms can easily be pulled off when an electric field is applied. They then move through the metal and form what we call an electric current.
When light is shone on the metal, some of the electrons can actually escape from the surface. The number that escape rises with the intensity of the light, but their energy of escape does not. Rather it depends on the colour of the light or, equivalently, its frequency nu (pronounced new).
The explanation of these properties by Albert Einstein in 1905 (the year in which he also produced the theory of relativity) was really the beginning of quantum theory. A beam of light can be thought of as a collection of particles, called photons. The number of photons is proportional to the intensity of the light, and the energy E of each photon is proportional to its frequency:
This formula had already been guessed in 1900 by the German physicist Max Planck, and the constant h is named after him. In ordinary units it is very small:
The electron is ejected from the metal when one of the photons hits it and gets absorbed by it, so that the photon's energy is transferred to the electron. The number of escaping electrons increases with the intensity of the light because when there are more photons there is a greater chance of one hitting an electron.
### How heavy are photons?
Photons move with the speed of light, so when we think of them as particles we must use special relativity instead of Newton's mechanics. The energy of a particle whose speed is v (not to be confused with the greek letter nu) and whose rest mass is m is, according to Einstein:
In this equation m is the particle's mass, v is its speed and c is the speed of light. For a particle at rest we can substitute 0 for v giving Einstein's famous formula:
For a particle moving with the speed of light, v = c, the denominator vanishes, so it can have finite energy only if the numerator vanishes too, that is m = 0. Experiments have tested that photons have zero mass to very high accuracy: we know that their mass is less than 10-18 times that of an electron.
Photons may not have mass but they do have momentum. In Newton's mechanics the momentum of a particle is simply its mass times its velocity; p = mv. However, in special relativity the momentum of a particle is:
Try it : the Earth's mass is approximately 6x1024 kg, and it moves around the Sun once per year at a distance of about 1.5x1011 metres. Use a calculator to compare Newton's classical value for the momentum of the Earth with Einstein's relativistic value. The speed of light is approximately 3x108 ms-1.
What do you notice about the two values?
Using the relativistic momentum equation we can express the energy of a particle in terms of its momentum as follows:
For a photon this is simply:
But as we have seen already the energy is also related to the frequency of the light by:
When we combine these two equations, we find that:
This tells us the momentum of a photon for light of frequency nu.
For waves whose speed is c, the wavelength is:
Putting these formulae together we find that:
This is one of the fundamental relations of quantum theory!
It turns that this equation is so fundamental that it applies to all particles, not just photons. (See "Quantum uncertainty" elsewhere in this issue.).
Try it : using an approximation for the momentum of the Earth in its orbit round the Sun, calculate its wavelengh, lambda. The the value of h, Planck's constant, is given above as approximately 6.626x10-34 Js.
What do you notice about the value?
Biographies of the mathematicians mentioned in the article are available from the "MacTutor history of mathematics archive":
### The author
Professor Peter Landshoff is doing research on quarks (what we and everything in the world are made of) and is a lecturer in quantum mechanics at the University of Cambridge. | 1,471 | 6,819 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2014-52 | latest | en | 0.951188 |
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# Tutor profile: Devan O.
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Devan O.
Finance Major graduating from Drury University with 3.94 GPA
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## Questions
### Subject:Basic Math
TutorMe
Question:
What is 30% of \$20?
Inactive
Devan O.
30% = 0.3 So, 30% of \$20 = 0.3 x \$20 0.3 x \$20 = \$6. So, 30% of \$20 is \$6.
### Subject:Finance
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Question:
Susan, the mother of John, wants to plan ahead for John's college. Susan figures that she will need \$100,000 to pay for John's college. Susan will need the \$100,000 when John turns 18 (John just turned 2). How much will Susan need to put into an account each month if the account returns 7% per year?
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Devan O.
N (# of payments) = 192 (16 years * 12 months) I/Y (return %) = 0.5833% (7% / 12) FV (future value) = \$100,000 Pmt (monthly payment) = ? Using a financial calculator, the payment is calculated at -\$283.89 This means Susan would need to put \$283.89 a month into the account earning 7% return per year.
### Subject:Algebra
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Question:
Solve for x: 4(x + 2) + 1 = 15 - 2x
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Devan O.
The first step to simplifying this equation is distributing the 4 to the (x + 2). To do this, we must multiply the 4 with x = 4x and with 2 = 8. The equation is now 4x + 8 + 1 = 15 - 2x. To simplify further, let's get all of the x's on the left side. To do this, we must add 2x to each side (-2x + 2x = 0 and eliminates x's on the right side). We now have 6x + 9 = 15. Now to get the 9 to the right side, let's subtract both sides by 9. This leaves us with 6x = 6. To get x by itself, let's divide both sides by 6 (6/6 = 1) leaving the equation as x = 1. The answer to the equation is x = 1.
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# Sputtering by Particle Bombardment: Experiments and Computer
Format: Hardcover
Language: English
Format: PDF / Kindle / ePub
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Before that he was an assistant research scientist at Johns Hopkins University, Md., which is where he also completed his postdoctoral research. D.) in Applied Physics: Applied Mathematics Complete requirements for the M. What is the plane’s new velocity with respect to the ground in standard position? 24.0° north of west at 190 km/h and encounters a wind from 15. assume that the plane’s new velocity is the vector sum of the plane’s original velocity and the wind velocity. 100 km north in 3.
Pages: 509
Publisher: Springer; 2007 edition (September 10, 2007)
ISBN: 3540445005
Through the controlled introduction of organic solvent, one can control the ordering of the phases during self-assembly http://1-million-link.com/lib/introduction-to-the-physics-of-gyrotrons-johns-hopkins-studies-in-applied-physics. This sum is the x-component of the resultant vector. 2. Find the x-component of R by finding and adding the x-components of A and B. graph and find the x.26 y y B R 2 2 B 4 R 6 A x (a) A 3 x (b) 5 x A+B=R Find the y-component of R by finding and adding the y-components of A and B.and y-components as signed numbers. 0R 0 2.27. graph and find the x. we find the resultant vector of several vectors as follows: 1. 3 epub. Laboratory topics vary by year, but include single-molecule fluorescence, fluorescence resonance energy transfer, confocal microscopy, two-photon microscopy, microendoscopy, and optical trapping , cited: http://istarestudi.com/?books/applied-physics-applied-physics. Sound would not be heard if some obstacle came between you and the stereo speakers and there were no nearby reflecting surfaces.74 s 5.00 ms 3.00 s 7 download. Sundqvist studies low-temperature condensed matter physics and electronics http://nickel-titanium.com/lib/magnetism-of-surfaces-interfaces-and-nanoscale-materials-volume-5-handbook-of-surface-science. Geometric frustration in Mott insulators permits perturbative electron fluctuations controlled by local spin configurations [1] ref.: http://nickel-titanium.com/lib/1985-workshop-on-the-physics-of-non-conventional-energy-sources-and-material-science-for-energy. The ability of a material to transfer heat by conduction is called its thermal conductivity http://nickel-titanium.com/lib/spectroscopy-and-optoelectronics-in-semiconductors-and-related-materials-proceedings-of-the. Generation, manipulations, propagation, and applications of coherent radiation. The basic theory of the interaction of electromagnetic radiation with resonant atomic transitions online. One is the observer and remains stationary.436 CHAPTER 16 WAVE MOTION AND SOUND (b) We simply change the sign from remain the same. where the movement of blood can be monitored using sound waves (Fig. (b) A three-dimensional ultrasound of a human fetus. ultrasound is also used to observe various human systems such as the nervous. reproductive online.
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based on 470 customer reviews | 2,625 | 10,761 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2017-34 | latest | en | 0.870365 |
https://9dok.org/document/yr3de747-control-theoretic-approach-for-covid-management.html | 1,675,817,670,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500664.85/warc/CC-MAIN-20230207233330-20230208023330-00529.warc.gz | 99,461,587 | 33,535 | # Control Theoretic Approach for COVID-19 Management
## Full text
(1)
The COVID-19 pandemic is one of the biggest challenges the world is currently facing. Until a vaccine and effective treatment are available, carefully planned measures are needed in every country to control the spread of the dis- ease. Choosing the right management policy is a sensitive task that requires several potentially contradicting objec- tives to be considered. The most impor- tant limiting constraint is the capacity of the healthcare system, which can easily be overwhelmed if the spread of the dis- ease is not controlled. It is clear that the transmission of the virus can be effi- ciently slowed down by appropriate restrictions (social distancing, lock- down), but these measures have nega- tive social and economic impacts that we can’t afford to overlook. At the moment, governments are continuously evaluating their control measures, trying to find a balance between public health concerns and the costs of social dis- tancing measures. This paper shows that control theory provides an appropriate framework for the support of decision- making through the systematic design of optimal management strategies.
A mathematical model of the epidemic spread
The computation of the control policy requires a mathematical model describing the relationship between important time-dependent quantities and capable of predicting the future behav- iour of the epidemic. The most common approach is to use compartmental models [1] for this purpose. In this mod- elling framework the total population is divided into groups (compartments) such that each compartment collects individ- uals of the same infection status. One possible grouping is obtained by intro- ducing the following compartments [2]:
Susceptible (S) collects individuals who can be infected ; Latent (L) contains those who have already contracted the disease, but do not show symptoms and are not infectious yet. Individuals who have just recently been infected and need
a few more days to develop symptoms are collected in class Pre-symptomatic (P). Depending on whether or not an infected individual develops symptoms he/she belongs to the compartment Symptomatic infected (I) or Asymptomatic infected (A). Three addi- tional groups are defined for the Hospitalised (H), Recovered (R) and Deceased (D) individuals. The transition diagram representing the interconnec- tions between the compartments is depicted in Figure 1. The transmission rates are given in the labels of the arrows. The model depends on several parameters (α,β,ρ,etc.) which can be determined and continuously updated by following the current literature and analysing the data registered worldwide on the active COVID-19 cases (e.g., L1).
Formulating COVID management as a control design task
In a control theory framework, dynam- ical systems are considered as operators mapping from an input signal (function) space to an output space. We distinguish between manipulable inputs which can be set (often between certain limits) by the user and disturbance inputs from the environment that cannot be directly influenced. The outputs are either directly measured quantities or they are computed from measurements. The inner variables representing the actual status of the model are the states. The control goals can be prescribed by
defining constraints and optimality cri- teria for the predicted future behaviour of the system. Possible examples for the former are (physical) bounds on the inputs and/or on the state variables and minimal control costs or operation time for the latter. Therefore, a complex con- trol problem can be expressed in the form of constrained optimisation [2]. In the compartmental model introduced above the control input is the scaling factor of coefficient βdetermining infection probability. By applying restrictions of varying stringency index (from mandatory mask wearing through closing of different institutions and lim- iting public gatherings to total lock- down) this factor can be varied between well-defined limits. Assuming that the number of hospitalised and deceased individuals can be reliably documented,
these two quantities are chosen for out- puts. The main goals of epidemic man- agement, such as protecting the health- care system and applying less stringent interventions to avoid social and eco- nomic crisis, can be formalised by defining a strict upper limit for the number of hospitalised individuals (e.g., H≤Hmax) and adding the control cost to the optimality criteria.
State estimation
In order to use the model to predict the future behaviour of the epidemic, infor- mation is needed about the non-meas- ured compartments. The state variables
ERCIM NEWS 124 January 2021
38
Special Theme
### Control Theoretic Approach for COVID-19 Management
by Gábor Szederkényi (Pázmány Péter Catholic University), Tamás Péni (SZTAKI) and Gergely Röst (University of Szeged)
A control theoretic approach can efficiently support the systematic design of strategies to suppress or mitigate the effects of the COVID-19 pandemic.
Figure1:Transitiondiagramofthecompartmentalmodeldescribingthetransmissiondynamics ofCOVID-19.
(2)
corresponding to these have to be esti- mated from the past measurements and the applied control actions using the nonlinear compartmental model.
Illustrative results
Figure 2 presents a simulation result obtained by performing the control design concept above for the Hungarian situation. In this specific scenario, we assumed that the capacity of the health- care system (Hmax) can be temporarily exceeded if needed, but only for a short time and by only a specific amount.
This scenario models the actual, real sit- uation, when there is an extra, but lim- ited and possibly costly reserve in the healthcare system that can be activated if necessary.
This research was partially supported by the Artificial Intelligence Hungarian National Laboratory, https://milab.hu/
[L1] COVID-NET: A weekly summary of US COVID-19 Hospitalization Data: https://kwz.me/h2X References:
[1] S. V. Rakovic, W. S. Levine (editors): “Handbook of Model Predictive Control”, Birkhauser 2019.
[2] F. Brauer, C. Castillo-Chavez, Z.
Feng (editors): “Mathematical Models in Epidemiology”, Texts in Applied Mathematics, 69, 1st ed.
2019.
[3] T. Péni, B. Csutak, G. Szederkényi, G. Röst: “Nonlinear model predictive control with logic constraints for COVID-19 management”, Nonlinear Dynamics, 2020.
Tamás Péni SZTAKI, Hungary peni@sztaki.hu
ERCIM NEWS 124 January 2021 39
Figure2:Simulationresultsobtainedbyapredictivecontrollercomputedbyconstrainedoptimisation.Thegoalistomitigatetheeffectofthe epidemicandprotectthefunctionalityofthehealthcaresystembytakinglessstringentmeasures.Thelimitationofthehealthcaresystemis modelledbyspecifyingtwoupperboundsHmax(1). andHmax(2). withHmax(1). <Hmax(2). forthenumberofhospitalisedpatients(H).Theprimarygoalisto keepHunderHmax(1). .Ifthisisnotpossible,thislimitcanbeexceeded,butonlyuptoHmax(2). andonlyforagiventimeperiod.Thecontrolinputcan varybetween“no-interventio”‘and“totallockdown”‘.Itcanbeseenthattherequiredcontrolgoalcanbeachievedbyapplyingstrictmeasuresat theverybeginningoftheepidemicandsystematicallyeasingtherestrictionsthereafter.Togetherwiththecontrolinput,thebottomfiguredepicts thetimedependentreplicationnumber(Rc)aswell.
Updating...
## References
Related subjects : | 1,625 | 7,423 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2023-06 | latest | en | 0.940132 |
https://izziswift.com/most-lightweight-way-to-create-a-random-string-and-a-random-hexadecimal-number/ | 1,638,546,705,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362891.54/warc/CC-MAIN-20211203151849-20211203181849-00167.warc.gz | 393,135,638 | 11,349 | Most lightweight way to create a random string and a random hexadecimal number
Question or problem about Python programming:
What is the most lightweight way to create a random string of 30 characters like the following?
And an hexadecimal number of 30 digits like the followin?
How to solve the problem:
Solution 1:
I got a faster one for the hex output. Using the same t1 and t2 as above:
>>> t1 = timeit.Timer("''.join(random.choice('0123456789abcdef') for n in xrange(30))", "import random")
>>> t2 = timeit.Timer("binascii.b2a_hex(os.urandom(15))", "import os, binascii")
>>> t3 = timeit.Timer("'%030x' % random.randrange(16**30)", "import random")
>>> for t in t1, t2, t3:
... t.timeit()
...
28.165037870407104
9.0292739868164062
5.2836320400238037
t3 only makes one call to the random module, doesn’t have to build or read a list, and then does the rest with string formatting.
Solution 2:
30 digit hex string:
>>> import os,binascii
>>> print binascii.b2a_hex(os.urandom(15))
The advantage is that this gets randomness directly from the OS, which might be more secure and/or faster than the random(), and you don’t have to seed it.
Solution 3:
In Py3.6+, another option is to use the new standard secrets module:
>>> import secrets
>>> secrets.token_hex(15)
>>> secrets.token_urlsafe(22) # may include '_-' unclear if that is acceptable
'teRq7IqhaRU0S3euX1ji9f58WzUkrg'
Solution 4:
import string
import random
lst = [random.choice(string.ascii_letters + string.digits) for n in xrange(30)]
str = "".join(lst)
print str
ocwbKCiuAJLRJgM1bWNV1TPSH0F2Lb
Solution 5:
Dramatically faster solution than those here:
timeit("'%0x' % getrandbits(30 * 4)", "from random import getrandbits")
0.8056681156158447
Solution 6:
Note: random.choice(string.hexdigits) is incorrect, because string.hexdigits returns 0123456789abcdefABCDEF (both lowercase and uppercase), so you will get a biased result, with the hex digit ‘c’ twice as likely to appear as the digit ‘7’. Instead, just use random.choice('0123456789abcdef').
Solution 7:
Another Method :
from Crypto import Random
import binascii
my_hex_value = binascii.hexlify(Random.get_random_bytes(30))
The point is : byte value is always equal to the value in hex.
Solution 8:
one-line function:
import random
import string
def generate_random_key(length):
return ''.join(random.choice(string.ascii_lowercase + string.digits) for _ in range(length))
print generate_random_key(30)
Solution 9:
In [1]: import random
In [2]: hex(random.getrandbits(16))
Out[2]: '0x3b19'
Solution 10:
Incidentally, this is the result of using timeit on the two approaches that have been suggested:
Using random.choice():
>>> t1 = timeit.Timer("''.join(random.choice(string.hexdigits) for n in xrange(30))", "import random, string")
>>> t1.timeit()
69.558588027954102
Using binascii.b2a_hex():
>>> t2 = timeit.Timer("binascii.b2a_hex(os.urandom(15))", "import os, binascii")
>>> t2.timeit()
16.288421154022217 | 813 | 2,976 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2021-49 | longest | en | 0.654381 |
https://www.physicsforums.com/threads/volume-question.8810/ | 1,670,541,117,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711368.1/warc/CC-MAIN-20221208215156-20221209005156-00075.warc.gz | 993,574,801 | 15,754 | # Volume question
hi :
i have this question, which it reads pretty simple but I am trying to see if someone can walk me through the answer...
A gas cylinder has two vents, V1 and V2, for
discharge. The cylinder empties in 7 minutes when
only vent V1 is open. The cylinder empties in 3
minutes when both vent V1 and vent V2 are open. About
how many minutes will the cylinder take to empty if
only vent V2 is open?
the answer is in actual minutes...
thanks
pedro
## Answers and Replies
arcnets
nampe,
obviously it doesn't matter what the vessel's volume is. So you could just assume it's 100 litres or something, and calculate how many litres per minute go thru each vent.
Homework Helper
You have to know how (why) it empties for the conditions of the vents being open. Is there a piston pushing the stuff out? Is it in an infinite negative pressure reservoir?
Mentor
turin is right, but my guess is that this is meant as an algebra problem, not a physics problem. Is that correct?
In that case, the rates are:
V1 = 1/7 (tanks per minute)
V1 + V2 = 1/3
Solve for V2 and invert to get the time for Valve 2 alone.
thanks every one for your replies... i found the answer and here it is...Think about it like this. For a given cylinder, the product (time to empty) x (total vent size) should be constant. Basically, if you open two vents of the same size, it should take half the time to empty. So let's say V1 and V2 are the sizes (areas) of vents 1 and 2. Then,
7 * V1 = 3 * (V1 + V2) = T * V2
From the first part of the equation, you can solve for V1 in terms of V2. That will then allow you to solve the second half of the equation for T, which is your answer.
so using 7 * V1 = 3 * (V1 + V2) = T * V2, i am solving for the first part of the equation...
7 * V1 = 3 * (V1 + V2)... 7V1=3V1+3V2...V1=3V2/4...now u got to place 7 * V1= T * V2 to obtain T value
7*3V2/4= T*V2...V2=7*3V2/4/V2 which V2=7*3/4=5.25
this was a gre practice question... pretty interesting ehhh?!
thanks again
Homework Helper
Originally posted by nampe
For a given cylinder, the product (time to empty) x (total vent size) should be constant.
Why? Did you find this in the solution guide or something? The rate that the cylinder empties is a lot more intimately related to why it empties, than how it empties.
Jess
Originally posted by turin
Why? Did you find this in the solution guide or something? The rate that the cylinder empties is a lot more intimately related to why it empties, than how it empties.
I assume that the tank empties because the gas inside is at greater than atmospheric pressure. When the valves are opened the tank pressure changes until the pressure in the tank is the same as the pressure outside the tank. This way it doesn't matter how the tank empties as long as the starting tank pressure is the same each time.
Jess
Homework Helper
If inactive pressure equalization is what causes the tank to empty, then it will never be empty (unless you assume that the ambient can be at a negative absolute pressure). Even if we assume that what is meant is pressure equalization with the ambient, then this will still take an infinite amount of time according to the proposed mechanism of strict baromotivation (I think it is a decaying exponential). | 860 | 3,260 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.9375 | 4 | CC-MAIN-2022-49 | latest | en | 0.933461 |
https://www.slideserve.com/ariana-dyer/geometry-powerpoint-ppt-presentation | 1,719,167,810,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862488.55/warc/CC-MAIN-20240623162925-20240623192925-00857.warc.gz | 864,412,814 | 20,285 | 1 / 25
# Geometry
Geometry. 9.1 An Introduction to Circles. Circle. The set of all points in a plane at a given distance from a given point. . P. 3. P is the center of the circle. The circle is the set of all points in the plane of the screen 3 away from P. Radius Plural is radii.
## Geometry
E N D
### Presentation Transcript
1. Geometry 9.1 An Introduction to Circles
2. Circle • The set of all points in a plane at a given • distance from a given point. . P 3 P is the center of the circle. The circle is the set of all points in the plane of the screen 3 away from P.
3. Radius Plural is radii • A segment that joins the center of the • circle to any point on the circle. . P radius
4. Chord • A segment whose endpoints lie on a circle. . P chord
5. Diameter • A chord that passes through the center • of the circle. twice The diameter is _________ as long as the radius. . P diameter
6. Secant • A line that contains a chord of a circle. . P SECANT
7. Tangent • A line in the plane of a circle that • intersects the circle in exactly one point. . P Tangent . Point of tangency
8. F B E A C D Name: 1) The center: A
9. F B E A C D Name: 2) Two diameters: DB and FC
10. F B E A C D Name: 3) A point of tangency: D
11. F B E A C D Name: 4) Four radii: AB and AF and AC AD and
12. F B E A C D Name: 5) A tangent: ED
13. F B E A C D Name: 6) A secant: FC
14. F B E A C D Name: 7) Six chords: FB and and DF DC and and and BC DB FC
15. F B E A C D 8) Why is AC not a Chord of A: . A chord is a segment with both endpoints on the circle. AC has only one endpoint on the circle.
