url stringlengths 14 2.42k | text stringlengths 100 1.02M | date stringlengths 19 19 | metadata stringlengths 1.06k 1.1k |
|---|---|---|---|
http://fricas.github.io/api/PartialFractionPackage.html | # PartialFractionPackage R¶
partialFraction(rf, var) returns the partial fraction decomposition of the rational function rf with respect to the variable var.
partialFraction(num, facdenom, var) returns the partial fraction decomposition of the rational function whose numerator is num and whose factored denominator is facdenom with respect to the variable var. | 2017-10-23 11:22:18 | {"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": 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": 0.7053033709526062, "perplexity": 898.5747154359461}, "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-2017-43/segments/1508187825900.44/warc/CC-MAIN-20171023111450-20171023131450-00110.warc.gz"} |
https://stats.stackexchange.com/questions/115571/back-transforming-regression-results-when-modeling-logy | # Back transforming regression results when modeling log(y)
I'm fitting a regression on the $\log(y)$. Is it valid to back transform point estimates (and confidence/prediction intervals) by exponentiation? I don't believe so, since $E[f(X)] \ne f(E[X])$ but wanted other's opinions.
My example below shows conflicts with back transforming (.239 vs .219).
set.seed(123)
a=-5
b=2
x=runif(100,0,1)
y=exp(a*x+b+rnorm(100,0,.2))
# plot(x,y)
### NLS Fit
f <- function(x,a,b) {exp(a*x+b)}
fit <- nls(y ~ exp(a*x+b), start = c(a=-10, b=15))
co=coef(fit)
# curve(f(x=x, a=co[1], b=co[2]), add = TRUE,col=2,lwd=1.2)
predict(fit,newdata=data.frame(x=.7))
[1] 0.2393773
### LM Fit
# plot(x,log(y))
# abline(lm(log(y)~x),col=2)
fit=lm(log(y)~x)
temp=predict(fit,newdata=data.frame(x=.7),interval='prediction')
exp(temp)
fit lwr upr
1 0.2199471 0.1492762 0.3240752
• Is this not one of the problems that is solved by log-linked gaussian GLMs? – generic_user Sep 16 '14 at 5:31
• @ARM Yes I believe so. Thanks for pointing that out. However using GLM it is harder to get prediction intervals but I think I can work it out. – Glen Sep 16 '14 at 6:14
• @Glen Do a search for Duan smearing on this site. – Dimitriy V. Masterov Sep 16 '14 at 17:32
It depends on what you want to obtain at the other end.
A confidence interval for a transformed parameter transforms just fine. If it has the nominal coverage on the log scale it will have the same coverage back on the original scale, because of the monotonicity of the transformation.
A prediction interval for a future observation also transforms just fine.
An interval for a mean on the log scale will not generally be a suitable interval for the mean on the original scale.
However, sometimes you can either exactly or approximately produce a reasonable estimate for the mean on the original scale from the model on the log scale.
However, care is required or you might end up producing estimates that have somewhat surprising properties (it's possible to produce estimates that don't themselves have a population mean for example; this isn't everyone's idea of a good thing).
So for example, in the lognormal case, when you exponentiate back, you have a nice estimate of $\exp(\mu_i)$, and you might note that the population mean is $\exp(\mu_i+\frac{1}{2}\sigma^2)$, so you may think to improve $\exp(\hat{\mu_i})$ by scaling it by some estimate of $\exp(\frac{1}{2}\sigma^2)$.
One should at least be able to get consistent estimation and indeed some distributional asymptotics via Slutsky's theorem (specifically the product-form) as long as one can consistently estimate the adjustment. The continuous mapping theorem says that you can if you can estimate $\sigma^2$ consistently... which is the case.
So as long as $\hat{\sigma}^2$ is a consistent estimator of $\sigma^2$, then $\exp(\hat{\mu_i})\cdot \exp(\frac{1}{2}\hat{\sigma}^2)$ converges in distribution to the distribution of $\exp(\hat{\mu_i})\cdot \exp(\frac{1}{2}\sigma^2)$ (which by inspection will then be asymptotically lognormally distributed). Since $\hat{\mu_i}$ will be consistent for $\mu_i$, bu the continuous mapping theorem, $\exp(\hat{\mu_i})$ will be consistent for $\exp(\mu_i)$, and so we have a consistent estimator of the mean on the original scale.
See here.
Some related posts:
Back transformation of an MLR model
Back Transformation
Back-transformed confidence intervals
• Thanks, I looked at the previous posts and, while enlightening, was still somewhat confused, hence my question. – Glen Sep 16 '14 at 6:15
• +1 Great answer! Just a quick clarification: Where did the $\frac{1}{2}$ came from as a scaler for $\hat{\sigma^2}$? I saw it in the definition of the lognormal in Wikipedia but it is not explained there either, is it just integrating out mean from the PDF? – usεr11852 Jul 15 '19 at 23:55
• You should be able to get it just by directly integrating: $E(Y) = \int_0^\infty y\, f(y)\, dy$ where $f$ is the density for the lognormal, but it is probably easier to do by calculating $E(e^X)$ for a normal (where $X=\log Y$), but then perhaps it is better to find the MGF for $X$ - which is no more difficult - and from which moments for $Y$ are very readily obtained (by replacing $t$ by $1,2,...$ in turn), essentially getting higher moments for free. – Glen_b Jul 16 '19 at 1:19
• @usεr11852 In either of the latter cases you take the $e^x$ or $e^{tx}$ into the $e^{...}$ term in the density, then complete the square in $x$, and bring additional constants (i.e. all except the normalizing constant for the normal) out the front of the integral (which has the $\frac12$ in it), leaving a Gaussian pdf integrated on the real line (with shifted mean from the original) which integrates to 1, leaving only the constants you brought out the front. This involves nothing more than very simple algebraic manipulations, ... ctd – Glen_b Jul 16 '19 at 2:12
• ctd... and from which the $t$-th raw moment of a lognormal is $e^{\mu t + \frac12 \sigma^2t^2}$. – Glen_b Jul 16 '19 at 2:24 | 2021-01-22 10:28:30 | {"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": 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": 0.9100063443183899, "perplexity": 546.5279343994058}, "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/1610703529179.46/warc/CC-MAIN-20210122082356-20210122112356-00773.warc.gz"} |
https://www.gamedev.net/forums/topic/322133-mass-file-conversion/ | Public Group
Mass File Conversion
This topic is 4748 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.
Recommended Posts
Does anyone know of any program that can convert many BMP images to PNG or TGAs at once. I have about 50 BMP textures , and I just converted my game to load TGAs and PNG. Do I have to manually open all these files up in paint and resave them or is there a faster way?
Share on other sites
If you have access to photoshop (I think 4 or 5.0 and above) you can batch process some file operations. Also, you may want to write a small utility that will do this for you in case you do not have access to photoshop.
Some sudo-code
Get dir listLoop through files in dirOpen file as bmpConvert data to TGASave as tgaNext
gl
Share on other sites
30 day trial:
?????
Or you could try googling for it. What you're looking for is a batch image converter. Photoshop will do it, if you have a copy of that.
**EDIT** UnderGamer, you beat me to it.
Thx
Share on other sites
Many free programs out there can do it. Xnview for example.
Share on other sites
This came up recently. If you're still looking for a solution, I think someone mentioned "ImageMagick"
Share on other sites
I think irfanview does it.
Share on other sites
Imagemagick's "convert" tool, and a shell script, should do it.
Even on win32, you might be able to do this with a batch file (On NT, you do have foreach, right?)
Mark
• 10
• 17
• 9
• 14
• 41
• Forum Statistics
• Total Topics
631067
• Total Posts
2997734
× | 2018-05-27 23:41:29 | {"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": 0.28268858790397644, "perplexity": 3658.692812578683}, "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-2018-22/segments/1526794870497.66/warc/CC-MAIN-20180527225404-20180528005404-00107.warc.gz"} |
https://www.ryanlevy.science/physics/Changing-Area/ | Consider the classic introductory problem where a bar is moving over a constant magnetic field. What is the generated EMF $$\varepsilon$$?
