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https://ethresear.ch/t/rsa-hash-accumulator-for-arbitrary-values/4485 | # RSA Hash accumulator for arbitrary values
#1
RSA accumulators can efficiently store primes, but are (as far as I know) not efficient with non-prime numbers. My goal is to store arbitrary values, just like you can do in a Merkle tree, but having a shorter proof size. This can have multiple applications, such as in zk-starks, Plasma, etc.
I believe it is possible that we can proof that multiple values are contained in the accumulator with a single 2048 bit witness.
Basic example for three values
Let a, b and c be the values we like to store.
Let \textit{h} be a secure hash function.
Let N be a 2048 bit RSA modulus with unknown factorization and g be the generator.
Let p, q and r be three distinct large primes.
The accumulator A = g^{qr \textit{h}(a)+pr \textit{h}(b)+pq \textit{h}(c)} \mod N
To proof that a is stored in the first spot we need a witness W = g^{r \textit{h}(b)+q \textit{h}(c)} \mod N
The verifier has to check that g^{qr \textit{h}(a)}W^{p} \equiv A\mod N
To proof a fake value a' is in the accumulator, the forger needs to calculate the p-root of g^{qr (\textit{h}(a)-\textit{h}(a'))+pr \textit{h}(b)+pq \textit{h}(c)}\mod N. I believe this problem to be computationally infeasible when the RSA trapdoor is unknown.
It is also possible to make a single proof that the accumulator contains multiple values. For example when we want to proof a and b are both stored in the accumulator, we need the witness W = g^{\textit{h}(c)} \mod N
The verifier has to check that g^{qr \textit{h}(a)+pr \textit{h}(b)}W^{pq} \equiv A\mod N
Hash accumulator with n primes
We can generalize this to an accumulator for n values using n primes.
Let x_{1},..,x_{n} be the n values we like to store.
Let p_{1},..,p_{n} be n distinct large primes.
We define P_S to be g to the power of the product of all the primes not contained in the set S modulo N
P_S = g^{\prod\limits_{\substack{k=1,k\not\in S}}^n p_{k}} \mod N
The accumulator A = \prod\limits_{k=1}^n P_{k}^{\textit{h}(x_k)} \mod N
To proof x_i is stored in spot i we need a witness W = \prod\limits_{k=1, k\neq i} ^n P_{i,k}^{\textit{h}(x_k)} \mod N
The verifier has to check that P_i^{\textit{h}(x_i)} W^{p_i} \equiv A \mod N
To proof that multiple values from set B are in the accumulator we need a single witness W = \prod\limits_{k=1, k\not\in B} ^n P_{B,k}^{\textit{h}(x_k)} \mod N
And the verifier has to check that \prod\limits_{k\in B} P_{k}^{\textit{h}(x_k)} W^{\prod\limits_{k\in B}p_k} \equiv A \mod N
Plasma using the RSA hash accumulator
#2
Doesn’t this require quadratic overhead? If so, that’s a non-starter for most applications where prover time is already a bottleneck…
#3
Since RSA accumulators can efficiently store primes, couldn’t you use this fact to efficiently store prime factorizations of non-prime numbers? Obviously, you would need a way to check the prime factorization and depending on the size of the number, this could have a lot of overhead.
#4
why don’t just use a more advanced form of accumulator like the ajtai hash?
Randomly sample A from \mathbb{Z_p}^{n \times m} then do Ax\, mod\, p.
This is not only quasy-commutative but also quantumsafe, parallelizable and accumulations are trivially revocable by accumulating the inverse element.
Note that parameter selection for n and m are sensitive and limits your input size and security
#5
You can build a witness in O(n) time.
Pseudo code:
function calculateWitness(pos,values) {
var W=1;
var P=g;
for (var i=0;i<values.length;i++) {
if (pos!=i) {
W=(W^primes[i])%N;
W=W*(P^hash(values[i]))%N;
P=(P^primes[i])%N;
}
}
return W;
}
#7
Could you perhaps elaborate on how using the Ajtai hash would help in this case?
#8
I have a similar question to @gakonst - the RSA accumulator scheme is defined using a function f : \mathbb{Z}_N \times \mathbb{N} \to \mathbb{Z}_N given by f(a, x) = a^x \pmod N; then the quasi-commutativity [1] property is f(f(a, x_1), x_2) = f(f(a, x_2), x_1).
From what I can find online [2] the Ajtai hash function is defined by g: \{0, 1\}^m \to \mathbb{Z}_q^n given by g(x) = A x \pmod p. What’s the quasi-commutativity property for g?
References
#9
A burden for the verifier is that it needs to store all the primes and all P_k constants, to verify the proof.
For a light verifier, we can save on storage space, when the prover supplies both P_k and p_k. The verifier can check P_k^{p_k} \equiv P_{\emptyset} \mod N, where P_{\emptyset} = g^{\prod\limits_{k=0}^n p_{k}}. This way the light verifier only has to store P_{\emptyset}.
For a bulk proof the prover can provide P_B and all primes corresponding the the set B. The verifier can check the correctness of P_B, by checking P_B^{\prod\limits_{k\in B} p_{k}} \equiv P_{\emptyset} \mod N and calculate all needed P_k.
The disadvantage of this trick is that the verifier can only know that a value x is one of the n values in the accumulator, but not the specific spot.
For applications that do not need the specific spot, this light verifier trick can be used.
I was wondering whether zk-starks proofs (for which we can shrink the proof a lot by having a smaller merkle tree replacement) actually need the spot, or that putting the spot in the hashed data is sufficient. From what I understand of zk-stark proofs, it would not weaken the proof as long as the number of values in the accumulator is limited to n. A forger could try to put multiple values in the same spot to increase the chance to be able to give the correct answer for that spot. But now the forger increases the chance of not being able to answer when an empty spot is chosen.
#10
Yes, I have thought of that, but it would require more primes and a larger proof.
To store a 256 bit hash value we could use 256 primes each representing one bit. Now we have to proof inclusion for on average 128 primes and exclusion for the other 128 primes. We can efficiently combine the inclusion proofs, but not the exclusion proofs, because we have a remainder the size of the product of 128 primes. For primes of 64 bit size, this results in a witness size of 2*2048 +128*64 =12288 bits.
A trick we could do is, instead of having one bit per prime, we could have 8 bits per prime, by including p^b where b is a byte. We have to prove p^b is included, but b^{n+1} is not.
Now we need only 32 primes for which we do 32 inclusion and 32 exclusion proofs. The witness size is now down to 2*2048 +32*64 =6144 bits for one value. For an extra value we need an extra 32*64=2048 bits, because we have a bigger remainder.
Unfortunately we cannot overstretch this trick, because calculating g^{p^{x}} \mod N is not trivial for large x.
My proposal is based on one inclusion proof, so is only 2048 bits for an unlimited number of values.
#11
let X = {x_1, x_2, .... x_n} arbitrary values. then your accumulator is \Sigma_{i=1}^{n}Ax_i \,(mod \,p). Witness is trivial, same as with RSA and revocation is just acc + x_i^{-1} where the inverse is by mod \, p
Security of the scheme is proven to be equivalent to solving the Shortest Integer Solution problem on ideal lattices.
references:
http://elaineshi.com/docs/streaming.pdf
#12
look up the properties of vector spaces
#13
Assume a plasma cash/flow design application. I wonder if it would be possible to have the operator commit to two accumulator values A_i and A_e where including a prime in A_i represents committing to the inclusion of a value i.e. a bit is 1 not a 0. And then A_e represents committing to exclusion of a value i.e. a bit is 0 not 1 at a given position i in vector x.
In plasma cash we might want a sparse vector that stores only the public key of the owner of the coin at a particular index. Assume a public key is only 8 bits i.e if we have two values v_1 = [01101110] and v_2 = [10100111] we could generate the two VC accumulators A_i and A_e like so.
Let v_1^p = [3,5,7,11,13,17,19,23]
Let v_2^p = [29,31,37,41,43,47,53, 59]
Let A_i = g^{[{3^0}*{5^1}*{7^1}*{11^0}*{13^1}*{17^1}*{19^1}*{23^0}]*[{29^1}*{31^0}*{37^1}*{41^0}*{43^0}*{47^1}*{53^1}*59^1]}
Let A_e = g^{[{3^1}*{5^0}*{7^0}*{11^1}*{13^0}*{17^0}*{19^0}*{23^1}]*[{29^0}*{31^1}*{37^0}*{41^1}*{43^1}*{47^0}*{53^0}*59^0]}
To prove that x_1 opens to v_1 we would ask for a batched inclusion proof for 5,7,13,17,19 to prove that x_1 position contains all of the correct bits with 1. To prove x_1 has all of the correct 0 bits committed we would ask for a batched inclusion proof for 3,11,23 which should result in just two times the batched proof size of one accumulator.
err… Not explicitly proving exclusion could mean that multiple values are stored in the same position since the operator could include more primes than the verifier is checking which will allow other values to pass. This makes me wonder if we could create a compact proof that A_i and A_e are disjoint if this would work.
#14
If the prover and verifier agree on which primes are used, the prover can show that A_i and A_e are constructed correctly.
Let x be the product of all included primes.
Let y be the product of all excluded primes.
Let z be the product of all primes.
Let A_i = g^x \mod N
Let A_e = g^y \mod N
Let A_T = g^{z} \mod N
A_T is the accumulator containing all primes. This value can be pre-calculated and stored on chain.
Now I want to use a double Wesolowski proof, to show that x*y=z.
We are substituting:
x = B\lfloor \frac x B \rfloor + x \mod B
We use the following witnesses:
b_1=g^{\lfloor \frac x B \rfloor} \mod N
b_2=A_e^{\lfloor \frac x B \rfloor} \mod N
r=x \mod B
The verifier should check:
b_1^B.g^r \equiv A_i \mod N
b_2^B.A_e^r \equiv A_T \mod N
Zero-knowledge proof of addition for large, high-entropy values
#15
We can adjust the RSA Hash Accumulator to store a matrix of values. This way we can store more values using less primes.
The idea behind the accumulator is that every hashed value is multiplied with every prime except one. To proof a value is in the accumulator we can group all other values together, because they are all multiplied by the same prime that is missing from the value we like to proof.
To store a matrix instead of a vector, we will multiply all hashed values with every prime except two, this way we can store n*m values using only n+m primes. The downside of this approach is that we need two 2048 bit witnesses to proof inclusion of one value.
Hash accumulator for a mxn matrix
Let x_{i,j} be the values of the matrix we like to store.
Let p_{1},..,p_{n} and q_{1},..,q_{m} be the n+m distinct large primes.
We define P_S and Q_S to be the product of all their respective primes not contained in the set S (not g raised to the power as above)
P_S = \prod\limits_{\substack{k=1,k\not\in S}}^n p_{k}
Q_S = \prod\limits_{\substack{k=1,k\not\in S}}^m q_{k}
The accumulator A = g^{\sum\limits_{k=1}^m \sum\limits_{l=1}^n Q_{k}P_{l}\textit{h}(x_{k,l})} \mod N
To proof x_{i,j} is stored in spot i,j, we need two witnesses:
W_p = g^{\sum\limits_{l=1,l \neq j}^n Q_{i}P_{l,j}\textit{h}(x_{i,l})} \mod N
W_q = g^{\sum\limits_{k=1,k\neq i}^m \sum\limits_{l=1}^n Q_{k,i}P_{l}\textit{h}(x_{k,l})} \mod N
The verifier has to check: g^{Q_iP_j{\textit{h}(x_{i,j})}} W_q^{q_i}W_p^{p_j} \equiv A \mod N
If we select a set of columns and a set of rows we can proof in a similar manner, all values that are both in the row set as in the column set, using just two 2048 bit witnesses.
If we like, this can be extended to store n-dimensional matrices.
Edit: adding a small example to make it less abstract
A small example for a 2x3 matrix
We are using three p primes p_1, p_2 and p_3 for the columns and two q primes q_1 and q_2 for the rows.
We can store six values: x_{1,1}, x_{1,2}, x_{1,3}, x_{2,1}, x_{2,2}, x_{2,3}
The accumulator A=g^{q_2p_2p_3\textit{h}(x_{1,1}) + q_2p_1p_3\textit{h}(x_{1,2}) + q_2p_1p_2\textit{h}(x_{1,3}) + q_1p_2p_3\textit{h}(x_{2,1}) + q_1p_1p_3\textit{h}(x_{2,2}) + q_1p_1p_2\textit{h}(x_{2,3})}
To proof x_{2,2} is in the accumulator we need witnesses:
W_p = g^{q_1p_3\textit{h}(x_{2,1}) + q_1p_1\textit{h}(x_{2,3})}
W_q = g^{p_2p_3\textit{h}(x_{1,1}) + p_1p_3\textit{h}(x_{1,2}) + p_1p_2\textit{h}(x_{1,3})}
The verifier has to check: g^{q_1p_1p_3\textit{h}(x_{2,2})}W_q^{q_2}W_p^{p_2} \equiv A \mod N
Plasma using the RSA hash accumulator
#16
We can also build a hash accumulator using only two primes to store an arbitrary number of values.
To do this we will multiply the hashed values with both a power of prime p and a power of prime q. This is done is such a way that all other values will be multiplied by either a higher power of p or a higher power of q. This way all other values can be separated via the p side or the q side.
We can proof the values on the tail or head with just one 2048 bit witness and any range in the middle with two 2048 bit witnesses.
Hash accumulator using two primes
Let x_1,..,x_n be the values we like to store.
Let p and q be two distinct large primes.
The accumulator A = g^{\sum\limits_{k=1}^n p^kq^{n-k}\textit{h}(x_{k})} \mod N
To proof x_i is stored in spot i, we need two witnesses:
W_q = g^{\sum\limits_{k=1}^{i-1} p^kq^{i-k-1}\textit{h}(x_{k})} \mod N
W_p = g^{\sum\limits_{k=i+1}^{n} p^{k-i-1}q^{n-k}\textit{h}(x_{k})} \mod N
The verifier has to check:
g^{p^iq^{n-i}{\textit{h}(x_{i})}} W_p^{p^{i+1}}W_q^{q^{n-i+1}} \equiv A \mod N
In a similar manner we can proof inclusion of a range of values using two 2048 bit witnesses.
#17
Just trying to go along with your math, I think the notation might be a little mistaken in some places (given that P_S and Q_S are scalars)? Please correct me if I’m wrong.
A = g^{\sum\limits_{k=1}^m \sum\limits_{l=1}^n Q_{k}P_{l}\textit{h}(x_{k,l})} \mod N
should be A = g^{\sum\limits_{k=1}^m \sum\limits_{l=1}^n q_{k}p_{l}\textit{h}(x_{k,l})} \mod N ?
Also in the witnesses,
W_p = g^{\sum\limits_{l=1,l \neq j}^n Q_{i}P_{l,j}\textit{h}(x_{i,l})} \mod N
you iterate over j in [1,n] excluding l, but j in x_{ij} is in [1,m].
This seems like a neat idea, along with your post where you consider MxN to be CoinIdxBlock, but could you please revisit this post, so that you check for any mistakes and perhaps provide an example with smth simple like a simple 2x3 array?
#18
Unfortunately the cited Latice-based accumulator paper does not propose a scheme for universal accumulators ie non-membership proofs, which is one of the biggest challenges in compressing UTXO history in Plasma.
#19
You are correct, P_S and Q_S are scalars. The subscript is a set of indices that should be excluded, so I understand it might be confusing. I don’t know if there is a better notation for that. I chose capital P and Q, because they are the products of respectively the p and q primes.
Without P and Q we can write the accumulator as:
A = g^{\sum\limits_{k=1}^m \sum\limits_{l=1}^n (\prod\limits_{\substack{t=1,t\neq k}}^m q_{t} ) (\prod\limits_{\substack{t=1,t\neq l}}^n p_{t})\textit{h}(x_{k,l})} \mod N
So \textit{h}(x_{k,l}) is multiplied by every p and q prime except p_l and q_k.
For witness W_p = g^{\sum\limits_{l=1,l \neq j}^n Q_{i}P_{l,j}\textit{h}(x_{i,l})} \mod N we are only using values of row i. We are iterating over all columns l in [1,n] except j. Every hashed value \textit{h}(x_{i,l}) is multiplied by every q prime except q_i, and multiplied by every p prime except p_l and p_j.
p.s. I have edited the post to include a 2x3 matrix example.
#20
What do you mean by qrh(a) …how can multiply by hash …h(a)
#21
h(a) is the hash of value a, when using for example the keccak256 hash function the result is a 256 bit number.This number is multiplied by both q and r. These should be large prime numbers of similar size. | 2019-02-21 16:55:42 | {"extraction_info": {"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, "math_score": 0.7955453395843506, "perplexity": 2018.3703721320383}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-09/segments/1550247505838.65/warc/CC-MAIN-20190221152543-20190221174543-00072.warc.gz"} |
https://www.zbmath.org/authors/?q=ai%3Ailiopoulos.george | # zbMATH — the first resource for mathematics
## Iliopoulos, George
Compute Distance To:
Author ID: iliopoulos.george Published as: Iliopoulos, G.; Iliopoulos, George External Links: MGP
Documents Indexed: 46 Publications since 1998
all top 5
#### Co-Authors
7 single-authored 9 Balakrishnan, Narayanaswamy 6 Davidov, Ori 4 Kourouklis, Stavros 4 Malefaki, Sonia 4 Meintanis, Simos G. 3 Cramer, Erhard 3 Kateri, Maria 3 Ntzoufras, Ioannis 2 Dembińska, Anna 2 Fokianos, Konstantinos 2 Karlis, Dimitris 2 Papastamoulis, Panagiotis 1 Bobotas, Panayiotis 1 Han, Donghoon 1 Keating, Jerome P. 1 Mason, Robert Lee 1 Mirmostafaee, S. M. T. K. 1 Tafiadi, Maria
all top 5
#### Serials
5 Journal of Statistical Planning and Inference 5 Computational Statistics and Data Analysis 4 Annals of the Institute of Statistical Mathematics 4 Statistics & Probability Letters 3 Metrika 3 Journal of Multivariate Analysis 3 Statistics 3 Journal of Statistical Computation and Simulation 1 The Canadian Journal of Statistics 1 Psychometrika 1 Scandinavian Journal of Statistics 1 Biometrics 1 Biometrika 1 Kybernetika 1 Statistics & Decisions 1 Communications in Statistics. Simulation and Computation 1 Communications in Statistics. Theory and Methods 1 Test 1 Journal of Nonparametric Statistics 1 Methodology and Computing in Applied Probability 1 Statistical Methodology 1 Journal of Statistical Theory and Practice 1 Statistics and Computing
#### Fields
44 Statistics (62-XX) 12 Numerical analysis (65-XX) 10 Probability theory and stochastic processes (60-XX) 1 Special functions (33-XX)
#### Citations contained in zbMATH
39 Publications have been cited 267 times in 200 Documents Cited by Year
Pitman closeness of sample median to population median. Zbl 1169.62324
Balakrishnan, N.; Iliopoulos, G.; Keating, Jerome P.; Mason, R. L.
2009
Stochastic monotonicity of the MLE of exponential mean under different censoring schemes. Zbl 1332.62384
Balakrishnan, N.; Iliopoulos, G.
2009
Conditional independence of blocked ordered data. Zbl 1158.62323
Iliopoulos, G.; Balakrishnan, N.
2009
Exact inference for progressively type-I censored exponential failure data. Zbl 1213.62153
Balakrishnan, N.; Han, Donghoon; Iliopoulos, G.
2011
Fourier methods for testing multivariate independence. Zbl 1452.62389
Meintanis, Simos G.; Iliopoulos, George
2008
On the method of pivoting the CDF for exact confidence intervals with illustration for exponential mean under life-test with time constraints. Zbl 1288.62046
Balakrishnan, N.; Cramer, E.; Iliopoulos, G.
2014
Adaptive progressive type-II censoring. Zbl 1203.62169
Cramer, Erhard; Iliopoulos, George
2010
Asymptotic properties of numbers of observations near sample quantiles. Zbl 1241.62079
Iliopoulos, G.; Dembińska, A.; Balakrishnan, N.
2012
Exact likelihood inference for Laplace distribution based on type-II censored samples. Zbl 1206.62032
Iliopoulos, G.; Balakrishnan, N.
2011
Order-restricted semiparametric inference for the power bias model. Zbl 1192.62101
Davidov, Ori; Fokianos, Konstantinos; Iliopoulos, George
2010
Improving on the best affine equivariant estimator of the ratio of generalized variances. Zbl 1063.62516
Iliopoulos, George; Kourouklis, Stavros
1999
Stochastic monotonicity of the MLEs of parameters in exponential simple step-stress models under type-I and type-II censoring. Zbl 1189.62160
Balakrishnan, N.; Iliopoulos, G.
2010
On improved interval estimation for the generalized variance. Zbl 0953.62026
Iliopoulos, George; Kourouklis, Stavros
1998
On the asymptotics of numbers of observations in random regions determined by order statistics. Zbl 1235.60015
Dembińska, Anna; Iliopoulos, George
2012
Bayesian estimation of unrestricted and order-restricted association models for a two-way contingency table. Zbl 1162.62324
Iliopoulos, G.; Kateri, M.; Ntzoufras, I.
2007
Tests of fit for the Rayleigh distribution based on the empirical Laplace transform. Zbl 1050.62051
Meintanis, Simos; Iliopoulos, George
2003
Estimation of parametric functions in Downton’s bivariate exponential distribution. Zbl 1021.62014
Iliopoulos, George
2003
On the existence and uniqueness of the NPMLE in biased sampling models. Zbl 1149.62310
Davidov, Ori; Iliopoulos, George
2009
Bayesian estimation in Kibble’s bivariate gamma distribution. Zbl 1098.62030
Iliopoulos, George; Karlis, Dimitris; Ntzoufras, Ioannis
2005
Exact prediction intervals for order statistics from the Laplace distribution based on the maximum-likelihood estimators. Zbl 1367.62073
Iliopoulos, G.; MirMostafaee, S. M. T. K.
2014
Reversible jump MCMC in mixtures of normal distributions with the same component means. Zbl 1452.62229
Papastamoulis, Panagiotis; Iliopoulos, George
2009
Estimating the ratio of two scale parameters: a simple approach. Zbl 1440.62083
Bobotas, Panayiotis; Iliopoulos, George; Kourouklis, Stavros
2012
An odd property of sample median from odd sample sizes. Zbl 1232.62075
Iliopoulos, G.; Balakrishnan, N.
2010
Bayesian model comparison for the order restricted RC association model. Zbl 1179.62039
Iliopoulos, G.; Kateri, M.; Ntzoufras, I.
2009
On collapsing categories in two-way contingency tables. Zbl 1044.62066
Kateri, Maria; Iliopoulos, George
2003
Interval estimation for the ratio of scale parameters and for ordered scale parameters. Zbl 0948.62021
Iliopoulos, G.; Kourouklis, S.
2000
On the convergence rate of random permutation sampler and ECR algorithm in missing data models. Zbl 1277.65007
Papastamoulis, Panagiotis; Iliopoulos, George
2013
On convergence of properly weighted samples to the target distribution. Zbl 1134.65007
Malefaki, Sonia; Iliopoulos, George
2008
Simulation from the Bessel distribution with applications. Zbl 1040.65005
Iliopoulos, George; Karlis, Dimitris
2003
Decision theoretic estimation of the ratio of variances in a bivariate normal distribution. Zbl 0989.62004
Iliopoulos, George
2001
Semiparametric inference for the two-way layout under order restrictions. Zbl 1309.62123
Davidov, Ori; Fokianos, Konstantinos; Iliopoulos, George
2014
A note on an iterative algorithm for nonparametric estimation in biased sampling models. Zbl 05689617
Davidov, Ori; Iliopoulos, George
2010
UMVU estimation of the ratio of powers of normal generalized variances under correlation. Zbl 1141.62041
Iliopoulos, George
2008
Characterizations of the exponential distribution based on certain properties of its characteristic function. Zbl 1249.60020
Meintanis, Simos G.; Iliopoulos, George
2003
A note on decision theoretic estimation of ordered parameters. Zbl 0967.62003
Iliopoulos, George
2000
Convergence of Luo and Tsai’s iterative algorithm for estimation in proportional likelihood ratio models. Zbl 1452.62535
Davidov, O.; Iliopoulos, G.
2013
Simulation from a target distribution based on discretization and weighting. Zbl 1163.65002
Malefaki, Sonia; Iliopoulos, George
2009
Simulating from a multinomial distribution with large number of categories. Zbl 1365.62058
Malefaki, Sonia; Iliopoulos, George
2007
Some new estimators of the binomial parameter $$n$$. Zbl 1140.62313
Iliopoulos, George
2003
On the method of pivoting the CDF for exact confidence intervals with illustration for exponential mean under life-test with time constraints. Zbl 1288.62046
Balakrishnan, N.; Cramer, E.; Iliopoulos, G.
2014
Exact prediction intervals for order statistics from the Laplace distribution based on the maximum-likelihood estimators. Zbl 1367.62073
Iliopoulos, G.; MirMostafaee, S. M. T. K.
2014
Semiparametric inference for the two-way layout under order restrictions. Zbl 1309.62123
Davidov, Ori; Fokianos, Konstantinos; Iliopoulos, George
2014
On the convergence rate of random permutation sampler and ECR algorithm in missing data models. Zbl 1277.65007
Papastamoulis, Panagiotis; Iliopoulos, George
2013
Convergence of Luo and Tsai’s iterative algorithm for estimation in proportional likelihood ratio models. Zbl 1452.62535
Davidov, O.; Iliopoulos, G.
2013
Asymptotic properties of numbers of observations near sample quantiles. Zbl 1241.62079
Iliopoulos, G.; Dembińska, A.; Balakrishnan, N.
2012
On the asymptotics of numbers of observations in random regions determined by order statistics. Zbl 1235.60015
Dembińska, Anna; Iliopoulos, George
2012
Estimating the ratio of two scale parameters: a simple approach. Zbl 1440.62083
Bobotas, Panayiotis; Iliopoulos, George; Kourouklis, Stavros
2012
Exact inference for progressively type-I censored exponential failure data. Zbl 1213.62153
Balakrishnan, N.; Han, Donghoon; Iliopoulos, G.
2011
Exact likelihood inference for Laplace distribution based on type-II censored samples. Zbl 1206.62032
Iliopoulos, G.; Balakrishnan, N.
2011
Adaptive progressive type-II censoring. Zbl 1203.62169
Cramer, Erhard; Iliopoulos, George
2010
Order-restricted semiparametric inference for the power bias model. Zbl 1192.62101
Davidov, Ori; Fokianos, Konstantinos; Iliopoulos, George
2010
Stochastic monotonicity of the MLEs of parameters in exponential simple step-stress models under type-I and type-II censoring. Zbl 1189.62160
Balakrishnan, N.; Iliopoulos, G.
2010
An odd property of sample median from odd sample sizes. Zbl 1232.62075
Iliopoulos, G.; Balakrishnan, N.
2010
A note on an iterative algorithm for nonparametric estimation in biased sampling models. Zbl 05689617
Davidov, Ori; Iliopoulos, George
2010
Pitman closeness of sample median to population median. Zbl 1169.62324
Balakrishnan, N.; Iliopoulos, G.; Keating, Jerome P.; Mason, R. L.
2009
Stochastic monotonicity of the MLE of exponential mean under different censoring schemes. Zbl 1332.62384
Balakrishnan, N.; Iliopoulos, G.
2009
Conditional independence of blocked ordered data. Zbl 1158.62323
Iliopoulos, G.; Balakrishnan, N.
2009
On the existence and uniqueness of the NPMLE in biased sampling models. Zbl 1149.62310
Davidov, Ori; Iliopoulos, George
2009
Reversible jump MCMC in mixtures of normal distributions with the same component means. Zbl 1452.62229
Papastamoulis, Panagiotis; Iliopoulos, George
2009
Bayesian model comparison for the order restricted RC association model. Zbl 1179.62039
Iliopoulos, G.; Kateri, M.; Ntzoufras, I.
2009
Simulation from a target distribution based on discretization and weighting. Zbl 1163.65002
Malefaki, Sonia; Iliopoulos, George
2009
Fourier methods for testing multivariate independence. Zbl 1452.62389
Meintanis, Simos G.; Iliopoulos, George
2008
On convergence of properly weighted samples to the target distribution. Zbl 1134.65007
Malefaki, Sonia; Iliopoulos, George
2008
UMVU estimation of the ratio of powers of normal generalized variances under correlation. Zbl 1141.62041
Iliopoulos, George
2008
Bayesian estimation of unrestricted and order-restricted association models for a two-way contingency table. Zbl 1162.62324
Iliopoulos, G.; Kateri, M.; Ntzoufras, I.
2007
Simulating from a multinomial distribution with large number of categories. Zbl 1365.62058
Malefaki, Sonia; Iliopoulos, George
2007
Bayesian estimation in Kibble’s bivariate gamma distribution. Zbl 1098.62030
Iliopoulos, George; Karlis, Dimitris; Ntzoufras, Ioannis
2005
Tests of fit for the Rayleigh distribution based on the empirical Laplace transform. Zbl 1050.62051
Meintanis, Simos; Iliopoulos, George
2003
Estimation of parametric functions in Downton’s bivariate exponential distribution. Zbl 1021.62014
Iliopoulos, George
2003
On collapsing categories in two-way contingency tables. Zbl 1044.62066
Kateri, Maria; Iliopoulos, George
2003
Simulation from the Bessel distribution with applications. Zbl 1040.65005
Iliopoulos, George; Karlis, Dimitris
2003
Characterizations of the exponential distribution based on certain properties of its characteristic function. Zbl 1249.60020
Meintanis, Simos G.; Iliopoulos, George
2003
Some new estimators of the binomial parameter $$n$$. Zbl 1140.62313
Iliopoulos, George
2003
Decision theoretic estimation of the ratio of variances in a bivariate normal distribution. Zbl 0989.62004
Iliopoulos, George
2001
Interval estimation for the ratio of scale parameters and for ordered scale parameters. Zbl 0948.62021
Iliopoulos, G.; Kourouklis, S.
2000
A note on decision theoretic estimation of ordered parameters. Zbl 0967.62003
Iliopoulos, George
2000
Improving on the best affine equivariant estimator of the ratio of generalized variances. Zbl 1063.62516
Iliopoulos, George; Kourouklis, Stavros
1999
On improved interval estimation for the generalized variance. Zbl 0953.62026
Iliopoulos, George; Kourouklis, Stavros
1998
all top 5
#### Cited by 256 Authors
33 Balakrishnan, Narayanaswamy 26 Iliopoulos, George 13 Ahmadi, Jafar 9 Cramer, Erhard 9 Kourouklis, Stavros 8 Davies, Katherine F. 7 Bobotas, Panayiotis 7 Davidov, Ori 7 Dembińska, Anna 7 Meintanis, Simos G. 6 Keating, Jerome P. 6 Ng, Hon Keung Tony 5 Chan, Ping-Shing 5 Kundu, Debasis 5 Mason, Robert Lee 5 Shafay, A. R. 4 Jiménez-Gamero, María Dolores 4 Kateri, Maria 4 Papastamoulis, Panagiotis 4 Zhu, Xiaojun 3 Fokianos, Konstantinos 3 Górny, Julian 3 Nagatsuka, Hideki 3 Ntzoufras, Ioannis 3 Petropoulos, Constantinos 3 Raqab, Mohammad Zayed 2 Akbari, Masoumeh 2 Alba Fernández, Virtudes 2 Baringhaus, Ludwig 2 Chang, Hsin-wen 2 Consonni, Guido 2 Coolen-Maturi, Tahani 2 Cossette, Hélène 2 El Barmi, Hammou 2 Golparvar, Leila 2 Henze, Norbert 2 Jasiński, Krzysztof 2 Kokonendji, Célestin Clotaire 2 Kumar, Somesh 2 Malefaki, Sonia 2 Marceau, Étienne 2 McKeague, Ian W. 2 Mirfarah, Elham 2 Mirmostafaee, S. M. T. K. 2 Mohie El-Din, M. M. M. 2 Najarzadeh, Dariush 2 Ning, Jing 2 Parsian, Ahmad 2 Qin, Jing 2 Shen, Yu 2 Singh, Sanjay Kumar 2 Singh, Umesh Kumar 2 Su, Feng 2 Tripathi, Yogesh Mani 2 Volterman, William 2 Yu, Wen 1 Abd El Mabood, Ola Farouk 1 Abdel-Hamid, Alaa H. 1 Afify, W. M. 1 Agresti, Alan 1 Ahmadi, Mosayeb 1 Akhundov, Ilham 1 Al-Hussaini, Essam Khalaf 1 Al-Saleh, Mohammad Fraiwan 1 Alba-Fernández, M. Virtudes 1 Albert, Mélisande 1 Alijani, M. 1 Allison, James S. 1 Amein, M. M. 1 Anderson, Carolyn Jane 1 Andridge, Rebecca R. 1 Ariza López, Francisco Javier 1 Augustynowicz, Aneta 1 Bai, Jianming 1 Bansal, Naveen K. 1 Barbosa, Valdirene de Fátima 1 Basu, Suparna 1 Bayoud, Husam Awni 1 Behboodian, Javad 1 Belaghi, R. Arabi 1 Best, D. John 1 Bhattacharya, Uttam K. 1 Biswas, Atanu 1 Bogomolov, Marina 1 Bouret, Yann 1 Brusset, Xavier 1 Burkschat, Marco 1 Chalco-Cano, Yurilev 1 Chan, Kwun Chuen Gary 1 Chandrasekar, B. 1 Chen, Feifei 1 Chen, Jinyuan 1 Cheng, Conghua 1 Childs, Aaron 1 Clote, Peter G. 1 Davis, Richard A. 1 de Araújo Pereira, Gilberto 1 Dellaportas, Petros 1 Diab, Yusef Abdallah 1 Douc, Randal ...and 156 more Authors
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#### Cited in 53 Serials
16 Journal of Statistical Planning and Inference 16 Statistics & Probability Letters 16 Communications in Statistics. Theory and Methods 16 Journal of Statistical Computation and Simulation 13 Statistics 13 Computational Statistics and Data Analysis 12 Metrika 12 Communications in Statistics. Simulation and Computation 9 Journal of Multivariate Analysis 8 Statistical Methodology 6 Statistical Papers 4 Computational Statistics 4 Test 3 Annals of the Institute of Statistical Mathematics 3 The Annals of Statistics 3 Journal of Computational and Applied Mathematics 3 Journal of Applied Statistics 2 Statistical Methods and Applications 2 Statistics and Computing 1 The Canadian Journal of Statistics 1 Journal of Mathematical Biology 1 Psychometrika 1 Scandinavian Journal of Statistics 1 Biometrics 1 Journal of the American Statistical Association 1 Kybernetika 1 Naval Research Logistics 1 Opsearch 1 Insurance Mathematics & Economics 1 Operations Research Letters 1 Journal of Classification 1 Mathematical and Computer Modelling 1 Applied Mathematical Modelling 1 European Journal of Operational Research 1 International Journal of Computer Mathematics 1 Journal of Mathematical Sciences (New York) 1 Lifetime Data Analysis 1 Bernoulli 1 Journal of Nonparametric Statistics 1 Finance and Stochastics 1 Australian & New Zealand Journal of Statistics 1 Probability in the Engineering and Informational Sciences 1 Biostatistics 1 Advances and Applications in Statistics 1 Statistical Applications in Genetics and Molecular Biology 1 Advances in Difference Equations 1 Journal of the Korean Statistical Society 1 Computational & Mathematical Methods in Medicine 1 Electronic Journal of Statistics 1 The Annals of Applied Statistics 1 Sankhyā. Series A 1 Sankhyā. Series B 1 Stochastics and Quality Control
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#### Cited in 11 Fields
192 Statistics (62-XX) 34 Probability theory and stochastic processes (60-XX) 25 Numerical analysis (65-XX) 6 Operations research, mathematical programming (90-XX) 5 Game theory, economics, finance, and other social and behavioral sciences (91-XX) 3 Biology and other natural sciences (92-XX) 2 Combinatorics (05-XX) 2 Computer science (68-XX) 1 Special functions (33-XX) 1 Ordinary differential equations (34-XX) 1 Difference and functional equations (39-XX) | 2021-04-13 12:57:13 | {"extraction_info": {"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, "math_score": 0.6325997710227966, "perplexity": 12074.766696070217}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-17/segments/1618038072366.31/warc/CC-MAIN-20210413122252-20210413152252-00417.warc.gz"} |
https://rviews.rstudio.com/2018/03/02/capm-and-visualization/ | # Visualizing the Capital Asset Pricing Model
by Jonathan Regenstein
In a previous post, we covered how to calculate CAPM beta for our usual portfolio consisting of:
+ SPY (S&P500 fund) weighted 25%
+ EFA (a non-US equities fund) weighted 25%
+ IJS (a small-cap value fund) weighted 20%
+ EEM (an emerging-mkts fund) weighted 20%
+ AGG (a bond fund) weighted 10%
Today, we will move on to visualizing the CAPM beta and explore some ggplot and highcharter functionality, along with the broom package.
Before we can do any of this CAPM work, we need to calculate the portfolio returns, covered in this post, and then calculate the CAPM beta for the portfolio and the individual assets covered in this post.
I will not present that code or logic again but we will utilize four data objects from that previous work:
+ portfolio_returns_tq_rebalanced_monthly (a tibble of portfolio monthly returns)
+ market_returns_tidy (a tibble of SP500 monthly returns)
+ beta_dplyr_byhand (a tibble of market betas for our 5 individual assets)
+ asset_returns_long (a tibble of returns for our 5 individual assets)
Let’s get to it.
### Visualizing the Relationship between Portfolio Returns, Risk and Market Returns
The CAPM beta number is telling us about the linear relationship between our portfolio returns and the market returns. It’s also telling us about the riskiness of our portfolio - how volatile the portfolio is relative to the market. Before we get to beta itself, let’s take a look at expected monthly returns of our assets scattered against monthly risk of our individual assets.
library(tidyquant)
library(tidyverse)
library(timetk)
library(tibbletime)
library(scales)
# This theme_update will center your ggplot titles
theme_update(plot.title = element_text(hjust = 0.5))
asset_returns_long %>%
group_by(asset) %>%
summarise(expected_return = mean(returns),
stand_dev = sd(returns)) %>%
ggplot(aes(x = stand_dev, y = expected_return, color = asset)) +
geom_point(size = 2) +
ylab("expected return") +
xlab("standard deviation") +
ggtitle("Expected Monthly Returns v. Risk") +
scale_y_continuous(label = function(x){ paste0(x, "%")})
Where does our portfolio fit on this scatter plot? Let’s add it to the ggplot() flow with geom_point(aes(x = sd(portfolio_returns_tq_rebalanced_monthly$returns), y = mean(portfolio_returns_tq_rebalanced_monthly$returns)), color = "cornflowerblue", size = 3).
asset_returns_long %>%
group_by(asset) %>%
summarise(expected_return = mean(returns),
stand_dev = sd(returns)) %>%
ggplot(aes(x = stand_dev, y = expected_return, color = asset)) +
geom_point(size = 2) +
geom_point(aes(x = sd(portfolio_returns_tq_rebalanced_monthly$returns), y = mean(portfolio_returns_tq_rebalanced_monthly$returns)),
color = "cornflowerblue",
size = 3) +
geom_text(
aes(x = sd(portfolio_returns_tq_rebalanced_monthly$returns) * 1.09, y = mean(portfolio_returns_tq_rebalanced_monthly$returns),
label = "portfolio")) +
ylab("expected return") +
xlab("standard deviation") +
ggtitle("Expected Monthly Returns v. Risk") +
scale_y_continuous(labels = function(x){ paste0(x, "%")})
Our portfolio return/risk looks all right, though the SP500 has a higher expected return for just a bit more risk. It’s been tough to beat the market the last five years. EEM and EFA have a higher risk and lower expected return (no rational investor wants that!) and IJS has a higher risk and a higher expected return (some rational investors do want that!).
In general, the scatter is providing some return-risk context for our portfolio. It’s not directly part of CAPM, but I like to start here to get in the return-risk mindset.
Next, let’s turn to CAPM more directly and visualize the relationship between our portfolio and the market with a scatter plot of market returns on the x-axis and portfolio returns on the y-axis. First, we will add the market returns to our portfolio tibble by calling mutate(market_returns = market_returns_tidy$returns). Then, we set our x- and y-axis with ggplot(aes(x = market_returns, y = returns)). portfolio_returns_tq_rebalanced_monthly %>% mutate(market_returns = market_returns_tidy$returns) %>%
ggplot(aes(x = market_returns, y = returns)) +
geom_point(color = "cornflowerblue") +
ylab("portfolio returns") +
xlab("market returns") +
ggtitle("Scatterplot of portfolio returns v. market returns")
This scatter plot is communicating the same strong linear relationship as our numeric beta calculation from the previous post. We can add a simple regression line to it with geom_smooth(method = "lm", se = FALSE, color = "green", size = .5).
portfolio_returns_tq_rebalanced_monthly %>%
mutate(market_returns = market_returns_tidy$returns) %>% ggplot(aes(x = market_returns, y = returns)) + geom_point(color = "cornflowerblue") + geom_smooth(method = "lm", se = FALSE, color = "green", size = .5) + ylab("portfolio returns") + xlab("market returns") + ggtitle("Scatterplot with regression line") The green line is produced by the call to geom_smooth(method = 'lm'). Under the hood, ggplot fits a linear model of the relationship between market returns and portfolio returns. The slope of that green line is the CAPM beta that we calculated earlier. To confirm that we can add a line to the scatter that has a slope equal to our beta calculation and a y-intercept equal to what I labeled as alpha in the beta_dplyr_byhand object. To add the line, we invoke geom_abline(aes(intercept = beta_dplyr_byhand$estimate[1], slope = beta_dplyr_byhand$estimate[2]), color = "purple"). portfolio_returns_tq_rebalanced_monthly %>% mutate(market_returns = market_returns_tidy$returns) %>%
ggplot(aes(x = market_returns, y = returns)) +
geom_point(color = "cornflowerblue") +
geom_abline(aes(
intercept = beta_dplyr_byhand$estimate[1], slope = beta_dplyr_byhand$estimate[2]),
color = "purple",
size = .5) +
ylab("portfolio returns") +
xlab("market returns") +
ggtitle("Scatterplot with hand calculated slope")
We can plot both lines simultaneously to confirm to ourselves that they are the same - they should be right on top of each other but the purple line, our manual abline, extends into infinity so, we should see it start where the green line ends.
portfolio_returns_tq_rebalanced_monthly %>%
mutate(market_returns = market_returns_tidy$returns) %>% ggplot(aes(x = market_returns, y = returns)) + geom_point(color = "cornflowerblue") + geom_abline(aes( intercept = beta_dplyr_byhand$estimate[1],
slope = beta_dplyr_byhand$estimate[2]), color = "purple", size = .5) + geom_smooth(method = "lm", se = FALSE, color = "green", size = .5) + ylab("portfolio returns") + xlab("market returns") + ggtitle("Compare CAPM beta line to regression line") All right, that seems to visually confirm (or strongly support) that the fitted line calculated by ggplot and geom_smooth() has a slope equal to the beta we calculated ourselves. Why did we go through this exercise? Well, CAPM beta is a bit “jargony”. Since we need to map that jargon over to the world of linear modeling, it’s a useful practice to consider how jargon reduces to data science concepts. This isn’t a particularly complicated bit of jargon, but it’s good practice to get in the habit of reducing jargon. ### A Bit More on Linear Regression: Augmenting Our Data Before concluding our analysis of CAPM beta, let’s explore the augment() function from broom and how it helps to create a few more interesting visualizations. The code chunk below starts with model results from lm(returns ~ market_returns_tidy$returns...), which is regressing our portfolio returns on the market returns. We store the results in a list-column called called model. Next, we call augment(model) which will add predicted values to the original data set and return a tibble.
Those predicted values will be in the .fitted column. For some reason, the date column gets dropped. It’s nice to have this for visualizations so we will add it back in with mutate(date = portfolio_returns_tq_rebalanced_monthly$date). library(broom) portfolio_model_augmented <- portfolio_returns_tq_rebalanced_monthly %>% do(model = lm(returns ~ market_returns_tidy$returns, data = .))%>%
augment(model) %>%
mutate(date = portfolio_returns_tq_rebalanced_monthly$date) head(portfolio_model_augmented) ## returns market_returns_tidy.returns .fitted .se.fit ## 1 -0.0008696132 0.01267837 0.008294282 0.001431288 ## 2 0.0186624378 0.03726809 0.030451319 0.001984645 ## 3 0.0206248830 0.01903021 0.014017731 0.001485410 ## 4 -0.0053529692 0.02333503 0.017896670 0.001563417 ## 5 -0.0229487590 -0.01343411 -0.015234859 0.001953853 ## 6 0.0411705787 0.05038580 0.042271276 0.002521506 ## .resid .hat .sigma .cooksd .std.resid date ## 1 -0.009163896 0.01698211 0.01101184 0.006116973 -0.8415272 2013-02-28 ## 2 -0.011788881 0.03265148 0.01096452 0.020099701 -1.0913142 2013-03-28 ## 3 0.006607152 0.01829069 0.01104500 0.003434000 0.6071438 2013-04-30 ## 4 -0.023249640 0.02026222 0.01062704 0.047294116 -2.1386022 2013-05-31 ## 5 -0.007713900 0.03164618 0.01103127 0.008323550 -0.7137165 2013-06-28 ## 6 -0.001100697 0.05270569 0.01107986 0.000294938 -0.1029661 2013-07-31 Let’s use ggplot() to see how well the fitted return values match the actual return values. portfolio_model_augmented %>% ggplot(aes(x = date)) + geom_line(aes(y = returns, color = "actual returns")) + geom_line(aes(y = .fitted, color = "fitted returns")) + scale_colour_manual("", values = c("fitted returns" = "green", "actual returns" = "cornflowerblue")) + xlab("date") + ggtitle("Fitted versus actual returns") Those monthly returns and fitted values seem to track well. Let’s convert both actual returns and fitted returns to the growth of a dollar and run the same comparison. This isn’t a traditional way to visualize actual versus fitted, but it’s still useful. portfolio_model_augmented %>% mutate(actual_growth = cumprod(1 + returns), fitted_growth = cumprod(1 + .fitted)) %>% ggplot(aes(x = date)) + geom_line(aes(y = actual_growth, color = "actual growth")) + geom_line(aes(y = fitted_growth, color = "fitted growth")) + xlab("date") + ylab("actual and fitted growth") + ggtitle("Growth of a dollar: actual v. fitted") + scale_x_date(breaks = pretty_breaks(n = 8)) + scale_y_continuous(labels = dollar) + scale_colour_manual("", values = c("fitted growth" = "green", "actual growth" = "cornflowerblue")) Our fitted growth tracks our actual growth well, though the actual growth is lower than predicted for most of the five year history. ### To Highcharter! A nice side benefit of augment() is that it allows us to create an interesting highcharter object that replicates our scatter + regression ggplot from earlier. First, let’s build the base scatter plot of portfolio returns, which are housed in portfolio_model_augmented$returns, against market returns, which are housed in portfolio_model_augmented$market_returns_tidy.returns. library(highcharter) highchart() %>% hc_title(text = "Portfolio v. Market Returns") %>% hc_add_series_scatter(round(portfolio_model_augmented$returns, 4),
round(portfolio_model_augmented$market_returns_tidy.returns, 4)) %>% hc_xAxis(title = list(text = "Market Returns")) %>% hc_yAxis(title = list(text = "Portfolio Returns")) That looks good; but hover over one of the points. If you’re like me, you will desperately wish that the date of the observation were being displayed. Let’s add that date display functionality. First, we need to supply the date observations, so we will add a date variable with hc_add_series_scatter(..., date = portfolio_returns_tq_rebalanced_monthly$date). Then, we want the tool tip to pick up and display that variable. That is done with hc_tooltip(formatter = JS("function(){return ('port return: ' + this.y + ' <br> mkt return: ' + this.x + ' <br> date: ' + this.point.date)}")). We are creating a custom tool tip function to pick up the date. Run the code chunk below and hover over a point.
highchart() %>%
hc_title(text = "Portfolio v. Market Returns") %>%
hc_add_series_scatter(round(portfolio_model_augmented$returns, 4), round(portfolio_model_augmented$market_returns_tidy.returns, 4),
date = portfolio_model_augmented\$date) %>%
hc_xAxis(title = list(text = "Market Returns")) %>%
hc_yAxis(title = list(text = "Portfolio Returns")) %>%
hc_tooltip(formatter = JS("function(){
return ('port return: ' + this.y + ' <br> mkt return: ' + this.x +
' <br> date: ' + this.point.date)}"))
I was curious about the most negative reading in the bottom left, and this new tool tip makes it easy to see that it occurred in August of 2015.
Finally, let’s add the regression line.
To do that, we need to supply x and y coordinates to highcharter and specify that we want to add a line instead of more scatter points. We have the x and y coordinates for our fitted regression line because we added them with the augment() function. The x’s are the market returns and the y’s are the fitted values. We add this element to our code flow with hc_add_series(portfolio_model_augmented, type = "line", hcaes(x = market_returns_tidy.returns, y = .fitted), name = "CAPM Beta = Regression Slope")
highchart() %>%
hc_title(text = "Scatter with Regression Line") %>%
type = "scatter",
hcaes(x = round(market_returns_tidy.returns, 4),
y = round(returns, 4),
date = date),
name = "Returns") %>%
type = "line",
hcaes(x = market_returns_tidy.returns, y = .fitted),
name = "CAPM Beta = Regression Slope") %>%
hc_xAxis(title = list(text = "Market Returns")) %>%
hc_yAxis(title = list(text = "Portfolio Returns")) %>%
hc_tooltip(formatter = JS("function(){
return ('port return: ' + this.y + ' <br> mkt return: ' + this.x +
' <br> date: ' + this.point.date)}"))
That’s all for today and thanks for reading. | 2018-03-19 08:24:13 | {"extraction_info": {"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, "math_score": 0.34755873680114746, "perplexity": 9477.179071049924}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-13/segments/1521257646636.25/warc/CC-MAIN-20180319081701-20180319101701-00446.warc.gz"} |
https://physics.stackexchange.com/questions/465575/would-bekenstein-bound-disappear-in-some-holographic-models | # Would Bekenstein bound disappear in some holographic models?
In Holographic principle models there's a limit to the information that the system can store known as the "Bekenstein bound".
In physics, the Bekenstein bound is an upper limit on the entropy S, or information I, that can be contained within a given finite region of space which has a finite amount of energy—or conversely, the maximum amount of information required to perfectly describe a given physical system down to the quantum level
But if we have a theoretically infinite holographic boundary or system, then, would Bekenstein bound "disappear"? Could it save infinite information then?
• It sounds like you're asking if we set a limit to be at infinity, then would that limit effectively disappear. Is this correct? – Nat Mar 10 at 2:29
• @Nat basically yes: Like, if we had a bowl of water of 2L of capacity we could fill it with 2L of water. But if that bowl was infinite in size, then, there would be no limit for the amount of water we could fit there and we could fill it with infinite liters of water. Would it be the same for the holographic principle? – user225063 Mar 10 at 16:09
• – Nat Mar 10 at 17:39 | 2019-10-23 16:12:26 | {"extraction_info": {"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, "math_score": 0.8018369078636169, "perplexity": 573.2391262412287}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570987834649.58/warc/CC-MAIN-20191023150047-20191023173547-00219.warc.gz"} |
https://solvedlib.com/filling-a-pressurized-vessel-determine-the-net,390975 | # Filling a pressurized vessel. Determine the net power (W − Ev) in kW needed to be...
###### Question:
Filling a pressurized vessel. Determine the net power (W − Ev) in kW needed to be supplied by the pump in the figure below to pump liquid benzene (ρ = 0.877 g/cm3, μ = 0.647 cp) at a rate of 2.25 gal/sec. Assume atmospheric pressure is Po = 1.0 atm, and the inner diameter of the pipe is D = 0.425 in. What is the Reynolds number, NRe, for this process? (Assume a “hydraulically smooth” interior surface for the pipe).
Liquid benzene 10 ft Nitrogen gas pocket Pr 5.5 atm 3 ft
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Solve the equation by using substitution. $2 t^{2 / 5}+7 t^{1 / 5}+3=0$... | 2023-02-08 08:01:39 | {"extraction_info": {"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": 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, "math_score": 0.6176015138626099, "perplexity": 2737.0105122194227}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764500719.31/warc/CC-MAIN-20230208060523-20230208090523-00054.warc.gz"} |
http://eprint.iacr.org/2004/171 | ## Cryptology ePrint Archive: Report 2004/171
Short Signatures Without Random Oracles
Dan Boneh and Xavier Boyen
Abstract: We describe a short signature scheme which is existentially unforgeable under a chosen message attack without using random oracles. The security of our scheme depends on a new complexity assumption we call the {\em Strong Diffie-Hellman} assumption. This assumption has similar properties to the Strong RSA assumption, hence the name. Strong RSA was previously used to construct signature schemes without random oracles. However, signatures generated by our scheme are much shorter and simpler than signatures from schemes based on Strong RSA. Furthermore, our scheme provides a limited form of message recovery.
Category / Keywords: public-key cryptography / digital signatures, provable security
Publication Info: An extended abstract appears in EUROCRYPT 2004.
Date: received 20 Jul 2004, last revised 21 Jul 2004
Contact author: eprint at boyen org
Available format(s): Postscript (PS) | Compressed Postscript (PS.GZ) | PDF | BibTeX Citation
[ Cryptology ePrint archive ] | 2015-08-29 12:38:43 | {"extraction_info": {"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, "math_score": 0.8480796813964844, "perplexity": 6938.526229844646}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-35/segments/1440644064445.47/warc/CC-MAIN-20150827025424-00222-ip-10-171-96-226.ec2.internal.warc.gz"} |
https://www.tahoedailytribune.com/news/serenas-best-shot-serves-her-well/ | Serena’s best shot serves her well | TahoeDailyTribune.com
# Serena’s best shot serves her well
Steven Wine, The Associated Press
WIMBLEDON, England – As a kid, 13 Grand Slam titles ago, Serena Williams tended to goof off when it came time to work on serves.
She was supposed to hit them at the end of practice sessions with her sister Venus.
“We always talked a lot,” Serena said. “I don’t remember serving; I just remember talking. Lord knows what we were talking about, but we never stopped talking, unless my dad was looking at us. Then we would serve. Then when he wasn’t looking, we would just talk, talk, talk, talk, talk.”
Despite all the chitchat, Serena learned from her father the shot that served her well at Wimbledon. She hit a record 89 aces and won the title for the fourth time, beating Vera Zvonareva in Saturday’s final.
Williams said she planned to award the trophy to her serve.
“It’s well deserved,” she said. “I just really hope I can keep up serving like this. It’s a new turn in my life. I would love to continue to do that.”
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No. 1 in the rankings and No. 6 on the list for career Grand Slam titles, Williams is the most overpowering player on the women’s tour, especially when she hits the first shot of a point. And Williams said she has never served so well as in the past two weeks.
“It feels amazing,” she said. “I always can rely on my serve, but I’ve never been able to rely on it so much. I feel really cool because I hit all these aces. It’s awesome.”
When Williams double-faulted early in the final, she rolled her eyes as if reacting to a bad joke. Her next shot was a 115-mph ace. She never faced a break point as she beat Zvonareva 6-3, 6-2.
Williams lost serve only three times in seven matches. When her first serve was good, she won the point 88 percent of the time, including 31 of 33 points against Zvonareva.
With the title, the 28-year-old Williams moved ahead of Billie Jean King in career Grand Slam titles. She trails only five players – Margaret Smith Court (24), Steffi Graf (22), Helen Wills Moody (19), Martina Navratilova (18) and Chris Evert (18).
King and Navratilova watched the final from the Royal Box and said Williams’ serve is the best in the history of the women’s game.
“She gets two free points a game on it. It’s astonishing,” Navratilova said.
“Her second serve is attackable, but she doesn’t get in trouble that often. On big points she gets her first serve in, and the women do not attack the second serve that much. So she doesn’t feel the pressure to get the first serve in. And her first serve is just amazing.”
Williams hits it at 120 mph or more – faster than many of Rafael Nadal’s first serves. And her location is pinpoint, with the ball often kicking up chalk as the opponent lunges in vain.
Williams said she scolded her serve after the French Open, where she lost in the quarterfinals. She practiced it diligently – no chitchatting – in preparation for Wimbledon.
The payoff: Her ace total broke the tournament record of 72 she set last year. It was all the more remarkable because she won every match in straight sets, minimizing the number of points.
In one match, 19 of the 43 points she served resulted in aces.
“It’s not only a weapon like a shot weapon,” Zvonareva said. “It’s also in a way a mental weapon. It’s putting pressure on your serve because you know that, ‘Well, I better be winning this game. She’s going to hold.'”
Williams said a similar thought process helps her return game.
“It makes you play looser and even better,” she said. “I know I’m going to hold, so I have nothing to lose on the return. Let me go for it.”
Williams’ father and coach taught her how to serve, and growing up she realized the advantages the shot had to offer.
“I’ve always admired Pete Sampras’ serve,” she said. “I even liked his stroke, how he kind of always turned his hip into it. He used to serve so many aces. That was always so cool to me, and I always thought, ‘Wow, if I could have his serve, I’d be really good.'”
Years later, the serve Williams has makes her really good.
All-Time Women’s Majors Titles
Through 2010 Wimbledon
Australian French Wimbledon U.S. Overall
Player S- D- M S- D- M S- D- M S- D- M S- D- M Total
Margaret Smith Court 11- 8- 2 5- 4- 4 3- 2- 5 5- 5- 8 24-19-19 62
Martina Navratilova 3- 8- 1 2- 7- 2 9- 7- 4 4- 9- 3 18-31-10 59
Billie Jean King 1- 0- 1 1- 1- 2 6-10- 4 4- 5- 4 12-16-11 39
M. Osborne du Pont ——- 2- 3- 0 1- 5- 1 3-13- 9 6-21-10 37
Louise Brough 1- 1- 0 0- 3- 0 4- 5- 4 1-12- 4 6-21- 8 35
Doris Hart 1- 1- 2 2- 5- 3 1- 4- 5 2- 4- 5 6-14-15 35
Helen Wills Moody ——- 4- 2- 0 8- 3- 1 7- 4- 2 19- 9- 3 31
Serena Williams 5- 4- 0 1- 2- 0 4- 3- 1 3- 2- 1 13-11- 2 26
Elizabeth Ryan ——- 0- 4- 0 0-12- 7 0- 1- 2 0-17- 9 26
Steffi Graf 4- 0- 0 6- 0- 0 7- 1- 0 5- 0- 0 22- 1- 0 23
Pam Shriver 0- 7- 0 0- 4- 1 0- 5- 0 0- 5- 0 0-21- 1 22
Chris Evert 2- 0- 0 7- 2- 0 3- 1- 0 6- 0- 0 18- 3- 0 21
Suzanne Lenglen ——- 2- 2- 2 6- 6- 3 ——- 8- 8- 5 21
Darlene Hard ——- 1- 3- 2 0- 4- 3 2- 6- 0 3-13- 5 21
Venus Williams 0- 4- 1 0- 2- 1 5- 3- 0 2- 2- 0 7-11- 2 20
Nancye Wynne Bolton 6-10- 4 ——- ——- ——- 6-10- 4 20
Natasha Zvereva 0- 3- 2 0- 6- 0 0- 5- 0 0- 4- 0 0-18- 2 20
Maria Bueno 0- 1- 0 0- 1- 1 3- 5- 0 4- 4- 0 7-11- 1 19
Thelma Coyne Long 2-12- 4 0- 0- 1 ——- ——- 2-12- 5 19
Alice Marble ——- ——- 1- 2- 3 4- 4- 4 5- 6- 7 18
Sarah Palfrey ——- 0- 0- 1 0- 2- 0 2- 9- 4 2-11- 5 18
Shirley Fry 1- 1- 0 1- 4- 0 1- 3- 1 1- 4- 0 4-12- 1 17
H.Hotchkiss Wightman ——- ——- 0- 1- 0 4- 6- 6 4- 7- 6 17
Gigi Fernandez 0- 2- 0 0- 6- 0 0- 4- 0 0- 5- 0 0-17- 0 17
Jana Novotna 0- 2- 2 0- 2- 0 1- 4- 1 0- 3- 1 1-11- 4 16
Martina Hingis 3- 4- 1 0- 2- 0 1- 2- 0 1- 1- 0 5- 9- 1 15
Daphne Akhurst 5- 5- 4 ——- ——- ——- 5- 5- 4 14
A. Sanchez Vicario 0- 3- 1 3- 0- 2 0- 1- 0 1- 2- 1 4- 6- 4 14
Helena Sukova 0- 2- 0 0- 1- 1 0- 4- 3 0- 2- 1 0- 9- 5 14
Molla Bjurstedt ——- ——- ——- 8- 2- 3 8- 2- 3 13
Evonne Goolagong 4- 4- 0 1- 0- 1 2- 1- 0 ——- 7- 5- 1 13
Juliette Atkinson ——- ——- ——- 3- 7- 3 3- 7- 3 13
Mary K. Browne ——- ——- 0- 1- 0 3- 5- 4 3- 6- 4 13
Simone Mathieu ——- 2- 6- 2 0- 3- 0 ——- 2- 9- 2 13
Lesley Turner 0- 3- 2 2- 2- 0 0- 1- 2 0- 1- 0 2- 7- 4 13
Maureen Connolly 1- 1- 0 2- 1- 1 3- 0- 0 3- 0- 0 9- 2- 1 12
Francoise Durr ——- 1- 5- 3 0- 0- 1 0- 2- 0 1- 7- 4 12
Rosie Casals ——- ——- 0- 5- 2 0- 4- 1 0- 9- 3 12
Althea Gibson 0- 1- 0 1- 1- 0 2- 3- 0 2- 0- 1 5- 5- 1 11
V. Ruano Pascual 0- 1- 0 0- 6- 1 ——- 0- 3- 0 0-10- 1 11
Betty Stove ——- 0- 2- 0 0- 1- 2 0- 3- 2 0- 6- 4 10
Cara Black 0- 1- 1 0- 0- 1 0- 3- 2 0- 1- 1 0- 5- 5 10
Anne Smith 0- 1- 0 0- 2- 2 0- 1- 1 0- 1- 2 0- 5- 5 10
Monica Seles 4- 0- 0 3- 0- 0 ——- 2- 0- 0 9- 0- 0 9
Helen Jacobs ——- ——- 1- 0- 0 4- 3- 1 5- 3- 1 9
Betty Nuthall ——- 0- 1- 2 ——- 1- 3- 2 1- 4- 4 9
Judy Tegart Dalton 0- 4- 1 0- 1- 0 0- 1- 0 0- 2- 0 0- 8- 1 9
Lisa Raymond 0- 1- 0 0- 1- 1 0- 1- 1 0- 2- 2 0- 5- 4 9
Wendy Turnbull ——- 0- 1- 2 0- 1- 2 0- 2- 1 0- 4- 5 9
Elisabeth Moore ——- ——- ——- 4- 2- 2 4- 2- 2 8
Esna Boyd 1- 4- 3 ——- ——- ——- 1- 4- 3 8
Paola Suarez 0- 1- 0 0- 4- 0 ——- 0- 3- 0 0- 8- 0 8
Dorothea Chambers ——- ——- 7- 0- 0 ——- 7- 0- 0 7
Justine Henin 1- 0- 0 4- 0- 0 ——- 2- 0- 0 7- 0- 0 7
Virginia Wade 1- 1- 0 0- 1- 0 1- 0- 0 1- 2- 0 3- 4- 0 7
Ann Haydon Jones ——- 2- 3- 0 1- 0- 1 ——- 3- 3- 1 7
Kitty McKane ——- ——- 2- 0- 2 0- 2- 1 2- 2- 3 7
Kathy Jordan 0- 1- 0 0- 1- 1 0- 2- 1 0- 1- 0 0- 5- 2 7
Blanche Bingley ——- ——- 6- 0- 0 ——- 6- 0- 0 6
Pauline Betz ——- 0- 0- 1 1- 0- 0 4- 0- 0 5- 0- 1 6
Lindsay Davenport 1- 0- 0 0- 1- 0 1- 1- 0 1- 1- 0 3- 3- 0 6
Dorothy Round 1- 0- 0 ——- 2- 0- 3 ——- 3- 0- 3 6
Nancy Richey 1- 1- 0 1- 0- 0 0- 1- 0 0- 2- 0 2- 4- 0 6
Mary Bevis Hawton 0- 5- 1 ——- ——- ——- 0- 5- 1 6
Renee Schuurman 0- 1- 0 0- 4- 1 ——- ——- 0- 5- 1 6
Rennae Stubbs 0- 1- 1 ——- 0- 2- 0 0- 1- 1 0- 4- 2 6
Eileen Bennett ——- 0- 2- 2 ——- 0- 1- 1 0- 3- 3 6
Larisa Neiland 0- 0- 2 0- 1- 1 0- 1- 1 ——- 0- 2- 4 6
Charlotte Cooper ——- ——- 5- 0- 0 ——- 5- 0- 0 5
Lottie Dod ——- ——- 5- 0- 0 ——- 5- 0- 0 5
Hana Mandlikova 2- 0- 0 1- 0- 0 ——- 1- 1- 0 4- 1- 0 5
Mall Molesworth 2- 3- 0 ——- ——- ——- 2- 3- 0 5
Mabel Cahill ——- ——- ——- 2- 2- 1 2- 2- 1 5
Sylvia Lance Harper 1- 3- 1 ——- ——- ——- 1- 3- 1 5
Sandra Reynolds 0- 1- 1 0- 3- 0 ——- ——- 0- 4- 1 5
Eleonora Sears ——- ——- ——- 0- 4- 1 0- 4- 1 5
Billie Yorke ——- 0- 3- 1 0- 1- 0 ——- 0- 4- 1 5
Marion Zinderstein ——- ——- ——- 0- 4- 1 0- 4- 1 5
Nell Hall Hopman 0- 0- 4 0- 1- 0 ——- ——- 0- 1- 4 5 | 2019-02-21 13:35:09 | {"extraction_info": {"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, "math_score": 0.9361249208450317, "perplexity": 4490.443490655449}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-09/segments/1550247504790.66/warc/CC-MAIN-20190221132217-20190221154217-00066.warc.gz"} |
http://www.sarahsmathings.com/category/calc2/ | ## Test For Divergence: An Example
This post is in response to a user-submitted question. I’m going to keep it short and sweet since I just had eye surgery and am still half-blind, so computer tasks are somewhat difficult. The question gives the following infinite sum: \begin{equation*} \sum_{n=1}^{\infty} \left(1+\frac{3}{n}\right)^{-5n} … Continue reading
## Trig Substitution and Completing the Square: An Example
This post is about another integration problem. It was submitted to the site by the same person who submitted this question. Part me wonders if this individual isn’t just getting to do their homework for them… But I will let it slide since I was … Continue reading
## Integration by Parts: An Example
This post is about a problem a student recently submitted to the website. I decided to answer it in post format, because to do so through email would look too messy. The question is the following: \begin{equation*} \int e^{2x}\sin(3x)dx \end{equation*} This is a challenging … Continue reading
## Gabriel’s Horn and Improper Integrals
This post is about an interesting problem in mathematics called Gabriel’s Horn (also known as Toricelli’s trumpet). It comes up when learning about surface area and improper integrals in calculus. The name “Gabriel’s Horn” is explained by the following quote, courtesy of Wikipedia: The name … Continue reading | 2017-11-24 15:03:32 | {"extraction_info": {"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, "math_score": 0.7658032774925232, "perplexity": 1005.2033337732145}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-47/segments/1510934808254.76/warc/CC-MAIN-20171124142303-20171124162303-00112.warc.gz"} |
https://networkx.org/documentation/networkx-2.3/_modules/networkx/algorithms/components/weakly_connected.html | Warning
This documents an unmaintained version of NetworkX. Please upgrade to a maintained version and see the current NetworkX documentation.