16. F B E A C D 9) Why is BD not a Chord of A: . A chord is a segment not a line.
17. G F H L I X K J Sphere • Sphere: The set of all points in space given distance from any given point is a sphere. • Many of the terms used with circles are also used with spheres. • For example, sphere X has a… center: radii: chords: diameter: secants: tangent: point of tangency:
18. 4 4 Congruent Circles congruent circles/spheres Circles (or spheres) are congruent if they have congruent radii.
19. Concentric Circles concentric circles: Circles that lie in the same plane and have the same center are concentric. concentric spheres: Concentric spheres have the same center.
20. Inscribed Polygon • A polygon is inscribed in a circle if each vertex of the polygon lies on the circle. • A circle is circumscribed about a polygon if each vertex of the polygon lies on the circle. The polygon is inscribed in the circle. Thus, the circle is circumscribed about the polygon. These two sentences have the same meaning.
21. A F C G D In sphere A, draw: 10. a diameter, BC 11. a chord, DE 12. a tangent, CF 13. a secant, DG B C E D
22. 14. Draw two concentric circles. Draw a tangent to one of the circles. Is it tangent to the other circle? • 15. Draw a large circle. Inscribe an isosceles triangle in the circle. • 16. Draw a rectangle. Circumscribe a circle about the rectangle. No.
23. O 6 x B Find the value of x. 6 6 60 3 90 3√3 30 3√3 3√3 x = 6√3
24. x 8 60 O x O 3 4 4 O x Find the value of x. O is the center of each circle. x = 8 x = 5 x x = 4√2 x 10 45 O x = 5√2
25. HW • P. 330-340 CE 1-11 WE 1-15
More Related | 982 | 3,212 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2024-26 | latest | en | 0.824122 |
https://discuss.leetcode.com/topic/26895/why-get-tle-error-if-i-comment-out-the-3rd-to-last-line | 1,516,200,963,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084886946.21/warc/CC-MAIN-20180117142113-20180117162113-00501.warc.gz | 645,687,555 | 10,129 | # Why get TLE error if I comment out the 3rd to last line??
• ``````def partition(self, head, x):
"""
:type x: int
:rtype: ListNode
"""
before = ListNode(0)
fake = before
after = ListNode(0)
before = before.next
else:
after = after.next
#after.next = None if commented out this line, I get TLE!! why??
return fake.next``````
• You get TLE because head.next was not null when you assigned head to after (in the else case of the while loop). This means that after.next is also not null and causes infinite loop.
• but "after.next = None" is out side the while loop? why would it cause infinite loop?
• Consider the input [2,1], 2. This means that at the beginning you have two nodes in the list:
``````head: {val = 2, next = node1}
node1: {val = 1, next = None}
``````
After the loop you get the following nodes:
``````before: {val = 1, next = None} //this is node1
after: {val = 2, next = hode1} //this is head
``````
Since node1 == before, this can be rewritten as:
``````before: {val = 1, next = None} //this is node1
after: {val = 2, next = before}
``````
Since after_head.next is equal to after, the line
``````before.next = after_head.next
``````
changes the state of the nodes to:
``````before: {val = 1, next = after}
after: {val = 2, next = before}
``````
As a result you return a list that contains a cycle and it causes TLE when the judge is trying to verify the answer.
• ohh that makes sense! thank you very much!
Looks like your connection to LeetCode Discuss was lost, please wait while we try to reconnect. | 446 | 1,552 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2018-05 | latest | en | 0.933883 |
https://smartconversion.com/ConversionExample/electrostatic-unit-of-charge-to-ESU-of-charge/1211/1214 | 1,695,476,551,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506481.17/warc/CC-MAIN-20230923130827-20230923160827-00314.warc.gz | 600,755,506 | 5,330 | # Convert electrostatic unit of charge to ESU of charge
Learn how to convert 1 electrostatic unit of charge to ESU of charge step by step.
## Calculation Breakdown
Set up the equation
$$1.0\left(electrostatic \text{ } unit \text{ } of \text{ } charge\right)={\color{rgb(20,165,174)} x}\left(ESU \text{ } of \text{ } charge\right)$$
Define the base values of the selected units in relation to the SI unit $$\left(coulomb\right)$$
$$\text{Left side: 1.0 } \left(electrostatic \text{ } unit \text{ } of \text{ } charge\right) = {\color{rgb(89,182,91)} 3.33564 \times 10^{-10}\left(coulomb\right)} = {\color{rgb(89,182,91)} 3.33564 \times 10^{-10}\left(C\right)}$$
$$\text{Right side: 1.0 } \left(ESU \text{ } of \text{ } charge\right) = {\color{rgb(125,164,120)} 3.33564 \times 10^{-10}\left(coulomb\right)} = {\color{rgb(125,164,120)} 3.33564 \times 10^{-10}\left(C\right)}$$
Insert known values into the conversion equation to determine $${\color{rgb(20,165,174)} x}$$
$$1.0\left(electrostatic \text{ } unit \text{ } of \text{ } charge\right)={\color{rgb(20,165,174)} x}\left(ESU \text{ } of \text{ } charge\right)$$
$$\text{Insert known values } =>$$
$$1.0 \times {\color{rgb(89,182,91)} 3.33564 \times 10^{-10}} \times {\color{rgb(89,182,91)} \left(coulomb\right)} = {\color{rgb(20,165,174)} x} \times {\color{rgb(125,164,120)} {\color{rgb(125,164,120)} 3.33564 \times 10^{-10}}} \times {\color{rgb(125,164,120)} \left(coulomb\right)}$$
$$\text{Or}$$
$$1.0 \cdot {\color{rgb(89,182,91)} 3.33564 \times 10^{-10}} \cdot {\color{rgb(89,182,91)} \left(C\right)} = {\color{rgb(20,165,174)} x} \cdot {\color{rgb(125,164,120)} 3.33564 \times 10^{-10}} \cdot {\color{rgb(125,164,120)} \left(C\right)}$$
$$\text{Cancel SI units}$$
$$1.0 \times {\color{rgb(89,182,91)} 3.33564 \times 10^{-10}} \cdot {\color{rgb(89,182,91)} \cancel{\left(C\right)}} = {\color{rgb(20,165,174)} x} \times {\color{rgb(125,164,120)} 3.33564 \times 10^{-10}} \times {\color{rgb(125,164,120)} \cancel{\left(C\right)}}$$
$$\text{Conversion Equation}$$
$$3.33564 \times 10^{-10} = {\color{rgb(20,165,174)} x} \times 3.33564 \times 10^{-10}$$
Cancel factors on both sides
$$\text{Cancel factors}$$
$${\color{rgb(255,204,153)} \cancel{3.33564}} \times {\color{rgb(99,194,222)} \cancel{10^{-10}}} = {\color{rgb(20,165,174)} x} \times {\color{rgb(255,204,153)} \cancel{3.33564}} \times {\color{rgb(99,194,222)} \cancel{10^{-10}}}$$
$$\text{Simplify}$$
$$1.0 = {\color{rgb(20,165,174)} x}$$
Switch sides
$${\color{rgb(20,165,174)} x} = 1.0$$
Solve $${\color{rgb(20,165,174)} x}$$
$${\color{rgb(20,165,174)} x} = 1$$
$$\text{Conversion Equation}$$
$$1.0\left(electrostatic \text{ } unit \text{ } of \text{ } charge\right) = {\color{rgb(20,165,174)} 1}\left(ESU \text{ } of \text{ } charge\right)$$ | 1,106 | 2,760 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2023-40 | latest | en | 0.30337 |
https://altoalsimce.org/which-three-lengths-cannot-be-the-lengths-of-the-sides-of-a-triangle/ | 1,660,414,557,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571982.99/warc/CC-MAIN-20220813172349-20220813202349-00524.warc.gz | 121,950,368 | 6,058 | Explanation:
Given the Triangle Inequality, the amount of any two political parties of a triangle should be higher than the third side.
Given the measurements
:
Therefore, this lengths cannot stand for a triangle.
Explanation:
Sketching abc in the xy-plane, together pictured here, we view that it has base 6 and also height 3. Since the formula for the area the a triangle is 1/2 * basic * height, the area of abc is 1/2 * 6 * 3 = 9.
You are watching: Which three lengths cannot be the lengths of the sides of a triangle?
The height,
, that triangle
in the figure is one-fourth the length of
. In terms of h, what is the area of triangle
?
Explanation:
If
*
, climate the size of
must be
.
Using the formula for the area that a triangle (
), through
, the area of the triangle must be
.
Explanation:
The area of a triangle = (1/2)bh whereby b is base and h is height. 18 = (1/2)4h which gives us 36 = 4h therefore h =9.
and
are similar triangles. The perimeter that Triangle A is 45” and the size of two of that is sides space 15” and 10”. If the perimeter the Triangle B is 135” and also what space lengths of 2 of the sides?
Explanation:
The perimeter is equal to the sum of the three sides. In comparable triangles, each side is in proportion to its correlating side. The perimeters are additionally in same proportion.
Perimeter A = 45” and perimeter B = 135”
The proportion of Perimeter A to Perimeter B is
This uses to the sides of the triangle. As such to obtain the any side the Triangle B, just multiply the correlating side by 3.
15” x 3 = 45”
10” x 3 = 30“
Explanation:
The sum of the lengths of 2 sides the a triangle can not be less than the size of the 3rd side. 8 + 4 = 12, i m sorry is much less than 13.
Two sides of a triangle are 20 and also 32. Which of the adhering to CANNOT be the 3rd side of this triangle.
Explanation:
Please remember the Triangle Inequality Theorem, which states that the sum of any kind of two sides of a triangle have to be better than the 3rd side. Therefore, the exactly answer is 10 since the amount of 10 and 20 would certainly not be higher than the 3rd side 32.
Explanation:
The sum of the lengths of any type of two political parties of a triangle have to exceed the size of the 3rd side; therefore, 5+7 > x, i beg your pardon cannot occur if x = 13.
The lengths of 2 sides that a triangle room 9 and 7. I m sorry of the following can be the length of the 3rd side?
Explanation:
Let us call the third side x. Follow to the Triangle Inequality Theorem, the sum of any kind of two political parties of a triangle need to be bigger than the various other two sides. Thus, all of the complying with must it is in true:
x + 7 > 9
x + 9 > 7
7 + 9 > x
We can solve these 3 inequalities to determine the possible values of x.
x + 7 > 9
Subtract 7 indigenous both sides.
x > 2
Now, we can look in ~ x + 9 > 7. Subtracting 9 from both sides, us obtain
x > –2
Finally, 7 + 9 > x, which means that 16 > x.
Therefore, x need to be better than 2, higher than –2, but additionally less 보다 16. The only number that satisfies every one of these needs is 12.
The prize is 12.
Report an Error
Example concern #5 : just how To discover The length Of The next Of an Acute / Obtuse Triangle
The lengths the a triangle are 8, 12, and also x. I beg your pardon of the adhering to inequalities shows every one of the possible values that x?
4 Explanation:
According to the Triangle Inequality Theorem, the sum of any kind of two political parties of a triangle should be higher (not higher than or equal) 보다 the continuing to be side. Thus, the adhering to inequalities have to all it is in true:
x + 8 > 12
x + 12 > 8
8 + 12 > x
Let"s fix each inequality.
x + 8 > 12
Subtract 8 from both sides.
x > 4
Next, let"s look in ~ the inequality x + 12 > 8
x + 12 > 8
Subtract 12 native both sides.
See more: The Holy Spirit Cannot Dwell In An Unclean Temple Scripture, Scripture Is Clear That We Are The Temple Of God:
x > –4
Lastly, 8 + 12 > x, which means that x –4.
To summarize, x have to be higher than 4 and also less than 20. We can write this together 4 Report an Error
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Charles Cohn Varsity Tutors llc 101 S. Hanley Rd, Suite 300 St. Louis, MO 63105 | 1,304 | 4,992 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.625 | 5 | CC-MAIN-2022-33 | latest | en | 0.930425 |
https://www.coursehero.com/tutors-problems/Mechanical-Engineering/536803-I-have-been-tasked-with-solving-y-3y2-0-using-the-technique-used/ | 1,542,557,449,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039744513.64/warc/CC-MAIN-20181118155835-20181118181835-00452.warc.gz | 823,648,696 | 20,714 | View the step-by-step solution to:
# I have been tasked with solving y'' - 3y^2 =0 using the technique used substituting v for y', therefore substituting v dv/dy for y''.
I have been tasked with solving y'' - 3y^2 =0 using the technique used substituting v for y', therefore substituting v dv/dy for y''. (Equation with "x" missing)
I broke it down as follows
Y'' -3y^2 =0
Y'' =3y^2
Substituting I get v dv/dy = 3y^2
Separating variables, I get v dv =3y^2dy
Integrating I get 1/2v^2 +c = y^3 +c
Solving for v, I get v = (2y^3-2c)^1/2 which back substituting, = dy/dx
From here, it is a blur!! Initial conditions are y(0) =2, y'(0) =4
Can anybody help me from here?
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Browse Documents | 315 | 1,052 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2018-47 | latest | en | 0.860311 |
https://apkcombo.com/learning-numbers-123-for-kids/suave.kidsnumbers.free/ | 1,601,017,773,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400222515.48/warc/CC-MAIN-20200925053037-20200925083037-00263.warc.gz | 269,857,677 | 18,349 | # Learning Numbers 123 for Kids
1.2.8
Aug 27, 2020 (1 month ago)
## Learning Numbers 123 for Kids APK
An easy and fun way, learning App for kids to identify and recognize Numbers.
Version 1.2.8 (19) Updated Aug 27, 2020 (1 month ago) Developer Suave Solutions Category Games, Education ID suave.kidsnumbers.free Installs 10,000+
It’s time to get this amazing numbers apps in your device for free which has been updated with more exciting features. Now, enjoy vast learning activity with fun in “Start Learning” and “Start Activity” options.
Tap the Learning button to get a lesson of 1 to 10 numbers which are designed also to display the objects with given numbers. Then tap on Activity button and explore lots of fun quiz into the main 6 features of complete sequence, identify numbers, guess the positions, count and Ans, let’s count and sequencing path. This educational app for kids is best to cater their learning and playing needs as well.
Features:
● Learning activity
● Quiz Activity
● Complete the sequence
● Identify the numbers
● Guess the positions
● Count and Ans
● Let’s count
● Sequencing path
● Interactive sounds
● Muting option
● Switching option-For switching one to another activity
● Verbal play guide
● Written play guide on screen
● Colorful graphics
● Social sharing buttons
How to play?
First download this app from Google play store for free and be ready to play with numbers. On first hand, there’s Start Learning feature which works like a complete reading book of numbers. There 1-10 numbers are available to learn and each number is linked with objects. The teaching background voice reads every number for the player and even it may also be muted. Then 2nd main feature of Start Activity is for bringing the user to play 6 activities in following ways.
1. Complete the sequence is a play to drag the given below numbers and drop into the blank boxes by following the correct sequence of counting.
2. Identify the numbers provides multiple stages activity to identify the written number from the given objects holding multiple digits.
3. Guess the positions is another exciting quiz of guessing the position of items by following the displayed written instructions. If it’s difficult for kids to read and understand the wording then teachers and parents can well perform their role here and help them.
4. This count and Ans activity is to count those objects which are equal to the given digit and then tap on them to answer.
5. Let’s count invites the player to count the displayed things and simply tap on the right number.
6. Sequencing path is truly amazing quiz which asks to complete a numeric path by dropping the correct digits in blank areas. Thus, a player completes the path for animals and bird, then they reach their destinations happily.
Dear parents and teachers!
The era of e-learning has brought many advance and effective ways to involve kid’s all 5 senses in learning because this involvement contains better learning. By considering this fact, Little Tree House offers best educational apps for kids to support them to prepare their academic lessons.
This learning Numbers 123 app is a commendable aiding tool for kids of 3-6 years old. It’s completely based on their syllabus and develops their interest to learn about numbers in playing and amusement.
So, try this entertaining and educational app for your preschoolers which is available free for all Android devices. App may have ads.
- Bug Fixes | 726 | 3,465 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2020-40 | latest | en | 0.91277 |
https://www.amelia.mn/presentations/TeachingRtoHS.html | 1,643,433,880,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320299927.25/warc/CC-MAIN-20220129032406-20220129062406-00598.warc.gz | 679,350,144 | 619,541 | July 1, 2014
## Goals and challenges
Goals
• \(\#\)useR2014
• "Authentic data-science experience." - Amelia
• "Make statisticians popular at parties." - James
Challenges
• Inexperienced teachers
• Limited time frames
• Curricular materials
• Learning to \(\#\)useR2014 can be hard
## 1. Simple syntax
`xyplot(mpg ~ wt | as.factor(cyl), data = mtcars)`
vs.
```par(mfrow = c(1,3))
plot(mtcars\$wt[mtcars\$cyl == 4], mtcars\$mpg[mtcars\$cyl == 4])
plot(mtcars\$wt[mtcars\$cyl == 6], mtcars\$mpg[mtcars\$cyl == 6])
plot(mtcars\$wt[mtcars\$cyl == 8], mtcars\$mpg[mtcars\$cyl == 8])```
```## null device
## 1```
## 2. Wrappers
`MakeWordBar`
```## function (text, min.freq = 2, top = 50, format = "count", col = "steelblue",
## color, ...)
## {
## if (!missing(color)) {
## stop("Remember to use the argument 'col' and not 'color'.")
## }
## if (class(text)[1] != "VCorpus") {
## stop("Remember to initialize text using initializeText()")
## }
## if (!missing(format) & (format != "count" & format != "percent")) {
## stop("invalid 'format' argument. 'format' must either be 'count' or 'percent'")
## }
## old.par <- par(no.readonly = TRUE)\$mar
## tdm <- TermDocumentMatrix(text)
## m <- as.matrix(tdm)
## v <- sort(rowSums(m), decreasing = T)
## d <- data.frame(word = names(v), freq = v, row.names = NULL)
## d <- d[d\$freq >= min.freq, ]
## if (format == "count") {
## if (!missing(top) & (top <= 0)) {
## stop("'top' must be a number greater than 0")
## }
## top.words <- head(d\$freq, n = top)
## top.words.names <- head(d\$word, n = top)
## largest.word <- max(nchar(as.character(top.words.names)))
## if (largest.word > 8) {
## adj = (largest.word - 7)/3
## par(mar = old.par + c(adj, 0, 0, 0))
## }
## else {
## par(mar = old.par)
## }
## }
## if (format == "percent") {
## if (missing(top)) {
## stop("'top' argument must be specificed when using format = 'percent'")
## }
## if (!missing(top) & (top <= 0 | top > 100)) {
## stop("'top' must be a number greater than 0 and less than or equal to 100")
## }
## n <- floor(nrow(d) * top/100)
## top.words <- head(d\$freq, n = n)
## top.words.names <- head(d\$word, n = n)
## largest.word <- max(nchar(as.character(top.words.names)))
## if (largest.word > 8) {
## adj = (largest.word - 7)/3
## par(mar = old.par + c(adj, 0, 0, 0))
## }
## else {
## par(mar = old.par)
## }
## }
## barplot(top.words, names.arg = top.words.names, las = 2,
## col = col, border = "white", ...)
## par(mar = old.par)
## }
## <environment: namespace:MobilizeSimple>```
```data("crude")
crude <- ProcessText(crude, removestopwords = TRUE)
MakeWordBar(crude, top=10)```
• MobilizeSimple package
• www.github.com/mobilizingcs | 937 | 3,057 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2022-05 | latest | en | 0.348881 |
https://www.archtoolbox.com/calculating-area/ | 1,709,098,879,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474697.2/warc/CC-MAIN-20240228044414-20240228074414-00154.warc.gz | 643,258,492 | 13,477 | # Calculating Area
Below are a series of diagrams that demonstrate how to calculate the area of two dimensional shapes.
## Area of a Circle
The area of a circle is the product of pi and the square of the radius.
Area = pi (r x r)
## Area of a Ellipse
An ellipse is flattened circle so the area calculation takes into account that there are two different radii.
Area = pi (r1 x r2)
## Area of a Triangle
The area of a triangle is half of the product of the base and height.
Area = (h x B)/2
## Area of an Equilateral Triangle
An equilateral triangle has 3 equal length sides. While the formula above will also work, there is an alternative method of calculating the area of an equilateral triangle.
Area = ((sqrt 3) (A x A)) / 4
## Area of a Rectangle
The area of a rectangle is the product of the long side and the short side. The area of a square is the length of a side squared.
Area = A x B
## Area of a Parallelogram
A parallelogram is rectangle that is skewed slightly. The area is a product of the base and the overall height.
Area = B x h
## Area of a Trapezoid
Trapezoids are half of a parallelogram. The diagram at left shows how 2 equal trapezoids make a parallelogram. Therefore, the area calculation is simply half of the area of a parallelogram.
Area = 1/2 (h (A + B))
A quadrilateral is simply two triangles put together, so the area can be calculated by summing the areas of the two triangles.
Area - (1/2 (h1 x B)) + (1/2 (h2 x B))
Article Updated: May 6, 2021
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https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(LibreTexts)/05%3A_The_Harmonic_Oscillator_and_the_Rigid_Rotor/5.01%3A_A_Harmonic_Oscillator_Obeys_Hooke's_Law | 1,726,296,500,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651559.58/warc/CC-MAIN-20240914061427-20240914091427-00482.warc.gz | 137,696,674 | 36,978 | # 5.1: A Harmonic Oscillator Obeys Hooke's Law
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The motion of two atoms in a diatomic molecule can be separated into translational, vibrational, and rotational motions. Both rotation and vibrational motions are internal motions that do not change the center of mass for the molecule (Figure 5.1.1 ), which is described by translational motion. Quantum translational motions can be modeled with the particle in a box model discussed previously and rotation and vibration can be modeled via the rigid rotor and harmonic oscillator models, respectively.