## The Standard Solution
Using Maxwell’s equations and Lenz’s law, EMF is given as $$\varepsilon = - \frac{d\Phi}{dt}$$
where $$\Phi=\int B\cdot dA=BA$$. Because $$B$$ is constant and the area is changing we get that
$\boxed{\varepsilon = -\ell B v }$
## Less Obvious Solution
Looking at Maxwell’s equations we have
$\nabla \times E= - \frac{\partial B}{\partial t}$
However, $$\frac{\partial B}{\partial t}=0$$, which seems to imply $$E=0$$ here! To rectify this, we notice that it implies the electric field is a constant not zero, which makes sense given the EMF given above. It’s also important to notice that you cannot move the time derivative past the integral sign without the chain rule, as the area part of the integral is time dependent.
Tags:
Categories:
Updated: | 2023-03-25 16:16:11 | {"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": 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": 0.982386589050293, "perplexity": 297.068938975932}, "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/1679296945368.6/warc/CC-MAIN-20230325161021-20230325191021-00076.warc.gz"} |
https://math.stackexchange.com/questions/1952882/least-squares-property-of-condtional-expectation/1952931 | # Least-Squares Property of Condtional Expectation
Let $Y$ be a vector-valued random element. Then then for any $\sigma$-field the conditional expectation satisfies the least-squares property $$argmin_{X \in L^2(\mathscr{F})}( \mathbb{E}[(X-Y)^2] )= \mathbb{E}[Y|\mathscr{F}].$$
My question is the following: if $g$ and $f$ are $\sigma(Y)$-measurable then what is the solution to the problem
$$argmin_{X \in L^2(\mathscr{F})}( \mathbb{E}[(g(X)-f(Y))^2])?$$
I expect something of the form $g(\mathbb{E}[f(Y)|\mathscr{F}])$ but I'm not certain, if and when that would be true.
• Your original statement does not seem to be correct to me; the argmin of $E[(X-Y)^2]$ over $L^2(F)$ is the conditional expectation of $Y$. As for your second question, it will depend quite drastically on what $g$ is. $g$ might very well be nonconvex for example... – Ian Oct 4 '16 at 2:41
• It's still wrong, by the way. – Ian Oct 4 '16 at 2:42
• Thanks for pointing that out I missed the crucial exponent in my typo. Ultimately I only need $g\in C^2$ and if necessary (but I would prefer not) I may assume it is convex. – AIM_BLB Oct 4 '16 at 2:52
If we make the added assumption that $g$ is injective and measurable wrt $\mathscr{F}$ then $$Z\triangleq f(Y)$$ then the least squares problem \begin{align} \hat{X}\triangleq \min_{X \in L^2(\mathscr{F})} \mathbb{E}[(X-Z)^2] & = \mathbb{E}[Z|\mathscr{F}] \\ &= \mathbb{E}[f(Y)|\mathscr{F}]. \end{align}
Now if $X$ in te minimization is of the form $g(\tilde{X})$ above then: $$\mathbb{E}[Z|\mathscr{F}] = \hat{X} = g(\tilde{X}),$$ the left-invertibility of $g$ allow us to conclude that $$g^{-1}\left(\mathbb{E}[Z|\mathscr{F}] \right) = \tilde{X} .$$ | 2019-07-24 00:20:29 | {"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": 1, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9962348937988281, "perplexity": 305.7711499602154}, "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-2019-30/segments/1563195530246.91/warc/CC-MAIN-20190723235815-20190724021815-00407.warc.gz"} |
https://msp.org/apde/2016/9-5/p08.xhtml | #### Vol. 9, No. 5, 2016
Download this article For screen For printing
Recent Issues
The Journal About the Journal Subscriptions Editorial Board Editors’ Interests Scientific Advantages Submission Guidelines Submission Form Editorial Login Contacts Author Index To Appear ISSN: 1948-206X (e-only) ISSN: 2157-5045 (print)
On the negative spectrum of the Robin Laplacian in corner domains
### Vincent Bruneau and Nicolas Popoff
Vol. 9 (2016), No. 5, 1259–1283
##### Abstract
For a bounded corner domain $\Omega$, we consider the attractive Robin Laplacian in $\Omega$ with large Robin parameter. Exploiting multiscale analysis and a recursive procedure, we have a precise description of the mechanism giving the bottom of the spectrum. It allows also the study of the bottom of the essential spectrum on the associated tangent structures given by cones. Then we obtain the asymptotic behavior of the principal eigenvalue for this singular limit in any dimension, with remainder estimates. The same method works for the Schrödinger operator in ${ℝ}^{n}$ with a strong attractive $\delta$-interaction supported on $\partial \Omega$. Applications to some Ehrling-type estimates and the analysis of the critical temperature of some superconductors are also provided.
##### Keywords
Robin Laplacian, eigenvalues estimates, corner domains
##### Mathematical Subject Classification 2010
Primary: 35J10, 35P15, 47F05, 81Q10
##### Milestones
Received: 18 January 2016
Revised: 30 March 2016
Accepted: 29 April 2016
Published: 29 July 2016
##### Authors
Vincent Bruneau Institut de Mathematiques Université de Bordeaux I 351 Cours de La Libération 33405 Talence France Nicolas Popoff Institut de Mathematiques Université de Bordeaux I 351 Cours de La Libération 33405 Talence France | 2018-01-17 15:01:06 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 5, "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": 0.2290077954530716, "perplexity": 2820.721334941602}, "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/1516084886946.21/warc/CC-MAIN-20180117142113-20180117162113-00403.warc.gz"} |
https://quomodocumque.wordpress.com/2014/12/16/how-many-points-does-a-random-curve-over-f_q-have/ | ## How many points does a random curve over F_q have?
So asks a charming preprint by Achter, Erman, Kedlaya, Wood, and Zureick-Brown. (2/5 Wisconsin, 1/5 ex-Wisconsin!) The paper, I’m happy to say, is a result of discussions at an AIM workshop on arithmetic statistics I organized with Alina Bucur and Chantal David earlier this year.
Here’s how they think of this. By a random curve we might mean a curve drawn uniformly from M_g(F_q). Let X be the number of points on a random curve. Then the average number of points on a random curve also has a geometric interpretation: it is
$|M_{g,1}(\mathbf{F}_q)|/|M_{g}(\mathbf{F}_q)|$
$|M_{g,2}(\mathbf{F}_q)|/|M_{g}(\mathbf{F}_q)|$?
That’s just the average number of ordered pairs of distinct points on a random curve; the expected value of X(X-1).
If we can compute all these expected values, we have all the moments of X, which should give us a good idea as to its distribution. Now if life were as easy as possible, the moduli spaces of curves would have no cohomology past degree 0, and by Grothendieck-Lefschetz, the number of points on M_{g,n} would be q^{3g-3+n}. In that case, we’d have that the expected value of X(X-1)…(X-n) was q^n. Hey, I know what distribution that is! It’s Poisson with mean q.
Now M_g does have cohomology past degree 0. The good news is, thanks to the Madsen-Weiss theorem (née the Mumford conjecture) we know what that cohomology is, at least stably. Yes, there are a lot of unstable classes, too, but the authors propose that heuristically these shouldn’t contribute anything. (The point is that the contribution from the unstable range should look like traces of gigantic random unitary matrices, which, I learn from this paper, are bounded with probability 1 — I didn’t know this, actually!) And you can even make this heuristic into a fact, if you want, by letting q grow pretty quickly relative to g.
So something quite nice happens: if you apply Grothendieck-Lefschetz (actually, you’d better throw in Kai Behrend’s name, too, because M_g is a Deligne-Mumford stack, not an honest scheme) you find that the moments of X still agree with those of a Poisson distribution! But the contribution of the tautological cohomology shifts the mean from q to q+1+1/(q-1).
This is cool in many directions!
• It satisfies one’s feeling that a “random set,” if it carries no extra structure, should have cardinality obeying a Poisson distribution — the “uniform distribution” on the groupoid of sets. (Though actually that uniform distribution is Poisson(1); I wonder what tweak is necessary to be able to tune the mean?)
• I once blogged about an interesting result of Bucur and Kedlaya which showed that a random smooth complete intersection curve in P^3 of fixed degree had slightly fewer than q+1 points; in fact, about q+1 – 1/q + o(q^2). Here the deviation is negative, rather than positive, as the new paper suggests is the case for general curves; what’s going on?