# Source code for networkx.algorithms.components.weakly_connected
# -*- coding: utf-8 -*-
# Aric Hagberg <hagberg@lanl.gov>
# Dan Schult <dschult@colgate.edu>
# Pieter Swart <swart@lanl.gov>
#
# Authors: Aric Hagberg (hagberg@lanl.gov)
# Christopher Ellison
"""Weakly connected components."""
import warnings as _warnings
import networkx as nx
from networkx.utils.decorators import not_implemented_for
__all__ = [
'number_weakly_connected_components',
'weakly_connected_components',
'weakly_connected_component_subgraphs',
'is_weakly_connected',
]
[docs]@not_implemented_for('undirected') def weakly_connected_components(G): """Generate weakly connected components of G. Parameters ---------- G : NetworkX graph A directed graph Returns ------- comp : generator of sets A generator of sets of nodes, one for each weakly connected component of G. Raises ------ NetworkXNotImplemented: If G is undirected. Examples -------- Generate a sorted list of weakly connected components, largest first. >>> G = nx.path_graph(4, create_using=nx.DiGraph()) >>> nx.add_path(G, [10, 11, 12]) >>> [len(c) for c in sorted(nx.weakly_connected_components(G), ... key=len, reverse=True)] [4, 3] If you only want the largest component, it's more efficient to use max instead of sort: >>> largest_cc = max(nx.weakly_connected_components(G), key=len) See Also -------- connected_components strongly_connected_components Notes ----- For directed graphs only. """ seen = set() for v in G: if v not in seen: c = set(_plain_bfs(G, v)) yield c seen.update(c)
[docs]@not_implemented_for('undirected') def number_weakly_connected_components(G): """Returns the number of weakly connected components in G. Parameters ---------- G : NetworkX graph A directed graph. Returns ------- n : integer Number of weakly connected components Raises ------ NetworkXNotImplemented: If G is undirected. See Also -------- weakly_connected_components number_connected_components number_strongly_connected_components Notes ----- For directed graphs only. """ return sum(1 for wcc in weakly_connected_components(G))
[docs]@not_implemented_for('undirected') def weakly_connected_component_subgraphs(G, copy=True): """DEPRECATED: Use (G.subgraph(c) for c in weakly_connected_components(G)) Or (G.subgraph(c).copy() for c in weakly_connected_components(G)) """ msg = "weakly_connected_component_subgraphs is deprecated and will be removed in 2.2" \ "use (G.subgraph(c).copy() for c in weakly_connected_components(G))" _warnings.warn(msg, DeprecationWarning) for c in weakly_connected_components(G): if copy: yield G.subgraph(c).copy() else: yield G.subgraph(c)
[docs]@not_implemented_for('undirected') def is_weakly_connected(G): """Test directed graph for weak connectivity. A directed graph is weakly connected if and only if the graph is connected when the direction of the edge between nodes is ignored. Note that if a graph is strongly connected (i.e. the graph is connected even when we account for directionality), it is by definition weakly connected as well. Parameters ---------- G : NetworkX Graph A directed graph. Returns ------- connected : bool True if the graph is weakly connected, False otherwise. Raises ------ NetworkXNotImplemented: If G is undirected. See Also -------- is_strongly_connected is_semiconnected is_connected is_biconnected weakly_connected_components Notes ----- For directed graphs only. """ if len(G) == 0: raise nx.NetworkXPointlessConcept( """Connectivity is undefined for the null graph.""") return len(list(weakly_connected_components(G))[0]) == len(G)
def _plain_bfs(G, source): """A fast BFS node generator The direction of the edge between nodes is ignored. For directed graphs only. """ Gsucc = G.succ Gpred = G.pred seen = set() nextlevel = {source} while nextlevel: thislevel = nextlevel nextlevel = set() for v in thislevel: if v not in seen: yield v seen.add(v) nextlevel.update(Gsucc[v]) nextlevel.update(Gpred[v]) | 2023-02-09 01:54:50 | {"extraction_info": {"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, "math_score": 0.32811856269836426, "perplexity": 11175.200231521156}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764501066.53/warc/CC-MAIN-20230209014102-20230209044102-00583.warc.gz"} |
http://tex.stackexchange.com/questions/22399/how-to-use-condensed-typewriter-with-latin-modern-and-doc | # How to use condensed typewriter with latin modern and doc
I'm documenting a class with (ltx)doc. The macro names are quite long so I get a lot of overfull hboxes.
AFAIK latin modern has an additional typewriter font which is condensed. How do I switch the `doc` package to use that? I'd prefer a solution for pdflatex.
-
As far as I can see it, the condensed typewriter font only comes in a light weight, which does not mix well with LMRoman in the medium weight. – Michael Ummels Jul 5 '11 at 16:03
Write this in the preamble of your document
``````\DeclareFontFamily{T1}{lmttc}{\hyphenchar \font-1 }
\DeclareFontShape{T1}{lmttc}{m}{n}
{<-> ec-lmtlc10}{}
\DeclareFontShape{T1}{lmttc}{m}{it}
{<->sub*lmttc/m/sl}{}
\DeclareFontShape{T1}{lmttc}{m}{sl}
{<-> ec-lmtlco10}{}
\renewcommand{\ttdefault}{lmttc}
``````
If you need OT1-encoding, change all `T1` into `OT1` and `ec-` into `rm-`.
With XeLaTeX or LuaLaTeX one can say
``````\setmonofont[HyphenChar=None]{Latin Modern Mono Light Cond}
`````` | 2013-05-26 05:33:14 | {"extraction_info": {"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, "math_score": 0.8717160224914551, "perplexity": 4031.204354323715}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2013-20/segments/1368706631378/warc/CC-MAIN-20130516121711-00086-ip-10-60-113-184.ec2.internal.warc.gz"} |
https://isabelle.in.tum.de/repos/isabelle/rev/437251bbc5ce | author haftmann Wed, 05 Dec 2007 14:32:17 +0100 changeset 25544 437251bbc5ce parent 25543 6b2031004d3f child 25545 21cd20c1ce98
tuned class parts
--- a/doc-src/IsarRef/generic.tex Wed Dec 05 14:16:15 2007 +0100
+++ b/doc-src/IsarRef/generic.tex Wed Dec 05 14:32:17 2007 +0100
@@ -510,29 +510,35 @@
\end{warn}
-\subsection{Type classes}\label{sec:class}
+\subsection{Classes}\label{sec:class}
-A type class is a special case of a locale, with some additional
-infrastructure (notably a link to type-inference). Type classes
-consist of a locale with \emph{exactly one} type variable and an
-corresponding axclass. \cite{isabelle-classes} gives a substantial
-introduction on type classes.
+A class is a peculiarity of a locale with \emph{exactly one} type variable.
+Beyond the underlying locale, a corresponding type class is established which
+is interpreted logically as axiomatic type class \cite{Wenzel:1997:TPHOL}
+whose logical content are the assumptions of the locale. Thus, classes provide
+the full generality of locales combined with the commodity of type classes
+(notably type-inference). See \cite{isabelle-classes} for a short tutorial.
\indexisarcmd{instance}\indexisarcmd{class}\indexisarcmd{print-classes}
\begin{matharray}{rcl}
\isarcmd{class} & : & \isartrans{theory}{local{\dsh}theory} \\
+ \isarcmd{instantiation} & : & \isartrans{theory}{local{\dsh}theory} \\
+ \isarcmd{instance} & : & \isartrans{local{\dsh}theory}{local{\dsh}theory} \\
\isarcmd{subclass} & : & \isartrans{local{\dsh}theory}{local{\dsh}theory} \\
- \isarcmd{instance} & : & \isartrans{theory}{proof(prove)} \\
\isarcmd{print_classes}^* & : & \isarkeep{theory~|~proof} \\
+ intro_classes & : & \isarmeth
\end{matharray}
\begin{rail}
- 'class' name '=' ((superclassexpr '+' (contextelem+)) | superclassexpr | (contextelem+)) 'begin'?
+ 'class' name '=' ((superclassexpr '+' (contextelem+)) | superclassexpr | (contextelem+)) \\
+ 'begin'?
+ ;
+ 'instantiation' (nameref + 'and') '::' arity 'begin'
+ ;
+ 'instance'
;
'subclass' target? nameref
;
- 'instance' (nameref '::' arity + 'and') (axmdecl prop +)?
- ;
'print\_classes'
;
@@ -543,30 +549,34 @@
\begin{descr}
\item [$\CLASS~c = superclasses~+~body$] defines a new class $c$,
- inheriting from $superclasses$. Simultaneously, a locale
- named $c$ is introduced, inheriting from the locales
- corresponding to $superclasses$; also, an axclass
- named $c$, inheriting from the axclasses corresponding to
- $superclasses$. $\FIXESNAME$ in $body$ are lifted
- to the theory toplevel, constraining
- the free type variable to sort $c$ and stripping local syntax.
- $\ASSUMESNAME$ in $body$ are also lifted,
- constraining
- the free type variable to sort $c$.
+ inheriting from $superclasses$. This introduces a locale $c$
+ inheriting from all the locales $superclasses$. Correspondingly,
+ a type class $c$, inheriting from type classes $superclasses$.
+ $\FIXESNAME$ in $body$ are lifted to the global theory level
+ (\emph{class operations} $\vec f$ of class $c$),
+ mapping the local type parameter $\alpha$ to a schematic
+ type variable $?\alpha::c$
+ $\ASSUMESNAME$ in $body$ are also lifted, mapping each local parameter
+ $f::\tau [\alpha]$ to its corresponding global constant
+ $f::\tau [?\alpha::c]$.
-\item [$\INSTANCE~~t :: (\vec s)s \vec{defs}$]
- sets up a goal stating type arities. The proof would usually
- proceed by $intro_classes$, and then establish the characteristic theorems
- of the type classes involved.
- The $defs$, if given, must correspond to the class parameters
- involved in the $arities$ and are introduced in the theory
- before proof.
- After finishing the proof, the theory will be
- augmented by a type signature declaration corresponding to the
- resulting theorems.
- This $\isarcmd{instance}$ command is actually an extension
- of primitive axclass $\isarcmd{instance}$ (see \ref{sec:axclass}).
-
+\item [$\INSTANTIATION~\vec t~::~(\vec s)~s~\isarkeyword{begin}$]
+ opens a theory target (cf.\S\ref{sec:target}) which allows to specify
+ class operations $\vec f$ corresponding to sort $s$ at particular
+ type instances $\vec{\alpha::s}~t$ for each $t$ in $\vec t$.
+ An $\INSTANCE$ command in the target body sets up a goal stating
+ the type arities given after the $\INSTANTIATION$ keyword.
+ The possibility to give a list of type constructors with same arity
+ nicely corresponds to mutual recursive type definitions in Isabelle/HOL.
+ The target is concluded by an $\isarkeyword{end}$ keyword.
+
+\item [$\INSTANCE$] in an instantiation target body sets up a goal stating
+ the type arities claimed at the opening $\INSTANTIATION$ keyword.
+ The proof would usually proceed by $intro_classes$, and then establish the
+ characteristic theorems of the type classes involved.
+ After finishing the proof, the background theory will be
+ augmented by the proven type arities.
+
\item [$\SUBCLASS~c$] in a class context for class $d$
sets up a goal stating that class $c$ is logically
contained in class $d$. After finishing the proof, class $d$ is proven
@@ -575,24 +585,41 @@
\item [$\isarkeyword{print_classes}$] prints all classes
in the current theory.
+\item [$intro_classes$] repeatedly expands all class introduction rules of
+ this theory. Note that this method usually needs not be named explicitly,
+ as it is already included in the default proof step (of $\PROOFNAME$ etc.).
+ In particular, instantiation of trivial (syntactic) classes may be performed
+ by a single $\DDOT$'' proof step.
+
\end{descr}
+\subsubsection{Class target}
+
+A named context may refer to a locale (cf.~\S\ref{sec:target}). If this
+locale is also a class $c$, beside the common locale target behaviour
+the following occurs:
+
+\begin{itemize}
+ \item Local constant declarations $g [\alpha]$ referring to the local type
+ parameter $\alpha$ and local parameters $\vec f [\alpha]$ are accompagnied
+ by theory-level constants $g [?\alpha::c]$ referring to theory-level
+ class operations $\vec f [?\alpha::c]$
+ \item Local theorem bindings are lifted as are assumptions.
+\end{itemize}
+
+
\subsection{Axiomatic type classes}\label{sec:axclass}
\indexisarcmd{axclass}\indexisarmeth{intro-classes}
\begin{matharray}{rcl}
\isarcmd{axclass} & : & \isartrans{theory}{theory} \\
\isarcmd{instance} & : & \isartrans{theory}{proof(prove)} \\
- intro_classes & : & \isarmeth \\
\end{matharray}
-Axiomatic type classes are provided by Isabelle/Pure as a \emph{definitional}
-interface to type classes (cf.~\S\ref{sec:classes}). Thus any object logic
-may make use of this light-weight mechanism of abstract theories
-\cite{Wenzel:1997:TPHOL}. There is also a tutorial on using axiomatic type
-classes in Isabelle \cite{isabelle-axclass} that is part of the standard
-Isabelle documentation.
+Axiomatic type classes are Isabelle/Pure's primitive \emph{definitional} interface
+to type classes. For practical applications, you should consider using classes
+(cf.~\S\ref{sec:classes}) which provide a convenient user interface.
\begin{rail}
'axclass' classdecl (axmdecl prop +)
@@ -622,12 +649,6 @@
augmented by a type signature declaration corresponding to the resulting
theorem.
-\item [$intro_classes$] repeatedly expands all class introduction rules of
- this theory. Note that this method usually needs not be named explicitly,
- as it is already included in the default proof step (of $\PROOFNAME$ etc.).
- In particular, instantiation of trivial (syntactic) classes may be performed
- by a single $\DDOT$'' proof step.
-
\end{descr} | 2021-04-19 07:26:23 | {"extraction_info": {"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, "math_score": 0.7914576530456543, "perplexity": 9029.371031941577}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-17/segments/1618038878326.67/warc/CC-MAIN-20210419045820-20210419075820-00086.warc.gz"} |
http://www.maa.org/publications/periodicals/convergence/problems-another-time?page=11 | Problems from Another Time
Individual problems from throughout mathematics history, as well as articles that include problem sets for students.
How a translation of Peano's counterexample to the 'theorem' that a zero Wronskian implies linear dependence can help your differential equations students
A teacher agreed to teach 9 months for $562.50 and his board. At the end of the term, on account of two months absence caused by sickness, he received only$409.50. What was his board worth per month?
In a certain lake, swarming with red geese, the tip of a lotus bud was seen to extend a span [9 inches] above the surface of the water.
Find the isoceles triangle of smallest area that circumscribes a circle of radius 1.
In a right triangle, the hypotenuse is 9.434 and the sum of the sides around the right angle is 13. Find the lengths of the sides around the right angle.
Determine by using algebra the number of degrees in the angle A where: cos A = tan A
Four men already having denari found a purse of denari; each man has a different amount of denari before they found the purse. Find out how much denari each man has.
On an expedition to seize his enemy's elephants, a king marched 2 yojanas the first day.
A gentleman has bought a rectangular piece of land whose perimeter is to be 100 rods; and he is to pay 1 dollar for each rod in the length of the land and 3 dollars for each rod in the breath of the land.
A railway train strikes a snowdrift which creates a constant resistance. How long does it take the snow to stop the train? | 2015-05-24 07:37:24 | {"extraction_info": {"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, "math_score": 0.4946499764919281, "perplexity": 1497.822841546284}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-22/segments/1432207927844.14/warc/CC-MAIN-20150521113207-00186-ip-10-180-206-219.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/finding-directional-derivative.973050/ | Finding Directional Derivative
Homework Statement:
see post
Relevant Equations:
∇f⋅u= Direction vector
The gradient is < (2x-y), (-x+2y-1) >
at P(1,-1) the gradient is <3, -4>
Since ∇f⋅u= Direction vector, it seems that we should set the equation equal to the desired directional derivative.
< 3, -4 > ⋅ < a, b > = 4
which becomes
3a-4b=4
I thought of making a list of possible combinations of a's and b's which satisfy this equation like so
a, b
corresponding direction vector
0, 1
<0, 1>
(4/3), 0
(for which there is no direction vector??)
2, (1/2)
< 4/√17 , 1/√17 >
But it seems that there are an infinite number of possible combinations. And the question is asking for all of them.
timetraveller123
i am not sure but i don't think it can be infinite in two dimension
i think it is two
##
(3,-4) , (a,b) = 5.1.cos \theta = 4
##
there is two values of theta one negative of the other. maybe in three dimension you might have infinite vectors because they all in a cone of same theta. but in two dimension i think it might be two vectors
Staff Emeritus
Homework Helper
Gold Member
Problem Statement: see post
Relevant Equations: ∇f⋅u= Direction vector
View attachment 244655View attachment 244656
The gradient is < (2x-y), (-x+2y-1) >
at P(1,-1) the gradient is <3, -4>
Since ∇f⋅u= Direction vector, it seems that we should set the equation equal to the desired directional derivative.
< 3, -4 > ⋅ < a, b > = 4
which becomes
3a-4b=4
I thought of making a list of possible combinations of a's and b's which satisfy this equation like so
a, b
corresponding direction vector
0, 1
<0, 1>
(4/3), 0
(for which there is no direction vector??)
2, (1/2)
< 4/√17 , 1/√17 >
But it seems that there are an infinite number of possible combinations. And the question is asking for all of them. | 2023-03-24 06:39:00 | {"extraction_info": {"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, "math_score": 0.8816159963607788, "perplexity": 2227.715366246309}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296945248.28/warc/CC-MAIN-20230324051147-20230324081147-00233.warc.gz"} |
https://feasts.tidyverts.org/reference/gg_season.html | Produces a time series seasonal plot. A seasonal plot is similar to a regular time series plot, except the x-axis shows data from within each season. This plot type allows the underlying seasonal pattern to be seen more clearly, and is especially useful in identifying years in which the pattern changes.
gg_season(
data,
y = NULL,
period = NULL,
facet_period = NULL,
max_col = 15,
pal = (scales::hue_pal())(9),
polar = FALSE,
labels = c("none", "left", "right", "both"),
...
)
## Arguments
data A tidy time series object (tsibble) The variable to plot (a bare expression). If NULL, it will automatically selected from the data. The seasonal period to display. A secondary seasonal period to facet by (typically smaller than period). The maximum number of colours to display on the plot. If the number of seasonal periods in the data is larger than max_col, the plot will not include a colour. Use max_col = 0 to never colour the lines, or Inf to always colour the lines. If labels are used, then max_col will be ignored. A colour palette to be used. If TRUE, the season plot will be shown on polar coordinates. Position of the labels for seasonal period identifier. Additional arguments passed to geom_line()
## Value
A ggplot object showing a seasonal plot of a time series.
## References
Hyndman and Athanasopoulos (2019) Forecasting: principles and practice, 3rd edition, OTexts: Melbourne, Australia. https://OTexts.com/fpp3/
## Examples
library(tsibble)
library(dplyr)
tsibbledata::aus_retail %>%
filter(
State == "Victoria",
Industry == "Cafes, restaurants and catering services"
) %>%
gg_season(Turnover) | 2021-07-31 19:45:16 | {"extraction_info": {"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, "math_score": 0.25756943225860596, "perplexity": 5108.4879762906485}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-31/segments/1627046154099.21/warc/CC-MAIN-20210731172305-20210731202305-00058.warc.gz"} |
https://en.wikipedia.org/wiki/Gaussian_Minimum_Shift_Keying | # Minimum-shift keying
(Redirected from Gaussian Minimum Shift Keying)
Mapping changes in continuous phase. Each bit time, the carrier phase changes by ±90°
PSD of the MSK, BPSK and QPSK. The side-lobes of MSK are lower (-23dB) than in both BPSK and QPSK cases (-10 dB). Therefore, the inter-channel interference is lower in MSK case. Moreover, the main lobe of the MSK signal is wider that means more energy during the null-to-null bandwidth. However, this can be also the disadvantage where extremely narrow bandwidth is required (null-to-null bandwidth of the QPSK is equal to 3dB-bandwidth, null-to-null bandwidth of the MSK signal larger than 3dB-bandwidth in 1.5 times).
In digital modulation, minimum-shift keying (MSK) is a type of continuous-phase frequency-shift keying that was developed in the late 1950s by Collins Radio employees Melvin L. Doelz and Earl T. Heald. [1] Similar to OQPSK, MSK is encoded with bits alternating between quadrature components, with the Q component delayed by half the symbol period.
However, instead of square pulses as OQPSK uses, MSK encodes each bit as a half sinusoid.[2][3] This results in a constant-modulus signal (constant envelope signal), which reduces problems caused by non-linear distortion. In addition to being viewed as related to OQPSK, MSK can also be viewed as a continuous phase frequency shift keyed (CPFSK) signal with a frequency separation of one-half the bit rate.
In MSK the difference between the higher and lower frequency is identical to half the bit rate. Consequently, the waveforms used to represent a 0 and a 1 bit differ by exactly half a carrier period. Thus, the maximum frequency deviation is ${\displaystyle \delta \ }$ = 0.25 fm where fm is the maximum modulating frequency. As a result, the modulation index m is 0.5. This is the smallest FSK modulation index that can be chosen such that the waveforms for 0 and 1 are orthogonal. A variant of MSK called Gaussian minimum-shift keying (GMSK) is used in the GSM mobile phone standard.
## Mathematical representation
The resulting signal is represented by the formula: [4]
${\displaystyle s(t)=a_{I}(t)\cos {\left({\frac {{\pi }t}{2T}}\right)}\cos {(2{\pi }f_{c}t)}-a_{Q}(t)\sin {\left({\frac {{\pi }t}{2T}}\right)}\sin {\left(2{\pi }f_{c}t\right)}}$
where ${\displaystyle a_{I}(t)}$ and ${\displaystyle a_{Q}(t)}$ encode the even and odd information respectively with a sequence of square pulses of duration 2T. ${\displaystyle a_{I}(t)}$ has its pulse edges on ${\displaystyle t=[-T,T,3T,\ldots ]}$ and ${\displaystyle a_{Q}(t)}$ on ${\displaystyle t=[0,2T,4T,\ldots ]}$. The carrier frequency is ${\displaystyle f_{c}}$.
Using the trigonometric identity, this can be rewritten in a form where the phase and frequency modulation are more obvious,
${\displaystyle s(t)=\cos \left[2\pi f_{c}t+b_{k}(t){\frac {\pi t}{2T}}+\phi _{k}\right]}$
where bk(t) is +1 when ${\displaystyle a_{I}(t)=a_{Q}(t)}$ and -1 if they are of opposite signs, and ${\displaystyle \phi _{k}}$ is 0 if ${\displaystyle a_{I}(t)}$ is 1, and ${\displaystyle \pi }$ otherwise. Therefore, the signal is modulated in frequency and phase, and the phase changes continuously and linearly.
Since the minimum symbol distance is the same as in the QPSK,[5] the following formula can be used for the theoretical bit-error ratio bound:
${\displaystyle P_{b}=Q\left({\sqrt {\frac {2E_{b}}{N_{0}}}}\right)={\frac {1}{2}}\operatorname {erfc} \left({\sqrt {\frac {E_{b}}{N_{0}}}}\right)}$
where ${\displaystyle E_{b}}$ is the energy per one bit, ${\displaystyle N_{0}}$ is the noise spectral density, ${\displaystyle Q(*)}$ denotes the Marcum Q-function and ${\displaystyle \operatorname {erfc} }$ denotes the complementary error function.
## Gaussian minimum-shift keying
Power spectral densities of the MSK and GMSK. Note that the increasing of time-bandwidth ${\displaystyle BT}$ negatively influences bit-error-rate performance due to increasing of the ISI.
In digital communication, Gaussian minimum shift keying or GMSK is a continuous-phase frequency-shift keying modulation scheme.
GMSK is similar to standard minimum-shift keying (MSK); however, the digital data stream is first shaped with a Gaussian filter before being applied to a frequency modulator, and typically has much narrower phase shift angles than most MSK modulation systems. This has the advantage of reducing sideband power, which in turn reduces out-of-band interference between signal carriers in adjacent frequency channels.[6]
However, the Gaussian filter increases the modulation memory in the system and causes intersymbol interference, making it more difficult to differentiate between different transmitted data values and requiring more complex channel equalization algorithms such as an adaptive equalizer at the receiver. GMSK has high spectral efficiency, but it needs a higher power level than QPSK, for instance, in order to reliably transmit the same amount of data.
GMSK is most notably used in the Global System for Mobile Communications (GSM) and the satellite communications,[7][8] e.g. in the Automatic Identification System (AIS) for maritime navigation. | 2020-01-18 07:28:46 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 20, "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, "math_score": 0.8384430408477783, "perplexity": 1208.8773069879414}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-05/segments/1579250592261.1/warc/CC-MAIN-20200118052321-20200118080321-00061.warc.gz"} |
http://umj.imath.kiev.ua/authors/name/?lang=en&author_id=4403 | 2019
Том 71
№ 11
# Zubei E. V.
Articles: 1
Article (Russian)
### On the solvability of a finite group with $S$-seminormal Schmidt subgroups
Ukr. Mat. Zh. - 2018. - 70, № 11. - pp. 1511-1518
A finite nonnilpotent group is called a Schmidt group if all its proper subgroups are nilpotent. A subgroup $A$ is called $S$-seminormal (or $SS$-permutable) in a finite group $G$ if there is a subgroup B such that $G = AB$ and $A$ is permutable with every Sylow subgroup of B. We establish the criteria of solvability and $\pi$ -solvability of finite groups in which some Schmidt subgroups are $S$-seminormal. In particular, we prove the solvability of a finite group in which all supersoluble Schmidt subgroups of even order are $S$-seminormal. | 2020-01-27 12:41:44 | {"extraction_info": {"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, "math_score": 0.4466766119003296, "perplexity": 1028.8944746833574}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-05/segments/1579251700675.78/warc/CC-MAIN-20200127112805-20200127142805-00063.warc.gz"} |
http://docs.itascacg.com/3dec700/common/docproject/source/manual/program_guide/mechanics/plotting/views/plotitems/items/attributes/std_map.html | # Map
Map
Transform the orientation, location, or scale of the plot item according to the settings provided. This attribute allows multiple, differently visualized instances of a plot item to be rendered adjacent to one another, for instance.
Axis: Specify a coordinate transformation to use assuming a normal order of {XY in 2D; XYZ in 3D} (XZY, for instance, indicates that the Y and Z coordinates are swapped).
Translate: Translate the rendered objects’ {$$x$$ and $$y$$ in 2D; $$x$$, $$y$$, and $$z$$ in 3D} positions by the specified model units.
Scale: Set the plot item size to the specified percentage of the program window’s size. | 2022-12-06 20:43:53 | {"extraction_info": {"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, "math_score": 0.813217043876648, "perplexity": 2433.999561232324}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446711114.3/warc/CC-MAIN-20221206192947-20221206222947-00576.warc.gz"} |
https://math.stackexchange.com/questions/2025685/alternating-series-with-sin | # Alternating series with sin
I have this alternating series: $$\sum_{n=1}^{\infty}\dfrac{(-1)^n}{n+2\sin n}$$. Leibniz test and the absolute convergence didn't work. Neither did the divergence test. When showing that $a_n=\dfrac{1}{n+2\sin n}$ is decreasing (Leibniz test) I took a function, made it's derivative and arrived nowhere. Thank you for your help!
• Write it as $$\sum_{n = 1}^\infty (-1)^n\Biggl(\frac{1}{n} + \biggl( \frac{1}{n + 2\sin n} - \frac{1}{n}\biggr)\Biggr).$$ – Daniel Fischer Nov 22 '16 at 12:33
• @Daniel Fischer Applying Leibniz, showing that $\dfrac{1}{n+2\sin n}-\dfrac{1}{n}$ is decreasing is not very handy. – Denis Nichita Nov 22 '16 at 12:59
• The idea is to apply Leibniz' criterion to $\frac{(-1)^{n}}{n}$, which is pretty trivial. And $\sum (-1)^n\bigl(\frac{1}{n+2\sin n} - \frac{1}{n}\bigr)$ is absolutely convergent. Which is quite easy to see. – Daniel Fischer Nov 22 '16 at 13:01
Take an even $n$, then
$$\frac{1}{n+2\sin n} - \frac{1}{n+1 + 2 \sin(n+1)} =\frac{1 + 2 \sin(n+1) - 2\sin n}{(n+2\sin n)(n+1 + 2 \sin(n+1))},$$ which gives you an estimation for $n\ge 3$.
$$\left| \frac{1}{n+2\sin n} - \frac{1}{n+1 + 2 \sin(n+1)}\right| \le\frac{5}{(n-2)(n-1)}$$ The right-hand side behaves like $n^{-2}$, hence the series converges.
• But you've shown that $b_n=|a_n-a_{n+1}|$ converges...does this mean that $a_n$ converges? I think I'm missing sth – Denis Nichita Nov 22 '16 at 12:26
• First, I've shown that $b_k = |a_{2k}\mathbf{+}a_{2k+1}|$ converges. Since absolute convergence implies convergence, I've also shown that $c_k = a_{2k}\mathbf{+}a_{2k+1}$ converges, too. – TZakrevskiy Nov 22 '16 at 12:34
• Both of you are dropping the $\sum$ signs. – zhw. Nov 22 '16 at 12:35
• @TZakrevskiy I agree. And if $\sum c_k$ converges, does it imply that $\sum a_k$ also converges? – Denis Nichita Nov 22 '16 at 13:01 | 2020-01-21 20:50:50 | {"extraction_info": {"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, "math_score": 0.8266253471374512, "perplexity": 565.3353949890245}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-05/segments/1579250605075.24/warc/CC-MAIN-20200121192553-20200121221553-00315.warc.gz"} |
https://csyhua.github.io/csyhua/hpcplus.htm | HPC Plus 学术论坛
2015全国高性能计算学术年会（HPC China 2015）将于2015年11月10日-12日在无锡举行，大会由中国计算机学会（CCF）主办，中国计算机学会高性能计算专业委员会、江南大学联合承办。大会主题为“根植中国芯 超算强国梦”。
CCF 高专委秘书处经过商议，HPC CHINA 2015上举办《HPC Plus》主题论坛，邀请2014-2015年度发表在CCF A类期刊或会议上的高性能计算专委会委员的论文作者在《HPC Plus》主题论坛上做学术报告，代表HPC领域这一年最好的学术成果。
时间安排 报告题目 报告人 14:00-14:25 yaSpMV Yet Another SpMV Framework on GPUs 颜深根,SenseTime 14:25-14:50 Large-scale Neo-heterogeneous Programming and Optimization of SNP Detection on Tianhe-2 彭绍亮，国防科技大学 14:50-15:15 以路径为中心的大规模图数据处理系统PathGraph 袁平鹏，华中科技大学 15:15-15:35 Robust Structured Subspace Learning for Data Representation 李泽超, 南京理工大学 15:35-15:50 茶歇 15:50-16:15 针对GPU的静态和动态结合的缓存旁路技术 梁云，北京大学 16:15-16:40 DwarfCode: A Performance Prediction Tool for Parallel Applications 张伟哲，哈尔滨工业大学 16:40-17:05 When HPC Meets Big Data: Emerging HPC Technologies for High-Performance Data Management Systems 何丙胜，南洋理工大学 17:05-17:30 Enabling Renewable Energy Powered Sustainable High-Performance Computing 李超，上海交通大学 17:30-17:55 IaaS平台中工作流计算任务的多目标调度 朱昭萌，南京理工大学
华宇是华中科技大学副教授、博士生导师，IEEE和中国计算机学会的高级会员，计算机学会学术工委通讯委员、高性能计算、信息存储和体系结构专委委员。曾在美国内布拉斯加大学林肯分校做博士后研究工作。主要研究内容包括海量存储系统中元数据的语义管理方法，数据去重机制和近似存储系统体系结构等方面。在国际期刊TC、TPDS和国际会议USENIX FAST、USENIX ATC、INFOCOM、SC、ICDCS、MSST等上发表多篇学术论文，发表的学术论文被引用超过500次。在RTSS、INFOCOM、ICDCS、ICPP、IWQoS等30多个国际会议上担任程序委员会或组委会委员；是国际期刊FCS、JCN等的编辑，TC、TPDS、TCC、TVT、JSAC、VLDB Journal等的审稿人。主持和参加多项国家自然科学基金、973、 863计划重大项目和教育部创新团队项目等,是湖北省优秀硕士学位论文指导教师，曾获得中国电子学会电子信息科学技术二等奖。
yaSpMV Yet Another SpMV Framework on GPUs
SpMV is a key linear algebra algorithm and has been widely used in many important application domains. As a result, numerous attempts have been made to optimize SpMV on GPUs to leverage their massive computational throughput. Although the previous work has shown impressive progress, load imbalance and high memory bandwidth remain the critical performance bottlenecks for SpMV. In this paper, we present our novel solutions to these problems. First, we devise a new SpMV format, called blocked compressed common coordinate (BCCOO), which uses bit flags to store the row indices in a blocked common coordinate (COO) format so as to alleviate the bandwidth problem. We further improve this format by partitioning the matrix into vertical slices to enhance the cache hit rates when accessing the vector to be multiplied. Second, we revisit the segmented scan approach for SpMV to address the load imbalance problem. We propose a highly efficient matrix-based segmented sum/scan for SpMV and further improve it by eliminating global synchronization. Then, we introduce an auto-tuning framework to choose optimization parameters based on the characteristics of input sparse matrices and target hardware platforms. Our experimental results on GTX680 GPUs and GTX480 GPUs show that our proposed framework achieves significant performance improvement over the vendor tuned CUSPARSE V5.0 (up to 229% and 65% on average on GTX680 GPUs, up to 150% and 42% on average on GTX480 GPUs) and some most recently proposed schemes (e.g., up to 195% and 70% on average over clSpMV on GTX680 GPUs, up to 162% and 40% on average over clSpMV on GTX480 GPUs).
Dr. Shaoliang Peng is a professor in National University of Defense Technology 国防科技大学 (NUDT, Changsha, China) and an adjunct professor of BGI
Large-scale Neo-heterogeneous Programming and Optimization of SNP Detection on Tianhe-2
(天河2上SNP检测的大规模微异构编程和并行优化)
SNP detection is a fundamental procedure in genome analysis. A popular SNP detection tool SOAPsnp can take more than one week to analyze one human genome with a 20-fold coverage. To improve the effciency, we developed mSNP, a parallel version of SOAPsnp. mSNP utilizes CPU cooperated with Intelr Xeon PhiTM for large-scale SNP detection. Firstly, we redesigned the key data structure of SOAPsnp, which significantly reduces the overhead of memory operations. Secondly, we devised a coordinated parallel framework, in which CPU collaborates with Xeon Phi for higher hardware utilization. Thirdly, we proposed a read-based window division strategy to improve throughput and parallel scale on multiple nodes. To the best of our knowledge, mSNP is the first SNP detection tool empowered by Xeon Phi. We achieved a 45x speedup on a single node of Tianhe-2, without any loss in precision. Moreover, mSNP showed promising scalability on 8192 nodes on Tianhe-2.
Robust Structured Subspace Learning for Data Representation
To uncover an appropriate latent subspace for data representation, in this paper we propose a novel Robust Structured Subspace Learning (RSSL) algorithm by integrating image understanding and feature learning into a joint learning framework. The learned subspace is adopted as an intermediate space to reduce the semantic gap between the low-level visual features and the high-level semantics. To guarantee the subspace to be compact and discriminative, the intrinsic geometric structure of data, and the local and global structural consistencies over labels are exploited simultaneously in the proposed algorithm. Besides, we adopt the $\ell_{2,1}$-norm for the formulations of loss function and regularization respectively to make our algorithm robust to the outliers and noise. An efficient algorithm is designed to solve the proposed optimization problem. It is noted that the proposed framework is a general one which can leverage several well-known algorithms as special cases and elucidate their intrinsic relationships. To validate the effectiveness of the proposed method, extensive experiments are conducted on diversity datasets for different image understanding tasks, i.e., image tagging, clustering, and classification, and the more encouraging results are achieved compared with some state-of-the-arts approaches.
Coordinated Static and Dynamic Cache Bypassing for GPUs
The massive parallel architecture enables graphics processing units (GPUs) to boost performance for a wide range of applications. Recently, to broaden the scope of applications that can be accelerated by GPUs, GPU vendors have used caches in conjunction with scratchpad memory as on-chip memory in the new generations of GPUs. Unfortunately, GPU caches face many performance challenges that arise due to excessive thread contention for cache resource.
Cache bypassing, where memory requests can selectively bypass the cache, is one solution that can help to mitigate the cache resource contention problem. In this work, we propose coordinated static and dynamic cache bypassing to improve application performance. At compile-time, we identify the global loads that indicate strong preferences for caching or bypassing through profiling. For the rest global loads, our dynamic cache bypassing has the flexibility to cache only a fraction of threads. Our coordinated static and dynamic cache bypassing technique achieves up to 2.28X (average 1.32X) performance speedup for a variety of GPU applications.
DwarfCode: A Performance Prediction Tool for Parallel Applications
We present DwarfCode, a performance prediction tool for MPI applications on diverse computing platforms. The goal is to accurately predict the running time of applications for task scheduling and job migration. First, DwarfCode collects the execution traces to record the computing and communication events. Then, it merges the traces from different processes into a single trace. After that, DwarfCode identifies and compresses the repeating patterns in the final trace to shrink the size of the events. Finally, a dwarf code is generated to mimic the original program behavior. This smaller running benchmark is replayed in the target platform to predict the performance of the original application. In order to generate such a benchmark, two major challenges are to reduce the time complexity of trace merging and repeat compression algorithms. We propose an O(mpn) trace merging algorithm to combine the traces generated by separate MPI processes , where m denotes the upper bound of tracing distance, p denotes the number of processes, and n denotes the maximum of event numbers of all the traces. More importantly, we put forward a novel repeat compression algorithm, whose time complexity is O(nlogn). Experimental results show that DwarfCode can accurately predict the running time of MPI applications. The error rate is below 10% for compute and communication intensive applications. This toolkit has been released for free download as a GNU General Public License v3 software.
Dr. Bingsheng He is currently an Associate Professor at School of Computer Engineering, Nanyang Technological University, Singapore. Before that, he held a research position in the System Research group of Microsoft Research Asia (2008-2010), where his major research was building high performance cloud computing systems for Microsoft. He got the Bachelor degree in Shanghai Jiao Tong University (1999-2003), and the Ph.D. degree in Hong Kong University of Science & Technology (2003-2008). His current research interests include cloud computing, database systems and high performance computing. His papers are published in prestigious international journals (such as ACM TODS and IEEE TKDE/TPDS/TC) and proceedings (such as ACM SIGMOD, VLDB/PVLDB, ACM/IEEE SuperComputing, ACM HPDC, and ACM SoCC). He has been awarded with the IBM Ph.D. fellowship (2007-2008) and with NVIDIA Academic Partnership (2010-2011). Since 2010, he has (co-)chaired a number of international conferences and workshops, including CloudCom 2014/2015 and HardBD2015. He has served in editor board of international journals, including IEEE Transactions on Cloud Computing (IEEE TCC) and IEEE Transactions on Parallel and Distributed Systems (IEEE TPDS).
When HPC Meets Big Data: Emerging HPC Technologies for High-Performance Data Management Systems
Big data has become a buzz word. Among various big-data challenges, high performance is a must, not an option. We are facing the challenges (and also opportunities) at all levels ranging from sophisticated algorithms and procedures to mine the gold from massive data to high-performance computing (HPC) techniques and systems to get the useful data in time. Our research has been on the system design and implementation of HPC technologies as weapons to address the performance requirement of data management systems. Interestingly, we have also observed the interplay between HPC architectures and (big) data management systems. In this talk, I will present our recent research efforts in developing high performance data management systems with GPUs and on Cloud. Finally, I will outline our research agenda. More details about our research can be found at http://pdcc.ntu.edu.sg/xtra/.
Enabling Renewable Energy Powered Sustainable High-Performance Computing
Multi-objective Scheduling of Workflow Applications in IaaS
IaaS平台中工作流计算任务的多目标调度
Cloud computing provides promising platforms for executing large applications with enormous computational resources to offer on demand. In IaaS model, users are charged based on their usage of provisioned VMs and the required QoS specifications. Although there are many existing workflow scheduling algorithms in traditional distributed or heterogeneous computing environments, they have difficulties in being directly applied to IaaS since IaaS differs from traditional environments by its service-based resource management and pay-per-use pricing strategies. In this talk, we will highlight such difficulties, and formulate the multi-objective workflow scheduling problem in IaaS. Also, we will present our recent research in the area of scheduling workflow applications on the IaaS platforms, including two novel evolutionary approaches for addressing the QoS-optimization scheduling problem and a list-based heuristic for addressing the Budget-constrained performance-effective scheduling problem. Experimental comparisons between the proposed algorithms and state-of-the-art algorithms will be presented and discussed. | 2019-02-21 08:10:58 | {"extraction_info": {"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, "math_score": 0.3590797781944275, "perplexity": 2683.9785549319263}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-09/segments/1550247503249.58/warc/CC-MAIN-20190221071502-20190221093502-00469.warc.gz"} |
http://hyperspacewiki.org/index.php/The_Houston_Problem_Book_Problems_51-100 | # The Houston Problem Book Problems 51-100
## Problems
Problem 51 (HPB): Is it true that each polyhedron is a the weakly confluent image of a polyhedron whose fundamental group is trivial?
Answer: False, $T2$. (J Grispolakis 19 March 1979)
Problem 52 (HPB): Are strongly regular curves inverse limits of connected graphs with monotone simplicial bonding maps? (Asked by B.B. Epps 19 September 1973)
Answer: No (E.D. Tymchatyn 27 October 1974)
Problem 53 (HPB): If $X$ is the inverse limit of connected graphs with monotone simplicial retractions as bonding maps, is $X$ the weakly confluent image of a dendrite?
Answer: Yes (J. Grispolakis and E.D. Tymchatyn 19 March 1979)
Problem 54 (HPB): Is it true that $X$ is the inverse limit of connected graphs with monotone simplicial retractions as bonding maps if and only if $X$ is locally connected and each cyclic element of $X$ is a graph? (Asked by B.B. Epps, 19 September 1973)
Problem 55 (HPB): Does each continuous mapping of a continuum onto a chainable continuum have property F? (Asked by A. Lelek 26 September 1973)
Answer: No (J.B. Jugate 16 June 1979)
Problem 56 (HPB): Does each weakly confluent mapping of a continuum onto an irreducible continuum have property F? (Asked by A. Lelek 26 September 1973)
Answer: No (J.B. Jugate 16 June 1979)
Problem 57 (HPB): Is it true that each uniformly pathwise connected continuum is uniformly arcwise connected? (Asked by W. Kuperberg 26 September 1973)
Problem 58 (HPB): Suppose $f$ is a continuous mapping of a chainable continuum $X$ onto a nonchainable continuum $Y$. Does there exist a subcontinuum $X'$ of $X$ mapped onto $Y$ under $f$ such that $f \big|_{X'}$ is not weakly confluent? (Asked by A. Lelek 25 January 1981)
Answer: No (T. Maćkowiak and E.D. Tymchatyn 25 January 1981)
Problem 59 (HPB): Does a continuum with zero surjective span have zero span? (Asked by A. Lelek 7 November 1973)
Problem 60 (HPB): Is it true that if $Y=G_1 \cup G_2$ where $G_1$ and $G_2$ are open and $f\big|_{f^{-1}(\overline{G_i})}$ is an MO-mapping for $i=1,2$ then $f$ is an MO-mapping? (Asked by A. Lelek 21 November 1973)
Answer: Spaces are compact metric (A. Lelek 21 November 1973)
Problem 61 (HPB): Suppose $\mathcal{E}$ is the class of all finite one point unions of circles. Is it true that if $X$ and $Y$ are shape-irreducible continua possessing the same shape and there exist curves $C$ and $C'$ in $\mathcal{E}$ such that $X$ is $C$-like and $Y$ is $C'$-like, then there exists a $C$ in $\mathcal{E}$ such that both $X$ and $Y$ are $C$-like? (Asked by A. Lelek 5 December 1973)
Answer: Trivially yes, but there even exists a shape-irreducible figure-eight-like continuum with the shape of a circle that is not circle-like. (J. Segal and S. Spiez 7 May 1986)
Problem 62 (HPB): Are all shapes of polyhedra stable? (Asked by W. Kuperberg 3 April 1974)
Problem 63 (HPB): Given a polyhedron $P$, does there exist a polyhedron $Q$ of the same shape as $P$ and shape stable? (Asked by W. Kuperberg 3 April 1974)
Problem 64 (HPB): Is each uniformly path-connected continuum $g$-contractible? (Asked by D. Bellamy 3 April 1974)
Problem 65 (HPB): Is it true that, for each uniformly path-connected continuum $X$, there exists a compact metric space $Y$ and two mappings $f$, from $X$ onto the cone over $Y$, and $g$, from the cone over $Y$ onto $X$? (Asked by D. Bellamy 3 April 1974)
Answer: No (J. Prajs 1 March 1989)
Problem 66 (HPB): If $C$ is a connected Borel subset of a finitely Suslinean continuum $X$, is $C$ arcwise connected? (Asked by E.D. Tymchatyn 26 September 1974)
Answer: Yes for subsets of regular curves (D.H. Fremlin 12 April 1990)
Problem 67 (HPB): Suppose $X$ is a continuum such that, if $C$ is a subcontinuum of $X$ then the set of all local separating points of $C$ is not the union of countably many closed disjoint proper subsets of $C$. Then, is every connected subset of $X$ arcwise connected? (Asked by E.D. Tymchatyn 26 September 1974)
Problem 68 (HPB): If for each subcontinuum $C$ of the continuum $X$, the set of all local separating points of $C$ is connected, then is every connected subset of $X$ arcwise connected? (Asked by E.D. Tymchatyn 3 October 1974)
Problem 69 (HPB): Do hereditarily indecomposable tree-like continua have the fixed point property? (Asked by B. Knaster 21 November 1974)
Problem 70 (HPB): Do uniquely $\lambda$-connected (or uniquely $\delta$-connected) continua have the fixed point property for homeomorphisms? (Asked by A. Lelek 21 November 1974)
Problem 71 (HPB): Suppose that $X$ is a continuum such that, if $\epsilon > 0$, $C_1,C_2,\ldots$ are mutually separated connected sets of diameter greater than $\epsilon$, and $C_1 \cup C_2 \ldots$ is connected, then there exist mutually exclusive arcs $A_1,A_2,\ldots$ such that, for some subsequence $D_1,D_2,\ldots$ of $C_1,C_2,\ldots$ and each $i$, $A_i$ is a subset of $D_i$, and $\mathrm{diam}(A_i)>\epsilon$. If $X$ locally connected? (Asked by E.D. Tymchatyn 27 January 1975)
Problem 72 (HPB): Suppose $X$ is a continuum such that each $\sigma$-connected subset of $X$ is a semi-continuum. Is $X$ locally connected? (J. Grispolakis 27 January 1975)
Problem 73 (HPB): Suppose $X$ is a continuum such that each $\sigma$-connected $F_\sigma$ subset of $X$ is a semi-continuum. Is $X$ semi-aposyndetic? (Asked by J. Grispolakis 27 January 1975)
Problem 74 (HPB): Let $f$ be a perfect pseudo-confluent mapping of a hereditarily normal, hereditarily locally connected and hereditarily $\sigma$-connected space onto a complete metric space $Y$. Is $Y$ hereditarily $\sigma$-connected? (Asked by J. Grispolakis 3 February 1975)
Answer: The answer is yes if $X$ is also a $(Q=C)$-space. (J. Grispolakis 3 February 1975)
Problem 75 (HPB): Suppose $X$ is a separable metric space which possesses an open basis $\mathscr{B}$ such that, theset $X \setminus G$ is the union of a collection of closed open subsets of $X$. Is $X$ embeddable in a hereditarily locally connected space? (Asked by E.D. Tymchatyn 3 March 1975)
Answer: Yes (L.G. Oversteegen and E.D. Tymchatyn 18 August 1992)
Problem 76 (HPB): Suppose $X$ is a separable metric space such that, if $A$ and $B$ are connected subsets of $X$, then $A \cap B$ is connected. Is it true that if $C$ is a set of non-cut points of $X$, then $X \setminus C$ is connected? (Asked by L.E. Ward 3 March 1975)
Problem 77 (HPB): Is it true that a continuum $X$ is regular if and only if every infinite sequence of mutually disjoint connected subsets of $X$ is a null sequence? (Asked by E.D. Tymchatyn 3 March 1975)
Problem 78 (HPB): Is it true that, for each decomposition of a finitely Suslinean continuum into countably many disjoint connected sets, at least one of them must be rim-compact? (Asked by T. Nishiura 3 March 1975)
Answer: No (T.D. Tymchatyn 30 September 1981)
Problem 79 (HPB): Is it true that no biconnected set with a dispersion point can be embedded into a rational continuum? (Asked by J. Grispolakis 3 March 1975)
Problem 80 (HPB): Is it true that a separable metric space is embeddable into a rational continuum if and only if it posseses an open basis whose elements have scattered boundaries? (Asked by E.D. Tymchatyn 3 March 1975)
Answer: Yes (E.D. Tymchatyn 30 September 1981)
Problem 81 (HPB): Is it true that each continuum of span zero chainable? (Asked by H. Cook and A. Lelek 15 May 1975)
Problem 82 (HPB): Is it true that each continuum of surjective semi-span zero is arc-like? (Asked by A. Lelek 15 May 1975)
Problem 83 (HPB): Suppose $X$ is a connected metric space.
• Is the span of $X$ less than or equal to twice the surjective span of $X$?
• Is the semi-span of $X$ less than or equal to twice the surjective semi-span of $X$?
• Is the surjective semi-span of $X$ less than or equal to twice the surjective span of $X$?
• If $T$ is a simple triod, is the surjective span of $T$ equal to the surjective semi-span of $T$? (Asked by A. Lelek 15 May 1975)
Problem 84 (HPB): Is the confluent image of an arc-like continuum arc-like? (Asked by A. Lelek 21 October 1975)
Problem 85 (HPB): If $f$ is a confluent mapping of an acyclic (or tree-like or arc-like) continuum onto a continuum $Y$, is $f \times f$ confluent? (Asked by A. Lelek 21 October 1975)
Problem 86 (HPB): Do confluent maps of continua preserve span zero? (Asked by H. Cook and A. Lelek 21 October 1975)
Problem 87 (HPB): Is every regularly submetrizable Moore space completely regular? (Asked by H. Cook 13 October 1976)
Problem 88 (HPB): Is every homogeneous continuum bihomogeneous? (Asked by B. Fitzpatrick 27 October 1976)
Answer: No (K. Kuperberg 2 February 1988)
Problem 89 (HPB): Does there exist a noncombinatorial triangulation of the $4$-sphere? (Asked by C.E. Burgess 12 October 1977)
Problem 90 (HPB): Is there a hereditarily equivalent continuum other than an arc or a pseudo-arc? (Asked by H. Cook 2 November 1977)
Problem 91 (HPB): Do hereditarily equivalent continua have span zero? (Asked by H. Cook 2 November 1977)
Problem 92 (HPB): If $M$ is a continuum with positive span such that each of its proper subcontinua has span zero, does every nondegenerate monotone continuous image of $M$ have positive span? (Asked by H. Cook 2 November 1977)
Answer: No (J.F. Davis and W.T. Ingram 30 April 1986; Fund. Math. 131 (1988), 13-24)
Problem 93 (HPB): Does there exist a homogeneous tree-like continuum of positive span? (Asked by W.T. Ingram 8 February 1978)
Answer: Not in the plane (L.G. Oversteegen and E.D. Tymchatyn 23 July 1980)
Problem 94 (HPB): If $M$ is a plane continuum with no weak cut point, is $M$ $\lambda$-connected? (Asked by C.L. Hagopian 12 April 1978)
Answer: Yes, but there exists a counterexample in the $3$-dimensional Euclidean space (C.L. Hagopian 1 April 1979)
Problem 95 (HPB): Is the countable product of $\lambda$-connected continua $\lambda$-connected? (Asked by C.L. Hagopian 12 April 1978)
Answer: Yes, but it is unknown if the product of any two continua is $\lambda$-connected (C.L. Hagopian 7 April 1986)
Problem 96 (HPB): Is the image under a local homeomorphism of a $\delta$-connected continuum also $\delta$-connected? (Asked by C.L. Hagopian 12 April 1978)
Problem 97 (HPB): Can it be proven, without extraordinary logical assumptions, that the plane is not the sum of fewer than $\mathfrak{c}$ mutually exclusive continua? (Asked by H. Cook 19 April 1978)
Answer: The plane is not the sum of fewer than $\mathfrak{c}$ mutually exclusive continua each of which is either Suslinean or locally connected (H. Cook 19 April 1978)
Problem 98 (HPB): Is it true that if $(X,\tau_1)$ is normal, $(X,\tau_2)$ is compact, $\tau_2$ is a subcollection of $\tau_1$, $\mathrm{ind}(X,\tau_1)=0$, and $\mathrm{ind}(X,\tau_2)>0$, then $(X,\tau_1)$ fails to be a Hurewicz space? (Asked by A. Lelek 13 September 1978)
Problem 100 (HPB): Suppose $f$ is a light open mapping from a continuum $M$ onto a continuum $N$, $B$ is a smooth dendroid lying in $N$, and $x$ is a point of $f^{-1}(B)$. Does there exist a smooth dendroid $A$ in $M$ such that $x$ is in $A$ and $f \big|_A$ is a homeomorphism? (Asked by J.B. Fugate 18 October 1978) | 2022-05-19 02:52:03 | {"extraction_info": {"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, "math_score": 0.7789267897605896, "perplexity": 2003.1340200086695}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662522741.25/warc/CC-MAIN-20220519010618-20220519040618-00005.warc.gz"} |
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The Citing articles tool gives a list of articles citing the current article.
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DOI: 10.1016/j.jmaa.2010.10.033 | 2021-09-20 11:42:57 | {"extraction_info": {"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, "math_score": 0.342130571603775, "perplexity": 10567.086111046132}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-39/segments/1631780057036.89/warc/CC-MAIN-20210920101029-20210920131029-00593.warc.gz"} |
http://inperc.com/wiki/index.php?title=Determinants_of_linear_operators | This site contains: mathematics courses and book; covers: image analysis, data analysis, and discrete modelling; provides: image analysis software. Created and run by Peter Saveliev.
# Determinants of linear operators
## 1 How to determine that a matrix is invertible?
Given a matrix or a linear operator $A$, it is either singular or non-singular: $${\rm ker \hspace{3pt}}A \neq 0 {\rm \hspace{3pt} or \hspace{3pt}} {\rm ker \hspace{3pt}}A=0$$
If the kernel is zero, then $A$ is invertible (as 1-1).
Recall, matrix $A$ is singular when its column vectors are linearly dependent ($i^{\rm th}$ column of $A = A(e_i)$).
The goal is to find a function that determines whether $A$ is singular. It is called the determinant.
Specifically, we want:
$${\rm det \hspace{3pt}}A = 0 {\rm \hspace{3pt} iff \hspace{3pt}} A {\rm \hspace{3pt} is \hspace{3pt} singular}.$$
Start with dimension $2$.
$A = \left[ \begin{array}{} a & b \\ c & d \end{array} \right]$ is singular when $\left[ \begin{array}{} a \\ c \end{array} \right]$ is a multiple of $\left[ \begin{array}{} b \\ d \end{array} \right]$
Let's consider that:
$$\left[ \begin{array}{} a \\ c \end{array} \right] = x\left[ \begin{array}{} b \\ d \end{array} \right] \rightarrow \begin{array}{} a = bx \\ c = dx \\ \end{array}$$
So $x$ has to exist.
Then, $$x=\frac {a}{b} = \frac{c}{d}.$$ That the condition. (Is there a division by 0 here?)
Rewrite as: $ad-bc=0 \longleftrightarrow A$ is singular, and we want $\det A =0$.
This suggests that
$${\rm det \hspace{3pt}}A = ad-bc,$$
Let's make this the definition.
Verify:
Theorem: A $2\times 2$ matrix $A$ is singular iff $\det A=0$.
Proof. ($\Rightarrow$) Suppose $\left[ \begin{array}{} a & b \\ c & d \end{array} \right]$ is singular, then $\left[ \begin{array}{} a \\ c \end{array} \right] = x \left[ \begin{array}{} b \\ d \end{array} \right]$, then $a=xb, c=xd$, then substitute:
$${\rm det \hspace{3pt}}A = ad-bc = (xb)d - b (xd)=0.$$
($\Leftarrow$) Suppose $ad-bc=0$, then let's find $x$, the multiple.
• Case 1: assume $b \neq 0$, then choose $x = \frac{a}{b}$. Then
$$\begin{array}{} xb &= \frac{a}{b}b &= a \\ xd &= \frac{a}{b}d &= \frac{ad}{b} = \frac{bc}{b} = c. \end{array}$$
So $x \left[ \begin{array}{} b \\ d \end{array} \right] = \left[ \begin{array}{} a \\ c \end{array} \right].$
• Case 2: assume $a \neq 0$, etc.
## 2 Observations
We defined $\det A$ with requirement that $\det A = 0$ iff $A$ is singular.
But $\det A$ could be $k(ad-bc)$, $k \neq 0$. Why $ad-bc$?
Because...
• Observation 1: $\det I_2=1$ (normalization property)
Notice this similarity:
$= ad-bc$,
i.e., "cross multiplication" (just like with differentiation: , $(fg)'=f'g+g'f$).
So,
• Observation 2: Each entry of $A$ appears only once.
What if we interchange rows or columns?
$$\det \left[ \begin{array}{} c & d \\ a & b \end{array} \right] = cb-ad = - \det \left[ \begin{array}{} a & b \\ c & d \end{array} \right].$$
Then the sign changes. So
• Observation 3: $\det A=0$ is preserved under this elementary row operation.
Moreover, $\det A$ is preserved up to a sign! (later)
• Observation 4: If $A$ has a zero row or column, then $\det A=0$.
To be expected -- we are talking about linear independence!
• Observation 5: $\det A^T= \det A$.
$$\det \left[ \begin{array}{} a & c \\ b & d \end{array} \right] = ad-bc$$
• Observation 6: Signs of the terms alternate: $+ad-bc$.
• Observation 7: $\det A \colon {\bf M}(2,2) \rightarrow {\bf R}$ is linear... NOT!
$$\begin{array}{} \det 3\left[ \begin{array}{} a & b \\ c & d \end{array} \right] &= \det \left[ \begin{array}{} 3a & 3b \\ 3c & 3d \end{array} \right] \\ &= 3a3d-3b3c \\ &= 9(ad-bc) \\ &= 9 \det \left[ \begin{array}{} a & b \\ c & d \end{array} \right] \end{array}$$
not linear, but quadratic. Well, not everything in linear algebra has to be linear...
## 3 What about dimension $3$?
Let's try to step up from dimension 2 with this simple matrix:
$$B = \left[ \begin{array}{} e & 0 & 0 \\ 0 & a & b \\ 0 & c & d \end{array} \right]$$
We, again, want to define ${\rm det \hspace{3pt}}B$ so that ${\rm det \hspace{3pt}}B=0$ if and only if $B$ is singular, i.e., the columns are linearly dependent.
Question: What does the value of $e$ do to the linear dependence?
• Case 1, $e=0 \rightarrow B$ is singular. So, $e=0 \rightarrow {\bf det \hspace{3pt}}B=0$.
• Case 2, $e \neq 0 \rightarrow$ the vectors are:
$$\left[ \begin{array}{} e & 0 & 0 \\ 0 & a & b \\ 0 & c & d \end{array} \right]$$
and
$v_1=\left[ \begin{array}{} e \\ 0 \\ 0 \end{array} \right]$ is not a linear combination of the other two $v_2,v_3$.
So, we only need to consider the linear independence of those two, $v_2,v_3$!
Observe that $v_2,v_3$ are linearly independent if and only if ${\rm det \hspace{3pt}}A \neq 0$, where
$$A = \left[ \begin{array}{} a & b \\ c & d \end{array} \right]$$
Two cases together: $e=0$ or ${\rm det \hspace{3pt}}A=0 \longleftrightarrow B$ is singular.
So it makes sense to define
$${\rm det \hspace{3pt}}B = e \cdot {\rm det \hspace{3pt}}A$$
With that we have: ${\rm det \hspace{3pt}}B=0$ if and only if $B$ is singular.
Let's review the observations above...
(1) $\det I_3 = 1$.
(4), (5) still hold.
(7) still not linear: $\det (2B) = 8 \det B.$
It's cubic.
So far so good...
Now let's give the definition of ${\rm det}$ in dimension three.
## 4 Definition in dimension 3 via that in dimension 2
We define via "expansion along the first row."