Before delving into the quantum mechanical harmonic oscillator, we will introduce the classical harmonic oscillator (i.e., involving classical mechanics) to build an intuition that we will extend to the quantum world. A classical description of the vibration of a diatomic molecule is needed because the quantum mechanical description begins with replacing the classical energy with the Hamiltonian operator in the Schrödinger equation. It also is interesting to compare and contrast the classical description with the quantum mechanical picture.
## The Classical Harmonic Oscillator
Simple harmonic oscillators about a potential energy minimum can be thought of as a ball rolling frictionlessly in a curved dish or a pendulum swinging frictionlessly back and forth (Figure 5.1.2 ). The restoring forces are precisely the same in either horizontal direction.
If we consider the bond to behave like a mass on a spring (Figure 5.1.2 ), then this restoring force ($$F$$) is proportional to the displacement ($$x$$) from the equilibrium length ($$x_o$$) - this is Hooke's Law:
$F = - kx \label {5.1.2}$
where $$k$$ is the force constant. Hooke's Law says that the force is proportional to, but in opposite direction to, the displacement ($$x$$). The force constant reflects the stiffness of the spring. The idea incorporated into the application of Hooke's Law to a diatomic molecule is that when the atoms move away from their equilibrium positions, a restoring force is produced that increases proportionally with the displacement from equilibrium. The potential energy for such a system increases quadratically with the displacement.
$V (x) = \dfrac {1}{2} k x^2 \label {5.1.3}$
Hooke's Law or the harmonic (i.e. quadratic) potential given by Equation $$\ref{5.1.3}$$ is an excellent approximation for the vibrational oscillations of molecules. The magnitude of the force constant $$k$$ depends upon the nature of the chemical bond in molecular systems just as it depends on the nature of the spring in mechanical systems. The larger the force constant, the stiffer the spring or the stiffer the bond. Since it is the electron distribution between the two positively charged nuclei that holds them together, a double bond with more electrons has a larger force constant than a single bond, and the nuclei are held together more tightly.
##### Caution
A stiff bond with a large force constant is not necessarily a strong bond with a large dissociation energy. A harmonic oscillator has no dissociation energy since it CANNOT be broken - there is always a restoring force to keep the molecule together. This is one of many deficiencies in using the harmonic oscillator model to describe molecular vibrations.
##### Two atoms or one?
You may have questioned the applicability of the harmonic oscillator model involving one moving mass bound to a fix wall via a spring like in Figure 5.1.2 for the vibration of a diatomic molecule with two moving masses like in Figure 5.1.1 . It turned out the two are mathematically the same with internal vibration motion described by a single reduced particle with a reduced mass $$μ$$.
For a diatomic molecule, Figure 5.1.3 , the vector $$\vec{r}$$ corresponds to the internuclear axis. The magnitude or length of $$r$$ is the bond length, and the orientation of $$r$$ in space gives the orientation of the internuclear axis in space. Changes in the orientation correspond to rotation of the molecule, and changes in the length correspond to vibration. The change in the bond length from the equilibrium bond length is the vibrational coordinate for a diatomic molecule.
##### Example 5.1.1
1. Show that minus the first derivative of the harmonic potential energy function in Equation $$\ref{5.1.3}$$ with respect to $$x$$ is the Hooke's Law force in Equation \ref{5.1.2}.
2. Show that the second derivative is the force constant, $$k$$.
3. At what value of $$x$$ is the potential energy a minimum; at what value of $$x$$ is the force zero?
4. Sketch graphs to compare the potential energy and the force for a system with a large force constant to one with a small force constant.
Solution a
Hooke's Law for a spring entails that the force applied on a spring $$F$$ is equal to the force constant, $$-k$$ times the distance compressed or stretched, $$x$$ (Equation \ref{5.1.2}). The derivative of $$V(x) = 0.5 k x^2$$ is
$V'(x) = (2)\left(\dfrac{1}{2}\right)kx = kx. \nonumber$
The negative of this is $$-V'(x) = -kx$$ which is exactly equal to Hooke's Law.
Solution b
The second derivative
$V"(x) = \dfrac{d}{dx} kx = k \nonumber$
Thus, the second derivative of this equation for potential energy is equal to the force constant, $$k$$.
Solution c
To find the minimum potential energy, it is easiest to set the first derivative equal to zero and solve for x. When $$V'(x) = kx = 0$$ then x must be equal to zero. Thus, the minimum potential energy is when x=0. Plugging this into Hooke's Law, $$F(0) = -k(0) = 0$$ so this is also the value for x when the force is zero.
Solution d
The force constant has a drastic effect on both the potential energy and the force. A system with a large force constant requires minimal change in $$x$$ to have a drastic change in potential energy or force, whereas a system with a small force constant is the exact opposite phenomenon.
## Solving the Harmonic Oscillator Model
The classical equation of motion for a one-dimensional simple harmonic oscillator with a particle of mass $$m$$ attached to a spring having spring constant $$k$$ is
$m \dfrac{d^2x(t)}{dt^2} = -kx(t) \label{5.1.4a}$
which can be rewritten in the standardform:
$\dfrac{d^2x(t)}{dt^2} + \dfrac{k}{m}x(t) = 0 \label{5.1.4b}$
Equation $$\ref{5.1.4a}$$ is a linear second-order differential equation that can be solved by the standard method of factoring and integrating. The resulting solution to Equation $$\ref{5.1.4a}$$ is
$x(t) = x_o \sin (\omega t + \phi) \label{5.1.5}$
with
$\omega = \sqrt{\dfrac{k}{m}} \label{5.1.6}$
and the momentum has time dependence
\begin{align*} p &= mv \\[4pt] &=mx_o \omega \cos (\omega t + \phi) \label{5.1.7} \end{align*}
Figure 5.1.4 show the displacement of the bond from its equilibrium length as a function of time. Such motion is called harmonic.
##### Example 5.1.2
Substitute the following functions into Equation $$\ref{5.1.4b}$$ to demonstrate that they are both possible solutions to the classical equation of motion.
1. $$x(t) = x_0 e^{i \omega t}$$
2. $$x(t) = x_0 e^{-i \omega t}$$
where
$\omega = \sqrt {\dfrac {k}{m}} \nonumber$
Note that the Greek symbol $$\omega$$ for frequency represents the angular frequency $$2π\nu$$.
Solution a
This requires simply placing the given function $$x(t) = x_0 e^{i \omega t}$$ into Equation $$\ref{5.1.4b}$$.
\begin{align*} \frac{d^2 x(t) }{dt^2} + \frac{k}{m} x(t) &= 0 \\[4pt] \frac{d^2 }{dt^2} \left( x_o e^{i \omega t} \right)+ \frac{k}{m}x_o e^{i \omega t} &= 0\\[4pt] x_o \frac{d^2 }{dt^2} \left( e^{i \omega t} \right)+ \frac{k}{m}x_o e^{i \omega t} &= 0 \\[4pt] x_o i^2 \omega^2 \left( e^{i \omega t} \right)+ \frac{k}{m}x_o e^{i \omega t} &= 0 \\[4pt] x_o i^2 \omega^2 e^{i \omega t} + \frac{k}{m}x_o e^{i \omega t} &= 0 \\[4pt] x_o i^2 \omega^2 + \frac{k}{m}x_o &= 0 \\[4pt] -x_o \frac{k}{m}+ \frac{k}{m}x_o = 0 \; \textrm{ with } \; \omega &= \sqrt{\frac{k}{m}} \end{align*} \nonumber
Solution b
This requires simply placing the given function $$x(t) = x_0 e^{-i \omega t}$$ into Equation $$\ref{5.1.4b}$$.
\begin{align*} \frac{d^2 x(t) }{dt^2} + \frac{k}{m} x(t) &= 0 \\[4pt] \frac{d^2 }{dt^2} \left( x_o e^{-i \omega t} \right)+ \frac{k}{m}x_o e^{-i \omega t}& = 0 \\[4pt] x_o \frac{d^2 }{dt^2} \left( e^{-i \omega t} \right)+ \frac{k}{m}x_o e^{-i \omega t} &= 0 \\[4pt] x_o i^2 \omega^2 \left( e^{-i \omega t} \right)+ \frac{k}{m}x_o e^{-i \omega t} &= 0 \\[4pt] x_o i^2 \omega^2 e^{-i \omega t} + \frac{k}{m}x_o e^{-i \omega t}& = 0\\[4pt] x_o i^2 \omega^2 + \frac{k}{m}x_o &= 0 \\[4pt] -x_o \frac{k}{m}+ \frac{k}{m}x_o = 0 \; \textrm{ with } \; \omega &= \sqrt{\frac{k}{m}} \end{align*} \nonumber
##### Exercise 5.1.1
Show that sine and cosine functions also are solutions to Equation $$\ref{5.1.4b}$$.
Using Equation $$\ref{5.1.4b}$$
$\frac{d^2x(t)}{dt^2} +\frac{k}{m}x(t)=0 \nonumber$
with $$\omega =\sqrt{\frac{k}{m}}$$.
For
\begin{align*} x(t)=x_{o} \sin(\omega t+\phi) \end{align*} \nonumber
Take the second derivative of $$x(t)$$
\begin{align*} \frac{d^2x(t)}{dt} &=-\omega^2x_{o}\sin(\omega t+\phi) \\[4pt] &= \omega^2 x_{o}\sin(\omega t+\phi) \\[4pt] &=-\frac{k}{m}x_{o} \sin \left(\sqrt{\frac{k}{m}}t+\phi \right) \end{align*} \nonumber
Plug in $$x(t)$$ and the second derivative of $$x(t)$$ into Equation $$\ref{5.1.4b}$$
\begin{align*} -\frac{k}{m}x_{o}\sin \left(\sqrt{ \frac{k}{m}} t + \phi \right) + \frac{k}{m} x_{o} \sin \left(\sqrt{\frac{k}{m}} t + \phi \right) =0 \end{align*} \nonumber
Hence, the sine equation is a solution to Equation $$\ref{5.1.4b}$$
We do the same for the cosine function
$x(t)=x_{o}\cos(\omega t + \phi) \nonumber$
Take the second derivative of $$x(t)$$
\begin{align*} \frac{d^2x(t)}{dt} &=-\omega^2x_{o}\cos(\omega t + \phi) \\[4pt] &= -\omega^2x_{o}\cos( \omega t +\phi) \\[4pt] &=-\frac{k}{m} x_{o} \cos\left( \sqrt{\frac{k}{m}} t + \phi \right ) \end{align*} \nonumber
Plug in $$x(t)$$ and the second derivative of $$x(t)$$ into Equation $$\ref{5.1.4b}$$
\begin{align*} -\frac{k}{m}x_{o} \cos\left( \sqrt{\frac{k}{m}} t + \phi \right) + \frac{k}{m}x_{o} \cos\left( \sqrt{\frac{k}{m}}t + \phi \right)=0 \end{align*} \nonumber
The cosine equation is also a solution to Equation $$\ref{5.1.4b}$$.
##### Exercise 5.1.2
Identify what happens to the frequency of the motion as the force constant increases in one case and as the mass increases in another case. If the force constant is increased 9-fold and the mass is increased by 4-fold, by what factor does the frequency change?
This is a simple application of Equation \ref{5.1.6}. As the force constant increases, the frequency of the motion increases, while as the mass increases, the frequency of the motion decreases. If the force constant increased 9-fold and the mass increased 4-fold,
$ω=\sqrt{\dfrac{9k}{4m}}= \dfrac{3}{2} \left(\dfrac{k}{m}\right) \nonumber$
The entire frequency of motion would increase by a factor of 3/2.
## Harmonic Oscillator Energies
The energy of the vibration is the sum of the kinetic energy and the potential energy. The momentum associated with the harmonic oscillator is
$p = m \dfrac {dx}{dt} \label {5.1.8}$
so combining Equations \ref{5.1.8} and \ref{5.1.3}, the total energy can be written as
\begin{align} E &= T + V \\[4pt] &= \dfrac {p^2}{2 m} + \dfrac {k}{2} x^2 \label {5.1.9} \end{align}
The total energy of the harmonic oscillator is equal to the maximum potential energy stored in the spring when $$x = \pm A$$, called the turning points (Figure 5.1.5 ). The total energy (Equation $$\ref{5.1.9}$$) is continuously being shifted between potential energy stored in the spring and kinetic energy of the mass.
The motion of a classical oscillator is confined to the region where its kinetic energy is nonnegative, which is what the energy relation Equation \ref{5.1.9} says. Physically, it means that a classical oscillator can never be found beyond its turning points, and its energy depends only on how far the turning points are from its equilibrium position. The energy of a classical oscillator changes in a continuous way. The lowest energy that a classical oscillator may have is zero, which corresponds to a situation where an object is at rest at its equilibrium position. The zero-energy state of a classical oscillator simply means no oscillations and no motion at all (a classical particle sitting at the bottom of the potential well in Figure 5.1.5 ). When an object oscillates, no matter how big or small its energy may be, it spends the longest time near the turning points, because this is where it slows down and reverses its direction of motion. Therefore, the probability of finding a classical oscillator between the turning points is highest near the turning points and lowest at the equilibrium position. (Note that this is not a statement of preference of the object to go to lower energy. It is a statement about how quickly the object moves through various regions.)
##### Example 5.1.3
1. What happens to the frequency of the oscillation as the vibration is excited with more and more energy?
2. What happens to the maximum amplitude of the vibration as it is excited with more and more energy?
###### Solution
a. Frequency
The energy of the harmonic oscillator can be written as
$E_{v}=h v\left(v+\dfrac{1}{2}\right) \nonumber$
and the frequency of oscillation is $$\omega=\sqrt{\frac{k}{m}}$$. Notice that the frequency depends only on the stiffness ($$k$$) and reduced mass ($$\mu$$) of the oscillator and not on the energy. Hence, increasing the energy of the vibrations does not affects its frequency.
b. Amplitude
The kinetic and potential terms for energy of the harmonic oscillator can be written as
\begin{align*} E &=K+V \\[4pt] &=\frac{1}{2} m \omega^{2} A^{2} \sin ^{2} \omega t+\frac{1}{2} k A^2 \cos^2 \omega t \end{align*} \nonumber
with $$\omega=\sqrt{\frac{k}{m}}$$ so
\begin{align*} E &=\frac{1}{2} k A^{2}\left(\sin ^{2} \omega t+\cos^2 \omega t\right) \\[4pt] &= \frac{1}{2} k A^2 \end{align*} \nonumber
The maximum amplitude of the vibration will increase as the energy increases.
##### Exercise 5.1.3
If a molecular vibration is excited by collision with another molecule and is given a total energy $$E_{hit}$$ as a result, what is the maximum amplitude of the oscillation? Is there any constraint on the magnitude of energy that can be introduced?
The equation that defines the energy of a molecular vibration can be approximated is:
$E_{h i t}=T+V=\frac{p^{2}}{2 m}+\frac{k}{2} x \nonumber$
The maximum amplitude of a harmonic oscillator is equal to x when the kinetic energy term of total energy equals zero
$E_{hit}=\frac{k}{2}x \nonumber$
Solving for x gives the maximum amplitude:
$x=\sqrt{\frac{2}{k} E_{h i t}} \nonumber$
The constraint for the energy that can be introduced cannot be greater than the energy required to break the bond between atoms. | 6,044 | 19,195 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2024-38 | latest | en | 0.198594 |
https://codegolf.stackexchange.com/questions/75129/rotate-every-2x2-block-in-a-matrix/207380#207380 | 1,719,042,303,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862252.86/warc/CC-MAIN-20240622045932-20240622075932-00439.warc.gz | 145,056,369 | 46,462 | # The Challenge
Given a n x m matrix with n > 1 and m > 1 filled with integers
1 2 3
4 5 6
and a list of integers with exactly as many values as 2x2 blocks in the matrix ((n-1)*(m-1) if you need the exact number)
[1, 2]
Output the matrix with every 2x2 block rotated by the current value in the list in the given order. The example above would yield
4 6 2
5 3 1
The first block gets rotated one time to the right and the second block gets rotated two to the right.
## Notes
• A positive integer means you rotate right by that many steps.
• A negative integer means you rotate left by that many steps.
• A zero means that you don't rotate.
• You rotate the blocks row-wise. That means that you start in the first row and go to the right. Once you rotated every block in that row you go to the next one. At the end every block was rotated exactly once.
• Keep in mind that the blocks overlap each other. The first matrix above has the blocks [[1,2],[4,5]] and [[2,3],[5,6]] for example.
• Each rotation of a block affects the rotation on the adjacent blocks. This is why you have to do the rotations in the pattern described above.
# Rules
• You may take the input in the most convenient format. Please specify in your answer which one you use. This does not allow you to read the matrix block-wise though.
• Function or full program allowed.
• Default rules for input/output.
• Standard loopholes apply.
• This is , so lowest byte-count wins. Tiebreaker is earlier submission.
# Test cases
Input format here is a list of lists for the matrix and a normal list for the values.
[[1,2],[3,4]], [-3] -> [[4,1],[3,2]]
[[1,1,1],[1,1,1]], [-333, 666] -> [[1,1,1],[1,1,1]]
[[1,2,3],[4,5,6]], [1,2] -> [[4,6,2],[5,3,1]]
[[1,2,3],[4,5,6],[7,8,9]], [4,0,12,-20] -> [[1,2,3],[4,5,6],[7,8,9]]
[[1,2,3,4,5],[5,4,3,2,1],[1,2,3,4,5]], [2,-3,4,1,6,24,21,-5] -> [[4,1,5,2,4],[2,1,3,5,5],[3,2,4,3,1]]
Happy Coding!
• first test wrong?
– ngn
Commented Jul 22, 2020 at 8:50
{(\@2$,(/.{@a@a+z(\@.{4,={+2/zW%~}*}z~}} Online demo Basically this applies the same technique twice in order to fold the block {4,={+2/zW%~}*} which operates on a 2x2 matrix and a number of times to rotate. [first row] [[second row] [third row] ... [last row]] [value_0 value_1 ... value_n] can be processed with .{block} and has the effect of [first row] [second row] value_0 {block}~ [third row] value_1 {block}~ ... because . (like % in CJam) doesn't gather the results into an array until it's finished. • You can save a byte by using 4,= for correct modulo (unless your block needs to run at least once?). Commented Mar 8, 2016 at 23:13 • Also what about zW% for the rotation? Commented Mar 8, 2016 at 23:20 • @MartinBüttner, I thought the rotation seemed too long, but I couldn't remember the shorter one. Nice trick on the modulo. Commented Mar 9, 2016 at 8:10 # APL (Dyalog Unicode), 37 35 bytes {⌽∘⍉@⍺⊢⍵}/⌽(⊂,{(4|⎕)/,⊢∘⊂⌺2 2⍳⍴⍵})⎕ Try it online! Switched to a more straightforward method after I realized @ also accepts a matrix of coordinates. Then we don't need to fiddle with the coordinate order; we extract the submatrix coordinates with ⊢∘⊂⌺2 2, and just rotate them directly using ⌽∘⍉. # APL (Dyalog Unicode), 37 bytes {1⌽@⍺⊢⍵}/⌽(⊂,{(4|⎕)/,2,∘⌽/2,⌿⊂¨⍳⍴⍵})⎕ Try it online! A full program that takes the matrix, then the vector of rotation amounts. Prints the resulting matrix with a leading space. @ can extract the elements at certain positions, manipulate them, and place them back into the original matrix, which is great for 2×2 rotation. In particular, 1⌽@(1 1)(2 1)(2 2)(1 2) extracts the top left submatrix [a b][c d] into a vector a c d b, rotates once to the left (1⌽) into c d b a, then puts the values back so that the submatrix becomes [c a][d b]. This achieves rotating the submatrix exactly once. {1⌽@⍺⊢⍵}/⌽(⊂,{(4|⎕)/,2,∘⌽/2,⌿⊂¨⍳⍴⍵})⎕ ⍝ Read from right: ⎕ ⍝ Take the matrix from stdin {...} ⍝ Pass to the dfn as ⍵ ⍳⍴⍵ ⍝ Matrix of 2D coordinates of ⍵ 2,⌿⊂¨ ⍝ Pair vertically adjacent coordinates 2,∘⌽/ ⍝ Catenate horizontally adjacent coordinate pairs, ⍝ flipping the right one so that it looks like (1 1)(2 1)(2 2)(1 2) , ⍝ Flatten the matrix of lists of coordinates (4|⎕)/ ⍝ Copy each (Rotations modulo 4) times ⌽(⊂,...) ⍝ Prepend the original matrix enclosed and reverse the entire array, ⍝ so that it is suitable for RTL reduce { }/ ⍝ RTL reduce by... 1⌽@⍺⊢⍵ ⍝ Take the matrix ⍵ and rotate once at coordinates ⍺ • a⊣⎕{a⊢←⌽∘⍉⍣⍺@⍵⊢a}¨,⊢∘⊂⌺2 2⍳⍴a←⎕ – ngn Commented Jul 22, 2020 at 9:27 ## CJam, 656360 55 bytes There must be a better way to do this... {_e_\z,:N(@/Ta*ee{~4,=,f{;1$,,\[XTN_)]f+_(+er\f=}~}/N/}
This is an unnamed function that expects the instructions and the matrix (in that order) on the stack and leaves the resulting matrix in their place.
Test it here.
### Explanation
I don't feel like writing the full breakdown for the code right now, so here is a rough overview:
• 2D array manipulation is a pain in CJam, so I'm instead unrolling the matrix, compute each rotation as a permutation of the elements at specific positions and then split the array into rows again at the end. The width of the matrix is stored in N.
• A rotation at position k in the unrolled array changes four indices: k <- k+1, k+1 <- k+1+N, k+N <- k, k+1+N <- k+1. For each index k along the instruction list, I compute a permutation corresponding to this, and apply it to the unrolled input array.
• This leaves the problem, that in the linear array some rotations will be positioned with its top left corner in the last column of the input. To skip these, I riffle zeroes into the instruction list, such that these invalid 2x2 sub-blocks are technically processed, but with a no-op.