• I have blogged about the question of average number of points on a random curve before. I’d be very interested to know whether the new heuristic agrees with the answer to the question proposed at the end of that post; if g is a large random matrix in GSp(Z_ell) with algebraic eigenvalues, and which multiplies the symplectic form by q, and you condition on Tr(g^k) > (-q^k-1) so that the “curve” has nonnegatively many points over each extension of F_q, does this give something like the distribution the five authors predict for Tr(g)? (Note: I don’t think this question is exactly well-formed as stated.)
## 6 thoughts on “How many points does a random curve over F_q have?”
1. Jon Awbrey says:
&latex \mathbb{F}_{q=2}?\$ ;)
2. Jon Awbrey says:
Arrgh … $\mathbb{F}_{q=2}?$ ;)
3. JSE says:
Not sure I get your point?
4. Qiaochu Yuan says:
You can start tuning the mean by taking a product of $k$ copies of the groupoid of finite sets, or equivalently looking at the groupoid of $k$-colored finite sets. The total size of such a finite set produces a Poisson distribution with parameter $k$.
5. AV says:
Two quick thoughts:
Firstly, applying the unitary matrix heuristic to deduce tiny trace would reasonably require that the Galois action on unstable cohomology is close to irreducible. Is that plausible? There may be natural “Hecke-type” correspondences between different M_gs, like “curve X is a covering of curve Y”, but I don’t know how far they obstruct that.
Secondly, concerning testability, can one sample a random curve from this measure, for moderately large g (or even not so moderately large)? One thought would be to take to look at a high degree branched cover of P^1 and pick the ones defined over F_q, but this sounds uncomputable.
6. JSE says:
The irreducibility is a good point! I’m sure you’re right that correspondences like those you mention could calve off some small pieces, but I think you would be OK (for large but fixed q) so long as the number of constituents was expected to be at worst exponential…? Re testability: that’s a great question, actually. Certainly e.g. for genus 3 you can sample a plane quartic and a hyperelliptic at random in the appropriate proportions but that’s probably much MORE moderate than you mean. Almost by definition it’s hard to imagine how you would do this once M_g is not unirational, right? | 2023-03-23 07:51:29 | {"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": 6, "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": 0.8508924245834351, "perplexity": 1281.2387195683368}, "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-2023-14/segments/1679296945030.59/warc/CC-MAIN-20230323065609-20230323095609-00449.warc.gz"} |
https://drake.mit.edu/doxygen_cxx/classdrake_1_1benchmarks_1_1free__body_1_1_free_body.html | Drake
FreeBody Class Reference
The purpose of the FreeBody class is to provide the data (initial values and gravity) and methods for calculating the exact analytical solution for the translational and rotational motion of a torque-free rigid body B with axially symmetric inertia, in a Newtonian frame (World) N. More...
#include <drake/multibody/benchmarks/free_body/free_body.h>
## Public Member Functions
FreeBody (const Eigen::Quaterniond &initial_quat_NB, const Eigen::Vector3d &initial_w_NB_B, const Eigen::Vector3d &initial_p_NoBcm_N, const Eigen::Vector3d &initial_v_NBcm_B, const Eigen::Vector3d &gravity_N)
Constructs a class that can be queried for exact values of orientation, position, and motion of a torque-free rigid body at time t. More...
~FreeBody ()=default
double get_I () const
Returns the body's moment of inertia about an axis perpendicular to its axis of rotation and passing through its center of mass. More...
double get_J () const
Returns body's moment of inertia about its axis of rotation. More...
const Eigen::Quaterniond & get_initial_quat_NB () const
const Eigen::Vector3d & get_initial_w_NB_B () const
Eigen::Vector3d CalcInitial_w_NB_N () const
const Eigen::Vector3d & get_initial_p_NoBcm_N () const
const Eigen::Vector3d & get_uniform_gravity_expressed_in_world () const
Eigen::Vector3d CalcInitial_v_NBcm_N () const
void set_initial_quat_NB (const Eigen::Quaterniond &quat_NB)
void set_initial_w_NB_B (const Eigen::Vector3d &w_NB_B)
void set_initial_p_NoBcm_N (const Eigen::Vector3d &p_NoBcm_N)
void set_initial_v_NBcm_B (const Eigen::Vector3d &v_NBcm_B)
void SetUniformGravityExpressedInWorld (const Eigen::Vector3d &gravity)
std::tuple< Eigen::Quaterniond, Eigen::Vector4d, Eigen::Vector3d, Eigen::Vector3d > CalculateExactRotationalSolutionNB (const double t) const
Calculates exact solutions for quaternion and angular velocity expressed in body-frame, and their time derivatives for torque-free rotational motion of axis-symmetric rigid body B in Newtonian frame (World) N, where torque-free means the moment of forces about B's mass center is zero. More...
std::tuple< Eigen::Vector3d, Eigen::Vector3d, Eigen::Vector3d > CalculateExactTranslationalSolution (const double t) const
Calculates exact solutions for translational motion of an arbitrary rigid body B in a Newtonian frame (world) N. More...
Implements CopyConstructible, CopyAssignable, MoveConstructible, MoveAssignable
FreeBody (const FreeBody &)=default
FreeBodyoperator= (const FreeBody &)=default
FreeBody (FreeBody &&)=default
FreeBodyoperator= (FreeBody &&)=default
## Detailed Description
The purpose of the FreeBody class is to provide the data (initial values and gravity) and methods for calculating the exact analytical solution for the translational and rotational motion of a torque-free rigid body B with axially symmetric inertia, in a Newtonian frame (World) N.
Examples of bodies with axially symmetric inertia include cylinders, rods or bars with a circular or square cross section and spinning tops. Since the only external forces on B are uniform gravitational forces, there exists an exact closed-form analytical solution for B's motion. The closed- form rotational solution is available since B is "torque-free", i.e., the moment of all forces about B's mass center is zero. This class calculates the body B's quaternion, angular velocity and angular acceleration expressed in B (body-frame) as well as the position, velocity, acceleration of Bcm (B's center of mass) in N (World). Algorithm from [Kane, 1983] Sections 1.13 and 3.1, Pages 60-62 and 159-169.
## ◆ FreeBody() [1/3]
FreeBody ( const FreeBody & )
default
## ◆ FreeBody() [2/3]
FreeBody ( FreeBody && )
default
## ◆ FreeBody() [3/3]
FreeBody ( const Eigen::Quaterniond & initial_quat_NB, const Eigen::Vector3d & initial_w_NB_B, const Eigen::Vector3d & initial_p_NoBcm_N, const Eigen::Vector3d & initial_v_NBcm_B, const Eigen::Vector3d & gravity_N )
inline
Constructs a class that can be queried for exact values of orientation, position, and motion of a torque-free rigid body at time t.
Parameters
[in] initial_quat_NB Value at time t = 0 of the quaternion relating right-handed orthonormal vectors Nx, Ny, Nz fixed in N (world) to right-handed orthonormal unit vectors Bx, By, Bz fixed in B (body). Note: The unit vector Bz is parallel to body B's symmetry axis. Note: The quaternion should already be normalized before it is passed. [in] initial_W_NB_B Value at time t = 0 of the angular velocity in N of body B, expressed in N. [in] initial_p_NoBcm_N Value at time t = 0 of the position vector from No (origin of world N) to Bcm (B's center of mass), expressed in N. [in] initial_v_NBcm_N Value at time t = 0 of the velocity in N of Bcm (B's center of mass), expressed in N. [in] gravity_N Local gravitational acceleration, expressed in N.
## ◆ ~FreeBody()
~FreeBody ( )
default
## ◆ CalcInitial_v_NBcm_N()
Eigen::Vector3d CalcInitial_v_NBcm_N ( ) const
inline
## ◆ CalcInitial_w_NB_N()
Eigen::Vector3d CalcInitial_w_NB_N ( ) const
inline
## ◆ CalculateExactRotationalSolutionNB()
std::tuple< Quaterniond, Vector4d, Vector3d, Vector3d > CalculateExactRotationalSolutionNB ( const double t ) const
Calculates exact solutions for quaternion and angular velocity expressed in body-frame, and their time derivatives for torque-free rotational motion of axis-symmetric rigid body B in Newtonian frame (World) N, where torque-free means the moment of forces about B's mass center is zero.