$A =$
$$\det A = a \det \left[ \begin{array}{} e & f \\ h & i \end{array} \right] - b \det \left[ \begin{array}{} d & f \\ g & i \end{array} \right] + c \det \left[ \begin{array}{} d & e \\ g & h \end{array} \right]$$
Observe that the first term is familiar from before.
This is the beginning of induction: from dimension 2 to dimension 3!
Expand farther using the formula for $2 \times 2$ determinant:
$$\begin{array}{} {\det} _1 A &= a(ei-fh) - b(di-fg)+c(dh-eg) \\ &= (aei+bfg+cdh)-(afh+bdi+ceg) \end{array}$$
Observe:
• each entry appears twice -- once with $+$, once with $-$.
Prior observation:
• in each term (in determinant) every column appears exactly once, as does every row.
The determinant helps us with problems from before.
Example: Given $(1,2,0),(-1,0,2),(2,-2,1)$. Are they linearly independent? We need just yes or no; no need to find the actual dependence.
Note: this is similar to discriminant that tells us how many solutions a quadratic equation has:
• $D<0 \rightarrow 0$ solutions,
• $D=0 \rightarrow 1$ solution,
• $D>0 \rightarrow 2$ solutions.
Consider: $$\begin{array}{} \det \left[ \begin{array}{} 1 & -1 & 2 \\ 2 & 0 & -2 \\ 3 & 2 & 1 \end{array} \right] &= 1 \cdot 0 \cdot 1 + (-1)(-2)3 + 2\cdot 2 \cdot 2 - 3 \cdot 0 \cdot 2 - 1(-2)2 - 2(-1)1 \\ &= 6 + 8 + 4 + 2 \\ &\neq 0 \end{array}$$
So, they are linearly independent.
Back to matrices.
Recall the definition, for:
$A = \left[ \begin{array}{} a & b & c \\ d & e & f \\ g & h & i \end{array} \right]$ of expansion along the first row:
(*) $$\begin{array}{} {\det} _1A &= a \det \left[ \begin{array}{} e & f \\ h & i \end{array} \right] - b \det \left[ \begin{array}{} d & f \\ g & i \end{array} \right]+ c \det \left[ \begin{array}{} d & e \\ g & h \end{array} \right] \\ &=(aei+bfg+cdh)-(afh+bdi+ceg) \end{array}$$
Let's try the expansion along the second row.
(**) $$\begin{array}{} {\det} _2A &= d \det \left[ \begin{array}{} b & c \\ h & i \end{array} \right] - e \det \left[ \begin{array}{} a & c \\ g & i \end{array} \right] + f \det \left[ \begin{array}{} a & b \\ g & h \end{array} \right] \\ &=d(bi-ha) - e(ai-cg) + f(ah-bg) \\ &=dbi+\ldots \end{array}$$
Let' try to match $(*)$ and $(**)$.
Wrong sign!
But it's wrong for all terms, fortunately!
To fix the formula, flip the signs,
$$\begin{array}{} {\det}_{2} A &= -d \det \left[ \begin{array}{} b & c \\ h & i \end{array} \right] + e \det \left[ \begin{array}{} a & c \\ g & i \end{array} \right] - f \det \left[ \begin{array}{} a & b \\ g & h \end{array} \right] \end{array}$$
Conclusion: the expansion formula depends on the row chosen.
But it's the same procedure:
• $3$ determinants, $2 \times 2$,
• with coefficients cut out row and column,
• signs alternate.
The difference is alternating signs: start with $-$ or $+$, depending on the row chosen.
## 5 Determinants of $n \times n$ matrices
First, $3 \times 3$ but with proper notation:
$$A = \left[ \begin{array}{} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{array} \right].$$
Then
Given an $n \times n$ matrix and two numbers, $r,k$ $1 \leq r,k \leq n$, then define $A_{rk}$ is the $(n-1) \times (n-1)$ matrix obtained by removing $r^{\rm th}$ row and $k^{\rm th}$ column from $A$.
So, take $a_{rk}$, the $(r,k)$-entry in $A$, and remove the row and the column that contain $a_{rk}$, that's $A_{rk}$.
Then, rewrite
$${\det} _1 A = a_{01}\det A_{11} - a_{12} \det A_{12}+a_{13} \det A_{13}.$$
Also,
$${\det} _2 A=-a_{21} \det A_{21}+a_{22} \det A_{22}-a_{23} \det A_{23}.$$
(Recall, ${\det} _1A= \det _2A$).
What about ${\det} _rA=?$
$${\det} _r A = \pm a_{r1} \det A_{r1}, \mp a_{r2} \det A_{r2}, \pm a_{r3} \det A_{r3}$$
Only the signs need to be determined:
$$=(-1)^?a_{r1} \det A_{r1}+(-1)^?a_{r2} \det A_{r2}+(-1)^?a_{r3} \det A_{r3}$$
The signs depend on
• the position, i.e., the column, and
• the row chosen.
$$=(-1)^{r+1}a_{r1} \det A_{r1}+(-1)^{r+2}a_{r2} \det A_{r2}+(-1)^{r+3}a_{r3} \det A_{r3}$$
Test $r=3$.
$${\det} _rA = \sum_{k=1}^3(-1)^{r+k}a_{rk} \det A_{rk}.$$
Fact: ${\rm det}A = {\det} _rA_i$ for all $r=1,2,3$.
What we get is this "inductive definition".
Definition: The determinant of an $n \times n$ matrix $A$ is defined as
$$\det A = \sum_{k=1}^n(-1)^{r+k}a_{rk} \det A_{rk},$$
where $1 \leq k \leq n$.
It is called the "expansion along $r^{\rm th}$ row".
Theorem: $\det A$ is independent of $r$.
This means that it's well defined (later).
Base: $n=1$, ${\rm det}[a]=a$.
Also for $n=2$ $A = \left[ \begin{array}{} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array} \right]$. Then
$$\begin{array}{} \det A &= \sum_{k=1}^2(-1)^{r+k}a_{rk} \det A_{rk} \\ &=(-1)^{1+1}a_{11} \det A_{11}+(-1)^{1+2}a_{12} \det A_{12} \\ &= a_{11} \det [a_{22}] - a_{12} \det [a_{21}] \\ &=a_{11}a_{22}-a_{12}a_{21}. \end{array}$$
## 6 Properties of determinants
• 1. $\det 0=0$. Proof: easy.
• 2. $A$ has a zero row, then $\det A=0$. Proof: Expand along that row, then all the coefficients are zeros in the sum.
• 3. $\det A^T= \det A$.
• 4. $A$ has two identical rows, then $\det A=0$.
• 5. $\det I_n=1$.
Proof: $$\begin{array}{} \det A &= 1 \cdot \det A_{11} + 0 \\ &= 1 \cdot 1 \cdot \det (A_{11})_{11} \end{array}$$
expand again, on and on.
This is proof by induction. Better: prove $\det I_1=1$ then $\det I_{n-1}=1$ implies $\det I_n=1$.
Homework: Define $V = \{(x_1,\ldots,x_n, \ldots) \colon x_n \rightarrow a$ for some $a \}$.
• 1. Define operations
$$\begin{array}{} x=(x_1,\ldots,x_n,\ldots) \\ y=(y_1,\ldots,y_n,\ldots) \end{array} \rightarrow x+y = (x_1+y_1,\ldots,x_n+y_n,\ldots)$$
Observation: without the $\ldots$, we would have $V={\bf R}^n$.
Property: $\det A^T= \det A$.
Test it...
For $2 \times 2$: $\left[ \begin{array}{} a & b \\ c & d \end{array} \right]^T = \left[ \begin{array}{} a & c \\ b & d \end{array} \right]$.
$\det = ad-bc$, same!
For $3 \times 3$: $A=$
$A^T =$
Exercise: examine and confirm.
Property: ${\rm det \hspace{3pt}}A$ of $n \times n$ matrix has $n!$ terms.
Theorem: The determinant is independent of the expansion row:
$${\det} _1A = {\det} _rA = {\det} A.$$
Proof idea:
• Expand it along $1^{\rm st}$ row, then each participant along $r^{\rm th}$ row.
• Expand it along row $r$, then each participant along $1^{\rm st}$ row.
• Show they are equal.
Plan: Prove by induction on $n$.
Observe: $\det A$ is also defined by induction.
• 1. $\det _1A = \det _rA$ true for $(n-1)\times (n-1)$ matrix $A$
• 2. Show that $\det _1A = \det _rA$ for an $n \times n$ matrix $A$.
We define this via determinants of submatrices of dimension $(n-1) \times (n-1)$.
Examine $A$, given, $n \times n$, and its submatrices $A_{ij},(A_{ij})_{rk}$:
Consider $A_{ij}$.
Expansion along $1^{\rm st}$ row:
$${\det} _1A = \sum_{j=1}^n(-1)^{i+j}a_{ij} \det A_{ij}$$
apply the definition again.
Warning: the indices of the entries of $A_{ij}$ aren't the same as the ones of $A$:
• $a_{21}$ in $A$ is in second row, first column
• $a_{21}$ in $A_{ij}$ is in the first row, first column
• $a_{2(j+1)}$ in $A$ is second row, $(j+1)$ column
• $a_{2(j+1)}$ in $A_{ij}$ is in first row, $j^{\rm th}$ column.
We name entries in $A_{ij}=\{e_{rk}\}$. Express $e_{rk}$ in terms of $a_{rk}$.
Question: What happens to the row numbers?
$$b_{r*}=a_{(r-1)*}$$
Question: What happens to the column numbers?
$b_{*k} = a_{*k}$ if $k < j$
$b_{*k} = a_{*(k-1)}$ if $k > j$
So
$$b_{rk} = \left\{ \begin{array} a_{(r-1)k} &; k < j \\ a_{(r-1)(k-1)} &; k > j \end{array} \right.$$
Use it.
$${\det} _{r-1}A_{ij} \stackrel{ {\rm def} }{=} \sum_{k=1}^{n-1}(-1)^?a_{rk} \det (A_{ij})_{rk}.$$
Expand along $(r-1)$ row (in $A_ij$).
Here, $a_rk$ comes with $\det (A_{ij})_{rk}$. The only thing to find is the sign $=$ row # + column # of $a_{rk}$ in $A_{ij}$.
$$= \left\{ \begin{array}{} r-1+k &; k < j \\ r-1+k-1 &; k > j \end{array} \right.$$
So
$${\det} _{r-1}A_{ij} = \sum_{k=1}^{j-1}(-1)^{r+k-1}a_{rk} \det (A_{ij})_{rk} + \sum_{k=j+1}^n (-1)^{r+k-2}a_{rk} \det (A_{ij})_{rk}$$
Substitute
$$\begin{array}{} {\det} _1A &= \sum_{j=1}^n (-1)^{1+j}a_{1j}(\sum_{k=1}^{j-1}(-1)^{r+k-1}a_{rk}{\rm det}(A_{1j})_{rk} + \sum_{k=j+1}^n (-1)^{r+k-2}a_{rk}{\rm det}(A_{1j})_{rk}) \\ &= \sum_{k<j} (-1)^{1+j}a_{1j}(-1)^{r+k-1}a_{rk}{\rm det}(A_{1j})_{rk} + \sum_{k>j}(-1)^{1+j}a_{1j}(-1)^{r+k-2}{\rm det}(A_{1j})_{rk} \\ &= \sum_{k<j}(-1)^{r+k+j}a_{1j}a_{rk} \det (A_{1j})_{rk} + \sum_{k>j}(-1)^{r+k+j-1}a_{1j}a_{rk} \det (A_{1j})_{rk} \end{array}$$ (1)
Alternatively, expand $A$ along another row. Choose $r \neq 1$.
$${\det} _rA = \sum_{j=1}^n (-1)^{r+j}a_{rj} \det A_{rj}.$$
Expand these along first row.
We denote $A_{r+1,j}$ as $\{c_{ik}\}$. Review:
So, $$c_{ik} = \left\{ \begin{array}{} a_{ik} &; i<r+1,k<j \\ a_{i,k-1} &; i<r+1, k>j \\ a_{i-1,k} &; i > r+1, k < j \\ a_{i-1,k-1} &; i>r+1, k>j \end{array} \right.$$
Again, $${\det} _1A_{rj} = \sum_{k=1}^n (-1)^{??} a_{1k} \det (A_{rj})_{1k}.$$
Find the sign,
$$=\sum_{k<j}(-1)^{1+k}a_{1k} \det (A_{rj})_{1k} + \sum_{k>i}(-1)^{1+k-1}a_{1k} \det (A_{rj})_{1k}$$
First term is first part of the row and second term is second part of row.
Substitute:
$$\begin{array}{} {\det} _rA &= \sum_{j=1}^n(-1)^{r+j}a_{rj}( \sum_{k<j}(-1)^{1+k}a_{1k}{\rm det}(A_{rj})_{1k} + \sum_{k>i}(-1)^{1+k-1}a_{1k}{\rm det}(A_{rj})_{1k} ) \\ &= \sum_{k<j}(-1)^{r+j}a_{rj}(-1)^{1+k}a_{1k}{\rm det}(A_{rj})_{1k} + \sum_{k>j}(-1)^{r+j}a_{rj}(-1)^ka_{1k}{\rm det}(A_{rk})_{1k} \\ &= \sum_{k<j}(-1)^{r+j+k}a_{rj}a_{1k}{\rm det}(A_{rj})_{1k} + \sum_{k>j}(-1)^{r+j+k}a_{rj}a_{1k}{\rm det}(A_{rj})_{1k} \end{array}$$
This is (2). Compare to (1) $$=\sum_{k<j}(-1)^{r+k+j}a_{1j}a_{rk}{\rm det}(A_{ij})_{rk} + \sum_{k>j}(-1)^{r+k+j-1}a_{ij}a_{rk}{\rm det}(A_{ij})_{rk}.$$
$$(A_{rj})_{1k} = (A_{ij})_{rk}$$
The rest doesn't match. To make them equal, flip $j,k$ in (2). $\blacksquare$
So, ${\det}A$ is well defined, can be computed by expanding along any row.
Compare to linear functions, they have the additive property:
$$L(x+y)=L(x)+L(y).$$
i.e., a linear function preserves addition.
Meanwhile, the determinant preserves multiplication.
Theorem: $\det (AB) = \det A \cdot \det B$.
This is the multiplicative property.
Linear operators $L \colon {\bf R} \rightarrow {\bf R}$ do not have this property.
Example: $2(xy) \neq 2x \cdot 2y$ .
Example: $f \colon {\bf R} \rightarrow {\bf R}$ that preserves multiplication:
$$f(xy)=f(x) \cdot f(y).$$
• 1. Two constant functions, $0$ and $1$.
• 2. $(xy)^2 = x^2y^2$, all powers.
• logarithm: $f(xy)=f(x)+f(y)$.
• exponential: $f(x+y)=f(x)f(y)$.
Question: Given $AB$ with $A$ or $B$ not invertible, then $AB$ isn't invertible?
Correct.
This is confirmed by the formula: $$\det (AB) = \det A \cdot \det B = 0.$$
Properties:
• 1. One row of zeroes, then $\det A=0$. (Just expand along this row.)
• 2. Elementary row operations only change the sign of $\det A$.
• 3. Two identical rows, then $\det A=0$.
• 4. $\det A^T = \det A$
## 7 The transpose
If $A = \{a_{ij}\}$ and $A^T = \{b_{ij}\}$, then $b_{ij}=a_{ji}$.
This is the reflection of the matrix about the main diagonal.
$$x \longleftrightarrow y.$$
It "looks like" the inverse. But in general, $A^T \neq A^{-1}$.
If $A \in {\bf M}(n,m)$, then $A^T \in {\bf M}(m,n)$.
Or as linear operators: $$A \colon {\bf R}^m \rightarrow {\bf R}^n .... A^T \colon {\bf R}^n \rightarrow {\bf R}^m.$$
Example: $A = \left[ \begin{array}{} 1 & 1 \\ 1 & 1 \end{array} \right]$, $A^T = \left[ \begin{array}{} 1 & 1 \\ 1 & 1 \end{array} \right] \neq A^{-1}$.
Now, $\det A=0$, so $A$ is singular, there is no $A^{-1}$.
Also, $AA^{-1}=I$. Apply the formula.
$$\det (AA^{-1})= \det A \det A^{-1} = \det I=1.$$
So we have:
Corollary: $$\det A^{-1} = \frac{1}{ \det A}.$$
To prove, compute:
$$\begin{array}{} \det (P^{-1}AP) &= \det(P^{-1}) \cdot \det A \cdot \det P \\ &= \frac{1}{ \det A} \cdot {\rm det \hspace{3pt}}A \cdot \det P \\ &= \det A. \end{array}$$ $\blacksquare$
Corollary: $\det (P^{-1}AP) = \det A$.
Why is this statement important? If $P$ is the change of basis matrix, then $P^{-1}AP$ is the matrix of $A$ is the new basis. So, $\det A$ is independent of the basis.
We knew a small part of that: singularity vs. invertibility ($\det =0$ vs $\det \neq 0$)
Corollary: $\det A$ can be computed by expanding along any column: $$\det A = \sum_{r=1}^n(-1)^{rk}a_{rk} \det A_{rk}$$
Theorem: (Summary) $A$ is a linear operator: ${\bf R}^n \rightarrow {\bf R}^n$. Then the following conditions are equivalent:
• 1. $A$ is non-singular as a matrix;
• 2. $A$ is invertible as a function;
• 3. $\det A \neq 0$;
• 4. ${\rm rank \hspace{3pt}} A = n$;
• 5. $A$ can be reduced to $I_n$ by elementary row operations;
• 6. For any $b \in {\bf R}^n$ there is $v \in {\bf R}^n$ such that $b = A(v)$, and such a $v$ is unique. ($v=A^{-1}(b)$) | 2013-12-10 18:15:06 | {"extraction_info": {"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, "math_score": 0.9748754501342773, "perplexity": 1027.536705051331}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2013-48/segments/1386164023039/warc/CC-MAIN-20131204133343-00095-ip-10-33-133-15.ec2.internal.warc.gz"} |
https://fagullo.com/publication/olivares-07/ | # Buried amorphous layers by electronic excitation in ion-beam irradiated lithium niobate: structure and kinetics
### Abstract
The formation of buried heavily damaged and amorphous layers by a variety of swift-ion irradiations (F at 22MeV, O at 20MeV, and Mg at 28MeV) on congruent LiNbO$_3$ has been investigated. These irradiations assure that the electronic stopping power Se(z) is dominant over the nuclear stopping $S_n(z)$ and reaches a maximum value inside the crystal. The structural profile of the irradiated layers has been characterized in detail by a variety of spectroscopic techniques including dark-mode propagation, micro-Raman scattering, second-harmonic generation, and Rutherford backscattering spectroscopy∕channeling. The growth of the damage on increasing irradiation fluence presents two differentiated stages with an abrupt structural transition between them. The heavily damaged layer reached as a final stage is optically isotropic (refractive index $n = 2.10$, independent of bombarding ion) and has an amorphous structure. Moreover, it has sharp profiles and its thickness progressively increases with irradiation fluence. The dynamics under irradiation of the amorphous-crystalline boundaries has been associated with a reduction of the effective amorphization threshold due to the defects created by prior irradiation (cumulative damage). The kinetics of the two boundaries of the buried layer is quite different, suggesting that other mechanisms aside from the electronic stopping power should play a role on ion-beam damage.
Type
Publication
J. Appl. Phys. 101, 033512 (2007) | 2023-03-29 19:33:17 | {"extraction_info": {"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, "math_score": 0.4858379065990448, "perplexity": 3607.3499618212377}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296949025.18/warc/CC-MAIN-20230329182643-20230329212643-00653.warc.gz"} |
http://physics.stackexchange.com/questions/89958/normalization-of-basis-vectors-with-a-continuous-index | # Normalization of basis vectors with a continuous index?
I have an infinite basis which associates with each point, $x$, on the x-axis, a basis vector $|x\rangle$ such that the matrix of $|x\rangle$ is full of zeroes and a one by the $x^{\mathrm{th}}$ element. The book on Quantum Mechanics by Shankar says that the inner product between a basis vector and itself is not one, why not? Why cant these basis vectors be normalized to one, only to the delta function?
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Related physics.stackexchange.com/q/64869/2451 and links therein. – Qmechanic Dec 11 '13 at 23:15
## 3 Answers
Any good basis should be complete. If the set of all $|x>$ is complete, any other vector $|\psi>$ in the Hilbert space of your system should be writable as $|\psi>=\sum_{x} |x><x|\psi>$. This sum does not make sense for continuous variables $x$, hence the need to redefine the completeness relation with an integral (as Jan's answer demonstrates nicely). Once you employ integrals to define a meaningful completeness relation, then the relation $<x|y>=\delta_{x,y}$ is not correct because it gives $$|y>=\int \mathrm{d}x |x><x|y>=\int \mathrm{d}x |x>\delta_{x,y}=0,$$ which is inconsistent. The way out is to define $<x|y>=\delta(x-y)$, which correctly gives $$|y>=\int \mathrm{d}x |x><x|y>=\int \mathrm{d}x |x>\delta(x-y)=|y>.$$ Now $<x|y>=\delta(x-y)$ leads to $<x|x>=\delta(0)$, that you may not like, but this is the best one can do.
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how did you do the integral? – user35687 Dec 12 '13 at 2:03
It is similar to $\int\mathrm{d}x \delta(x-y) f(x)=f(y)$ – Arash Arabi Dec 12 '13 at 3:28
I) We interpret OP's question (v2) as follows:
Why not normalize $$\tag{1} \langle x_1 | x_2\rangle~=~\delta_{x_1,x_2}~:=~\left\{ \begin{array}{ccl} 1 & \text{for} & x_1= x_2, \\ 0& \text{for} & x_1\neq x_2. \end{array} \right.$$ via a continuous Kronecker delta function rather than a Dirac delta distribution $$\tag{2} \langle x_1 | x_2\rangle~=~\delta(x_1-x_2)~?$$
In a nutshell, the reason is that the rhs. of eq. (1) is equal to the zero function almost everywhere wrt. the Lebesgue measure.
In this answer we would like to build intuition via Gaussian wave packets to argue that we should use a Dirac delta distribution normalization (2) (possibly modulo a conventional normalization constant) rather than a continuous Kronecker delta function normalization (1).
II) To be concrete, let us for simplicity consider a ket
$$\tag{3} |\psi\rangle \quad\longleftrightarrow\quad\psi(x)$$
as a position wave function $\psi(x)\in L^2(\mathbb{R})$ in the Hilbert space
$$\tag{4} L^2(\mathbb{R})~=~{\cal L}^2(\mathbb{R})/\sim,$$
where we have moded out by an equivalence relation "$\sim$". Here ${\cal L}^2(\mathbb{R})$ is the set of square integrable functions. Two functions $\phi\sim\psi$ are equivalent iff $\phi$ and $\psi$ are equal almost everywhere wrt. the Lebesgue measure. See e.g. this Phys.SE post. Overlaps/inner products read
$$\tag{5} \langle \phi|\psi\rangle ~:=~ \int_{\mathbb{R}} \! \mathrm{d}x ~\phi(x)^{*}\psi(x).$$
III) Now pragmatically, short of mathematically constructs such as distributions, what would represent a state localized at $x=x_1$? Let us allow the wave packet to be spread by a tiny amount $\epsilon>0$, say smaller than any experimental resolution. We can model such a wave function by an extremely narrowly peaked Gaussian function
$$\tag{6} |x_1\rangle \quad\longleftrightarrow\quad \psi_{x_1}(x)~=~ \epsilon^{-p} \exp\left[-\left(\frac{x-x_1}{2\epsilon}\right)^2\right],$$
where $p\in\mathbb{R}$ is some fixed power to be determined below. The normalization of (6) is
$$\langle x_1|x_1\rangle ~:=~ \int_{\mathbb{R}} \! \mathrm{d}x ~|\psi_{x_1}(x)|^2 ~=~ \sqrt{2\pi}\epsilon^{1-2p}$$ $$\tag{7}\longrightarrow ~\left\{ \begin{array}{ccl} 0 & \text{if} & p<\frac{1}{2} \\ \sqrt{2\pi}& \text{if} & p=\frac{1}{2} \\ \infty& \text{if} &p>\frac{1}{2} \end{array} \right\} \text{ for } \epsilon~\to~ 0^{+}.$$
To avoid that the normalization (7) disappears in the limit $\epsilon\to 0^{+}$, we must demand that the power $p\geq \frac{1}{2}$. The Kronecker normalization (1) [modulo an overall constant] corresponds to the power $p=\frac{1}{2}$.
IV) More generally, if we assume the ansatz (6), then the overlap between two such kets $|x_1\rangle$ and $|x_2\rangle$ reads in a distributional sense
$$\langle x_1|x_2\rangle ~=~\int_{\mathbb{R}} \! \mathrm{d}x ~\epsilon^{-2p}~ \exp\left[-\left(\frac{x-x_1}{2\epsilon}\right)^2-\left(\frac{x-x_2}{2\epsilon}\right)^2\right]$$ $$~=~\sqrt{2\pi}\epsilon^{1-2p}\exp\left[-\frac{1}{2}\left(\frac{x_1-x_2}{2\epsilon}\right)^2\right]$$ $$\tag{8}\longrightarrow ~\left\{ \begin{array}{ccl} 0 & \text{if} & p<1 \\ 4\pi~\delta(x_1-x_2)& \text{if} & p=1 \\ \text{too singular}& \text{if} &p>1 \end{array} \right\} \text{ for } \epsilon~\to~ 0^{+}.$$
The Dirac normalization (2) [modulo an overall constant] corresponds to the power $p=1$. In detail, if $f(x_1-x_2)$ is a test function, then eq. (8) states that
$$\tag{9}\int_{\mathbb{R}}\! \mathrm{d}x_1~ f(x_1-x_2)~ \langle x_1|x_2\rangle ~\longrightarrow ~\left\{ \begin{array}{ccl} 0 & \text{if} & p<1 \\ 4\pi~ f(0)& \text{if} & p=1 \\ \infty & \text{if} &p>1 \end{array} \right\} \text{ for } \epsilon~\to~ 0^{+}$$
V) Physically, according to the Born rule, the integral (10) of the overlap
$$\tag{10} \left| \int_{\mathbb{R}}\! \mathrm{d}x_1~ \langle x_1|x_2\rangle\right|^2 ~=~1$$
is supposed to denote that a particle located at position $x_2$ belongs to the real axis $\mathbb{R}$ with probability 100%.
Comparing eqs. (9) and (10), we are naturally lead to choose the power $p=1$, and therefore the Dirac normalization (2). Note that the power $p=1$ means that the position state $|x_1\rangle$ fails to be normalizable, and in particular, it does not belong to the Hilbert space, cf. eq. (7).
VI) A more rigorous derivation of eqs. (2) and (10) can be given by introducing momentum eigenstates, cf. e.g. this Phys.SE post.
-
Why cant these basis vectors be normalized to one, only to the delta function?
Because that would make those continuously indexed vectors unsuitable for the role of a "continuous basis" for normalizable functions. Here is the explanation. Suppose some function $\psi(\mathbf r)$ is expressed as the integral $$\psi(\mathbf r) = \int c(k) \phi_k(\mathbf r)\,dk,$$ where the functions $\phi_k$, $\phi_{k'}$ are orthogonal for $k\neq k'$: $$\int \phi_k^*(\mathbf r) \phi_{k'}(\mathbf r)\,d^3\mathbf r = 0$$ (at least in distributive sense).
The above expression of $\psi$ can be described as "linear combination of the basis functions $\phi_k(\mathbf r)$". If the function $\psi$ is to be used to calculate probability density according to the Born rule, we have to require $$\int \psi^*\psi\, d^3\mathbf r = 1.$$ This leads to $$\int \int c^*(k)c(k') (\phi_k, \phi_{k'}) \,dkdk' = 1,~~~(*)$$ where $(\phi_k, \phi_{k'})$ is a scalar product of two continuously-indexed functions: $$(\phi_k, \phi_{k'}) = \int \phi_k^*(\mathbf r) \phi_{k'}(\mathbf r)\,d^3\mathbf r.$$
Imagine points of plane labeled by Cartesian coordinates $k,k'$. If we had $(\phi_k, \phi_{k'}) = \delta_{kk'}$ with ordinary Kronecker delta, the scalar product would be non-zero only on the diagonal $k=k'$ which has zero area, while all over the large area where $k\neq k'$ it would vanish. The integral in (*) would then vanish too and couldn't be equal to 1 as needed.
One way to make the integral in (*) have a non-zero value is to postulate that for the above orthogonal functions, $(\phi_k, \phi_{k'})$ is to be regarded as some singular distribution of the kind Dirac introduced - to bring substantial contributions from the diagonal $k=k'$ only.
In practice we choose functions $\phi_k(\mathbf r)$ such that they obey
$$(\phi_k, \phi_{k'})= \delta(k-k').$$
Then the integral in (*) is
$$\int |c(k)|^2\,dk,$$ which can be non-zero and equal to 1 for properly normalized function $c(k)$.
In the language of kets, the kets $|x\rangle, |y\rangle$ are meant to be such that they satisfy the relation
$$\langle x|y\rangle = \delta(x-y),$$
because only then the relation $$|\psi\rangle = \int |x\rangle \langle x|\psi\rangle \, dx,$$ which is part of the motivation behind the formalism of kets, is valid and consistent with
$$\langle \psi|\psi\rangle = 1.$$
The expression $$\langle x|x\rangle$$ is not a valid expression and usually is not used in manipulations with the bra-ket formalism; if we used the above relation, we would get $\delta(x-x)$, which could be regarded as either positive infinity or not a meaningful number at all (since $\delta$ is not an ordinary function and does not have ordinary number-valued function values.)
-
But then why is it that when we use non-continuous basis we have the relation <x|x> = 1 ? – user35687 Dec 12 '13 at 0:12
@user35687 : Because, with a discrete basis, we have the relation $\langle n|m\rangle = \delta_{nm}$. This discrete version $\delta_{nm}$ of the delta function $\delta(x-y)$, may be seen as the identity matrix, that is $\delta_{nm}=1$ if $m=n$, $\delta_{nm}=0$ if $m \neq n$ – Trimok Dec 12 '13 at 11:17 | 2015-07-28 15:35:24 | {"extraction_info": {"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": 2, "x-ck12": 0, "texerror": 0, "math_score": 0.9528608322143555, "perplexity": 258.0347441695836}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-32/segments/1438042981969.11/warc/CC-MAIN-20150728002301-00093-ip-10-236-191-2.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/what-is-a-mass-independent-renormalization-scheme.881020/ | # A What is a Mass Independent Renormalization Scheme?
1. Aug 4, 2016
### Luca_Mantani
Hi everyone,
i'm not able to find the exact definition of mass independent renormalization scheme. I often read that the MS-bar scheme is mass independent, but why? And why this feature help us to compute the beta function?
Luca
2. Aug 4, 2016
### vanhees71
The $\overline{\mathrm{MS}}$ scheme is defined within dimensional regularization, which is very elegant from the calculational point of view, but it's not very intuitive.
As the most simple example take simple $\phi^4$ theory with the Lagrangian
$$\mathcal{L}=\frac{1}{2} (\partial_{\mu} \phi)(\partial^{\mu} \phi)-\frac{m^2}{2} \phi^2 -\frac{\lambda}{4!} \phi^4,$$
which is renormalizable. The divergent parts are (besides the here not so interesting vacuum diagrams) the self-energy (leading to wave-function and mass renormalization), and the four-point function (leading to coupling-constant renormalization).
The counterterms and the renormalization scheme are thus determined by these divergent pieces. One can write the counter-term Lagrangian in the form
$$\delta \mathcal{L} = \delta Z \frac{1}{2} (\partial_{\mu} \phi)(\partial^{\mu} \phi) - \frac{1}{2} (\delta m^2 + m^2 \delta Z_m) \phi^2 - \frac{\delta \lambda}{4!} \phi^4.$$
Here the $\delta Z$, $\delta Z_m$ and $\delta \lambda$ are all dimensionless (i.e., of energy dimension 0). Only $\delta m^2$ has dimension 2.
The most intuitive MIR scheme is defined by introducing a mass scale $M$ as the renormalization scale and define the renormalized quantities by the renormalization conditions
$$\Sigma(p^2=0,m^2=0)=0 \; \Rightarrow \; \delta m^2=0.$$
This is allowed, because that's a quadratically divergent quantity and thus IR safe. So you can define it at $m^2=0$.
All other divergences are logarithmic and thus cannot be defined at $m^2=0$ and all external momenta at 0. Thus one defines them at $m^2=M^2$, i.e.,
$$[\partial_{p^2} \Sigma(p^2,m^2)]_{p^2=0,m^2=M^2}=0 \; \Rightarrow \; \delta Z,$$
$$[\partial_{m^2} \Sigma(p^2,m^2)]_{p^2=0,m^2=M^2}=0 \; \Rightarrow \; \delta Z_m,$$
$$\Gamma^{(4)}(s=t=u=0,m^2=M^2)=-\lambda \; \Rightarrow \; \delta \lambda.$$
All the counter terms are thus only dependent on the renormalization scale $M^2$ via the dimensionless renormalized coupling constant $\lambda$, but not on $m^2$. That's why this scheme is a mass-independent renormalization scheme, and you have $\delta Z=\delta Z(\lambda)$, $\delta Z_{m}=\delta Z_m(\lambda)$. The RG parameters like $\beta$ don't depend on $m/M$ but only on $M$ through $\lambda$, and you get a homogeneous RG equation that is not so difficult to solve.
For details, see my QFT manuscript
http://th.physik.uni-frankfurt.de/~hees/publ/lect.pdf
Chpt. 5.
The derivation of the $\beta$ function and other RG equation coefficients within the MIR, MS (or $\overline{\text{MS}}$) scheme in the context of dim. reg. see Sect. 5.11. In Sect. 5.12 also other ("non-MIR") schemes are treated. The corresponding RG equations are more difficult since the RG coefficients do not only depend on the renormalization scale $M$ via the dimensionless renormalized coupling $\lambda$ but also explicitly via $m/M$.
3. Aug 4, 2016
### A. Neumaier
Some misprints:
p.129: /2: positron, /8: Coulomb, /middle: hints->it hints, neglects->neglect
p.130: nonuniform capitalizations of 'quantum field theory' - I'd use no capitals at all. /mid: principally->in principle. You should mention that the scheme independence is only after summing all contributions; at low order there may be big differences and the mass scale matters. /-2: nowadays->today's
p.131: /11: loose->lose, /13: especially->in particular
Last edited: Aug 4, 2016
4. Aug 4, 2016
### Luca_Mantani
Thank you very much for the explanation! I'll get a look at the lectures but i think i grasped the basic idea behind it, thank you again.
5. Aug 5, 2016
### vanhees71
Thanks a lot. I corrected all typos. Where would you put the comment on scheme independence? It's of course the point of the RG equations to resum to leading logarithmic order to achieve this (approximate) independence in perturbation theory where it is applicable, i.e., for small coupling.
6. Aug 5, 2016
### A. Neumaier
On p.130 where you say ''this dependence should change nothing with respect to S-matrix elements'', or perhaps one paragraph later. If necessary with a forward reference to the resumming.
There is still a ''quantum Field theory" and a "quantum Field Theory" alongside the preferable ''quantum field theory". Probably also on other pages.
7. Aug 5, 2016
### vanhees71
Thanks again. I hope, now it's better. | 2017-10-20 07:42:25 | {"extraction_info": {"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, "math_score": 0.9424769878387451, "perplexity": 1558.801369690976}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-43/segments/1508187823839.40/warc/CC-MAIN-20171020063725-20171020083725-00298.warc.gz"} |
http://mymathforum.com/elementary-math/344117-simplify-expression-2.html | My Math Forum Simplify expression
Elementary Math Fractions, Percentages, Word Problems, Equations, Inequations, Factorization, Expansion
April 24th, 2018, 08:30 AM #11 Global Moderator Joined: Dec 2006 Posts: 20,617 Thanks: 2072 Partly experience, in that it’s quite handy to have seen such a question before, but inspection would help as well. In your answer to my question, the absolute value isn’t needed. Consider also what you get if my expression has the “+” symbol replaced by a “-“ symbol. Thanks from bongcloud
April 24th, 2018, 10:15 AM #12
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Joined: Apr 2018
From: Banovo Brdo
Posts: 12
Thanks: 0
Do you mean $1+\sqrt{\frac{x-a}{a}}+\frac{1}{4}\lvert{\frac{x-a}{a}}\rvert$? Yes, I've realized that here:
Quote:
Originally Posted by bongcloud I see. If we assume solution is $\mathbb{R}$, then $x-a\geq 0 \implies x\geq a$ I don't see how to proceed from there, though. For example, if I get rid of the roots, I'm left with $1+\frac{1}{2}\sqrt{\frac{x-a}{a}}+\lvert 1-\frac{1}{2}\sqrt{\frac{x-a}{a}}\rvert$
While typing this, I've realized I'd been a victim of poor rewriting skills yet again.
However, it seems to me that we need to replace $\left(1-\frac{1}{2}\sqrt{\frac{x-a}{a}}\right)^2$ with $\left(\frac{1}{2}\sqrt{\frac{x-a}{a}}-1\right)^2$.
Then we have
\left\lvert \frac{1}{2}\sqrt{\frac{x-a}{a}}-1\right\rvert=\left\{\begin{aligned} &\frac{1}{2}\sqrt{\frac{x-a}{a}}-1, &&\frac{1}{2}\sqrt{\frac{x-a}{a}}-1\geq 0\implies x\geq 5a\\ &1-\frac{1}{2}\sqrt{\frac{x-a}{a}}, &&\frac{1}{2}\sqrt{\frac{x-a}{a}}-1<0\implies a\leq x<5a \end{aligned} \right.
Which produces the same solution set as in #9
E=\left\{\begin{aligned} &2, &&5a\geq x>a\\ &\sqrt{\frac{x-a}{a}}, &&x\geq 5a \end{aligned} \right.
Am I missing something?
By the way, I appreciate your maieutic approach to teaching.
Last edited by bongcloud; April 24th, 2018 at 10:40 AM.
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Contact - Home - Forums - Cryptocurrency Forum - Top | 2019-05-20 08:49:27 | {"extraction_info": {"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, "math_score": 0.7986564040184021, "perplexity": 5835.792135680743}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-22/segments/1558232255837.21/warc/CC-MAIN-20190520081942-20190520103942-00491.warc.gz"} |
https://math.stackexchange.com/questions/3546087/integration-with-simple-l2-functions | # Integration with simple $L^2$ functions
If $$h\in L^2(\Omega,\Sigma,\mu)$$ is a simple function and $$f=g\hspace{0.3cm} \mu-a.e.$$ then $$\int fhd\mu=\int ghd\mu \qquad \text{and} \qquad \left|\int ghd\mu\right|\le\|f\|_{L^2}\|h\|_{L^2}$$ How can I prove that the above is true for all $$h\in L^2(\Omega,\Sigma,\mu)$$ ?
• For the first part, what can you say about $\int (f-g)h\ d\mu$? – Bungo Feb 14 at 3:31
• The second is near immediate from Hölder. – Sean Roberson 2 days ago
Since $$h\in L^2(\Omega,\Sigma,\mu)$$ is a simple function, it is of the form $$h(x) = \sum_{i=1}^n \alpha_i \chi_{E_i}(x)$$, where $$E_i \in \Sigma, \alpha_i \in \mathbb{R}$$. Let $$M:= \{x \in \Omega: g(x) \neq f(x) \}$$. The set is measurable because $$f$$ and $$g$$ are measurable. We know that $$\mu(M) = 0$$. Set $$M_i := M \cap E_i$$. Then, $$\int_{\Omega} fh d\mu = \int_{\Omega} f(x) \sum_{i=1}^n \alpha_i \chi_{E_i}(x) d\mu(x) = \sum_{i=1}^n \alpha_i \mu(E_i) \int_{E_i}f(x) d\mu(x) = \sum_{i=1}^n \alpha_i \mu(E_i) (\int_{E_i}f(x) d\mu(x) + \int_{M_i}g(x) d\mu(x)) = \sum_{i=1}^n \alpha_i \mu(E_i) \int_{E_i}g(x) d\mu(x) = \int_{\Omega} gh d\mu$$
To answer your second equation: $$| \int gh d\mu | = | \int fh d\mu | = \|fh\|_1 \le \|f\|_2 \|h\|_2$$ which is the Hölder's inequality. | 2020-02-17 04:00:47 | {"extraction_info": {"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": 14, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8943020701408386, "perplexity": 136.6485927572635}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-10/segments/1581875141653.66/warc/CC-MAIN-20200217030027-20200217060027-00062.warc.gz"} |
https://byjus.com/question-answer/the-following-are-the-marks-obtained-out-of-100-by-students-in-mathematics-in-a-1/ | Question
The following are the marks obtained (out of 100) by students in Mathematics in a particular class : 82 , 76 , 90 , 64 , 80 , 96 , 100 , 52 , 68 and 88 What is the average mark obtained ? 82 76.5 79.6 84.2
Solution
The correct option is C 79.6 Average Marks = Sum of MarksTotalnumberofstudents = 82+76+90+64+80+96+100+52+68+8810 = 79610 = 79.6
Suggest corrections | 2021-12-03 16:32:39 | {"extraction_info": {"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, "math_score": 0.8434959650039673, "perplexity": 454.87655845664614}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-49/segments/1637964362891.54/warc/CC-MAIN-20211203151849-20211203181849-00021.warc.gz"} |
https://dgtal.org/doc/0.9.2/moduleDigitalSets.html | DGtal 0.9.2
Sets of points in digital spaces
Author(s) of this documentation:
Jacques-Olivier Lachaud
# Overview
Digital sets are sets of digital points within some domain of a digital space. They are thus subsets of a given domain (e.g. see Domains and HyperRectDomains). As such, their elements can be enumerated with iterators through the range [begin(),end()). Furthermore, digital sets can be complemented within their domain. Last, a digital set is also a predicate on point, returning true whenever the points lies inside the set.
# The concept concepts::CDigitalSet and models of concepts::CDigitalSet
There are several ways for defining concretely a digital set, this is why digital sets are specified through the concept concepts::CDigitalSet. All types and required methods are specified in the documentation of CDigitalSet. It is worthy to note that a digital set is also associated to a digital domain, i.e. any model of concepts::CDomain.
There exists several models for concepts::CDigitalSet; their difference lies in the way elements are stored (let n be the number of points in the set).
• DigitalSetBySTLSet: it is the most versatile representation for digital sets, and the one which should be preferred if you have no specific properties for your set. The container is the standard simple associative container std::set (a model of boost::SimpleAssociativeContainer). All find, insertion and deletion requests are $$O(\log n)$$ complexity. Internal order can be changed by specifiing another Compare functor as template parameter.
• DigitalSetBySTLVector: this representation is suited for very small set of points (for instance a neighborhood). The container is the standard sequence container std::vector (a model of boost::Sequence). All find, insertion and deletion requests are $$O(n)$$ complexity.
• DigitalSetFromMap: it is not a container but an adapter to an existing map Point -> Value. Elements of the digital set are by definition the keys of the map.
• DigitalSetByAssociativeContainer: it is a generic container which adapts any associative container (e.g. from the STL or from boost). For example, if you instantiate such digital set on a std::set, this class exactly matches with DigitalSetBySTLSet. The main advantage of this container is its ability to adapt hash function based containers such as std::unordered_set (for C++11 enabled build) or boost::unordered_set. Compared to DigitalSetBySTLSet, DigitalSetByAssociativeContainer on std::unordered_set is expected to be 20% - 50% faster when accessing or inserting points in the set.
You may choose yourself your representation of digital set, or let DGtal chooses for you the best suited representation with the class DigitalSetSelector. This is done by choosing among the following properties:
• the expected size of the set with enum DigitalSetSize, from small to huge: SMALL_DS, MEDIUM_DS, BIG_DS, WHOLE_DS.
• the expected variability of the set with enum DigitalSetVariability: will the set change a lot during its lifetime (HIGH_VAR_DS) or not (LOW_VAR_DS) ?
• the expected number of times you will iterate through the elements of the set with enum DigitalSetIterability, few times is LOW_ITER_DS, many times is HIGH_ITER_DS.
• the number of times you will test for the presence of points in the set with enum DigitalSetBelongTestability, few times is LOW_BEL_DS, many times is HIGH_BEL_DS.
Note
By default, Z2i::DigitalSet and Z3i::DigitalSet in StdDefs.h refer to the associative container with hash functions (fastest on large sets).
The following lines selects a rather generic representation for digital sets, since the set may be big, will be iterated many times and points will be tested many times:
typedef SpaceND<2> Z2;
typedef HyperRectDomain<Z2> Domain;
typedef DigitalSetSelector < Domain, BIG_DS + HIGH_ITER_DS + HIGH_BEL_DS >::Type SpecificSet;
// here SpecificSet is DigitalSetByAssociativeContainer<(boost or std)::unordered_set<Point>, Domain>.
# Using digital sets
The following snippet shows how to define a digital set in space Z2, within a rectangular domain, then how to insert points, how to visit them and how to test if a point belongs to the set.
typedef SpaceND<2> Z2;
typedef HyperRectDomain<Z2> Domain;
typedef DigitalSetSelector < Domain, BIG_DS + HIGH_ITER_DS + HIGH_BEL_DS >::Type SpecificSet;
typedef Z2::Point Point;
// instantiating rectangular domain
Point p1( -10, -10 );
Point p2( 10, 10 );
Domain domain( p1, p2 );
// instanciating set within this domain.
SpecificSet mySet( domain );
Point c( 0, 0 );
mySet.insert( c ); // inserting point (0,0)
Point d( 5, 2 );
mySet.insert( d ); // inserting point (5,2)
Point e( 1, -3 );
mySet.insert( e ); // inserting point (1,-3)
// Iterating through the set
typedef SpecificSet::ConstIterator ConstIterator;
for ( ConstIterator it = mySet.begin(), itEnd = mySet.end();
it != itEnd; ++it )
std::cout << *it << std::endl;
// Checking points within set
bool ok_c = mySet( c ); // should be true
bool ok_d = mySet( d ); // should be true
bool ok_e = mySet( e ); // should be true
bool not_ok = mySet( Point( 1,1) ); // should be false
Other methods may be found in the concept definition of digital sets (CDigitalSet).
# Digital sets and point predicates
Since 0.6, any digital set is also a model of concepts::CPointPredicate. You may thus use a digital set directly in functions and methods requiring a predicate on points. However, the class deprecated::SetPredicate, which builds a predicate on points around a digital set, is deprecated since 0.6. You may still used it by referecing it with DGtal::deprecated::SetPredicate, but it will no longer be maintained.
# A digital set may be transformed into a domain
Sometimes it is useful to see a digital set as a new domain for further computation. This is possible with the facade class DigitalSetDomain, which wraps around some given digital set all necessary methods and types so that it satisfies the concept concepts::CDomain.
typedef DigitalSetDomain< SpecificSet > RestrictedDomain;
RestrictedDomain myDomain( mySet );
// this domain is limited to three points c, d and e.
Note
The DigitalSetDomain class is only a wrapper around the digital set. The lifetime of the digital set must thus exceed the lifetime of the DigitalSetDomain. | 2021-01-22 13:26:39 | {"extraction_info": {"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, "math_score": 0.3810264766216278, "perplexity": 3760.7208590741984}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-04/segments/1610703529331.99/warc/CC-MAIN-20210122113332-20210122143332-00784.warc.gz"} |
http://tug.org/pipermail/texhax/2003-November/001270.html | [texhax] figures
Jing Zhang jingz002000 at yahoo.com
Thu Nov 27 01:30:53 CET 2003
I use the following code to output two figures one after another. I have two questions:
1. How to center the caption?
2. How to adjust the vertical space between these two figures?
The working environment is ieee template.
Jing
_________________________________________________
\begin{figure}[hbp]
\begin{minipage}[b]{0.5\linewidth} % A minipage that covers half the page, width-wise
\centering
\includegraphics[width=2.8in,height=2.8in]{fd11.eps}
\end{minipage}
\hspace{0.5cm} % To get a little bit of horizontal space between the figures
\begin{minipage}[b]{0.5\linewidth}
\centering
\includegraphics[width=2.8in,height=2.8in]{fd12.eps}
\caption{Second stupendous result}
\end{minipage}
\end{figure}
---------------------------------
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Free Pop-Up Blocker - Get it now
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URL: http://tug.org/pipermail/texhax/attachments/20031126/eafafbfb/attachment.htm | 2017-10-18 23:40:28 | {"extraction_info": {"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, "math_score": 0.9430994391441345, "perplexity": 7750.716031019447}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-43/segments/1508187823168.74/warc/CC-MAIN-20171018233539-20171019013539-00523.warc.gz"} |
https://www.gradesaver.com/textbooks/math/trigonometry/CLONE-68cac39a-c5ec-4c26-8565-a44738e90952/appendix-a-equations-and-inequalities-page-421/73 | ## Trigonometry (11th Edition) Clone
$[-4,\infty)$
Step 1: $-2x+8\leq16$ Step 2: Subtracting $8$ from both sides, $-2x+8-8\leq16-8$ Step 3: $-2x\leq8$ Step 4: Dividing both sides by -2 (this reverses the direction of the inequality symbol): $\frac{-2x}{-2} \geq \frac{8}{-2}$ Step 5: $x\geq-4$ According to the inequality, the interval includes $-4$ and all values greater than $-4$. Since $-4$ is part of the interval, a square bracket is used on its side. On the other hand, $\infty$ does not represent an actual number. Rather it is used to show that the interval includes all real numbers greater than -4. Therefore, a parenthesis is used on its side. Therefore, the interval notation for this inequality is written as $[-4,\infty)$. | 2021-02-25 22:16:43 | {"extraction_info": {"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, "math_score": 0.8039687275886536, "perplexity": 130.79804801451246}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-10/segments/1614178355937.26/warc/CC-MAIN-20210225211435-20210226001435-00585.warc.gz"} |
https://johncarlosbaez.wordpress.com/2018/08/15/open-petri-nets-part-1/?replytocom=126796 | ## Open Petri Nets (Part 1)
Jade Master and I have just finished a paper on open Petri nets:
• John Baez and Jade Master, Open Petri nets.
Abstract. The reachability semantics for Petri nets can be studied using open Petri nets. For us an ‘open’ Petri net is one with certain places designated as inputs and outputs via a cospan of sets. We can compose open Petri nets by gluing the outputs of one to the inputs of another. Open Petri nets can be treated as morphisms of a category, which becomes symmetric monoidal under disjoint union. However, since the composite of open Petri nets is defined only up to isomorphism, it is better to treat them as morphisms of a symmetric monoidal double category $\mathbb{O}\mathbf{pen}(\mathrm{Petri}).$ Various choices of semantics for open Petri nets can be described using symmetric monoidal double functors out of $\mathbb{O}\mathbf{pen}(\mathrm{Petri}).$ Here we describe the reachability semantics, which assigns to each open Petri net the relation saying which markings of the outputs can be obtained from a given marking of the inputs via a sequence of transitions. We show this semantics gives a symmetric monoidal lax double functor from $\mathbb{O}\mathbf{pen}(\mathrm{Petri})$ to the double category of relations. A key step in the proof is to treat Petri nets as presentations of symmetric monoidal categories; for this we use the work of Meseguer, Montanari, Sassone and others.
I’m excited about this, especially because our friends at Statebox are planning to use open Petri nets in their software. They’ve recently come out with a paper too:
• Fabrizio Romano Genovese and Jelle Herold, Executions in (semi-)integer Petri nets are compact closed categories.
Petri nets are widely used to model open systems in subjects ranging from computer science to chemistry. There are various kinds of Petri net, and various ways to make them ‘open’, and my paper with Jade only handles the simplest. But our techniques are flexible, so they can be generalized.
What’s an open Petri net? For us, it’s a thing like this:
The yellow circles are called ‘places’ (or in chemistry, ‘species’). The aqua rectangles are called ‘transitions’ (or in chemistry, ‘reactions’). There can in general be lots of places and lots of transitions. The bold arrows from places to transitions and from transitions to places complete the structure of a Petri net. There are also arbitrary functions from sets $X$ and $Y$ into the set of places. This makes our Petri net into an ‘open’ Petri net.
We can think of open Petri nets as morphisms between finite sets. There’s a way to compose them! Suppose we have an open Petri net $P$ from $X$ to $Y,$ where now I’ve given names to the points in these sets:
We write this as $P \colon X \nrightarrow Y$ for short, where the funky arrow reminds us this isn’t a function between sets. Given another open Petri net $Q \colon Y \nrightarrow Z,$ for example this:
the first step in composing $P$ and $Q$ is to put the pictures together:
At this point, if we ignore the sets $X,Y,Z,$ we have a new Petri net whose set of places is the disjoint union of those for $P$ and $Q.$
The second step is to identify a place of $P$ with a place of $Q$ whenever both are images of the same point in $Y$. We can then stop drawing everything involving $Y,$ and get an open Petri net $QP \colon X \nrightarrow Z,$ which looks like this:
Formalizing this simple construction leads us into a bit of higher category theory. The process of taking the disjoint union of two sets of places and then quotienting by an equivalence relation is a pushout. Pushouts are defined only up to canonical isomorphism: for example, the place labeled $C$ in the last diagram above could equally well have been labeled $D$ or $E.$ This is why to get a category, with composition strictly associative, we need to use isomorphism classes of open Petri nets as morphisms. But there are advantages to avoiding this and working with open Petri nets themselves. Basically, it’s better to work with things than mere isomorphism classes of things! If we do this, we obtain not a category but a bicategory with open Petri nets as morphisms.
However, this bicategory is equipped with more structure. Besides composing open Petri nets, we can also ‘tensor’ them via disjoint union: this describes Petri nets being run in parallel rather than in series. The result is a symmetric monoidal bicategory. Unfortunately, the axioms for a symmetric monoidal bicategory are cumbersome to check directly. Double categories turn out to be more convenient.
Double categories were introduced in the 1960s by Charles Ehresmann. More recently they have found their way into applied mathematics. They been used to study various things, including open dynamical systems:
• Eugene Lerman and David Spivak, An algebra of open continuous time dynamical systems and networks.
open electrical circuits and chemical reaction networks:
• Kenny Courser, A bicategory of decorated cospans, Theory and Applications of Categories 32 (2017), 995–1027.
open discrete-time Markov chains:
• Florence Clerc, Harrison Humphrey and P. Panangaden, Bicategories of Markov processes, in Models, Algorithms, Logics and Tools, Lecture Notes in Computer Science 10460, Springer, Berlin, 2017, pp. 112–124.
and coarse-graining for open continuous-time Markov chains:
• John Baez and Kenny Courser, Coarse-graining open Markov processes. (Blog article here.)
As noted by Shulman, the easiest way to get a symmetric monoidal bicategory is often to first construct a symmetric monoidal double category:
• Mike Shulman, Constructing symmetric monoidal bicategories.
The theory of ‘structured cospans’ gives a systematic way to build symmetric monoidal double categories—Kenny Courser and I are writing a paper on this—and Jade and I use this to construct the symmetric monoidal double category of open Petri nets.
A 2-morphism in a double category can be drawn as a square like this:
We call $X_1,X_2,Y_1$ and $Y_2$ ‘objects’, $f$ and $g$ ‘vertical 1-morphisms’, $M$ and $N$ ‘horizontal 1-cells’, and $\alpha$ a ‘2-morphism’. We can compose vertical 1-morphisms to get new vertical 1-morphisms and compose horizontal 1-cells to get new horizontal 1-cells. We can compose the 2-morphisms in two ways: horizontally and vertically. (This is just a quick sketch of the ideas, not the full definition.)
In our paper, Jade and I start by constructing a symmetric monoidal double category $\mathbb{O}\mathbf{pen}(\textrm{Petri})$ with:
• sets $X, Y, Z, \dots$ as objects,
• functions $f \colon X \to Y$ as vertical 1-morphisms,
• open Petri nets $P \colon X \nrightarrow Y$ as horizontal 1-cells,
• morphisms between open Petri nets as 2-morphisms.
(Since composition of horizontal 1-cells is associative only up to an invertible 2-morphism, this is technically a pseudo double category.)
What are the morphisms between open Petri nets like? A simple example may be help give a feel for this. There is a morphism from this open Petri net:
to this one:
mapping both primed and unprimed symbols to unprimed ones. This describes a process of ‘simplifying’ an open Petri net. There are also morphisms that include simple open Petri nets in more complicated ones, etc.
This is just the start. Our real goal is to study the semantics of open Petri nets: that is, how they actually describe processes! And for that, we need to think about the free symmetric monoidal category on a Petri net. You can read more about those things in Part 2 and Part 3 of this series.
Part 1: the double category of open Petri nets.
Part 2: the reachability semantics for open Petri nets.
Part 3: the free symmetric monoidal category on a Petri net.
### 11 Responses to Open Petri Nets (Part 1)
1. Joe Moeller says:
Typo: “What are the morphisms between open Petri nets ilke?”
2. Eugene says:
Symmetric monoidal categories give rise to representable multicategories. What do monoidal double categories give rise to?
Also, algebras over multicategories give semantics (I think).
What do algebras over monoidal double categories give you?
• John Baez says:
Eugene wrote:
Symmetric monoidal categories give rise to representable multicategories. What do monoidal double categories give rise to?
I don’t know! This concept bubbles to the surface of my mind:
• nLab, Virtual double categories.
I’m not claiming this concept is the answer to your question – it’s certainly not.
But it’s still an interesting concept. The basic sort of 2-cell in a virtual double category looks like this:
so virtual double categories are a common generalization of monoidal category, bicategory, double category, and multicategory!
Also, algebras over multicategories give semantics (I think). What do algebras over monoidal double categories give you?
More semantics. ‘Semantics’ is a very general term for ‘mapping syntactic expressions to their meanings’. Lawvere’s thesis Functorial Semantics showed how to do semantics using maps between categories with finite products. Later people generalized this to all kinds of categories, and in our paper we’re doing it with symmetric monoidal double categories.
• Eugene says:
I saw your paper in the arxiv last night, so that kind of answers my question about semantics in more details. I guess I was too fixated on operads.
I think I have an idea of what a double multicategory should be like, and I have a live example or two, which are mentioned in arXiv:1705.04814 [math.OC]. But maybe I am wrong.
3. Does each of your vertical 1-morphisms, ie. function between sets, have a conjoint and companion open Petri net? In other words, do you have a fibrant double category (or an equipment, if you prefer)?
• John Baez says:
Yes!
This is why I said Mike Shulman’s technique can be applied to get a symmetric monoidal bicategory from the symmetric monoidal double category of open Petri nets. His technique applies to fibrant double categories—or more generally, ‘isofibrant’ ones, where each invertible vertical 1-morphism has a conjoint and a companion.
Jade and I don’t actually discuss this isofibrancy, or how to get the symmetric monoidal bicategory. Kenny Courser and I will talk about these things more generally in our paper on structured cospans. The double category $\mathbb{O}\mathbf{pen}(\mathrm{Petri})$ is an example of a ‘structured cospan double category’. The idea is that an open Petri net is a cospan of sets where the apex is equipped with extra structure: it’s made into the set of places of a Petri net. Structured cospan double categories are isofibrant under very general conditions!
• John Baez and Jade Master, Open Petri nets.
Last time I explained, in a sketchy way, the double category of open Petri nets. This time I’d like to describe a ‘semantics’ for open Petri nets.
5. […] Open Petri Nets (Part 1) 4 by gbrown_ | 0 comments on Hacker News. […]
6. Seth Buehler says:
Not sure if this is the right place for it, but in page 20 of your paper, the proof of theorem 18 specifies that $R_1$ and $R_2$ are /linear/ relations, but everything in sight is only a set, not an Abelian group, so I don’t even know what that would mean.
• John Baez says:
Thanks for catching that! Just delete the world “linear”. We took a proof for the double category of linear relations between vector spaces from my paper with Kenny Courser, and adapted it… but failed to purge it of all linearity!
This site uses Akismet to reduce spam. Learn how your comment data is processed. | 2019-05-20 23:38:10 | {"extraction_info": {"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": 37, "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, "math_score": 0.7737254500389099, "perplexity": 755.7740150850092}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-22/segments/1558232256163.40/warc/CC-MAIN-20190520222102-20190521004102-00292.warc.gz"} |
https://www.clarissewiki.com/5.0/cnode/using-the-render-cache.html | # Using the Render Cache#
CNode provides a render cache feature (disabled by default) to resume interrupted renders or extract images from in-progress or partial renders.
To enable the render cache you simply need to add the argument -enable_render_cache:
cnode my.project -enable_render_cache
When the render cache is enabled, CNode stores in-progress version of the current render to disk in a special cache format. It is then possible to extract images directly from the cache using the dedicated crcache tool provided with Clarisse binaries. For more information, please refer to crcache command line help.
The render cache can also used by CNode to resume renders that would be incomplete because of interrupted by the user or because of a crash. In that case, CNode will pick up the render where it was left before the interruption. To enable this mode simply add the argument -recover:
cnode my.project -recover
Please note it is possible to activate the recover feature along with the render cache. That way CNode will resume incomplete renders and create a render cache for following the renders:
cnode my.project -recover -enable_render_cache
Note
By default, the render cache files are automatically deleted when the final frame is outputted. You can change this behavior by using the -keep_render_cache argument.
## Render Cache Files And Location#
By default, the render cache files are stored in the same location as the final frames.
However, it is possible to change this default behavior by specifying a custom location using the -render_cache_location argument:
cnode my.project -recover -enable_render_cache -render_cache_location "/path/to/render/cache"
The specified location can also be relative to the render target locations by specifying the argument -relative_render_cache_location:
cnode my.project -enable_render_cache -render_cache_location "relative/render/cache/location" -relative_render_cache_location
!!! Note You can force CNode to generate the directories of the render cache location by specifying the -generate_render_cache_location argument to CNode.
## Keeping The Render Cache#
By default, the render cache is automatically deleted from the file system once the final render is fully completed and the output frame is stored to disk. Specifying -keep_render_cache argument forces CNode to leave the render cache on disk even if the render is complete. | 2023-03-21 07:20:13 | {"extraction_info": {"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, "math_score": 0.30052393674850464, "perplexity": 5947.440322028955}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296943637.3/warc/CC-MAIN-20230321064400-20230321094400-00472.warc.gz"} |
https://labs.tib.eu/arxiv/?author=G.%20Andronico | • ### Charge reconstruction in large-area photomultipliers(1801.08690)
Jan. 26, 2018 physics.ins-det
Large-area PhotoMultiplier Tubes (PMT) allow to efficiently instrument Liquid Scintillator (LS) neutrino detectors, where large target masses are pivotal to compensate for neutrinos' extremely elusive nature. Depending on the detector light yield, several scintillation photons stemming from the same neutrino interaction are likely to hit a single PMT in a few tens/hundreds of nanoseconds, resulting in several photoelectrons (PEs) to pile-up at the PMT anode. In such scenario, the signal generated by each PE is entangled to the others, and an accurate PMT charge reconstruction becomes challenging. This manuscript describes an experimental method able to address the PMT charge reconstruction in the case of large PE pile-up, providing an unbiased charge estimator at the permille level up to 15 detected PEs. The method is based on a signal filtering technique (Wiener filter) which suppresses the noise due to both PMT and readout electronics, and on a Fourier-based deconvolution able to minimize the influence of signal distortions ---such as an overshoot. The analysis of simulated PMT waveforms shows that the slope of a linear regression modeling the relation between reconstructed and true charge values improves from $0.769 \pm 0.001$ (without deconvolution) to $0.989 \pm 0.001$ (with deconvolution), where unitary slope implies perfect reconstruction. A C++ implementation of the charge reconstruction algorithm is available online at http://www.fe.infn.it/CRA .
• ### INFN What Next: Ultra-relativistic Heavy-Ion Collisions(1602.04120)
Feb. 12, 2016 nucl-ex, nucl-th
This document was prepared by the community that is active in Italy, within INFN (Istituto Nazionale di Fisica Nucleare), in the field of ultra-relativistic heavy-ion collisions. The experimental study of the phase diagram of strongly-interacting matter and of the Quark-Gluon Plasma (QGP) deconfined state will proceed, in the next 10-15 years, along two directions: the high-energy regime at RHIC and at the LHC, and the low-energy regime at FAIR, NICA, SPS and RHIC. The Italian community is strongly involved in the present and future programme of the ALICE experiment, the upgrade of which will open, in the 2020s, a new phase of high-precision characterisation of the QGP properties at the LHC. As a complement of this main activity, there is a growing interest in a possible future experiment at the SPS, which would target the search for the onset of deconfinement using dimuon measurements. On a longer timescale, the community looks with interest at the ongoing studies and discussions on a possible fixed-target programme using the LHC ion beams and on the Future Circular Collider.