# Python 2, 166 159 bytes
A,R=input();m=~-len(A[0])
for j,r in enumerate(R):exec r%4*"a,b,c,d=A[j/m][j%m:][:2]+A[j/m+1][j%m:][:2];A[j/m][j%m:j%m+2]=c,a;A[j/m+1][j%m:j%m+2]=b,d;"
print A
Try it online! | 2,046 | 5,896 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2024-26 | latest | en | 0.897756 |
https://nrich.maths.org/public/leg.php?code=-420&cl=3&cldcmpid=756 | 1,524,248,713,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125944677.39/warc/CC-MAIN-20180420174802-20180420194802-00076.warc.gz | 684,946,627 | 9,788 | Search by Topic
Resources tagged with smartphone similar to Cayley:
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Two and Two
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How many solutions can you find to this sum? Each of the different letters stands for a different number.
Number Daisy
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Can you find six numbers to go in the Daisy from which you can make all the numbers from 1 to a number bigger than 25?
Ben's Game
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Ben passed a third of his counters to Jack, Jack passed a quarter of his counters to Emma and Emma passed a fifth of her counters to Ben. After this they all had the same number of counters.
Difference Sudoku
Stage: 4 Challenge Level:
Use the differences to find the solution to this Sudoku.
Product Sudoku
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The clues for this Sudoku are the product of the numbers in adjacent squares.
Consecutive Numbers
Stage: 2 and 3 Challenge Level:
An investigation involving adding and subtracting sets of consecutive numbers. Lots to find out, lots to explore.
Consecutive Negative Numbers
Stage: 3 Challenge Level:
Do you notice anything about the solutions when you add and/or subtract consecutive negative numbers?
Weights
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Different combinations of the weights available allow you to make different totals. Which totals can you make?
M, M and M
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If you are given the mean, median and mode of five positive whole numbers, can you find the numbers?
Squares in Rectangles
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A 2 by 3 rectangle contains 8 squares and a 3 by 4 rectangle contains 20 squares. What size rectangle(s) contain(s) exactly 100 squares? Can you find them all?
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Stage: 3 Challenge Level:
Many numbers can be expressed as the sum of two or more consecutive integers. For example, 15=7+8 and 10=1+2+3+4. Can you say which numbers can be expressed in this way?
Stage: 3 Challenge Level:
How many different symmetrical shapes can you make by shading triangles or squares?
One and Three
Stage: 4 Challenge Level:
Two motorboats travelling up and down a lake at constant speeds leave opposite ends A and B at the same instant, passing each other, for the first time 600 metres from A, and on their return, 400. . . .
Stage: 3 Challenge Level:
A game for 2 or more people, based on the traditional card game Rummy. Players aim to make two `tricks', where each trick has to consist of a picture of a shape, a name that describes that shape, and. . . .
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Children at Large
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There are four children in a family, two girls, Kate and Sally, and two boys, Tom and Ben. How old are the children?
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Make 37
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Four bags contain a large number of 1s, 3s, 5s and 7s. Pick any ten numbers from the bags above so that their total is 37.
Eight Hidden Squares
Stage: 2 and 3 Challenge Level:
On the graph there are 28 marked points. These points all mark the vertices (corners) of eight hidden squares. Can you find the eight hidden squares?
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My two digit number is special because adding the sum of its digits to the product of its digits gives me my original number. What could my number be?
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Stage: 4 Challenge Level:
A spider is sitting in the middle of one of the smallest walls in a room and a fly is resting beside the window. What is the shortest distance the spider would have to crawl to catch the fly?
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If you have only 40 metres of fencing available, what is the maximum area of land you can fence off?
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Find a cuboid (with edges of integer values) that has a surface area of exactly 100 square units. Is there more than one? Can you find them all?
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Play the divisibility game to create numbers in which the first two digits make a number divisible by 2, the first three digits make a number divisible by 3...
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What size square corners should be cut from a square piece of paper to make a box with the largest possible volume?
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Stage: 3 Challenge Level:
Think of two whole numbers under 10, and follow the steps. I can work out both your numbers very quickly. How?
Dozens
Stage: 2 and 3 Challenge Level:
Do you know a quick way to check if a number is a multiple of two? How about three, four or six?
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Stage: 3 Challenge Level:
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Explore when it is possible to construct a circle which just touches all four sides of a quadrilateral.
Funnel
Stage: 4 Challenge Level:
A plastic funnel is used to pour liquids through narrow apertures. What shape funnel would use the least amount of plastic to manufacture for any specific volume ?
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Stage: 4 Challenge Level:
Chris and Jo put two red and four blue ribbons in a box. They each pick a ribbon from the box without looking. Jo wins if the two ribbons are the same colour. Is the game fair?
Days and Dates
Stage: 3 Challenge Level:
Investigate how you can work out what day of the week your birthday will be on next year, and the year after...
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Liam's house has a staircase with 12 steps. He can go down the steps one at a time or two at time. In how many different ways can Liam go down the 12 steps?
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Explore the effect of combining enlargements.
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Stage: 3 Challenge Level:
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Curvy Areas
Stage: 4 Challenge Level:
Have a go at creating these images based on circles. What do you notice about the areas of the different sections? | 2,036 | 8,746 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2018-17 | latest | en | 0.914592 |
https://numbermatics.com/n/1738600/ | 1,723,657,358,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641118845.90/warc/CC-MAIN-20240814155004-20240814185004-00715.warc.gz | 321,997,001 | 6,834 | 1738600
1,738,600 is an even composite number composed of three prime numbers multiplied together.
What does the number 1738600 look like?
This visualization shows the relationship between its 3 prime factors (large circles) and 24 divisors.
1738600 is an even composite number. It is composed of three distinct prime numbers multiplied together. It has a total of twenty-four divisors.
Prime factorization of 1738600:
23 × 52 × 8693
(2 × 2 × 2 × 5 × 5 × 8693)
See below for interesting mathematical facts about the number 1738600 from the Numbermatics database.
Names of 1738600
• Cardinal: 1738600 can be written as One million, seven hundred thirty-eight thousand, six hundred.
Scientific notation
• Scientific notation: 1.7386 × 106
Factors of 1738600
• Number of distinct prime factors ω(n): 3
• Total number of prime factors Ω(n): 6
• Sum of prime factors: 8700
Divisors of 1738600
• Number of divisors d(n): 24
• Complete list of divisors:
• Sum of all divisors σ(n): 4042710
• Sum of proper divisors (its aliquot sum) s(n): 2304110
• 1738600 is an abundant number, because the sum of its proper divisors (2304110) is greater than itself. Its abundance is 565510
Bases of 1738600
• Binary: 1101010000111011010002
• Base-36: 119IG
Squares and roots of 1738600
• 1738600 squared (17386002) is 3022729960000
• 1738600 cubed (17386003) is 5255318308456000000
• The square root of 1738600 is 1318.5598204103
• The cube root of 1738600 is 120.2448703515
Scales and comparisons
How big is 1738600?
• 1,738,600 seconds is equal to 2 weeks, 6 days, 2 hours, 56 minutes, 40 seconds.
• To count from 1 to 1,738,600 would take you about four weeks!
This is a very rough estimate, based on a speaking rate of half a second every third order of magnitude. If you speak quickly, you could probably say any randomly-chosen number between one and a thousand in around half a second. Very big numbers obviously take longer to say, so we add half a second for every extra x1000. (We do not count involuntary pauses, bathroom breaks or the necessity of sleep in our calculation!)
• A cube with a volume of 1738600 cubic inches would be around 10 feet tall.
Recreational maths with 1738600
• 1738600 backwards is 0068371
• 1738600 is a Harshad number.
• The number of decimal digits it has is: 7
• The sum of 1738600's digits is 25
• More coming soon!
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The information we have on file for 1738600 includes mathematical data and numerical statistics calculated using standard algorithms and methods. We are adding more all the time. If there are any features you would like to see, please contact us. Information provided for educational use, intellectual curiosity and fun!
Keywords: Divisors of 1738600, math, Factors of 1738600, curriculum, school, college, exams, university, Prime factorization of 1738600, STEM, science, technology, engineering, physics, economics, calculator, one million, seven hundred thirty-eight thousand, six hundred.
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Some bits of this website may not work unless you switch it on. | 949 | 3,472 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2024-33 | latest | en | 0.871449 |
https://voyance-telephone-avis.com/the-foolproof-what-is-free-fall-physics-strategy/ | 1,585,736,384,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370505730.14/warc/CC-MAIN-20200401100029-20200401130029-00010.warc.gz | 781,136,703 | 14,475 | # What Is Free Fall Physics at a Glance
## The Importance of What Is Free Fall Physics
While this occurs, you’re probably in the telogen phase. Accelerometers are employed in vehicle Electronic stability control systems to gauge the vehicle’s actual movement. Equally important is that a high specific impulse can generate a large enough increase in a satellite’s momentum to allow the spacecraft to modify orbitsa feature unavailable on currently orbiting CubeSats. If you’re only asked for positions and velocities, you may likewise be able to work the issue using Conservation of Energy. However, you ought to be in a position to cancel the real units to get there at seconds for time.
This is something which is mutually beneficial, he explained, and something that we would like to encourage. They like the notion of being a true one-person shop. Thank you for everyone who helps. I still had enough time with my kids, which was among the reasons I wished to work from house in the very first spot. 1 way to receive high quality measurementsof position for a role of time is by employing video taken with a stationary camera.
The neural network won’t ever learn how to predict the other faces. Among the most dangerous details of the passion trap is the subtle, illusory way it requires hold. It’s defined as the reversal of position of a point in a specific direction. In the very first area of the ride, force is put on the car to lift it to the cap of the free-fall tower. Specifically, we’re likely to examine the important things you will need to know to address problems involving objects in free fall.
Your introduction should use a main source. http://en.wikipedia.com/wiki/Pensiero A comprehension of acceleration, for instance, is vital to the study of force. Symmetry is apparent in nature in a wide selection of situations.
Read about the fantastic advances in our comprehension of the processes that surround us. The rest of The information has to be extracted from the problem statement based upon your comprehension of the aforementioned principles. In some instances, you might feel compelled to finish an assignment to the customer’s exact specifications as opposed to exploring different solutions that may offer superior outcomes. Obviously, there are several distinctive procedures which can be used, and yours may differ than the one listed here.
Interpreting this as anything else is extremely challenging to do. It is crucial to bear in mind the indicators of the numbers you plug into the equations. Check Definition and Ratio problems to see whether you’re able to locate a practical example.
So I give them 2 more minutes to finish their towers and let them talk. If you become in the pattern of answering emails or responding to texts at midnight, it is going to be difficult to discontinue this behavior farther down the line. You will undoubtedly recognize that I’ve not answered my leading question, but that is going to have to wait. The means to do it is to consider which answer you desire.
It’s deemed satisfactory in the event the business outperforms the projections. However, it’s more probable that you’re not the market share leader. If you are aware that your costing systems inflate the true expenses, perhaps you use market-based pricing. Sounds simple enough, but you would not be aware of how many people (even seasoned small business owners) fall for a sales pitch and never start looking into the business or claims being provided. The record of applying price discounts to fulfill short-term sales objectives isn’t promising. So it is with managers who should make the numbers.
## The What Is Free Fall Physics Stories
This calculator can help you to solve all kinds of uniform acceleration difficulties. You’re able to realize that the algorithms do well exceptfor little faces. For many difficulties, you will only should work in 1 dimension.
In addition, I want students to look carefully at the picture and I ask them how they know that it is accelerating. The Nokia 5500 sport includes a 3D accelerometer that may be obtained from software. It didn’t act as a pile driver since it never gave up its momentum. This acceleration is called the gravitational acceleration. This totally free fall acceleration is also referred to as acceleration due to gravity.
So we’re likely to should use gravity to address these equations. To carry out kinematics calculations, all we should do is plug the first conditions into the suitable equation of motion and after that read out the answer. If you’d like position at a specified velocity, pick equation 2. Generally, in a Physics class you will be requested to solve two kinds of equations when calculating free fall. The equation that’s used to figure distance and velocity is provided below.
Lastly, higher thrust density will make it possible for satellites to achieve complex fuel-optimized orbits in a fair moment. But gravity is a strong force. A thruster will fire to modify orbits as soon as the satellite passes the planet’s equator. There wouldn’t be any effects of gravity inside the room.
## The Hidden Gem of What Is Free Fall Physics
The designer must earn a compromise between sensitivity and the most acceleration that may be measured. This ought to help determine the accuracy of my results. But recently I realized that it’s wrong. The value looks reasonable enough.
Increased cross-sectional areas bring about a higher amount of air resistance. The apparatus was made to record the exact time of flight for the ball. Aluminum melts at a far lower temperature, so even if it glowed it wouldn’t be the exact same color. That happens due to air resistance. It is the result of collisions of the object’s leading surface with air molecules.
These absolutely free fall problems will reveal to you just how to address an assortment of word problems associated with objects which are falling from a particular height. This type of sensor can be utilized to measure or monitor people. Be aware that at the exact same distance under the point of release, the rock has the exact same velocity in both situations. The experimentally determined price of this acceleration varies slightly from one spot to another on the face of the earth. | 1,219 | 6,259 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2020-16 | latest | en | 0.947714 |
http://mathhelpforum.com/discrete-math/273645-induction.html | 1,529,909,991,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267867493.99/warc/CC-MAIN-20180625053151-20180625073151-00340.warc.gz | 201,594,391 | 12,296 | 1. Induction
I'm trying to prove that for $n>13$ , there exists $x, y \in{Z_+}$ such that $n=3x+5y$. I'm having trouble with the extended inductive step. I would appreciate it if you don't just outright solve this for me. But if you do, use spoilers.
2. Re: Induction
not true for $\displaystyle n=15$
should be true for $\displaystyle n>15$
are we required to use induction?
3. Re: Induction
Originally Posted by Idea
not true for $\displaystyle n=15$
should be true for $\displaystyle n>15$
are we required to use induction?
The OP is true on non-negative integers,$\displaystyle \mathbb{Z}\setminus\mathbb{Z}^-$
4. Re: Induction
if we include $\displaystyle 0$ in $\displaystyle \mathbb{Z}_+$ then it is true for $\displaystyle n>7$
otherwise it is true for $\displaystyle n>15$
5. Re: Induction
Hey VonNemo19.
I'd try and show that for certain numbers n = 2z and n = 2z + 1 that you can find integer solutions to that expression (i.e. look at different cases of numbers by divisibility and show they hold in all cases).
If you can't do odd and even then try looking at enough prime numbers for some period and use the inductive step on that.
Could you show us the attempts you make?
6. Re: Induction
I would not use "induction" at all but rather use "Euclid's algorithm". Start from the obvious fact that 3(2)+ 5(-1)= 1 and multiply each side by n: 3(2n)+ 5(-n)= n. So, for every n, one solution is x= 2n, y= -n. But then it is clear that x= 2n- 5k, y= -n+ 3k is also a solution for any integer, k: 3(2n- 5k)+ 5(-n+ 3k)= 6n- 15k- 5n+ 15k= n.
Requiring that the solutions be positive means that we must have 2n- 5k> 0 and -n+ 3k> 0. That is, 3k> n and 5k< 2n. That is equivalent to n/3< k< 2n/5. For what values of n is there at least one integer between n/3 and 2n/5?
7. Re: Induction
Originally Posted by VonNemo19
I'm trying to prove that for $n>13$ , there exists $x, y \in{Z_+}$ such that $n=3x+5y$. I'm having trouble with the extended inductive step.
(I went ahead and gave my full attempt, because 1) it's the next day, and 2) I didn't see around giving hints.)
Starting with n = 16, inclusive, you can show that n = 3x + 5y, for positive integers x and y.
Case where y = 1, x > 3
- - - - - - - - - - - - - - - -
Base case
x = 4, y = 1
n = 3(4) + 5(1)
n = 12 + 5
n = 17
n + 1 =
3(4) + 5(1) + 1 =
3(1 + 3) + 5(3 - 2) + 1 =
3(1) + 3(3) + 5(3) - 5(2) + 1 =
3(1) + 5(3) + 9 - 10 + 1 =
3(1) + 5(3) =
3 + 15 =
18
Inductive step
Recall that y = 1.
n = 3x + 5(1)
n = 3x + 5
n + 1 = 3x + 5 + 1
n + 1 = 3x + 5 + (-9 + 10)
n + 1 = 3x - 9 + 5 + 10
n + 1 = 3(x - 3) + 5(1 + 2)
n + 1 = 3(x - 3) + 5(3)
Case where y > 1
- - - - - - - - - - - -
Base case
x = 2, y = 2
n = 3(2) + 5(2)
n = 6 + 10
n = 16
n + 1 =
3(2) + 5(2) + 1 =
3(4 - 2) + 5(1 + 1) + 1 =
3(4) - 3(2) + 5(1) + 5(1) + 1 =
3(4) + 5(1) - 3(2) + 5(1) + 1 =
3(4) + 5(1) - 6 + 5 + 1 =
3(4) + 5(1) =
12 + 5 =
17
Inductive step
n = 3x + 5y
n + 1 = 3x + 5y + 1
n + 1 = 3x + 5y + (6 - 5)
n + 1 = 3x + 6 + 5y - 5
n + 1 = 3(x + 2) + 5(y - 1)
Thus, by the Principle of Mathematical Induction, I have shown that the proposition, as I have restated it, is true.
8. Re: Induction
Hey. Sorry about the absence. Yeah. We're talking about non negative solutions. I've made some progress. If I assume k, then I can always add 1 to both sides and then sub back for k, right? So, $k =3x_0+5y_0\rightarrow{}k +1=6+k-5=3(x_0+2)+5(y_0-1)$. But then $y -1\geq{0}$. So I get confused. | 1,408 | 3,461 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2018-26 | latest | en | 0.903155 |
https://sites.google.com/site/gibsonridgerepository/gr2analyst/reference/algorithms/posh-mehs | 1,516,695,650,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084891791.95/warc/CC-MAIN-20180123072105-20180123092105-00710.warc.gz | 801,003,855 | 10,490 | GR2Analyst > Reference > Algorithms >
### POSH/MEHS
Probability of Severe Hail (POSH) and Maximum Estimate Hail Size (MEHS) are the hail algorithm outputs from GR2Analyst. The algorithm is described below, and also in more detail in the powerpoint presentation GR2Analyst Algorithms in WFO Operations (Lincoln, 2012), which is the source of this page's images/tables.
The hail algorithm is based upon:
Witt, Arthur, Michael D. Eilts, Gregory J. Stumpf, J. T. Johnson, E. De Wayne Mitchell, Kevin W. Thomas, 1998: An Enhanced Hail Detection Algorithm for the WSR-88D. Wea. Forecasting, 13, 286–303.
## Overview
Compares radar representation to the environment, from that creates Severe Hail Index (SHI)
SHI used as basis for Probability of Severe Hail (POSH) and Maximum Estimated Hail Size (MEHS)
## Severe Hail Index Technique from Witt et al (1998)
The Severe Hail Index (SHI) is calculated by the equation:
Where:
SHI is the Severe Hail Index in J/m/s
WT is temp weighting function; hail forms at <0°C, most severe hail forms at <-20°C
H is height above radar level (km)
E is energy flux based on radar reflectivity; weighted based on transition zone from water (40dbZ) and ice (50dbZ). E is closely related to hail damage potential.
H0 and HT are height of freezing level and height of storm top, respectively
SHI values were then compared to actual hail events. A strong correlation was found between SHI, freezing level (H0), and hail occurrence. A warning threshold for SHI was created from value with highest critical success index (CSI). Correlation between warning threshold and H0 was ~0.80, although this varied based upon the verification time window (how much time before/after radar indication of hail to look for reports).
The Warning Threshold is calculated by the equation:
Where:
WT is the Warning Threshold in J/m/s
H0 is the height of the freezing level
### Probability of Severe Hail from SHI
Probability of severe hail (POSH) function created from warning threshold and SHI such that when SHI=WT, POSH=50%. POSH is calculated by the equation:
Where:
POSH is the Probability of Severe Hail (as a dimensionless ratio)
SHI is the Severe Hail Index in J/m/s
WT is the Warning Threshold in J/m/s
The reliability diagrams for POSH are shown below. The plot on the left is from the cases used to develop the POSH equation. The plot on the right is from the set of verification
cases. The table below the reliability diagrams shows the statistics for the algorithm in different parts of the United States.
### Maximum-Estimate Hail Size from SHI
Maximum Estimated Hail Size (MEHS) function created from SHI such that 75% of hail reports will be smaller than MEHS. MEHS is calculated by the equation:
Where:
MEHS is the Maximum Estimated Hail Size in mm
SHI is the Severe Hail Index in J/m/s
A reliability diagrams for MEHS is shown below. Due to the substantial uncertainty in estimating hail size, further statistics were not calculated in the paper.
Hail icons have their color and size dependent upon the max. estimated hail size. Icons are centered on the highest MEHS value.
Icons can be configured in:
File -> Configure Hail Icons
Increment determines usage of icon sizes and icons colored based upon color from MEHS color table.
## Sources of Uncertainty
### SHI in Witt et al (1998)
Environment Data
Equation-fitting to data with issues (StormData)
POSH algorithm worked best in S. Plains and was worse in Florida test region. Paper authors attribute some of this to lower quality of reports due to less frequent severe weather and lower population density. Due to large uncertainty in maximum hail size reports, MEHS errors were not quantified.
### GR2Analyst Implementation of SHI
GR2Analyst does not implement the algorithm based on the storm-centroid, but it is instead implemented on a gridded cell-by-cell basis. The integration is done for each radar pixel, stacked vertically. Note: Because of this, storm motion is important!
Environment data comes from latest RUC analysis profile valid for radar location. If no profile available, the values are set to defaults of: 0°C = 10,000ft and -20°C = 20,000ft
Note: This might cause issue when looking at archived data if the profiles were not downloaded to that computer either automatically (during the event) or manually (after the event). The RUC analysis is applied to entire radar range; causes most error in situations with strong fronts or varying airmasses. | 1,052 | 4,481 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2018-05 | latest | en | 0.897503 |
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# More on Slot Cycles
28 February 2005
John,
My wife and I like to play reel slots with a bonus round such as Elvis, Wheel of Fortune and most of all IGT's Pinball.