The quaternion characterizes the orientation between right-handed orthogonal unit vectors Nx, Ny, Nz fixed in N and right-handed orthogonal unit vectors Bx, By, Bz fixed in B, where Bz is parallel to B's symmetry axis.
Note
CalculateExactRotationalSolutionABInitiallyAligned() implements the algorithm from [Kane, 1983] Sections 1.13 and 3.1, Pages 60-62 and 159-169.
Parameters
t Current value of time.
Returns
Machine-precision values at time t are returned as defined below.
Note
This function allows for initial misalignment of Nx, Ny, Nz and Bx, By, Bz.
std::tuple Description
quat_NB Quaternion relating frame N to frame B: [e0, e1, e2, e3]
| Note: quat_NB is analogous to the rotation matrix R_NB. quatDt | Time-derivative of `quat_NB', i.e., [ė0, ė1, ė2, ė3]. w_NB_B | B's angular velocity in N, expressed in B. alpha_NB_B | B's angular acceleration in N, expressed in B.
## ◆ CalculateExactTranslationalSolution()
std::tuple< Vector3d, Vector3d, Vector3d > CalculateExactTranslationalSolution ( const double t ) const
Calculates exact solutions for translational motion of an arbitrary rigid body B in a Newtonian frame (world) N.
Algorithm from high-school physics.
Parameters
t Current value of time.
Returns
Machine-precision values at time t are returned as defined below.
std::tuple Description
xyz Vector3d [x, y, z], Bcm's position from No, expressed in N.
xyzDt Vector3d [ẋ, ẏ, ż] Bcm's velocity in N, expressed in N.
xyzDDt Vector3d [ẍ ÿ z̈], Bcm's acceleration in N, expressed in N.
## ◆ get_I()
double get_I ( ) const
inline
Returns the body's moment of inertia about an axis perpendicular to its axis of rotation and passing through its center of mass.
For example, for a cylinder of radius r, length h and uniformly distributed mass m with its rotational axis aligined along its body frame z-axis this would be: I = Ixx = Iyy = m / 12 (3 r² + h²)
## ◆ get_initial_p_NoBcm_N()
const Eigen::Vector3d& get_initial_p_NoBcm_N ( ) const
inline
## ◆ get_initial_quat_NB()
const Eigen::Quaterniond& get_initial_quat_NB ( ) const
inline
## ◆ get_initial_w_NB_B()
const Eigen::Vector3d& get_initial_w_NB_B ( ) const
inline
## ◆ get_J()
double get_J ( ) const
inline
Returns body's moment of inertia about its axis of rotation.
For example, for a cylinder of radius r, length h and uniformly distributed mass m with its rotational axis aligined along its body frame z-axis this would be: J = Izz = m r² / 2
## ◆ get_uniform_gravity_expressed_in_world()
const Eigen::Vector3d& get_uniform_gravity_expressed_in_world ( ) const
inline
## ◆ operator=() [1/2]
FreeBody& operator= ( FreeBody && )
default
## ◆ operator=() [2/2]
FreeBody& operator= ( const FreeBody & )
default
## ◆ set_initial_p_NoBcm_N()
void set_initial_p_NoBcm_N ( const Eigen::Vector3d & p_NoBcm_N )
inline
## ◆ set_initial_quat_NB()
void set_initial_quat_NB ( const Eigen::Quaterniond & quat_NB )
inline
## ◆ set_initial_v_NBcm_B()
void set_initial_v_NBcm_B ( const Eigen::Vector3d & v_NBcm_B )
inline
## ◆ set_initial_w_NB_B()
void set_initial_w_NB_B ( const Eigen::Vector3d & w_NB_B )
inline
## ◆ SetUniformGravityExpressedInWorld()
void SetUniformGravityExpressedInWorld ( const Eigen::Vector3d & gravity )
inline
The documentation for this class was generated from the following files: | 2018-12-15 03:30:09 | {"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": 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": 0.5101610422134399, "perplexity": 8145.224965049536}, "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-2018-51/segments/1544376826686.8/warc/CC-MAIN-20181215014028-20181215040028-00151.warc.gz"} |
https://www.sangakoo.com/en/unit/complement-of-a-set/problems | # Problems from Complement of a set
Let $$A = \{a,e,i,o,u\}$$ and $$B = \{b,c,d,f,g,\ldots,z\}$$ be the sets of vowels and consonants of the alphabet.
a) What is the complementary of $$A$$?
b) And $$B$$?
See development and solution
### Development:
We realize that sets $$A$$ and $$B$$ are, respectively, the vowels and the consonants of the alphabet. Therefore, we can easily answer the proposed questions.
a) The complement of $$A$$ is $$B$$ since the letters that are not vowels are consonants.
b) Reciprocally, the letters that are not consonant are vowels, since the letters of the alphabet can divided into consonants and vowels.
### Solution:
a) $$A^c=B$$
b) $$B^c=A$$
Hide solution and development | 2020-06-05 04:04:18 | {"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": 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": 0.8782778382301331, "perplexity": 713.5990750393984}, "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/1590348492427.71/warc/CC-MAIN-20200605014501-20200605044501-00237.warc.gz"} |
https://pypi.org/project/sass-cli/ | CLI for SASS compliation
Project description
I don’t want to install Ruby, but I want to compile *.scss from the command line.
Same here.
If you use the Ruby CLI, you will find some things missing. This command accepts two arguments, which may either be a pair of filenames or a pair of directory names.
Examples
Two files:
$sass assets/scss/file.scss build/output.css Two dirs: $ sass assets/scss build/css
Project details
Uploaded source | 2022-12-05 21:42:17 | {"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": 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": 0.24016453325748444, "perplexity": 9756.527886987153}, "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-2022-49/segments/1669446711045.18/warc/CC-MAIN-20221205200634-20221205230634-00673.warc.gz"} |
http://www.physicsforums.com/showthread.php?t=153210 | # Electric Field of Earth problem
by stylez03
Tags: earth, electric, field
P: 139 1. The problem statement, all variables and given/known data Electric Field of the Earth. The earth has a net electric charge that causes a field at points near its surface equal to 153 N/C and directed in toward the center of the earth. 1. What charge would a human with a mass of 58.0 kg have to acquire to overcome his or her weight by the force exerted by the earth's electric field? 2. What would be the magnitude of the repulsive force between two people each with the charge calculated in part (a) and separated by a distance of 120 m? 2. Relevant equations $$E = \frac{1}{4*pi*e_o}*\frac{q}{r^2*\vec{r}}$$ $$G = 9.81 & m/s$$ 3. The attempt at a solution I couldn't find any relevant information in the electric field section that I could map to this problem to help me solve it. Does anyone have a suggestion on where to start?
Sci Advisor HW Helper P: 1,772 Your looking at the wrong equation here. This is a simple problem. Just a matter of balancing two forces. How are charge, electric field, and electric force related?
P: 139 Electric force on a charged body is exherted by the electric field created by other charged bodies?
Sci Advisor HW Helper P: 1,772 Electric Field of Earth problem I meant what is the (very simple) equation relating the three quantities?
P: 139 I can't seem to find the equation that relates to all three, I'm looking at the index of current chapter of all the equations in this chapter and I can't find one that does?
Sci Advisor HW Helper P: 1,772 I'm surprised. Well it's easy: F=qE . Notice how it is similar to the equation for weight: F=mg . In both cases the force is found as the product of a quantity of matter times the strength of a field. Balance those forces.
P: 139 So it should be q = F/E because for part 1 it's looking for the charge. q = 58.0 kg / 153 ??
PF Gold
P: 2,080
Quote by Chi Meson I'm surprised. Well it's easy: F=qE . Notice how it is similar to the equation for weight: F=mg .
Hmm, I gave stylez the same equation in another thread about 12 hours earlier
HW Helper
P: 1,772
Quote by stylez03 So it should be q = F/E because for part 1 it's looking for the charge. q = 58.0 kg / 153 ??