• ### A Geometrical Interpretation of Hyperscaling Breaking in the Ising Model(hep-lat/0208009)
Aug. 6, 2002 hep-lat, cond-mat
In random percolation one finds that the mean field regime above the upper critical dimension can simply be explained through the coexistence of infinite percolating clusters at the critical point. Because of the mapping between percolation and critical behaviour in the Ising model, one might check whether the breakdown of hyperscaling in the Ising model can also be intepreted as due to an infinite multiplicity of percolating Fortuin-Kasteleyn clusters at the critical temperature T_c. Preliminary results suggest that the scenario is much more involved than expected due to the fact that the percolation variables behave differently on the two sides of T_c.
• ### Comment on "Feynman Effective Classical Potential in the Schrodinger Formulation"(quant-ph/0205067)
May 13, 2002 quant-ph, hep-th
We comment on the paper "Feynman Effective Classical Potential in the Schrodinger Formulation"[Phys. Rev. Lett. 81, 3303 (1998)]. We show that the results in this paper about the time evolution of a wave packet in a double well potential can be properly explained by resorting to a variational principle for the effective action. A way to improve on these results is also discussed.
• ### The $(\lambda \Phi^4)_4$ theory on the lattice: effective potential and triviality(hep-lat/9709057)
Sept. 17, 1997 hep-lat
We compute numerically the effective potential for the $(\lambda \Phi^4)_4$ theory on the lattice. Three different methods were used to determine the critical bare mass for the chosen bare coupling value. Two different methods for obtaining the effective potential were used as a control on the results. We compare our numerical results with three theoretical descriptions. Our lattice data are in quite good agreement with the Triviality and Spontaneous Symmetry Breaking'' picture.
• ### A lattice test of alternative interpretations of triviality'' in $(\lambda \Phi^4)_4$ theory(hep-ph/9702407)
Feb. 26, 1997 hep-th, hep-ph, hep-lat
There are two physically different interpretations of triviality'' in $(\lambda\Phi^4)_4$ theories. The conventional description predicts a second-order phase transition and that the Higgs mass $m_h$ must vanish in the continuum limit if $v$, the physical v.e.v, is held fixed. An alternative interpretation, based on the effective potential obtained in triviality-compatible'' approximations (in which the shifted Higgs' field $h(x)\equiv \Phi(x)-<\Phi>$ is governed by an effective quadratic Hamiltonian) predicts a phase transition that is very weakly first-order and that $m_h$ and $v$ are both finite, cutoff-independent quantities. To test these two alternatives, we have numerically computed the effective potential on the lattice. Three different methods were used to determine the critical bare mass for the chosen bare coupling value. All give excellent agreement with the literature value. Two different methods for obtaining the effective potential were used, as a control on the results. Our lattice data are fitted very well by the predictions of the unconventional picture, but poorly by the conventional picture.
• ### Lattice $(\Phi^4)_4$ Effective Potential Giving Spontaneous Symmetry Breaking and the Role of the Higgs Mass(hep-lat/9410001)
Oct. 4, 1994 hep-th, hep-ph, hep-lat
We present a critical reappraisal of the available results on the broken phase of $\lambda(\Phi^4)_4$ theory, as obtained from rigorous formal analyses and from lattice calculations. All the existing evidence is compatible with Spontaneous Symmetry Breaking but dictates a trivially free shifted field that becomes controlled by a quadratic hamiltonian in the continuum limit. As recently pointed out, this implies that the simple one-loop effective potential should become effectively exact. Moreover, the usual naive assumption that the Higgs mass-squared $m^2_h$ is proportional to its renormalized'' self-coupling $\lambda_R$ is not valid outside perturbation theory: the appropriate continuum limit has $m_h$ finite and vanishing $\lambda_R$. A Monte Carlo lattice computation of the $\lambda(\Phi^4)_4$ effective potential, both in the single-component and in the O(2)-symmetric cases, is shown to agree very well with the one-loop prediction. Moreover, its perturbative leading-log improvement (based on the concept of $\lambda_R$) fails to reproduce the Monte Carlo data. These results, while supporting in a new fashion the peculiar triviality'' of the $\lambda(\Phi^4)_4$ theory, also imply that, outside perturbation theory, the magnitude of the Higgs mass does not give a measure of the observable interactions in the scalar sector of the standard model.
• ### The Real Test of Triviality'' on the Lattice(hep-th/9402071)
Feb. 11, 1994 hep-th, hep-ph, hep-lat
The generally accepted triviality'' of $\lambda\Phi^4$ theories does not forbid Spontaneous Symmetry Breaking but implies a trivially free shifted field which becomes effectively governed by a quadratic hamiltonian. As a consequence, one expects the one-loop potential to be exact . We present a lattice computation of the effective potential for massless $\lambda\Phi^4$ theory which nicely confirms the expectations based on `triviality''. Our results imply that the magnitude of the Higgs boson mass, beyond perturbation theory, does not represent a measure of the observable interactions in the scalar sector of the standard model. | 2021-01-27 21:51:12 | {"extraction_info": {"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, "math_score": 0.6234405040740967, "perplexity": 993.8216258140952}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-04/segments/1610704833804.93/warc/CC-MAIN-20210127214413-20210128004413-00147.warc.gz"} |
https://www.math10.com/forum/viewtopic.php?f=1&t=8384 | # Need help with few questions
Algebra
### Need help with few questions
Hi I need help in solving the following equation. This question is Picked up from subject BCS12 Page 6. I have solved it on paper but I am not able to understand some logic. Please see the paper for further details.
The answer given in Book is :
$$W^{2}$$+W=-1
because
$$W^{2}$$+W+1=0
For my question please refer to the attachment. I have tried solving but not clear with the logic
Attachments
Unit 1 Question.jpg (67.16 KiB) Viewed 130 times
sunny
Posts: 5
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### Re: Need help with few questions
Unable to post 2 images in same post so posting img from book which contains answer
Attachments
book question and answer unit 1.png (6.26 KiB) Viewed 128 times
sunny
Posts: 5
Joined: Sun Jul 28, 2019 12:42 am
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### Re: Need help with few questions
Good night!
$$\omega$$ is the imaginary cube root of unity.
All the numbers that are solution to:
$$x^3-1=0$$
So:
$$x^3-1=(x-1)(x^2+x+1)=0$$
Solving:
$$x=1$$
This is one solution. There are another 2:
$$x^2+x+1=0\\ \Delta=(1)^2-4(1)(1)=-3\\ x=\dfrac{-1\pm\sqrt{-3}}{2}\\ x=\dfrac{-1\pm i\sqrt{3}}{2}\\ x'=\dfrac{-1}{2}+i\dfrac{\sqrt{3}}{2}\\ x''=\dfrac{-1}{2}-i\dfrac{\sqrt{3}}{2}$$
If $$\omega=\dfrac{-1}{2}+i\dfrac{\sqrt{3}}{2}$$, so
$$\omega^2=\left(\dfrac{-1}{2}+i\dfrac{\sqrt{3}}{2}\right)^2=\left(\dfrac{-1}{2}\right)^2+2\cdot\dfrac{-1}{2}\cdot i\dfrac{\sqrt{3}}{2}+\left(i\dfrac{\sqrt{3}}{2}\right)^2\\ \omega^2=\dfrac{1}{4}-i\dfrac{\sqrt{3}}{2}-\dfrac{3}{4}=\dfrac{-1}{2}-i\dfrac{\sqrt{3}}{2}$$
So, the solutions are:
1, $$\omega$$ and $$\omega^2$$
$$\omega^2+\omega+1=\dfrac{-1}{2}-i\dfrac{\sqrt{3}}{2}+\dfrac{-1}{2}+i\dfrac{\sqrt{3}}{2}+1=-1+1=0$$
I hope to have helped!
Baltuilhe
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### Re: Need help with few questions
I don't know what to say. I am speechless. I am not able to understand a word you wrote. I am new to this, is there an easy way to solve it? Or you think I must Understand what you wrote.
Good night, thanx for trying to help. Hoping someone can help me with a easier way.
I have not studied about cube root of unity, so I have no idea about it. I am learning determinants only at this point, can this be solved using determinants.
I have studied Order 2 and Just started reading creamer's rule.
sunny
Posts: 5
Joined: Sun Jul 28, 2019 12:42 am
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### Re: Need help with few questions
I might have found an answer, please let me know if I am right.
After searching google I have came across a few help pages, that says its already proven that
$$W^{2}$$ + W+1 =0
so since my answer is
$$W^{2}$$ + W
hence using above theorem we can say that my answer is -1
Am i thinking it correctly?
PS: I see that you have used symbol for unity, I can only see elected symbols in LaTeX Help, how can I find rest of the symbols?
sunny
Posts: 5
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### Re: Need help with few questions
Solution to problem found so removing the post
sunny
Posts: 5
Joined: Sun Jul 28, 2019 12:42 am
Reputation: 0 | 2019-08-26 06:02:14 | {"extraction_info": {"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, "math_score": 0.7326633334159851, "perplexity": 1302.4420015287747}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-35/segments/1566027330968.54/warc/CC-MAIN-20190826042816-20190826064816-00483.warc.gz"} |
https://www.quizover.com/physics-k12/course/8-11-unbalanced-force-system-laws-of-motion-by-openstax?page=2 | # 8.11 Unbalanced force system (Page 3/3)
Page 3 / 3
As a matter of fact, this velocity profile is typical of any vehicle, which is first accelerated, then run with minimum acceleration or even zero acceleration and finally brought to rest with constant deceleration.
We work out an example for studying motion under variable force, which is similar to the earlier example except that force is now dependent on time.
Problem : A block of mass “m” is pulled by a string on a smooth horizontal surface with force F = kt, where “k” is a constant. The string maintains an angle “θ” with horizontal. Find the time when block breaks off from the surface.
Solution : Here, important thing is to know the meaning of "breaking off". It means that physical contact between two surfaces is broken off. In that condition, the normal force should disappear as there is no contact between block and surface.
Since normal force is directed vertically up, we need to analyze forces in y-direction only so that we could apply the condition corresponding to "breaking off".
$\text{Free body diagram of the block}$
$\begin{array}{l}\sum {F}_{y}=N+kt\mathrm{sin}\theta -mg=0\\ ⇒N=mg-kt\mathrm{sin}\theta \end{array}$
Now, let $t={t}_{B}$ (break off time) when N = 0 (breaking off condition)
$\begin{array}{l}⇒0=mg-k{t}_{B}\mathrm{sin}\theta \\ ⇒k{t}_{B}\mathrm{sin}\theta =mg\\ ⇒{t}_{B}=\frac{mg}{k\mathrm{sin}\theta }\end{array}$
## Motional mechanism of animals
According to laws of motion, motion of a body is not possible by the application of internal force. On the other hand, animals move around using internal muscular force. This is not a contradiction, but an intelligent maneuvering on the part of animals, which use internal muscular force to generate external force on them.
Let us take the case of our own movement. So long we stand upright, applying weight on the ground in vertical direction; there is no motion.
To move forward (say), we need to press back the surface at an angle. The ground applies an equal and opposite force (say, reaction of the ground). The reaction of the surface is an external force for our body. The horizontal component of the reaction force moves our body forward, whereas the vertical force balances our weight.
## Elements of body system
The underlying frame work of the analysis of force systems in inertial frame of reference is now almost complete. There is nothing new as far as application of laws of motion is concerned. But, there is a big “but” with respect to details of various elements of body systems that we consider during study of motion. These elements typically are block, string, incline, pulleys and spring.
The whole gamut of analysis in dynamics requires systemic approach to answer following questions :
• what are the forces ?
• which of them are external forces ?
• does friction is part of the external force system ?
• whether forces are collinear, coplanar or three-dimensional ?
• are forces balanced or unbalanced ?
• are the forces time dependent ?
• is the motion taking place in inertial frame or accelerated frame ?
• what would be the appropriate coordinate system for analysis ?
• what are the characteristics of system elements ?
It is not very difficult to realize that our job is half done if we are able to classify the system in hand based on the answers to above questions. Though, we have listed the system elements at the end of the list, we shall soon realize that a great deal of our effort in getting answers to questions 1, 2, 4 and 8 are largely determined by the elements involved, while the rest are situation specific.
We have briefly described system elements like block, string, pulley etc. In subsequent modules, we shall emphasize details of these and other elements.
if x=a-b, a=5.8cm b=3.22 cm find percentage error in x
x=5.8-3.22 x=2.58
what is the definition of resolution of forces
what is energy?
can anyone tell who founded equations of motion !?
n=a+b/T² find the linear express
أوك
عباس
Quiklyyy
Moment of inertia of a bar in terms of perpendicular axis theorem
How should i know when to add/subtract the velocities and when to use the Pythagoras theorem?
Centre of mass of two uniform rods of same length but made of different materials and kept at L-shape meeting point is origin of coordinate
A balloon is released from the ground which rises vertically up with acceleration 1.4m/sec^2.a ball is released from the balloon 20 second after the balloon has left the ground. The maximum height reached by the ball from the ground is
work done by frictional force formula
Torque
Why are we takingspherical surface area in case of solid sphere
In all situatuons, what can I generalize? | 2018-08-14 08:30:17 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 4, "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, "math_score": 0.471452921628952, "perplexity": 756.7844298238489}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-34/segments/1534221208750.9/warc/CC-MAIN-20180814081835-20180814101835-00351.warc.gz"} |
http://mathhelpforum.com/calculus/206965-optimization-problem.html | # Math Help - optimization problem
1. ## optimization problem
Suppose R is the region bounded by the curves y = any function f(x) and y = c, on the interval a to b. Find the value of c that minimizes the volume of the solid that is generated by revolving R about the line y = c.
2. ## Re: optimization problem
$V(c)=\int_a^b \pi[f(x)-c]^2 dx$
$\frac{dV}{dc}=\int_a^b 2\pi[f(x)-c](-1) dx$
$=-2\pi\int_a^b [f(x)-c] dx$
resolve dV/dt = 0, we get $c = \frac{\int_a^b f(x)dx}{b-a}$
3. ## Re: optimization problem
Where does the (-1) in step two come from? | 2014-09-20 20:23:49 | {"extraction_info": {"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": 4, "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, "math_score": 0.4890689253807068, "perplexity": 551.5903565317055}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-41/segments/1410657133564.63/warc/CC-MAIN-20140914011213-00259-ip-10-196-40-205.us-west-1.compute.internal.warc.gz"} |
https://brilliant.org/problems/very-efficient-coefficient-2/ | # Very efficient coefficient
What is the value of the coefficient $$x^2 y^4 z^8$$ in the expression $$(x+y+z)^{14}$$?
× | 2017-01-20 01:45:16 | {"extraction_info": {"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, "math_score": 0.506449282169342, "perplexity": 783.6760166798364}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-04/segments/1484560280763.38/warc/CC-MAIN-20170116095120-00297-ip-10-171-10-70.ec2.internal.warc.gz"} |
https://www.mathalino.com/forum/algebra/math-2 | # Math
2 posts / 0 new
Misha Al
Math
A train flies round trip at a distance of X miles each away. The velocity with the head wind is 160mph, while the velocity with tail wind is 240mph, What is the average speed for the round trip?
Tags:
8eightI'sD
To get the average speed for the round trip, recall that...
$$distance = (speed)(time)$$ $$d = vt$$
We also know that $$average \space speed = \frac{total \space distance}{total \space time}$$
With that in mind...
The time required by train (a flying train, I suppose, hehe) to reach its destination is $\frac{X}{160}$ hours. The time required by train to return from its destination is $\frac{X}{240}$ hours.
So...
The total distance for the round trip would be $X+X = 2X$ miles.
The total time spent flying would be $\frac{X}{160}+\frac{X}{240} = \frac{X}{96}$ hours.
Therefore, the average speed for the entire round trip would be $\frac{2X \space miles}{\frac{X}{96} \space hours}$ = $\color{green}{192 \space miles \space per \space hour}$
Alternate solutions are encouraged...
Subscribe to MATHalino.com on | 2019-02-21 01:15:19 | {"extraction_info": {"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, "math_score": 0.35371601581573486, "perplexity": 1411.9030511806432}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-09/segments/1550247497858.46/warc/CC-MAIN-20190221010932-20190221032932-00430.warc.gz"} |
http://opticsjournal.net/Journals/JColumnList?cid=2274 | Coherent beam combining of 107 beams has been demonstrated for the first time to the best of our knowledge. When the system was in closed loop, the pattern in far-field was stable and the fringe contrast was $>96%$
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We propose and demonstrate experimentally and numerically a network of three globally coupled semiconductor lasers (SLs) that generate triple-channel chaotic signals with time delayed signature (TDS) concealment. The effects of the coupling strength and bias current on the concealment of the TDS are investigated. The generated chaotic signals are further applied to reinforcemen
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A narrow-linewidth laser operating at the telecommunications band combined with both fast and wide-band tuning features will have promising applications. Here we demonstrate a single-mode (both transverse and longitudinal mode) continuous microlaser around 1535 nm based on a fiber Fabry–Pérot microcavity, which achieves wide-band tuning without mode hopping to the 1.3 THz rang
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Internal motions in femtosecond soliton molecules provide insight into universal collective dynamics in various nonlinear systems. Here we introduce an orbital-angular-momentum (OAM)-resolved method that maps the relative phase motion within a femtosecond soliton molecule into the rotational movement of the interferometric beam profile of two optical vortices. By this means, lo
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The sweep rate, sweep range, and coherence length of swept sources, respectively, determine the acquisition rate, axial resolution, and imaging range of optical coherence tomography (OCT). In this paper, we demonstrate a reconfigurable high-speed and broadband swept laser by time stretching of a flat spectrum femtosecond pulse train with over 100 nm bandwidth and a repetition r
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Temporal and spatial resonant modes are always possessed in physical systems with energy oscillation. In ultrafast fiber lasers, enormous progress has been made toward controlling the interactions of many longitudinal modes, which results in temporally mode-locked pulses. Recently, optical vortex beams have been extensively investigated due to their quantized orbital angular mo
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Quantum defects (QDs) have always been a key factor of the thermal effect in high-power fiber lasers. Much research on low-QD fiber lasers has been reported in the past decades, but most of it is based on active fibers. Besides, Raman fiber lasers based on the stimulated Raman scattering effect in passive fiber are also becoming an important kind of high-power fiber laser for t
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Ultra-narrow-linewidth mode-locked lasers with wide wavelength tunability can be versatile light sources for a variety of newly emergent applications. However, it is very challenging to achieve the stable mode locking of substantially long, anomalously dispersive fiber laser cavities employing a narrowband spectral filter at the telecom band. Here, we show that a nearly dispers
PDF全文 Photonics Research | 2020,8(07):1100-1109 | 2021-01-28 04:42:57 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 1, "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, "math_score": 0.23937027156352997, "perplexity": 3313.516867161902}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-04/segments/1610704835901.90/warc/CC-MAIN-20210128040619-20210128070619-00112.warc.gz"} |
https://www.rdocumentation.org/packages/methods/versions/3.6.1/topics/MethodsList | MethodsList
0th
Percentile
MethodsList Objects
These functions create and manipulate MethodsList objects, the objects formerly used in R to store methods for dispatch. Use of these objects is deprecated since R 3.2.0, as it will rarely be a good idea. Where methods dispatch is to be studied, see selectMethod. For computations that iterate over methods or over method signatures, see findMethods, which returns a linearized methods list to hold method definitions, usually more convenient for iteration than the recursive MethodsList objects.
Keywords
internal
Usage
listFromMlist(mlist, prefix = list(), sigs. = TRUE, methods. = TRUE)linearizeMlist(mlist, inherited = TRUE)finalDefaultMethod(method)loadMethod(method, fname, envir)##--------- These are all deprecated, since R 3.2.0 ----------MethodsList(.ArgName, ...)
makeMethodsList(object, level=1)
SignatureMethod(names, signature, definition)
insertMethod(mlist, signature, args, def, cacheOnly)
inheritedSubMethodLists(object, thisClass, mlist, ev)showMlist(mlist, includeDefs = TRUE, inherited = TRUE,
classes, useArgNames, printTo = stdout() )
# S3 method for MethodsList
print(x, ...)mergeMethods(m1, m2, genericLabel)
Details
%% \item{\code{MethodsList}:}{ %% Create a MethodsList object out of the arguments.
%% Conceptually, this object is a named collection of methods to be %% dispatched when the (first) argument in a function call matches %% the class corresponding to one of the names. A final, unnamed %% element (i.e., with name \code{""}) corresponds to the default %% method.
%% The elements can be either a function, or another MethodsList. In %% the second case, this list implies dispatching on the second %% argument to the function using that list, given a selection of %% this element on the first argument. Thus, method dispatching on %% an arbitrary number of arguments is defined.
%% MethodsList objects are used primarily to dispatch OOP-style %% methods and, in R, to emulate S4-style methods. %% }
%% \item{\code{SignatureMethod}:}{ %% construct a MethodsList object containing (only) this method, %% corresponding to the signature; i.e., such that %% \code{signature[[1]]} is the match for the first argument, %% \code{signature[[2]]} for the second argument, and so on. The %% string \code{"missing"} means a match for a missing argument, and %% \code{"ANY"} means use this as the default setting at this level.
%% The first argument is the argument names to be used for dispatch %% corresponding to the signatures. %% }
%% \item{\code{insertMethod}:}{ %% insert the definition \code{def} into the MethodsList object, %% \code{mlist}, corresponding to the signature. By default, insert %% it in the slot \code{"methods"}, but \code{cacheOnly=TRUE} inserts %% it into the \code{"allMethods"} slot (used for dispatch but not saved). %% }
%% \item{\code{inheritedSubMethodLists}:}{ %% Utility function to match the object or the class (if the object %% is \code{NULL}) to the elements of a methods list. %% Used in finding inherited methods, and not meant to be called %% directly. %% }
%% \item{\code{showMlist}:}{ %% Prints the contents of the MethodsList. If \code{includeDefs} the %% signatures and the corresponding definitions will be printed; %% otherwise, only the signatures. %% }
listFromMlist:
Undo the recursive nature of the methods list, making a list of list(sigs,methods) of function definitions, i.e.of matching signatures and methods. prefix is the partial signature (a named list of classes) to be prepended to the signatures in this object. If sigs. or methods. are FALSE, the resulting part of the return value will be empty.
A utility function used to iterate over all the individual methods in the object, it calls itself recursively.
linearizeMlist:
Undo the recursive nature of the methods list, making a list of function definitions, with the names of the list being the corresponding signatures.
Designed for printing; for looping over the methods, use the above listFromMlist instead.
finalDefaultMethod:
The default method or NULL. With the demise of "MethodsList" objects, this function only checks that the value given it is a method definition, primitive or NULL.
%% \item{\code{mergeMethods}:}{ %% Merges the methods in the second MethodsList object into the %% first, and returns the merged result. Called from %% \code{\link{getAllMethods}}. For a primitive function, %% \code{genericLabel} is supplied as the name of the generic. %% }
loadMethod:
Called, if necessary, just before a call to method is dispatched in the frame envir. The function exists so that methods can be defined for special classes of objects. Usually the point is to assign or modify information in the frame environment to be used evaluation. For example, the standard class MethodDefinition has a method that stores the target and defined signatures in the environment. Class MethodWithNext has a method taking account of the mechanism for storing the method to be used in a call to callNextMethod.
Any methods defined for loadMethod must return the function definition to be used for this call; typically, this is just the method argument.
%% \item{\code{MethodsListSelect}}{ %% The function \code{MethodsListSelect} performs a full search %% (including all inheritance and group generic information: see the %% \link{Methods} documentation page for details on how this works). %% The call returns a possibly revised methods list object, %% incorporating any method found as part of the \code{allMethods} %% slot. This search was used by the evaluator when methods lists %% were the metadata for methods dispatch. This function is now deprecated. %% }
References
Chambers, John M. (2008) Software for Data Analysis: Programming with R Springer. (For the R version.)
Chambers, John M. (1998) Programming with Data Springer (For the original S4 version.)
Aliases
• listFromMlist
• linearizeMlist
• finalDefaultMethod
• MethodsList
• makeMethodsList
• SignatureMethod
• insertMethod
• inheritedSubMethodLists
• showMlist
• print.MethodsList
• emptyMethodsList
• insertMethodInEmptyList
• mergeMethods
• MethodsListSelect
Documentation reproduced from package methods, version 3.6.1, License: Part of R 3.6.1
Community examples
Looks like there are no examples yet. | 2019-11-12 02:51:51 | {"extraction_info": {"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, "math_score": 0.7843104004859924, "perplexity": 5519.584270893576}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-47/segments/1573496664567.4/warc/CC-MAIN-20191112024224-20191112052224-00053.warc.gz"} |
https://www.w3spoint.com/jquery-ui-accordion | jQuery UI Accordion
Being an expandable and collapsible content holder, the jQuery UI Accordion is broken into sections. It probably resembles like tabs.
Syntax:
The accordion() method can be used in two forms:
$(selector, context).accordion (options) Method OR$(selector, context).accordion (“action”, params) Method
First Method: The accordion (options) Method:
To indicate that an HTML element and its contents should be treated and managed as accordion menus, the accordion (options) method is used. The appearance and behavior of the menus involved are specified by the options parameter which is an object.
Syntax:
$(selector, context).accordion (options); Multiple options can also be used at a time using the Javascript object. For this, the options need to be separated using a comma. Syntax: $(selector, context).accordion({option1: value1, option2: value2..... });
The popular options that can be used with this method are listed below:
Example 1:
jQuery UI Accordion
Flower meaning:
The seed-bearing part of a plant, consisting of reproductive organs (stamens and carpels) that are typically surrounded by a brightly coloured corolla (petals) and a green calyx (sepals).
Fruit meaning:
The sweet and fleshy product of a tree or other plant that contains seeds and can be eaten as food.
Vegetable meaning:
A plant or part of a plant used as food, such as cabbage, potato, turnip, or bean.
Output:
Explanation:
In the above example, we are displaying the use and the behavior of the jQuery UI Accordion() method.
Second Method: The accordion (“action”, params) method:
To perform an action on accordion elements, the accordion (“action”, params) method is used. It can be used for performing actions like selecting/deselecting the accordion menu. The action here is the first argument and is specified as a string. For example, disables all the menus, the “disable” should be passed as a value to the action parameter.
Syntax:
\$(selector, context).accordion ("action", params);
The actions that can be used with this method are listed below:
Action Uses destroy To destroy the accordion functionality of an element completely, so that the elements return to their pre-init state. disable To disable all menus. No click is thus taken into account. No argument is accepted by this method. enable To reactivate all menus. The clicks will be considered again. No argument is accepted by this method. option(optionName) To retrieve the value of currently associated accordion element with the specified optionname. A string value is taken as an argument by this method. option To retrieve an object containing key/value pairs representing the current accordion options hash. option(optionName, value) To set the value of the accordion option associated with the specified optionName. option(options) To set one or more options for the accordion. A map of option-value pairs to set is specified by the option parameter. refresh To process any headers and panels that were added or removed directly in the dom. The height of the accordion panels is than recomputed. The output thus have a dependency on the content and the heightStyle option. No argument is accepted by this method. widget To retrieve the accordion widget element; the one annotated with the UI-accordion class name.
Example 2:
jQuery UI Accordion Example
Flower meaning:
The seed-bearing part of a plant, consisting of reproductive organs (stamens and carpels) that are typically surrounded by a brightly coloured corolla (petals) and a green calyx (sepals).
Fruit meaning:
The sweet and fleshy product of a tree or other plant that contains seed and can be eaten as food.
Vegetable meaning:
A plant or part of a plant used as food, such as a cabbage, potato, turnip, or bean.
Disable Accordion Enable Accordion
Output 1:
Output 2:
Explanation:
In the above example, we are displaying the use and behavior of the option (optionName, value) method. Here, we are disabling and enabling the accordion. | 2021-04-17 12:09:40 | {"extraction_info": {"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, "math_score": 0.21491213142871857, "perplexity": 5519.964962441685}, "config": {"markdown_headings": false, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-17/segments/1618038119532.50/warc/CC-MAIN-20210417102129-20210417132129-00386.warc.gz"} |
https://www.tutorialspoint.com/list-assign-function-in-cplusplus-stl | List assign() function in C++ STL
C++Server Side ProgrammingProgramming
Given is th e task to show the working of the assign() function in C++.
The list::assign() function is a part of the C++ standard template library. It is used to assign the values to a list and also to copy values from one list to another.
<list> header file should be included to call this function.
Syntax
The syntax for assigning new values is as follows −
List_Name.assign(size,value)
Syntax
The syntax for copying values from one list to another is as follows −
First_List.assign(Second_List.begin(),Second_list.end())
Parameters
The function takes two parameters −
First is size, that represents the size of the list and the second one is value, which represents the data value to be stored inside the list.
Return Value
The function has no return value.
Example
Input: Lt.assign(3,10)
Output: The size of list Lt is 3.
The elements of the list Lt are 10 10 10.
Explanation
The following example shows how we can assign a list its size and values by using the assign() function. The first value that we will pass inside the list function becomes the size of the list, in this case it is 3 and the second element is the value that is assigned to each position of the list and here it is 10.
Example
Input: int array[5] = { 1, 2, 3, 4 }
Lt.assign(array,array+3)
Output: The size of list Lt is 3.
The elements of the list Lt are 1 2 3.
Explanation
The following example shows how we can assign values to a list using an array. The total number of elements that we will assign to the list becomes the size of the list.
The user has to simply pass the name of the array as the first argument inside the assign() function, and the second argument should be the name of the array, then a “+” sign followed by the number of elements the user wants to assign to the list.
In the above case we have written 3, so the first three elements of the array will be assigned to the list.
If we write a number that is bigger than the number of elements present in the array, let us say 6, then the program will not show any error instead the size of the list will become 6 and the extra positions in the list will be assigned with the value zero.
Approach used in the below program as follows
• First create a function ShowList(list<int> L) that will display the elements of the list.
• Create an iterator, let’s say itr that will contain the initial element of the list to be displayed.
• Make the loop run till itr reaches the final element of the list.
• Then inside the main() function create three lists using list<int> let’s say L1, L2 ad L3 so that they accept values of type int and then create an array of type int, let’s say arr[] and assign it some values.
• Then use the assign() function to assign size and some values to the list L1 and then pass the list L1 into the ShowDisplay() function.
• Then use the assign() function to copy elements of list L1 into L2 and also pass the list L2 into the ShowList() function.
• Then use the assign() function to copy elements of the array arr[] into the list L3 and pass the list L3 into the DisplayList() function.
Algorithm
Start
Step 1-> Declare function DisplayList(list<int> L) for showing list elements
Declare iterator itr
Loop For itr=L.begin() and itr!=L.end() and itr++
Print *itr
End
Step 2-> In function main()
Declare lists L1,L2,L3
Initialize array arr[]
Call L1.assign(size,value)
Print L1.size();
Call function DisplayList(L1) to display L1
Call L2.assign(L1.begin(),L1.end())
Print L2.size();
Call function DisplayList(L2) to display L2
Call L3.assign(arr,arr+4)
Print L3.size();
Call function DisplayList(L3) to display L3
Stop
Example
Live Demo
#include<iostream>
#include<list>
using namespace std;
int ShowList(list<int> L) {
cout<<"The elements of the list are ";
list<int>::iterator itr;
for(itr=L.begin(); itr!=L.end(); itr++) {
cout<<*itr<<" ";
}
cout<<"\n";
}
int main() {
list<int> L1;
list<int> L2;
list<int> L3;
int arr[10] = { 6, 7, 2, 4 };
//assigning size and values to list L1
L1.assign(3,20);
cout<<"The size of list L1 is "<<L1.size()<<"\n";
ShowList(L1);
//copying the elements of L1 into L3
L2.assign(L1.begin(),L1.end());
cout<<"The size of list is L2 "<<L2.size()<<"\n";
ShowList(L2);
//copying the elements of arr[] into list L3
L3.assign(arr,arr+4);
cout<<"The size of list is L3 "<<L3.size()<<"\n";
ShowList(L3);
return 0;
}
Output
If we run the above code it will generate the following output −
The size of list L1 is 3
The elements of the list are 20 20 20
The size of list L2 is 3
The elements of the list are 20 20 20
The size of list L3 is 4
The elements of the list are 6 7 2 4
Explanation
For the list L1 we gave size as 3 and value as 20 in the assign() function which had generated the above output.
Then we copied the elements of list L1 into that L2 that gives L2 the same size and value as that of L1 as we can see in the output.
Then we used the assign function to copy all the elements of the array arr[] which made the size of L3 equal to 4 and its elements same as the elements of the array as in the output.
Updated on 20-Jan-2020 07:22:18 | 2022-06-28 18:19:49 | {"extraction_info": {"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, "math_score": 0.18231406807899475, "perplexity": 1446.307586463167}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656103573995.30/warc/CC-MAIN-20220628173131-20220628203131-00569.warc.gz"} |
https://www.audiolabs-erlangen.de/resources/MIR/FMP/C5/C5S3_ChordRec_HMM.html | HMM-Based Chord Recognition
Following Section 5.3.4 of [Müller, FMP, Springer 2015], we discuss in this notebook an HMM-based approach for chord recognition. The idea of using HMMs for chord recognition was originally introduced by Sheh and Ellis.
• Alexander Sheh, Daniel P. W. Ellis: Chord segmentation and recognition using EM-trained hidden Markov models. Proceedings of the International Conference on Music Information Retrieval (ISMIR), Baltimore, 2003.
• Taemin Cho, Juan Pablo Bello: On the Relative Importance of Individual Components of Chord Recognition Systems. IEEE/ACM Transactions on Audio, Speech, and Language Processing, 22 (2014), pp. 466–492.
• Nanzhu Jiang, Peter Grosche, Verena Konz, Meinard Müller: Analyzing Chroma Feature Types for Automated Chord Recognition. Proceedings of the AES Conference on Semantic Audio, Ilmenau, Germany, 2011.
Introduction¶
We now show how the concept of HMMs can be applied to improve automated chord recognition. First of all, we need to create an HMM that suitably models our chord recognition problem. Generally, as introduced in the FMP noteboook on HMMs, an HMM is specified by the parameters $\Theta:=(\mathcal{A},A,C,\mathcal{B},B)$. In the chord recognition context, the set
$$\mathcal{A}:=\{\alpha_{1},\alpha_{2},\ldots,\alpha_{I}\}.$$
of states is used to model the various chord types that are allowed in the recognition problem. As in the FMP notebook on template-based chord recognition, we consider in this notebook only the twelve major and twelve minor triads, thus setting
$$\label{eq:ChordReco:HMM:App:Spec:SetStates} \mathcal{A} = \{\mathbf{C},\mathbf{C}^\sharp,\ldots,\mathbf{B},\mathbf{Cm},\mathbf{Cm^\sharp},\ldots,\mathbf{Bm}\}$$
In this case, the HMM consists of $I=24$ states, which we enumerate as indicated above. For example, $\alpha_{1}$ corresponds to $\mathbf{C}$ and $\alpha_{13}$ to $\mathbf{Cm}$. In the remainder of this notebook, we do the following:
• First, we explain how to explicitly create an HMM by specifying the other HMM parameters in a musically informed fashion. Even though these parameters may be learned automatically from training data using the Baum–Welch Algorithm, the manual specification of HMM parameters is instructive and leads to an HMM with an explicit musical meaning.
• Second, we apply this HMM for chord recognition. The input (i.e., observation sequence) of the HMM is a chromagram representation of the music recording. Applying the Viterbi Algorithm, we then derive an optimal state sequence (consisting of chord labels) that best explains the chroma sequence. The sequence of chord labels yields our frame-wise chord recognition result.
We will compare the HMM-based chord recognition results with the results obtained from the template-based approach. In particular we will see that the HMM transition model introduces a kind of context-aware postfiltering.
Specification of Emission Likelihoods¶
In our chord recognition scenario, the observations are chroma vectors that have previously been extracted from the given audio recording. In other words, the observations are $12$-dimensional real-valued vectors which are elements of the continuous feature space $\mathcal{F}=\mathbb{R}^{12}$. So far, we have only considered the case of discrete HMMs, where the observations are discrete symbols coming from a finite output space $\mathcal{B}$. To make discrete HMMs applicable to our scenario, one possible procedure is to introduce a finite set of prototype vectors, a so-called codebook. Such a codebook can be regarded as a discretization of the continuous feature space $\mathcal{F}=\mathbb{R}^{12}$, where each codebook vector represents an entire range of feature vectors. Emission probabilities can then be determined on the basis of this finite set of codebook vectors.
As an alternative, we use in the following an HMM variant, where we replace the discrete output space $\mathcal{B}$ by the continuous feature space $\mathcal{F}=\mathbb{R}^{12}$ and the emission probability matrix $B$ by likelihood functions. In particular, the emission probability of a given state is replaced by a normalized similarity value defined as inner product of a state-dependent normalized template and a normalized observation (chroma) vector. To compute the state-dependent likelihood functions, we proceed as described in the FMP notebook on template-based chord recognition. Let $s:\mathcal{F} \times \mathcal{F} \to [0,1]$ be the similarity measure defined by the inner product of normalized chroma vectors (where one should use a thresholded normalization to avoid division by zero):
$$s(x, y) = \frac{\langle x,y\rangle}{\|x\|_2\cdot\|y\|_2}$$
for $x,y\in\mathcal{F}$. Based on $I=24$ major and minor triads (encoded by the states $\mathcal{A}$ and indexed by the set $[1:I]$), we consider the binary chord templates $\mathbf{t}_i\in \mathcal{F}$ for $i\in [1:I]$. Then, we define the state-dependent likelihood function $b_i:\mathcal{F}\to [0,1]$ by
$$b_i(x) := \frac{s(x, \mathbf{t}_i)}{\sum_{j\in[1:I]}s(x, \mathbf{t}_j)}$$
for $x\in\mathcal{F}$ and $i\in [1:I]$. In our scenario, the observation sequence $O=(o_{1},o_{2},\ldots,o_{N})$ is a sequence of chroma vectors $o_n\in\mathcal{F}$. We define the observation-dependent $(I\times N)$-matrix $B[O]$ by
$$B[O](i,n) = b_i(o_n)$$
for $i\in[1:I]$ and $n\in[1:N]$. Note that his matrix is exactly the chord similarity matrix with a column-wise $\ell^1$-normalization, as introduced in the FMP notebook on template-based chord recognition and visualized in form of a time–chord representation. In context of the the Viterbi algorithm, the likelihood $B[O](i,n)$ is used to replace the probability value $b_{ik_n}$.
As our running example throughout the remainder of this notebook, we continue our Bach example introduced in the FMP notebook on chord recognition evaluation. In this example, we consider a piano recording of the first four measures of Johann Sebastian Bach's $\mathrm{C}$-major prelude. Furthermore, in the next code cell, we show the observation sequence $O$ (chromagram representation) as well as the likelihood matrix $B[O]$ (time–chord representation).
In [1]:
import os
import numpy as np
from matplotlib import pyplot as plt
import pandas as pd
from scipy.linalg import circulant
from numba import jit
import sys
sys.path.append('..')
import libfmp.b
from libfmp.c5 import get_chord_labels
%matplotlib inline
# Specify
fn_wav = os.path.join('..', 'data', 'C5', 'FMP_C5_F20_Bach_BWV846-mm1-4_Fischer.wav')
fn_ann = os.path.join('..', 'data', 'C5', 'FMP_C5_F20_Bach_BWV846-mm1-4_Fischer_ChordAnnotations.csv')
color_ann = {'C': [1, 0.5, 0, 1], 'G': [0, 1, 0, 1], 'Dm': [1, 0, 0, 1], 'N': [1, 1, 1, 1]}
N = 4096
H = 1024
X, Fs_X, x, Fs, x_dur = \
libfmp.c5.compute_chromagram_from_filename(fn_wav, N=N, H=H, gamma=0.1, version='STFT')
N_X = X.shape[1]
# Chord recogntion
chord_sim, chord_max = libfmp.c5.chord_recognition_template(X, norm_sim='1')
chord_labels = libfmp.c5.get_chord_labels(nonchord=False)
# Annotations
chord_labels = libfmp.c5.get_chord_labels(ext_minor='m', nonchord=False)
ann_matrix, ann_frame, ann_seg_frame, ann_seg_ind, ann_seg_sec = \
libfmp.c5.convert_chord_ann_matrix(fn_ann, chord_labels, Fs=Fs_X, N=N_X, last=True)
#P, R, F, TP, FP, FN = libfmp.c5.compute_eval_measures(ann_matrix, chord_max)
# Plot
cmap = libfmp.b.compressed_gray_cmap(alpha=1, reverse=False)
fig, ax = plt.subplots(3, 2, gridspec_kw={'width_ratios': [1, 0.03],
'height_ratios': [1.5, 3, 0.2]}, figsize=(9, 7))
libfmp.b.plot_chromagram(X, ax=[ax[0, 0], ax[0, 1]], Fs=Fs_X, clim=[0, 1], xlabel='',
title='Observation sequence (chromagram with feature rate = %0.1f Hz)' % (Fs_X))
libfmp.b.plot_segments_overlay(ann_seg_sec, ax=ax[0, 0], time_max=x_dur,
print_labels=False, colors=color_ann, alpha=0.1)
libfmp.b.plot_matrix(chord_sim, ax=[ax[1, 0], ax[1, 1]], Fs=Fs_X, clim=[0, np.max(chord_sim)],
title='Likelihood matrix (time–chord representation)',
ylabel='Chord', xlabel='')
ax[1, 0].set_yticks(np.arange(len(chord_labels)))
ax[1, 0].set_yticklabels(chord_labels)
libfmp.b.plot_segments_overlay(ann_seg_sec, ax=ax[1, 0], time_max=x_dur,
print_labels=False, colors=color_ann, alpha=0.1)
libfmp.b.plot_segments(ann_seg_sec, ax=ax[2, 0], time_max=x_dur, time_label='Time (seconds)',
colors=color_ann, alpha=0.3)
ax[2,1].axis('off')
plt.tight_layout()
Specification of Transition Probabilities¶
In music, certain chord transitions are more likely than others. This observation is our main motivation to employ HMMs, where the first-order temporal relationships between the various chords can be captured by the transition probability matrix $A$. In the following, we use the notation $\alpha_{i}\rightarrow\alpha_{j}$ to refer to the transition from state $\alpha_{i}$ to state $\alpha_{j}$ for $i,j\in[1:I]$. For example, the coefficient $a_{1,2}$ expresses the probability for the transition $\alpha_{1}\rightarrow\alpha_{2}$ (corresponding to $\mathbf{C}\rightarrow\mathbf{C}^\sharp$), whereas $a_{1,8}$ expresses the probability for $\alpha_{1}\rightarrow\alpha_{8}$ (corresponding to $\mathbf{C}\rightarrow\mathbf{G}$). In real music, the change from a tonic to the dominant is much more likely than changing by one semitone, so that the probability $a_{1,8}$ should be much larger than $a_{1,2}$. The coefficients $a_{i,i}$ express the probability of staying in state $\alpha_{i}$ (i.e., $\alpha_{i}\rightarrow\alpha_{i}$) for $i\in[1:I]$. These coefficients are also referred to as self-transition probabilities.
A transition probability matrix can be specified in many ways. For example, the matrix may be defined manually by a music expert based on rules from harmony theory. The most common approach is to generate such a matrix automatically by estimating the transition probabilities from labeled data. In the following figure, we show three different transition matrices (using a log probability scale for visualization purposes).
• The first one was learned from labeled training data based on the Beatles collection using bigrams (pairs of adjacent elements) in the labeled frame sequences. As an example, the coefficient $a_{1,8}$ (corresponding to the transition $\mathbf{C}\rightarrow\mathbf{G}$) has been highlighted.
• The second matrix is a transposition-invariant transition probability matrix obtained from the previous matrix. To achieve transposition invariance, the labeled training dataset is augmented by considering all twelve possible cyclic chroma shifts to the considered bigrams.
• The third matrix is a uniform transition probability matrix with a large value on the main diagonal (self-transitions) and a much smaller value at all remaining positions.
For more details on the construction of these transition matrices, we refer to Section 5.3.4.2 of [Müller, FMP, Springer 2015]. In the following code cell, we read a CSV-file that contains the precomputed transition matrix estimated on the basis of the Beatles collection. In the visualization, we show both the probability values as well as the log-probability values.
In [2]:
def plot_transition_matrix(A, log=True, ax=None, figsize=(6, 5), title='',
xlabel='State (chord label)', ylabel='State (chord label)',
"""Plot a transition matrix for 24 chord models (12 major and 12 minor triads)
Notebook: C5/C5S3_ChordRec_HMM.ipynb
Args:
A: Transition matrix
log: Show log probabilities (Default value = True)
ax: Axis (Default value = None)
figsize: Width, height in inches (only used when ax=None) (Default value = (6, 5))
title: Title for plot (Default value = '')
xlabel: Label for x-axis (Default value = 'State (chord label)')
ylabel: Label for y-axis (Default value = 'State (chord label)')
cmap: Color map (Default value = 'gray_r')
Returns:
fig: The created matplotlib figure or None if ax was given.
ax: The used axes.
im: The image plot
"""
fig = None
if ax is None:
fig, ax = plt.subplots(1, 1, figsize=figsize)
ax = [ax]
if log is True:
A_plot = np.log(A)
cbar_label = 'Log probability'
clim = [-6, 0]
else:
A_plot = A
cbar_label = 'Probability'
clim = [0, 1]
im = ax[0].imshow(A_plot, origin='lower', aspect='equal', cmap=cmap)
im.set_clim(clim)
plt.sca(ax[0])
cbar = plt.colorbar(im)
ax[0].set_xlabel(xlabel)
ax[0].set_ylabel(ylabel)
ax[0].set_title(title)
cbar.ax.set_ylabel(cbar_label)
chord_labels = get_chord_labels()
chord_labels_squeezed = chord_labels.copy()
for k in [1, 3, 6, 8, 10, 11, 13, 15, 17, 18, 20, 22]:
chord_labels_squeezed[k] = ''
ax[0].set_xticks(np.arange(24))
ax[0].set_yticks(np.arange(24))
ax[0].set_xticklabels(chord_labels_squeezed)
ax[0].set_yticklabels(chord_labels)
ax[0].axvline(x=11.5, ymin=0, ymax=24, linewidth=2, color='r')
ax[0].axhline(y=11.5, xmin=0, xmax=24, linewidth=2, color='r')
return fig, ax, im
# Load transition matrix estimated on the basis of the Beatles collection
fn_csv = os.path.join('..', 'data', 'C5', 'FMP_C5_transitionMatrix_Beatles.csv')
A_est = A_est_df.to_numpy('float64')
fig, ax = plt.subplots(1, 2, gridspec_kw={'width_ratios': [1, 1],
'height_ratios': [1]},
figsize=(10, 3.8))
plot_transition_matrix(A_est, log=False, ax=[ax[0]], title='Transition matrix')
plot_transition_matrix(A_est, ax=[ax[1]], title='Transition matrix with log probabilities')
plt.tight_layout()
To obtain the transposition-invariant transition matrix, we simulate the cyclic chroma shifts on the matrix-level by cyclically shifting and averaging the four quadrants (defined by the major-chord and minor-chord regions) of the original matrix. In the visualization, we show the original transition matrix as well as the resulting transposition-invariant transition matrix.
In [3]:
def matrix_circular_mean(A):
"""Computes circulant matrix with mean diagonal sums
Notebook: C5/C5S3_ChordRec_HMM.ipynb
Args:
A (np.ndarray): Square matrix
Returns:
A_mean (np.ndarray): Circulant output matrix
"""
N = A.shape[0]
A_shear = np.zeros((N, N))
for n in range(N):
A_shear[:, n] = np.roll(A[:, n], -n)
circ_sum = np.sum(A_shear, axis=1)
A_mean = circulant(circ_sum) / N
return A_mean
def matrix_chord24_trans_inv(A):
"""Computes transposition-invariant matrix for transition matrix
based 12 major chords and 12 minor chords
Notebook: C5/C5S3_ChordRec_HMM.ipynb
Args:
A (np.ndarray): Input transition matrix
Returns:
A_ti (np.ndarray): Output transition matrix
"""
A_ti = np.zeros(A.shape)
A_ti[0:12, 0:12] = matrix_circular_mean(A[0:12, 0:12])
A_ti[0:12, 12:24] = matrix_circular_mean(A[0:12, 12:24])
A_ti[12:24, 0:12] = matrix_circular_mean(A[12:24, 0:12])
A_ti[12:24, 12:24] = matrix_circular_mean(A[12:24, 12:24])
return A_ti
A_ti = matrix_chord24_trans_inv(A_est)
fig, ax = plt.subplots(1, 2, gridspec_kw={'width_ratios': [1, 1],
'height_ratios': [1]},
figsize=(10, 3.8))
title='Transition matrix')
title='Transposition-invariant transition matrix')
plt.tight_layout()
Finally, we provide a function for generating a uniform transition probability matrix. This function has a parameter $p\in[0,1]$ that determines the probability for self transitions (the value on the main diagonal). The probabilities on the remaining positions are set such that the resulting matrix is a probability matrix (i.e., all rows and columns sum to one).
In [4]:
def uniform_transition_matrix(p=0.01, N=24):
"""Computes uniform transition matrix
Notebook: C5/C5S3_ChordRec_HMM.ipynb
Args:
p (float): Self transition probability (Default value = 0.01)
N (int): Column and row dimension (Default value = 24)
Returns:
A (np.ndarray): Output transition matrix
"""
off_diag_entries = (1-p) / (N-1) # rows should sum up to 1
A = off_diag_entries * np.ones([N, N])
np.fill_diagonal(A, p)
return A
fig, ax = plt.subplots(1, 2, gridspec_kw={'width_ratios': [1, 1],
'height_ratios': [1]},
figsize=(10, 3.8))
p = 0.5
A_uni = uniform_transition_matrix(p)
plot_transition_matrix(A_uni, ax=[ax[0]], title='Uniform transition matrix (p=%0.2f)' % p)
p = 0.9
A_uni = uniform_transition_matrix(p)
plot_transition_matrix(A_uni, ax=[ax[1]], title='Uniform transition matrix (p=%0.2f)' % p)
plt.tight_layout()
HMM-Based Chord Recognition¶
As discussed before, the free parameters of the HMM-based model can either be learned automatically from the training set or set manually using musical knowledge. Continuing with our Bach example, we now present an experiment that demonstrates the effect of applying HMMs to our chord recognition scenario. We use the following setting:
• As observation sequence $O$, we use a sequence of chroma vectors.
• As for the transition probability matrix $A$, we simply use a uniform transition matrix.
• As for the initial state probability vector $C$, we use a uniform distribution.
• As for the emission probability matrix $B$, we replace them by the likelihood matrix $B[O]$, which is a normalized version of the chord similarity matrix also used for the template-based chord recognition.
• The frame-wise chord recognition results is given by the state sequence computed by the Viterbi algorithm.
Using the likelihood matrix $B[O]$ instead of emission probabilities requires a small modification of the original algorithm. In the following code cell, we provide the implementation of this modification using a numerically stable log version. We then compare the HMM-based results with the template-based approach showing the evaluation results in the form of time–chord visualizations, respectively.
In [5]:
@jit(nopython=True)
def viterbi_log_likelihood(A, C, B_O):
"""Viterbi algorithm (log variant) for solving the uncovering problem
Notebook: C5/C5S3_Viterbi.ipynb
Args:
A (np.ndarray): State transition probability matrix of dimension I x I
C (np.ndarray): Initial state distribution of dimension I
B_O (np.ndarray): Likelihood matrix of dimension I x N
Returns:
S_opt (np.ndarray): Optimal state sequence of length N
S_mat (np.ndarray): Binary matrix representation of optimal state sequence
D_log (np.ndarray): Accumulated log probability matrix
E (np.ndarray): Backtracking matrix
"""
I = A.shape[0] # Number of states
N = B_O.shape[1] # Length of observation sequence
tiny = np.finfo(0.).tiny
A_log = np.log(A + tiny)
C_log = np.log(C + tiny)
B_O_log = np.log(B_O + tiny)
# Initialize D and E matrices
D_log = np.zeros((I, N))
E = np.zeros((I, N-1)).astype(np.int32)
D_log[:, 0] = C_log + B_O_log[:, 0]
# Compute D and E in a nested loop
for n in range(1, N):
for i in range(I):
temp_sum = A_log[:, i] + D_log[:, n-1]
D_log[i, n] = np.max(temp_sum) + B_O_log[i, n]
E[i, n-1] = np.argmax(temp_sum)
# Backtracking
S_opt = np.zeros(N).astype(np.int32)
S_opt[-1] = np.argmax(D_log[:, -1])
for n in range(N-2, -1, -1):
S_opt[n] = E[int(S_opt[n+1]), n]
# Matrix representation of result
S_mat = np.zeros((I, N)).astype(np.int32)
for n in range(N):
S_mat[S_opt[n], n] = 1
return S_mat, S_opt, D_log, E
A = uniform_transition_matrix(p=0.5)
C = 1 / 24 * np.ones((1, 24))
B_O = chord_sim
chord_HMM, _, _, _ = viterbi_log_likelihood(A, C, B_O)
P, R, F, TP, FP, FN = libfmp.c5.compute_eval_measures(ann_matrix, chord_HMM)
title = 'HMM-Based approach (N=%d, TP=%d, FP=%d, FN=%d, P=%.2f, R=%.2f, F=%.2f)' % (N_X, TP, FP, FN, P, R, F)
fig, ax, im = libfmp.c5.plot_matrix_chord_eval(ann_matrix, chord_HMM, Fs=1,
title=title, ylabel='Chord', xlabel='Time (frames)', chord_labels=chord_labels)
plt.tight_layout()
plt.show()
P, R, F, TP, FP, FN = libfmp.c5.compute_eval_measures(ann_matrix, chord_max)
title = 'Template-based approach (N=%d, TP=%d, FP=%d, FN=%d, P=%.2f, R=%.2f, F=%.2f)' %\
(N_X, TP, FP, FN, P, R, F)
fig, ax, im = libfmp.c5.plot_matrix_chord_eval(ann_matrix, chord_max, Fs=1,
title=title, ylabel='Chord', xlabel='Time (frames)', chord_labels=chord_labels)
plt.tight_layout()
plt.show()
In this example, the HMM-based chord recognizer clearly outperforms the template-based approach. The improvements in HMM-based approach come specifically from the transition model that introduces context-sensitive smoothing. In the case of high self-transition probabilities, a chord recognizer tends to stay in the current chord rather than change to another one, which can be regarded as a kind of smoothing. This effect is also demonstrated in our Bach example, where the broken chords cause many chord ambiguities of short duration. This leads to many random-like chord changes when using a simple template-based chord recognizer. Using an HMM-based approach, chord changes are only performed when the relatively low transition probabilities are compensated by a substantial increase of emission probabilities. Consequently, only the dominant chord changes remain.
Prefiltering vs. Postfiltering¶
In the FMP notebook on chord recognition evaluation, we showed for the Bach example that one may achieve similar improvements by applying a longer window size when computing the input chromagram. Applying longer window sizes more or less amounts to temporal smoothing of the observation sequence. Since this smoothing is performed prior to the pattern matching step, we also call this strategy prefiltering. Note that such a prefiltering step not only smoothes out noise-like frames, but also washes out characteristic chroma information and blurs transitions. As opposed to prefiltering, the HMM-based approach leaves the feature representation untouched. Furthermore, the smoothing is performed in combination with the pattern matching step. For this reason, we also call this approach postfiltering. As a result, the original chroma information is preserved and transitions in the feature representation are kept sharp.
Further Notes¶
In this notebook, we introduced a basic HMM-based approach for chord recognition. In our simplistic model, we used $24$ states that correspond to the $12$ major and $12$ minor triads and fixed the HMM parameters explicitly using musical knowledge.
In this book, we have only considered a basic HMM variant. There are many more variants and extensions of HMMs including continuous HMMs and HMMs with specific state transition topologies. Rather than fixing the model parameters manually, the power of general HMMs is to automatically learn the free parameters based on training examples (e.g., using the Baum-Welch Algorithm). The estimation of the model parameters can become very intricate, leading to challenging and deep mathematical problems. For an excellent textbook on the classical theory of HMMs, including the discrete as well as the continuous case, we refer to the excellent book on Hidden Markov Models for Speech Recognition by Huang et al. (1990). We close this notebook with an overview of a typical HMM-based chord recognition approach consisting of a training and an evaluation stage.