Two questions:
1. Are bonus round games programmed to hit the bonus symbol at some independent frequency, or is hitting the bonus symbol controlled by the RNG, just like any other possible symbol on the pay line?
2. Is a separate RNG used during the bonus round to control the possible payouts (since there are typically a smaller number of potential bonus round payouts than there are reel symbol combinations)?
Thanks,
Mike & Chris
Dear Mike & Chris,
The bonus symbol is just another symbol on the reels, as far as determining which symbols land on the payline is concerned. The program running the slot uses the output from the RNG to choose the stops on the reels that will land on the payline -- and if some of those symbols happen to be symbols that launch a bonus round, all the better for the player.
The frequency with which the bonus round occurs is determined solely by the number of times the bonus round symbols appear on the reels.
The same RNG is used to determine the results of bonus rounds. It's a piece of cake to convert random numbers from 1 to 1,000,000,000 to random numbers from 1 to 10.
Best of luck in and out of the casinos,
John
John,
I was told that the machine plays games when idle, and when a coin is inserted, it decided the outcome of the game before the second coin is inserted. I was also told that the manufacturers overlap the cycles thus preventing counting the number of games that are in a cycle. I also read that the RNG was like a string a lights, when you pushed the button and that combination is what you received. I was wondering if you could put to rest the theory of operation.
Thanks,
Kevin
Dear Kevin,
In the past, machines locked in the result of a spin when a coin was inserted. But anytime a machine is sitting with a result locked in, it is vulnerable to being cheated, so today's machines wait until the last possible moment to poll the RNG and lock in the result. It happens sometime after the player pulls the handle or hits the Spin button.
Machines do "play games" while idle. A more technical (and better) way to say that is to say that the RNG is constantly generating numbers, even while the machine is not being played.
Now, this business about cycles has to be one of the most misunderstood areas of slot operations. There are actually two cycles in the slot and many people just do not understand what the cycles are and why they're irrelevant.
One think sometimes referred to as a cycle is the total number of outcomes possible on the machine. This is found by multiplying together the number of stops on each reel. I read somewhere that some slot directors like machines with smaller cycles because they make the jackpots more likely to hit. Nonsense. The jackpot can hit with the same frequency on a machine with 2,000 stops and on a machine with 200,000 stops. It all depends on the number of times the jackpot symbol appears on the virtual reels.
The other cycle is the RNG's cycle or period. It tells how many numbers the RNG can generate before it starts repeating. This cycle, like the one before, is also mostly irrelevant.
Now, it is true the the RNG is really a psuedo-RNG. The numbers aren't truly generated at random. A deterministic function, which takes one or more prior results as parameters, is used to determine the next number. The numbers will come out in a definite pattern, and if you knew the pattern you could cheat the machine -- as cheats have done in the past.
Manufacturers have taken a number of steps to make RNG cheating very difficult. One we already mentioned, having the RNG run constantly. Another is a consequence of using more powerful processors in machines. RNGs can generate tens, hundreds, maybe even thousands of numbers per second. Even if you knew the jackpot was going to be the result in 2.53 seconds, it would be very difficult for you to start the spin while that is the result when results are only current for a fraction of a second.
Another anti-cheating step is to alter the speed at which the RNG runs. It may run in turbo mode for a few seconds and then fall back to turtle mode for a length of time. The last anti-cheating step I want to mention is periodically reseeding the RNG, that is, changing the parameters that are used as input to it. This has the effect of sending the RNG to a different point in its cycle.
And that brings us to an aspect of a cycle that we haven't discussed yet--moving predictably from one step to another, just as the minutes on a digital clock will predictably go from 00 to 59 and then repeat. I think that's what someone was referring to when he spoke of "overlapping cycles", but I really don't know what he meant. I don't know which cycles are supposed to be overlapped.
John
Send your slot and video poker questions to John Robison, Slot Expert, at slotexpert@comcast.net. Because of the volume of mail I receive, I regret that I can't send a reply to every question. Also be advised that it may take several months for your question to appear in my column.
John Robison
John Robison is an expert on slot machines and how to play them. John is a slot and video poker columnist and has written for many of gaming’s leading publications. He holds a master's degree in computer science from the prestigious Stevens Institute of Technology.
You may hear John give his slot and video poker tips live on The Good Times Show, hosted by Rudi Schiffer and Mike Schiffer, which is broadcast from Memphis on KXIQ 1180AM Friday afternoon from from 2PM to 5PM Central Time. John is on the show from 4:30 to 5. You can listen to archives of the show on the web anytime.
#### Books by John Robison:
The Slot Expert's Guide to Playing Slots
John Robison
John Robison is an expert on slot machines and how to play them. John is a slot and video poker columnist and has written for many of gaming’s leading publications. He holds a master's degree in computer science from the prestigious Stevens Institute of Technology.
You may hear John give his slot and video poker tips live on The Good Times Show, hosted by Rudi Schiffer and Mike Schiffer, which is broadcast from Memphis on KXIQ 1180AM Friday afternoon from from 2PM to 5PM Central Time. John is on the show from 4:30 to 5. You can listen to archives of the show on the web anytime.
#### Books by John Robison:
The Slot Expert's Guide to Playing Slots | 1,436 | 6,611 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2024-10 | latest | en | 0.963233 |
https://nl.mathworks.com/matlabcentral/cody/problems/233-reverse-the-vector/solutions/211710 | 1,590,586,848,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347394074.44/warc/CC-MAIN-20200527110649-20200527140649-00109.warc.gz | 471,405,591 | 15,542 | Cody
# Problem 233. Reverse the vector
Solution 211710
Submitted on 3 Mar 2013 by Mário
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
%% x = 1; y_correct = 1; assert(isequal(reverse(x),y_correct))
2 Pass
%% x = -10:1; y_correct = 1:-1:-10; assert(isequal(reverse(x),y_correct))
3 Pass
%% x = 'able was i ere i saw elba'; y_correct = 'able was i ere i saw elba'; assert(isequal(reverse(x),y_correct)) | 162 | 530 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2020-24 | latest | en | 0.747803 |
https://www.brianthicks.com/guide/functional-sets/8/ | 1,696,112,809,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510730.6/warc/CC-MAIN-20230930213821-20231001003821-00629.warc.gz | 741,448,538 | 3,242 | # Functional Sets, Part 8: Map
Elm 0.18 · Updated February 11, 2018 · 3 Minute Read · ∞ Permalink
Now that we have filter and friends, we’re almost done with our Set implementation. Once we have map, we’ll be done!
map applies a function to every item in a collection. Say we have a list containing the numbers one through five. We can use List.map to double every number:
List.map (\i -> i * 2) [1, 2, 3] -- [2, 4, 6]
We’re going to make map for our sets. So, given the same list (but converted to a set) we should get:
map (\i -> i * 2) (fromList [1, 2, 3]) -- {2, 4, 6}
Let’s do this… What if we deconstruct the set piece by piece like we did with member? That way, we’d apply the function exactly once to each item, and get a new set!
Not so fast!
What about functions that return the same value, like always 1? When applied to the set from earlier we would get a set with 1, five times. That won’t work, it’s not a valid set! Sets always have unique values, so we need a different approach.
foldl comes to our rescue here again. Since it moves values through a function to an accumulator piece by piece, we can build a new Set as we go. We’ll start with an empty list as our accumulator value, then insert the mapped values one at a time. If we get any duplicates, they’ll be removed as we go.
Here’s how it looks:
map : (comparable -> comparable2) -> Set comparable -> Set comparable2
map fn set =
foldl
(\item acc -> insert (fn item) acc)
empty
set
It works just like we said above:
map (\i -> i * 2) (fromList [1, 2, 3]) -- {2, 4, 6}
And if we use a function that returns duplicate values, they’re deduplicated automatically:
map (always 1) (fromList [1, 2, 3]) -- {1}
And… that’s it! That’s the whole Set API! If you’ve been following along this whole time, you should know a lot more about how to implement data structures in Elm. We’ll be back next week to sum up and sneak a peek at how we can extend our Sets to be Dicts. | 546 | 1,956 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2023-40 | latest | en | 0.852667 |
https://studentshare.org/physics/1623334-physics-project-2 | 1,547,627,460,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583657097.39/warc/CC-MAIN-20190116073323-20190116095323-00012.warc.gz | 669,364,066 | 20,272 | Search
# Physics project 2 - Essay Example
Summary
In a uniform gravitational field, it coincides with the objects centre of mass. (In modern Britain the spelling centre is standard. Both spellings originated in England; centre is now…
## Extract of sample"Physics project 2"
Download file to see previous pages In work on floating bodies he demonstrated that the orientation of a floating objects is the one that makes its center of gravity as low as possible. He developed mathematical techniques for finding the centers of gravity of objects of uniform density of various well-defined shapes, in particular a triangle, a hemisphere, and a frustum of a circular parabolic.
The center of gravity is an important point on an aircraft, as it defines the amount of mass forward or behind the center of gravity that needs to be moved in order to pitch the plane up or down without applying any external forces. In conventional designs the Co G is often located very near the line 1/3rd back from the front of the wing. That is the line where most wings generate their lift, known as the center of pressure (Co P), so by balancing the plane at that point, the lift and weight balance out with no net torque. The Co is sometimes moved slightly to the rear of this line in order to provide the plane with a natural "nose up" tendency when lift increases (like when applying more power).
If the balance of the plane is moved too far from the Cog, the control surfaces may have trouble controlling the plane. The actual force generated by the surfaces is typically quite small (a few pounds) but due to their location at the end of the tail (typically) they generate considerable torque to pitch the plane. If the Co starts to move away from the Co P there will be an increasing amount of constant torque they have to counteract, and if it moves too far, it may be more than the controls can counter.
The center of gravity is a geometric property of any object. The center of gravity is the average location of the weight of an object. We can completely describe the motion of any object through space in terms of the translation of the center of gravity of the object from one place to another and the rotation of the ...Download file to see next pagesRead More
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+16312120006 | 1,465 | 6,829 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2019-04 | latest | en | 0.949576 |
https://www.syfy.com/syfywire/two-of-earths-moons-in-one-picture | 1,579,949,056,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251672440.80/warc/CC-MAIN-20200125101544-20200125130544-00510.warc.gz | 1,096,693,319 | 39,320 | # Two of Earth’s Moons in One Picture
Contributed by
Jan 25, 2013
Photographer Lauren Hartnett took a fantastic picture: Two Moons literally passing in the night!
Hartnett was in Houston, Texasâat NASAâs Johnson Space Center, appropriately enoughâon Jan. 4, 2012 when she snapped this. It shows the Moon, just a few days past half full, with the International Space Station sailing past it to the upper left. The station is in low-Earth orbit, so it's much closer than the Moon; it's just a little perspective and orbital mechanics that made them look so close together.
Most people arenât aware the ISS is easily visible to the naked eye when it passes overhead. In fact, if the geometry is right it can outshine Venus and be the third brightest object in the sky (after the Sun and Moon). Iâve seen it many, many times myself, and a few times Iâve also seen incredibly bright flashes of light off the solar panels as they catch the Sun just right. Itâs amazing.
What fascinates me about this picture are the relative sizes of the two moons. I can do some math with this!
The Moon was about 404,000 kilometers (250,000 miles) away when this picture was taken, and is about 3470 km (2170 miles) in diameter.
The ISS orbits the Earth at a height of 400 km (250 miles). Thatâs how far away it would be if it passed straight overhead, but for Hartnett it was at a 30° angle from the zenith (given the Moon's altitude above the horizon at the time). If you remember your trig, that multiplies the distance by a factor of 1.15 (the distance roughly scales as the secant of the angle away from the zenith, if youâre playing along at home), so the ISS was actually about 460 km (290 miles) away at the moment this shot was snapped.
The ISS is about 110 meters across. So letâs see. That makes the Moon about 31,500 times wider physically than the ISS, but 880 times farther away. That means it should be 31,500 / 880 = 36 times the size of the ISS in the picture. I measured it on my screen, and that's almost exactly the ratio I got.
That's so cool. Math!
And of course, it took some pretty sophisticated math to predict the location of the ISS at that time. Happily, if you want to see it, you donât have to crunch the numbers: There are lots of websites thatâll just tell you. I use Heavens Above the most, but others can be found with a flick of your Googling wrist. Give it a try! Seeing the space station is amazing. And just think: There are people up there, living on board. And one, Commander Chris Hadfield, takes lots of pictures from orbit and puts them on Twitter.
So maybe, the next time you see the ISS and try to get a picture, someone up there is snapping one of you, too.
Tip oâ the lens cap to astronaut Ron Garan on Twitter. | 685 | 2,776 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2020-05 | longest | en | 0.952369 |
https://community.fabric.microsoft.com/t5/Desktop/TOPN-Question/td-p/2379309 | 1,718,901,628,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861957.99/warc/CC-MAIN-20240620141245-20240620171245-00853.warc.gz | 154,166,089 | 115,568 | cancel
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Frequent Visitor
## TOPN Question
Hi, I am new to this community and am trying to advance my DAX knowledge for my job. Could someone tell me what the numbers below the columns represent?
Thank you!
VAR __SomeVariable =
TOPN(
502,
__SomeTable,
[Column1],
0,
[Column2],
0,
'Product'[Column3],
1,
'Location'[Column4],
1,
'Location'[Column5],
1
)
2 ACCEPTED SOLUTIONS
Community Champion
Hi @TechoGirl0 ,
If you look at the syntax of TopN ,
- 502 is the number of rows to return
- 1 or 0 is the order to be applied. 0/FALSE/DESC – descending; 1/TRUE/ASC – ascending.
``````Syntax =
TOPN(<n_value>, <table>, <orderBy_expression>, [<order>[, <orderBy_expression>, [<order>]]…]) ``````
Thanks,
Samarth
Best Regards,
Samarth
If this post helps, please consider accepting it as the solution to help the other members find it more quickly.
Community Support
Hi, @TechoGirl0
Best Regards,
Community Support Team _Charlotte
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.
2 REPLIES 2
Community Support
Hi, @TechoGirl0
Best Regards,
Community Support Team _Charlotte
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.
Community Champion
Hi @TechoGirl0 ,
If you look at the syntax of TopN ,
- 502 is the number of rows to return
- 1 or 0 is the order to be applied. 0/FALSE/DESC – descending; 1/TRUE/ASC – ascending.
``````Syntax =
TOPN(<n_value>, <table>, <orderBy_expression>, [<order>[, <orderBy_expression>, [<order>]]…]) ``````
Thanks,
Samarth
Best Regards,
Samarth
If this post helps, please consider accepting it as the solution to help the other members find it more quickly.
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Top Solution Authors
Top Kudoed Authors | 659 | 2,593 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2024-26 | longest | en | 0.781009 |
https://math.stackexchange.com/questions/1999791/inverse-of-a-symmetric-tridiagonal-almost-toeplitz-matrix/2001363 | 1,618,317,840,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038072366.31/warc/CC-MAIN-20210413122252-20210413152252-00511.warc.gz | 476,426,048 | 39,823 | # Inverse of a symmetric tridiagonal almost-Toeplitz matrix
I'm trying to analytically find the inverse of the following $N \times N$ tridiagonal matrix:
$$T = \begin{bmatrix} 1 & -c \\ -c & 2 & -c \\ & -c & 2 & \ddots \\ & & \ddots & \ddots & \ddots \\ & & & \ddots & 2 & -c \\ & & & & -c & 2 & -c \\ & & & & & -c & 1 \end{bmatrix}$$
That is, all elements just off the main diagonal are $-c$, where $c \in (0, 1)$; the main diagonal has a one in the top-left and in the bottom-right, and twos in between; and everything else is a zero.
This comes from a statistical model I'm using (a CAR spatial model for a path graph) where, by construction, $T^{-1}$ is a covariance matrix. Thus I know $T^{-1}$ is symmetric and the elements of its main diagonal are all positive. I'd happily settle for only getting formulas for those main diagonal elements.
Copying from the Wikipedia page on tridiagonal matrices (and doing a little bit of work), the $i$th diagonal element of $T^{-1}$ is $(T^{-1})_{ii} = \theta_{i-1} \theta_{N-i} / \theta_N$, where
$$\theta_i = a_i \theta_{i-1} - c^2 \theta_{i-2} \text{ for } i = 2,3,\dots,N$$
with initial conditions $\theta_0 = \theta_1 = 1$. The $a_i$ are the main diagonal elements of $T$, so e.g. if $N = 5$ then $\{a_i\} = \{1,2,2,2,1\}$. One way to answer my question would be to solve the above recurrence relation, i.e. find a closed-form for $\theta_i$ for any $N \ge 2$. (I tried this using a generating function, but got stuck due to the pesky non-constant diagonal of $T$, i.e. the $a_i$.)
In the following related question, the person who answered it used a very different, very involved approach: Inverse of a symmetric tridiagonal matrix.. There, the matrix to be inverted had a constant main diagonal, so it was easier.
• Another approach you may like is the Woodbury matrix identity. In particular, take $A$ to be the Toeplitz matrix, $C = -I$, and $U = V^T$ to be the correct $2 \times n$ matrix. – Ben Grossmann Nov 4 '16 at 22:32
The inverse can be analytically computed using the recurrence relations mentioned in the OP's link or here. I give below some details.
To apply those results for the given matrix, we need to solve two sets of constant coefficient linear recurrence relations. First set is
$$\theta_{i} = 2\theta_{i-1} - c^{2}\theta_{i-1}, \; i=2,...,N-1; \quad\text{and}\quad\theta_{N} = \theta_{N-1} - c^{2}\theta_{N-2};$$
subject to initial conditions $\theta_{0}=1, \theta_{1}=1$. To ease notation, let $x:=\sqrt{1-c^{2}}$, and note that $x\in(0,1)$ since $c\in(0,1)$. The first recurrence equation of the first set, can be easily solved to get $\theta_{i} = \frac{1}{2}\left[(1+x)^{i} + (1-x)^{i}\right]$ for $i=2,...,N-1$ (notice from initial conditions that this solution is also valid for $i=0,1$). Combining this solution with the second recurrence equation of the first set, we get $\theta_{N} = \frac{x}{2}\left[(1+x)^{N-1} - (1-x)^{N-1}\right]$. This completes the solution of the first set of recurrence relations.
The second set is
$$\phi_{j} = 2\phi_{j+1} - c^{2}\phi_{j+2}, \; j=N-1,...,2; \quad\text{and}\quad\phi_{1} = \phi_{2} - c^{2}\phi_{3};$$
subject to terminal conditions $\phi_{N+1}=1, \phi_{N}=1$. Solving the first recurrence equation of the second set, we get $\phi_{j} = \frac{1}{2}\left[(1-x)^{N-j+1} + (1+x)^{N-j+1}\right]$ (notice from terminal conditions that this solution is also valid for $j=N,N+1$). Combining this solution with the second recurrence relation of the second set yields $\phi_{1} = \frac{x}{2}\left[(1+x)^{N-1} - (1-x)^{N-1}\right]$. This completes the solution of the second set of recurrence relations.
Edit: As noted in the comment below, the $\phi_{j}$ sequence is $\theta_{i}$ sequence in reverse, i.e., $\theta_{i} = \phi_{N-i+1}$. So solving any one set of recurrence equation suffices.
Now the $i,j$-th element of the inverse of the $N\times N$ matrix $T_{N}$, denoted as $\left(T_{N}^{-1}\right)_{ij}$, can be written explicitly by substituting the solution of these recurrence equations into the formula of the above links. For instance, for $i \leq j$, we have
$$\left(T_{N}^{-1}\right)_{ij} = (-1)^{i+j}\:(-c)^{j-i}\: \displaystyle\frac{\theta_{i-1}\phi_{j+1}}{\theta_{N}} \\ = c^{j-i}\: \displaystyle\frac{\left[(1+x)^{i-1} + (1-x)^{i-1}\right] \left[(1+x)^{N-j} + (1-x)^{N-j}\right]}{2x \left[(1+x)^{N-1} - (1-x)^{N-1}\right]}.$$
As an example, for $N=5$, $c=0.3$, we get $\left(T_{5}^{-1}\right)_{23} = 0.0824$, etc.
• This is great! A few notes: The $\phi_j$ sequence is just the $\theta_i$ sequence in reverse, i.e. $\theta_i = \phi_{N-i+1}$, so I guess we really only need one of them (I'm thinking this is because $T$ is symmetric along the anti-diagonal, aka persymmetric). Two, $\theta_N$ has a minus in it, i.e. $\theta_N = \phi_1 = \frac{x}{2}\left[(1+x)^{N-1} - (1-x)^{N-1}\right]$, which affects the final equation. Lastly, trivially, a small simplification: $(-1)^{i+j}(-c)^{j-i} = (-1)^{2 j}c^{j-i} = c^{j-i}$. – Dagremu Nov 6 '16 at 19:15
• You are right in all three points. Edited. – Abhishek Halder Nov 7 '16 at 5:20 | 1,698 | 5,081 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2021-17 | latest | en | 0.824719 |
https://discuss.pytorch.org/t/inconsistent-behavior-in-torch-repeat-s-backprop/88021 | 1,653,811,606,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652663048462.97/warc/CC-MAIN-20220529072915-20220529102915-00193.warc.gz | 254,316,618 | 5,682 | # Inconsistent behavior in torch.repeat()'s backprop
Hi,
When having two tensors: A and X where A is constant (requires_grad=False) and has the shape [100, 100, 1,1], and X is a trainable parameter (requires_grad=True) and has the shape [1, 100, 1, 1], I would expect these 3 expressions to produce the same result:
1. (A * X)
2. (A * X.expand(100, 100, 1, 1))
3. (A * X.repeat(100, 1, 1, 1))
Since the sizes of the tensors do not match, the first dimension of X should be increased by copying the data, and then the multiplication is performed.
As expected, all these expressions do in fact produce the same result, but thereās a tiny issue during back-propagation.
I would expect Xās gradients in all 3 cases, since they are mathematically identical, to be the same, but it turns out lines 1&2 consistently get the exact same result, while line 3 always has minor changes.