What's the difference betwen mass and weight?
P: 139 Mass is a measurement of the amount of matter something where as weight is the measurement of the pull of gravity on an object.
P: 139 EDIT: Thanks I found the right equation. For part B it asks: What would be the magnitude of the repulsive force between two people each with the charge calculated in part (a) and separated by a distance of 120 m? Is there a separate to find the repulsive force?
P: 139 $$\frac{1} {4*pi*8.85*10^-12} * \frac{-3.72*10^-9^2} {120^2}$$ Though the online program says I'm off by an additive constant.
P: 98 where did the 10^-9 come from?
P: 139
Quote by mace2 where did the 10^-9 come from?
For some of the problems when they mention charge, the example I had to build off of was to take it and multiply it by 10^-9. The equation I'm using is the following:
$$\frac{1} {4*pi*8.85*10^-12} * \frac{q1*q2}{r^2}$$
Should q1 and q2 just be -3.72 and since q1 = q2 it should just be q^2 ? Is this the equation you would use to find the repulsive force?
P: 98 Yeah, that will give you the electrostatic force. Since they're both negatively charged they will repel. Don't guess about 10^-9, just work with what you have. The math doesn't lie. q1 & q2 & so on are all measured in Coloumbs.
P: 139
Quote by mace2 Yeah, that will give you the electrostatic force. Since they're both negatively charged they will repel. Don't guess about 10^-9, just work with what you have. The math doesn't lie. q1 & q2 & so on are all measured in Coloumbs.
I tried:
$$\frac{1} {4*pi*8.85*10^-12} * \frac{-3.72^2}{120^2}$$ | 2014-08-27 17:08:44 | {"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": 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": 0.41674256324768066, "perplexity": 636.4657698751392}, "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-2014-35/segments/1408500829661.96/warc/CC-MAIN-20140820021349-00056-ip-10-180-136-8.ec2.internal.warc.gz"} |
http://fany.savina.net/2010/01/replicating-a-scientific-experiment/ | # Replicating a Scientific Experiment
Since I was in my molecules and atoms period, I also tried doing this experiment: Measuring the Diameter of a Molecule. The experiment was found on a school website. I read the essay summarising the experiment and I thought the concept was pretty simple. It was basically filling out a rectangular straight-sided pan with water and to make sure the water bulged over the side of the pan. Then a wire rod was to be placed over the water near the middle of the pan. It had to just lightly touch the water. A single drop of oil was to be dropped into the middle of 1 section created by the rod. The oil was then to spread out over the surface of the water and when it had covered the whole surface between the side of the pan and the side of rod, the rod was to be moved another couple of inches to allow the oil to spread even more. It continued liked this until you were supposed to see the oil starting to break up and see fissures in what should look like a very thin blanket of oil over the water. As soon as this happened the rod must be left alone and the length and width of the oil blanket must be measured. The idea was that all the molecules started spreading out and just before the cracks could come, they would all be flatly the one beside the other, making the oil one molecule thick. A calculation was then given out to find out the height of these molecules. Volume=Length*Width*Height
$V=L$×$W$×$H$
Assuming the drop was 1 cubic millimetre and the oil's length was 400 mm and the width was 300 mm:
$1=400$×$300$×$H$
$1=120000$×$H$
$120000$×$H=1$
$H=\frac{1}{120000}$
$H=0.00000833$
0.00000833mm is the supposed height of one oil molecule. If you want more details on this experiment, visit the site, but details were my exact problem.
When I tried this experiment I did everything the experiment told me to do. The thing is, it doesn't tell you a lot. A lot of variables that need to be known are not explained, for example, the type of oil, the amount of time the experiment takes, or even the water temperature. I used olive oil because it was the only oil I had. I wasn't able to reproduce this experiment even though I tried three times. The first time I used ambient temperature water and a tiny drop of oil from the tip of a needle. It didn't work. The second time I used ambient temperature water again but this time with a slightly larger drop which I think was roughly 1 cubic mm. I used a pipette for the drop this time. This time it seemed to actually spread but it spread out into a bigger drop VERY SLOWLY. I left it overnight and in the morning the drop had clung on the edge of my pan and it had broken down into several bubbles. The third time I used the same amount of oil but I used hot water. This time the drop didn't even bother spreading.
My conclusion is that I have no idea if the proposed answer they gave could be right or if it is downright wrong. If it is wrong then the experiment was probably found somewhere else and it was probably not carried out before putting it on their website. If it is right, then whoever wrote out the experiment did not do it properly because I could not carry it out due to missing details.
This entry was posted in Chemistry, Science and tagged , , . Bookmark the permalink. | 2017-04-25 20:32:16 | {"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": 0, "img_math": 12, "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": 0.5139205455780029, "perplexity": 407.8228935563037}, "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-2017-17/segments/1492917120878.96/warc/CC-MAIN-20170423031200-00377-ip-10-145-167-34.ec2.internal.warc.gz"} |
https://proofwiki.org/wiki/Definition:Convolution_of_Real_Sequences | # Definition:Convolution of Real Sequences
### Definition 1
Let $\sequence f$ and $\sequence g$ be real sequences.
The convolution of $f$ and $g$ is defined as:
$\ds \sequence {f_i} * \sequence {g_i} := \sum_{j \mathop = 0}^i f_j g_{i - j}$
### Definition 2
Let $f: \Z \to \R$ and $g: \Z \to \R$ be mappings from the integers to the real numbers.
The convolution of $f$ and $g$ is defined as:
$\ds \map f i * \map g i := \sum_{j \mathop \in \Z} f_j g_{i - j}$ | 2021-09-21 02:24:43 | {"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": 0.8737154006958008, "perplexity": 277.4625051564444}, "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/1631780057131.88/warc/CC-MAIN-20210921011047-20210921041047-00405.warc.gz"} |
https://www.physicsforums.com/threads/fractions-of-all-numbers.350791/ | # Fractions of all numbers
1. Nov 1, 2009
### marcusmath
I really didn't know how to word the title so sorry if it's a little confusing.
And I didn't know whether to post this in number theory or not but ah well.
The other day, I started thinking about this and I was just wondering if it had been done before or if it's even correct;
Half of all positive integers are divisible by 2,
A third are divisible by 3,
A quarter by 4 etc.
Now, of the third that are divisible by 3, a half of those numbers are also divisible by 2.
So 1/6 of positive integers are divisible by 3 and not 2.
I continued this trend to find;
$$\frac{1}{2}$$ of p. ints are divisible by 2
$$\frac{1}{3}\left(1-\frac{1}{2}\right)$$ are divisible by 3 and not 2
$$\frac{1}{5}\left(1-\left(\frac{1}{2}+\frac{1}{3}\left(1-\frac{1}{2}\right)\right)\right)$$ are divisible by 5 and not 3 or 2
$$\frac{1}{7}\left(1-\left(\frac{1}{2}+\frac{1}{3}\left(1-\frac{1}{2}\right)+\frac{1}{5}\left(1-\left(\frac{1}{2}+\frac{1}{3}\left(1-\frac{1}{2}\right)\right)\right)\right)\right)$$ are divisible by 7 and not 5, 3 or 2
If I sum all these to infinity, it should equal 1 as it will then cover all positive integers.
So could you by any chance verify the truth of this statement,
$$\sum^{\infty}_{p_{0}=2}\frac{1}{p_{0}}\left(1-\sum^{p_{0}-1}_{p_{1}=2}\frac{1}{p_{1}}\left(1-\sum^{p_{1}-1}_{p_{2}=2}\frac{1}{p_{2}}\left(...\right)\right)\right) = 1$$
I know this is quite awkward looking and I don't even know if I've used correct notation, I just sort of scrawled it down as I was thinking it.