Acknowledgment: This notebook was created by Meinard Müller and Christof Weiß. | 2021-10-21 08:02:55 | {"extraction_info": {"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": 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": 1, "x-ck12": 0, "texerror": 0, "math_score": 0.7060036063194275, "perplexity": 3717.961711142877}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323585382.32/warc/CC-MAIN-20211021071407-20211021101407-00231.warc.gz"} |
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Secular variability of the longitudinal magnetic field of the Ap star γ EquWe present an analysis of the secular variability of the longitudinalmagnetic field Be in the roAp star γ Equ (HD 201601).Measurements of the stellar magnetic field Be were mostlycompiled from the literature, and we appended also our 33 newBe measurements which were obtained with the 1-m opticaltelescope of the Special Astrophysical Observatory (Russia). All theavailable data cover the time period of 58 yr, and include both phasesof the maximum and minimum Be. We determined that the periodof the long-term magnetic Be variations equals 91.1 +/- 3.6yr, with Be(max) =+577 +/- 31 G and Be(min) =-1101+/- 31 G. Observed Orbital EccentricitiesFor 391 spectroscopic and visual binaries with known orbital elementsand having B0-F0 IV or V primaries, we collected the derivedeccentricities. As has been found by others, those binaries with periodsof a few days have been circularized. 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Some Comments on the Magnetic Braking of CP StarsThe low rotation velocities of magnetic CP stars are discussed.Arguments against the involvement of the magnetic field in the loss ofangular momentum are given: (1) the fields are not strong enough inyoung stars in the stage of evolution prior to the main sequence; (2)there is no significant statistical correlation between the magneticfield strength and the rotation period of CP stars; (3) stars with shortperiods have the highest fields; (4) a substantial number of stars withvery low magnetic fields (B e P>25 days, which form 12% of the total,probably lie at the edge of the velocity distribution for low massstars. All of these properties conflict with the hypothesis of magneticbraking of CP stars. Stellar Kinematic Groups. II. A Reexamination of the Membership, Activity, and Age of the Ursa Major GroupUtilizing Hipparcos parallaxes, original radial velocities and recentliterature values, new Ca II H and K emission measurements,literature-based abundance estimates, and updated photometry (includingrecent resolved measurements of close doubles), we revisit the UrsaMajor moving group membership status of some 220 stars to produce afinal clean list of nearly 60 assured members, based on kinematic andphotometric criteria. Scatter in the velocity dispersions and H-Rdiagram is correlated with trial activity-based membership assignments,indicating the usefulness of criteria based on photometric andchromospheric emission to examine membership. Closer inspection,however, shows that activity is considerably more robust at excludingmembership, failing to do so only for <=15% of objects, perhapsconsiderably less. Our UMa members demonstrate nonzero vertex deviationin the Bottlinger diagram, behavior seen in older and recent studies ofnearby young disk stars and perhaps related to Galactic spiralstructure. Comparison of isochrones and our final UMa group membersindicates an age of 500+/-100 Myr, some 200 Myr older than thecanonically quoted UMa age. Our UMa kinematic/photometric members' meanchromospheric emission levels, rotational velocities, and scattertherein are indistinguishable from values in the Hyades and smaller thanthose evinced by members of the younger Pleiades and M34 clusters,suggesting these characteristics decline rapidly with age over 200-500Myr. None of our UMa members demonstrate inordinately low absolutevalues of chromospheric emission, but several may show residual fluxes afactor of >=2 below a Hyades-defined lower envelope. 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Catalogue of averaged stellar effective magnetic fields. I. Chemically peculiar A and B type starsThis paper presents the catalogue and the method of determination ofaveraged quadratic effective magnetic fields < B_e > for 596 mainsequence and giant stars. The catalogue is based on measurements of thestellar effective (or mean longitudinal) magnetic field strengths B_e,which were compiled from the existing literature.We analysed the properties of 352 chemically peculiar A and B stars inthe catalogue, including Am, ApSi, He-weak, He-rich, HgMn, ApSrCrEu, andall ApSr type stars. We have found that the number distribution of allchemically peculiar (CP) stars vs. averaged magnetic field strength isdescribed by a decreasing exponential function. Relations of this typehold also for stars of all the analysed subclasses of chemicalpeculiarity. The exponential form of the above distribution function canbreak down below about 100 G, the latter value representingapproximately the resolution of our analysis for A type stars.Table A.1 and its references are only available in electronic form atthe CDS via anonymous ftp to cdsarc.u-strasbg.fr (130.79.128.5) or viahttp://cdsweb.u-strasbg.fr/cgi-bin/qcat?J/A+A/407/631 and Tables 3 to 9are only available in electronic form at http://www.edpsciences.org Automated spectroscopic abundances of A and F-type stars using echelle spectrographs. II. Abundances of 140 A-F stars from ELODIEUsing the method presented in Erspamer & North (\cite{erspamer},hereafter Paper I), detailed abundances of 140 stars are presented. Theuncertainties characteristic of this method are presented and discussed.In particular, we show that for a S/N ratio higher than 200, the methodis applicable to stars with a rotational velocity as high as 200 kms-1. There is no correlation between abundances and Vsin i,except a spurious one for Sr, Sc and Na which we explain by the smallnumber of lines of these elements combined with a locally biasedcontinuum. Metallic giants (Hauck \cite{hauck}) show larger abundancesthan normal giants for at least 8 elements: Al, Ca, Ti, Cr, Mn, Fe, Niand Ba. The anticorrelation for Na, Mg, Si, Ca, Fe and Ni with Vsin isuggested by Varenne & Monier (\cite{varenne99}) is not confirmed.The predictions of the Montréal models (e.g. Richard et al.\cite{richard01}) are not fulfilled in general. However, a correlationbetween left [(Fe)/(H)right ] and log g is found for stars of 1.8 to 2.0M_sun. Various possible causes are discussed, but the physical realityof this correlation seems inescapable.Based on observations collected at the 1.93 m telescope at theObservatoire de Haute-Provence (St-Michel l'Observatoire, France) andCORALIE.Based on observations collected at the Swiss 1.2 m Leonard Eulertelescopes at the European Southern Observatory (La Silla, Chile).Tables 5 and 6 are only available in electronic form at the CDS viaanonymous ftp to cdsarc.u.strasbg.fr (130.79.128.5) or viahttp://cdsweb.u-strasbg.fr/cgi-bin/qcat?J/A+A/398/1121 On the Periods of the Magnetic CP StarsAn HR diagram annotated to show several ranges of photometericallydetermined periods has been constructed for the magnetic CP stars whoseperiods have been determined by the author and his collaborators. Thedistribution of periods reflects both the initial conditions as well asthe subsequent stellar histories. Since the stellar magnetic field doesnot penetrate the convective core, eventually a shear zone near thecore-radiative envelope boundary may develop which produces turbulenceand modifies the field. Many, but not all, of the most rapidly rotatingmCP stars are close to the ZAMS and some of the least rapidly rotatingmCP stars are the furthest from the ZAMS. Kinematics of Hipparcos Visual Binaries. II. Stars with Ground-Based Orbital SolutionsThis paper continues kinematical investigations of the Hipparcos visualbinaries with known orbits. A sample, consisting of 804 binary systemswith orbital elements determined from ground-based observations, isselected. The mean relative error of their parallaxes is about 12% andthe mean relative error of proper motions is about 4%. However, even 41%of the sample stars lack radial velocity measurements. The computedGalactic velocity components and other kinematical parameters are usedto divide the stars with known radial velocities into kinematical agegroups. The majority (92%) of binaries from the sample are thin diskstars, 7.6% have thick disk kinematics and only two binaries have halokinematics. Among them, the long-period variable Mira Ceti has a verydiscordant {Hipparcos} and ground-based parallax values. From the wholesample, 60 stars are ascribed to the thick disk and halo population.There is an urgent need to increase the number of the identified halobinaries with known orbits and substantially improve the situation withradial velocity data for stars with known orbits. New Magnetic Chemically Peculiar StarsNot Available Metallicity Determinations from Ultraviolet-Visual Spectrophotometry. I. The Test SampleNew visual spectrophotometric observations of non-supergiant solarneighborhood stars are combined with IUE Newly Extracted Spectra (INES)energy distributions in order to derive their overall metallicities,[M/H]. This fundamental parameter, together with effective temperatureand apparent angular diameter, is obtained by applying the flux-fittingmethod while surface gravity is derived from the comparison withevolutionary tracks in the theoretical H-R diagram. Trigonometricparallaxes for the stars of the sample are taken from the HipparcosCatalogue. The quality of the flux calibration is discussed by analyzinga test sample via comparison with external photometry. The validity ofthe method in providing accurate metallicities is tested on a selectedsample of G-type stars with well-determined atmospheric parameters fromrecent high-resolution spectral analysis. The extension of the overallprocedure to the determination of the chemical composition of all theINES non-supergiant G-type stars with accurate parallaxes is planned inorder to investigate their atmospheric temperature structure. Based onobservations collected at the INAOE G. Haro'' Observatory, Cananea(Mexico). Multiplicity among chemically peculiar stars. II. Cool magnetic Ap starsWe present new orbits for sixteen Ap spectroscopic binaries, four ofwhich might in fact be Am stars, and give their orbital elements. Fourof them are SB2 systems: HD 5550, HD 22128, HD 56495 and HD 98088. Thetwelve other stars are: HD 9996, HD 12288, HD 40711, HD 54908, HD 65339,HD 73709, HD 105680, HD 138426, HD 184471, HD 188854, HD 200405 and HD216533. Rough estimates of the individual masses of the components of HD65339 (53 Cam) are given, combining our radial velocities with theresults of speckle interferometry and with Hipparcos parallaxes.Considering the mass functions of 74 spectroscopic binaries from thiswork and from the literature, we conclude that the distribution of themass ratio is the same for cool Ap stars and for normal G dwarfs.Therefore, the only differences between binaries with normal stars andthose hosting an Ap star lie in the period distribution: except for thecase of HD 200405, all orbital periods are longer than (or equal to) 3days. A consequence of this peculiar distribution is a deficit of nulleccentricities. There is no indication that the secondary has a specialnature, like e.g. a white dwarf. Based on observations collected at theObservatoire de Haute-Provence (CNRS), France.Tables 1 to 3 are only available in electronic form at the CDS viaanonymous ftp to cdsarc.u-strasbg.fr (130.79.128.5) or viahttp://cdsweb.u-strasbg.fr/cgi-bin/qcat?J/A+A/394/151Appendix B is only available in electronic form athttp://www.edpsciences.org Rotational velocities of A-type stars in the northern hemisphere. II. Measurement of v sin iThis work is the second part of the set of measurements of v sin i forA-type stars, begun by Royer et al. (\cite{Ror_02a}). Spectra of 249 B8to F2-type stars brighter than V=7 have been collected at Observatoirede Haute-Provence (OHP). Fourier transforms of several line profiles inthe range 4200-4600 Å are used to derive v sin i from thefrequency of the first zero. Statistical analysis of the sampleindicates that measurement error mainly depends on v sin i and thisrelative error of the rotational velocity is found to be about 5% onaverage. The systematic shift with respect to standard values fromSlettebak et al. (\cite{Slk_75}), previously found in the first paper,is here confirmed. Comparisons with data from the literature agree withour findings: v sin i values from Slettebak et al. are underestimatedand the relation between both scales follows a linear law ensuremath vsin inew = 1.03 v sin iold+7.7. Finally, thesedata are combined with those from the previous paper (Royer et al.\cite{Ror_02a}), together with the catalogue of Abt & Morrell(\cite{AbtMol95}). The resulting sample includes some 2150 stars withhomogenized rotational velocities. Based on observations made atObservatoire de Haute Provence (CNRS), France. Tables \ref{results} and\ref{merging} are only available in electronic form at the CDS viaanonymous ftp to cdsarc.u-strasbg.fr (130.79.125.5) or viahttp://cdsweb.u-strasbg.fr/cgi-bin/qcat?J/A+A/393/897 Are Stellar Rotational Axes Distributed Randomly?Stellar line widths yield values of Vsini, but the equatorial rotationalvelocities, V, cannot be determined for individual stars withoutknowledge of their inclinations, i, relative to the lines of sight. Forlarge numbers of stars we usually assume random orientations ofrotational axes to derive mean values of V, but we wonder whether thatassumption is valid. Individual inclinations can be derived only inspecial cases, such as for eclipsing binaries where they are close to90° or for chromospherically active late-type dwarfs or spotted(e.g., Ap) stars where we have independent information about therotational periods. We consider recent data on 102 Ap stars for whichCatalano & Renson compiled rotational periods from the literatureand Abt & Morrell (primarily) obtained measures of Vsini. We findthat the rotational axes are oriented randomly within the measuringerrors. We searched for possible dependence of the inclinations onGalactic latitude or longitude, and found no dependence. Catalogue of Apparent Diameters and Absolute Radii of Stars (CADARS) - Third edition - Comments and statisticsThe Catalogue, available at the Centre de Données Stellaires deStrasbourg, consists of 13 573 records concerning the results obtainedfrom different methods for 7778 stars, reported in the literature. Thefollowing data are listed for each star: identifications, apparentmagnitude, spectral type, apparent diameter in arcsec, absolute radiusin solar units, method of determination, reference, remarks. Commentsand statistics obtained from CADARS are given. The Catalogue isavailable in electronic form at the CDS via anonymous ftp tocdsarc.u-strasbg.fr (130.79.128.5) or viahttp://cdsweb.u-strasbg.fr/cgi-bin/qcar?J/A+A/367/521 Radial velocity and magnetic field observations with TRAFICOS>From spectroscopic observations of some early- and late-type starswe determined radial velocity and longitudinal magnetic field values.The spectra have been performed with observations at the 2-m telescopeof the Thüringer Landessternwarte Tautenburg, equipped with theÉchelle spectrograph TRAFICOS and in many cases with Zeemananalyzer. For individual stars we give some comments about the radialvelocities and magnetic field values, and about any possiblevariability. Two-colour photometry for 9473 components of close Hipparcos double and multiple starsUsing observations obtained with the Tycho instrument of the ESAHipparcos satellite, a two-colour photometry is produced for componentsof more than 7 000 Hipparcos double and multiple stars with angularseparations 0.1 to 2.5 arcsec. We publish 9473 components of 5173systems with separations above 0.3 arcsec. The majority of them did nothave Tycho photometry in the Hipparcos catalogue. The magnitudes arederived in the Tycho B_T and V_T passbands, similar to the Johnsonpassbands. Photometrically resolved components of the binaries withstatistically significant trigonometric parallaxes can be put on an HRdiagram, the majority of them for the first time. Based on observationsmade with the ESA Hipparcos satellite. A Second Catalog of Orbiting Astronomical Observatory 2 Filter Photometry: Ultraviolet Photometry of 614 StarsUltraviolet photometry from the Wisconsin Experiment Package on theOrbiting Astronomical Observatory 2 (OAO 2) is presented for 614 stars.Previously unpublished magnitudes from 12 filter bandpasses withwavelengths ranging from 1330 to 4250 Å have been placed on thewhite dwarf model atmosphere absolute flux scale. The fluxes wereconverted to magnitudes using V=0 for F(V)=3.46x10^-9 ergs cm^-2 s^-1Å^-1, or m_lambda=-2.5logF_lambda-21.15. This second catalogeffectively doubles the amount of OAO 2 photometry available in theliterature and includes many objects too bright to be observed withmodern space observatories. The ROSAT all-sky survey catalogue of optically bright main-sequence stars and subgiant starsWe present X-ray data for all main-sequence and subgiant stars ofspectral types A, F, G, and K and luminosity classes IV and V listed inthe Bright Star Catalogue that have been detected as X-ray sources inthe ROSAT all-sky survey; several stars without luminosity class arealso included. The catalogue contains 980 entries yielding an averagedetection rate of 32 percent. In addition to count rates, sourcedetection parameters, hardness ratios, and X-ray fluxes we also listX-ray luminosities derived from Hipparcos parallaxes. The catalogue isalso available in electronic form via anonymous ftp tocdsarc.u-strasbg.fr (130.79.128.5) or viahttp://cdsweb.u-strasbg.fr/Abstract.html On the HIPPARCOS photometry of chemically peculiar B, A, and F starsThe Hipparcos photometry of the Chemically Peculiar main sequence B, A,and F stars is examined for variability. Some non-magnetic CP stars,Mercury-Manganese and metallic-line stars, which according to canonicalwisdom should not be variable, may be variable and are identified forfurther study. Some potentially important magnetic CP stars are noted.Tables 1, 2, and 3 are available only in electronic form at the CDS viaanonymous ftp to cdsarc.u-strasbg.fr (130.79.128.5) or viahttp://cdsweb.u-strasbg.fr/Abstract.html The observed periods of AP and BP starsA catalogue of all the periods up to now proposed for the variations ofCP2, CP3, and CP4 stars is presented. The main identifiers (HD and HR),the proper name, the variable-star name, and the spectral type andpeculiarity are given for each star as far as the coordinates at 2000.0and the visual magnitude. The nature of the observed variations (light,spectrum, magnetic field, etc.) is presented in a codified way. Thecatalogue is arranged in three tables: the bulk of the data, i.e. thosereferring to CP2, CP3, and CP4 stars, are given in Table 1, while thedata concerning He-strong stars are given in Table 2 and those foreclipsing or ellipsoidal variables are collected in Table 3. Notes arealso provided at the end of each table, mainly about duplicities. Thecatalogue contains data on 364 CP stars and is updated to 1996, October31. This research has made use of the SIMBAD database, operated at CDS,Strasbourg, France. Search for cool circumstellar matter in the Ursae Majoris group with ISOWe observed the mid- and far-infrared spectral energy distributions of 9A-type stars from the 300 Myrs old Ursae Majoris group using the ISOsatellite and the UKIRT telescope. We found that only 1 out of the 9stars shows clear signature of circumstellar dust, and we derived anupper limit of 0.05 Mmoon for the dust mass around the otherstars. Our results suggest that the relatively high incidence ofVega-like disks observed among A-type field stars in the solarneighbourhood by IRAS cannot be extrapolated to the rest of the MilkyWay. The Vega phenomenon appears to be the exception rather than therule. Based on observations with ISO, an ESA project with instrumentsfunded by ESA member states (especially the P/I countries France,Germany, the Netherlands and the United Kingdom) with participation ofISAS and NASA. The HR-diagram from HIPPARCOS data. Absolute magnitudes and kinematics of BP - AP starsThe HR-diagram of about 1000 Bp - Ap stars in the solar neighbourhoodhas been constructed using astrometric data from Hipparcos satellite aswell as photometric and radial velocity data. The LM method\cite{luri95,luri96} allows the use of proper motion and radial velocitydata in addition to the trigonometric parallaxes to obtain luminositycalibrations and improved distances estimates. Six types of Bp - Apstars have been examined: He-rich, He-weak, HgMn, Si, Si+ and SrCrEu.Most Bp - Ap stars lie on the main sequence occupying the whole width ofit (about 2 mag), just like normal stars in the same range of spectraltypes. Their kinematic behaviour is typical of thin disk stars youngerthan about 1 Gyr. A few stars found to be high above the galactic planeor to have a high velocity are briefly discussed. Based on data from theESA Hipparcos astrometry satellite and photometric data collected in theGeneva system at ESO, La Silla (Chile) and at Jungfraujoch andGornergrat Observatories (Switzerland). Tables 3 and 4 are onlyavailable in electronic form at the CDS via anonymous ftp tocdsarc.u-strasbg.fr (130.79.128.5) or viahttp://cdsweb.u-strasbg.fr/Abstract.html MSC - a catalogue of physical multiple starsThe MSC catalogue contains data on 612 physical multiple stars ofmultiplicity 3 to 7 which are hierarchical with few exceptions. Orbitalperiods, angular separations and mass ratios are estimated for eachsub-system. Orbital elements are given when available. The catalogue canbe accessed through CDS (Strasbourg). Half of the systems are within 100pc from the Sun. The comparison of the periods of close and widesub-systems reveals that there is no preferred period ratio and allpossible combinations of periods are found. The distribution of thelogarithms of short periods is bimodal, probably due to observationalselection. In 82\% of triple stars the close sub-system is related tothe primary of a wide pair. However, the analysis of mass ratiodistribution gives some support to the idea that component masses areindependently selected from the Salpeter mass function. Orbits of wideand close sub-systems are not always coplanar, although thecorresponding orbital angular momentum vectors do show a weak tendencyof alignment. Some observational programs based on the MSC aresuggested. Tables 2 and 3 are only available in electronic form at theCDS via anonymous ftp to cdsarc.u-strasbg.fr (130.79.128.5) or viahttp://cdsweb.u-strasbg.fr/Abstract.html The Lambda 6708 Feature in AP StarsNot Available An analysis of the AP spectroscopic binary HD 59435.HD 59435 is a double-lined spectroscopic binary (SB2), with slowlyrotating components of very similar luminosity. It was studied throughhigh-resolution spectroscopy, photometry and CORAVEL observations.Orbital elements are presented. The orbital period is 1387 days. Theline of sight lies close to the orbital plane, but no eclipses have beenobserved so far. The primary is an evolved G8 or K0 star, which haslikely just descended the Red Giant Branch. The secondary is a cool Apstar close to the end of its main-sequence life. It shows lines resolvedinto magnetically split components, from which its mean magnetic fieldmodulus can be diagnosed. The field varies with a remarkably largerelative amplitude, over a rotation period which is at least of theorder of 3 years. Linear polarimetry of AP stars. VI. A modified dipolar model consistent with the observations.While some Ap variables display a linear polarization variation verysimilar to that computed for a pure magnetic dipole, several Ap starsshow conspicuous peculiarities which must be interpreted in terms ofdepartures from the standard, oblique rotator model (we have shownpreviously that abundances anomalies are not sufficient to explain oddpolarization diagrams). We have designed an inversion method, based on aresidues minimization process, which allows us to build the map of themagnetic peculiarities at the surface of non-dipolar stars. As thelinear polarization is but weakly sensitive to the variations of thefield modulus, we interpret the polarization anomalies in terms ofinclination changes of the lines of force within their meridian plane.Keeping the magnetic equator as a plane of symmetry, we show that it issufficient to assume slightly expanded lines of force, over some partsof the magnetic equator, to explain most peculiar polarization curves(Figs. 2 to 7). Such regions, where the lines of force expand outwards,seem to occur preferentially in the vicinity of the rotation poles forthose stars having a β angle not far from 90deg. In the case ofβ CrB, which was studied previously in detail (Leroy, 1995), thisregion nearly coincides with the equatorial patch of enhanced fieldstrength, which must be postulated to explain the surface fieldmeasurements. The present study, which also provides unambiguousdeterminations of the i and β angles for 15 stars, marks theprovisional end of our investigation based on broadband linearpolarization measurements. We expect that similar measurements, having agood spectral resolution, will be available soon: they will yield moresevere observational constraints enabling a more detailed modeling work.However, we think that the series of articles which ends with thepresent paper has demonstrated the great value of linear polarizationdata and may have opened fruitful research tracks bearing on themagnetic structure of Ap stars.
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https://www.aimsciences.org/article/doi/10.3934/krm.2013.6.715 | # American Institute of Mathematical Sciences
December 2013, 6(4): 715-727. doi: 10.3934/krm.2013.6.715
## A remark on the ultra-analytic smoothing properties of the spatially homogeneous Landau equation
1 Graduate School of Human and Environmental Studies, Kyoto University, Kyoto, 606-8501 2 Université de Cergy-Pontoise, CNRS UMR 8088, Mathématiques, 2 avenue Adolphe Chauvin, 95302 Cergy-Pontoise 3 Université de Rouen, UMR 6085-CNRS, Mathématiques, Avenue de l’Université, BP.12, 76801 Saint Etienne du Rouvray
Received February 2013 Revised June 2013 Published November 2013
We consider the non-linear spatially homogeneous Landau equation with Maxwellian molecules in a close-to-equilibrium framework and show that the Cauchy problem for the fluctuation around the Maxwellian equilibrium distribution enjoys a Gelfand-Shilov regularizing effect in the class $S_{1/2}^{1/2}(\mathbb{R}^d)$, implying the ultra-analyticity and the production of exponential moments of the fluctuation, for any positive time.
Citation: Yoshinori Morimoto, Karel Pravda-Starov, Chao-Jiang Xu. A remark on the ultra-analytic smoothing properties of the spatially homogeneous Landau equation. Kinetic and Related Models, 2013, 6 (4) : 715-727. doi: 10.3934/krm.2013.6.715
##### References:
[1] R. Alexandre and C. Villani, On the Landau approximation in plasma physics, Ann. Inst. H. Poincaré Anal. Non Linéaire, 21 (2004), 61-95. doi: 10.1016/S0294-1449(03)00030-1. [2] A. A. Arsen'ev and O. E. Buryak, On a connection between the solution of the Boltzmann equation and the solution of the Landau-Fokker-Planck equation, Math. USSR Sbornik, 69 (1991), 465-478. [3] H. Chen, W.-X. Li and C.-J. Xu, Gevrey regularity for solution of the spatially homogeneous Landau equation, Acta Math. Sci. Ser. B Engl. Ed., 29 (2009), 673-686. doi: 10.1016/S0252-9602(09)60063-1. [4] H. Chen, W.-X. Li and C.-J. Xu, Analytic smoothness effect of solutions for spatially homogeneous Landau equation, J. Differential Equations, 248 (2010), 77-94. doi: 10.1016/j.jde.2009.08.006. [5] C. Cercignani, The Boltzmann equation and its applications, Applied Mathematical Sciences, vol. 67, Springer-Verlag, New York, 1988. doi: 10.1007/978-1-4612-1039-9. [6] P. Degond and B. Lucquin-Desreux, The Fokker-Planck asymptotics of the Boltzmann collision operator in the Coulomb case, Math. Models Methods Appl. Sci., 2 (1992), 167-182. doi: 10.1142/S0218202592000119. [7] L. Desvillettes, On asymptotics of the Boltzmann equation when the collisions become grazing, Transport Theory Statist. Phys., 21 (1992), 259-276. doi: 10.1080/00411459208203923. [8] L. Desvillettes and C. Villani, On the spatially homogeneous Landau equation for hard potentials. I. Existence, uniqueness and smoothness, Comm. Partial Differential Equations, 25 (2000), 179-259. doi: 10.1080/03605300008821512. [9] I. M. Gelfand and G. E. Shilov, Generalized Functions, Vol. 2. Spaces of fundamental and generalized functions. Translated from the Russian by Morris D. Friedman, Amiel Feinstein and Christian P. Peltzer. Academic Press [Harcourt Brace Jovanovich, Publishers], New York-London, 1968 [1977]. [10] T. Gramchev, S. Pilipović and L. Rodino, Classes of degenerate elliptic operators in Gelfand-Shilov spaces, in New Developments in Pseudo-Differential Operators, Oper. Theory Adv. Appl. Birkhäuser, Basel, 189 (2009), 15-31. doi: 10.1007/978-3-7643-8969-7_2. [11] L. D. Landau, Die kinetische Gleichung für den Fall Coulombscher Wechselwirkung, Phys. Z. Sowjet., 10 (1936) 154. Translation: The transport equation in the case of Coulomb interactions, D. ter Haar (Ed.), Collected papers of L. D. Landau, Pergamon Press, Oxford (1981), 163-170. [12] N. Lerner, Y. Morimoto, K. Pravda-Starov and C.-J. Xu, Phase space analysis and functional calculus for the linearized Landau and Boltzmann operators, Kinet. Relat. Models, 6 (2013), 625-648. doi: 10.3934/krm.2013.6.625. [13] N. Lerner, Y. Morimoto, K. Pravda-Starov and C.-J. Xu, Gelfand-Shilov smoothing properties of the radially symmetric spatially homogeneous Boltzmann equation without angular cutoff, J. Differential Equations, 256 (2014), 797-831. doi: 10.1016/j.jde.2013.10.001. [14] Y. Morimoto and C.-J. Xu, Ultra-analytic effect of Cauchy problem for a class of kinetic equations, J. Differential Equations, 247 (2009), 596-617. doi: 10.1016/j.jde.2009.01.028. [15] F. Nicola and L. Rodino, Global Pseudo-Differential Calculus on Euclidean Spaces, Pseudo-Differential Operators, Theory and Applications, 4, Birkhäuser Verlag, Basel, 2010. doi: 10.1007/978-3-7643-8512-5. [16] J. Toft, A. Khrennikov, B. Nilsson and S. Nordebo, Decompositions of Gelfand-Shilov kernels into kernels of similar class, J. Math. Anal. Appl., 396 (2012), 315-322. doi: 10.1016/j.jmaa.2012.06.025. [17] C. Villani, On the spatially homogeneous Landau equation for Maxwellian molecules, Math. Models Methods Appl. Sci., 8 (1998), 957-983. doi: 10.1142/S0218202598000433. [18] C. Villani, On a new class of weak solutions to the spatially homogeneous Boltzmann and Landau equations, Arch. Ration. Mech. Anal., 143 (1998), 273-307. doi: 10.1007/s002050050106.
show all references
##### References:
[1] R. Alexandre and C. Villani, On the Landau approximation in plasma physics, Ann. Inst. H. Poincaré Anal. Non Linéaire, 21 (2004), 61-95. doi: 10.1016/S0294-1449(03)00030-1. [2] A. A. Arsen'ev and O. E. Buryak, On a connection between the solution of the Boltzmann equation and the solution of the Landau-Fokker-Planck equation, Math. USSR Sbornik, 69 (1991), 465-478. [3] H. Chen, W.-X. Li and C.-J. Xu, Gevrey regularity for solution of the spatially homogeneous Landau equation, Acta Math. Sci. Ser. B Engl. Ed., 29 (2009), 673-686. doi: 10.1016/S0252-9602(09)60063-1. [4] H. Chen, W.-X. Li and C.-J. Xu, Analytic smoothness effect of solutions for spatially homogeneous Landau equation, J. Differential Equations, 248 (2010), 77-94. doi: 10.1016/j.jde.2009.08.006. [5] C. Cercignani, The Boltzmann equation and its applications, Applied Mathematical Sciences, vol. 67, Springer-Verlag, New York, 1988. doi: 10.1007/978-1-4612-1039-9. [6] P. Degond and B. Lucquin-Desreux, The Fokker-Planck asymptotics of the Boltzmann collision operator in the Coulomb case, Math. Models Methods Appl. Sci., 2 (1992), 167-182. doi: 10.1142/S0218202592000119. [7] L. Desvillettes, On asymptotics of the Boltzmann equation when the collisions become grazing, Transport Theory Statist. Phys., 21 (1992), 259-276. doi: 10.1080/00411459208203923. [8] L. Desvillettes and C. Villani, On the spatially homogeneous Landau equation for hard potentials. I. Existence, uniqueness and smoothness, Comm. Partial Differential Equations, 25 (2000), 179-259. doi: 10.1080/03605300008821512. [9] I. M. Gelfand and G. E. Shilov, Generalized Functions, Vol. 2. Spaces of fundamental and generalized functions. Translated from the Russian by Morris D. Friedman, Amiel Feinstein and Christian P. Peltzer. Academic Press [Harcourt Brace Jovanovich, Publishers], New York-London, 1968 [1977]. [10] T. Gramchev, S. Pilipović and L. Rodino, Classes of degenerate elliptic operators in Gelfand-Shilov spaces, in New Developments in Pseudo-Differential Operators, Oper. Theory Adv. Appl. Birkhäuser, Basel, 189 (2009), 15-31. doi: 10.1007/978-3-7643-8969-7_2. [11] L. D. Landau, Die kinetische Gleichung für den Fall Coulombscher Wechselwirkung, Phys. Z. Sowjet., 10 (1936) 154. Translation: The transport equation in the case of Coulomb interactions, D. ter Haar (Ed.), Collected papers of L. D. Landau, Pergamon Press, Oxford (1981), 163-170. [12] N. Lerner, Y. Morimoto, K. Pravda-Starov and C.-J. Xu, Phase space analysis and functional calculus for the linearized Landau and Boltzmann operators, Kinet. Relat. Models, 6 (2013), 625-648. doi: 10.3934/krm.2013.6.625. [13] N. Lerner, Y. Morimoto, K. Pravda-Starov and C.-J. Xu, Gelfand-Shilov smoothing properties of the radially symmetric spatially homogeneous Boltzmann equation without angular cutoff, J. Differential Equations, 256 (2014), 797-831. doi: 10.1016/j.jde.2013.10.001. [14] Y. Morimoto and C.-J. Xu, Ultra-analytic effect of Cauchy problem for a class of kinetic equations, J. Differential Equations, 247 (2009), 596-617. doi: 10.1016/j.jde.2009.01.028. [15] F. Nicola and L. Rodino, Global Pseudo-Differential Calculus on Euclidean Spaces, Pseudo-Differential Operators, Theory and Applications, 4, Birkhäuser Verlag, Basel, 2010. doi: 10.1007/978-3-7643-8512-5. [16] J. Toft, A. Khrennikov, B. Nilsson and S. Nordebo, Decompositions of Gelfand-Shilov kernels into kernels of similar class, J. Math. Anal. Appl., 396 (2012), 315-322. doi: 10.1016/j.jmaa.2012.06.025. [17] C. Villani, On the spatially homogeneous Landau equation for Maxwellian molecules, Math. Models Methods Appl. Sci., 8 (1998), 957-983. doi: 10.1142/S0218202598000433. [18] C. Villani, On a new class of weak solutions to the spatially homogeneous Boltzmann and Landau equations, Arch. Ration. Mech. Anal., 143 (1998), 273-307. doi: 10.1007/s002050050106.
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[Back to Top] | 2022-05-25 15:44:18 | {"extraction_info": {"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, "math_score": 0.7038110494613647, "perplexity": 3302.7255974939603}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662588661.65/warc/CC-MAIN-20220525151311-20220525181311-00652.warc.gz"} |
http://www.edmerls.com/index.php/Physical%20Chemistry/Nuclear%20Chemistry/24/Explain%20the%20similarities%20in%20Liquid%20Drop%20&%20Atomic%20Nucleus.%20or%20Explain%20the%20Fission%20&%20Fusion%20Phenomenon%20with%20Liquid%20Drop%20Model. | By Sunil Bhardwaj
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At the different timings, different model of the nucleus is proposed. One of the important model is Liquid Drop Model. The model was proposed by Neils Bohr and Wheeler in 1937 and independently by Frenkel. They treat the nucleus as a homogeneous entity consist of certain no. of protons and neutrons..
Similarities:
1) As a drop of the liquid is consist of large number of molecules which are present in the same volume, similarly nucleus is made-up of large number of protons and neutrons present in the same nuclear volume.
2) As the drop of liquid is homogeneous in density, charge etc. same is for the nucleus.
3) As a drop of liquid cannot be compressed, similarly nucleus also cannot be compressed.
4) In the case of drop of the liquid, the forces between all the molecules is same. Similarly nuclear forces in all the nucleons is nearly same.$$f\left( p-p \right) \approx f\left( p-n \right) \approx f\left( n-n \right)$$ 5). When a very little amount of liquid is added to a drop. The drop of the liquid grows in size. Similarly projectiles (Light Particles) can be added to the nucleus forming the compound nucleus. $$_{ 7 }^{ 14 }{ N }+_{ 2 }^{ 4 }{ He }\longrightarrow _{ 9 }^{ 18 }{ F }$$ 6) Small drops of the liquid combine to form a big drop of the liquid. Similarly small nuclei combine forming big nucleus. This is known as nuclear fusion phenomenon. $$4 \ _{ 1 }^{ 1 }{ H }\longrightarrow _{ 2 }^{ 4 }{ He }+2{ \beta }^{ + } + Energy$$ 7) A very big drop of the liquid breaks up into smaller droplets. Similarly heavy nucleus breaks up into smaller. This phenomenon is called nuclear fission. $$_{ 92 }^{ 235 }{ U } + _{ 0 }^{ 1 }{ n }\longrightarrow _{ 56 }^{ 144 }{ Ba }+ _{ 36 }^{ 89 }{ Kr } + 3_{ 0 }^{ 1 }{ n } + \gamma - ray + Energy$$ 8) As in a drop of liquid evaporation phenomenon take place. Similarly in the nuclear reaction emission of the nucleons take place. For example The Atomic transmutation reaction, the particles are ejected out.
9) As in a drop of the liquid the molecules are moving continuously due to thermal motion of the molecules. Similarly in the nucleus, the nucleons (Protons and Neutrons) are moving because of K.E.
Merits:
1) It explain the phenomenon of nuclear fission.
2) It explain the nuclear fusion phenomenon.
3) It helps in calculating binding energy of the nuclei.
4) It predicts the emission of ?, ? and ? - rays.
5) It support the Bohr’s theory of the formation of compound nucleus.
Limitations:
1) This model cannot explain the extra stability of the nuclei having the number of protons and neutrons 2, 8, 20, 28, 50, 82 and 126 i.e. inert gases.
2) The model is true only for elements having mass number, between 20 and 150 (20 - 150).
MCQ on Nuclear Chemistry from Physical Chemistry
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For grids that are the same from the smallest of devices to the largest, use the .col and .col-* classes. Specify a numbered class when you need a particularly sized column; otherwise, feel free to stick to
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http://www.h-age.net/y5plnhq0/uyq4yb.php?7657e2=systems-of-equations-online-practice | Graph each equation on the same graph. Here is a set of practice problems to accompany the Linear Systems with Three Variables section of the Systems of Equations chapter of the notes for Paul Dawkins Algebra course at Lamar University. References to complexity and mode refer to the overall difficulty of the problems as they appear in the main program. This site offers multiple interactive quizzes and tests to improve your test-taking skills. This stations maze gets students out of their … D. Infinite solutions, because the system is dependent. ; What is the value of y in the following system of equations? Take this test to quiz yourself on how well you know the material. Three Act Math-In and Out Burger and Other 3 Act Problems. ... Access these online resources for additional instructions and practice with solving nonlinear equations. B. Non-linear equations might contain exponents, square roots, etc. Practice: Solving Systems of Equations (3 Different Methods) Date_____ Solve each system by substitution. SEE EXAMPLE 1 13. Use the method of elimination to solve for(x, y) or (x, y, z) in the following systems of {e1, e2,...} equations. Identify what we are looking for. You can solve a system of equations through addition, subtraction, multiplication, or substitution. Identify the solution of the given graphs. System of Equations Substitution - Sample Math Practice Problems The math problems below can be generated by MathScore.com, a math practice program for schools and individual families. Click on the concept that you need help with or follow each lesson in the order presented for a complete study of Systems of Equations. Select the answer from the 4 answer choices: A,B,C, or D. If you have any questions or you need help, feel free to ask and contact me. Math Games Online. This website uses cookies to ensure you get the best experience. Access these online resources for additional instruction and practice with solving systems of equations by substitution. Solving Systems Of Equations It might seem obvious, but to meaningfully solve a system of equations, they must share one or more variables. One number is 3 less than the other. Manga High Online Practice-Elimination. Systems of Equations Game If you want to solve systems of equations and score tones of points, we have the perfect game for you. A practical Algebra Simultaneous Equations Solver: Solves System of Equations; Simultanous equation calculator is an online tool that solves systems of equations step by step. c. Find the point, or points, or intersection. Graphing Systems of Equations Practice Problems. Recall that a linear equation can take the form $$Ax+By+C=0$$. How To Solve Applications with Systems of Equations. Read the problem. Students get to practice graphing lines in a computer enhanced way. A system of equations in three variables is dependent if it has an infinite number of solutions. ©8 HKeuhtmac uSWoofDtOwSaFrKej RLQLPCC.3 z hAHl5lW 2rZiigRhct0s7 drUeAsqeJryv3eTdA.k p qM4a0dTeD nweiKtkh1 RICnDfbibnji etoeK JAClWgGefb arkaC n17.8-3-Worksheet by Kuta Software LLC Answers to Practice: Solving Systems of Equations (3 Different Methods) (ID: 1) Linear Equations Worksheets Standard Form to Slope Intercept Form Worksheets Finding the Slope of an Equation of a Line Worksheets Find Slope From Two Points Worksheets Finding Slope Quizzes: Combining Like Terms Straight Line Graph Slope Formula - Finding slope of a line using point-point method System of Linear Equations Linear Equations Quizzes: Solving systems of linear equations online. This unit will teach you three different methods for solving a system of equations. You'll learn to solve first-order equations, autonomous equations, and nonlinear differential equations. Access this online resource for additional instruction and practice with systems of equations. A system of equations can be solved using several different methods. Systems of Equations; Section 4.2 Exercises Practice Makes Perfect. Systems of equations are a set of two (or more) equations that have two (or more) variables. System of Equations - Factorization on Brilliant, the largest community of math and science problem solvers. System of Equations Addition - Sample Math Practice Problems The math problems below can be generated by MathScore.com, a math practice program for schools and individual families. Systems of linear equations are a common and applicable subset of systems of equations. In the case of two variables, these systems can be thought of as lines drawn in two-dimensional space. Nonlinear Systems of Equations; Solve a System of Nonlinear Equations; Most of the time, a systems of equations question on the ACT will involve two equations and two variables.It is by no means unheard of to have three or more equations and variables, but systems of equations are rare enough already and ones with more than two equations are even rarer than that. Practice questions. If the graphs extend beyond the small grid with x and y both between −10 and 10, graphing the lines may be cumbersome. Here’s the link: Desmos – Linear Systems Bundle. Systems of Equations Game - a fun and interactive way to practice your math skills! In this post we’ll take a look at some activities and how they might work in your classroom. {y = 3x 2 − 2x + 7 y + 5 = 1__ 2 x Consider the system of equations {y = x 2 y = mx + b. No solution, because the system is independent. Solve systems of two linear equations in two variables algebraically, and estimate solutions by graphing the equations. Solve one of the equations for either variable. This website uses cookies to ensure you get the best experience. The equations relate to one another, and each can be solved only with the information that the other provides. Access this online resource for additional instruction and practice with systems of equations. The main difference is that we’ll usually end up getting two (or more!) A solution to the system is the values for the set of variables that can simultaneously satisfy all equations of the system. Here is a set of practice problems to accompany the Linear Systems with Three Variables section of the Systems of Equations chapter of the notes … The system of equations has: A. I love this practice activity from Math Games soooo much. 72. Lesson Quiz. Systems of equations with substitution: potato chips, Systems of equations with substitution: -3x-4y=-2 & y=2x-5, Practice: Systems of equations with substitution, Substitution method review (systems of equations), Solving systems of equations with elimination, Solving systems of equations with substitution. The correct answer is Choice (E). A system of nonlinear equations is a system of two or more equations in two or more variables containing at least one equation that is not linear. Ex 1: (6, 5) is the solution to the system. You can expand on this knowledge with MIT's 2x2 Systems course, designed to introduce coupled differential equations. Find the numbers. The sum of two number is 15. 10 questions2. Systems of equations can have one unique solution (intersecting lines), no solution (non-overlapping parallel lines), or infinite solutions (completely overlapping lines). I really emphasize with students that you can see how many solutions a system of equations has. Brilliant. System of Equations - Substitution on Brilliant, the largest community of math and science problem solvers. Card Sort: Linear Systems – Practice solving, graphing, and analyzing systems; Polygraph works best with a device for each student, but with a touch of creativity the rest can work with a single projector and whiteboard for the class. Make sure all the words and ideas are understood. Paul's Online Notes Practice Quick Nav Download This unit will teach you three different methods for solving a system of equations. Systems of Equations Game . Algebra - System of Equations is a test on solving System of Equations. Let’s dive in. Select the answer from the 4 answer choices: A,B,C, or D. If you have any questions or you need help, feel free to ask and contact me. 73. 3 problems include identifying the solution from a graph. Systems of equations have a lot of moving parts, so we have a lot of activities to look at. Three Methods for Solving Systems of Equations 1. Solve with elimination. Solve a system of equations by substitution. What is the value of a in the (a, b) solution to the following system of equations? References to complexity and mode refer to the overall difficulty of the problems as they appear in the main program. Practice System of Equations, receive helpful hints, take a quiz, improve your math skills. This activity also practices identifying how many solutions a pair of equations has. The sum of two number is 30. Algebra - System of Equations is a test on solving System of Equations. Improve your math knowledge with free questions in "Solve a system of equations by graphing" and thousands of other math skills. No solution, because the system is inconsistent. Systems of equations word problem (coins) Example: A man has 14 coins in his pocket, all of which are dimes and quarters. 13 - Systems of Equations Word Problems Stations Maze - Students need LOTS of practice with word problems! Learn how to Solve Systems of 3 Equations using the Elimination Method in this free math video tutorial by Mario's Math Tutoring. Practice System of Equations, receive helpful hints, take a quiz, improve your math skills. We can use either Substitution or Elimination, depending on what’s easier. Online Practice Game-Substitution. System of Linear Equations - Problem Solving on Brilliant, the largest community of math and science problem solvers. Finding the slopes of graphs, and graphing proportional relationships and equations Completing and writing rules for function tables With Math Games, students can work at a suitable level for their individual abilities, in a format that’s simple to use and lots of fun to engage with! My Dashboard; math-mooc; Pages; Practice Problems - Systems of Linear Equations and Inequalities in Two Variables Algebra 1 Test Practice. You can find some great ideas for using task cards here and here. Substitution method review (systems of equations) Next lesson. Donate or volunteer today! The point of intersection is the solution to the system of equations. Systems Of Equations - Introduction In algebra, a system of equations is a group of two or more equations that contain the same set of variables. You can play this basketball game alone, against another player, or against the computer. If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. By using this website, you agree to our Cookie Policy. The question asks you to solve for a, so you need to eliminate the b values from the system. Students will practice solving systems of linear equations using the substitution method to complete this 20 problems crossword puzzle. Though it is best to know how to solve any systems question in multiple ways, it is completely okay to pick one solving method and stick with it each time. Solve simple cases by inspection. Systems of equations in three variables that are inconsistent could result from three parallel planes, two parallel planes and one intersecting plane, or three planes that intersect the other two but not at the same location. Linear Equations Worksheets Standard Form to Slope Intercept Form Worksheets Finding the Slope of an Equation of a Line Worksheets Find Slope From Two Points Worksheets Finding Slope Quizzes: Combining Like Terms Straight Line Graph Slope Formula - Finding slope of a line using point-point method System of Linear Equations Linear Equations Quizzes: To log in and use all the features of Khan Academy, please enable JavaScript in your browser. Here is a set of practice problems to accompany the Nonlinear Systems section of the Systems of Equations chapter of the notes for Paul Dawkins Algebra course at Lamar University. Graphing a. Graph one equation b. Graph the other equation on the same plane. Our mission is to provide a free, world-class education to anyone, anywhere. Students are to find the x-coordinate of the solution to each system of equations, locate their answer in the answers bank and complete the puzzle accordingly. Below you will find all lessons and practice problems for Solving Systems of Equations. C. One solution, because the system is independent. If all lines converge to a common point, the system is said to be consistent and has a solution at this point of intersection. ***To solve a system of equations by graphing simply graph both equations on the same coordinate plane and find where they intersect. Graph each equation on the same graph. The difference is that linear equations yield straight lines and only contain only variables, coefficients and constants. 1) −21 x − 3y = 7 7x + y = 3 2) 8x + 4y = −24 5x + y = −18 3) x − 2y = 10 8x + 2y = 8 4) x + 3y = −7 8x + 7y = −5 Solve each system by elimination. Name what we are looking for. System of Equations Addition - Sample Math Practice Problems The math problems below can be generated by MathScore.com, a math practice program for schools and individual families. Now that you've completed the Graphing Systems of Equations lesson, you must be ready to practice a few on your own. Once a solution to a variable is found, working backwards through equations to find other variable values is possible. Practice: Systems of equations with substitution. Solving systems of linear equations by graphing is a good way to visualize the types of solutions that may result. www.math10.ca Systems of Equations Practice … Solving Systems of Equations by Graphing. Select one of the links below to get started. By using this website, you agree to our Cookie Policy. You’ll discover the solving method that suits you the best when it comes to systems of equations once you practice on multiple problems. There are 5 questions in this test and it shouldn't be too hard to do. Answer questions and then view immediate feedback. Tips to Remember When Graphing Systems of Equations. SEE EXAMPLE 1 11. High School Math Solutions – Systems of Equations Calculator, Elimination A system of equations is a collection of two or more equations with the same set of variables. If you're seeing this message, it means we're having trouble loading external resources on our website. ; Answers and explanations. Instructional Video-Solve Linear Systems by Substitution; Instructional Video-Solve by Substitution; Key Concepts. System of Linear Diophantine Equations on Brilliant, the largest community of math and science problem solvers. This is the currently selected item. Systems of equations are extremely useful in applications where there is more than one unknown. Choose variables to represent those quantities. Learn more Accept. Click on the above link to check it out. The decision is accompanied by a detailed description, you can also determine the compatibility of the system of equations, that is the uniqueness of the solution. This online calculator allows you to solve a system of equations by various methods online. If all lines converge to a common point, the system is said to be consistent and has a … B. If the total value of his change is \$2.75, how many dimes and how many quarters does he have? Direct Translation Applications. Solving & Graphing Systems of Equations Chapter Exam Take this practice test to check your existing knowledge of the course material. Systems of linear equations are a common and applicable subset of systems of equations. Below you will find all lessons and practice problems for Solving Systems of Equations. Tips to Remember When Graphing Systems of Equations. Mathematics | L U Decomposition of a System of Linear Equations Last Updated: 02-04-2019 L U decomposition of a matrix is the factorization of a given square matrix into two triangular matrices, one upper triangular matrix and one lower triangular matrix, such that the product of these two matrices gives the original matrix. The following steps will be useful to solve system of equations using substitution. By using this website, you agree to our Cookie Policy. answers for a variable (since we may be dealing with quadratics or higher degree polynomials), and we need to plug in answers to get the other variable. Show Step-by-step Solutions. Interactive Equation Game In this game, students must match different equations with their solutions as fast as possible. Sometimes we need solve systems of non-linear equations, such as those we see in conics. {y = 3 y = x 2 − 4x + 7 12. Systems of Equations ; Key Concepts. In the case of two variables, these systems can be thought of as lines drawn in two-dimensional space. Test and improve your knowledge of OAE Mathematics: Systems of Equations with fun multiple choice exams you can take online with Study.com Step 1 : In the given two equations, solve one of the equations either for x or y. Improve your math knowledge with free questions in "Solve a system of equations using substitution" and thousands of other math skills. See what lessons you have mastered and what lessons you still need further practice on. References to complexity and mode refer to the overall difficulty of the problems as they appear in the main program. For example, 3x + 2y = 5 and 3x + 2y = 6 have no solution because 3x + 2y cannot simultaneously be 5 and 6 . It is consistent and independent. Systems of equations with elimination: potato chips Our mission is to provide a free, world-class education to anyone, anywhere. Solving a system of equations requires you to find the value of more than one variable in more than one equation. This is a systems of equations quiz by graphing.Included in this quiz:1. It is … There are 5 questions in this test and it shouldn't be too hard to do. You appear to be on a device with a "narrow" screen width (, Derivatives of Exponential and Logarithm Functions, L'Hospital's Rule and Indeterminate Forms, Substitution Rule for Indefinite Integrals, Volumes of Solids of Revolution / Method of Rings, Volumes of Solids of Revolution/Method of Cylinders, Parametric Equations and Polar Coordinates, Gradient Vector, Tangent Planes and Normal Lines, Triple Integrals in Cylindrical Coordinates, Triple Integrals in Spherical Coordinates, Linear Homogeneous Differential Equations, Periodic Functions & Orthogonal Functions, Heat Equation with Non-Zero Temperature Boundaries, Absolute Value Equations and Inequalities, \begin{align*}y & = {x^2} + 6x - 8\\ y & = 4x + 7\end{align*}, \begin{align*}y & = 1 - 3x\\ \frac{{{x^2}}}{4} + {y^2} & = 1\end{align*}, \begin{align*}xy & = 4\\ \frac{{{x^2}}}{4} + \frac{{{y^2}}}{{25}} & = 1\end{align*}, \begin{align*}y & = 1 - 2{x^2}\\ {x^2} - \frac{{{y^2}}}{9} & = 1\end{align*}. A perfect simultanous equations solver that helps you solve simultatious equations online. Solve systems of equations where one of the equations is solved for one of the variables. Take this test to quiz yourself on how well you know the material. We discuss what systems of equations are and how to transform them into matrix notation. #3: Three Act Upgrades. This website uses cookies to ensure you get the best experience. A system of nonlinear equations is a system where at least one of the equations is not linear. System of Equations Substitution - Sample Math Practice Problems The math problems below can be generated by MathScore.com, a math practice program for schools and individual families. If the equation is written in standard form, you can either find the x and y intercepts or rewrite the equation in slope intercept form. Khan Academy is a 501(c)(3) nonprofit organization. PRACTICE & PROBLEM SOLVING UNDERSTAND PRACTICE Additional Exercises Available Online Practice Tutorial Determine how many solutions each system of equations has by graphing them. Solving a System of Nonlinear Equations Using Substitution. If you want to know how to solve a system of equations… | 2021-03-03 17:50:34 | {"extraction_info": {"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, "math_score": 0.3250569999217987, "perplexity": 526.2442945172619}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-10/segments/1614178367183.21/warc/CC-MAIN-20210303165500-20210303195500-00604.warc.gz"} |
https://mathematica.stackexchange.com/questions/207188/modify-real-part-and-leaves-imaginary-part-unchanged | # Modify real part and leaves imaginary part unchanged
How can I flip the sign of the real part but not affect the imaginary part of a complex number:
a+bi => -a + bi
Example list:
list = {{-0.282095 + 0.282095 I, -0.27254 + 0.291336 I,
-0.262018 + 0.300835 I, -0.250437 + 0.310542 I}}
expected:
{{0.282095 + 0.282095 I, 0.27254 + 0.291336 I,
0.262018 + 0.300835 I, 0.250437 + 0.310542 I}}
So it's "similar" to conjugate but works on the real not imaginary.
-Conjugate[list]
(* {{0.282095 + 0.282095 I, 0.27254 + 0.291336 I,
0.262018 + 0.300835 I, 0.250437 + 0.310542 I}} *)
• This is not even a Mathematica but a Math answer. I like it – infinitezero Oct 2 '19 at 12:06
list /. Complex[x_, y_] :> Complex[-x, y]
{{0.282095 + 0.282095 I, 0.27254 + 0.291336 I, 0.262018 + 0.300835 I, 0.250437 + 0.310542 I, 2, 3 I}}
• This assumes that all entries are truly Complex[_,_]. But for a list {{-0.282095 + 0.282095 I, -0.3445}}, which also contains only complex numbers (although one of them happens to be purely real, and even have Real head), this won't work correctly. – Ruslan Oct 3 '19 at 5:20
• Agreed. This solution is for complex numbers of the form a + b i that was asked in the question. To include real numbers it needs to be modified to {Complex[x_,y_]:>Complex[-x,y], x_:>-x}. – Suba Thomas Oct 3 '19 at 14:21
• Actually this would also break for "complex" number like e.g. 1.234+0I, which actually collapses to a Real – Ruslan Oct 3 '19 at 14:23
f[z_] = -Re[z] + I Im[z]
f[list]
(* {{0.282095 + 0.282095 I, 0.27254 + 0.291336 I, 0.262018 + 0.300835 I, 0.250437 + 0.310542 I}} *) | 2021-01-21 06:03:30 | {"extraction_info": {"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, "math_score": 0.229335218667984, "perplexity": 2323.3217817602053}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-04/segments/1610703522242.73/warc/CC-MAIN-20210121035242-20210121065242-00476.warc.gz"} |
https://xianblog.wordpress.com/tag/urinals/ | ## the riddle of the stands
Posted in Books, Kids, R with tags , , , , , on May 11, 2018 by xi'an
The simple riddle of last week on The Riddler, about the minimum number of urinals needed for n men to pee if the occupation rule is to stay as far as possible from anyone there and never to stand next to another man, is quickly solved by an R code:
```ocupee=function(M){
ok=rep(0,M)
ok[1]=ok[M]=1
ok[trunc((1+M/2))]=1
while (max(diff((1:M)[ok!=0])>2)){
i=order(-diff((1:M)[ok!=0]))[1]
ok[(1:M)[ok!=0][i]+trunc((diff((1:M)[ok!=0])[i]/2))]=1
}
return(sum(ok>0))
}
```
with maximal occupation illustrated by the graph below:
Meaning that the efficiency of the positioning scheme is not optimal when following the sequential positioning, requiring $N+2^{\lceil log_2(N-1) \rceil}$ urinals. Rather than one out of two, requiring 2N-1 urinals. What is most funny in this simple exercise is the connection exposed in the Riddler with an Xkcd blag written a few years go about the topic. | 2023-02-08 16:10:08 | {"extraction_info": {"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": 1, "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, "math_score": 0.29007095098495483, "perplexity": 1642.0293111948151}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764500837.65/warc/CC-MAIN-20230208155417-20230208185417-00575.warc.gz"} |
https://www.geraldbelton.com/post/getting-started-with-r-a-beginners-guide-part-2/getting-started-with-r-a-beginner-s-guide-part-2/ | # Getting Started With R: A Beginner's Guide, Part 2
Last week, we installed R and R Studio, and we tried out a few simple R commands in the console. But using R in interactive mode, while powerful, has some limits. Today we are going to learn how to use R as a programming language, and we will write our first R Script. But first, let’s look at how we can use R Studio to keep our work organized.
A lot of tutorials introduce these topics much later, if at all. I think it’s very important to learn how to use these organizational tools from the very beginning. Eventually, you are going to need to leave R to go do something else, and you’ll want to be able to come back to R and continue what you were doing. You will have multiple R projects going at the same time, and you’ll want to be able to keep them separated.
You’ve probably closed R studio since last week’s lesson. When you quit R, a box popped up asking “Save workspace image to ~/.Rdata?” If you choose “Yes” at this prompt, when you restart R Studio, you will see in the Environment pane the objects you created in your previous session. In that same pane, you can select the “History” tab, and see all of the commands you ran in that last session. This is not the ideal way to start, stop, and re-start your work in R, but it’s a start.
## Working Directory
Your “working directory” is where R will look (by default) for any files you want to load, and where R (again, by default) will save any files that you write to disk. You can check your working directory with:
getwd()
It’s also displayed at the top of the R Studio console.
You can change your working directory directly with the command:
setwd("~/MyNewDirectory")
The above command assumes that there is already a directory called “MyNewDirectory,” and it is a subdirectory of you home directory. You can also change your working directory by navigating to it in the Files pane of R Studio, and then selecting “More” and “Set as Working Directory” from the Files menu.
Note well that I said you can do these things, not that you should do them. As we will see, there is a better way:
## R Studio Projects
As a general rule, it’s a very good idea to keep all the files associated with a project in one place. That would include data files, R scripts, figures, analytical results, etc. And R Studio makes it very easy to accomplish this via its support for projects.
To demonstrate, let’s make a project to use for the rest of this series of tutorials. In the menu bar at the top of R Studio, click “File” then “New Project.” You’ll see this:
As you can see, you can create a new directory, or choose one that already exists on you computer. The third option, Version Control, is something we will talk about later.
If you choose “New Directory,” you will get an additional menu with three choices: Empty Project, R Package, and Shiny Web Application. Choose Empty Project. Then give your new project a name. I called mine “tutorials.”
Now let’s create an R script. An R script is a file containing a series of commands that can be executed by R; in other words, a computer program.
In R Studio, click the File menu item at the top left of the screen, then select New File, and then R Script. Or you can use the keyboard shortcut, Ctrl-Shift-N. Now the console window no longer takes up the entire left side of your window; it has been split in half. The top left pane is now labeled “Untitled1.” Click on the little picture of a floppy disk, and a dialogue box will pop up, allowing you to name script. Let’s name this one “iris.R.” By convention, the file names of R scripts end with “.R” or “.r,” and you should follow this convention unless you have a good reason to do otherwise.
Since R is primarily a tool for analyzing data, we are going to need some data! Fortunately, there are a lot of ways to get data into R, and we will look at those later. But R also has some very convenient datasets built-in. For this project, we are going to use the iris dataset which is included in R. This dataset contains four measurements of 150 flowers representing three different species of iris.
Let’s inspect the data. Type “iris” in the console window, and press Enter. You’ll see… well, you’ll see a bunch of data scroll by faster than you can tell what it is. Try this instead:
head(iris)
## Sepal.Length Sepal.Width Petal.Length Petal.Width Species
## 1 5.1 3.5 1.4 0.2 setosa
## 2 4.9 3.0 1.4 0.2 setosa
## 3 4.7 3.2 1.3 0.2 setosa
## 4 4.6 3.1 1.5 0.2 setosa
## 5 5.0 3.6 1.4 0.2 setosa
## 6 5.4 3.9 1.7 0.4 setosa
That’s better, now we can see how the iris data is organized. Each row is an observation, and each column is a variable.
Since “head” shows us the first six rows of our data, what do you suppose would happen if you typed “tail(iris)?” Try it and see!
You can learn more about the iris data by typing “?iris”, and you will learn that iris is a data frame containing a famous dataset created by a researcher named Edgar Anderson.
But wait… we typed these commands in the console, not in our new R script. Let’s fix that! Look at your Environment window, and you’ll see another tab labeled “History.” Click that, and you’ll see all of the commands you have run during this R session, in the order that you ran them. You can select a command by clicking on it, and you can select multiple commands using Ctrl-click. Select “head(iris)” and all of the subsequent commands, then click “To Source” in the menu bar. Now the commands are there in your “iris.R” script.
Let’s plot the iris data. In the iris.R window, type plot(iris$Petal.Length, iris$Petal.Width, main="Edgar Anderson's Iris Data") (or copy and paste it from here). When you hit Enter, the cursor moves to a new line and… nothing happens. That’s because you’ve edited the script, but not sent the command to R to be executed. To execute the command, you can put the cursor anywhere in that line and press Ctrl-R, or put the cursor in that line and click “Run” at the top of the window. You can also use your mouse to select multiple commands and then click Run, and the commands will execute in order.
Once you’ve executed that command, you’ll see the File window (in the bottom right corner of R Studio) change to the Plot window. Depending on your screen settings, you might need to click the “Zoom” button to get a good look at your plot. It’s a simple scatter plot, with petal length on the x axis, and petal width on the y axis. You can already see that there seems to be some clustering of the data. Let’s make the plots for each iris species a different color:
plot(iris$Petal.Length, iris$Petal.Width, pch=21, bg=c("red","green3","blue")
[unclass(iris\$Species)], main="Edgar Anderson's Iris Data")
We’ve added some stuff to our basic plot, but don’t worry about those details right now; we are going to go in depth on plotting later. But do notice that the color-coding allows us to instantly see the relationship of petal width to length for the three different species of iris. Also notice that the above two lines are a single command. R doesn’t mind if a command is broken across multiple lines in a script, it uses the () to know when it gets to the end. It’s generally a good idea to break very long commands into multiple lines to make your code easier to read.
Let’s do one more thing before we call it a day. We’ll output our nice plot to a pdf file:
dev.print(pdf, "iris_plot.pdf")
You’ll see some cryptic text in the console screen, and if you click the tab to change the Plots window to the Files window, you’ll see that there is a new file called “iris_plot.pdf” in that window. Make sure your script file is saved. Now you can exit R Studio, and when you come back, you can easily re-run the same script to recreate the same plot. Even better, you have your input data, your processing script, and your output, all in the same folder. The could be very helpful when you come back to a project months later, look at the plot, and say to yourself, “Self, how did you make that plot?”
I strongly recommend you adopt this workflow for all of your projects:
1. Create an R Project.
2. Keep your inputs in the project folder.
3. Keep your processing scripts there, and run them in pieces or all at once.
4. Save your outputs in that folder.
You can do things in R studio using your mouse, such as importing a data file by clicking on it, or saving a plot using the menu in the plot window. Don’t do that! Get in the habit of doing all of your loading, processing, and saving, in your script file. You’ll make it much easier for someone else (or even yourself, months later) to understand how a table was created, how a figure was generated, and what transformations and calculations were done to your data.
Last week I said we would get to version control, and how to share your data and code, but we didn’t quite get there. So that will be our topic for next week. | 2021-10-27 10:22:36 | {"extraction_info": {"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, "math_score": 0.38508161902427673, "perplexity": 1113.9604268476355}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323588113.25/warc/CC-MAIN-20211027084718-20211027114718-00544.warc.gz"} |
https://math.stackexchange.com/questions/3181444/help-proving-that-the-order-topology-is-a-topology | # Help proving that the order topology is a topology
I have been looking into topology recently and have run into a brick wall with the order topology. A lot of the sources that I have been looking at have brushed over the proof that the order topology meets the axioms of a topology. I can prove that the whole space and the empty set are in the topology, but I have trouble proving the finite intersection and arbitrary union are in the topology. It seems easy with the properties of a metric space but without this idea of a metric I can't visualize how this would work. Any help would be greatly appreciated.
• Please try to make the titles of your questions more informative. For example, Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here. – Shaun Apr 9 at 19:24
• Thanks for the advice is there any way to edit my question? @shaun – spazmferret Apr 9 at 19:25
• Use the edit button. – Shaun Apr 9 at 19:26
• found it thanks – spazmferret Apr 9 at 19:26
• What definition of order topology are you using? – Chessanator Apr 9 at 19:31
Check that the set $$\mathcal{B}$$ is a base for a topology on $$(X,<)$$, having at least two points, where $$\mathcal{B}$$ is given by $$\{(x,y)\mid x < y, x,y \in X\} \cup \{[\min(X),x)\mid x \in X\} \cup \{(x,\max(X)]\mid x \in X\}$$
where the latter two are only used if $$\min(X)$$ resp. $$\max(X)$$ exist.
So suppose both exist (that's the most "work", if neither exist we only have the open intervals to deal with, which is a bit easier), and let $$p \in X$$. If $$p=\min(X)$$ then we have some $$p\neq q \in X$$, and we know that $$p \in [\min(X),q) \in \mathcal{B}$$. If $$p=\max(X)$$ then we have some $$p \neq q \in X$$ and we have $$p \in (q,\max(X)]\in \mathcal{B}$$. If neither of these cases holds, $$p$$ is neither the minimal nor the maximum of $$(X,<)$$ and so we have $$q_1,q_2 \in X$$ with $$q_1 < p$$ and $$p < q_2$$. But then $$p \in (q_1, q_2) \in \mathcal{B}$$.
This shows that $$\bigcup \mathcal{B}=X$$, which is the first condition to be a base.
Now, let $$B_1,B_2$$ be two members of $$\mathcal{B}$$ and let $$p \in B_1 \cap B_2$$. If $$p = \min(X)$$ then $$B_1$$ must be of the form $$[\min(X), q_1)$$ (as no other types in $$\mathcal{B}$$ can contain $$\min(X)$$, as those imply an element strictly smaller than $$\min(X)$$ which cannot be), and likewise $$B_2$$ has to be of the form $$[\min(X), q_2)$$. But then $$x \in B_3:=[\min(X), \min(q_1,q_2)) \subseteq B_1 \cap B_2$$ and we are done checking the condition for intersections in this case. The $$p=\max(X)$$ is quite similar, check it.
So we can assume $$p \notin \{\min(X),\max(X)\}$$ and $$B_1 = (q_1, q_2)$$, say, and $$B_2 = (r_1, r_2)$$ for some $$q_1 < q_2, r_1 < r_2$$ in $$X$$ (even either is of one of the two special types, we could safely remove the min/max endpoint and have an open interval instead). Then take $$B_3 = (\max(q_1,r_1), \min(q_2,r_2))$$ and verify that $$p \in B_3 \subseteq B_1 \cap B_2$$, finishing the intersection condition check.
This makes the order topology's definition well-defined.
I usually prefer to take all sets $$\mathcal{S} = \{(x,\rightarrow):=\{y \in X: y > x\}\mid x \in X\}\cup \{(\leftarrow,x):=\{y \in X: y < x\}\mid x \in X\}$$
and define that as a subbase (so that the order topology is the smallest topology that contains $$\mathcal{S}$$) and the induced base then becomes exactly $$\mathcal{B}$$, as can easily be checked as well.
• Thanks a bunch, this is super helpful. – spazmferret Apr 9 at 21:56 | 2019-07-21 13:09:38 | {"extraction_info": {"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": 43, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8473140001296997, "perplexity": 164.7722010365704}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-30/segments/1563195527000.10/warc/CC-MAIN-20190721123414-20190721145414-00076.warc.gz"} |
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Kindly specify | 2023-01-31 10:56:59 | {"extraction_info": {"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, "math_score": 0.9145994782447815, "perplexity": 3423.737244434521}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764499857.57/warc/CC-MAIN-20230131091122-20230131121122-00700.warc.gz"} |
http://physics.stackexchange.com/questions/33679/as-the-universe-expands-why-do-some-things-stretch-but-not-others/33684 | # As the universe expands, why do some things stretch but not others?
I got into watching a video on Olbers' Paradox a few days ago, and from there read about the origins of the universe, its expansion, and so on... it's always fascinated me, but now something about it bothers me. I've heard many analogies about it (the dot-balloon, the raisin bread loaf, and others), but none really seem to explain this question. (This comes close, but dances around the answer more than explain it.)
At the beginning of the universe as we know it, the universe itself was very small, so all the stars giving off light would have made it very bright (16:29 in this video). Since that time, the wavelength of that light has been stretched (17:01, same video). I found a few explanations saying that space itself stretched (here; described as "ether" in the article), which would stretch out the wavelengths.
But here's what bothers me: If space is stretching out, redshifting all the light soaring around our universe, why are we not stretching? Theoretically, the universe is expanding an incredible amount faster than the speed of light, and the edge of the universe is an unimaginably large number of megaparsecs away from us. But should we not notice some of the stretch here, too?
That is to say, if the light in space (the "ether", though I'm not fond of that term) is stretching out, why is everything on Earth still the same size as it was a hundred years ago? Is it stretching uniformly, but we are just unable to notice such a small stretch? Or does mass have some property that space and light do not, that prevents it from stretching out? I've also heard about time stretching, too; does this have an impact on it?
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Possible duplicate: physics.stackexchange.com/q/2110/2451 – Qmechanic Aug 8 '12 at 3:58
@Qmechanic Good link! (I searched for quite a while and didn't find anything like that...) This paper seemed to be the most informative, however it answered the question as being "if the cosmological expansion is greater than the atomic force initially, then atoms will expand, otherwise they will not." Is there an answer as to whether the cosmological or atomic force was larger initially? – Eric Aug 8 '12 at 5:25
This is not my field but the way I understand it is that the expansion involves unbound states. It does not affect bound states. For example protons, bound by the strong interaction, once generated, during the expansion, and decoupled, i.e. the quark gluon plasma has stopped existing, remain protons with the dimensions we know them. Incorporating your comment question:
Is there an answer as to whether the cosmological or atomic force was larger initially?
Decoupling means that as expansion progresses locally the cosmological force becomes smaller than the strong force ( in the case of protons decoupling) and therefore there is no longer a dissolution and recreation of protons from the energy soup of the Big Bang, in this case the quark gluon plasma which should exist before protons can appear.
The same is true for galaxies, which are a gravitationally bound state and separate between each other due to the expansion but remain bound internally.
However the only locally visible effect of the accelerating expansion is the disappearance (by runaway redshift) of distant galaxies; gravitationally bound objects like the Milky Way do not expand.
Photons (and neutrinos) are not bound states, and therefore follow the expansion of space changing their wavelength due to it. Always keep in mind that this expansion happens locally at every spacetime point of what we define as space time for usual physics studies.
This is a field which is researched still, but this model seems to fit observations up to now.
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Excellent explanation. I'm going to keep the question open for a bit longer to see if other answers arrive, but this is definitely a more thorough explanation than I had expected. Thanks a lot! – Eric Aug 8 '12 at 5:45
The expansion of the universe is due to the expansion of spacetime. There's a good article on this here.
Suppose you take two non-interacting particles, put them some distance apart and make them stationary with respect to each other. If you now watch them for a few billion years you'll see the particles start to accelerate away from each other. This happens because the spacetime between the two particles is stretching i.e. there is more "space" between them.
One way of interpreting the acceleration is to say there is a force between the two particles repelling them. This is a slightly dodgy description because there isn't really a force; it's just expansion of space. Nevertheless, if you tied the two particles together with a rope and watch for a few billion years there would be a tension in the rope so the force is real in this sense.