I understand that these operations will have to add multiple values along the first dimension during back-propagation, which are probably translated to CUDA āatomicAddā operations that are known to be nondeterminism as explained here, but isnāt it weird that lines 1 and 2 always get the same results, and line 3 is the only non-deterministic calculation?
I donāt understand how can torch.repeat() be different than torch.expand() during back-propagation, given that torch.expand() in this case is deterministic and gives the exact same results as (A*X).
The code below assumes that line 1 (that calculated (A*X)) has the correct result, and first compares the gradient values of it to the gradients produced by using torch.expand(), then uses torch.repeat() to show that it gives different results.
Code:
``````import torch
device = 'cuda'
size = 100
dtype = torch.float
a = torch.rand(size, size, 1, 1, device=device, dtype=dtype)
x = torch.ones(1, size, 1, 1, dtype=dtype, device=device, requires_grad=True)
y = torch.ones(1, size, 1, 1, dtype=dtype, device=device, requires_grad=True)
(a * x).sum().backward() # Calculating the expected result, expression #1
(a * y.expand(size,size,1,1)).sum().backward() # Expression #2 - using torch.expand()
(a * y.repeat(size,1,1,1)).sum().backward() # Expression #3 - using torch.repeat()
torch.equal(x.grad,y.grad) # This always returns False even though it performs the exact same operation!!!
``````
1. When running on CPU (device = ācpuā) everything works as expected (torch.expand() gives the same result as (A*X) and torch.repeat()).
2. The behavior stays the same when using dtype=torch.double.
3. Everything works as expected when changing X and Y to torch.double, but generating A in floating-point, and only after the generation moving it to double (by using .double()).
4. When reducing the size to below 5, everything works as expected (worked more than 20 times in a row), when using size=5, it has about 50% chance of working.
Thanks.
Hi,
There is one major difference between these 3 cases: `.repeat()` is the only one that actually allocates memory for the ālarger version of Xā.
One possible explanation for what you see is that it has more memory to read when doing the multiplication and reductions, a different kernel launching configuration needs to be used. And so operation can happen in a different order which can lead to small differences Iām afraid.
Also checking the implementation, it seems like they donāt call the exact same sum method when doing the reduction, which might explain the 1-bit derror as well.
Hi @albanD,
The backward of expand is defined here where `sum_to()` is a nice wrapper that will call sum with a list of dimensions to reduce.
On the other hand, the backward of repeat in this same file is `repeat_backward()` which is defined here (most likely was implemented before we added support for multiple dim to sum). | 942 | 3,833 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2022-21 | latest | en | 0.889415 |
https://captaincalculator.com/health/calorie/calories-burned-shoveling-snow-calculator/ | 1,726,524,643,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651714.51/warc/CC-MAIN-20240916212424-20240917002424-00308.warc.gz | 133,480,154 | 16,231 | # Calories Burned Shoveling Snow | Calculator & Formula
LAST UPDATE: September 24th, 2020
The average person burns 370-715 calories per hour shoveling snow.
The number of calories burned will depend on your weight and the intensity of your shoveling. A 150-pound (68kg) person shoveling with moderate effort will burn 379 calories per hour. A 200-pound (90.8kg) person shoveling with vigorous effort will burn 716 calories per hour.
## How many calories are burned shoveling snow?
### Formula
Calories burned per minute = (MET x body weight in Kg x 3.5) ÷ 200
“MET” is a measurement of the energy cost of physical activity for a period of time. You can find an activity’s MET on the chart above.
A task with a MET of 1 is roughly equal to a person’s energy expenditure from sitting still at room temperature not actively digesting food.
A task with a MET of 2 uses twice as much energy as a task with a MET of 1. A task with a MET of 10 uses 10 times as much energy as a task with a MET of 1.
MET values “do not estimate the energy cost of physical activity in individuals in ways that account for differences in body mass, adiposity, age, sex, efficiency of movement, geographic and environmental conditions in which the activities are performed. Thus, individual differences in energy expenditure for the same activity can be large and the true energy cost for an individual may or may not be close to the stated mean MET level as presented in the Compendium.” (as quoted from the main page of the Compendium of Physical Activities).
### Example
A person weighs 180 pounds (81.65kg) and shovels snow with moderate effort (a task that has a MET value of 5.4) for 1 hour (60 minutes).
Calories Burned from shovelling snow with moderate effort (per minute) = (5.4 x 81.65 x 3.5) ÷ 200 = 7.72
Calories Burned from shovelling snow with moderate effort (for 60 minutes) = 7.72 x 60 = 463 | 475 | 1,897 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2024-38 | latest | en | 0.928981 |
http://www.convertit.com/Go/BusinessConductor/Measurement/Converter.ASP?From=dram+fl&To=static+moment+of+area | 1,632,177,242,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057119.85/warc/CC-MAIN-20210920221430-20210921011430-00424.warc.gz | 72,901,826 | 3,698 | New Online Book! Handbook of Mathematical Functions (AMS55)
Conversion & Calculation Home >> Measurement Conversion
Measurement Converter
Convert From: (required) Click here to Convert To: (optional) Examples: 5 kilometers, 12 feet/sec^2, 1/5 gallon, 9.5 Joules, or 0 dF. Help, Frequently Asked Questions, Use Currencies in Conversions, Measurements & Currencies Recognized Examples: miles, meters/s^2, liters, kilowatt*hours, or dC.
Conversion Result: ```fluid dram = 3.6966911953125E-06 volume (volume) ``` Related Measurements: Try converting from "dram fl" to acre foot, barrel, coomb, drop, freight ton, gallon, hekat (Israeli hekat), jigger, koku (Japanese koku), last, load, methuselah, oil arroba (Spanish oil arroba), petroleum barrel, pint (fluid pint), quart (fluid quart), tou (Chinese tou), tun (English tun), UK bushel (British bushel), wine arroba (Spanish wine arroba), or any combination of units which equate to "length cubed" and represent capacity, section modulus, static moment of area, or volume. Sample Conversions: dram fl = .10026042 bath (Israeli bath), .00079995 beer gallon (English beer gallon), .00156657 board foot, .00000102 cord (of wood), .00000816 cord foot (of wood), .015625 cup, .00000373 displacement ton, .0033569 dry quart, .00010851 firkin, .0000155 hogshead, .00005425 kilderkin, .00002052 koku (Japanese koku), .00369669 liter, .00195313 magnum, .16666667 pony, .00003497 sack, .00000388 tun (English tun), .00010164 UK bushel (British bushel), .00040658 UK peck (British peck), .00650527 UK pint (British pint).
Feedback, suggestions, or additional measurement definitions?
Please read our Help Page and FAQ Page then post a message or send e-mail. Thanks! | 470 | 1,708 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2021-39 | latest | en | 0.676969 |
http://nrich.maths.org/public/leg.php?code=5039&cl=3&cldcmpid=1975 | 1,500,799,864,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549424296.90/warc/CC-MAIN-20170723082652-20170723102652-00517.warc.gz | 231,417,617 | 10,574 | # Search by Topic
#### Resources tagged with Interactivities similar to Hello Again:
Filter by: Content type:
Stage:
Challenge level:
### There are 155 results
Broad Topics > Information and Communications Technology > Interactivities
### Khun Phaen Escapes to Freedom
##### Stage: 3 Challenge Level:
Slide the pieces to move Khun Phaen past all the guards into the position on the right from which he can escape to freedom.
### Conway's Chequerboard Army
##### Stage: 3 Challenge Level:
Here is a solitaire type environment for you to experiment with. Which targets can you reach?
### When Will You Pay Me? Say the Bells of Old Bailey
##### Stage: 3 Challenge Level:
Use the interactivity to play two of the bells in a pattern. How do you know when it is your turn to ring, and how do you know which bell to ring?
### Cogs
##### Stage: 3 Challenge Level:
A and B are two interlocking cogwheels having p teeth and q teeth respectively. One tooth on B is painted red. Find the values of p and q for which the red tooth on B contacts every gap on the. . . .
### Matching Fractions, Decimals and Percentages
##### Stage: 3 Challenge Level:
An activity based on the game 'Pelmanism'. Set your own level of challenge and beat your own previous best score.
### Picturing Triangle Numbers
##### Stage: 3 Challenge Level:
Triangle numbers can be represented by a triangular array of squares. What do you notice about the sum of identical triangle numbers?
### Diagonal Dodge
##### Stage: 2 and 3 Challenge Level:
A game for 2 players. Can be played online. One player has 1 red counter, the other has 4 blue. The red counter needs to reach the other side, and the blue needs to trap the red.
### You Owe Me Five Farthings, Say the Bells of St Martin's
##### Stage: 3 Challenge Level:
Use the interactivity to listen to the bells ringing a pattern. Now it's your turn! Play one of the bells yourself. How do you know when it is your turn to ring?
### Rolling Around
##### Stage: 3 Challenge Level:
A circle rolls around the outside edge of a square so that its circumference always touches the edge of the square. Can you describe the locus of the centre of the circle?
### An Unhappy End
##### Stage: 3 Challenge Level:
Two engines, at opposite ends of a single track railway line, set off towards one another just as a fly, sitting on the front of one of the engines, sets off flying along the railway line...
### Volume of a Pyramid and a Cone
##### Stage: 3
These formulae are often quoted, but rarely proved. In this article, we derive the formulae for the volumes of a square-based pyramid and a cone, using relatively simple mathematical concepts.
### Cubic Net
##### Stage: 4 and 5 Challenge Level:
This is an interactive net of a Rubik's cube. Twists of the 3D cube become mixes of the squares on the 2D net. Have a play and see how many scrambles you can undo!
### Balancing 2
##### Stage: 3 Challenge Level:
Meg and Mo still need to hang their marbles so that they balance, but this time the constraints are different. Use the interactivity to experiment and find out what they need to do.
### Muggles Magic
##### Stage: 3 Challenge Level:
You can move the 4 pieces of the jigsaw and fit them into both outlines. Explain what has happened to the missing one unit of area.
### Online
##### Stage: 2 and 3 Challenge Level:
A game for 2 players that can be played online. Players take it in turns to select a word from the 9 words given. The aim is to select all the occurrences of the same letter.
### Balancing 3
##### Stage: 3 Challenge Level:
Mo has left, but Meg is still experimenting. Use the interactivity to help you find out how she can alter her pouch of marbles and still keep the two pouches balanced.
### Balancing 1
##### Stage: 3 Challenge Level:
Meg and Mo need to hang their marbles so that they balance. Use the interactivity to experiment and find out what they need to do.
### Instant Insanity
##### Stage: 3, 4 and 5 Challenge Level:
Given the nets of 4 cubes with the faces coloured in 4 colours, build a tower so that on each vertical wall no colour is repeated, that is all 4 colours appear.
### Poly-puzzle
##### Stage: 3 Challenge Level:
This rectangle is cut into five pieces which fit exactly into a triangular outline and also into a square outline where the triangle, the rectangle and the square have equal areas.
### Shuffle Shriek
##### Stage: 3 Challenge Level:
Can you find all the 4-ball shuffles?
### Lost
##### Stage: 3 Challenge Level:
Can you locate the lost giraffe? Input coordinates to help you search and find the giraffe in the fewest guesses.
### Interactive Spinners
##### Stage: 3 Challenge Level:
This interactivity invites you to make conjectures and explore probabilities of outcomes related to two independent events.
### Drips
##### Stage: 2 and 3 Challenge Level:
An animation that helps you understand the game of Nim.
### Diamond Mine
##### Stage: 3 Challenge Level:
Practise your diamond mining skills and your x,y coordination in this homage to Pacman.
### Isosceles Triangles
##### Stage: 3 Challenge Level:
Draw some isosceles triangles with an area of $9$cm$^2$ and a vertex at (20,20). If all the vertices must have whole number coordinates, how many is it possible to draw?
### Subtended Angles
##### Stage: 3 Challenge Level:
What is the relationship between the angle at the centre and the angles at the circumference, for angles which stand on the same arc? Can you prove it?
### Nine Colours
##### Stage: 3 Challenge Level:
Can you use small coloured cubes to make a 3 by 3 by 3 cube so that each face of the bigger cube contains one of each colour?
### Shuffles Tutorials
##### Stage: 3 Challenge Level:
Learn how to use the Shuffles interactivity by running through these tutorial demonstrations.
### Teddy Town
##### Stage: 1, 2 and 3 Challenge Level:
There are nine teddies in Teddy Town - three red, three blue and three yellow. There are also nine houses, three of each colour. Can you put them on the map of Teddy Town according to the rules?
### A Tilted Square
##### Stage: 4 Challenge Level:
The opposite vertices of a square have coordinates (a,b) and (c,d). What are the coordinates of the other vertices?
### Rollin' Rollin' Rollin'
##### Stage: 3 Challenge Level:
Two circles of equal radius touch at P. One circle is fixed whilst the other moves, rolling without slipping, all the way round. How many times does the moving coin revolve before returning to P?
### Number Pyramids
##### Stage: 3 Challenge Level:
Try entering different sets of numbers in the number pyramids. How does the total at the top change?
### Bow Tie
##### Stage: 3 Challenge Level:
Show how this pentagonal tile can be used to tile the plane and describe the transformations which map this pentagon to its images in the tiling.
### Konigsberg Plus
##### Stage: 3 Challenge Level:
Euler discussed whether or not it was possible to stroll around Koenigsberg crossing each of its seven bridges exactly once. Experiment with different numbers of islands and bridges.
### Partitioning Revisited
##### Stage: 3 Challenge Level:
We can show that (x + 1)² = x² + 2x + 1 by considering the area of an (x + 1) by (x + 1) square. Show in a similar way that (x + 2)² = x² + 4x + 4
### Shear Magic
##### Stage: 3 Challenge Level:
What are the areas of these triangles? What do you notice? Can you generalise to other "families" of triangles?
### Changing Places
##### Stage: 4 Challenge Level:
Place a red counter in the top left corner of a 4x4 array, which is covered by 14 other smaller counters, leaving a gap in the bottom right hand corner (HOME). What is the smallest number of moves. . . .
### Sliding Puzzle
##### Stage: 1, 2, 3 and 4 Challenge Level:
The aim of the game is to slide the green square from the top right hand corner to the bottom left hand corner in the least number of moves.
### Icosian Game
##### Stage: 3 Challenge Level:
This problem is about investigating whether it is possible to start at one vertex of a platonic solid and visit every other vertex once only returning to the vertex you started at.
### Speeding Up, Slowing Down
##### Stage: 3 Challenge Level:
Experiment with the interactivity of "rolling" regular polygons, and explore how the different positions of the red dot affects its speed at each stage.
### Factor Lines
##### Stage: 2 and 3 Challenge Level:
Arrange the four number cards on the grid, according to the rules, to make a diagonal, vertical or horizontal line.
### Disappearing Square
##### Stage: 3 Challenge Level:
Do you know how to find the area of a triangle? You can count the squares. What happens if we turn the triangle on end? Press the button and see. Try counting the number of units in the triangle now. . . .
### Nim-interactive
##### Stage: 3 and 4 Challenge Level:
Start with any number of counters in any number of piles. 2 players take it in turns to remove any number of counters from a single pile. The winner is the player to take the last counter.
### More Number Pyramids
##### Stage: 3 Challenge Level:
When number pyramids have a sequence on the bottom layer, some interesting patterns emerge...
### Jam
##### Stage: 4 Challenge Level:
To avoid losing think of another very well known game where the patterns of play are similar.
### Fifteen
##### Stage: 2 and 3 Challenge Level:
Can you spot the similarities between this game and other games you know? The aim is to choose 3 numbers that total 15.
### Inside Out
##### Stage: 4 Challenge Level:
There are 27 small cubes in a 3 x 3 x 3 cube, 54 faces being visible at any one time. Is it possible to reorganise these cubes so that by dipping the large cube into a pot of paint three times you. . . .
### Square Coordinates
##### Stage: 3 Challenge Level:
A tilted square is a square with no horizontal sides. Can you devise a general instruction for the construction of a square when you are given just one of its sides?
### Flippin' Discs
##### Stage: 3 Challenge Level:
Identical discs are flipped in the air. You win if all of the faces show the same colour. Can you calculate the probability of winning with n discs?
### Multiplication Tables - Matching Cards
##### Stage: 1, 2 and 3 Challenge Level:
Interactive game. Set your own level of challenge, practise your table skills and beat your previous best score. | 2,432 | 10,447 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2017-30 | longest | en | 0.864037 |
https://stats.stackexchange.com/questions/31718/pca-with-on-data-with-unequal-length/335257 | 1,618,597,364,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038088245.37/warc/CC-MAIN-20210416161217-20210416191217-00459.warc.gz | 629,708,603 | 39,433 | # PCA with on data with unequal length
I am wondering how to perform PCA on data with unequal lengths. In the simplest case you could split up the data matrix so that one block would be $T_{1}Xn$ and the rest is $T_{2}Xn$. The second block has effectively missing data in the first $T_{1}-T_{2}Xn$ block.
Many of the applications of PCA on missing data are variants of the EM algorithm methodology (these typically will assume the distribution is multivariate Gaussian) that are more suited to applications where the data is missing at random. Oftentimes, these techniques do not particularly apply to data with uneven lengths. However, there are techniques to estimate the mean and covariance matrix in these cases and it is possible to apply PCA to the resulting covariance matrix. However, when the number of variables becomes large (absolutely or relative to the number of observations), then it may no longer be convenient to estimate the covariance matrix in this fashion.
• Excuse me, what does this mean data is T1Xn and T2Xn? And did by "unequal length" you just mean that there are missing values? – ttnphns Jul 5 '12 at 19:24
• So the data is $TXn$, but you could break it into the respective panels that would be of different length. The first $T_{2}-T_{1}Xn$ block would be missing in the second panel. – John Jul 5 '12 at 19:45
• I'm sorry if this comes a bit too late... I'm currently looking for a solution to the same problem. You mention that there are techniques to estimate the mean and covariance matrix in these cases and it is possible to apply PCA to the resulting covariance matrix... would you mind kindly giving me some references? Thank you in advance! – Imaco Mar 18 at 15:56
• This is a common one: ideas.repec.org/p/fth/pennfi/05-96.html – John Mar 18 at 22:10 | 435 | 1,794 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2021-17 | latest | en | 0.93089 |
https://stromcv.com/doing-divide-long-division-61 | 1,674,938,110,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499654.54/warc/CC-MAIN-20230128184907-20230128214907-00862.warc.gz | 570,502,528 | 5,928 | How to divide long division
We will explore How to divide long division can help students understand and learn algebra.
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Divide each digit of the dividend with the divisor starting from left to right. Bring down the next digit after each step as shown below: 1. Divide 2 by 2. Write the remainder after subtracting the
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How To Do Long Division? Definition, Steps, Method, Examples
Step 1: D for Divide. How many times can you divide 5 into 15. The answer is 3. So you put 3 on the quotient line. Step 2: M for Multiply. You multiply your answer from step 1 and your divisor: 3 x 5 = 15. Write this underneath the 15. Step 3: S
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To divide larger numbers, use long division. Unlike the other Big Four operations, long division moves from left to right. For each digit in the dividend (the number you’re
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To perform long division, first identify the dividend and divisor. To divide 100 by 7, where 100 is the dividend and 7 is the divisor, set up the long division problem by writing the dividend under
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Michael Smith | 603 | 2,681 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2023-06 | latest | en | 0.911317 |
http://gambaswiki.org/wiki/cat/trigo | 1,620,381,015,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988775.80/warc/CC-MAIN-20210507090724-20210507120724-00114.warc.gz | 19,868,303 | 2,250 | # Trigonometric Functions
`ACos` Computes the arc cosine of an angle. `ACosh` Computes the hyperbolic arc cosine of an angle. `Ang` Computes the angle polar coordinate from two rectangular coordinates. `ASin` Computes the arc sine of an angle. `ASinh` Computes the hyperbolic arc sine of an angle. `ATan` Computes the arc tangent of one number. `ATan2` Computes the arc tangent of two numbers. `ATanh` Computes the hyperbolic arc tangent of an angle. `Cos` Computes the cosine of an angle. `Cosh` Computes the hyperbolic cosine of an angle. `Deg` Converts radians to degrees. `Hyp` Calculate the hypotenuse of a triangle. `Mag` Computes the distance polar coordinate from two rectangular coordinates. `Pi` Returns π or a multiple of π. `Sin` Computes the sine of an angle. `Sinh` Computes the hyperbolic sine of an angle. `Tan` Computes the tangent of an angle. `Tanh` Computes the hyperbolic tangent of an angle. `Rad` Converts degrees to radians. | 239 | 950 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2021-21 | longest | en | 0.61943 |
https://stats.stackexchange.com/questions/338804/hypothesis-testing-on-coefficients-in-two-subsets-of-data-after-stepwise-regress/338807 | 1,638,690,049,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363149.85/warc/CC-MAIN-20211205065810-20211205095810-00031.warc.gz | 600,988,734 | 36,039 | # Hypothesis Testing on coefficients in two subsets of data after Stepwise Regression
Is it a reasonable approach to run a hypothesis test to test whether the coefficients of a variable in two regressions on two different subsets of the same population are different if you have used stepwise regression to build the model in each subset? So assuming the variable you want to test is included in the regression for each of the subsets but other independent variables are included for one and not for the other, can you still test for a significant difference in the variable you want to examine?
If so, what approach would be optimal? I was thinking about using an interaction term between the variable which defines the two subset variables (in this case gender) and the variable I want to test for differences in the coefficient in (in this case years of education) however I'm not sure exactly how to go about this and which regression I'd use the interaction term in (i.e whether it would be the male or female regression).
No.
Stepwise model building is not reasonable, but is cargo cult science. The $$p$$-values produced from stepwise regression model building do not have the typical meaning of "probability of observing the estimate/test statistic assuming the null hypothesis is true," but rather "probability of observing the estimate/test statistic based on a series of unstated conditionals that are almost certainly predicated on some number of variables not included in the presented model."
Use stepwise regression if you want to appear to be performing meaningful statistical analysis, while providing results (estimates, "noise" variables, missing "real" variables, $$p$$-values, $$R^{2}$$, etc.) that are very likely to be biased.