2. Nov 1, 2009
### CRGreathouse
To make it a little easier: the sum of "divisible by 2" + "not divisible by 2 but divisible by 3" + "not divisible by 2 or 3 but divisible by 5" is "divisible by 2, 3, or 5" which is "not relatively prime to 2, 3, and 5":
$$f_1 = 1/2$$
$$f_2 = 1/2 + 1/2\cdot1/3 = 1 - 1/2\cdot2/3$$
$$f_3 = 1/2 + 1/2\cdot1/3 + 1/2\cdot2/3\cdot1/5 = 1 - 1/2\cdot2/3\cdot4/5$$
. . .
$$f_k = 1 - \sum_{q\in\mathbb{P},q\le p_k}\left(1-\frac1p\right)$$
Now instead of lots of sums, you only have one. All you need to show is that the sum in question tends to 0 as k goes to infinity (which is true). | 2018-07-20 05:54:09 | {"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": 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": 0.49952539801597595, "perplexity": 319.5493998659261}, "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-2018-30/segments/1531676591497.58/warc/CC-MAIN-20180720041611-20180720061611-00564.warc.gz"} |
http://www.conwaylife.com/forums/viewtopic.php?p=42474 | For discussion of specific patterns or specific families of patterns, both newly-discovered and well-known.
drc wrote:
Ethanagor wrote:I may or may not have just discovered a new spaceship. I checked through the wiki and couldn't find it, but tell me if this is already a thing. It is speed 2c/20 and orthogonal. Here is the RLE:
That ship is already known, other rules aren't on the wiki, and it's not in life.
Sorry, I didn't realize I wan't using the life rules... Oops! I can delete the response if you want...
"It's not easy having a good time. Even smiling makes my face ache." - Frank N. Furter
Ethanagor
Posts: 78
Joined: March 15th, 2017, 7:34 pm
Location: the Milky Way galaxy
Reporting negative result for c/8 diagonal for width 6 asymmetry, width 11 bilateral symmetry, width 12 glide-reflective symmetry, width 13 gutter symmetry, and width 14 skew gutter symmetry.
Also negative result for c/11 width 6 asymmetry and width 13 gutter symmetry.
I could not work out how to get partials from the c/8d search (gfind-pt), but here are some of the more "promising" partials from the c/11 search:
x = 341, y = 33, rule = B3/S2329bo$3b2o8bo14bobo12bo3bo13bo3bo$2bo2bo6bobo13bobo11bobobobo11bobobobo17bo3bo$2bo2bo5bo2bo14bo12bobobobo11bobobobo16bobobobo40bo3bo45bo3bo47bo5bo$12b2o27b2obobob2o9b2obobob2o15bobobobo19bo3bo15bobobobo21bo3bo17bobobobo45bobo3bobo20bo5bo$2bo38b2obobob2o9b2obobob2o14b2obobob2o17bobobobo14bobobobo20bobobobo16bobobobo13bo3bo26bo2bo3bo2bo18bobo3bobo26bo5bo$bobo25b2o10b2obobob2o9b2obobob2o14b2obobob2o17bobobobo13b2obobob2o19bobobobo15b2obobob2o12bo3bo27b2o5b2o19bobo3bobo26bo5bo30bo5bo$4bo9bo13b3o13bobo15bobo17b2obobob2o16b2obobob2o12b2obobob2o18b2obobob2o14b2obobob2o12bo3bo56bo5bo26b3o3b3o29bo5bo$b2o11bo13bobo12b2ob2o37bobo19b2obobob2o12b2obobob2o18b2obobob2o14b2obobob2o145bo3bo$2bo9bo2bo13bo14bobo14bo3bo41b2obobob2o15bobo21b2obobob2o17bobo13b3o3b3o27bo3bo54bo3bobo3bo$2b2o8b2o28bo5bo10b3o3b3o16bo3bo21bobo45bobo36bo7bo26bobobobo22b2ob2o27bobo3bobo28b2o5b2o$bob2o7bobo12bobo11bo2bobo2bo10bo5bo15b3o3b3o39bo3bo45bo3bo12b3o3b3o25bo7bo19bo7bo62b2o5b2o$2obo23b2o54bo5bo19bo3bo14b3o3b3o20bo3bo16b3o3b3o11bobobobo26b2o5b2o18b2o2bobo2b2o24b3o3b3o30bo3bo$13bobo11bobo77b3o3b3o13bo5bo19b3o3b3o15bo5bo45b2obobob2o18b2o2bobo2b2o22b2ob2o3b2ob2o25bobobobobobo$3b2o7bobo13bo28b2o9b2o38bo5bo41bo5bo34b2o5b2o27b2ob2o20b5ob5o22b2obo5bob2o25bo3bobo3bo$2bo7b2o2bo12bobo28bo9bo189b4o3b4o24bo7bo29b3ob3o$29bo29b2obobob2o129bobo3bobo51b3obo3bob3o23b3o3b3o30b2ob2o$28bobo29bo5bo15bo7bo15bo9bo10bob2o3b2obo42bo3bo12bo7bo26bo5bo20bo7bo24bo3bobo3bo27bo7bo$28b2o51b3o5b3o13bo11bo9bo9bo16b3o5b3o11bo2bo5bo2bo7b2o7b2o24b3o3b3o20b2o3b2o28b2ob2o29bo2bo3bo2bo$60bobobobo14bo2bo3bo2bo14bo2b2ob2o2bo38b2o5b2o68b3ob3o21bo5bo25b2ob2ob2ob2o25b2o9b2o$60b2o3b2o14b3o5b3o15b2o5b2o12b2obobob2o18bo7bo33b2o5b2o26b3ob3o18bo4bobo4bo22b2obo3bob2o25b2o9b2o$58bo9bo13bo7bo16bo7bo10b2o3bobo3b2o17b2o3b2o36bo3bo53bobo7bobo21b2ob2o3b2ob2o25bo3bobo3bo$60bo5bo15b3o3b3o15b[code][/code]2o2bobo2b2o9b2o3bobo3b2o92bo7bo20bo5bo63bo2bobo2bo$58b3o5b3o14bo5bo17bobo3bobo11bob3ob3obo18bobobobo68bobo3bobo20bobobobo63bob2ob2obo$60bo5bo16bo5bo18b2o3b2o11b2o3bobo3b2o19bobo69bo2b2ob2o2bo16b3o2bobo2b3o62bo3bo$60b3ob3o16b2o3b2o18b2o3b2o14b3ob3o96b2o3b2o21b2o3b2o64bo5bo$60b3ob3o16bo5bo17bobo3bobo13b2o3b2o21bo3bo98b3ob3o63b3o3b3o$59bo7bo14b3o3b3o15bobo5bobo9bo3bo3bo3bo17b3ob3o94bob2o5b2obo61b2o3b2o$59bo2bobo2bo38bo9bo10b3o5b3o15b6ob6o162b2o2bo3bo2b2o$57bo4bobo4bo10b3o7b3o14b2obobob2o12b2o5b2o18bo2bobo2bo$57b2o3bobo3b2o36bo3bobo3bo10bo3bobo3bo15b2obobobobob2o$58b2o7b2o14b2o3b2o18b3ob3o$81b5ob5o!