Anyhow, now we have everything we need to understand why the Earth isn't stretching. The expansion of spacetime creates a stretching force, but this will only have an effect if there is no other force to oppose it. For example you are indeed being stretched by the expansion, but the interatomic forces between the atoms in your body are vastly stronger than the stretching due to expansion, so you remain the same size. Likewise the gravitational force between the Sun and Earth is vastly greater than the stretching force so the Earth's orbit doesn't change.
The stretching force is vanishingly small at small distances, but it gets greater and greater with increasing distance so at some point it wins. Galaxies and indeed galaxy clusters are still too small to be stretched, but at greater sizes than this the stretching wins. That's why galaxy clusters are the largest objects observed in the universe. At greater sizes spacetime expansion wins.
A footnote: if anyone's still interested in this subject, there's a paper Local cosmological effects of order H in the orbital motion of a binary system just out claiming that the effect of the expansion on the Solar System might be measurable.
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It's such a darn shame I can't accept two answers... very well-written, and a great explanation of all the forces in play. Thanks so much! – Eric Aug 8 '12 at 6:22
When you model the expanding universe in cosmology, you do so with a particular solution to the Einstein field equations called the FRW metric. The defining feature of this metric is, of course, metric expansion. This means that distances will increase over time. One assumption that goes into the FRW metric is homogeneity. Since the universe is homogeneous on large scales, this works excellently for very large portions of the universe. However, galaxies are certainly not homogeneous. So, you need to use a different metric inside of galaxies - and because of this, space inside of galaxies is totally unaffected by metric expansion. It's not even that the effect is too small to be noticed, galaxies are totally unaffected by expansion. So, we can generalize this to say that expansion occurs in between bound systems. There is a good entry on this at the Usenet FAQ:
http://math.ucr.edu/home/baez/physics/Relativity/GR/expanding_universe.html
Dark energy, however, is a bit trickier. Since it is a negative pressure vacuum energy, it exerts an extremely small force everywhere. So, it has a small effect inside of bound systems. This is because dark energy is a cosmological constant - which is also a term in the Einstein field equations. Since these still govern gravitational interactions inside of galaxies, dark energy has an effect there. The easiest way to see by is by looking at attractive gravitational force between two objects with a cosmological constant in the Newtonian limit: $$F = {GMm \over r^2} - {\Lambda m c^2 \over 3} r$$ However, this effect is utterly negligible.
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I do not think anyone at this time can answer this question with certainty because there could be a number of explanations of what others are observing. I think we need to learn from nature for some of these more difficult questions. My theory could be; that the whole universe is expanding, including you, me and everything that consists of the universe. For example instead of the big bang or the big stretch, we could have the big revolve. This option would follow the course of nature maybe a little more closely. Most things that we know of in the universe revolve and or orbit. Everything in space that we see is moving and orbiting something or another. It appears that by measuring light we can see a shift of some moving away and others moving closer. Now you read that they say that the space in between is stretching. Maybe it is expanding as we are expanding and then the observation could be similar. If it were true and we are all expanding then it can explain some parts of the laws of gravity as well but not all of them. Maybe they can be explained under a different law. In nature the earth itself is revolving and it is renewing itself by the movement if its plates. Over time all of our records of existence will be wiped off the face of the earth. Maybe our sun and our stars known as the white light is part of a dimension that we can see, and all of the particles from this source are travelling and expanding through space and then will eventually collect into the black holes to be spit out the other side ( another dimension ) to again be returned to our dimension through the sun and stars. Thus we have the big revolve. We have a related view point, it will take billions of years for this to happen. Maybe from another view it will take a split second by their time. Maybe our universe is inside and is a part of another universe, maybe someone in the other universe is cold and will throw another log on the fire, and thus our universe goes up in a puff of smoke. All I know for sure is we have a long way to go to get closer to the truth, the guys at NASA are on the right track, one step at a time and by using only proven science to build upon the next mission.
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Hmm... interesting take with a lot of neat ideas. I'm anxious to get some more opinions on this, too. – Eric Aug 8 '12 at 3:13 | 2014-12-25 19:20:00 | {"extraction_info": {"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, "math_score": 0.6941980123519897, "perplexity": 318.77784398042803}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-52/segments/1419447548024.117/warc/CC-MAIN-20141224185908-00041-ip-10-231-17-201.ec2.internal.warc.gz"} |
https://cs.stackexchange.com/questions/78142/is-there-a-natural-example-of-a-total-computable-but-non-primitive-recursive | # Is there a "natural" example of a total, computable but non-primitive recursive function?
Every example of a total, computable but non-primitive recursive function seems to be explicitly constructed for proof theory, or in Godelian proofs of "what is the name of this book?" kind. But is there a non-primitive recursive function or algorithm that occurs naturally, like in physics or number theory or even in industry software?
• The standard example is the Ackermann function. My (possibly incorrect) understanding is that it was "contrived" for this purpose. However, its occurrence in other contexts such as in the asymptotic complexity of union-find and its relationship to things like arrow notations suggests that it is somewhat "natural" despite its origins. Jul 21 '17 at 0:27
• I don't know if you call it natural or not, but you can always just compute $R(x) = P_x(x) + 1$ where $P$ is an enumeration of your computing model, for any computing model with only total machines. By that definition, $\forall y ~:~ R \ne P_y$. Maybe that is what you mean by Godelian though. Jul 21 '17 at 6:06 | 2021-10-18 15:06:54 | {"extraction_info": {"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, "math_score": 0.8180183172225952, "perplexity": 493.86360629507135}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323585203.61/warc/CC-MAIN-20211018124412-20211018154412-00713.warc.gz"} |
http://www.maa.org/publications/periodicals/convergence/problems-another-time?page=25&device=mobile | # Problems from Another Time
Individual problems from throughout mathematics history, as well as articles that include problem sets for students.
An official asks a woman why she has so many bowls to wash. The woman explains that she had dinner guests who ate meat, rice, and soup. Judging by the number of bowls, how many guests were there?
A man agreed to pay for 13 valuable houses worth $5000 each, what the last would amount to, reckoning 7 cents for the first, 4 times 7 cents for the second, and so on, increasing the price 4 times on each to the last. A father left$20,000 to be divided among his four sons ages 6, 8, 10, and 12 years respectively so that each share placed at 4 1/2 compounded interest should amount to the same value when its possessor becomes the age 21.
Two bicyclists travel in opposite directions around a quarter-mile track and meet every 22 seconds. When they travel in the same direction on this track, the faster passes the slower once every 3 minutes and 40 seconds. Find the rate of each rider
Having been given the perimeter and perpendicular of a right angled triangle, it is required to find the triangle.
Given a right triangle where you know the length of the base and the sum of the perpendicular side and the hypotenuse.
If 40 oranges are worth 60 apples, and 75 apples are worth 7 dozen peaches, and 100 peaches are worth 1 box of grapes and three boxes of grapes are worth 40 pounds of pecans, how many peaches can be bought for 100 oranges?
A water tub holds 73 gallons; the pipe which fills it usually admits 7 gallons in 5 minutes; and the tap discharges 20 gallons in 17 minutes.
In a square box that contains 1000 marbles, how many will it take to reach across the bottom of the box in a straight row?
One hundred men besieged in a castle, have sufficient food to allow each one bread to the weight of 14 lot a day for ten months. | 2015-08-03 22:32:04 | {"extraction_info": {"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, "math_score": 0.3721281886100769, "perplexity": 1327.77158327408}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-32/segments/1438042990114.79/warc/CC-MAIN-20150728002310-00204-ip-10-236-191-2.ec2.internal.warc.gz"} |
https://icesp.org.br/mom-and-yunr/open-set-in-topology-with-examples-387871 | ( Extreme topologies) On any set Xwe can define the following: 1. , is also called distance function or simply distance. {\displaystyle S\rightarrow X} The answer to the normal Moore space question was eventually proved to be independent of ZFC. Adams, Colin Conrad, and Robert David Franzosa. For metric spaces second-countability, separability, and the Lindelöf property are all equivalent. Many examples with applications to physics and other areas of math include fluid dynamics, billiards and flows on manifolds. to any topological space T are continuous. A space in which all components are one-point sets is called totally disconnected. Let X be a set and let τ be a family of subsets of X. (See Heine–Borel theorem). In mathematics, general topology is the branch of topology that deals with the basic set-theoretic definitions and constructions used in topology. d However, if their number is infinite, this might not be the case; for instance, the connected components of the set of the rational numbers are the one-point sets, which are not open. In particular, if X is a metric space, sequential continuity and continuity are equivalent. In topology and related areas of mathematics, a metrizable space is a topological space that is homeomorphic to a metric space. If a set is given a different topology, it is viewed as a different topological space. Euclidean Examples The most basic example is the space R with the order topology. Ivanov, V.M. I am a Physics undergrad, and just started studying Topology. The components of any topological space X form a partition of X: they are disjoint, nonempty, and their union is the whole space. {\displaystyle (X,\tau )} In Example 9 mentioned above, it is clear that is a -open set; thus it is --open, -preopen, and --open. 1. This topology on R is strictly finer than the Euclidean topology defined above; a sequence converges to a point in this topology if and only if it converges from above in the Euclidean topology. Topology. where X is a topological space and S is a set (without a specified topology), the final topology on S is defined by letting the open sets of S be those subsets A of S for which f−1(A) is open in X. For non first-countable spaces, sequential continuity might be strictly weaker than continuity. Each choice of definition for 'open set' is called a topology. Viro, O.A. The product topology is sometimes called the Tychonoff topology. Every sequence of points in a compact metric space has a convergent subsequence. (the empty set… If we change the definition of 'open set', we change what continuous functions, compact sets, and connected sets are. x (Every open set in the usual topology is a union of sets/intervals from the first collection in the union above.) Open Sets in a Metric Space. A subset of X may be open, closed, both (clopen set), or neither. Set-theoretic topology is a subject that combines set theory and general topology. Hence these last two topologies cannot arise from a metric. is omitted and one just writes Upper Saddle River: Prentice Hall, 2000. Skip navigation Sign in. Instead of specifying the open subsets of a topological space, the topology can also be determined by a closure operator (denoted cl), which assigns to any subset A ⊆ X its closure, or an interior operator (denoted int), which assigns to any subset A of X its interior. In other words, the sets {pi−1(U)} form a subbase for the topology on X. Conversely, any function whose range is indiscrete is continuous. ∈ A continuum (pl continua) is a nonempty compact connected metric space, or less frequently, a compact connected Hausdorff space. A compact set is sometimes referred to as a compactum, plural compacta. Title of his doctoral dissertation ( 1931 ) of a nonempty topological that! A coarser topology and/or τX is replaced by a finer topology often spaces... Every continuous image of a space in which every subset is open 4 subbase the. Compactness is Tychonoff 's theorem: the ( arbitrary ) product of open sets are the! Algebra a over a topological space the original space sometimes we may refer to a topological algebra over... Check that the topology τY is replaced by open set in topology with examples finer topology concepts also have several ;. Every open set Usuch that a2U a the ( arbitrary ) product of theory... The individual articles term was coined by David van Dantzig ; it appears in union... When an equivalence relation is defined on it are the fundamental building blocks of topology algebra a over a space... Connected if it takes limits of sequences in general, the set of real numbers additionally, connectedness and are. ( 2 points ) let X and Y we want to get an appropriate topology on X, ). In many situations gp -Closed, gp open set. discrete metric lastly, open sets are the building. And Robert David Franzosa identify them at every point homeomorphic to a topological space to from... Between two topological spaces, this means that for every open set. S a theorem we ’ ll later! A union of open sets are open members of τ are called Tychonoff. Preserve sequential limits '' there will probably be a family of subsets of X X ; d!! ) of a such that open in X most other branches of topology set! The rational numbers Q, and sets that are open and closed finite... Indiscrete is continuous the normal Moore space question, a set, Pre open set in Topology.Wikipedia a! An approach to topology that deals with the subspace topology of S, as... What continuous functions τ are called sequential spaces. open, closed, and algebraic topology, but for products!, sequential continuity and continuity are equivalent path-connected ( or pathwise connected or 0-connected ) if there a. Component is also an open set Usuch that a2U a the continuous image of a topological space that is used... Plural compacta function f: ( X, Ttriv: the topology on R, the basic open sets an. Often difficult to use directly definition and examples of topologies ) on any infinite set. Cartesian. X may be open, closed, and algebraic topology elements xand yof a set X, Ttriv the... Viewed as a subset of a such that the space X in R 2 as the open sets in X! Proofs, and thus there are equivalent are called the Tychonoff topology explicitly this. Also holds: any function whose range is indiscrete is continuous only if it is easy to check that basis. Have many topologies on them and continuity are equivalent are called the connected components of space! Introduced earlier complex numbers, Defining topologies via continuous functions preserve sequential ''! At 19:22 and bounded least likely to be more precise, one can \recover '' all the sets... Τ ) is a topological structure exist and thus there are equivalent are called the box topology is than. Following definitions, X, Tdis: the topology ˝is implicit continuity are equivalent are sequential. Definitions for a topology on any normed vector space this topology converges to every point zero! Whole space R with the quotient topology, which is a subject that combines set and! Hausdorff spaces where limit points are unique is injective, this does not imply that it is union... Of the topologies of each Xi forms a basis zero, one can \recover all! \Recover '' all the open and closed sets consist of Xitself and all nite subsets of.! B } and let = { a, b ) ( 2 points ) let X Y. Property are all equivalent integers is open 4 one path-component, i.e later... Point of the following conditions are equivalent are called the connected components of the space is paracompact Hausdorff... Their intersections are cylinder sets here, the set of all products of open sets in spaces X have neccessary., \tau ) $where$ \tau \$ is the empty set and let τ a... Include fluid dynamics, billiards and flows on manifolds δ-ε definition of 'open set ', we shall give! That it is easy to check that the space X is said be... All open intervals are open balls defined by the open sets in X called totally disconnected the strongest on. ) let Xbe a topological space codomain is Hausdorff, then the pair X! Numbers, Defining topologies via continuous functions for 'open set ' is due to John von Neumann eventually proved be! ; however, often topological spaces, the sets whose complement is finite, component. Can check that these open sets are the fundamental building blocks of topology devoted the! Is defined on it also holds: any function whose range is indiscrete is continuous number of examples which do! It is viewed as a counterexample in many situations in metric spaces the! Show that a subset of a Hausdorff space is paracompact and Hausdorff, then natural... '' in R of finite length is compact if and only if it is a finite set is a... Spaces where limit points keywords: Pre- closed set, gp -Closed, gp set! In the nite complement topology on a finite-dimensional vector space examples ( Part ). On preimages are often difficult to use directly a subset of X was the subject of intense research open. Continuous map is an introduction to algebraic topology definition is equivalent to the concept of continuity with open sets weak. '', and Robert David Franzosa probably be a family of subsets of.. With … example 2.3 as well questions that are eventually constant Hausdorff spaces limit! Except zero fact this property characterizes continuous functions preserve sequential limits.... Continuum theory is the normal Moore space question, a question in general, the {... Function f−1 need not be continuous each component is also an open set. eventually constant characterizes functions. Called open sets hence these last two topologies can not arise from a metric topology, which associates algebraic such... Family of subsets of X is said to be more precise, one sees the! Change the definition of 'open set ', we shall instead give a meaning to which UˆXare... Topology whose open sets are the same for finite products, a Xwe. Have the following properties: 1 a finite set is not open, this topology converges to every point zero. Compact sets, by taking complements a subject that combines set theory ( ZFC.... Does not imply that it is easy to check that these open sets in.., often topological spaces. a neighborhood of every point except zero with continuous inverse function f−1 need be! Time when subjected to continuous change a set X endowed with the usual topology subsets of is. Give a meaning to which subsets UˆXare \open '' thus there are equivalent are sequential. Lecture on it appears in the title of his doctoral dissertation ( 1931.... Algebra a over a topological space conditions are equivalent: the continuous image of a space. Keywords: Pre- closed set, gp open set if it takes limits of nets and. ⊆ τ2 ( see also comparison of topologies ) on any normed vector space question in general spaces. Topologies ) on any infinite set. be examinable unless I actually lecture on.... The context of metric spaces second-countability, separability, and so on set ; and whole! Specified in terms of limit points are unique many ways to define a topology \tau } X a! Space question, a set and the whole space R are open and closed sets consist of and... Basic open sets is open, as is the branch of general assumed! The topological space X is said to be ambiguous is then the pair ( X ; d ) the. Colin Conrad, and in fact this property characterizes continuous functions preserve sequential limits '' metric spaces )! The topologies of each Xi forms a topology on a set GˆR is 4... That makes it an algebra over K. a unital associative topological algebra is a finite set is ). On preimages are often difficult to use directly a ) ( 2 points ) let be! To every point except zero ) is an approach to topology that avoids mentioning points can also given...: 1:17:06 importantly the following: general topology family of subsets of X \basic closed sets a. Let = { a } } this gives back the above δ-ε definition of continuity with open is... Has a convergent subsequence open 4 products they coincide any normed vector space the particular topology τ now the... Cofinite topology questions that are neither open nor closed, and many of the theory ways! A quotient topology, in the topology induced by d is τ { \displaystyle \tau.! The first collection in the case where their number is finite, each component is neighborhood! Is easy to check that the only metric possible on a finite set is closed and bounded nets instead sequences... And sets that are independent of ZFC spaces must be Hausdorff spaces where limit points are unique it. By inclusion ) of a number of examples which you do not have the conditions. Such 6 i.e., its complement is open, it ’ S a theorem we ’ do... As groups to topological spaces are metric spaces, open set in topology with examples of sequences need be! | 2021-05-18 04:52:02 | {"extraction_info": {"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, "math_score": 0.93380206823349, "perplexity": 550.5766992520416}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-21/segments/1620243989820.78/warc/CC-MAIN-20210518033148-20210518063148-00329.warc.gz"} |
https://jmodelica.org/assimulo/tutorial_sundials.html | # Sundials¶
Assimulo comes with some “own” solvers, while most solvers are solvers provided from elsewhere and wrapped into Assimulo.
The most central solver group are those collected in the SUNDIALS Sundials package.
Note
• The SUNDIALS code is left unchanged.
• Not all of SUNDIALS parameters are currently lifted to Python.
Examples for the use of SUNDIALS within Assimulo are given in tutorialCVode.py and For the complete example, tutorialIDA.py | 2019-01-16 03:32:49 | {"extraction_info": {"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, "math_score": 0.5508444309234619, "perplexity": 10553.44151754627}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-04/segments/1547583656665.34/warc/CC-MAIN-20190116031807-20190116053807-00380.warc.gz"} |
https://itectec.com/ubuntu/ubuntu-change-ubuntu-server-18-04-lts-bionic-console-screen-resolution/ | # Ubuntu – Change Ubuntu Server 18.04 LTS Bionic Console Screen Resolution
18.04consolegrub2servervirtualbox
• Running Ubuntu 18.04 sever on a physical machine or VirtualBox?
• Do you need/want to use the full native resolution of your screen? Higher than 640×480 / 800×600?
Then you may be stuck as I've been, because the solution that used to work on 14.04 and 16.04 doesn't work on 18.04.
Problems to solve (goals)
1. hwinfo --framebuffer gives an empty output -> find another solution
2. adding the usual lines to /etc/default/grub, only helps at the first stage of the boot. After GRUB2 has done its work, the resolution switches back to a lower value -> resolve this too
3. exclude xorg based tools like xrandr (this is a server without GUI by default)
4. increase VT1-7 (Ctrl+Alt+F1, F2F7 ) resolution in case of Desktop systems with a GUI
5. set the resolution to 1280×1024
• 1. Get supported video mode (use vbeinfo instead of hwinfo)
• reboot
• hold down SHIFT after the BIOS/UEFI finished
• press c´ for the GRUB command line
• type set pager=1, then hit ENTER
• type vbeinfo, then hit ENTER
• take a note about the supported video mode you need Mode 0x031b: 1280x1024 (+3840), 24 bits
• reboot
2. Modify / add the following lines to /etc/default/grub to match the ones below
> GRUB_CMDLINE_LINUX_DEFAULT="video=0x0345 gfxpayload=true"
> ...
> # The resolution used on graphical terminal
> # note that you can use only modes which your graphic card supports via VBE
> # you can see them in real GRUB with the command vbeinfo'
> GRUB_GFXMODE=1280x1024x24
3. Update GRUB
sudo update-grub
4. Reboot
sudo reboot
Note1: I've tested the above solution with Ubuntu 18.04 Server and Desktop on VirtualBox.
Note2: Didn't include linux terminal (emulator), putty, cygwin, conemu and other fancy tools, because this is about the bare console.
None of the sources I used had a complete solution, but putting the parts together, solved the problem for me.
Change Ubuntu Server 14.04 Screen Resolution
https://ubuntuforums.org/archive/index.php/t-1468789.html | 2021-08-03 19:24:26 | {"extraction_info": {"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, "math_score": 0.2633954882621765, "perplexity": 11262.51395110786}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-31/segments/1627046154471.78/warc/CC-MAIN-20210803191307-20210803221307-00544.warc.gz"} |
https://m.hanspub.org/journal/paper/611 | # 含权双调和椭圆型问题的特征值不等式Eigenvalue Inequality for a Weighted Biharmonic Elliptic Problem
Abstract: In this paper, we study the relation between the first and the second eigenvalue of a weighted biharmonic elliptic problem with Dirichlet boundary. By some variational technique we obtain the corresponding inequality, and some evaluations are put forward in low dimension space.
[1] L. E. Payne, G. Polya and H. F. Weinberger. On the ratio of consecutive eigenvalues. Journal of Math and Physics, 1956, 35: 289-298.
[2] G. N. Hile, M. H. Protter. Inequalities for eigenvalues of the Laplacian. Indiana University Mathematic Journal, 1980, 29: 523-538.
[3] H. C. Yang. Estimates of the difference between consecutive eigenvalues. International Centre for Theoretical Physics, 1995, (3): 47-63.
[4] E. M. Harrell II, J. Stubbe. On trace identities and universal eigenvalue estimates for some partial differential operators. Transactions of the American Mathematical Society, 1997, 349(5): 1797-1809.
[5] G. N. Hile, R. Z. Yeh. Inequalities for eigenvalues of the Biharmonic operator. Pacific Journal of Mathematics, 1984, 112(1): 115-133.
[6] M. S. Ashbaugh, L. Hermi. A unified approach to universal inequalities for eigenvalues of elliptic operators. Pacific Journal of Mathematics, 2004, 217(2): 201-219.
[7] P. Li. Eigenvalue estimates on homogeneous manifolds. Comment Mathematic Helvetic, 1980, 55(1): 347-363.
[8] 屈长征, 崔尚斌. 复Monge-Ampere方程的特征值问题[J]. 纯粹数学与应用数学, 1995, 11(2): 37-40.
Top | 2021-09-18 15:51:37 | {"extraction_info": {"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, "math_score": 0.8775097727775574, "perplexity": 2251.520480254198}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-39/segments/1631780056548.77/warc/CC-MAIN-20210918154248-20210918184248-00016.warc.gz"} |
https://mathoverflow.net/questions/352925/applying-analytic-coordinate-changes-to-singular-function-germs/368103 | # Applying analytic coordinate changes to singular function germs [closed]
Suppose we are given a function germ \begin{align} f = \sum a_{ijk}x^iy^jz^k \end{align} such that $$f\in \mathfrak{m}^2$$, where $$\mathfrak{m}$$ is the ideal in $$\mathbb{C}\{x,y,z\}$$ of holomorphic functions vanishing at 0. I am currently reading an expository text on the du Val surface singularities in which the author sometimes simplifies a germ like this by using analytic coordinate changes. Stuff like: "If the 2-jet of $$f$$ is $$x^2$$, an analytic coordinate change can be used to remove any further appearances of $$x$$ in $$f$$" or "if $$J_2(f)=x^2 + y^2$$ then the existence of an $$a_{ijk}\neq 0$$ with $$i + j <2$$ implies that at least one term of the form $$z^m$$ or $$xz^m$$ or $$yz^m$$ appears in $$f$$, and then an analytic coordinate change can be used to make $$f = x^2 + y^2 + z^{n+1}$$."
I don't really understand which coordinate changes are applied here. Does anyone have a reference explaining techniques like this in further detail?
Analytic change of coordinates in $$(\mathbb{C}^3,0)$$ must take $$0$$ to $$0$$. So they are induced by (local) $$\mathbb{C}$$-automorphisms of $$\mathcal{O}_{\mathbb{C}^3,0}=\mathbb{C}\{x,y,z\}$$ taking $$\mathfrak{m}=(x,y,z)$$ to itself. It is not difficult to check that those maps are always of type $$x=a_{1,1}x_1+a_{2,1}y_1+a_{3,1}z_1+\ldots$$ $$y=a_{1,2}x_1+a_{2,2}y_1+a_{3,2}z_1+\ldots$$ $$z=a_{1,3}x_1+a_{2,3}y_1+a_{3,3}z_1+\ldots$$ where $$A=(a_{i,j})\in GL_3$$ and stand $$\ldots$$ for 'higher order terms'.
Being very practical, in each case you should express a given function $$f(x,y,z)$$ in the new coordinates $$(x_1,y_1,z_1)$$ in a way that simplify its form. Most of the times it reduces to a linear algebra question...
Of course this also works for every dimension $$n$$. | 2023-02-04 22:24:09 | {"extraction_info": {"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": 30, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9217686057090759, "perplexity": 82.33402304938069}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764500154.33/warc/CC-MAIN-20230204205328-20230204235328-00184.warc.gz"} |
https://mathematica.stackexchange.com/questions/51138/finding-integral-bounds | # Finding integral bounds
i have this integral shown below is equal to 1, and i need to find "a" on mathematica, but I'm not sure how.
integral (from 0 to a) sqrt((-50.8938 sin(8.4823 t))^2+(4-11.3097 sin(11.3097 t))^2) dt = 1
any suggestions?
• If you are asking a question about Mathematica why don't you write your code in a Mathematica-friendly way? – Öskå Jun 19 '14 at 13:54
• Start with the free-form input, see closely related: Symbolic Definite Integration – Artes Jun 19 '14 at 15:03
f[t_?NumericQ] =
Sqrt[(-50.8938 Sin[8.4823 t])^2 + (4 - 11.3097 Sin[11.3097 t])^2];
Looking at a plot of f[t] to find an initial value for t in FindRoot
Plot[f[t], {t, 0, .1}]
Clear[a]
a = a /. FindRoot[
NIntegrate[f[t], {t, 0, a}] == 1, {a, 0.08}] //
Quiet
0.0680318
Check
NIntegrate[f[t], {t, 0, a}]
1.
Show[
RegionPlot[0 <= t <= a && y <= f[t],
{t, 0, .1}, {y, 0, 40},
PlotPoints -> 50,
AspectRatio -> 1/GoldenRatio],
Plot[f[t], {t, 0, .1}]] | 2019-11-22 21:03:00 | {"extraction_info": {"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, "math_score": 0.17792829871177673, "perplexity": 4960.929358505733}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-47/segments/1573496671548.98/warc/CC-MAIN-20191122194802-20191122223802-00448.warc.gz"} |
https://www.albert.io/ie/sat-chemistry-subject-test/mass-of-an-ammonium-compound | Free Version
Difficult
# Mass of an Ammonium Compound
SATCHM-2YLE4P
What is the mass of $6.022 × 10^{23}$ formula units of $(NH_4)_2SO_4$?
A
$228.34g$
B
$114.11g$
C
$210.29g$
D
$342.14g$
E
$132.16g$ | 2017-03-01 20:31:56 | {"extraction_info": {"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, "math_score": 0.6344687938690186, "perplexity": 13033.521117681272}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-09/segments/1487501174276.22/warc/CC-MAIN-20170219104614-00095-ip-10-171-10-108.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/proof-by-induction-involving-divisibility.799058/ | # Proof by Induction Involving Divisibility
1. Feb 20, 2015
### Colleen G
1. The problem statement, all variables and given/known data
Let P(n): 7|(34n+1-52n-1. Prove that P(n) is true for every natural number n.
2. Relevant equations
*I know that proving by induction requires a proving P(1) true, and then proving P(k+1) true.
*If a|b, then b=a*n, for some n∈ℤ
3. The attempt at a solution
I have proved the "base case" for P(1). Long story short, for n=1, you wind up with 7|238 by definition of divisibility since 238=7(34). So P(1) is true.
The next part is where it gets tricky - Assuming P(k) is true, that is, 7|(34k+1-52k-1, which is the inductive hypothesis. Then proving for P(k+1). What I have so far is...
7|7|(34(k+1)+1-52(k+1)-1
=34k+5-52k+1
=34 * 34k+1-51*52k
=34 * 34k+1-(3*52k+2*52k)
=34 * 34k+1-3*52k-2*52k
=..............
Now I don't know what to do! I know I have to get it to the point where I can use the inductive hypothesis, which is what I'm trying to do, but I've hit a wall, or taken a wrong turn when split the 5 into a two and a three. Any ideas?
2. Feb 20, 2015
### Dick
Here's a short hint 3^4=81=56+25=56+5^2 and 56 is divisible by 7.
3. Feb 21, 2015
### Colleen G
Ok yes I see that, but am having trouble using it. Are the steps that I have taken so far correct? Or have I done more than necessary. What I'm saying is, can I use this information about 3^4 from the step that I left off at?
4. Feb 21, 2015
### Dick
Convince yourself that $P(k+1)=81*3^{4k+1}-25*5^{2k-1}$. Use that $81=56+25$. See how you can express that in terms of $P(k)$?
5. Feb 21, 2015
### Ray Vickson
This is badly written: the "equation" you wrote, namely
$$7|7|(3^{4(k+1)+1} - 5^{2(k+1)-1} \\ = 3^{4k+5} - 5^{2k+1} \\ \vdots$$
does not even make sense.
To say it properly, here is a hint: letting $F(n) = 3^{4n+1} - 5^{2n-1}$, prove that for positive integer $k$, $F(k) = 7j$ for some positive integer $j$ implies $F(k+1) = 7 m$ for some positive integer $m$.
6. Feb 21, 2015
### Colleen G
Thank you for you help, Dick! I understand now.
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Draft saved Draft deleted | 2018-01-21 17:04:12 | {"extraction_info": {"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, "math_score": 0.753143846988678, "perplexity": 939.9447616237699}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-05/segments/1516084890795.64/warc/CC-MAIN-20180121155718-20180121175718-00184.warc.gz"} |
http://quant.stackexchange.com/tags/arbitrage/hot | # Tag Info
24
Consider the standard error, and in particular the distance between the upper and lower limits: $$\Delta = (\bar{x} + SE \cdot \alpha) - (\bar{x} - SE \cdot \alpha) = 2 \cdot SE \cdot \alpha$$ Using the formula for standard error, we can solve for sample size: n = \left(\frac{2 \cdot s \cdot ...
11
Setting aside, that it's not pure riskless arbitrage, but rather statistical arbitrage: You can extract the profit by performing continuous delta hedging. If you constantly adjust your hedge position you gain/lose money by delta hedging. Being long option (gamma long), you sell at higher prices and buy at lower ones. Over the course of time you realize ...
10
There are really a few issues here: 1) When do I turn off a model because I believe the model is invalid? This is a subjective call and depends on many things such as how strong the economic reasoning behind the model is, how crowded the space is, and how poorly the model is performing relative to backtest. With regard to the last point, as a rule of ...
9
You ought to have pre-determined "kill" switches, like a maximum allowable drawdown or time-from-high. Ideally you should get an idea of what these values would be from your backtests. When you do shutdown a model, don't just throw away the code. The strategy might not be working at the moment, but it could come back in the future. I just heard that models ...
8
An index is just an abstract concept and does not hold securities. Hence no source of revenue from lending them. A portfolio mirroring an index holds the securities and can in fact generate revenue by loaning the securities to others wanting to short the stocks. This provides a positive bias. That is often offset by a negative bias when the index ...
8
The "price protection" refers to RegNMS in the US. A stock exchange that does not have the best price must route all order flow to the exchange that does. The SIP in the figure is a consolidated feed that lists the best price among all exchanges. Consider this example: a broker sends a market order to buy JNJ to NYSE where the best offer is \$86.97. ... 7 Neither. Black--Scholes says nothing about the parameter values:$\mu$and$\sigma.$A very large$\mu$and very small$\sigma$is very unlikely to actually occur in the market and if it did you could make money with high probability without using option contracts. BS simply says that if the market follows a certain process then a certain option price is ... 6 You can use a equity based model. Stop trading when your equity drops below your "X-day" equity moving average, and resume trading when your equity crosses above the "X day" equity moving average. You could also do this by measuring the slope of the curve, and not trading when the slope is statistically below 0. I like this method because it does not tie ... 6 In my mind, there are two questions here: 1) How does DB make money given a zero expense ratio? This is covered by Dirk and Lliane. Basically, DB gets cheap funding and stock loan fees in return for paying marketing / index / hedging costs. The ETF investor gets zero expense ratio in return for taking DB credit risk. 2) Why does it look like the etf ... 6 Are there any other mechanisms at play here which might explain this kind of tracking error? Dirk is right, you often lend the titles internally or not, etc. You can also write calls for your index, this is not orthodox, but it's ETF, there is no orthodoxy there... Edit : With the graph and given the outperforming is seasonnal (around May), I think we ... 6 This depends a little bit on your definition of volatility arbitrage but in general what is meant is a strategy that takes advantage of the difference between implied volatility and realized volatility. Normally you receive implied variance and pay realized variance. This strategy is the classical example of picking up nickles in front of a steamroller ... 6 The main problem is that you cannot achieve Libor in the markets. So the old-fashioned method of discounting at Libor doesn't work any more. As an example, if you compound up the 3m Libor with today's price on a 3x6 FRA, you won't get 6m Libor. Traditionally, that would mean arbitrage, but these days it's just a fact of life. You cannot achieve 3m Libor for ... 6 This is called on the run/off the run arbitrage, a type of convergence trade. The basic idea is that as the liquidity premium disappears for the on-the-run issue, the price will fall and converge to the price of previous issues. Here are a couple papers - http://people.stern.nyu.edu/lpederse/courses/LAP/papers/SearchBargaining/VayanosWeill.pdf ... 5 I am not sure why your question had so many upvotes because in currency markets anything else but triangular arbitrage does not exist. What is a quadrangular arb, I have never heard of it despite having traded fx among other asset classes for over ten years now. Think about it: Lets say you observe the price of EUR/USD. You can build triangular arbs by ... 5 Short answer: yes. Long answer: the challenge in trading these things, like you mentioned, is that each contract is not perfectly hedgable. This is an intentional choice made by the exchanges that list these products, so that they can provide an incentive to trading firms(locals) to provide liquidity for these new products and help boost trading volume. ... 5 I typically have several tests. Like Ralph Winters said, one calculation to use is the equity curve. I use "...equity curve slope must be greater than x1..." to continue using a model. Another test is, "...%wins must be greater than x2...". Another is, "...average %return per winning transaction must be greater than x3...". Another is "...%return per ... 5 To see the connection between put-call parity and option price you should read this highly insightful paper by Espen Gaarder Haug & Nassim Nicholas Taleb: Option traders use (very) sophisticated heuristics, never the Black– Scholes–Merton formula It shows how you can heuristically derive option pricing formulas by adapting the tails and skewness ... 5 Both premiums are actually always positive by definition. The difference will be positive when the forward price exceeds the strike and vice versa. 5 Yes, there is a software application that you can purchase for$39.99 which stores all your tick data in a highly compressed format while still allowing maximum throughput and lowest latency data queries that I have ever seen. The package provides APIs to all languages under the sun but because they have a special sale going on it comes with the complete ...
5
Not sure why Python is recommended when you clearly ask for a .Net solution (well you may look at IronPython but I do not recommend it given there are much better options, see below), aside the fact that Python is horribly slow even when performing non-mission-critical data analysis and research. Even C# easily runs circles around most python scripts, given ...
5
The option is a contract that gives you the right to buy the stock in one year for 18. Today people are trading the stock for 20, so you can sell the stock short for 20 today. Selling the stock short means someone will give you 20 cash today in return for a stock IOU, where you are obligated to deliver the stock to them on a later date. So you get 20 cash ...
5
In three bullet points: Efficiency: the obtained prices maximize assumed utilities of different agents. In their paper "The Valuation of Option Contracts and a Test of Market Efficiency", Cohen, Black and Scholes compare the theoretical value of options to their market price. The efficiency is in this sense: can agents obtain more or less in practice than ...
4
Fatih Yilmaz, formerly of Bank of America (currently BlueGold), has a piece called "Imaginal Spreads and Pairs Trading" on exactly this topic, if you can find it (I couldn't find a copy on the public internet), originally published April 17, 2009. He writes: Academics and industry practitioners generally concentrate on time series aspects of currency ...
4
Here's the relationship of rho on calls and puts. When you buy call options instead of the the underlying, you are effectively buying an indirect leveraged position in the underlying. A simple way to see this is buy re-arranging the terms of the Put-Call parity equation solving for the call price. The value of the call is equal to a synthetic position ...
4
Interesting question! I don't think you will get very far just using mid prices, though...any sufficiently sensitive test will flag nearly every situation as an arbitrage since $A_\text{mid}+B_\text{mid} \neq (A+B)_\text{mid}$ in most cases. Instead, what about viewing each price set as a dimension in $n$-dimensional space? The arbitrages occur if the ...
4
On Bloomberg. Go to ETF -> holdings and type "97 Enter".
4
While triangular arbitrages exists, they are a rare, short lived, and shallow. In several academic datasets they are very rarely seen, mainly for two reasons, market efficiency aside: (1) the time resolution of the data is not tick by tick but aggregated at some level (for example at 1 second intervals), (2) the dataset doesn't include all available quotes ...
4
You could compute index dividend yield from ATM options using linearized put-call parity (assuming index options are European.) The present value of the dividend payment is: $PV(div) = P - C + (S - K) + K(e^{rT} - 1)$ where $r$ is interest rate to the option expiration and $T$ is time to maturity in years. Then the implied dividend is: $d = ... 4 Is there anything which can be done to account for the underlyings with no listed option contracts? Classical options pricing theory relies on the idea that any option contract can be simulated with the appropriate dynamic hedging strategy. Options pricing practice indicates that this is sort-of true. So one thing you can do is synthesize the given ... 4 In the derivatives context, "arbitrage free" means almost surely for the probability measure under consideration. This is in opposition with statistical arbitrage used at high frequencies for example. More precisely the assumption is that there is no$T\geq 0$and self-financed portfolio$V$such that$V_0 = 0$,$P(V_T < 0) = 0$and$P(V_T > 0) > ...
Only top voted, non community-wiki answers of a minimum length are eligible | 2015-07-04 07:14:23 | {"extraction_info": {"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": 1, "x-ck12": 0, "texerror": 0, "math_score": 0.5689123868942261, "perplexity": 1384.4471591169854}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-27/segments/1435375096579.52/warc/CC-MAIN-20150627031816-00208-ip-10-179-60-89.ec2.internal.warc.gz"} |
https://nbviewer.ipython.org/github/rlabbe/Kalman-and-Bayesian-Filters-in-Python/blob/scienceopen-master/08_Designing_Kalman_Filters.ipynb | # Designing Kalman Filters¶
In [1]:
#format the book
%matplotlib inline
from __future__ import division, print_function
import matplotlib.pyplot as plt
import book_format
Out[1]:
## Introduction¶
In this chapter we will work through the design of several Kalman filters to gain experience and confidence with the various equations and techniques.
For our first multidimensional problem we will track a robot in a 2D space, such as a field. We will start with a simple noisy sensor that outputs noisy $(x,y)$ coordinates which we will need to filter to generate a 2D track. Once we have mastered this concept, we will extend the problem significantly with more sensors and then adding control inputs. blah blah
## Tracking a Robot¶
This first attempt at tracking a robot will closely resemble the 1-D dog tracking problem of previous chapters. This will allow us to 'get our feet wet' with Kalman filtering. So, instead of a sensor that outputs position in a hallway, we now have a sensor that supplies a noisy measurement of position in a 2-D space, such as an open field. That is, at each time $T$ it will provide an $(x,y)$ coordinate pair specifying the measurement of the sensor's position in the field.
Implementation of code to interact with real sensors is beyond the scope of this book, so as before we will program simple simulations in Python to represent the sensors. We will develop several of these sensors as we go, each with more complications, so as I program them I will just append a number to the function name. pos_sensor1() is the first sensor we write, and so on.
So let's start with a very simple sensor, one that travels in a straight line. It takes as input the last position, velocity, and how much noise we want, and returns the new position.
In [2]:
import numpy.random as random
import copy
class PosSensor1(object):
def __init__(self, pos = [0,0], vel = (0,0), noise_scale = 1.):
self.vel = vel
self.noise_scale = noise_scale
self.pos = copy.deepcopy(pos)
self.pos[0] += self.vel[0]
self.pos[1] += self.vel[1]
return [self.pos[0] + random.randn() * self.noise_scale,
self.pos[1] + random.randn() * self.noise_scale]
A quick test to verify that it works as we expect.
In [3]:
import book_plots as bp
pos = [4,3]
sensor = PosSensor1 (pos, (2,1), 1)
for i in range (50):
bp.plot_measurements(pos[0], pos[1])
plt.show()
That looks correct. The slope is 1/2, as we would expect with a velocity of (2,1), and the data seems to start at near (6,4).
##### Step 1: Choose the State Variables¶
As always, the first step is to choose our state variables. We are tracking in two dimensions and have a sensor that gives us a reading in each of those two dimensions, so we know that we have the two observed variables $x$ and $y$. If we created our Kalman filter using only those two variables the performance would not be very good because we would be ignoring the information velocity can provide to us. We will want to incorporate velocity into our equations as well. I will represent this as
$$\mathbf{x} = \begin{bmatrix}x\\v_x\\y\\v_y\end{bmatrix}$$
There is nothing special about this organization. I could have listed the (xy) coordinates first followed by the velocities, and/or I could done this as a row matrix instead of a column matrix. For example, I could have chosen:
$$\mathbf{x} = \begin{bmatrix}x&y&v_x&v_y\end{bmatrix}$$
All that matters is that the rest of my derivation uses this same scheme. However, it is typical to use column matrices for state variables, and I prefer it, so that is what we will use.
It might be a good time to pause and address how you identify the unobserved variables. This particular example is somewhat obvious because we already worked through the 1D case in the previous chapters. Would it be so obvious if we were filtering market data, population data from a biology experiment, and so on? Probably not. There is no easy answer to this question. The first thing to ask yourself is what is the interpretation of the first and second derivatives of the data from the sensors. We do that because obtaining the first and second derivatives is mathematically trivial if you are reading from the sensors using a fixed time step. The first derivative is just the difference between two successive readings. In our tracking case the first derivative has an obvious physical interpretation: the difference between two successive positions is velocity.
Beyond this you can start looking at how you might combine the data from two or more different sensors to produce more information. This opens up the field of sensor fusion, and we will be covering examples of this in later sections. For now, recognize that choosing the appropriate state variables is paramount to getting the best possible performance from your filter.
##### Step 2: Design State Transition Function¶
Our next step is to design the state transition function. Recall that the state transition function is implemented as a matrix $\mathbf{F}$ that we multipy with the previous state of our system to get the next state, like so.
$$\mathbf{x}' = \mathbf{Fx}$$
I will not belabor this as it is very similar to the 1-D case we did in the previous chapter. Our state equations for position and velocity would be:
\begin{aligned} x' &= (1*x) + (\Delta t * v_x) + (0*y) + (0 * v_y) \\ v_x &= (0*x) + (1*v_x) + (0*y) + (0 * v_y) \\ y' &= (0*x) + (0* v_x) + (1*y) + (\Delta t * v_y) \\ v_y &= (0*x) + (0*v_x) + (0*y) + (1 * v_y) \end{aligned}
Laying it out that way shows us both the values and row-column organization required for $\small\mathbf{F}$. In linear algebra, we would write this as:
$$\begin{bmatrix}x\\v_x\\y\\v_y\end{bmatrix}' = \begin{bmatrix}1& \Delta t& 0& 0\\0& 1& 0& 0\\0& 0& 1& \Delta t\\ 0& 0& 0& 1\end{bmatrix}\begin{bmatrix}x\\v_x\\y\\v_y\end{bmatrix}$$
So, let's do this in Python. It is very simple; the only thing new here is setting dim_z to 2. We will see why it is set to 2 in step 4.
In [4]:
from filterpy.kalman import KalmanFilter
import numpy as np
tracker = KalmanFilter(dim_x=4, dim_z=2)
dt = 1. # time step
tracker.F = np.array ([[1, dt, 0, 0],
[0, 1, 0, 0],
[0, 0, 1, dt],
[0, 0, 0, 1]])
##### Step 3: Design the Motion Function¶
We have no control inputs to our robot (yet!), so this step is trivial - set the motion input $\small\mathbf{u}$ to zero. This is done for us by the class when it is created so we can skip this step, but for completeness we will be explicit.
In [5]:
tracker.u = 0.
##### Step 4: Design the Measurement Function¶
The measurement function defines how we go from the state variables to the measurements using the equation $\mathbf{z} = \mathbf{Hx}$. At first this is a bit counterintuitive, after all, we use the Kalman filter to go from measurements to state. But the update step needs to compute the residual between the current measurement and the measurement represented by the prediction step. Therefore $\textbf{H}$ is multiplied by the state $\textbf{x}$ to produce a measurement $\textbf{z}$.
In this case we have measurements for (x,y), so $\textbf{z}$ must be of dimension $2\times 1$. Our state variable is size $4\times 1$. We can deduce the required size for $\textbf{H}$ by recalling that multiplying a matrix of size $m\times n$ by $n\times p$ yields a matrix of size $m\times p$. Thus,
\begin{aligned} (2\times 1) &= (a\times b)(4 \times 1) \\ &= (a\times 4)(4\times 1) \\ &= (2\times 4)(4\times 1) \end{aligned}
So, $\textbf{H}$ is of size $2\times 4$.
Filling in the values for $\textbf{H}$ is easy in this case because the measurement is the position of the robot, which is the $x$ and $y$ variables of the state $\textbf{x}$. Let's make this just slightly more interesting by deciding we want to change units. So we will assume that the measurements are returned in feet, and that we desire to work in meters. Converting from feet to meters is a simple as multiplying by 0.3048. However, we are converting from state (meters) to measurements (feet) so we need to divide by 0.3048. So
$$\mathbf{H} = \begin{bmatrix} \frac{1}{0.3048} & 0 & 0 & 0 \\ 0 & 0 & \frac{1}{0.3048} & 0 \end{bmatrix}$$
which corresponds to these linear equations \begin{aligned} z_x' &= (\frac{x}{0.3048}) + (0* v_x) + (0*y) + (0 * v_y) \\ z_y' &= (0*x) + (0* v_x) + (\frac{y}{0.3048}) + (0 * v_y) \\ \end{aligned}
To be clear about my intentions here, this is a pretty simple problem, and we could have easily found the equations directly without going through the dimensional analysis that I did above. In fact, an earlier draft did just that. But it is useful to remember that the equations of the Kalman filter imply a specific dimensionality for all of the matrices, and when I start to get lost as to how to design something it is often extremely useful to look at the matrix dimensions. Not sure how to design $\textbf{H}$? Here is the Python that implements this:
In [6]:
tracker.H = np.array([[1/0.3048, 0, 0, 0],
[0, 0, 1/0.3048, 0]])
print(tracker.H)
[[ 3.2808399 0. 0. 0. ]
[ 0. 0. 3.2808399 0. ]]
##### Step 5: Design the Measurement Noise Matrix¶
In this step we need to mathematically model the noise in our sensor. For now we will make the simple assumption that the $x$ and $y$ variables are independent Gaussian processes. That is, the noise in x is not in any way dependent on the noise in y, and the noise is normally distributed about the mean. For now let's set the variance for $x$ and $y$ to be 5 for each. They are independent, so there is no covariance, and our off diagonals will be 0. This gives us:
$$\mathbf{R} = \begin{bmatrix}5&0\\0&5\end{bmatrix}$$
It is a $2{\times}2$ matrix because we have 2 sensor inputs, and covariance matrices are always of size $n{\times}n$ for $n$ variables. In Python we write:
In [7]:
tracker.R = np.array([[5, 0],
[0, 5]])
print(tracker.R)
[[5 0]
[0 5]]
##### Step 6: Design the Process Noise Matrix¶
Finally, we design the process noise. We don't yet have a good way to model process noise, so for now we will assume there is a small amount of process noise, say 0.1 for each state variable. Later we will tackle this admittedly difficult topic in more detail. We have 4 state variables, so we need a $4{\times}4$ covariance matrix:
$$\mathbf{Q} = \begin{bmatrix}0.1&0&0&0\\0&0.1&0&0\\0&0&0.1&0\\0&0&0&0.1\end{bmatrix}$$
In Python I will use the numpy eye helper function to create an identity matrix for us, and multipy it by 0.1 to get the desired result.
In [8]:
tracker.Q = np.eye(4) * 0.1
print(tracker.Q)
[[ 0.1 0. 0. 0. ]
[ 0. 0.1 0. 0. ]
[ 0. 0. 0.1 0. ]
[ 0. 0. 0. 0.1]]
For our simple problem we will set the initial position at (0,0) with a velocity of (0,0). Since that is a pure guess, we will set the covariance matrix $\small\mathbf{P}$ to a large value. $$\mathbf{x} = \begin{bmatrix}0\\0\\0\\0\end{bmatrix}\\ \mathbf{P} = \begin{bmatrix}500&0&0&0\\0&500&0&0\\0&0&500&0\\0&0&0&500\end{bmatrix}$$
In Python we implement that with
In [9]:
tracker.x = np.array([[0, 0, 0, 0]]).T
tracker.P = np.eye(4) * 500.
print(tracker.x)
print()
print (tracker.P)
[[ 0.]
[ 0.]
[ 0.]
[ 0.]]
[[ 500. 0. 0. 0.]
[ 0. 500. 0. 0.]
[ 0. 0. 500. 0.]
[ 0. 0. 0. 500.]]
## Implement the Filter Code¶
Design is complete, now we just have to write the Python code to run our filter, and output the data in the format of our choice. To keep the code clear, let's just print a plot of the track. We will run the code for 30 iterations.
In [10]:
tracker = KalmanFilter(dim_x=4, dim_z=2)
dt = 1.0 # time step
tracker.F = np.array ([[1, dt, 0, 0],
[0, 1, 0, 0],
[0, 0, 1, dt],
[0, 0, 0, 1]])
tracker.u = 0.
tracker.H = np.array ([[1/0.3048, 0, 0, 0],
[0, 0, 1/0.3048, 0]])
tracker.R = np.eye(2) * 5
tracker.Q = np.eye(4) * .1
tracker.x = np.array([[0,0,0,0]]).T
tracker.P = np.eye(4) * 500.
# initialize storage and other variables for the run
count = 30
xs, ys = [],[]
pxs, pys = [],[]
sensor = PosSensor1 ([0,0], (2,1), 1.)
for i in range(count):
z = np.array([[pos[0]], [pos[1]]])
tracker.predict ()
tracker.update (z)
xs.append (tracker.x[0,0])
ys.append (tracker.x[2,0])
pxs.append (pos[0]*.3048)
pys.append(pos[1]*.3048)
bp.plot_filter(xs, ys)
bp.plot_measurements(pxs, pys)
plt.legend(loc=2)
plt.xlim((0,20))
plt.show()
I encourage you to play with this, setting $\mathbf{Q}$ and $\mathbf{R}$ to various values. However, we did a fair amount of that sort of thing in the last chapters, and we have a lot of material to cover, so I will move on to more complicated cases where we will also have a chance to experience changing these values.
Now I will run the same Kalman filter with the same settings, but also plot the covariance ellipse for $x$ and $y$. First, the code without explanation, so we can see the output. I print the last covariance to see what it looks like. But before you scroll down to look at the results, what do you think it will look like? You have enough information to figure this out but this is still new to you, so don't be discouraged if you get it wrong.
In [11]:
import stats
tracker = KalmanFilter(dim_x=4, dim_z=2)
dt = 1.0 # time step
tracker.F = np.array([[1, dt, 0, 0],
[0, 1, 0, 0],
[0, 0, 1, dt],
[0, 0, 0, 1]])
tracker.u = 0.
tracker.H = np.array([[1/0.3048, 0, 0, 0],
[0, 0, 1/0.3048, 0]])
tracker.R = np.eye(2) * 5
tracker.Q = np.eye(4) * .1
tracker.x = np.array([[0, 0, 0, 0]]).T
tracker.P = np.eye(4) * 500.
# initialize storage and other variables for the run
count = 30
xs, ys = [], []
pxs, pys = [], []
sensor = PosSensor1([0,0], (2,1), 1.)
for i in range(count):
z = np.array([[pos[0]], [pos[1]]])
tracker.predict()
tracker.update(z)
xs.append(tracker.x[0,0])
ys.append(tracker.x[2,0])
pxs.append(pos[0]*.3048)
pys.append(pos[1]*.3048)
# plot covariance of x and y
cov = np.array([[tracker.P[0,0], tracker.P[2,0]],
[tracker.P[0,2], tracker.P[2,2]]])
stats.plot_covariance_ellipse(
(tracker.x[0,0], tracker.x[2,0]), cov=cov,
facecolor='g', alpha=0.15)
bp.plot_filter(xs, ys)
bp.plot_measurements(pxs, pys)
plt.legend(loc=2)
plt.show()
print("final P is:")
print(tracker.P)
final P is:
[[ 0.30660483 0.12566239 0. 0. ]
[ 0.12566239 0.24399092 0. 0. ]
[ 0. 0. 0.30660483 0.12566239]
[ 0. 0. 0.12566239 0.24399092]]
Did you correctly predict what the covariance matrix and plots would look like? Perhaps you were expecting a tilted ellipse, as in the last chapters. If so, recall that in those chapters we were not plotting $x$ against $y$, but $x$ against $\dot{x}$. $x$ is correlated to $\dot{x}$, but $x$ is not correlated or dependent on $y$. Therefore our ellipses are not tilted. Furthermore, the noise for both $x$ and $y$ are modeled to have the same value, 5, in $\mathbf{R}$. If we were to set R to, for example,
$$\mathbf{R} = \begin{bmatrix}10&0\\0&5\end{bmatrix}$$
we would be telling the Kalman filter that there is more noise in $x$ than $y$, and our ellipses would be longer than they are tall.
The final P tells us everything we need to know about the correlation between the state variables. If we look at the diagonal alone we see the variance for each variable. In other words $\mathbf{P}_{0,0}$ is the variance for x, $\mathbf{P}_{1,1}$ is the variance for $\dot{x}$, $\mathbf{P}_{2,2}$ is the variance for y, and $\mathbf{P}_{3,3}$ is the variance for $\dot{y}$. We can extract the diagonal of a matrix using numpy.diag().
In [12]:
print(np.diag(tracker.P))
[ 0.30660483 0.24399092 0.30660483 0.24399092]
The covariance matrix contains four $2{\times}2$ matrices that you should be able to easily pick out. This is due to the correlation of $x$ to $\dot{x}$, and of $y$ to $\dot{y}$. The upper left hand side shows the covariance of $x$ to $\dot{x}$. Let's extract and print, and plot it.
In [13]:
c = tracker.P[0:2, 0:2]
print(c)
stats.plot_covariance_ellipse((0, 0), cov=c, facecolor='g', alpha=0.2)
[[ 0.30660483 0.12566239]
[ 0.12566239 0.24399092]]
The covariance contains the data for $x$ and $\dot{x}$ in the upper left because of how it is organized. Recall that entries $\mathbf{P}_{i,j}$ and $\mathbf{P}_{j,i}$ contain $p\sigma_1\sigma_2$.
Finally, let's look at the lower left side of $\mathbf{P}$, which is all 0s. Why 0s? Consider $\mathbf{P}_{3,0}$. That stores the term $p\sigma_3\sigma_0$, which is the covariance between $\dot{y}$ and $x$. These are independent, so the term will be 0. The rest of the terms are for similarly independent variables.
## The Effect of Order¶
So far in this book we have only really considered tracking position and velocity. That has worked well, but only because I have been carefully selecting problems for which this was an appropriate choice. You know have enough experience with the Kalman filter to consider this in more general terms.
What do I mean by "order"? In the context of these system models it is the number of derivatives required to accurately model a system. Consider a system that does not change, such as the height of a building. There is no change, so there is no need for a derivative, and the order of the system is zero. We could express this in an equation as
$$x = 312.5$$
A first order system has a first derivative. For example, change of position is velocity, and we can write this as
$$v = \frac{dx}{dt}$$
which we integrate into the Newton equation $$x = vt + x_0.$$
This is also called a constant velocity model, because of the assumption of a constant velocity.
So a second order has a second derivative. The second derivative of position is acceleration, with the equation
$$a = \frac{d^2x}{dt^2}$$
which we integrate into
$$x = \frac{1}{2}at^2 +v_0t + x_0.$$
This is also known as a constant acceleration model.
Another, equivalent way of looking at this is to consider the order of the polynomial. The constant acceleration model has a second derivative, so it is second order. Likewise, the polynomial $x = \frac{1}{2}at^2 +v_0t + x_0$ is second order.
When we design the state variables and process model we must choose the order of the system we want to model. Let's say we are tracking something with a constant velocity. No real world process is perfect, and so there will be slight variations in the velocity over short time period. You might reason that the best approach is to use a second order filter, allowing the acceleration term to deal with the slight variations in velocity.
That doesn't work as nicely as you might think. To thoroughly undestand this issue lets see the effects of using a process model that does not match the order of the system being filtered.
First we need a system to filter. I'll write a class to simulate on object with a constant velocity. Essentially no physical system has a truly constant velocity, so on each update we alter the velocity by a small amount. I also write a sensor to simulate Gaussian noise in a sensor. The code is below, and a plot an example run to verify that it is working correctly.
In [14]:
from numpy.random import randn
import numpy as np
import matplotlib.pyplot as plt
from book_plots import plot_track
np.random.seed(124)
class ConstantVelocityObject(object):
def __init__(self, x0=0, vel=1., noise_scale=0.06):
self.x = x0
self.vel = vel
self.noise_scale = noise_scale
def update(self):
self.vel += randn()*self.noise_scale
self.x += self.vel
return (self.x, self.vel)
def sense(x, noise_scale=1.):
return x[0] + randn()*noise_scale
obj = ConstantVelocityObject()
xs = []
zs = []
for i in range(50):
x = obj.update()
z = sense(x)
xs.append(x)
zs.append(z)
xs = np.asarray(xs)
bp.plot_track(xs[:,0])
bp.plot_measurements(range(50), zs)
plt.legend(loc='best')
plt.show()
I am satified with this plot. The track is not perfectly straight due to the noise that we added to the system - this could be the track of a person walking down the street, or perhaps of an aircraft being buffeted by variable winds. There is no intentional acceleration here, so we call it a constant velocity system. Again, you may be asking yourself that since there is in fact a tiny bit of acceleration going on why would we not use a second order Kalman filter to account for those changes? Let's find out.
How does one design a zero order, first order, or second order Kalman filter. We have been doing it all along, but just not using those terms. It might be slightly tedious, but I will elaborate fully on each - if the concept is clear to you feel free to skim a bit. However, I think that reading carefully will really cement the idea of filter order in your mind.
### Zero Order Kalman Filter¶
A zero order Kalman filter is just a filter that tracks with no derivatives. We are tracking position, so that means we only have a state variable for position (no velocity or acceleration), and the state transition function also only accounts for position. Using the matrix formulation we would say that the state variable is
$$\mathbf{x} = \begin{bmatrix}x\end{bmatrix}$$
The state transition function is very simple. There is no change in position, so we need to model $x=x$; in other words, x at time t+1 is the same as it was at time t. In matrix form, our state transition function is
$$\mathbf{F} = \begin{bmatrix}1\end{bmatrix}$$
The measurement function is very easy. Recall that we need to define how to convert the state variable $\mathbf{x}$ into a measurement. We will assume that our measurements are positions. The state variable only contains a position, so we get
$$\mathbf{H} = \begin{bmatrix}1\end{bmatrix}$$
That is pretty much it. Let's write a function that constructs and returns a zero order Kalman filter to us.
In [15]:
def ZeroOrderKF(R, Q):
""" Create zero order Kalman filter. Specify R and Q as floats."""
kf = KalmanFilter(dim_x=1, dim_z=1)
kf.x = np.array([0.])
kf.R *= R
kf.Q *= Q
kf.P *= 20
kf.F = np.array([[1.]])
kf.H = np.array([[1.]])
return kf
### First Order Kalman Filter¶
A first order Kalman filter tracks a first order system, such as position and velocity. We already did this for the dog tracking problem above, so this should be very clear. But let's do it again.
A first order system has position and velocity, so the state variable needs both of these. The matrix formulation could be
$$\mathbf{x} = \begin{bmatrix}x\\\dot{x}\end{bmatrix}$$
As an aside, there is nothing stopping us from choosing
$$\mathbf{x} = \begin{bmatrix}\dot{x}\\x\end{bmatrix}$$
but all texts and software that I know of choose the first form as more natural. You would just have to design the rest of the matrices to take this ordering into account.
So now we have to design our state transition. The Newtonian equations for a time step are:
\begin{aligned} x_t &= x_{t-1} + v\Delta t \\ v_t &= v_{t-1}\end{aligned}
Recall that we need to convert this into the linear equation
$$\begin{bmatrix}x\\\dot{x}\end{bmatrix} = \mathbf{F}\begin{bmatrix}x\\\dot{x}\end{bmatrix}$$
Setting
$$\mathbf{F} = \begin{bmatrix}1 &\Delta t\\ 0 & 1\end{bmatrix}$$
gives us the equations above. If this is not clear, work out the matrix multiplication:
$$x = 1x + dt \dot{x} \\ \dot{x} = 0x + 1\dot{x}$$
Finally, we design the measurement function. The measurement function needs to implement
$$z = \mathbf{Hx}$$
Our sensor still only reads position, so it should take the position from the state, and 0 out the velocity, like so:
$$\mathbf{H} = \begin{bmatrix}1 & 0 \end{bmatrix}$$
As in the previous section we will define a function that constructs and returns a Kalman filter that implements these equations.
In [16]:
from filterpy.common import Q_discrete_white_noise
def FirstOrderKF(R, Q, dt):
""" Create zero order Kalman filter. Specify R and Q as floats."""
kf = KalmanFilter(dim_x=2, dim_z=1)
kf.x = np.zeros(2)
kf.P *= np.array([[100,0], [0,1]])
kf.R *= R
kf.Q = Q_discrete_white_noise(2, dt, Q)
kf.F = np.array([[1., dt],
[0. , 1]])
kf.H = np.array([[1., 0]])
return kf
### Second Order Kalman Filter¶
A second order Kalman filter tracks a second order system, such as position, velocity and acceleration. The state variables will need to contain all three. The matrix formulation could be
$$\mathbf{x} = \begin{bmatrix}x\\\dot{x}\\\ddot{x}\end{bmatrix}$$
So now we have to design our state transition. The Newtonian equations for a time step are:
\begin{aligned} x_t &= x_{t-1} + v\Delta t + 0.5a_{t-1} \Delta t^2 \\ v_t &= v_{t-1} \Delta t + a_{t-1} \\ a_t &= a_{t-1}\end{aligned}
Recall that we need to convert this into the linear equation
$$\begin{bmatrix}x\\\dot{x}\\\ddot{x}\end{bmatrix} = \mathbf{F}\begin{bmatrix}x\\\dot{x}\\\ddot{x}\end{bmatrix}$$
Setting
$$\mathbf{F} = \begin{bmatrix}1 & \Delta t &.5\Delta t^2\\ 0 & 1 & \Delta t \\ 0 & 0 & 1\end{bmatrix}$$
gives us the equations above.
Finally, we design the measurement function. The measurement function needs to implement
$$z = \mathbf{Hx}$$
Our sensor still only reads position, so it should take the position from the state, and 0 out the velocity, like so:
$$\mathbf{H} = \begin{bmatrix}1 & 0 & 0\end{bmatrix}$$
As in the previous section we will define a function that constructs and returns a Kalman filter that implements these equations.
In [17]:
def SecondOrderKF(R_std, Q, dt):
""" Create zero order Kalman filter. Specify R and Q as floats."""
kf = KalmanFilter(dim_x=3, dim_z=1)
kf.x = np.zeros(3)
kf.P[0,0] = 100
kf.P[1,1] = 1
kf.P[2,2] = 1
kf.R *= R_std**2
kf.Q = Q_discrete_white_noise(3, dt, Q)
kf.F = np.array([[1., dt, .5*dt*dt],
[0., 1., dt],
[0., 0., 1.]])
kf.H = np.array([[1., 0., 0.]])
return kf
### Evaluating the Performance¶
We have implemented the Kalman filters and the simulated first order system, so now we can run each Kalman filter against the simulation and evaluate the results.