Some relevant citations
Babyak, M. A. (2004). What you see may not be what you get: A brief, nontechnical introduction to overfitting in regression-type models. Psychosomatic Medicine, 66:411–421.
Flom, P. L. and Cassell, D. L. (2007). Stopping stepwise: Why stepwise and similar selection methods are bad, and what you should use.
Henderson, D. A. and Denison, D. R. (1989). Stepwise regression in social and psychological research. Psychological Reports, 64:251–257.
Huberty, C. J. (1989). Problems with stepwise methods—better alternatives. Advances in Social Science Methodology, 1:43–70.
Hurvich, C. M. and Tsai, C.-L. (1990). The impact of model selection on inference in linear regression. The American Statistician, 44(3):214–217.
McIntyre, S. H., Montgomery, D. B., Srinivasan, V., and Weitz, B. A. (1983). Evaluating the statistical significance of models developed by stepwise regression. Journal of Marketing Research, 20(1):1–11.
Pope, P. T. and Webster, J. T. (1972). The use of an $$F$$-statistic in stepwise regression procedures. Technometrics, 14(2):327–340.
Rencher, A. C. and Pun, F. C. (1980). Inflation of $$R^{2}$$ in best subset regression. Technometrics, 22(1):49–53.
Romano, J. P. and Wolf, M. (2005). Stepwise multiple testing as formalized data snooping. Econometrica, 73(4):1237–1282.
Sribney, B., Harrell, F., and Conroy, R. (2011). Problems with stepwise regression.
Steyerberg, E. W., Eijkemans, M. J., and Habbema, J. D. F. (1999). Stepwise selection in small data sets: a simulation study of bias in logistic regression analysis. Journal of Clinical Epidemiology, 52(10):935–942.
Thompson, B. (1995). Stepwise regression and stepwise discriminant analysis need not apply here: A guidelines editorial. Educational and Psychological Measurement, 55(4):525–534.
Whittingham, M., Stephens, P., Bradbury, R., and Freckleton, R. (2006). Why do we still use stepwise modelling in ecology and behaviour? Journal of Animal Ecology, 75(5):1182–1189.
Wilkinson, L. (1979). Tests of significance in stepwise regression. Psychological Bulletin, 86(1):168–174.
• recently developed methods do provide correct p-values in the stepwise regression context if that's what you want arxiv.org/abs/1401.3889 (me, I'm a Bayesian, so i don't, but methods exist for those that do)
– Cyan
Jan 6 '19 at 15:38
• Fair point, Thank you @Cyan Sadly, that's not what the vast majority of published applications or taught applications of stepwise model building are. My suspicion is that the structural push towards soft funding of academic researchers + a confirmation-bias oriented aversion to 'negative findings' pushes the majority of researchers away from substantive science, and into algorithmic rituals permitting publication, and coincidentally enough, justification of the researcher's next paycheck. Hmm... apparently I am in a dark mood this morning. Jan 6 '19 at 17:19 | 1,145 | 4,653 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 5, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2021-49 | latest | en | 0.933523 |
https://stefczykradzi.pl/crushing-machine/11-06_18118.html | 1,642,995,604,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304471.99/warc/CC-MAIN-20220124023407-20220124053407-00356.warc.gz | 532,786,892 | 8,403 | [email protected]
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# Cement Sand And Crush Formulas
• ### How to Calculate Cement Sand and Coarse Aggregate
Density of Cement 1440 kg/ m 3 0.3636 X 1440 523 kg 10.5 Approx. Bags . Sand Quantity ( Sand Part / Concrete Parts ) * Concrete Volume (1.5/5.5) * 2 0.5454 m 3. Coarse Aggregate (Coarse Aggregate Part / Concrete Parts ) *
• ### How To Calculate Cement Sand Aggregates Quantity in
Aug 03, 2018 Consider volume of concrete 1m3. Dry Volume of Concrete 1 x 1.54 1.54 m3. Sand (1.5/5.5) x 1.54 0.42 m3 1.5 is a part of Sand, 5.5 is sum of ratio. Density of Sand is 1450/m3. For KG 0.42 x 1450 609 kg. As we know that 1m3 35.31 CFT. For Calculation in Cubic Feet 0.42 x 35.31 14.83 Cubic Feet.
• ### cement sand and crush formulas
cement sand crush calculation formula. How to calculate cement sand quantity in 13 mortar Quora Thanks for A2A The process of calculation to ascertain the quantity of constituents in a concrete mix or mortar with defined proportion is a part of
• ### How to Calculate Quantities of Cement Sand and Aggregate
Feb 07, 2017 The ratio of dry volume to the wet volume of concrete is 1.54. So 1.54 Cum of dry materials (cement, sand and aggregate) is required to produce 1 Cum of concrete. Volume of Cement required 1/ (124) X 1.54 1/7 X 1.54 0.22 Cum. Volume of Sand required 2/7 X 1.54 0.44 Cum or 15.53 cft.
• ### Make Excel sheet for Cement Sand and Crushed Stone
(one part cement two part sand four part aggregates). Volume of concrete work is taken as 500 cft. Now, in order to determine the quantity of cement in cft., put the cursor on the specific cell of concrete volume and apply the formula i.e. (D2*1.54*A2)/(A2B2C2). Here 1.54 is taken to convert wet volume of concrete to dry volume of concrete.
• ### Calculate Quantities of Materials for Concrete Cement
Assuming bulk densities of materials per cubic meter, cubic feet and cubic yards as follows Cement 1500 kg/m3 93.642 lb/ft3 2528.332 lb/cubic yards Sand 1700 kg/m3 105 lb/ft3 2865.443 lb/cubic yards Coarse aggregates 1650 kg/m3 103 lb/ft3 2781.166 lb/cubic yards Specific gravities of concrete materials are as follows Cement 3.15 Sand 2.6
• ### What are the Correct Concrete Mixing Ratios Ratio Chart
Some basic mixing ratios for concrete are 123, 133, 124. These mixing ratios are based on the proportions of cement sand stone in that order. The ratio you use will depend on what psi strength you need. To make concrete there are four basic materials you need Portland cement - You can buy this in a 94lb bag
• ### MOISTURE CONTENT OF SAND Concrete Product
Example No. 1 (A) equals weight of damp sand as received or as weighed into weigh batcher. (B) equals saturated surface dry condition of the same sand. (A) equals 520 grams. (B) equals 500 grams. 520 500 X 20 x 100 4.00% of Moisture in Sand by Weight 500 500
• ### Concrete Calculator Estimate Cement Sand Gravel
On this page, you can calculate material consumption viz., cement, sand, stone gravel for the following concrete mix ratios - 11.53, 124, 136, 148, 1510. Once, the quantities are determined, it is easy to estimate the cost of a concrete block, driveway, patio, yard or any other structure with the price prevailing in your area.
• ### THE QUANTITIES OF MATERIALS IN ONE CUBIC METER
Target mean strength of concrete is derived from the below formula . f. t f. ck ... the Vm was not the simple addition of the unit volume ratio of the cement, sand, and water. ...
• ### Basic formula for finding cement sand and crush by 148
The portable concrete crusher is a basic crushing system. ... formula classifier efficiency in cement - Crusher South Africa . ... cement plants, sand and gravel ... Chat Online aci concrete mix design excel in description - Water/cement ratio- This component is
• ### Calculate Quantities of Materials for Concrete Cement
Thus, the quantity of cement required for 1 cubic meter of concrete 0.98/0.1345 7.29 bags of cement. The quantities of materials for 1 m3 of concrete production can be calculated as follows The weight of cement required 7.29 x 50 364.5 kg. Weight of fine aggregate (sand) 1.5 x 364.5 546.75 kg.
• ### PCC Concrete Calculation PCC Concrete Ratio Plain
Water Clean water that is safe to drink should be utilized in PCC. The following formula is applied for this -. Material ratio of material/sum of ratio x dry volume. Suppose the ratio of cement, sand and crushed stone is 124. To calculate dry volume, the formula will be wet volume x 1.54. Wet volume Length x Breadth x Thickness. YouTube.
• ### What are the Correct Concrete Mixing Ratios Ratio Chart
When making your own concrete its important to use the correct concrete mixing ratios to produce a strong, durable concrete mix. Some basic mixing ratios for concrete are 123, 133, 124. These mixing ratios are based on the proportions of cement sand stone in that order. The ratio you use will depend on what psi strength you need.
• ### Mixing Your Own Concrete JLC Online
Dec 19, 2019 Mixing concrete from scratch involves the combination of portland cement, sand, coarse aggregate (crushed and graded stone), and water. The industry standard uses a 1-2-3 formula with one part portland cement, two parts sand, and three parts aggregate for a
• ### MOISTURE CONTENT OF SAND Concrete Product
The simple formula shown above illustrates the method of checking moisture content of sand, gravel, crushed rock, and /or fresh concrete. We list these few materials because it is the usual situation that may occur in a block plant. In all computations used in batch
• ### Aggregates for Concrete Memphis
concrete volume (70% to 85% by mass) and strongly influ-ence the concretes freshly mixed and hardened proper-ties, mixture proportions, and economy. Fine aggregates (Fig. 5-1) generally consist of natural sand or crushed stone with most particles smaller than 5 mm (0.2 in.). Coarse aggregates (Fig. 5-2) consist of one or a com-CHAPTER 5
• ### Materials
Concrete Sand. McCars concrete sand is an outstanding component for a variety of concrete formulas used in Central Texas, including cement and asphalt mixtures. We mine, crush, wash, and sift at our quarry so your mixture is smooth.
• ### What is Concrete Mixing Formula Concrete Information
Concrete Mixing Formula. Lets imagine that we have to prepare the dough for a floor 2 meters long (200cm), 2.5 meters wide (250cm) and 0.2 meters high (20cm), so it will be enough to multiply the length by the width, finally, the result by the height, in this case we will do 2 x 2.5 5 x 0.2 1mt (volume of mix ) From the calculation ...
• ### Formulas for Lightweight Concrete Green Home Building
I realize if the formula is a little porous because of less sand than the plaster and mineral paint should make up for any porosity on the exterior. We also did some formulas using portland cement and lime, and one using lime with magnesium phosphate.
• ### Concrete Calculator Estimate Cement Sand Gravel
Example calculation Estimate the quantity of cement, sand and stone aggregate required for 1 cubic meter of 124 concrete mix. Ans. Materials required are 7 nos. of 50 kg bag of cement, 0.42 m 3 of sand and 0.83 m 3 of stone aggregate.
• ### FINENESS MODULUS OF FINE AGGREGATE
1.1 This method determines the fineness modulus of concrete fine aggregate used in evaluation of natural and manufactured sands for portland cement concrete. 1.2 The values given in parentheses (if provided) are not standard and may not be exact mathematical conversions. Use each system of units separately.
• ### Cement and Sand ratio for brickwork How to Calculate
To calculate the sand quantity Given that the sand cement ratio is 15. So, sand required is (.305)/6.25cum. When converted into kg .251440360Kg. If you are looking for construction cost calculator then have a look on it here. With this simple formula, it is easy to estimate the amount of cement consumption required for brick masonry ...
• ### Methods of Proportioning Cement Sand and Aggregates in
Methods of Proportioning Concrete Arbitrary Method of Proportioning Concrete The general expression for the proportions of cement, sand and coarse aggregate is 1 n 2n by volume. 1 1 2 and 1 1.2 2.4 for very high strength. 1 1.5 3 and 1 2 4 for normal works. 1 3 6 and 1 4 8 for foundations and mass concrete works.
• ### How To Calculate Number Of Bricks Cement And Sand For
Required amount of Sand 0.306565 X 6/7 0.26277 Cubic metre (m3) Therefore, For 1 cum of brickwork, we need. 500 Numbers of bricks. 63 kg of cement. 0.263 m 3 of sand. Feel free to use the below calculator.
• ### CITY OF HOUSTON CEMENT STABILIZED SAND PART 1
sand-cement mixture within 4 hours after add ition of water to mix at plant. B. Do not place or compact sand-cement mixture in standing or free water. C. Where potable water lines cross wastewater line, embed wastewater line with cement stabilized sand in accordance with Texas Administrative Code 290.44(e)(4)(B) 1.
• ### Density of Cement Sand and Aggregate Cement Density
The Density of Sand utility returns sand density based on sand conditions (wet/dry in bulk / packaged).. The density of the sand is affected if the sand is compacted (bulged) or loose and if it is wet or dry.When packed, the grains of sand are forced to form a narrower formation, and more matter is in the volume.. Natural sand i.e. locally available River sand having a bulk
• ### Fly Ash Bricks Mixing Proportion 3 Important Formulas
Aug 15, 2013 River Sand or Stone Dust 30 to 40%. Cement 8 to 12%. In this Mix design, gypsum and lime are replaced with cement. This mix ratio is not so popular and not a profitable design for entrepreneurs. Cement is more expensive than Gypsum and Lime, hence this formula is useful only during non-availability of gypsum and lime.
• ### RCC Calculator Estimate Cement Sand RCC Online Calculator
Reinforced cement concrete or RCC calculator can be used to calculate M15, M20 and M25 mix ratio of cement, sand and jelly. At materialtree.com, you can also shop for superior quality cement bags at best market price.
• ### How to Calculate Cement Sand and Aggregate required for
Cement Sand Aggregate (in Kgs) is 50 kgs 115 kgs 209 kgs (by weight) Water required for the mixture 27.5 kgs. Total weight of concrete ingredients 5011520927.5 401.5 say 400 kg. Density of concrete 2400 kg/cum. So, 1 bag of cement produces 400/2400 0.167 cum
• ### Pemanfaatan pasir sungai dan batu pecah asal Sukadana
The result of this research the modulus of elasticity of concrete formula E 4791.7 f c . The ratio of concrete compressive strength 3rd, 7th, and 28th days age was 68%, 89% and 100%. Finally, the conclusion is that river sand and crushed aggregate from Sukadana appropriate as normal concrete material.
• ### HIGHSTRENGTH STRUCTURAL LIGHTWEIGHT CONCRETE
Conventional cellular concrete produced with a pre-formed foam mixture is produced by discharging a stream of preformed foam into a mixing unit on site or a transit mix load of sand-cement grout or cement-water slurry. This foam surfactant resembles shaving cream or the foam used for firefighting.
• ### SIEVE ANALYSIS OF FINE AND COARSE AGGREGATE
Crushed Coarse Aggregate for Concrete TR 112 TR 113 Method B Lightweight Aggregate for Concrete TR 113 Method A Recycled PCC Base Course TR 112 TR 113 Method B Sand Clay Gravel Base Course Aggregate TR 112 TR 113 Method C Sand Base Course Aggregate TR 112 TR 113 Method B Stone Base Course Aggregate TR 112 TR 113
• ### Formulas for Lightweight Concrete Green Home Building
Formulas for Lightweight Concrete. Fernando Martinez Lewels has a M.S.C.E degree from the University of Texas at El Paso. He is now working with the Agartif company in Chihuahua, Mexico (about 170 miles from El Paso Texas).
• ### Concrete Mix Ratio What Is Concrete Mix Ratio Types of
Cement Mixture Formula. The safest bet for any concrete mix is four-two-one four parts crushed rock two parts sand and one part cement. The four-two-one mix, obviously, has seven parts. Conveniently, when mixing concrete, the ratio can be mixed on any range of scales. Concrete Mix Ratios Table | 3,047 | 12,191 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2022-05 | latest | en | 0.759251 |
https://www.lidolearning.com/questions/m-bb-ncert8-ch14-ex14p2-q5-sum2/factorise-/ | 1,685,372,303,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224644867.89/warc/CC-MAIN-20230529141542-20230529171542-00724.warc.gz | 930,522,368 | 12,761 | # NCERT Solutions Class 8 Mathematics Solutions for Exercise 14.2 in Chapter 14 - Factorisation
Question 26 Exercise 14.2
Factorise:
q^{2}-10 q+121
Solution:
The middle term is split into two elements using the "Splitting the Middle Term by Factorization" approach and further by grouping the pairs, given quadratic polynomial can be factorized.
Observed that, 21 = -7 x -3 and -7 + (-3) = -10
\begin{array}{l} q^{2}-10 q+21=q^{2}-3 q-7 q+21 \\ =q(q-3)-7(q-3) \end{array}
\begin{array}{l} =(q-7)(q-3) \\ \text { This implies } q^{2}-10 q+21=(q-7)(q-3) \end{array}
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Connect with us on social media! | 280 | 858 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2023-23 | latest | en | 0.755582 |
http://mathhelpforum.com/calculus/115842-question-about-inflection-points.html | 1,480,914,081,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541525.50/warc/CC-MAIN-20161202170901-00049-ip-10-31-129-80.ec2.internal.warc.gz | 166,785,995 | 10,046 | 1. ## a question about inflection points
If there is a maximum and a minimum on a curve, will the inflection point between them be exactly in the middle? Or can it be located anywhere within the distance between the maximum and the minimum?
2. I don't think so.
Let $y=ax^3+bx^2$
Then $y'=3ax^2+2bx$ giving maxs/mins at x=0,-2b/(3a)
And $y''=6ax+2b$ giving infl pts at x=-b/(3a)
which is half way inbetween.
Now even if we change that to $y=ax^3+bx^2+cx+d$
the inflection pt is the same (x that is)
while the max/min is the solution to $3ax^2+2bx+c=0$
which gives $x={-2b\pm\sqrt{4b^2-12ac}\over 6a}$ and that average is once again -b/(3a)
So it is true for polynomials of degree 3.
Moving onto the next power...
Try $y'=x^3-3x^2+2x$ giving mins at x=0 and 2 and a max at 1
while $y''=3x^2-6x+2$ giving infl pts at $x=1\pm 1/\sqrt{3}$
and those two inflection pts are not at .5 and 1.5, the midpoints of 0 and 1, and of 1 and 2. | 336 | 938 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 9, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2016-50 | longest | en | 0.919325 |
https://www.kdbfaq.com/tag/milliseconds/ | 1,591,519,842,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590348526471.98/warc/CC-MAIN-20200607075929-20200607105929-00568.warc.gz | 787,639,303 | 14,491 | ## How do I extract hours, minutes, or seconds from a time?
Short answer: Times are integers (think milliseconds).
```q)(`hh`mm`ss \$ .z.T), `int \$ .z.T mod 1000
16 49 2 233
q)
```
You can extract the hours, minutes, and seconds from a time by passing `hh, `mm, and `ss, respectively, as left arguments to \$ (cast):
```q)now: .z.T
q)now
16:49:02.233
q)`hh \$ now
16
q)`mm \$ now
49
q)`ss \$ now
2
q)`hh`mm`ss \$ now
16 49 2
q)
```
Getting the milliseconds from a time is slightly less obvious. Times (type -19) are represented internally by q as 32-bit integers; typically the value counts the number of milliseconds since midnight, but it can also represent a span of time. We can cast freely back and forth between the two types and the values are preserved:
```q)`time \$ 0
00:00:00.000
q)`int \$ 00:00:00.000
0q)`time \$ 24 * 60 * 60 * 1000
24:00:00.000
q)`int \$ 24:00:00.000
86400000
q)`int \$ now
60542233
q)`time \$ 60542233
16:49:02.233
q)
```
Not only is the internal representation of time simply an integer, we can mix integers and times in integer arithmetic operations, and the result is always a time:
```q)01:00:00.000 + 00:01:00.000
01:01:00.000
q)01:00:00.000 + 60000
01:01:00.000
q)01:00:00.000 * 4
04:00:00.000
q)now - 01:30:20.123
15:18:42.110
q)
```
By using div and mod, then, we have an alternative means to calculate the components of a time:
```q)now div 3600000 // milliseconds per hour
00:00:00.016
q)now mod 1000 // just the milliseconds, please
00:00:00.233
q)
```
Although extracting milliseconds from a time while keeping the time type (as in the second example above) is sometimes useful, we normally want to get back these components of a time as integers, so let’s cast it:
```q)`int \$ now mod 1000
233
q)
```
By the way, there is a shortcut for getting hours, minutes, and seconds from global variables that hold times: dot notation.
```q)x: .z.T
q)x.hh, x.mm, x.ss
16 49 2
q)
```
However, we rarely use global variables to hold time values.