EDIT:
negative result for c/11 width 12 even bilateral symmetry. Here are the most "promising" partials:
x = 181, y = 28, rule = B3/S23119b2o20b2o31b2o$99b2o14bo8bo14b6o11b2o15b4o$50b2o4b2o8b2o4b2o8b2o4b2o7b6o11bo10bo13b2o2b2o9b6o13bo2bo$bo3b2o3bo6b2o4b2o6b2o4b2o10b3o4b3o5b4o4b4o4b4o4b4o5b2o2b2o12bo8bo13bobo2bobo8b2o2b2o13b4o$obo6bobo4b4o2b4o5b3o2b3o10b3o4b3o36bo2bo2bo2bo11b3o2b3o14bobo2bobo7bobo2bobo11bob2obo$o2bo4bo2bo3bobob4obobo6bo2bo27bo3bo2bo3bo4bo3bo2bo3bo3bob2o2b2obo10b2o6b2o13bo6bo7bobo2bobo10bobo2bobo$15bobob4obobo3b3o4b3o25bo2bo2bo2bo6bo2bo2bo2bo8b2o14bobo4bobo13b2o4b2o7bo6bo10bob4obo$2b3o2b3o6b2obo2bob2o4bo8bo9bo8bo37bo6bo10bob2o4b2obo11b3o4b3o6bobo2bobo9b2o6b2o$3b2o2b2o9bob2obo7bob4obo10bo3b2o3bo6bo2bo2bo2bo21b2ob2ob2o12bo6bo13b3o4b3o5b3o4b3o8b2o6b2o$2bo6bo8bo4bo6b2ob4ob2o9b4o2b4o5bo10bo5bobo4bobo3b3ob4ob3o9b2o6b2o27b2o6b2o7b2o8b2o$o10bo4bo2bo2bo2bo5b3o2b3o12bo4bo8b2obo2bob2o5bo2bo4bo2bo2b2o8b2o34bo2bo7b2o2b4o2b2o6b2obo4bob2o$2bobo2bobo6b2o6b2o6bo4bo30bo2bo9bobo4bobo4b2o6b2o11bo6bo30bob2obo11b2o4b2o$o2b2o2b2o2bo3bob2ob2ob2obo23bobo2bobo7bob2o2b2obo7bo6bo6b2o4b2o10b3o6b3o13b2o2b2o8b2o4b2o11bo4bo$2bobo2bobo6b3ob2ob3o5bo2b2o2bo10b3o4b3o24b2o2b2o7bo6bo34b2o4b2o8b2o2b2o$29bobobo2bobobo8bobo4bobo6b3o4b3o7bobo2bobo6bo6bo10b2o8b2o44bob2o2b2obo$49bobo4bobo6bobo4bobo9b4o7b4o2b4o33bo6bo7b3o2b3o9bob2o2b2obo$66bob4obo7b2ob4ob2o4bo2bo2bo2bo12b2o2b2o18b2o10bobo2bobo9bob6obo$65b2o2b2o2b2o8bo4bo6b4o2b4o11b2o4b2o13b3o4b3o6bo6bo$82bo6bo6b2o4b2o13bob2obo14b2obo2bob2o6b2ob2ob2o$83b2o2b2o6b3o4b3o10b2o2b2o2b2o48b4o$80bo10bo6b4o14bo6bo15b2o2b2o7bo2bo2bo2bo12b2o$97bo4bo14bo4bo15b8o5bo3bo2bo3bo6b4o4b4o$137b3o4b3o5b3o4b3o7b2obo4bob2o$136bo2bo4bo2bo22bo8bo$138b3o2b3o9bo2bo12b2o4b2o$138bo6bo7bo2b2o2bo$153bo6bo$170b3o4b3o! EDIT 2: Negative result for c/11 width 11 odd bilateral symmetry. Things to work on: • Work on the snowflakes orthogonoid Goldtiger997 Posts: 446 Joined: June 21st, 2016, 8:00 am Location: 11.329903°N 142.199305°E ### Re: Spaceship Discussion Thread I have confirmed the absence of even-symmetric full-period 3c/9 ships at width 12 using gfind, and I will likely soon do the same for 2c/8 glide-symmetric full-period width-16. My knightt width-11 asymmetric c/6 search is still going strong at #D 63 after 635 CPU-hours; the last two lines output were #D 11 45 104 161 216 220 241 267 283 298 317 331 348 359 373 380 385 399 405 431 469 480 515 535 551 572 618 655 780 872 929 1003 1110 1235 1383 1759 2291 2751 2970 3232 3621 4096 4534 4884 5307 5757 6191 6622 7135 7661 8169 8722 9363 10123 10756 11410 12301 13160 14200 15230 16439#D 63 37 223244119 27 913689402 129 109613574 88 791463671 11 973599409 825355 0 0 0 825783 1 17288 . Could someone explain what they mean? x₁=ηx V ⃰_η=c²√(Λη) K=(Λu²)/2 Pₐ=1−1/(∫^∞_t₀(p(t)ˡ⁽ᵗ⁾)dt) $$x_1=\eta x$$ $$V^*_\eta=c^2\sqrt{\Lambda\eta}$$ $$K=\frac{\Lambda u^2}2$$ $$P_a=1-\frac1{\int^\infty_{t_0}p(t)^{l(t)}dt}$$ http://conwaylife.com/wiki/A_for_all Aidan F. Pierce A for awesome Posts: 1634 Joined: September 13th, 2014, 5:36 pm Location: 0x-1 ### Re: Spaceship Discussion Thread A for awesome wrote:the last two lines output were #D 11 45 104 161 216 220 241 267 283 298 317 331 348 359 373 380 385 399 405 431 469 480 515 535 551 572 618 655 780 872 929 1003 1110 1235 1383 1759 2291 2751 2970 3232 3621 4096 4534 4884 5307 5757 6191 6622 7135 7661 8169 8722 9363 10123 10756 11410 12301 13160 14200 15230 16439#D 63 37 223244119 27 913689402 129 109613574 88 791463671 11 973599409 825355 0 0 0 825783 1 17288 . Could someone explain what they mean? The search space for knightt is a tree where each node represents a "row" of a potential spaceship. I'm not exactly sure what knightt considers to be a single "row", but it is probably similar to gfind: one node in the graph represents 2p actual rows of the spaceship, where p is the period. The root node then represents an empty row (this is the front of the spaceship). If a node in the tree only leads to dead ends, then it is removed from the search. The first output line shows how many nodes are left at each level in the search. That is, at a depth of 1, there are only 11 possible nodes; at depth 2 there are 45 nodes; etc. The second output line gives some statistics, but I'm not sure what they all are. You would need to check the source code to figure that out. Obviously, 63 is the current depth of the search. One of the numbers is the current memory usage, but I'm not sure which one. -Matthias Merzenich Sokwe Moderator Posts: 1264 Joined: July 9th, 2009, 2:44 pm ### Re: Spaceship Discussion Thread Sokwe wrote:The second output line gives some statistics, but I'm not sure what they all are. You would need to check the source code to figure that out. Obviously, 63 is the current depth of the search. One of the numbers is the current memory usage, but I'm not sure which one. It's in lines 2392-94 of the source code, it appears: printf("#D %d %d %d %d %d %d %d %d %d %d %d %d %d %d %d %d %d %d\n", rownum, numtriesb, numtries, contriesb, contries, conmissqb, conmissq, contritsb, contrits, contramissesb, contramisses, sortop - sortroads, inrealship, inship, openship, offroad - allroads, trims, numhits); Unfortunately, while knightt's variable names are much more helpful than, say, those in the original version of zfind, they're still not transparent enough for me to figure out what all they mean (apart from rownum, of course). x₁=ηx V ⃰_η=c²√(Λη) K=(Λu²)/2 Pₐ=1−1/(∫^∞_t₀(p(t)ˡ⁽ᵗ⁾)dt) $$x_1=\eta x$$ $$V^*_\eta=c^2\sqrt{\Lambda\eta}$$ $$K=\frac{\Lambda u^2}2$$ $$P_a=1-\frac1{\int^\infty_{t_0}p(t)^{l(t)}dt}$$ http://conwaylife.com/wiki/A_for_all Aidan F. Pierce A for awesome Posts: 1634 Joined: September 13th, 2014, 5:36 pm Location: 0x-1 ### Re: Spaceship Discussion Thread I'd be surprised if these aren't known, but they're not listed anywhere that I can see: x = 77, y = 29, rule = B3/S2348b2o14bo10b2o$2ob2o16b2o4b2o13b2o2bo7b2o9bo3b2o2bo$14bob2ob3o6b3obo3b2o2bo4bo3bo5b3obo3b2o2bo4bo3bo$o3bo5b2o3bo4b2o6b2o3bo7bobo3b2o6b2o3bo7bobo3b2o$b2obo8b2obobob2o6b2obobo3bo5bobobo7b2obobo3bo5bobobo$2bobobo3bo5bobo12bobob2o3b2obob2o11bobob2o3b2obob2o$3b2obob3o2b2obob2o10b2obobo7bobo11b2obobo7bobo$4bobo6b2ob2o14b2o7bob2o14b2o7bob2o$7bo6bobo16bo5bob2o17bo5bob2o$5bo6bo19b2o2bo4bo17b2o2bo4bo$5bo3bobob2o18b2o3bob2o18b2o3bob2o$6bo2b5o20b3ob2o21b3ob2o$7b3o25bo26bo5$2b2o3bo2bo3bobo2b2ob2o$6bo4bobo3bo2bobo$2bo3bob2obobobobobob2o$3b2obo4bobo3bobob2o$4bobobo2bo2b2obobo$5b2obob2o3bobob2o$6bobo8b2o$9bo7bo$7bo6bo2b2o$7bo4bo3b2o$8bo3bob3o$9b3o3bo! EDIT: Another: x = 22, y = 13, rule = B3/S234b2o$2o3bo2b2o2bobo2b2ob2o$4bob3o2bo3bo2bobo$o3bob2obobobobobob2o$b2obo4bobo3bobob2o$2bobobo2bo2b2obobo$3b2obob2o3bobob2o$4bobo8b2o$7bo7bo$5bo6bo2b2o$5bo4bo3b2o$6bo3bob3o$7b3o3bo! x₁=ηx V ⃰_η=c²√(Λη) K=(Λu²)/2 Pₐ=1−1/(∫^∞_t₀(p(t)ˡ⁽ᵗ⁾)dt) $$x_1=\eta x$$ $$V^*_\eta=c^2\sqrt{\Lambda\eta}$$ $$K=\frac{\Lambda u^2}2$$ $$P_a=1-\frac1{\int^\infty_{t_0}p(t)^{l(t)}dt}$$ http://conwaylife.com/wiki/A_for_all Aidan F. Pierce A for awesome Posts: 1634 Joined: September 13th, 2014, 5:36 pm Location: 0x-1 ### Re: Spaceship Discussion Thread Remember this "double loafer"? x = 14, y = 35, rule = B3/S232bo8bo$bobo6bobo$o2bo6bo2bo$b2o8b2o$6b2o$4b2o2b2o$4bo4bo$4bo4bo$3b8o$2b4o2b4o$bo2bo4bo2bo$o3bo4bo3bo$bo2bo4bo2bo3$b3o6b3o3$2bo8bo$b3o6b3o$2ob3o2b3ob2o$obo2bo2bo2bobo$b2o3b2o3b2o$5bo2bo$bo3b4o3bo2$bo2bo4bo2bo$2b3o4b3o$2bo8bo$b2o2b4o2b2o$14o$3o8b3o$6b2o$2o3b4o3b2o$o12bo!