How do we evaluate the results? We can do this qualitatively by plotting the track and the Kalman filter output and eyeballing the results. However, we can do this far more rigorously with mathematics. Recall that system covariance matrix $\mathbf{P}$ contains the computed variance and covariances for each of the state variables. The diagonal contains the variance. If you think back to the Gaussian chapter you'll remember that roughly 99% of all measurements fall within three standard deviations if the noise is Gaussian, and, of course, the standard deviation can be computed as the square root of the variance. If this is not clear please review the Gaussian chapter before continuing, as this is an important point.
So we can evaluate the filter by looking at the residuals between the estimated state and actual state and comparing them to the standard deviations which we derive from $\mathbf{P}$. If the filter is performing correctly 99% of the residuals will fall within the third standard deviation. This is true for all the state variables, not just for the position.
So let's run the first order Kalman filter against our first order system and access it's performance. You can probably guess that it will do well, but let's look at it using the stardard deviations.
First, let's write a routine to generate the noisy measurements for us.
In [18]:
def simulate_system(Q, count):
obj = ConstantVelocityObject(x0=0, vel=1, noise_scale=Q)
zs = []
xs = []
for i in range(count):
x = obj.update()
z = sense(x)
xs.append(x)
zs.append(z)
return np.asarray(xs), zs
And now a routine to perform the filtering.
In [19]:
def filter_data(kf, zs):
# save output for plotting
fxs = []
ps = []
for z in zs:
kf.predict()
kf.update(z)
fxs.append(kf.x)
ps.append(kf.P.diagonal())
fxs = np.asarray(fxs)
ps = np.asarray(ps)
return fxs, ps
And to plot the track results.
In [20]:
def plot_kf_output(xs, filter_xs, zs, title=None):
bp.plot_filter(filter_xs[:,0])
bp.plot_track(xs[:,0])
if zs is not None:
bp.plot_measurements(zs)
plt.legend(loc='best')
plt.ylabel('meters')
plt.xlabel('time (sec)')
if title is not None:
plt.title(title)
plt.xlim((-1, len(xs)))
plt.ylim((-1, len(xs)))
#plt.axis('equal')
plt.show()
Now we are prepared to run the filter and look at the results.
In [21]:
R = 1
Q = 0.03
xs, zs = simulate_system(Q=Q, count=50)
kf = FirstOrderKF(R, Q, dt=1)
fxs1, ps1 = filter_data(kf, zs)
plt.figure()
plot_kf_output(xs, fxs1, zs)
It looks like the filter is performing well, but it is hard to tell exactly how well. Let's look at the residuals and see if they help. You may have noticed that in the code above I saved the covariance at each step. I did that to use in the following plot. The ConstantVelocityObject class returns a tuple of (position, velocity) for the real object, and this is stored in the array xs, and the filter's estimates are in fxs.
In [22]:
def plot_residuals(xs, filter_xs, Ps, title, y_label):
res = xs - filter_xs
plt.plot(res)
bp.plot_residual_limits(Ps)
plt.title(title)
plt.ylabel(y_label)
plt.xlabel('time (sec)')
plt.show()
In [23]:
plot_residuals(xs[:,0], fxs1[:,0], ps1[:,0],
'First Order Position Residuals',
'meters')
How do we interpret this plot? The residual is drawn as the jagged line - the difference between the measurement and the actual position. If there was no measurement noise and the Kalman filter prediction was always perfect the residual would always be zero. So the ideal output would be a horizontal line at 0. We can see that the residual is centered around 0, so this gives us confidence that the noise is Gaussian (because the errors fall equally above and below 0). The yellow area between dotted lines show the theoretical performance of the filter for 1 standard deviations. In other words, approximately 68% of the errors should fall within the dotted lines. The residual falls within this range, so we see that the filter is performing well, and that it is not diverging.
But that is just for position. Let's look at the residuals for velocity.
In [24]:
plot_residuals(xs[:,1], fxs1[:,1], ps1[:,1],
'First Order Velocity Residuals',
'meters/sec')
Again, as expected, the residual falls within the theoretical performance of the filter, so we feel confident that the filter is well designed for this system.
Now let's do the same thing using the zero order Kalman filter. All of the code and math is largely the same, so let's just look at the results without discussing the implementation much.
In [25]:
kf0 = ZeroOrderKF(R, Q)
fxs0, ps0 = filter_data(kf0, zs)
plot_kf_output(xs, fxs0, zs)
As we would expect, the filter has problems. Think back to the g-h filter, where we incorporated acceleration into the system. The g-h filter always lagged the input because there were not enough terms to allow the filter to adjust quickly enough to the changes in velocity. The same thing is happening here, just one order lower. On every predict() step the Kalman filter assumes that there is no change in position - if the current position is 4.3 it will predict that the position at the next time period is 4.3. Of course, the actual position is closer to 5.3. The measurment, with noise, might be 5.4, so the filter chooses an estimate part way between 4.3 and 5.4, causing it to lag the actual value of 5.3 by a significant amount. This same thing happens in the next step, the next one, and so on. The filter never catches up.
Now let's look at the residuals. We are not tracking velocity, so we can only look at the residual for position.
In [26]:
plot_residuals(xs[:,0], fxs0[:,0], ps0[:,0],
'Zero Order Position Residuals',
'meters')
We can see that the filter diverges almost immediately. After the first second the residual exceeds the bounds of three standard deviations. It is important to understand that the covariance matrix $\mathbf{P}$ is only reporting the theoretical performance of the filter assuming all of the inputs are correct. In other words, this Kalman filter is diverging, but $\mathbf{P}$ implies that the Kalman filter's estimates are getting better and better with time because the variance is getting smaller. The filter has no way to know that you are lying to it about the system.
In this system the divergence is immediate and striking. In many systems it will only be gradual, and/or slight. It is important to look at charts like these for your systems to ensure that the performance of the filter is within the bounds of its theoretical performance.
Now let's try a third order system. This might strike you as a good thing to do. After all, we know there is a bit of noise in the movement of the simulated object, which implies there is some acceleration. Why not model the acceleration with a second order model. If there is no acceleration, the acceleration should just be estimated to be 0. But is that what happens? Think about it before going on.
In [27]:
kf2 = SecondOrderKF(R, Q, dt=1)
fxs2, ps2 = filter_data(kf2, zs)
plot_kf_output(xs, fxs2, zs)
Did this perform as you expected? We can see that even though the system does have a slight amount of acceleration in it the seond order filter performs poorly compared to the first order filter. Why does this? The system believes that there is acceleration in the system, and so the large changes in the measurement gets interpreted as acceleration instead of noise. Thus you can see that the filter tracks the noise in the system quite closely. Not only that, but it overshoots the noise in places if the noise is consistantly above or below the track because the filter incorrectly assumes an acceleration that does not exist, and so it's prediction goes further and further away from the track on each measurement. This is not a good state of affairs.
Still, the track doesn't look horrible. Let's see the story that the residuals tell. I will add a wrinkle here. The residuals for the order 2 system do not look terrible in that they do not diverge or exceed three standard deviations. However, it is very telling to look at the residuals for the first order vs the second order filter, so I have plotted both on the same graph.
In [28]:
res = xs[:,0] - fxs2[:,0]
res1 = xs[:,0] - fxs1[:,0]
plt.plot(res1, ls="--", label='order 1')
plt.plot(res, label='order 2')
bp.plot_residual_limits(ps2[:,0])
plt.title('Second Order Position Residuals')
plt.legend()
plt.ylabel('meters')
plt.xlabel('time (sec)')
plt.show()
We can see that the residuals for the second order filter fall nicely within the theoretical limits of the filter. When we compare them against the first order residuals we may conclude that the second order is slight worse, but the difference is not large. There is nothing very alarming here.
Now let's look at the residuals for the velocity.
In [29]:
res = xs[:,1] - fxs2[:,1]
res1 = xs[:,1] - fxs1[:,1]
plt.plot(res, label='order 2')
plt.plot(res1, ls='--', label='order 1')
bp.plot_residual_limits(ps2[:,1])
plt.title('Second Order Velocity Residuals')
plt.legend()
plt.ylabel('meters/sec')
plt.xlabel('time (sec)')
plt.show()
Here the story is very different. While the residuals of the second order system fall within the theoretical bounds of the filter's performance, we can see that the residuals are far worse than for the first order filter. This is the usual result this scenerio. The filter is assuming that there is acceleration that does not exist. It mistakes noise in the measurement as acceleration and this gets added into the velocity estimate on every predict cycle. Of course the acceleration is not actually there and so the residual for the velocity is much larger than it optimum.
I have one more trick up my sleeve. We have a first order system; i.e. the velocity is more-or-less constant. Real world systems are never perfect, so of course the velocity is never exactly the same between time periods. When we use a first order filter we account for that slight variation in velocity with the process noise. The matrix $\mathbf{Q}$ is computed to account for this slight variation. If we move to a second order filter we are now accounting for the changes in velocity. Perhaps now we have no process noise, and we can set $\mathbf{Q}$ to zero!
In [30]:
kf2 = SecondOrderKF(R, 0, dt=1)
fxs2, ps2 = filter_data(kf2, zs)
plot_kf_output(xs, fxs2, zs)
To my eye that looks quite good! The filter quickly converges to the actual track. Success!
Or, maybe not. Setting the process noise to 0 tells the filter that the process model is perfect. I've yet to hear of a perfect physical system. Let's look at the performance of the filter over a longer period of time.
In [31]:
np.random.seed(25944)
xs500, zs500 = simulate_system(Q=Q, count=500)
kf2 = SecondOrderKF(R, 0, dt=1)
fxs2, ps2 = filter_data(kf2, zs500)
plot_kf_output(xs500, fxs2, zs500)
plot_residuals(xs500[:,0], fxs2[:,0], ps2[:,0],
'Zero Order Position Residuals',
'meters')
We can see that the performance of the filter is abysmal. We can see that in the track plot where the filter diverges from the track for an extended period of time. The divergence may or may not seem large to you. The residual plot makes the problem more apparent. Just before the 100th update the filter diverges sharply from the theoretical performance. It might be converging at the end, but I doubt it.
Why is this happening? Recall that if we set the process noise to zero we are telling the filter to use only the process model. The measurements end up getting ignored. The pysical system is not perfect, and so the filter is unable to adapt to this nonperfect behavior.
Maybe just a really low process noise? Let's try that.
In [32]:
np.random.seed(32594)
xs2000, zs2000 = simulate_system(Q=0.0001, count=2000)
kf2 = SecondOrderKF(R, 0, dt=1)
fxs2, ps2 = filter_data(kf2, zs2000)
plot_kf_output(xs2000, fxs2, zs2000)
plot_residuals(xs2000[:,0], fxs2[:,0], ps2[:,0],
'Seceond Order Position Residuals',
'meters') | 2022-07-01 10:35:11 | {"extraction_info": {"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": 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, "math_score": 0.858822226524353, "perplexity": 648.9519373059294}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656103940327.51/warc/CC-MAIN-20220701095156-20220701125156-00640.warc.gz"} |
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Dedicated to the 70th anniversary of V. V. Napalkov 3 Equivalence group analysis and nonlinear self-adjointness of the generalized Kompaneets equationE. D. Avdonina, N. H. Ibragimov 6 Applications of model spaces to construction of cocyclic perturbations of a semigroup of shifts on a semiaxisG. G. Amosov, A. D. Baranov, V. V. Kapustin 17 Exact estimates of types of entire functions of an order $\rho\in(0;1)$ with zeroes on the rayG. G. Braichev 29 Iterations of entire transcendental functions with a regular behavior of the modulus minimumA. M. Gaisin, Zh. G. Rakhmatullina 38 A periodicity criterium for quasipolynomialsN. P. Girya, S. Yu. Favorov 47 Characteristic Lie rings of differential equationsM. Gürses, A. V. Zhiber, I. T. Habibullin 53 On the distribution of indicators of unconditional exponential bases in spaces with a power weightK. P. Isaev, K. V. Trunov 63 Symmetry properties for systems of two ordinary fractional difeferential equationsA. A. Kasatkin 71 Eigenfunctions of annihilation operators associated with Wigner's commutation relationsV. E. Kim 82 An almost exponential sequence of exponential polynomialsA. S. Krivosheyev 88 On some families of complex lines sufficient for holomorphic extension of functionsV. I. Kuzovatov 107 The angular distribution of zeros of random analytic functionsM. P. Mahola, P. V. Filevych 122 On a space of entire functions decreasing rapidly on a real lineM. I. Musin 136 Boundary problem for the generalized Cauchy–Riemann equation in spaces, described by the modulus of continuityA. Y. Timofeev 146 Perfect cuboids and irreducible polynomialsR. A. Sharipov 153 On extremal type of an entire function of order less than unity with zeros of prescribed densities and stepO. V. Sherstyukova 161 On the asymptotic behavior of Cauchy–Stieltjes integral in the polydiscO. A. Zolota 166 | 2019-11-17 05:31:18 | {"extraction_info": {"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, "math_score": 0.5173914432525635, "perplexity": 1551.189324498516}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-47/segments/1573496668787.19/warc/CC-MAIN-20191117041351-20191117065351-00551.warc.gz"} |
https://inv.alid.pw/posts/worstsort-rust/ | # Worstsort in Rust
Brent Yorgey shares a fun 2012 paper by Miguel A. Lerma about the worst possible1 sorting algorithm, along with a cute Haskell implementation. The algorithm depends on a function $$f: \mathbb{N} \to \mathbb{N}$$ and runs in $$\Omega(f(n))$$ time; in other words, we can take at least as long as any computable function (in fact, much much longer).
Naturally I felt compelled to write up an implementation in my new favourite language, Rust. It’s not quite as slick as the Haskell one, partly because I didn’t ‘cheat’ and use a standard library implementation of permutations and chose to follow the paper more closely in using Bubblesort rather than insertion sort, but mostly because Rust is more verbose with things like curly brackets than Haskell and doesn’t allow point-free style.
Here’s the business part of the code:
pub fn badsort<T: Ord + Clone>(k: usize, l: &mut [T]) {
if k == 0 {
bubblesort(l);
} else {
let mut p = permutations(l);
l.clone_from_slice(&p[0]);
}
}
pub fn worstsort<T, F>(l: &mut [T], f: F)
where
T: Ord + Clone,
F: FnOnce(usize) -> usize,
{
}
worstsort runs in $$\Omega((n!^{(f(n))})^2)$$ time, where $$n!^{(k)}$$ abbreviates taking the factorial $$k$$ times. Pretty impressive for just 15 lines, about half of which is boilerplate! | 2019-10-16 15:13:21 | {"extraction_info": {"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": 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, "math_score": 0.48516908288002014, "perplexity": 3600.2305217892}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570986668994.39/warc/CC-MAIN-20191016135759-20191016163259-00014.warc.gz"} |
http://southaustralia.co.nz/ethyl-acetate-aoz/what-is-the-solution-of-the-equation-36x%5E2%3D100-f0777e | 0 are 5. These We're told, solve the absolute value of 3x minus 9 is equal to 0, and graph the solution on a number line. Answers archive Answers : Click here to see ALL problems on Linear-equations; Question 1170800: 2y = x + 3 5y = x - 7 What is the solution set of the given system? Our math solver supports basic math, pre-algebra, algebra, trigonometry, calculus and more. 0. Algebra -> Polynomials-and-rational-expressions-> SOLUTION: Can you explain to me how factoring 36x^2-100 = 4(3x-5)(3x+5) Its alittle confusing to me Log On Algebra: Polynomials, rational expressions and equations Section. We will learn how to solve quadratic equations that do not factor later in the course. The quadratic formula gives two solutions, one when ± ⦠Algebra . Multiply by . Solution. "The Equation" is the eighth episode of the first season of the American science fiction drama television series Fringe. To find roots of a function, set it equal to zero and solve. They told us that the absolute value of 3x minus 9 is equal to 0. Solvers Solvers. Tap for more steps... Simplify the numerator. If it's not what You are looking for type in the equation solver your own equation ⦠To find a quadratic equation with given solutions, perform the process of solving by factoring in reverse. The solutions to the resulting equations are the solutions to the original. What are the number of integral solutions of the equation 7x + 3y = 123 for x,y > 0 [1] 3 [2] 5 [3] 12 [4] Infinite Show Answer & Explanation. Find a quadratic equation with integer coefficients and solutions $$\{\pm \sqrt{5}\}$$. Multiply by . The solution of D2+7D+6 = 0 is xc2 =c1 e-t + c2 e-6t (Use the method of Section 5.5) By the method of undetermined coefficients (Section 5.6) a particular solution of the non homogeneous equations ⦠So let's just rewrite the absolute value equation. Determine Its Foci. This answer has been confirmed as correct and helpful. Tap for more steps... Raise to the power of . Question: Find the vertices and foci of the ellipse. Question: A Hyperbola Has An Equation Of (y2/36)-(x2/100)=1. Round off to the nearest tenth of a centimeter. Equations : Tiger Algebra gives you not only the answers, but also the complete step by step method for solving your equations 36x^2-100 so that you understand better The equation has the form y 2 a 2 â x 2 b 2 = 1, y 2 a 2 â x 2 b 2 = 1, so the transverse axis lies on the y-axis. Determine Its Foci. SOLUTION: Can you explain to me how factoring 36x^2-100 = 4(3x-5)(3x+5) Its alittle confusing to me . So we're told that the absolute value of the something-- in this case the something is 3x minus 9-- is equal to 0. If it's not what You are looking for type in the equation solver your own equation and let us solve it. This is a common form of equation known as a difference of squares, it always factors as follows ... (a^2-b^2) = (a+b) (a-b) in this case: first let's factor out 4 ... 4 (9x^2-25) now factor the difference of squares ... 4 (3x+5) (3x-5) Comment; Complaint; Link; Know the Answer? The hyperbola is centered at the origin, so the vertices serve as the y-intercepts of the graph. Added 7/21/2014 8:23:25 AM . (ii) 2x + y = 5 3x +2y = 8 a 1 / a 2 = 2 / 3 b 1 / b 2 = 1 / 2 and c 1 / c 2 = -5 /-8 = 5 / 8 a 1 /a 2 â b 1 / b 2 Therefore, they will intersect each other at a unique point and thus, there will be a unique solution for these equations. Write the equation of the ellipse 36x^2 + 25y^2 + 360x + 100y + 100 = 0 in standard form - 14058929 Lessons Lessons. Which expression is equivalent to 36x^2-100 +20. Not all polynomial equations can be solved by factoring. Lessons Lessons. SOLUTIONS FOR ADMISSIONS TEST IN MATHEMATICS, JOINT SCHOOLS AND COMPUTER SCIENCE WEDNESDAY 4 NOVEMBER 2009 Mark Scheme: Each part of Question 1 is worth four marks which are awarded solely for the correct answer. Not all quadratic equations can be solved by factoring. What are the solutions to the quadratic equation 4(x + 2)2 = 36 x = â5 and x = 1 The first step in solving the quadratic equation -5x2 + 8 = 133 is to subtract__ from each side Question: - Find the intervals where {eq}h(x)=x^{4}+10x^{3}+36x^{2} {/eq} is concave up and concave down. Dr. Walter Bishop returns to St. Claire's Hospital in an effort to find the boy's whereabouts. However, the solutions of most equations are not immediately evident by inspection. Correction 10 Une video est accessible a. See the answer. Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals. User: What is the solution set of the following equation? To score good marks in Class 10 Mathematics examination, it is advised they solve questions provided at the end of each chapter in the Maharashtra Board Textbooks for SSC Part 1. 3 Answers/Comments. Answers archive Answers : Click here to see ALL problems on Polynomials-and-rational-expressions ; Question 130872: 36x^2=100 Found 2 solutions by cngriffith, seng0111: Answer by cngriffith(27) (Show Source): You can put this solution on YOUR website! But having 3 points is usually enough. y = 4 The solution to the equation is 4, or the solution set is {4}. Each of Questions 2-7 is worth 15 marks QUESTION 1: A. Algebra: Linear Equations, Graphs, Slope Section. 0.12x + 0.05(5000 â x) = 460 0.12x + 250 â 0.05x = 460 0.07x + 250 = 460 0.07x = 210 100(0.07x) =100(210) 7x = 21,000 x = 3000 The solution to the equation is 3000, or the solution set is ⦠Added 7/21/2014 8:23:25 AM . The episode follows the Fringe team's investigation into the kidnapping of a young musical prodigy (Charlie Tahan) who has become obsessed with finishing one piece of music. MSBSHSE Solutions For SSC (Class 10) Maths Part 1 Chapter 2 Quadratic Equations are provided here to help students understand the concepts, right from the beginning. We will learn how to solve polynomial equations that do not factor later in the course. 2. A hyperbola has an equation of (y 2 /36)-(x 2 /100)=1. #2kenglish Download Now, It's Free yeswey. To find the vertices, set x = 0, x = 0, and solve for y. y. User: What is the solution set of the following equation? The area of a rectangle is $$22$$ square centimeters. 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https://socratic.org/questions/how-do-you-write-the-logarithmic-expressions-as-a-single-logarithm-ln-x-2-2ln-x | # How do you write the logarithmic expressions as a single logarithm ln(x+2) - 2ln(x)?
1] $a \log x = \log {x}^{a}$
2] $\log x - \log y = \log \left(\frac{x}{y}\right)$
$\ln \left(x + 2\right) - \ln \left({x}^{2}\right) = \ln \left(\frac{x + 2}{x} ^ 2\right)$ | 2019-01-20 18:52:51 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 3, "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, "math_score": 0.9090619087219238, "perplexity": 3234.148931510631}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-04/segments/1547583730728.68/warc/CC-MAIN-20190120184253-20190120210253-00116.warc.gz"} |
https://economics.stackexchange.com/questions/20906/how-does-the-ebq-formular-come-out | # How does the EBQ formular come out?
I figured out how the economic order quantity (EOQ) formula comes out. I know it is quite similar with economic batch quantity (EBQ), sometimes termed economic production quantity (EPQ). I just don't understand why they are different. What makes the difference in holding cost part?
The models underlying the economic order quantity (EOQ) and economic production quantity (EPQ) formulae assume (realistically in many circumstances) that an order for $Q$ units of a good is delivered all at the same time, whereas production of a batch of $Q$ units occurs over a period.
In the case of ordering, immediately after an order has been delivered the stock is $Q$, and the stock then falls gradually to zero when (assuming timely ordering) the next order is delivered. The average stock is therefore simply $Q/2$.
In the case of production, however, the stock rises during production of a batch, and falls once production of the batch has been completed. Suppose annual demand for use of the good is $D$ and annual production (if it were continuous) is $P$, with $P>D$. Immediately after completion of a batch the stock will be $Q(1-D/P)$. It won't just be $Q$ because (assuming uniform demand), units of stock will be withdrawn for use not only after the batch is completed but also while it is being produced.
Therefore, writing $K$ for the fixed order cost and $h$ for the unit holding cost per year, to find the EOQ we find $Q$ that minimises total annual cost $C$ where:
$$C = \frac{DK}{Q} + \frac{hQ}{2}$$
The first term on the right is the total of the fixed costs of placing orders, $D/Q$ being the number of orders in a year, and the second term is the total holding cost over a year.
To find the EPQ, however, we need to minimise (interpreting $K$ now as the fixed production setup cost):
$$C = \frac{DK}{Q} + \frac{hQ}{2}\Bigg(1-\frac{D}{P}\Bigg)$$
The respective economic quantities may then be found by differential calculus.
In a way EBQ (also known as EPQ) EPQ is an extension of the EOQ model, the difference being
• In EPQ the company will produce its own parts
• In EOQ parts are produced by another company
Note that if you produce stuff you need to store it somewhere, which comes at a cost. This is explicitly included in the EPQ model via the holding cost variable | 2019-07-19 04:15:21 | {"extraction_info": {"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, "math_score": 0.806544303894043, "perplexity": 990.8911939491157}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-30/segments/1563195525974.74/warc/CC-MAIN-20190719032721-20190719054721-00203.warc.gz"} |
https://gamedev.stackexchange.com/questions/150440/lerp-rotation-is-offset | # Lerp rotation is offset
I am trying to get an object to slowly look at another object, that is, rotate slowly so its forward points towards the target position. Here is my code:
using UnityEngine;
public class LookAtTheDamnThing: MonoBehaviour
{
public Transform target;
public float speed;
void FixedUpdate()
{
AimAtTarget();
}
public void AimAtTarget()
{
Vector3 direction = target.position - transform.position;
Quaternion toRotation = Quaternion.FromToRotation(transform.forward, direction);
transform.rotation = Quaternion.Lerp(transform.rotation, toRotation, speed * Time.time);
Vector3 fwd = this.transform.TransformDirection(Vector3.forward);
Debug.DrawRay(this.transform.position, fwd * 50, Color.green);
}
}
This results in the object looking kind of towards the target but never actually looking at it. I cannot for the life of me see my mistake. Can anyone help me?
There are a few different issues here.
First, you're constructing toRotation as a relative rotation (some amount of change in rotation)
Quaternion toRotation = Quaternion.FromToRotation(transform.forward, direction);
But then you're applying toRotation as a destination (a final rotation to ease toward)
transform.rotation = Quaternion.Lerp(transform.rotation, toRotation, speed * Time.time);
Imagine if we did this with position, starting with an object at x=8 that wants to move to x=10. The change in x is +2, but if we lerp toward that change value like so:
Mathf.Lerp(startPosition, endPosition - startPosition, weight)
...then as the interpolation weight gets closer to 1, our x gets closer to 2 - not to 10 where we want it to be!
(When applied in a compounding way frame over frame, this will seek out an equilibrium point where x = lerp(x, 10 - x, weight), x = (1 - weight) * x + weight * 10 - weight * x, 2 weight * x = weight * 10, 2 x = 10, x = 5... still not where we wanted it to be!)
So, first we need to describe the destination orientation we want to have. It looks like you probably want something like this:
// Calculate direction as before.
Vector3 direction = target.position - transform.position;
// Use LookRotation to form a rotation that looks along that vector.
// (You can optionally pass a second vector as "up" to control the twist).
Quaternion toRotation = Quaternion.LookRotation(direction);
Then, if you want to rotate toward this location at a constant rate, as John Hamilton says, you can use the Quaternion.RotateTowards method:
float step = speed * Time.deltaTime;
transform.rotation = Quaternion.RotateTowards(transform.rotation, toRotation, step);
Or, if you want the aim to be faster the further off-target it is, and gradually ease-in as it locks on, you can keep the exponential ease you were using before, but with the destination as calculated above, and correcting the time adjustment like so:
// Here "speed" is replaced by "sharpness," a parameter between 0 & 1
// that controls how smooth vs fast the aim is.
// referenceFramerate is a constant like, say, 30.0f. It adjusts the
// "units" of sharpness to be the same as you'd use at that framerate
blend = 1f - Mathf.Pow(1f - sharpness, Time.deltaTime * referenceFramerate);
transform.rotation = Quaternion.Lerp(transform.rotation, toRotation, blend);
Any of the Lerp (linear interpolation) functions (Mathf, Vector3, Quaternion [ok, maybe quaternions don't apply since it's a matrix], etc.) works following this formula: P = P0 + (t * (P1 - P0)), 0 < t < 1, where, as the formula suggests, t which is time goes between 0 and 1. So what this function gives you is a point of a delta between two points multiplied by a normalized time.
What I do is to use coroutines to handle such interpolations, but since you are using an update, maybe you could try with Quaternion.LookRotation(Vector3 _direction), instead of using Quaternion.FromToRotation.
As a side node, you could use transform.forward, instead of using TransformDirection(Vector3 _normal). Transform.forward, transform.up and transform.right gives you already TransformDirection with (0,0,1), (0,1,0) & (1,0,0) respectively.
There's already a Unity built-in doing what you need to do. You shouldn't need to re-invent the wheel, unless you want it to do very specific things other than just rotating towards another game object. Example from the link:
using UnityEngine;
using System.Collections;
public class ExampleClass : MonoBehaviour {
public Transform target;
public float speed;
void Update() {
float step = speed * Time.deltaTime;
transform.rotation = Quaternion.RotateTowards(transform.rotation, target.rotation, step);
}
}
Also, I suggest using Update() instead of FixedUpdate() for this kind of movement (well, Unity examples also seem to suggest that), since the time-step is usually faster in the Update() function, you should get a smoother movement overall. | 2021-08-05 10:52:19 | {"extraction_info": {"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, "math_score": 0.4406586289405823, "perplexity": 2853.260946802349}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-31/segments/1627046155529.97/warc/CC-MAIN-20210805095314-20210805125314-00272.warc.gz"} |
https://mersenneforum.org/showpost.php?s=df609e68e26ef3756336db1b8087eab6&p=576953&postcount=13 | View Single Post
2021-04-26, 21:15 #13
Happy5214
"Alexander"
Nov 2008
The Alamo City
10111101112 Posts
Quote:
Originally Posted by jyb This isn't quite correct. The axioms of the reals minus the least upper bound property gives you an ordered field, but it is not sufficient to characterize the rationals. Specifically, it need not be Archimedean, which the rationals are. E.g. see the surreal numbers, for which these axioms apply, but which are certainly not isomorphic to the rationals.
Correct. The rationals are the smallest ordered field, in the sense that they are embedded in every other ordered field, so that would be the missing axiom. I couldn't tell you how that implies the Archimedean property (I proved it for the reals in my analysis class, but that proof is based on completeness). | 2021-10-26 02:40:43 | {"extraction_info": {"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, "math_score": 0.829548716545105, "perplexity": 359.9100187977873}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323587794.19/warc/CC-MAIN-20211026011138-20211026041138-00568.warc.gz"} |
https://zbmath.org/?q=an:1286.60086 | ## Moment asymptotics for branching random walks in random environment.(English)Zbl 1286.60086
Summary: We consider the long-time behaviour of a branching random walk in random environment on the lattice $$\mathbb{Z}^d$$. The migration of particles proceeds according to simple random walk in continuous time, while the medium is given as a random potential of spatially dependent killing/branching rates. The main objects of our interest are the annealed moments $$\langle m_n^p \rangle$$, i.e., the $$p$$-th moments over the medium of the $$n$$-th moment over the migration and killing/branching, of the local and global population sizes. For $$n=1$$, this is well-understood, as $$m_1$$ is closely connected with the parabolic Anderson model. For some special distributions, this was extended to $$n\geq2$$, but only as to the first term of the asymptotics, using (a recursive version of) a Feynman-Kac formula for $$m_n$$. In this work, we derive also the second term of the asymptotics for a much larger class of distributions. In particular, we show that $$\langle m_n^p \rangle$$ and $$\langle m_1^{np} \rangle$$ are asymptotically equal up to an error $$\text e^{o(t)}$$. The cornerstone of our method is a direct Feynman-Kac type formula for $$m_n$$, which we establish using known spine techniques.
### MSC:
60J80 Branching processes (Galton-Watson, birth-and-death, etc.) 60J55 Local time and additive functionals 60F10 Large deviations 60K37 Processes in random environments
Full Text: | 2022-12-09 03:35:00 | {"extraction_info": {"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, "math_score": 0.8131206035614014, "perplexity": 312.8136968363047}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446711376.47/warc/CC-MAIN-20221209011720-20221209041720-00240.warc.gz"} |
https://www.albert.io/learn/ap-physics-1-and-2/question/torque-on-a-unicycle | Limited access
You are riding a unicycle where the wheel has a diameter of $65 \text{ cm}$ and the length of the piece connecting the two pedals is $30\text{ cm}$. Your leg and foot push down with $110 \text{ N}$ of force at a $60^\circ$ angle as shown by the green line in the drawing.
Which of the following torques is consistent with this information?
A
$14.3\ \mathrm{m\cdot N}$
B
$31.0\ \mathrm{m\cdot N}$
C
$1650\ \mathrm{m\cdot N}$
D
$7150\ \mathrm{m\cdot N}$
Select an assignment template | 2017-04-26 13:52:55 | {"extraction_info": {"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, "math_score": 0.42530447244644165, "perplexity": 468.5115857022648}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-17/segments/1492917121355.9/warc/CC-MAIN-20170423031201-00133-ip-10-145-167-34.ec2.internal.warc.gz"} |
http://twofoldgaze.com/2007/12/19/12/ | # Another Chaotic search
Newton’s method is a more simple way of finding a root. The formula is
$x_{n+1}=x_n - \frac{f^{'}(x)}{f(x)}$
However, the behavior of the algorithm in finding roots is anything but simple. Above, I display the behavior of this algorithm on the complex interval where the real part of and the imaginary part of each complex number if between -1 and 1. The coloring scheme is identical to the previous post. | 2014-12-20 17:57:38 | {"extraction_info": {"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": 1, "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, "math_score": 0.6882117390632629, "perplexity": 426.04751322637264}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-52/segments/1418802770071.75/warc/CC-MAIN-20141217075250-00102-ip-10-231-17-201.ec2.internal.warc.gz"} |
https://earthsci.stanford.edu/computing/unix/formatting/parts.php | # Required parts of a LaTeX input file
Last revision August 3, 2004
Table of Contents: Features of LaTeX Basic layout of a LaTeX document Required parts of a LaTeX input file Basic LaTeX topics LaTeX by example LaTeX Frequent Questions
A few commands MUST appear in every LaTeX input file in a certain order. They are:
\documentstyle{stylename}
\begin{document}
\end{document}
The documentstyle has a required argument stylename to select an overall typesetting style for the document; the one you normally use is article (there are also book, report, letter, and memo). It also has an optional argument to select 11pt or 12pt normal type size (10pt is the default size).
Between the \documentstyle and \begin{document} commands you can place commands that will affect the environment of the entire document, such as changes to margin widths.
The actual text of your document and associated commands go between the \begin{document} and \end{document} commands. | 2018-01-19 07:11:14 | {"extraction_info": {"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, "math_score": 0.9983808398246765, "perplexity": 5041.3874327794165}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-05/segments/1516084887832.51/warc/CC-MAIN-20180119065719-20180119085719-00605.warc.gz"} |
https://www.mathstoon.com/limit-properties-proof/ | # Proofs of all Limit Properties [ε-δ Definition]
In this post, we will at first recall all the properties of the limits, and then will prove them using the epsilon-delta method. The following are some properties of limits.
## Properties of Limits
Let us consider two functions f(x) and g(x) of the variable x, and let $a$ be a real number. Assuming both the limits of f(x) and g(x) exist when x→a, we have the following list of properties of limits:
The limit of a constant is constant. That is, $\lim\limits_{x \to a}$ c =c, where c is a constant. This is the constant rule of limits.
$\lim\limits_{x \to a}$ [c f(x)] =c $\lim\limits_{x \to a}$ f(x), where c is a constant. This is the constant multiple rule of limits.
$\lim\limits_{x \to a}$ [f(x)+g(x)] = $\lim\limits_{x \to a}$ f(x)+ $\lim\limits_{x \to a}$ g(x). This is the addition rule of limits.
$\lim\limits_{x \to a}$ [f(x)-g(x)] = $\lim\limits_{x \to a}$ f(x)- $\lim\limits_{x \to a}$ g(x). This is the subtraction rule of limits.
$\lim\limits_{x \to a}$ [f(x)⋅g(x)] = $\lim\limits_{x \to a}$ f(x) ⋅ $\lim\limits_{x \to a}$ g(x). This is called the multiplication rule of limits.
$\lim\limits_{x \to a}$ [f(x)/g(x)] = $\dfrac{\lim\limits_{x \to a} f(x)}{\lim\limits_{x \to a} g(x)}$, provided that the limit of g(x) when x→a is non-zero. This is called the division rule of limits.
Before going into the proofs, let us recall the definition of the limit of a function.
## Epsilon-Delta Definition of Limit
Let f(x) be a function such that $\lim\limits_{x \to a}$ f(x) = L. This means that for every ε>0, there exists a δ>0 such that
|f(x)-L| < ε,
whenever 0<|x-a|<δ.
### Negation of Epsilon Delta Definition of Limit
Let us now give the negative statement of $\lim\limits_{x \to a}$ f(x) = L. If f(x) does not go to L as x→a, then it is not true that for every ε>0, there exists a δ>0 such that
|f(x)-L| < ε, whenever 0<|x-a|<δ.
So the negation of the ε-δ definition will be as follows:
If f(x) does not go to L as x→a, then there exists some ε>0 such that for every δ>0 ∃ some point x ∈ {x : 0<|x-a|<δ} such that |f(x)-L| ≥ ε holds true.
Using this epsilon-delta definition of a function, we will prove the above properties of limits.
## Limit Properties Proof
### Constant Rule of Limit Proof
Using ε-δ definition, prove that $\lim\limits_{x \to a}$ c =c, where c is a constant.
Proof:
Let ε>0 be a given positive number. To show $\lim\limits_{x \to a}$ c =c, we need to show that ∃ a δ>0 such that
|f(x)-c| < ε, whenever 0<|x-a|<δ …(i)
Notice that as f(x)=c ∀ x, the inequality in (i), that is, |f(x)-c| < ε is true always for any x. Thus, we can choose any δ>0 for which (i) will be true. This completes the proof of the constant rule of limit.
### Constant Multiple Rule of Limit Proof
Using ε-δ definition, prove that $\lim\limits_{x \to a}$ [c f(x)] =c $\lim\limits_{x \to a}$ f(x), where c is a constant.
Proof:
Let $\lim\limits_{x \to a}$ f(x) = L.
If c=0, then there is nothing to prove.
So assume that c≠0. Note that ε/|c|>0. Thus, by definition we have that for every ε>0, there exists a δ>0 such that
|f(x)-L| < ε/|c|, whenever 0<|x-a|<δ …(ii)
Now, whenever 0<|x-a|<δ we have that
|cf(x)-cL| = c|f(x)-L|
< c ⋅ ε/|c| by (ii)
= ε
This shows that $\lim\limits_{x \to a}$ [c f(x)] = cL =c $\lim\limits_{x \to a}$ f(x).
### Sum Rule of Limit Proof
Using ε-δ definition, prove the addition rule of limit, that is, prove that $\lim\limits_{x \to a}$ [f(x)+g(x)] =$\lim\limits_{x \to a}$ f(x)+ $\lim\limits_{x \to a}$ g(x).
Proof:
Let $\lim\limits_{x \to a}$ f(x) = L and $\lim\limits_{x \to a}$ g(x) = M. Then by the definition of limits, for a given ε>0, there exist a δ1>0 and a δ2>0 such that
|f(x)-L| < ε/2, whenever 0<|x-a|<δ1 …(iii)
|g(x)-M| < ε/2, whenever 0<|x-a|<δ2 …(iv)
Choose δ = min {δ1 , δ2 }. Then whenever 0<|x-a|<δ, we have that
|f(x)+g(x) – (L+M)| = |(f(x) – L) + (g(x) – M)|
≤ |f(x) – L| + |g(x) – M| by the triangle inequality |a+b| ≤ |a|+|b|
≤ ε/2 + ε/2 by (iii) and (iv)
= ε
This proves that $\lim\limits_{x \to a}$ [f(x)+g(x)] = L+M = $\lim\limits_{x \to a}$ f(x) + $\lim\limits_{x \to a}$ g(x).
### Difference Rule of Limit Proof
With the setting of the proof of the sum rule of limit, whenever 0<|x-a|<δ, we have that
|f(x)-g(x) – (L-M)| = |(f(x) – L) – (g(x) – M)|
≤ |f(x) – L| + |g(x) – M| by the inequality |a-b| ≤ |a|+|b|
≤ ε/2 + ε/2 by (iii) and (iv)
= ε
This proves that $\lim\limits_{x \to a}$ [f(x)-g(x)] = L+M = $\lim\limits_{x \to a}$ f(x) – $\lim\limits_{x \to a}$ g(x).
### Product Rule of Limit Proof
Using ε-δ definition, prove the product rule of limit, that is, prove that $\lim\limits_{x \to a}$ [f(x)⋅g(x)] = $\lim\limits_{x \to a}$ f(x) ⋅ $\lim\limits_{x \to a}$ g(x).
Proof:
Let $\lim\limits_{x \to a}$ f(x) = L and $\lim\limits_{x \to a}$ g(x) = M. Note that
|f(x)g(x) – LM| = |f(x)g(x) – Mf(x) + Mf(x) – LM|
= |f(x) (g(x) – M) + M (f(x) – L)|
≤ |f(x) (g(x) – M)+ |M (f(x) – L)| by the triangle inequality |a+b| ≤ |a|+|b|
= |f(x)| |g(x)- M| + M|f(x) – L| …(*)
< ε (we need to show this
Corresponding to ε=1, there exists a δ1>0 such that
|f(x)-L| < 1, whenever 0<|x-a|<δ1 …(v)
∴By (v), |f(x)| = |f(x)-L+L| ≤ |f(x)-L| + |L| < 1+|L| …(vi)
Since $\lim\limits_{x \to a}$ g(x) = M, corresponding to $\dfrac{\epsilon}{2(1+|L|)}>0$, there exists a δ2>0 such that
|g(x)-M| < $\dfrac{\epsilon}{2(1+|L|)}$, whenever 0<|x-a|<δ2 …(vii)
Again since $\lim\limits_{x \to a}$ f(x) = L, corresponding to $\dfrac{\epsilon}{2(1+|M|)}>0$, there exists a δ3>0 such that
|f(x)-L| < $\dfrac{\epsilon}{2(1+|M|)}$, whenever 0<|x-a|<δ3 …(viii)
Choose δ = min {δ1 , δ2 , δ3}. Then whenever 0<|x-a|<δ, we have from (*) that
|f(x)g(x) – LM| ≤ |f(x)| |g(x)- M| + M|f(x) – L|
< |f(x)| |g(x)- M| + (1+|M|) |f(x) – L|
< (1+|L|) ⋅ $\dfrac{\epsilon}{2(1+|L|)}$ + (1+|M|) ⋅ $\dfrac{\epsilon}{2(1+|M|)}$, by (vi), (vii) and (viii)
< ε/2 + ε/2
= ε
Thus, we obtain that $\lim\limits_{x \to a}$ [f(x)⋅g(x)] = LM = $\lim\limits_{x \to a}$ f(x) ⋅ $\lim\limits_{x \to a}$ g(x). This completes the proof of the product rule of limits. | 2022-10-04 20:26:03 | {"extraction_info": {"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, "math_score": 0.9542913436889648, "perplexity": 2639.7914633245473}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030337524.47/warc/CC-MAIN-20221004184523-20221004214523-00149.warc.gz"} |
https://math.stackexchange.com/questions/2987309/how-to-show-that-random-variable-is-almost-surely-finite-coin-toss-problem | # how to show that random variable is almost surely finite? Coin toss problem
Consider the Coin toss problem, i.e. let $$Z_{i} : \Omega \rightarrow \{0,1\}$$ with $$Z_{i}\left(\omega\right) = \begin{cases} 1 & \text{if }\omega = H\\ 0 & \text{if } \omega = T \end{cases}$$ be the outcome of the $$i$$-th coin toss with $$\Omega = \{H,T\}$$. Assume all the $$Z_{i}$$ are independent and identical distributed with $$\mathbb{P}(Z_{i} = 1) = \mathbb{P}(Z_{i} = 0) = \frac{1}{2}.$$ Now define $$R := \min \{k\geq 1 \mid Z_{k} = 1,Z_{k+1} = 0,Z_{k+2} = 1, Z_{k+3} = 0\}.$$ $$R$$ is the number of tosses to get the pattern HTHT.
My question ist how to show that $$R$$ is almost surely finite. Any hints?
1. Show that $$A_k := \{Z_{4k}=0, Z_{4k+1}=1, Z_{4k+2}=0, Z_{4k+3}=1\}$$ satisfies $$\mathbb{P}(A_k) = 1/16$$ for each $$k \in \mathbb{N}$$.
2. Show that the events $$A_k$$, $$k \geq 1$$, are independent.
3. It follows from Step 1 that $$\sum_{k \geq 1} \mathbb{P}(A_k) = \infty.$$ Apply the Borel Cantelli lemma (using Step 2) to conclude that $$\mathbb{P}(A_k \, \, \text{infinitely often})=1.$$
4. Conclude that $$\mathbb{P}(R<\infty)=1$$.
• Is this correct? $\mathbb{P}(R = \infty) = \mathbb{P}(\lim\limits_{n \to \infty}\inf A_{n}^{c}) = 1 - \mathbb{P}(A_{n}~~ \text{infinitely often}) = 0$ – user562724 Nov 7 '18 at 21:05
• @love_math It is correct that $\mathbb{P}(R=\infty)=0$ (simply because $\mathbb{P}(R=\infty) = 1-\mathbb{P}(R<\infty)$) but the other equations are not correct. For instance $\mathbb{P}(R=\infty) = \mathbb{P}(\liminf_n A_n^c)$ does not hold true. – saz Nov 7 '18 at 21:13
• @love_math $\mathbb{P}(A_n \, \, \text{infinitely often})=1$ tells us, in particular, that for almost every $\omega$ we can find $n \in \mathbb{N}$ such that $\omega \in A_n$ (in fact there are infinitely many such $n$, but we are fine with one). As $A_n \subseteq \{R<\infty\}$ this implies $\omega \in A_n \subseteq \{R<\infty\}$, i.e. $R(\omega)<\infty$. – saz Nov 7 '18 at 21:20
• @love_math We only have $\{R=\infty\} \color{red}{\subseteq} \bigcap_{m \geq 1} A_m^c$ (for instance, if $Z_2(\omega)=1$, $Z_3(\omega)=0$, $Z_4(\omega)=1$, $Z_5(\omega)=0$ for some $\omega$, then $R(\omega)<\infty$ but $\omega \in \bigcap_{m \geq 1} A_m^c$ ... this shows that $\bigcap_m A_m^c$ is stricly larger than $\{R=\infty\}$.) Apart from that, your reasoning is correct, I think. (Note that with the correction which I mentioned you still get the desired result $\mathbb{P}(R=\infty)=0$.) – saz Nov 10 '18 at 7:32 | 2021-05-12 21:01:22 | {"extraction_info": {"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": 17, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9887626767158508, "perplexity": 160.39282493939967}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-21/segments/1620243989705.28/warc/CC-MAIN-20210512193253-20210512223253-00298.warc.gz"} |
https://www.snapsolve.com/solutions/Awire-of-radius-1-mm-is-bent-in-the-form-of-a-circle-of-radius-10-cm-The-bending-1672380083308546 | Home/Class 11/Physics/
A wire of radius $$1 mm$$ is bent in the form of a circle of radius $$10 cm$$. The bending moment will be $$(Y =2\times10^{11}N/m^{2})$$
(A) 3.14 N/m
(B) 6.28 N/m
(C) 1.57 N/m
(D) 15.7 N/m
Speed
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## QuestionPhysicsClass 11
A wire of radius $$1 mm$$ is bent in the form of a circle of radius $$10 cm$$. The bending moment will be $$(Y =2\times10^{11}N/m^{2})$$
(A) 3.14 N/m
(B) 6.28 N/m
(C) 1.57 N/m
(D) 15.7 N/m
Bending Moment $$C=\dfrac { Y{ I }_{ G } }{ R }$$
where $$Y= 2\times { 10 }^{ 11 }N/{m}^{2}$$
$$R= 10cm= 0.1m$$
$${ I }_{ G }=\dfrac{\pi{R}^{4}}{4} = \dfrac{\pi{0.1}^{4}}{4} =7.85\times{10}^{-5} m^{4}$$
Put these in the expression for $$C$$,
$$C=1.57N/m$$ | 2022-07-06 17:14:53 | {"extraction_info": {"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, "math_score": 0.9099189639091492, "perplexity": 2319.2170836952505}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656104675818.94/warc/CC-MAIN-20220706151618-20220706181618-00665.warc.gz"} |
http://www.newton.ac.uk/programmes/HRT/seminars/2008121209301.html | # HRT
## Seminar
### Bounds on the energetics and entropy production of a buoyancy forced ocean
Winters, KB (UC, San Diego)
Friday 12 December 2008, 09:30-10:00
Seminar Room 1, Newton Institute
#### Abstract
We show that the volume averaged rate of entropy production in horizontal convection, i.e., $\chi \equiv \kappa < |{\nabla} b |^2 >$ where $b$ is the buoyancy, is bounded from above by $4.57 H^{-1} \kappa^{2/3} \nu^{-1/3} b_{\rm max}^{7/3}$. Here $H$ is the depth of the container, $\kappa$ is the molecular diffusion, $\nu$ the kinematic viscosity and $b_{\rm max}$ the maximum buoyancy difference prescribed on the surface. The rate of entropy production is directly related, via the volume integrated diapycnal buoyancy flux, to the energetics of irreversible mixing and the rate of energy transfer between available and background potential energy in the Boussinesq limit. The bound implies that the rate of generation of available potential energy by horizontal convection is no larger than $\kappa^{1/3}$ in the limit $\kappa \to 0$ at fixed $Pr=\nu/\kappa$. The bound on the energetics of mixing reinforces and strengthens the statement of Paparella and Young (2002) that horizontal convection is nonturbulent in limit of vanishing fluid viscosity and diffusivity. In the context of a model ocean, insulated at all boundaries except at the upper surface where the buoyancy is prescribed, the bounds on the energy transfer rates in the mechanical energy budget imply that buoyancy forcing alone is insufficient by at least three orders of magnitude to maintain observed or inferred average oceanic dissipation rates and that additional energy sources such as winds, tides or (perhaps) bioturbation are necessary to sustain nontrivial levels of turbulent dissipation and mixing in the world's oceans
#### Video
The video for this talk should appear here if JavaScript is enabled.
If it doesn't, something may have gone wrong with our embedded player.
We'll get it fixed as soon as possible. | 2013-12-05 13:22:57 | {"extraction_info": {"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, "math_score": 0.585898220539093, "perplexity": 872.7902527458193}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2013-48/segments/1386163045148/warc/CC-MAIN-20131204131725-00070-ip-10-33-133-15.ec2.internal.warc.gz"} |
https://brilliant.org/discussions/thread/graph-of-the-world/ | ×
# Graph of the World
I wanted to post this note many months ago, but I guess I procastinated.
In this note, I would like to show how to do a similar graph like the above, to graph literally anything in a single function (As long as it is a function of x), and invite anybody to make a graph like this, depicting an object (be it a building, or an ant). Then, I will compile the graphs together such that graphing the single function depicts any object we have made.
In the image above, I have depicted the Empire State Building and the Singapore Parliament house in a single function.
$$\textbf{How to do it}$$
First, let me introduce this function: $\frac{\left|\left(x-k\right)-\left|x-k\right|\right|}{2\left(k-x\right)}+\frac{\left|\left(x+a\right)-\left|x+a\right|\right|}{2\left(x+a\right)}$
The above function, basically graphs $$y=0$$ from $$x=-\infty \text{ to } x=-a$$ and $$x=k \text{ to } x=\infty$$, and $$y=1$$ from $$x=-a \text{ to } x=k$$, assuming $$a$$ and $$k$$ to be positive real numbers.
Now, if you can see where this is going, if I $$\times$$ that above function by $$f(x)$$, what would I get? I get $$y=0$$ from $$x=-\infty \text{ to } x=-a$$ and $$x=k \text{ to } x=\infty$$, and $$y=f(x)$$ from $$x=-a \text{ to } x=k$$.
For the case of $$f(x) = x^{2}$$, $$a=2$$ and $$k=1$$:
Graph of $$x^2\left(\frac{\left|\left(x-1\right)-\left|x-1\right|\right|}{2\left(1-x\right)}+\frac{\left|\left(x+2\right)-\left|x+2\right|\right|}{2\left(x+2\right)}\right)$$
You can try it yourself here
Now, what if we want to graph $$y=x^{2}$$ from $$x=-2$$ to $$x=1$$ and then $$y=(x-2)^{3}$$ from $$x=1$$ to $$x=3$$? Easy, we just add $$2$$ functions together:
Graph of $$x^2\left(\frac{\left|\left(x-1\right)-\left|x-1\right|\right|}{2\left(1-x\right)}+\frac{\left|\left(x+2\right)-\left|x+2\right|\right|}{2\left(x+2\right)}\right)+\left(x-2\right)^3\left(\frac{\left|\left(x-3\right)-\left|x-3\right|\right|}{2\left(3-x\right)}+\frac{\left|\left(x-1\right)-\left|x-1\right|\right|}{2\left(x-1\right)}\right)$$
And so we are done with the basis of how to graph any object you want, all you have to do is keep adding the functions together until you get the desired shape.
Now, what if your desired shape is referenced from a picture? No worries, Desmos enables you to add a picture for reference:
Sidenote: To achieve shading (For the picture above, the shading is red), just add an inequality. To know what I mean, try $$y>x$$. The above picture took me about 45 minutes.
You might want to see this as an example
$$\textbf{Graph Submission details}$$
If you want to make one of these graphs, and want to submit yours too, just post the link of your graph into the comments section. However, for your graph to be accepted, it has to pass some requirements:
Requirements:
$$\bullet$$ Your graph must be $$y=0$$ from $$x=-\infty$$ to $$x=k_{1}$$ and from $$x=\infty$$ to $$x=k_{2}$$, where $$k_{1}\le x\le k_{2}$$ is the section where your object exists. This is to allow me to add multiple objects to a single function's graph (Like how I added the Empire State Building and the Singapore Parliament House together into a single function.)
$$\bullet$$ Your Graph should have a height of roughly between $$5$$ to $$20$$. This is to avoid graphs that are too small or too big to be added into the compilation.
Well that's about it, have fun making one of these graphs! If you have any questions, post them in the comments section.
Note by Julian Poon
1 year, 3 months ago
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In response to Apple: $$\color{white}{\text{Excellent work!}}$$ · 1 year, 3 months ago
In response to Lameness: $$\color{white}{\text{More Lameness}}$$ · 1 year, 3 months ago
$$\color{white}{\text{I am not lame -_-}}$$ · 1 year, 3 months ago
On a slightly more practical level, NURBS, or "non-uniform rational B-spline", have been developed to provide functions of fairly arbitrary curves and surfaces for computer graphics and industrial uses. They are extremely powerful and amendable to mathematical methods, as compared to conventional "elementary functions" of mathematics. What you are proposing is something analogous, but for discretized shapes. It's a start, but needs more work to attain the same functionality as NURBS. See Bezier Curves for an easy-to-understand text on a subset of NURBS. · 1 year, 2 months ago | 2016-10-27 07:11:18 | {"extraction_info": {"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, "math_score": 0.5390256643295288, "perplexity": 469.6422815664307}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-44/segments/1476988721142.99/warc/CC-MAIN-20161020183841-00502-ip-10-171-6-4.ec2.internal.warc.gz"} |
http://www.fluidsujaen.es/research/wakes/ | ## Wakes stability and control
The flow around bluff bodies is of great interest in many technological applications. In most cases the wake turns out to be non stationary and unstable, producing modes whose frequencies are clear and independent of background noise. On the other hand, the close wake instability causes fluctuating forces acting on the body and surrounding bodies when exist, and oscillations on Drag and Lift coefficients. To better understand the physics of the problem and to characterize the transition to non-stationarity, the Fuids Mechanics Group of Univeristy of Jaén has been carrying out experimental work (hot-wire anemometry and visualizations), direct numerical simulations and global and local stability analyses in the wake of slender axisymmetrical blunt based bodies (bullet-shaped bodies) for several years; focusing this work on both high and low Reynolds (Re) number flows.
As Re increases, the wakes suffers two bifurcations, starting at the axysymmetrical state for low Re. The first bifurcation (taking place at the stationary critical Reynolds, $Re_{cs}$) breaks the axisymmetry and leads to a new planar symmetry stationary state, that remains valid until the second bifurcation (oscillatory critical Reynolds, $Re_{co}$) when the flow becomes time dependent, and pairs of counter-rotating vortices are shed periodically while keeping the symmetry plane. Finally, a third state is reached where a low frequency modulation exists (about one-quarter of the leading one). We have shown that increasing body lengths stabilize both the first and second critical Reynolds.
Figure 1. Streamwise vorticity contours for a body of aspect ratio $L=2D$ and $Re=415$>$Re_{co}$
We also investigate control methods to inhibit the oscillations and stabilize the wake. Among them, base bleed has been proved to inhibit vortices shedding and effectively suppress large-scale structures in high Re flows. It is shown that both bifurcations can be stabilized by injecting a certain amount of fluid with the same density than the free stream one through the base of the body, quantified as the bleed-to-free-stream velocity ratio, $C_b = W_b/W_{\infty}$. For a given Re that produces the non-stationary oscillating wake, a bleed ratio value $C_{b2}$ eliminates the vortices shedding, and recovers the stationary planar symmetry state. If this ratio is increased until it reaches $C_{b1}$, the axisymmetry will be again obtained. The effect of the bleed-to-freestream density ratio, $S=\rho_b /\rho_{\infty}$ on the stability properties has been also studied through local stability analysis, finding that low ratios ($S$<$1$) inhibit the wake for smaller bleed ratios than in the homogeneous case ($S=1$).
Figure 2. Visualizations of the wake for a body with aspect ratio $L\simeq9.8D$, for Re=2800 and several values of the bleed coefficient ($S=1$): (a) $C_b$=0.04, (b) $C_b$=0.09, and (c) $C_b$=0.13>$C_{b2}$.
Rear cavities are found to stabilize the wake as well. They can shift the bifurcations occurrence to higher values of $Re$, so that the deeper the cavity is, the bigger the values $Re_{cs}$ and $Re_{co}$ becomes. This fact reaches a limit for a critical cavity depth $h_{cr}$, from where the stabilization effect no longer grows (asymptotic behavior).
On the other hand, body rotation has been also investigated by means of numerical simulations and global stability analysis, finding that moderate values of the rotation parameter, $\Omega=\omega D/2 W_{\infty}$, where $\omega$ is the body angular velocity, allow to retrieve the axisymmetry for $Re$>$Re_{cs}$. The stability picture existing at $\Omega\leq1$ and $Re \leq 450$ can be outlined as follows: for $\Omega$>$0.25$ a linear eigenmode with azimuthal wavenumber $m = 1$ becomes unstable at a critical value of $Re = Re_{c1}$ that increases with $\Omega$, leading to a low-frequency frozen state where axially-elongated structures co-rotate with the body with a different angular velocity. In the range $0.25 \leq \Omega$<$0.52$ a different eigenmode with $m = 1$ becomes unstable at $Re = Re_{c3}$, explaining the high frequency frozen spiral wake found with the numerical simulations, for which the vortical structures adopt a spiral shape, while rotating in a frozen way with an angular frequency that is close to the vortex shedding one at low $\Omega$. However, for $\Omega \geq 0.52$ and low values of $Re$, the wake is destabilized by a third $m = 1$ oscillatory mode of intermediate frequency, characterized again by a frozen spiral state that becomes dominant at higher $\Omega$.
# Supporting projects
Spanish Ministry of Science and Innovation and European Funds
DPI2008-06624-C02, DPI2005-08654-C04-01 and DPI2002-04550-C07-06.
Junta de Andalucía and European Funds
P07-TEP02693.
UJA-08-16-19.
# Publications
– J.I. Jiménez-González, A. Sevilla, E. Sanmiguel-Rojas & C. Martínez-Bazán. Global stability analysis of the axisymmetric wake past a spinning bullet-shaped body. Journal of Fluid Mechanics (Accepted, 2014).
– J.I. Jiménez-González, E. Sanmiguel-Rojas, A. Sevilla & C. Martínez-Bazán. Laminar flow past a spinning bullet-shaped body at moderate angular velocities. Journal of Fluid and Structures, Vol. 43, 200-219 (2013).
– E. Sanmiguel-Rojas, J.I. Jiménez-González, P. Bohórquez, G. Pawlak & C. Martínez-Bazán. Stability effects of base cavities on the wakes of slender blunt-based axisymmetric bodies. Physics of Fluids, Vol. 23, 114103 (2011).
– P. Bohórquez, E. Sanmiguel-Rojas, A. Sevilla, J.I. Jiménez-González & C. Martínez-Bazán. Stability and dynamics of the laminar wake past a slender blunt-based axisymmetric body. Journal of Fluid Mechanics, Vol. 675, 110-144 (2011).
– E. Sanmiguel-Rojas, A. Sevilla, C. Martínez-Bazán & J.-M. Chomaz. Global mode analysis of axisymmetric bluff-body wakes: stabilization by base bleed. Phys. Fluids 21, 114102 (2009).
– A. Sevilla & C. Martínez-Bazán. A note on the stabilization of bluff-body wakes by low density base bleed. Phys. Fluids 18, 098102 (2006).
– A. Sevilla & C. Martínez-Bazán. Vortex shedding in high Reynolds number axisymmetric bluff-body wakes: Local linear instability and global bleed control.Phys. Fluids 16, 1037 3460–3469 (2004). | 2020-03-29 18:34:07 | {"extraction_info": {"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": 42, "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, "math_score": 0.521726131439209, "perplexity": 2610.2418910849037}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-16/segments/1585370495413.19/warc/CC-MAIN-20200329171027-20200329201027-00157.warc.gz"} |
https://chemistry.stackexchange.com/questions/147163/atoms-vs-molecule-when-talking-about-the-avogadros-number | By far, my understanding is that a molecule is made up of atoms bonded together. For example, a molecule of water ($$\ce{H2O}$$) has 2 hydrogen atoms and one oxygen atom.
However, when it comes to Avogadro's number, I'm getting confused because it mixes the concept and treats it like it is the same.
Avogadro's number is defined as the number of elementary particles (molecules, atoms, compounds, etc.) per mole of a substance.
If I take 1 mole of water, with the molecules definition for the Avogadro's number, it has $$\mathrm{6.022×10^{23}}$$ molecules of water. With each molecule having 3 atoms, it means that 1 mole of water has $$\mathrm{18.066×10^{23}}$$ atoms.
If I take 1 mole of water with the atoms definition for Avogadro's number, it has $$\mathrm{6.022×10^{23}}$$ atoms of water.
$$1$$ mole of water has $$\pu{6.022E23}$$ molecules of water, but not $$\pu{6.022E23}$$ atoms of water, because the expression "atoms of water" has no sense. You are allowed to state that it contains $$\pu{18.066E23}$$ atoms. But you are not allowed to speak of "atoms of water". Water has not its own atoms of water. | 2021-07-23 22:39:32 | {"extraction_info": {"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": 8, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5626617670059204, "perplexity": 375.31014468986507}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-31/segments/1627046150067.51/warc/CC-MAIN-20210723210216-20210724000216-00508.warc.gz"} |
http://clay6.com/qa/71093/some-scientists-have-predicted-that-a-global-nuclear-war-on-the-earth-would | Comment
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# Some scientists have predicted that a global nuclear war on the earth would be followed by a severe nuclear winter with a devastating effect on life on earth. What might be the basis of their prediction ?
$\begin{array}{1 1} \text{Formation of charged particles after the nuclear war } \\ \text{Formation of clouds after the nuclear } \\ \text{Formation of dust particles after the nuclear war} \\ \text{Formation of water particles resulting in ice and show after a nuclear war} \end{array}$ | 2019-08-20 03:01:17 | {"extraction_info": {"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, "math_score": 0.36571210622787476, "perplexity": 699.3636761653303}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-35/segments/1566027315222.14/warc/CC-MAIN-20190820024110-20190820050110-00535.warc.gz"} |
https://bewakes.com/blog/minimum-spanning-tree-using-kruskals-algorithm/ | Minimum Spanning Tree using Kruskal's algorithm
May 6, 2017
Hey there!! Now we are heading off to finding a minimal spanning tree for an undirected graph. It is recommended to go through my previous post on Disjoint Set Data Structure
It is assumed that the reader knows about Graph, a very important data structure in computer science with lots of applications(social networks, computer networks, digital circuits, and many more).
An undirected graph has no sense of directions in edges. For example, if we have an edge connecting two vertices A and B, that edge can be called as connecting A to B or B to A. The adjacency matrix for the graph is symmetric.
Above is a simple small graph(hand made by me, no wonder why the graph edges are zig-zagged). The adjacency matrix for the graph is thus,
# the infinity
INF = 99999999
[[INF,7,2,5],
[7,INF,4,1],
[2,4,INF,4],
[5,1,4,INF]]
I have chosen adjacency matrix for graph representation here. And as mentioned earlier, the adjacency matrix is symmetric because it is an undirected graph. INF means infinity and is the weight of the vertices that are not connected. Ignore the blue edges for now.
Kruskal's algorithm goes like this:
• List out the edges of the graph.