## How do I extract the milliseconds from a time?
mod 1000:
q)now: .z.T
q)now
00:15:00.812
q)now mod 1000
00:00:00.812
If you want the milliseconds as an integer, you can simply follow the mod with a conversion to int:
q)`int\$ now mod 1000
812
You don’t want to pass a datetime to mod; the result you’ll get is not what you had in mind. You must convert it to a time first:
q)now: .z.Z
q)now
2011.03.26T09:51:26.624
q)now mod 1000
102.4107
q)`int\$ (`time\$ now) mod 1000
624
q) | 831 | 2,485 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2020-24 | latest | en | 0.715569 |
https://www.lmfdb.org/L/8/570e4/1.1/c3e4/0/0 | 1,623,876,403,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487626008.14/warc/CC-MAIN-20210616190205-20210616220205-00037.warc.gz | 795,106,079 | 8,138 | # Properties
Label 8-570e4-1.1-c3e4-0-0 Degree $8$ Conductor $105560010000$ Sign $1$ Analytic cond. $1.27927\times 10^{6}$ Root an. cond. $5.79923$ Motivic weight $3$ Arithmetic yes Rational yes Primitive no Self-dual yes Analytic rank $0$
# Origins of factors
## Dirichlet series
L(s) = 1 − 4·2-s + 6·3-s + 4·4-s + 10·5-s − 24·6-s + 36·7-s + 16·8-s + 9·9-s − 40·10-s − 22·11-s + 24·12-s + 92·13-s − 144·14-s + 60·15-s − 64·16-s − 105·17-s − 36·18-s + 136·19-s + 40·20-s + 216·21-s + 88·22-s − 56·23-s + 96·24-s + 25·25-s − 368·26-s − 54·27-s + 144·28-s + ⋯
L(s) = 1 − 1.41·2-s + 1.15·3-s + 1/2·4-s + 0.894·5-s − 1.63·6-s + 1.94·7-s + 0.707·8-s + 1/3·9-s − 1.26·10-s − 0.603·11-s + 0.577·12-s + 1.96·13-s − 2.74·14-s + 1.03·15-s − 16-s − 1.49·17-s − 0.471·18-s + 1.64·19-s + 0.447·20-s + 2.24·21-s + 0.852·22-s − 0.507·23-s + 0.816·24-s + 1/5·25-s − 2.77·26-s − 0.384·27-s + 0.971·28-s + ⋯
## Functional equation
\begin{aligned}\Lambda(s)=\mathstrut &\left(2^{4} \cdot 3^{4} \cdot 5^{4} \cdot 19^{4}\right)^{s/2} \, \Gamma_{\C}(s)^{4} \, L(s)\cr=\mathstrut & \,\Lambda(4-s)\end{aligned}
\begin{aligned}\Lambda(s)=\mathstrut &\left(2^{4} \cdot 3^{4} \cdot 5^{4} \cdot 19^{4}\right)^{s/2} \, \Gamma_{\C}(s+3/2)^{4} \, L(s)\cr=\mathstrut & \,\Lambda(1-s)\end{aligned}
## Invariants
Degree: $$8$$ Conductor: $$2^{4} \cdot 3^{4} \cdot 5^{4} \cdot 19^{4}$$ Sign: $1$ Analytic conductor: $$1.27927\times 10^{6}$$ Root analytic conductor: $$5.79923$$ Motivic weight: $$3$$ Rational: yes Arithmetic: yes Character: induced by $\chi_{570} (1, \cdot )$ Primitive: no Self-dual: yes Analytic rank: $$0$$ Selberg data: $$(8,\ 2^{4} \cdot 3^{4} \cdot 5^{4} \cdot 19^{4} ,\ ( \ : 3/2, 3/2, 3/2, 3/2 ),\ 1 )$$
## Particular Values
$$L(2)$$ $$\approx$$ $$0.3311578304$$ $$L(\frac12)$$ $$\approx$$ $$0.3311578304$$ $$L(\frac{5}{2})$$ not available $$L(1)$$ not available
## Euler product
$$L(s) = \displaystyle \prod_{p} F_p(p^{-s})^{-1}$$
$p$$\Gal(F_p)$$F_p(T)$
bad2$C_2$ $$( 1 + p T + p^{2} T^{2} )^{2}$$
3$C_2$ $$( 1 - p T + p^{2} T^{2} )^{2}$$
5$C_2$ $$( 1 - p T + p^{2} T^{2} )^{2}$$
19$C_2^2$ $$1 - 136 T + 783 p T^{2} - 136 p^{3} T^{3} + p^{6} T^{4}$$
good7$D_{4}$ $$( 1 - 18 T + 382 T^{2} - 18 p^{3} T^{3} + p^{6} T^{4} )^{2}$$
11$D_{4}$ $$( 1 + p T + 236 p T^{2} + p^{4} T^{3} + p^{6} T^{4} )^{2}$$
13$C_2^2$ $$( 1 - 46 T - 81 T^{2} - 46 p^{3} T^{3} + p^{6} T^{4} )^{2}$$
17$D_4\times C_2$ $$1 + 105 T - 691 T^{2} + 198450 T^{3} + 58961262 T^{4} + 198450 p^{3} T^{5} - 691 p^{6} T^{6} + 105 p^{9} T^{7} + p^{12} T^{8}$$
23$D_4\times C_2$ $$1 + 56 T + 2658 T^{2} - 1335936 T^{3} - 185963117 T^{4} - 1335936 p^{3} T^{5} + 2658 p^{6} T^{6} + 56 p^{9} T^{7} + p^{12} T^{8}$$
29$D_4\times C_2$ $$1 - 153 T - 9565 T^{2} + 2418012 T^{3} - 78460746 T^{4} + 2418012 p^{3} T^{5} - 9565 p^{6} T^{6} - 153 p^{9} T^{7} + p^{12} T^{8}$$
31$D_{4}$ $$( 1 - 222 T + 65743 T^{2} - 222 p^{3} T^{3} + p^{6} T^{4} )^{2}$$
37$D_{4}$ $$( 1 + 582 T + 185602 T^{2} + 582 p^{3} T^{3} + p^{6} T^{4} )^{2}$$
41$D_4\times C_2$ $$1 + 228 T - 43414 T^{2} - 9677232 T^{3} + 767884899 T^{4} - 9677232 p^{3} T^{5} - 43414 p^{6} T^{6} + 228 p^{9} T^{7} + p^{12} T^{8}$$
43$D_4\times C_2$ $$1 + 212 T - 119146 T^{2} + 1076112 T^{3} + 17030802443 T^{4} + 1076112 p^{3} T^{5} - 119146 p^{6} T^{6} + 212 p^{9} T^{7} + p^{12} T^{8}$$
47$D_4\times C_2$ $$1 + 273 T - 143953 T^{2} + 2958228 T^{3} + 29217344628 T^{4} + 2958228 p^{3} T^{5} - 143953 p^{6} T^{6} + 273 p^{9} T^{7} + p^{12} T^{8}$$
53$D_4\times C_2$ $$1 - 299 T - 138207 T^{2} + 20973654 T^{3} + 16331287858 T^{4} + 20973654 p^{3} T^{5} - 138207 p^{6} T^{6} - 299 p^{9} T^{7} + p^{12} T^{8}$$
59$D_4\times C_2$ $$1 - 453 T - 235195 T^{2} - 13429638 T^{3} + 109629530364 T^{4} - 13429638 p^{3} T^{5} - 235195 p^{6} T^{6} - 453 p^{9} T^{7} + p^{12} T^{8}$$
61$D_4\times C_2$ $$1 - 457 T - 216379 T^{2} + 13131438 T^{3} + 81893722754 T^{4} + 13131438 p^{3} T^{5} - 216379 p^{6} T^{6} - 457 p^{9} T^{7} + p^{12} T^{8}$$
67$D_4\times C_2$ $$1 - 1648 T + 1441562 T^{2} - 1108800768 T^{3} + 716645716043 T^{4} - 1108800768 p^{3} T^{5} + 1441562 p^{6} T^{6} - 1648 p^{9} T^{7} + p^{12} T^{8}$$
71$D_4\times C_2$ $$1 + 1239 T + 630425 T^{2} + 234014886 T^{3} + 140937876264 T^{4} + 234014886 p^{3} T^{5} + 630425 p^{6} T^{6} + 1239 p^{9} T^{7} + p^{12} T^{8}$$
73$D_4\times C_2$ $$1 + 4 p T - 215126 T^{2} - 1910576 p T^{3} - 94269219677 T^{4} - 1910576 p^{4} T^{5} - 215126 p^{6} T^{6} + 4 p^{10} T^{7} + p^{12} T^{8}$$
79$D_4\times C_2$ $$1 + 15 T - 80293 T^{2} - 13583400 T^{3} - 236715240972 T^{4} - 13583400 p^{3} T^{5} - 80293 p^{6} T^{6} + 15 p^{9} T^{7} + p^{12} T^{8}$$
83$D_{4}$ $$( 1 - 313 T + 1165660 T^{2} - 313 p^{3} T^{3} + p^{6} T^{4} )^{2}$$
89$D_4\times C_2$ $$1 + 229 T - 1139511 T^{2} - 49918794 T^{3} + 886799000014 T^{4} - 49918794 p^{3} T^{5} - 1139511 p^{6} T^{6} + 229 p^{9} T^{7} + p^{12} T^{8}$$
97$D_4\times C_2$ $$1 + 1050 T - 979606 T^{2} + 269598000 T^{3} + 2495971407807 T^{4} + 269598000 p^{3} T^{5} - 979606 p^{6} T^{6} + 1050 p^{9} T^{7} + p^{12} T^{8}$$
$$L(s) = \displaystyle\prod_p \ \prod_{j=1}^{8} (1 - \alpha_{j,p}\, p^{-s})^{-1}$$ | 2,785 | 5,094 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2021-25 | latest | en | 0.250203 |
https://medium.com/analytics-vidhya/the-fundamentals-of-image-recognition-5ff241c63a25?source=user_profile---------9---------------------------- | 1,660,372,774,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571909.51/warc/CC-MAIN-20220813051311-20220813081311-00015.warc.gz | 370,473,962 | 37,701 | Published in
Analytics Vidhya
# The Fundamentals of Image Recognition
What is this? These are just strange symbols placed in a particular order that hold no meaning if not digested correctly, yet your mind is continuously making sense of this collection of nonsense.
Cool. How?
Well, the most common method thought to facilitate this understanding is the periodic movement of the eyes from left to right, deciphering the meaning of each word by analyzing the order of the symbols and their subsequent pronunciation.
Strangely, but rationally enough, this is the basis for all efficient image recognition algorithms. Let’s get into it.
# A Practical Example
Imagine that you are walking through an art museum. Additionally, imagine that you take five minutes to analyze the meaning, and objects in a given painting. The only catch: You must understand the painting to the above extent by only observing one square-inch of the canvas.
What would the result of this situation be? Well, it would not consist of much, as the observer would be unable to decipher any true meaning from simply analyzing the designated square-inch. If, however, 30 observers were given the same challenge, notified of the other observer’s analysis location, and the result of each square-inch’s analysis, would they collectively be able to decipher the meaning of the already mysterious piece of art? While the answer of the question is subjective, it would most commonly be ‘yes’. This model of pixel isolation, and subsequent analysis and identification through joint ‘effort’ pixel analysis is the approach many of the most popular algorithms currently take.
Abnormalities and similarities in pixel shade and position are taken as input to curate a value for that given pixel. Patterns in these values suggest an object or unifying characteristic in the image that can then be group as an independent entity.
# Pixel Isolation and Analysis
The principal way algorithms detect objects in a given image is through the use of pixel isolation and subsequent analysis of the values given by that pixel. A computer cannot simply be given an image and expected to make sense of it through collective observation alone. That is similar to asking a baby to take a look at a block of text from The Odyssey, and make sense of it.
The theoretical approach cannot work, but a systematic one can. Similar to the natural human tendency to glance one’s eyes from left to right to understand words on the page, the computer can use a variety of methods to isolate a given set of pixelated dimensions and analyze them for pattern.
This technique allows for a greater depth of analysis for the image, and level precision for the software, as seen by the workflow below.
Although a larger ‘pixel grid’ is used to isolate specific parts of a photo in more conventional use cases, for our purposes, the above 7-by-7 grid is sufficient for explanation. The image on the far-left of the workflow model shows a 7-by-7 grid being overlaid onto the original image. Within each grid box, somewhat of an analysis takes place. In this case, the analyses yielded a ‘common consensus’ if you will, on the location of unique items on the image. Overlaying this with additional data, the algorithm is able to output, with precision, the approximate location on various object in the image, who’s locations had not been fully estimated in the initial image analysis.
Once locations of object in an image can be rendered, the image of the objects themselves can be referenced with a larger dataset of images to receive a clear, English labeling as to what the object is.
# My Program
To test the convenience and accuracy of this image analysis method, I created my own program using many of the same factors.
Unlike the above model, my program was trained using the labeled NVIDIA Fashion dataset. The distinction ‘labeled’ means essentially what it sounds like, data that is given a predetermined value. 50,000 of these images were selected to be used for the training of the algorithm, while the remaining 10,000 were used for testing of the accuracy of the algorithm.
To arrive with the desired functionality of the algorithm, I implemented a three sequential layer in my network with three distinct activation functions: the first having the ‘flatten’ function, the second with ‘ReLU’, and the third with ‘softmax’.
The first layer’s function is to format the image such that the pixelated dimensions of the image are ready for subsequent analysis and identification from the following three layers.
While this exact structure is not consistent through all programs, the altering of image format is vital in creating an image recognition program with any minor degree of functionality.
# What does this mean?
The exact repercussions of advancement in this domain are still somewhat unknown. While the potential level for positive impact is quite high, the level of the negative impact seems to be at a drastically darker level then initially intended.
We have seen, and will continue to see an increased level of hyper-supervised government rule if we allow for image recognition technology to increase at its current rate. In China, the technology is being used to identify and target religious minorities for deportation and brash treatment. Protestors in the infamous Hong Kong protests are being identified, and statutorily targeted through the use of facial recognition posts that have found a convenient home in the administrative region. This activity is a perfect example of policy control gone wrong. If the world wants to experience the positive effects of this technology at even a minor degree, measures must be taken to prevent against its extreme manipulation.
From a conventional use standpoint, the effects of algorithmic advancement are tremendous. Imagine world in which doctors would no longer need to spend hours analyzing the scans of patients looking for abnormalities, but could rather delegate that task to a narrowly designed image recognition tool that could understand the root of this medical outlier, and curate a diagnosis in response to it. I believe with the proper policy in place, this technology, partnered with more generalize AI applications, has the ability to change the world at a grand scale.
# Contact:
Email: mmcd.jack@gmail.com
--
--
## More from Analytics Vidhya
Analytics Vidhya is a community of Analytics and Data Science professionals. We are building the next-gen data science ecosystem https://www.analyticsvidhya.com | 1,253 | 6,546 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2022-33 | latest | en | 0.9528 |
https://encode.su/threads/1951-lightning-fast-lossless-image-compression-(heads-up) | 1,611,603,641,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703644033.96/warc/CC-MAIN-20210125185643-20210125215643-00006.warc.gz | 325,090,187 | 12,570 | Thread: lightning-fast lossless image compression (heads up)
1. lightning-fast lossless image compression (heads up)
I've been talking to some graphics guys, and it seems that you can losslessly compress or decompress images by about a factor of two at hundreds of megabytes per second, single core, using SIMD instructions (e.g., SSE4). (8-bit grayscale images from the imagecompression.info gray8 benchmark.)
Unfortunately, the people who know how to do this are not publishing how or releasing the code because it's not a standard compression format and their companies are pursuing standardized compression formats like JPEG-LS instead.
On the other hand, it seems that the techniques are not patented and it doesn't appear that they will be, so this might be a good thing for a free software person who's good at images and SIMD to look into, if they're so inclined.
This posting at Jan Wassenberg's blog is particularly intriguing:
http://wassenberg.dreamhosters.com/b...e-compression/
Jan did some very cool work for his dissertation on symmetrical compression of 16-bit images, but that code did not work well for 8-bit color channels; the newer variant he describes does, at a cost in compression ratio of a bit under 20 percent.
On his blog he says:
The most interesting part is fixed-to-variable coding for entire vector registers of bits. For the image residuals above, it compresses to within 5-20% of the entropy at speeds comparable to or exceeding state of the art integer coders (http://arxiv.org/abs/1209.2137). A single core can compress/decompress between 4600 MB/s (sparse bit arrays with 0.05% ones) and 1000 MB/s (relatively dense, 12% ones). Of course this scales to multiple cores, as there is no shared dictionary to update.
Although the instruction set improved with SSE4, writing vector code is still an exercise in improvisation. Finding 1-bits is accomplished by a hash function based on de Bruijn sequences (http://supertech.csail.mit.edu/papers/debruijn.pdf) and using the FPU normalization hardware. Vectorized shifts are emulated with the PSHUFB universal shuffle instruction.
I'm not an expert on either images or SIMDfying algorithms, but that sounded very promising.
2. Thanks:
avitar (15th May 2014)
3. Simple/fast lossless image compression is scanning line by line using some simple predictor for pixel value like average of 2 neighboring pixels (or a bit better predictor from LOCO-I) - and encoding difference from the real pixel value using some entropy coder for two-sided exponential distribution (exp(-|x-y|)).
This way you can get about 50% reduction for a photography, and much better for simpler pictures.
Using FSE as entropy coder would give you about 500MB/s per core here, what could be improved using SIMD.
4. Thanks (2):
avitar (15th May 2014),Paul W. (15th May 2014)
5. Originally Posted by Jarek
Simple/fast lossless image compression is scanning line by line using some simple predictor for pixel value like average of 2 neighboring pixels (or a bit better predictor from LOCO-I) - and encoding difference from the real pixel value using some entropy coder for two-sided exponential distribution (exp(-|x-y|)).
I've done sort-of that myself, long ago, in an experimental compressor for window system save areas, just averaging the pixel immediately to the left with the one immediately above and using a cheesy fixed Huffman-ish code for the difference. My cheesy version doesn't work as well as it used to, though, now that people use more colors, gradients, thumbnails, and (especially) photographs in web pages, docs, and desktop backgrounds.
Maybe it can be fixed, though, with a slightly better predictor, plugging in FSE, and reasonable speed optimizations. (Or better yet, finding a well-written free library that already does that.)
This way you can get about 50% reduction for a photography, and much better for simpler pictures.
Using FSE as entropy coder would give you about 500MB/s per core here, what could be improved using SIMD.
How much potential do you see for speeding up FSE with SIMD?
6. The prediction with subtraction should be SIMD vectorized, getting ~ x8 speedup. So the speed depends mainly on the speed of entropy coder. tANS/FSE decoding step is e.g.
t = decodingTable[x];
useSymbol[t.symbol];
x = t.newX + readbits[t.nbBits];
where x is below 16 bits - we could use 8x SIMD. The table use has to be done one by one, but all the rest should get a speedup.
I don't know the exact values that can be achieved, dnd claimed getting about 1.5x speedup, but without anything to confirm that ... I think png level of compression can be obtain with about 500MB/s here.
7. Thanks:
Paul W. (16th May 2014)
8. Jarek;
I don't know the exact values that can be achieved, dnd claimed getting about 1.5x speedup, but without anything to confirm that ... I think png level of compression can be obtain with about 500MB/s here.
1.5x seems really low to me, but I don't know much, and I'm not sure what the issues are. As I understood what Wassenberg did, and I may well not understand it at all, he was picking one quantization for a whole SIMD register worth of pixels, where you have a fixed choice of 4/4 byte split, 6/2 byte split, or exact match of whole registers. Nastily high-frequency detailed images like mandrill would have to require you to encode a lot of (aligned) 8-byte sequences 4/4 way, easier (mostly-gentle-gradient) images like lena with minor color noise would mostly allow you to use 6/2, and really easy (non-photo, synthetic) images with a lot of flat color areas would often use exact (8/0).
(Basically, the encoding per aligned 8-byte unit is determined by the worst-matching byte of each 8. If I've got this all wrong, being not an image guy or particularly good at any aspect of this, I'd appreciate being gently straightened out.)
If that's roughly right, it seems there should be a tradeoff where you can get significantly better compression at a significant cost in speed, if you pick a smaller-than-8-byte unit to do this across, like 4 bytes instead, or you have more choices of splits (like 3 and 1 in addition to 4 and 2). Crudeness and assuming stretches of regularity should get you more vectorizability, right?
One question I have is about the split vs. unified treatment of color channels. Is there a screamingly fast way to interleave/deinterleave the bits of the bytes for different colors? Given the typical correlations between color channels, I'd think it'd be beneficial to just interleave 3 x 8 bits into 24, then do the magic, with all the higher bits together and all the lower bits together. I wouldn't have thought that possible, but the (Leiserson et al.) paper that Wassenberg cites about using magical DeBruijn sequences for indexing high (or low) bits (on hardware without hardware support for it) makes me a bit more optimistic that maybe there's a fast bit-fiddly trick that fills the bill.
9. Originally Posted by Paul W.
One question I have is about the split vs. unified treatment of color channels. Is there a screamingly fast way to interleave/deinterleave the bits of the bytes for different colors? Given the typical correlations between color channels, I'd think it'd be beneficial to just interleave 3 x 8 bits into 24, then do the magic, with all the higher bits together and all the lower bits together. I wouldn't have thought that possible, but the (Leiserson et al.) paper that Wassenberg cites about using magical DeBruijn sequences for indexing high (or low) bits (on hardware without hardware support for it) makes me a bit more optimistic that maybe there's a fast bit-fiddly trick that fills the bill.
http://graphics.stanford.edu/~seander/bithacks.html
The DeBruijn sequence trick is neat. I've been thinking about DeBruijn sequences lately.
10. Thanks:
Paul W. (16th May 2014)
11. Look at LZ4.
Hardware / compiler support turned out to be much better than DeBruijn.
Overall, LZ4 gives you an implementation that's portable, yet about optimal on most platforms.
12. Originally Posted by m^2
Look at LZ4.
Hardware / compiler support turned out to be much better than DeBruijn.
Overall, LZ4 gives you an implementation that's portable, yet about optimal on most platforms.
My take is that the reason for using the DeBruijn method is that it agrees well with SIMD. I'm pretty sure Intel doesn't offer that instruction in hardware with vectorization.
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• | 1,979 | 8,567 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2021-04 | latest | en | 0.910738 |
https://my.oschina.net/idreamo/blog/684525 | 1,611,000,280,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703515235.25/warc/CC-MAIN-20210118185230-20210118215230-00250.warc.gz | 472,927,786 | 16,616 | # C Primer Plus 第6章 C控制语句：循环 6.7 逗号运算符
2016/05/31 06:32
//postage.c -- 一类邮资费率
#include <stdio.h>
int main (void)
{
const int FIRST_OZ = 37;
const int NEXT_OZ = 23;
int ounces,cost;
printf(" ounces cost\n");
for (ounces=1,cost=FIRST_OZ;ounces<=16;ounces++,cost+=NEXT_OZ)
printf("%5d $4.2f\n ",ounces,cost/100.0); return 0; } /*输出的前4行看上去是这样的 ounces cost 1$0.37
2 $0.60 3$0.83
4 \$1.06
*/
houseprice=249,500;
houseprice=249;
500;
houseprice=（249，500）；
/*zeno.c -- 求序列的和*/
#include <stdio.h>
int main (void)
{
int t_ct; //项计数
double time,x;
int limit;
printf("Enter the number of terms you want:");
scanf("%d",&limit);
for(time=0,x=1,t_ct=1;t_ct<=limit;t_ct++,x*=2.0)
{
time+=1.0/x;
printf("time = %f when terms = %d.\n",time,t_ct);
}
return 0;
}
/*下面是前几项的输出
Enter the number of terms you want:15
time = 1.000000 when terms = 1.
time = 1.500000 when terms = 2.
...
time = 1.999939 when terms = 15.
*/
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0 | 388 | 928 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2021-04 | latest | en | 0.281896 |
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