• For each edge, make a set containing only the edge. If there are n edges then, there will be n sets(partitions) at this step.
• Take an edge with smallest weight value and check if the vertices connected by it are in same set. If not, union them, else continue with the next smallest edge.
• Repeat till all vertices are contained in a single set.
## Code
Lets directly go into the code for the implementation is very very easy.
# number of vertices
# now get the edges
# an edge is a tuple (vertex 1, vertex 2, weight)
edges = [(i, j, adjacency[i][j]) for i in range(v) for j in range(i, v) if adjacency[i][j]!=INF]
# sort edges
# because we need to take out smallest edges first.
edges.sort(key=lambda x: x[2])
# CREATE a disjoint set datastructure
# The constructor will automaticall create separate set for each vertex
disset = DisjointSet(v) # v is number of vertices
c = 0 # counter
wts = 0 # total weight of the final spanning treee
final_edges = []
while c < len(edges):
edge = edges[c]
i,j, w = edge[0], edge[1], edge[2]
if disset.find(i) != disset.find(j):
wts+=w
disset.union(i, j)
final_edges.append(edge)
c+=1
print(final_edges)
Running the above code for the graph shown above, we get the output: [(1, 3, 1), (0, 2, 2), (1, 2, 4)] which are the blue edges in the figure.
That's all about the Kruskal's Algorithm. Easy, ain't it?? | 2019-12-11 21:35:36 | {"extraction_info": {"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, "math_score": 0.49009597301483154, "perplexity": 1099.8079958521114}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-51/segments/1575540533401.22/warc/CC-MAIN-20191211212657-20191212000657-00150.warc.gz"} |
https://math.stackexchange.com/questions/3188169/splitting-field-of-sqrt-vphantom-sum1-sqrt2-and-galois-group | # Splitting field of $\sqrt{\vphantom{\sum}1+{\sqrt2}}$ and Galois group
Let $$\alpha= \sqrt{\vphantom{\sum}1+{\sqrt2}}$$.
(a) Let $$p(x)$$ be the minimal polynomial of $$\alpha$$. Find $$p(x)$$. Let K be the splitting field of $$p(x)$$
(b)Let $$E= \mathbb{Q}(i, \sqrt2)$$ Show that $$E \subseteq K$$
(c) Show that $$K=E\sqrt{\vphantom{\sum}1+{\sqrt2}}$$
(d) Show that there exists $$j:E \longrightarrow E$$ such that $$j(i)=-i$$ and $$j(\sqrt2)=-(\sqrt2)$$
(e) Show that there exists $$f:K \longrightarrow K$$ which extends $$j$$ such that $$f(\sqrt{\vphantom{\sum}1+{\sqrt2}})=\sqrt{\vphantom{\sum}1-{\sqrt2}}$$
(f)Show that $$f$$ has order 4
(g) Let $$g \in$$ Gal$$(K/ \mathbb{Q})$$ be a complex conjugation. Then show that $$fg=gf^3$$
(h) Conclude that Gal $$(K/ \mathbb{Q})$$ is isomorphic to $$D_8$$
Attempt
Finding the minimal polynomial is the only easy part here $$(x^2 -1)^2 -2= x^4 -2x^2 -1$$
Also I calculated the splitting field which I got to be the Galois closure as well which is done by computing $$x=-i\sqrt{\vphantom{\sum}-1+{\sqrt2}}$$
$$x=i\sqrt{\vphantom{\sum}-1+{\sqrt2}}$$
$$x=-\sqrt{\vphantom{\sum}1+{\sqrt2}}$$
$$x=\sqrt{\vphantom{\sum}1+{\sqrt2}}$$
So we just get $$\mathbb{Q}(i,\sqrt{\vphantom{\sum}1+{\sqrt2}})$$
(h) I have no idea how to proceed but I have found the following facts in the dummit and foote
1) The Galois group of the splitting field of an irreducible polynomial f acts transitively on the roots of f
2)The order of the Galois group of the splitting field of an irreducible polynomial f is divisible by degree of f
Not sure how I'm supposed to use these hints
(b) Look at this it's obvious to see that the 'elements' that K and E have in common is $$i$$ but what is the mechanism for showing such? I'm not sure how to proceed here.
(c) For this part is it okay to just take $$\mathbb{Q}(i,\sqrt{\vphantom{\sum}1+{\sqrt2}})$$ = $$\mathbb{Q}(i,\sqrt2)$$ x $$\sqrt{\vphantom{\sum}1+{\sqrt2}})$$ and proceed by normal multiplication?
(f) It's not hard to see from the splitting field $$x^4 -2x^2 -1$$ that it is irreducible over $$\mathbb{Q}$$ and the degree of the extension is 4.
for (d) and (e) i'm drawing a blank but i'm thinking they are related and somehow the minimal polynomial of $$\sqrt{\vphantom{\sum}1+{\sqrt2}})$$ and $$\sqrt{\vphantom{\sum}1-{\sqrt2}})$$ are some how related by the function $$j$$ but I just can't quite see it.
Please note I don't have much experience in this field and I'm currently using the dummit and foote.
Can anyone assist me.
• If the hints don't help you, you can look at the tower of quadratic extensions $\Bbb{Q}(ai,b\sqrt{2},c\sqrt{1+b\sqrt{2}})/\Bbb{Q}(ai,b\sqrt{2})/\Bbb{Q}(ai)/\Bbb{Q}$ where $a = \pm 1,b = \pm 1,c=\pm 1$ and see how the order $2$ Galois groups affect $a,b,c$. – reuns Apr 15 at 3:31
(b) take products and sums of the generators in $$K$$ to get elements in $$E$$. E.g $$\sqrt{1 + \sqrt{2}}$$ is contained in $$K$$, so $$1 + \sqrt{2}$$ is. Similarly $$1 - \sqrt{2}$$ is. Subtracting these two elements $$1$$ and $$\sqrt{2}$$ are both contained in $$K$$.
(c) When you say $$E \sqrt{1 + \sqrt{2}}$$ you must mean $$L = E ( \sqrt{ 1 + \sqrt{2}})$$. Since $$E \subset K$$, we can use degrees here to show they are the same (if a finite field extension is contained in another finite field extension of the same degree then they are the same. $$L/E$$ has degree $$2$$ and $$E/\mathbb{Q}$$ has degree $$4$$, as $$E / \mathbb{Q}(\sqrt{2})$$ has degree $$2$$ and $$\mathbb{Q}(\sqrt{2}) / \mathbb{Q}$$ has degree $$2$$. Now show that $$K$$ has degree $$8$$ using the tower $$K / \mathbb{Q}(\sqrt{1 + \sqrt{2}})/\mathbb{Q}(\sqrt{2}) / \mathbb{Q}$$.
(d) For a simple extension $$L(a)/L$$ of fields, with $$a$$ having minimal polynomial $$f(x) \in L[x]$$, there is always an automorphism permuting any two roots of $$f$$ contained in $$L[x]$$. To get one of the desired maps in the problem, take the map $$\mathbb{Q}(\sqrt{2}) \rightarrow \mathbb{Q}(\sqrt{2})$$ sending the root $$\sqrt{2}$$ of the minimal polynomial $$x^2 - 2$$ to the other root $$-\sqrt{2}$$ of the minimal polynoimal. This gives an automorphism of $$\mathbb{Q}(\sqrt{2})$$. This gives a map $$\mathbb{Q}(\sqrt{2}) \rightarrow \mathbb{Q}(\sqrt{2}, i)$$, which extends to a map $$\mathbb{Q}(\sqrt{2}, i)$$ sending $$i$$ to $$i$$. The reason this extension exists is that $$\mathbb{Q}(\sqrt{2}, i) / \mathbb{Q}(\sqrt{2})$$ is a simple extension generated by $$i$$, with minimal polynomial $$x^2 + 1$$, and $$x^2 + 1$$ has a root in the target of the embedding $$\mathbb{Q}(\sqrt{2}) \rightarrow \mathbb{Q}(\sqrt{2}, i)$$.
So the two general facts we used were
1. Let $$L(a)/L$$ be a finite simple extension of fields. For any root $$b$$ of the minimal polynomial $$f$$ in $$L(a) / L$$, there is an $$L$$-automorphism of $$L(a)$$ sending $$a$$ to $$b$$.
2. Let $$L(a)$$ be a finite simple extension of fields. If $$M/L$$ contains a root $$b$$ of the minimal polynoimal $$f(x)$$ of $$a$$ in $$L[x]$$, then there is an embedding $$L(a) \rightarrow M$$ sending $$a$$ to $$b$$.
(e) this uses the same methods discussed in the last one. Here $$K/E$$ is a simple extension with $$a = \sqrt{1 + \sqrt{2}}$$ its generator. To proceed, calculate the minimal polynomial $$f(x)$$ of $$a$$ over $$E$$. There is an $$E$$-automorphism of $$K$$ sending $$a$$ to the other root of $$f(x)$$ in $$K$$.
(f) you got it!
(g) $$fg$$ and $$gf^3$$ are determined by what they do to the roots of certain polynomials. This is nice for us, since we have identified finitely many of these, and automorphisms send roots to other roots. So it is simply a matter of evaluating what they do to the roots. E.g. $$f g ( \sqrt{1 + \sqrt{2} } ) = f ( \sqrt{1 + \sqrt{2}} ) = \sqrt{1 - \sqrt{2}}$$. Do all the calculations of this kind, observe that they are the same for $$fg$$ and $$gf^3$$, and you'll be done.
When we found roots of the minimal polynomials and observed how each automorphism permutes them, we have really reduced an "intractible" question of automorphisms of fields to a tractible question of a finite group acting on the roots; automorphisms are determined by how they act on the roots of certain polynomials, and they permute those roots. In this way, instead of automorphisms acting on a field we have automorphisms acting on a finite set. The second should seem a lot easier to answer, since it is about something finite- in the worst case, we have a brute force approach.
(h) This is potentially the most tricky part, but it's also where all this computation has been leading us. $$D_8$$ is given by generators and relations as $$\langle x, y | x^4, y^2, xy = yx^3 \rangle$$. To give an isomorphism, let's start with a map from the free group $$\langle x, y \rangle$$ to the galois group in question sending $$x$$ to $$f$$ and $$y$$ to $$g$$. We must show a few things to establish an isomorphism:
1. $$x^4$$ is sent to $$\text{Id}$$, and so are $$y^2$$ and $$x^{-3}y^{-1}xy$$. You showed the last one already, but the first two are not so different!
2. All automorphisms in the galois group are some combination of $$f$$ and $$g$$.
3. The order of the galois group is $$8$$.
I have to go for now, but I hope this is enough. | 2019-04-25 19:41:32 | {"extraction_info": {"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": 123, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9043008685112, "perplexity": 113.5352660028851}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-18/segments/1555578733077.68/warc/CC-MAIN-20190425193912-20190425215912-00252.warc.gz"} |
https://stats.stackexchange.com/questions/177252/how-to-compare-two-different-rankings-of-same-set-of-objects | # How to compare two different rankings of same set of objects?
I have $N$ objects $A_1,\dots,A_N$. I have two algorithms that assign two different scores to each of this objects. Thus sorting on this scores gives me two different rankings of these same objects. Now how do I measure the similarity (or dissimilarity) of these different rankings in a visually understandable manner. Are their known approaches to this?
• A Spearman correlation would capture the ordinality in the agreement of the rankings. Visualizing that relationship could be as simple as creating deciles for each set of scores and cross-classifying them (i.e., a crosstab). Next, by highlighting with different colors the agreement on the diagonal versus the diagonal plus one or two deciles off and summarizing the percentages of each with a few numbers should communicate what you need. – Mike Hunter Oct 16 '15 at 13:20
• What is the purpose of the comparison? Are these rankings according to different criteria, or are they supposed to use the same criterion, but made by different judges? ... Probably these matters. – kjetil b halvorsen Oct 16 '15 at 13:20
• These rankings are according to different criterion, but there should be certain natural things I expect in them. For instance, certain objects are really good and I naturally expect to them to score higher. However Algorithm 1's criterion doesn't reflect this whereas Algorithm 2 does. I do have a rigorous explanation of this effect. However, I don't have a nice way of putting the numbers. – dineshdileep Oct 16 '15 at 13:44 | 2020-06-06 05:29:52 | {"extraction_info": {"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, "math_score": 0.5269224047660828, "perplexity": 770.8716122161311}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-24/segments/1590348509972.80/warc/CC-MAIN-20200606031557-20200606061557-00191.warc.gz"} |
https://stats.stackexchange.com/questions/360639/computing-the-cdf-of-the-minimum-of-particular-dependent-random-variables | # Computing the CDF of the minimum of particular dependent random variables
For each $i=1,\dots,n$ let $Z_i\sim\text{Poisson}(\lambda_i)$, and suppose $\{Z_i\}$ are independent. Also for each $i=1,\dots,n$, let $\{Y_{ij}\}_{j\in\mathbb{N}}$ be an infinite sequence of iid random variables, where $Y_{ij}\sim\text{Bernoulli}(p_i)$. Define
$$X_1=\sum_{i=1}^n\sum_{j=1}^{Z_i}Y_{ij}\quad\text{and}\quad{X_2}=\sum_{i=1}^n\nu_iZ_i$$
for some strictly positive scalars $\nu_i$. It's easy to show that
$$X_1\sim\text{Poisson}\bigg(\sum_{i=1}^n\lambda_ip_i\bigg),$$
but that $X_2$ does not follow any well-known distribution (as a linear combination of Poisson random variables). How can I numerically compute $\mathbb{P}[\min\{X_1-X_2-3,X_1-2X_2\}\leqslant0]$ without using Monte Carlo?
What I have
For fixed scalars $\alpha,\beta\in\mathbb{R}_+$, I computed the characteristic function of $X_1-\alpha X_2-\beta$ to be
$$\varphi(t)=\exp\bigg[-\iota\beta{t}+\sum_{i=1}^n\lambda_ip_i\big(\varphi_i(t)-1\big)\bigg]$$
where
$$\varphi_i(t)=(1-p_i)e^{-\iota\alpha\nu_it}+p_ie^{\iota(1-\alpha\nu_i)t}.$$
for $\iota=\sqrt{-1}$. I can then use the Gil-Pelaez inversion formula to numerically calculate $\mathbb{P}[X_1-\alpha{X_2}-\beta\leqslant0]$ (this works quite well in practice). To calculate the minimum, I need to work out something about the joint distribution, which is where I'm stuck.
• Hvae you tried computing $E[e^{i(t_1X_1+t_2X_2)}]$? By inverting this you'll get $P(X_1\leq a,X_2\leq b)$, and then just use inclusion-exclusion to calculate the minimum. – Alex R. Aug 8 '18 at 0:22
• @AlexR. That would be the characteristic function of the vector-valued random variable $X=(X_1,X_2)$, right? Is there an obvious or well-known extension of the GP inversion formula to the vector-valued case? – David M. Aug 8 '18 at 0:41
• Is this a homework question? – Sextus Empiricus Aug 8 '18 at 6:23
• @MartijnWeterings No, it’s a stylized version of a more complex problem I encountered. If you just want to give a hint or tip, I would appreciate that too – David M. Aug 8 '18 at 12:11
• Not that it helps much but we could write the problem more compact (and somewhat more general) as a sum of random vectors: $$R_i \sim \text{Poisson}(\lambda_i)$$ $$\vec{V} = \begin{bmatrix} V_1 \\ V_2 \end{bmatrix} = \sum_{i=1}^n \begin{bmatrix} \nu_i \\ \xi_i \end{bmatrix} R_i$$ compute $P[V_1 \leq a \text{ or } V_2 \leq b] \equiv 1-P[V_1>a \text{ and } V_2 >b]$. I guess that you should be able to express the characteristic function for $\vec{V}$ and then use some inversion formula to perform the numeric computation. – Sextus Empiricus Aug 9 '18 at 9:54 | 2021-05-06 22:48:51 | {"extraction_info": {"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, "math_score": 0.9593656659126282, "perplexity": 305.89571817410007}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-21/segments/1620243988763.83/warc/CC-MAIN-20210506205251-20210506235251-00579.warc.gz"} |
http://ywov.siberianhuskyadozioni.it/lesson-2-fun-with-functions-linear-ones-answer-key.html | # Lesson 2 Fun With Functions Linear Ones Answer Key
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They are 1) substitution, and 2) elimination. Card sort answer key, one per student Assessments 1. Show which of these possibilities is the case by successively transforming the given equation into simpler forms, until an equivalent equation of the form x = a, a = a, or a = b results (where a and b are different numbers). Defining Functions. Each input has exactly one output. Common Core Algebra I Unit 8 Lesson 1 Introduction To Quadratic Functions. Great Valley School District is an equal opportunity education institution and will not discriminate on the basis of actual or perceived race, color, age, creed religion, gender, sexual orientation, gender identity, gender expression, ancestry, national origin, marital status, pregnancy or handicap/disability in its programs and activities, or employment practices as required by Title VI. Lesson Plan. The Chapter 4 Resource Mastersincludes the core materials needed for Chapter 4. Graph the equation 3 x + 2 y = 6 by using the x and y. Some of the worksheets for this concept are Graphing linear inequalities in two variables, Grades mmaise salt lake city, Unit 4 analyze and graph linear equations functions and, 3 graphing linear functions, Algebra 1 standard lesson plan overview, Work 2 2. 3: Expressing a Number in Many Ways; Lesson 2. Be sure to example if there are no solutions, one solution or infinite solutions. Graphically, where the line crosses the $x$-axis, is called a zero, or root. Common Core Algebra I Unit 8 Lesson 5 Stretching Parabolas And Completing The Square. $y = 2{x^2} - 5x + 3$ Using function notation, we can write this as any of the following. These units are very broad, and it would be impossible to cover all aspects of these topics given space constraints of this book. You must get 5 in a row correct. Practice and Problem Solving: D 1. Algebra 2 Common Core answers to Chapter 2 - Functions, Equations, and Graphs - 2-3 Linear Functions and Slope-Intercept Form - Lesson Check - Page 78 2 including work step by step written by community members like you. Lesson 1 Homework- Slope Solutions. Solving Systems of Linear Equations Make this Foldable to record information about solving systems. com and read and learn about trigonometry, multiplying and dividing fractions and several additional algebra topics. Download free printable worksheets Linear Equations pdf of CBSE and kendriya vidyalaya Schools as per latest syllabus in pdf, CBSE Class 8 Mathematics Worksheet - Linear Equations in One Variable - Practice worksheets for CBSE students. Questions then ask students to sort the graphs based on the form of the equations. Textbook Authors: Hall, Prentice, ISBN-10: 0133186024, ISBN-13: 978--13318-602-4, Publisher: Prentice Hall. If ever you have to have service with math and in particular with algebra 2 plato answers or solving linear equations come visit us at Polymathlove. In this lesson students learn to solve two-step equations using inverse operations and a graphic organizer. Worksheets are Work, Writing linear equations, Notes linear nonlinear functions, Linear or nonlinear 1, Function table t1l1s1, Graphing lines, Comparing linear and nonlinear functions, Real world applications of linear equations. Access lesson materials for Free Gizmos. Linear Functions and Systems (Algebra 2 Curriculum - Unit 2)This bundle includes notes, homework assignments, two quizzes, a study guide and a unit test that cover the following topics:• Domain and Range of a Relation• Relations vs. Standard: A1. Complete the following exercise. Find a Function - Find an example of a function in the media. Lesson 5 – Introduction to Exponential Functions Mini-Lesson Page 172 This next example is long but will illustrate the key difference between EXPONENTIAL FUNCTIONS and LINEAR FUNCTIONS. Use polynomial expressions and functions to model and to answer questions about data patterns and graphs that cannot be represented with linear, quadratic, inverse variation, or exponential functions As they work through the Problems of this Unit, it will be helpful to keep asking questions such as:. MOTION WORKSHEET 2 ANSWERS PDF. 2 (Part 1) Solve One Variable Equations - Lesson 7. Unit 2 Linear Expressions, Equations & Inequalities. Students graph linear, quadratic, absolute value, and square root functions using transformations. Transforming Linear Functions Write the correct answer. y = 2x − 4 Equation 1 7x − 2y = 5 Equation 2 Step 1: Equation 1 is already solved for y. If ever you have to have service with math and in particular with algebra 2 plato answers or solving linear equations come visit us at Polymathlove. The slope of a line is the change of x Sample answer: No, it is not a function because one of the elements of the domain, 3, is paired with two. Lesson 1 Homework- Slope Solutions. Answer to Question 3 is ”C”. These guided notes include solving 1-step equations, 2- step equations, multi-step equations and equations with variable on both sides. Free Gizmos change each semester The next change will be on July 1, 2020. 4 Linear Functions and Models 95 Solution (a) If we let represent the value of each car after x years,then represents the original value of each car, so The y-intercept of the linear function is$28,000. FSA Mathematics Practice Test Answer Key Session 2 8. For any number of quarts, there is only one possible number of pints. The points on the line are solutions of the equation. The first worksheet is 2 step equations, the harder one involves brackets and variables on both sides. Idea here is to express one variable using the other variable in one equation, and use it in the second equation, where we would get a linear equation with one variable. solving equations maze worksheet. Functions can be classified in two different categories: linear or nonlinear. 1a Show that linear functions grow by equal differences over equal intervals and that exponential functions grow by equal factors over equal intervals. 7 Solve a simple system consisting of a linear equation and a quadratic equation in two variables algebraically and graphically. Any quadratic expression can be written as a perfect square by a method called completing the square. A linear inequality divides the coordinate plane into two halves by a boundary line where one half represents the solutions of the inequality. Common Core Algebra I Unit 8 Lesson 1 Introduction To Quadratic Functions. help is an online resource designed to help people learn algebra. Go back 3 4. write the equations of circular motion analogous to the equations of linear motion. 2 (Part 1) Solve One Variable Equations - Lesson 7. Textbook Module 2: Exploring Constant Change, Textbook Topic 2: Solving Linear Equations and Inequalities, Textbook Lesson 1: Strike a Balance:. 2 Multiple Choice Identify the choice that best completes the statement or answers the question. 1 – Select and/or use appropriate strategies to solve or represent number sentences. Did you enjoy the activity? You have seen that in a linear function, equal differences in x produce equal differences in y. Come to Algebra-equation. This quiz is about rational function graphs and their specific components. In particular, the lesson will help you identify and help students who have the following difficulties: • Solving equations with one variable. Find the amplitude and period of the function. Worksheets for each activity 3. solving equations maze dodge the monsters. Kinematics (Description of Motion) Problems. Tom Carol, NY This program laid the foundation for the most successful step-by-step solution to teaching algebra that Ive ever seen or had the pleasure of implementing inside the classroom. For this card, write two equations. LESSON 5-2. Unit 2 Review (lessons 1-7) Answers to Unit 2 Review (lessons 1-7) HW pages 81-82 Answers. Algebra 2 -25 - Functions, Equations, and Graphs WARM UP Solve each equation for y. Plug 17 in for x in the original equation. Put the table #1 transparency on the overhead and ask students to copy it on their graph paper. 5 Interpret the equation y = mx + b as defining a linear function, whose graph is a straight line; give examples of functions that are not linear. • MGSE9-12. In earlier modules, students analyze the process of solving equations and developing fluency in writing, interpreting, and translating between various forms of linear equations (Module 1) and linear and exponential functions (Module 3). Notice that the graph of y = sin 2 x is compressed horizontally and one period is completed in half the original period. Exercises 10 2. 1 Ready Answer Key f. Graph Linear Equations The graph of a linear equations represents all the solutions of the equation. HS-AT-S-EI17 Students will write and solve linear sentences, describing real-world. Go back 4 5. | 2020-08-04 13:28:58 | {"extraction_info": {"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": 3, "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, "math_score": 0.2532587945461273, "perplexity": 1232.4982558725762}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-34/segments/1596439735867.94/warc/CC-MAIN-20200804131928-20200804161928-00066.warc.gz"} |
https://brilliant.org/problems/an-electricity-and-magnetism-problem-by-nishant-3/ | # Moving wire over a Square Frame!
Electricity and Magnetism Level 4
$$ABCD$$ is a square frame made from different wires of same length and each having different uniform resistance per unit length. Resistances of wires forming sides $$AB, BC, CD$$ and $$DA$$ are $$100W, 400 W, 500 W$$ and $$200 W$$ respectively. An ideal cell is connected across $$B$$ and $$D$$. A straight conducting wire containing a resistance and a galvanometer in series starts rotating about pivoted point $$A$$ from initial position as shown with uniform angular velocity $$\omega = \frac{\pi}{360} rad/sec$$. One end of the straight wire (rotating) is pivoted at $$A$$ and other end always in sliding contact with a side of the square. The time (in seconds) after start when galvanometer shows zero deflection is?
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http://ljosidrafverktakar.is/y7xql75/how-to-find-equivalent-expressions-with-exponents-aba920 | # how to find equivalent expressions with exponents
From simplify exponential expressions calculator to division, we have got every aspect covered. It is often simpler to work directly from the definition and meaning of exponents… You should learn to find the lowest common denominator. First, the Laws of Exponentstell us how to handle exponents when we multiply: So let us try that with fractional exponents: What is another way to write 5 3 Franklin Bradley, AK Powers Complex Examples. 9^1=3^2. On the left-hand side above is the exponential statement "y = b x".On the right-hand side above, "log b (y) = x" is the equivalent logarithmic statement, which is pronounced "log-base-b of y equals x"; The value of the subscripted "b" is "the base of the logarithm", just as b is the base in the exponential expression … Order of Operations Factors & Primes Fractions Long Arithmetic Decimals Exponents & Radicals Ratios & Proportions Percent Modulo Mean, Median & Mode Scientific Notation Arithmetics Algebra Equations Inequalities System of Equations System of Inequalities Basic Operations Algebraic Properties Partial Fractions Polynomials Rational Expressions Sequences Power Sums Induction Logical Sets Convert expressions with rational exponents to their radical equivalent Square roots are most often written using a radical sign, like this, $\sqrt{4}$. Then I bought this software. b 4 = b × b × b × b. Exponents. Some examples are 8^3=2^9,8^6=4^9,8^1=2^3, and 8^2=4^3. There are five types of problems in this exercise: 1. 8^3=2^9. Linda Taylor, KY. My daughter is in 10th grade and son is in 7th grade. The user is asked to fill in the blanks with the correct exponent to make true statements. 8^2=4^3. There are 56 combinations of answers that include one to the anything equals something to the zero. Exponents are also called Powers or Indices. Exponents are also called powers or indices. Search. Any time you might need advice with math and in particular with equivalent expression calculator or inverse functions come visit us at Solve-variable.com. For example, 72 1 = 72; 72 0 = 1. Stage Design - A Discussion between Industry Professionals. All rights reserved. Because raising a power to a power means that you multiply exponents (as long as the bases are the same), you can simplify the following expressions: Leave the exponent as … In 8 2 the "2" says to use 8 twice in a multiplication, so 8 2 = 8 × 8 = 64. To simplify an expression with a negative exponent, you just flip the base number and exponent to the bottom of a fraction with a on top. How can you use the fact that anything to the zero power equals one? In the event you actually seek assistance with math and in particular with Algebra Exponents Calculator or solution come pay a visit to us at Mathfraction.com. Get Free Negative Exponents Lesson 5 now and use Negative Exponents Lesson 5 immediately to get % off or $off or free shipping. The LCD = 6x^2y^2 2) Your next step is to convert each fraction to an equivalent fraction where the denominator is the LCD. 4 5 = 4 × 4 × 4 × 4 × 4 = 1,024. Come to Algebra-equation.com and read and learn about operations, mathematics and … For example, 32 3–5 = 3– 3 = 1/33 = 1/27. Basic Simplifying With Neg. When you do see an exponent that is a decimal, you need to convert the decimal to a fraction. Improve your math knowledge with free questions in "Identify equivalent expressions involving exponents I" and thousands of other math skills. To log in and use all the features of Khan Academy, please enable JavaScript in your browser. Use this worksheet and quiz to find out how well you understand the properties of exponents and equivalent expressions. $$5\cdot 5=5^{2}$$ An expression that represents repeated multiplication of the same factor is called a power. Some of the worksheets displayed are Sample work from, Evaluating expressions date period, A guide to exponents, Radicals and rational exponents, Fractional exponents and radical expressions, Using order of operations, Name date, Exponents es1. Algebraic expressions (6th grade) Read and write equivalent expressions with variables and exponents An updated version of this instructional video is available. When raising a power to a power, you multiply the exponents, but the bases have to be the same. 8^1=2^3. An exponent, also called a power, is a compressed way of expressing repeated multiplications of a number by itself. Source: Annie DeAngelo and Maeve O’Connell. Certified Information Systems Security Professional (CISSP) Remil ilmi. Complete the equation by putting it back together. I find the program to be of such great use! Like, the property a^n • a^m = a^ (m+n) lets you simplify (3x^2y^3) (2x^5y^6) to … Showing top 8 worksheets in the category - Equivalent Expressions Using Exponents. I used to spend hours teaching them arithmetic, equations and algebraic expressions. The exponent of a number says how many times to use the number in a multiplication.. Let's see why in an example. Directions: Using the digits 0-9 only once each, create as many true equations as possible. Your email address will not be published. Purplemath. Apply exponent rules: This problem provides a set of three expressions that can be simplified via exponent rules. We read am as a raised to the power m or just a to the power m. Problem 1: Students learn how to compute using integer exponents building on their earlier experiences with adding and subtracting integers. Source: Annie DeAngelo and Maeve O’Connell, Tags 6.EE.1 Annie DeAngelo DOK 2 Maeve O'Connell, Directions: Using the digits 1 to 9 at most once each, fill in the boxes …. Arithmetic w/ Polynomials & Rational Expressions, Reasoning with Equations and Inequalities, Linear, Quadratic, and Exponential Models, Similarity, Right Triangles, and Trigonometry, Expressing Geometric Properties with Equations, Interpreting Categorical and Quantitative Data, Making Inferences and Justifying Conclusions, Conditional Probability & the Rules of Probability, Comparing Hundredths and Tenths 2 Open Middle, Comparing Hundredths and Tenths 1 Open Middle, Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License. Improve your math knowledge with free questions in "Identify equivalent expressions involving exponents II" and thousands of other math skills. Open Middle is the registered trademark of the Open Middle Partnership. Khan Academy is a 501(c)(3) nonprofit organization. If you're seeing this message, it means we're having trouble loading external resources on our website. Rarely do you see them as decimals. As one example which you probably already know, they are used to simplify or reduce expressions. Donate or volunteer today! Thank you! a m = a × a × a × a…m times. You can write the two expressions to show equality: 2x (3y + 2) = 6xy + 4x. This means taking the two new numbers and keeping the function in the middle the same: 6xy + 4x. We know how to calculate the expression 5 x 5. Porque seria el mismo dijito sumado porque como si lo miras biennes el mismo dijito porque algunos son ejemplos, Your email address will not be published. Recognize the pattern: This problem has a table with various powers of a particular base. Usually you see exponents as whole numbers, and sometimes you see them as fractions. The number a is known as base and m is said to be exponent and am is said to be the exponent form of the number. © 2016-2020 Open Middle Partnership. Then, there are a number of rules and laws regarding exponents you can use to calculate the expression. Writing negative exponents as fractions will make it easier for you to understand how to work with them in an equation. Save my name, email, and website in this browser for the next time I comment. integer exponents to generate equivalent numerical expressions. Simplifying Exponent Expressions. Since order doesnt matter for multiplication you will often find that you and a friend or you and the teacher have worked out the same problem with completely different steps but have gotten the same answer in the end. Some more examples: Play this game to review Pre-algebra. 9^4=3^8. We have got a large amount of excellent reference information on subjects starting from syllabus to equations and inequalities To simplify with exponents, don't feel like you have to work only with, or straight from, the rules for exponents. Any number raised to the power of 1 keeps the same value; any number with an exponent of 0 is equal to 1. Required fields are marked *. 9^2=3^4. Aside from that, I have found. This is your equivalent expression. What is another way to write 5 3 8^6=4^9. Examples, solutions, videos, worksheets, and activities to help Algebra 1 students learn how to simplify expressions with exponents. This expression can be written in a shorter way using something called exponents. Rewriting exponential expressions as A⋅Bᵗ, Practice: Rewrite exponential expressions, Equivalent forms of exponential expressions, Practice: Equivalent forms of exponential expressions, Solving exponential equations using properties of exponents. But there is another way to represent them. By filling in the values the user will be able to find a pattern and place it in the provided space. We carry a lot of excellent reference information on subject areas varying from algebra ii to subtracting rational Write the expression: This problem has a single exponential expression that … Note that rational exponents are subject to all of the same rules as other exponents when they appear in algebraic expressions. Learn how to simplify expressions using the power rule and the negative exponent rule of exponents. 5 3 = 5 × 5 × 5. In words: 8 2 could be called "8 to the power 2" or "8 to the second power", or simply "8 squared" . Polymathlove.com provides insightful advice on Equivalent Expressions Calculator, operations and adding and subtracting rational expressions and other math topics. Now this algebra tutor teaches my children and they are improving at a better pace. 2. Our mission is to provide a free, world-class education to anyone, anywhere. The following diagram shows the law of exponents: product, quotient, power, zero exponent and negative exponent. If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. The number 5 is called the base, and the number 2 is called the exponent. Just in case you have to have assistance on adding fractions or value, Polymathlove.com is the ideal site to pay a visit to! 3. Play this game to review Pre-algebra. How To Find Equivalent Expressions With Exponents To simplify with exponents dont feel like you have to work only with or straight from the rules for exponents. Brought to you by Sciencing. Equivalent Expressions Using Exponents. 6.EE.1 Annie DeAngelo DOK 2 Maeve O'Connell 2020-03-19. We do this by multiplying both the numerator and denominator by the same value. Equivalent forms of exponential expressions Equivalent forms of exponential expressions CCSS.Math: HSA.SSE.A.2 , HSN.RN.A.2 , HSN.RN.A , or straight from, the rules for exponents and denominator by the same value ; number. To an equivalent fraction where the denominator is the ideal site to pay a visit to n't feel like have! Visit us at Solve-variable.com as one example which you probably already know, they are improving a. 4 × 4 how to find equivalent expressions with exponents 4 × 4 × 4 = 1,024 certified Information Systems Professional. A pattern and place it in the blanks with the correct exponent to make true statements you multiply the,. An expression that represents repeated multiplication of the open Middle is the LCD = 2!, anywhere to 1 problem provides a set of three expressions that can be written in a shorter using! Hsn.Rn.A 8^1=2^3 the ideal site to pay a visit to exponent of 0 is equal to 1 and... Trouble loading external resources on our website world-class education to anyone, anywhere ) = +. Expressions calculator to division, we have got every aspect covered the exponent... Once each, create as many true equations as possible, we have got every aspect covered to find pattern... See exponents as fractions will make it easier for you to understand how to simplify expressions using the digits only. Anything to the power of 1 keeps the same: 6xy + 4x learn to find the lowest denominator!.Kasandbox.Org are unblocked i used to spend hours teaching them arithmetic, equations and algebraic expressions:! That can be written in a shorter way using something called exponents the registered trademark of open. And keeping the function in the Middle the same it means we 're having trouble loading external resources our... In algebraic expressions web filter, please enable JavaScript in Your browser are! Include one to the zero = 6xy + 4x value ; any number with an exponent of number. Enable JavaScript in Your browser for you to understand how to simplify expressions with variables and exponents an version..., worksheets, and sometimes you see them as fractions convert each fraction to an equivalent fraction where denominator. Use the fact that anything to the anything equals something to the power of 1 keeps the factor. Many true equations as possible expressions ( 6th grade ) Read and write equivalent expressions using the power rule the! You multiply the exponents, but the bases have to work only,... Exponent and negative exponent category - equivalent expressions with variables and exponents an updated version this. The exponent of 0 is equal to 1 same: 6xy + 4x multiply the exponents, do n't like! In 10th grade and son is in 7th grade you see them as fractions will make it easier for to! Them arithmetic, equations and algebraic expressions as one example which you already... Following diagram shows the law of exponents fractions will make it easier for you to understand to! The negative exponent decimal, you need to convert each fraction to an fraction. Exponents an updated version of this instructional video is available - equivalent expressions with exponents we got. And write equivalent expressions with variables and exponents an updated version of this instructional is... Denominator is the ideal site to pay a visit to fact that anything to anything! Filling in the provided space equals something to the power rule and the negative exponent rule of.... Calculator or inverse functions come visit us at Solve-variable.com the numerator and denominator the... Expressions equivalent forms of exponential expressions equivalent forms of exponential expressions calculator to division, have! Fill in the category - equivalent expressions with variables and exponents an updated version this... Algebraic expressions how can you use the number 5 is called the exponent of 0 is equal to.! To the zero power equals one 6x^2y^2 2 ) = 6xy +.. Keeping the function in the Middle the same rules as other exponents they. Factor is called the base, and activities to help Algebra 1 students learn how to simplify with.... Can write the two expressions to show equality: 2x ( 3y 2. Mission is to convert the decimal to a power, you need to convert the decimal to a power a. Examples, solutions, videos, worksheets, and sometimes you see them as fractions Middle the:. Use all the features of Khan Academy, please make sure that the *. Math and in particular with equivalent expression calculator or inverse functions come visit us Solve-variable.com! I comment external resources on our website value ; any number with an exponent that is a,. Us at Solve-variable.com by the same value 5\cdot 5=5^ { 2 }$ \$ 5\cdot 5=5^ 2... As whole numbers, and the number in a shorter way using something called.... And website in this browser for the next time i comment 3– 3 = =. And in particular with equivalent expression calculator or inverse functions come visit at! To pay a visit to you see them as fractions O ’.. We 're having trouble loading external resources on our website the same rules as other exponents when appear! Step is to convert the decimal to a fraction each fraction to an equivalent fraction where the is... With various powers of a particular base c ) ( 3 ) nonprofit organization 're having trouble loading resources! Fraction to an equivalent fraction where the denominator is the registered trademark of the open Middle is the ideal to. Middle Partnership find the lowest common denominator only once each, create as many true equations as possible the.. Function in the category - equivalent expressions using the digits how to find equivalent expressions with exponents only each... Do see an exponent that is a 501 ( c ) ( )! Have got every aspect covered negative exponents as fractions negative exponents as fractions integer building! 1 students learn how to simplify with exponents, but the bases have be... Video is available open Middle is the ideal site to pay a visit to exponential expressions to... Exponents an updated version of this instructional video is available, HSN.RN.A 8^1=2^3 be in! Updated version of this instructional video is available find a pattern and place it the. Will make it easier for you to understand how to work with them in an equation represents repeated of! The provided space of answers that include one to the power rule and the number 5 is called the,. Visit us at Solve-variable.com exponent rules: this problem provides a set of three expressions that can be simplified exponent. To compute using integer exponents building on their earlier experiences with adding and subtracting integers or reduce expressions can... The exponent Annie DeAngelo and Maeve O ’ Connell denominator by the same rules as other exponents when appear... Sometimes you see them as fractions, 72 1 = 72 ; 72 0 = 1 with exponents our is! Subtracting integers are improving at a better pace sure that the domains *.kastatic.org *! The expression to all of the same value 5 is called the,. The registered trademark of the same: 6xy + 4x b × b × b b..., HSN.RN.A.2, HSN.RN.A 8^1=2^3 and in particular with equivalent expression calculator or inverse functions visit... To convert each fraction to an equivalent fraction where the denominator is the =. = 1/33 = 1/27 next step is to convert each fraction to an fraction! B × b × b expressions to show equality: 2x ( 3y + 2 ) next! Algebra 1 students learn how to work only with, or straight from the! To anyone, anywhere i comment user is asked to fill in the space... Read and write equivalent expressions using exponents 56 combinations of answers that include one the... You see them as fractions will make it easier for you to understand how simplify... An expression that represents repeated multiplication of how to find equivalent expressions with exponents open Middle Partnership know, they used. Base, and activities to help Algebra 1 students learn how to compute using exponents. Expression can be simplified via exponent rules now this Algebra tutor teaches children... Domains *.kastatic.org and *.kasandbox.org are unblocked anything to the zero power equals one is in 10th and! Use the fact that anything to the power of 1 keeps the same value any. Law of exponents: product, quotient, power, zero exponent and exponent... To help Algebra 1 students learn how to simplify expressions with variables and exponents an updated version this... Earlier experiences with adding and subtracting integers are improving at a better pace Professional ( ). Have to be the same value for the next time i comment power, zero exponent and negative exponent two... You should learn to find the lowest common denominator convert each fraction to an equivalent fraction where the is! Be written in a shorter way using something called exponents find the lowest denominator! Expressions to show equality: 2x ( 3y + 2 ) Your next step is to provide a free world-class... Power to a fraction sure that the domains *.kastatic.org and *.kasandbox.org are unblocked of. Via exponent rules 4 5 = 4 × 4 × 4 × 4 =.! One example which you probably already know, they are used to simplify expressions using.... Open Middle Partnership = 4 × 4 = b × b an updated version of instructional! An expression that represents repeated multiplication of the same factor is called a power to fraction. Them arithmetic, equations and algebraic expressions ( 6th grade ) Read and write equivalent expressions with variables and an! Simplify or reduce expressions create as many true equations as possible any time you need... Daughter is in 7th grade.kastatic.org and *.kasandbox.org are unblocked able to find a pattern place.
0 replies | 2021-07-25 10:35:22 | {"extraction_info": {"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": 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, "math_score": 0.5123899579048157, "perplexity": 1170.5095000595218}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-31/segments/1627046151641.83/warc/CC-MAIN-20210725080735-20210725110735-00210.warc.gz"} |
http://www.solutioninn.com/you-want-to-estimate-the-average-sat-score-for-all | # Question
You want to estimate the average SAT score for all students who took the Ethan-Davies SAT Preparation course during the past 2 years. You select a simple random sample of 100 such students from a comprehensive list of all Davies students who took the course over the last two years and find that the average SAT score for the sample was 1940. Assume you know the population standard deviation here is 83 points.
a. Produce the 95% confidence interval estimate of the mean SAT score for the population of Ethan-Davies students.
b. Carefully interpret the interval.
c. Identify the standard error and the margin of error terms in your interval.
Sales0
Views68 | 2016-10-28 02:55:06 | {"extraction_info": {"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, "math_score": 0.8324854969978333, "perplexity": 461.16997836349395}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-44/segments/1476988721555.36/warc/CC-MAIN-20161020183841-00371-ip-10-171-6-4.ec2.internal.warc.gz"} |
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## concentration
Chrystelle Junior | Next Rank: 30 Posts
Joined
07 Dec 2006
Posted:
19 messages
#### concentration
Tue Jan 02, 2007 5:43 am
hi
Does someone have an idea?
1, The concentration of a certain chemical in a full water tank depends on the depth of the water. At a depth that is x feet below the top of the tank, the concentration is parts per million, where 0 < x < 4. To the nearest 0.1 foot, at what depth is the concentration equal to 6 parts per million?
(A) 2.4 ft
(B) 2.5 ft
(C) 2.8 ft
(D) 3.0 ft
(E) 3.2 ft
aim-wsc Legendary Member
Joined
20 Apr 2006
Posted:
2470 messages
Followed by:
14 members
85
Target GMAT Score:
801-
Wed Jan 03, 2007 3:47 am
absolutely stumped.
or maybe i am not been able to see the question some problem with my browser...
you hv not attached any file with it, right?
_________________
Getting started @BTG?
Please do not PM me, (not active anymore) contact Eric.
### GMAT/MBA Expert
Stacey Koprince GMAT Instructor
Joined
27 Dec 2006
Posted:
2228 messages
Followed by:
681 members
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GMAT Score:
780
Sat Jan 06, 2007 3:54 pm
Hi, Chrystelle
I think your transcription of the problem might be missing some info - possibly at the line that says "the concentration is parts per million" - is there supposed to be a value between "is" and "parts"? As written, there is not enough information given in the problem to solve it. Please do post again if you figure out what is missing.
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See More Top Beat The GMAT Experts | 2018-04-24 01:18:06 | {"extraction_info": {"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, "math_score": 0.24075157940387726, "perplexity": 13123.000534357794}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-17/segments/1524125946314.70/warc/CC-MAIN-20180424002843-20180424022843-00636.warc.gz"} |
https://socratic.org/questions/how-do-you-find-the-zeros-of-y-4x-4-11x-2-3 | How do you find the zeros of y=4x^4-11x^2-3?
Sep 22, 2016
$x = \pm \frac{1}{2} i$
$x = \pm \sqrt{3}$
Explanation:
$4 {x}^{4} - 11 {x}^{2} - 3 = 0$ can be factored
$\left(4 {x}^{2} + 1\right) \left({x}^{2} - 3\right) = 0$
So we have
$4 {x}^{2} + 1 = 0$ or ${x}^{2} - 3 = 0$
Proceeding solve each factor for $x$
$4 {x}^{2} = - 1$
${x}^{2} = - \frac{1}{4}$
$x = \pm \sqrt{\frac{1}{4}} i$
$x = \pm \frac{1}{2} i$
${x}^{2} = 3$
$x = \pm \sqrt{3}$ | 2020-01-17 19:24:24 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 13, "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, "math_score": 0.5763767957687378, "perplexity": 2698.1989507091353}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-05/segments/1579250590107.3/warc/CC-MAIN-20200117180950-20200117204950-00185.warc.gz"} |
https://www.physicsforums.com/threads/complex-exponential.301299/ | # Complex exponential
1. Mar 20, 2009
### tommyhakinen
what is e$$^{jw\infty}$$ ?
2. Mar 20, 2009
### WWGD
You cannot define it in the complex plane, since oo is not a complex number. You may
want to work on the Riemann sphere, or something, if you want to work with the log,
tho I don't see how to do so at this point. e^z must be defined in terms of a branch
of logz , to be well-defined, in the sense of being single-valued, among other things.
Since logz is not even defined in the whole complex plane, you cannot define a logz
to be analytic in the extended complex plane | 2018-02-23 00:59:20 | {"extraction_info": {"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, "math_score": 0.891369640827179, "perplexity": 637.8172525768477}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-09/segments/1518891814300.52/warc/CC-MAIN-20180222235935-20180223015935-00283.warc.gz"} |
https://juliet.readthedocs.io/en/latest/user/quicktest.html | Getting started¶
Two ways of using juliet¶
In the spirit of accomodating the code for everyone to use, juliet can be used in two different ways: as an imported library and also in command line mode. Both give rise to the same results because the command line mode simply calls the juliet libraries in a python script.
To use juliet as an imported library, inside any python script you can simply do:
import juliet
dataset = juliet.load(priors = priors, t_lc=times, y_lc=flux, yerr_lc=flux_error)
results = dataset.fit()
In this example, juliet will perform a fit on a lightcurve dataset defined by a dictionary of times times, relative fluxes flux and error on those fluxes flux_error given some prior information priors which, as we will see below, is also defined through a dictionary.
In command line mode, juliet can be used through a simple call in any terminal. To do this, after installing juliet, you can from anywhere in your system simply do:
juliet -flag1 -flag2 --flag3
In this example, juliet is performing a fit using different inputs defined by -flag1, -flag2 and --flag3. There are several flags that can be used to accomodate your juliet runs through command-line which we’ll explore in the tutorials. There is a third way of using juliet, which is by calling the juliet.py code and applying these same flags (as it is currently explained in project’s wiki page). However, no further updates will be done for that method, and the ones defined above should be the preferred ones to use.
A first fit to data with juliet¶
To showcase how juliet works, let us first perform an extremely simple fit to data using juliet as an imported library. We will fit the TESS data of TOI-141 b, which was shown to host a 1-day transiting exoplanet by Espinoza et al. (2019). Let us first load the data corresponding to this object, which is hosted in MAST. For TESS data, juliet has already built-in functions to load the data arrays directly given a web link to the data — let’s load it and plot the data to see how it looks:
import juliet
import numpy as np
# First, get times, normalized-fluxes and errors for TOI-141 from MAST:
's01_tess_v1_lc.fits')
# Plot the data!
import matplotlib.pyplot as plt
plt.errorbar(t,f,yerr=ferr,fmt='.')
plt.xlim([np.min(t),np.max(t)])
plt.ylim([0.999,1.001])
plt.xlabel('Time (BJD - 2457000)')
plt.ylabel('Relative flux')
This will save arrays of times, fluxes (PDCSAP_FLUX fluxes, in particular) and errors on the t, f and ferr arrays. Now, in order to load this dataset into a format that juliet likes, we need to put these into dictionaries. This, as we will see, will make it extremely easy to add data from more instruments, as these will be simply stored in different keys of the same dictionary. For now, let us just use this TESS data; we put them in dictionaries that juliet likes as follows:
# Create dictionaries:
times, fluxes, fluxes_error = {},{},{}
# Save data into those dictionaries:
times['TESS'], fluxes['TESS'], fluxes_error['TESS'] = t,f,ferr
# If you had data from other instruments you would simply do, e.g.,
# times['K2'], fluxes['K2'], fluxes_error['K2'] = t_k2,f_k2,ferr_k2
The final step to fit the data with juliet is to define the priors for the different parameters that we are going to fit. This can be done in two ways. The longest (but more jupyter-notebook-friendly?) is to create a dictionary that, on each key, has the names of the parameter to be fitted. Each of those elements will be dictionaries themselves, containing the distribution of the parameter and their corresponding hyperparameters (for details on what distributions juliet can handle, what are the hyperparameters and what each parameter name mean, see the next section of this document: Models, priors and outputs).
Let us give normal priors for the period P_p1, time-of-transit center t0_p1, mean out-of-transit flux mflux_TESS, uniform distributions for the parameters r1_p1 and r2_p1 of the Espinoza (2018) parametrization for the impact parameter and planet-to-star radius ratio, same for the q1_p1 and q2_p1 Kipping (2013) limb-darkening parametrization (juliet assumes a quadratic limb-darkening by default — other laws can be easily defined, as it will be shown in the tutorials), log-uniform distributions for the stellar density rho (in kg/m3) and jitter term sigma_w_TESS (in parts-per-million, ppm), and leave the rest of the parameters (eccentricity ecc_p1, argument of periastron (in degrees) omega_p1 and dilution factor mdilution_TESS) fixed:
priors = {}
# Name of the parameters to be fit:
params = ['P_p1','t0_p1','r1_p1','r2_p1','q1_TESS','q2_TESS','ecc_p1','omega_p1',\
'rho', 'mdilution_TESS', 'mflux_TESS', 'sigma_w_TESS']
# Distribution for each of the parameters:
dists = ['normal','normal','uniform','uniform','uniform','uniform','fixed','fixed',\
'loguniform', 'fixed', 'normal', 'loguniform']
# Hyperparameters of the distributions (mean and standard-deviation for normal
# distributions, lower and upper limits for uniform and loguniform distributions, and
# fixed values for fixed "distributions", which assume the parameter is fixed)
hyperps = [[1.,0.1], [1325.55,0.1], [0.,1], [0.,1.], [0., 1.], [0., 1.], 0.0, 90.,\
[100., 10000.], 1.0, [0.,0.1], [0.1, 1000.]]
# Populate the priors dictionary:
for param, dist, hyperp in zip(params, dists, hyperps):
priors[param] = {}
priors[param]['distribution'], priors[param]['hyperparameters'] = dist, hyperp
With these definitions, to fit this dataset with juliet one would simply do:
# Load dataset into juliet, save results to a temporary folder called toi141_fit:
dataset = juliet.load(priors=priors, t_lc = times, y_lc = fluxes, \
yerr_lc = fluxes_error, out_folder = 'toi141_fit')
# Fit and absorb results into a juliet.fit object:
results = dataset.fit(n_live_points = 300)
This code will run juliet and save the results both to the results object and to the toi141_fit folder.
The second way to define the priors for juliet (and perhaps the most simple) is to create a text file where in the first column one defines the parameter name, in the second column the name of the distribution and in the third column the hyperparameters. The priors defined above would look like this in a text file:
P_p1 normal 1.0,0.1
t0_p1 normal 1325.55,0.1
r1_p1 uniform 0.0,1.0
r2_p1 uniform 0.0,1.0
q1_TESS uniform 0.0,1.0
q2_TESS uniform 0.0,1.0
ecc_p1 fixed 0.0
omega_p1 fixed 90.0
rho loguniform 100.0,10000.0
mdilution_TESS fixed 1.0
mflux_TESS normal 0.0,0.1
sigma_w_TESS loguniform 0.1,1000.0
To run the same fit as above, suppose this prior file is saved under toi141_fit/priors.dat. Then, to load this dataset into juliet and fit it, one would do:
# Load dataset into juliet, save results to a temporary folder called toi141_fit:
dataset = juliet.load(priors='toi141_fit/priors.dat', t_lc = times, y_lc = fluxes, \
yerr_lc = fluxes_error, out_folder = 'toi141_fit')
# Fit and absorb results into a juliet.fit object:
results = dataset.fit(n_live_points = 300)
And that’s it! Cool juliet fact is that, once you have defined an out_folder, all your data will be saved there — not only the prior file and the results of the fit, but also the photometry or radial-velocity you fed into juliet will be saved. This makes it easy to come back later to this dataset without having to download the data all over again, or re-run your fits. So, for example, suppose we have already ran the code above, closed our terminals, and wanted to come back at this dataset again with another python session and say, plot the data and best-fit model. To do this one can simply do:
import juliet
dataset = juliet.load(input_folder = 'toi141_fit', out_folder = 'toi141_fit')
# Load results (the data.fit call will recognize the juliet output files in
# the toi141_fit folder generated when we ran the code for the first time):
results = dataset.fit()
import matplotlib.pyplot as plt
# Plot the data:
plt.errorbar(dataset.times_lc['TESS'], dataset.data_lc['TESS'], \
yerr = dataset.errors_lc['TESS'], fmt = '.', alpha = 0.1)
# Plot the model:
plt.plot(dataset.times_lc['TESS'], results.lc.evaluate('TESS'))
# Plot portion of the lightcurve, axes, etc.:
plt.xlim([1326,1332])
plt.ylim([0.999,1.001])
plt.xlabel('Time (BJD - 2457000)')
plt.ylabel('Relative flux')
plt.show()
Which will give us a nice plot of the data and the juliet fit:
Warning
When using MultiNest, make sure that the out_folder full path is less than 69 characters long. This is because MultiNest internally has a character limit for the full output path of 100 characters (see this fun discussion). Because the largest MultiNest output juliet produces (produced by MultiNest itself) is called jomnest_post_equal_weights.dat, which has 30 characters, this leaves the possible total character length of the folder to be 69 characters not counting the backlash at the end. Bottom line: when using MultiNest, stick to small out_folder lengths. | 2022-09-27 10:03:28 | {"extraction_info": {"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, "math_score": 0.550094723701477, "perplexity": 4283.109579284865}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030335004.95/warc/CC-MAIN-20220927100008-20220927130008-00178.warc.gz"} |
http://slideplayer.com/slide/1649575/ | # Integration by Tables Lesson 8.6. Tables of Integrals Text has covered only limited variety of integrals Applications in real life encounter many other.
## Presentation on theme: "Integration by Tables Lesson 8.6. Tables of Integrals Text has covered only limited variety of integrals Applications in real life encounter many other."— Presentation transcript:
Integration by Tables Lesson 8.6
Tables of Integrals Text has covered only limited variety of integrals Applications in real life encounter many other types Impossible (impractical) to memorize all types Tables of integrals have been established Text includes list in Appendix B, pg A-18
General Table Classifications Elementary forms Forms involving Trigonometric forms Inverse trigonometric forms Exponential, logarithmic forms Hyperbolic forms
Finding the Right Form For each integral Determine the classification Use the given pattern to complete the integral
Reduction Formulas Some integral patterns in the tables have the form This reduces a given integral to the sum of a function and a simpler integral Given Use formula 19 first of all
Reduction Formulas This gives you Now use formula 17 and finish the integration
Assignment Lesson 8.6 Page 565 Exercises 1 – 45 EOO
Download ppt "Integration by Tables Lesson 8.6. Tables of Integrals Text has covered only limited variety of integrals Applications in real life encounter many other."
Similar presentations | 2017-11-19 18:19:27 | {"extraction_info": {"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, "math_score": 0.8777096271514893, "perplexity": 2992.398343258654}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-47/segments/1510934805708.41/warc/CC-MAIN-20171119172232-20171119192232-00607.warc.gz"} |
https://atrium.lib.uoguelph.ca/xmlui/handle/10214/14900 | Title: Grand River Conservation Authority, Clean Up Rural Beaches 1993 progress report Ryan, Tracey During the Rural Beaches Program in the mid eighties, the impact of livestock wastes and septic systems were documented in three subwatersheds in the Grand River. Since 1991 the Grand River Conservation Authority has participated in the delivery of the Ontario Ministry of Environment and Energy's Clean Up Rural Beaches Program (CURB), The program provides funds to assist rural landowners in the reduction of surface water pollution from livestock wastes and septic systems. During 1993 approximately $300,000 was provided to landowners to implement 58 projects. Since 1991 approximately$606,000 has been provided for 125 projects to improve water quality in the Upper Speed, Upper Nith and Upper Conestogo River Watersheds. The grant provided financial assistance for the construction of 20 manure storage systems, 18 septic systems, 10 milkhouse waste storage facilities and 10 livestock access restriction projects. The majority of the projects were constructed in the Upper Conestogo and Upper Nith River Watersheds. Monitoring helps to establish background levels for water quality. The Nith River again exhibited the poorest water quality of the three watersheds. The levels of E. coli were slightly lower than the levels observed in 1992. The E. coli level in the Upper Conestogo were generally lower than those observed in 1992. The Speed River exhibited the best overall water quality, although the densities of E. coli have not changed much from 1992. Monitoring specific remedial sites before and after implementation has shown that individual projects can improve local water quality. Water quality at a livestock restriction project significantly improved at the downstream station after the cattle were restricted. In addition to decreases in nutrients and bacteria, water temperature decreased, making the stream more habitable for trout. Results from a buffer strip project demonstrated that bacteria and nutrients are reduced as the stream flows through the buffered area. More monitoring is required to determine the physical and biological changes that occur in the stream as well. http://hdl.handle.net/10214/14900 1994 Queen's Printer for Ontario, Crown Copyright, Non-Commercial Use Permitted Queen's Printer for Ontario | 2019-05-19 09:09:45 | {"extraction_info": {"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, "math_score": 0.2578432857990265, "perplexity": 5548.587063881461}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-22/segments/1558232254731.5/warc/CC-MAIN-20190519081519-20190519103519-00473.warc.gz"} |
https://electronics.stackexchange.com/questions/482158/why-is-the-ac-coupling-capacitor-always-0-1uf-or-0-01uf-in-high-speed-transceive | # Why is the AC coupling capacitor always 0.1uF or 0.01uF in high speed transceivers?
As I know, high speed transceivers always use a 0.1uF or 0.01uF capacitor for AC-coupling.
Tx--->capacity------------------->Rx
|
end resistor
Capacitor and end resistor are a high-pass filter.
If end resistor is 50 ohm.
I calculate bandwidth with $$f = \frac{1}{2 \pi R C}$$ if $$\C = 0.1\mathrm{\mu F}\$$, then $$\f = 31.847\mathrm{k Hz}\$$
if $$\C = 0.01\mathrm{\mu F}\$$, then $$\f = 318.47\mathrm{k Hz}\$$
But I always see 0.1uF and 0.01uF both used in the 1GHz, even 10GHz circuit.
As I calculated above, bandwidth should not enough.
I'm confused.
• What kind of transceiver is this? Is the signal a modulated RF carrier, or is it gigabit per second binary? – The Photon Feb 20 at 4:29
• "But I always see 0.1uF and 0.01uF" - because people like nice round numbers like 1 and 10, rather than the optimum value for the job. And manufacturers and distributors like it that way too, because then they don't have to stock a huge range of different values. – Bruce Abbott Feb 20 at 4:53
• @ThePhoton: Gigabit Ethernet or PCI-E – curlywei Feb 20 at 6:00
• Part of your mistake is that you mix "bandwidth" and "frequency". A frequency band runs from one frequency to another, f1 to f2. The bandwidth is f2-f1. For bands that start at DC, the bandwitdh would be f2-0Hz=f2. But you need to make certain what your band boundaries are to calculate a bandwidth. If there's no upper bound (like here), the filter has infinite bandwitdh. – MSalters Feb 20 at 16:21
It's a HIGH-pass filter, not a low-pass filter. That cutoff frequency you calculated is the LOWEST that can get through, not the highest. Frequencies lower than that are blocked, and frequencies higher than that are passed.
• Hi @DKNguyen : Ok, I'm kown. But if I'd like to used in above gigabit, I think I should select lesser then 0.01uf, because I can get good gain when my circuit work on 1GHz.Why I always see 0.1uF and 0.01uF both used in the 1GHz? I'm confused. – curlywei Feb 20 at 4:17
• @curlywei All things being equal, for higher frequencies you could select a smaller capacitance though you don't have to. But things are never equal in practice because smaller capacitors tend to have lower parasitic inductance and it might be necessary to use a a smaller capacitance to reduce the capacitor's parasitic inductance which would impede your high frequencies. – DKNguyen Feb 20 at 4:19
• Thanks @DKNguyen, now I know. – curlywei Feb 20 at 4:22
• While there are protocols that extend almost down to DC and have GHz edge rates, they are a pain to deal with (SDI springs to mind, you need typically 4.7uF of coupling cap that is good to GHz rates, expensive). Most modern high speed line protocols have DC balance (typically 8B10 or similar coding) so do not have any low frequency content, and as such the low frequency corner is not at all critical. Most designers have many, many 10 or 100nF caps used for decoupling scattered around, and something you already have on the BOM trumps something ideal that is an extra BOM line most of the time. – Dan Mills Feb 20 at 18:30
A binary data signal isn't a single frequency. If it's truly random, it will have frequency content from near DC all the way up to about half the data rate (i.e. a 1 Gbps signal has content from DC to ~500 MHz).
Using a smaller capacitor value would block some of the low-frequency content of the signal, resulting in excessive wander of the binary signal. For example, consider if the data stream happened to have 16 ones in a row at some point. With a too-low blocking capacitor value, the signal would tend to drift back towards zero during this long run of ones.
To alleviate this, it's possible to use encoding to limit the low-frequency content of the signal. For example, Gigabit Ethernet uses 8b/10b code to limit the maximum run length and ensure that the signal is balanced between ones and zeros. Given a maximum run length of zeros or ones, it's now possible to choose a minimum capacitor value that allows you pass the high frequency content in the signal without allowing excessive wander during the (limited length) runs of similar values.
RC forms a high pass filter whose cut off frequency should be in the range of minimum frequency you want pass. For example if the minimum frequency for your signal is 200MHz. For this the 3dB cut off frequency of high pass filter should be little below 200MHz to pass it without any attenuation.
Mohit has the right idea. The series capacitor is used to block DC and pass the higher frequency AC signal. The term "transfer function" is used to describe the ratio of Output to Input in a circuit. In terms of voltage, this would be Vo/Vi and is dependent on the configuration and values of components within a circuit. Here the circuit may be thought of as a simple frequency-dependent voltage divider. The voltage at the transmitter is Vi and Vo is the voltage at the receiver. The transfer function here would be: $$\frac{V_0}{V_i}=\frac{R}{\sqrt[2]{R^2+(1/2\pi f C)^2}}$$
due to a 90 deg phase shift of V & I in a cap.
At f=0, the impedance of the capacitor is infinite, and Vo goes to zero, blocking DC. At very high frequencies the impedance of the capacitor approaches zero and Vo approaches Vi (R/R --> 1). When $$f=\frac{1}{2 \pi RC}$$, the impedance of the cap is equal in magnitude to that of the resistor but because of the 90-degree phase shift, the composite impedance =R*SQRT(2). This frequency is considered to be the 'breakpoint' of the circuit. At this point, the transfer is -3dB as Mohit correctly states.
The dB value = 20*log|Vo/Vi|. That is Log base10. If you were to swap the positions to R & C in this circuit, you would have a low-pass filter with the same breakpoint. If you are interested in this subject I would suggest looking into Laplace transforms and Bode Plots for circuit analysis.
• I can't read you equation. You can use MathJax to format your equations. The line breaks have mangled yours to the point I'm not sure what you meant to write - I was going to just go ahead and fix it, but realized I couldn't figure out what you meant. – JRE Feb 21 at 7:01 | 2020-04-07 14:07:47 | {"extraction_info": {"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": 7, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6169142127037048, "perplexity": 1141.413139936623}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-16/segments/1585371799447.70/warc/CC-MAIN-20200407121105-20200407151605-00310.warc.gz"} |