I tried to put two front ends side by side and search for a continuation between,and here are all partials whose length > 150:
x = 1456, y = 61, rule = B3/S232bo8bo11bo8bo11bo8bo18bo8bo11bo8bo11bo8bo18bo8bo11bo8bo11bo8bo18bo8bo11bo8bo11bo8bo18bo8bo11bo8bo11bo8bo18bo8bo11bo8bo11bo8bo18bo8bo11bo8bo11bo8bo18bo8bo11bo8bo11bo8bo18bo8bo11bo8bo11bo8bo18bo8bo11bo8bo11bo8bo18bo8bo11bo8bo11bo8bo18bo8bo11bo8bo11bo8bo18bo8bo11bo8bo11bo8bo18bo8bo11bo8bo11bo8bo18bo8bo11bo8bo11bo8bo18bo8bo11bo8bo11bo8bo18bo8bo11bo8bo11bo8bo18bo8bo11bo8bo11bo8bo18bo8bo11bo8bo11bo8bo18bo8bo11bo8bo11bo8bo18bo8bo11bo8bo11bo8bo$bobo6bobo9bobo6bobo9bobo6bobo16bobo6bobo9bobo6bobo9bobo6bobo16bobo6bobo9bobo6bobo9bobo6bobo16bobo6bobo9bobo6bobo9bobo6bobo16bobo6bobo9bobo6bobo9bobo6bobo16bobo6bobo9bobo6bobo9bobo6bobo16bobo6bobo9bobo6bobo9bobo6bobo16bobo6bobo9bobo6bobo9bobo6bobo16bobo6bobo9bobo6bobo9bobo6bobo16bobo6bobo9bobo6bobo9bobo6bobo16bobo6bobo9bobo6bobo9bobo6bobo16bobo6bobo9bobo6bobo9bobo6bobo16bobo6bobo9bobo6bobo9bobo6bobo16bobo6bobo9bobo6bobo9bobo6bobo16bobo6bobo9bobo6bobo9bobo6bobo16bobo6bobo9bobo6bobo9bobo6bobo16bobo6bobo9bobo6bobo9bobo6bobo16bobo6bobo9bobo6bobo9bobo6bobo16bobo6bobo9bobo6bobo9bobo6bobo16bobo6bobo9bobo6bobo9bobo6bobo16bobo6bobo9bobo6bobo9bobo6bobo$o2bo6bo2bo7bo2bo6bo2bo7bo2bo6bo2bo14bo2bo6bo2bo7bo2bo6bo2bo7bo2bo6bo2bo14bo2bo6bo2bo7bo2bo6bo2bo7bo2bo6bo2bo14bo2bo6bo2bo7bo2bo6bo2bo7bo2bo6bo2bo14bo2bo6bo2bo7bo2bo6bo2bo7bo2bo6bo2bo14bo2bo6bo2bo7bo2bo6bo2bo7bo2bo6bo2bo14bo2bo6bo2bo7bo2bo6bo2bo7bo2bo6bo2bo14bo2bo6bo2bo7bo2bo6bo2bo7bo2bo6bo2bo14bo2bo6bo2bo7bo2bo6bo2bo7bo2bo6bo2bo14bo2bo6bo2bo7bo2bo6bo2bo7bo2bo6bo2bo14bo2bo6bo2bo7bo2bo6bo2bo7bo2bo6bo2bo14bo2bo6bo2bo7bo2bo6bo2bo7bo2bo6bo2bo14bo2bo6bo2bo7bo2bo6bo2bo7bo2bo6bo2bo14bo2bo6bo2bo7bo2bo6bo2bo7bo2bo6bo2bo14bo2bo6bo2bo7bo2bo6bo2bo7bo2bo6bo2bo14bo2bo6bo2bo7bo2bo6bo2bo7bo2bo6bo2bo14bo2bo6bo2bo7bo2bo6bo2bo7bo2bo6bo2bo14bo2bo6bo2bo7bo2bo6bo2bo7bo2bo6bo2bo14bo2bo6bo2bo7bo2bo6bo2bo7bo2bo6bo2bo14bo2bo6bo2bo7bo2bo6bo2bo7bo2bo6bo2bo14bo2bo6bo2bo7bo2bo6bo2bo7bo2bo6bo2bo$b2o8b2o9b2o8b2o9b2o8b2o16b2o8b2o9b2o8b2o9b2o8b2o16b2o8b2o9b2o8b2o9b2o8b2o16b2o8b2o9b2o8b2o9b2o8b2o16b2o8b2o9b2o8b2o9b2o8b2o16b2o8b2o9b2o8b2o9b2o8b2o16b2o8b2o9b2o8b2o9b2o8b2o16b2o8b2o9b2o8b2o9b2o8b2o16b2o8b2o9b2o8b2o9b2o8b2o16b2o8b2o9b2o8b2o9b2o8b2o16b2o8b2o9b2o8b2o9b2o8b2o16b2o8b2o9b2o8b2o9b2o8b2o16b2o8b2o9b2o8b2o9b2o8b2o16b2o8b2o9b2o8b2o9b2o8b2o16b2o8b2o9b2o8b2o9b2o8b2o16b2o8b2o9b2o8b2o9b2o8b2o16b2o8b2o9b2o8b2o9b2o8b2o16b2o8b2o9b2o8b2o9b2o8b2o16b2o8b2o9b2o8b2o9b2o8b2o16b2o8b2o9b2o8b2o9b2o8b2o16b2o8b2o9b2o8b2o9b2o8b2o$6b2o19b2o19b2o26b2o19b2o19b2o26b2o19b2o19b2o26b2o19b2o19b2o26b2o19b2o19b2o26b2o19b2o19b2o26b2o19b2o19b2o26b2o19b2o19b2o26b2o19b2o19b2o26b2o19b2o19b2o26b2o19b2o19b2o26b2o19b2o19b2o26b2o19b2o19b2o26b2o19b2o19b2o26b2o19b2o19b2o26b2o19b2o19b2o26b2o19b2o19b2o26b2o19b2o19b2o26b2o19b2o19b2o26b2o19b2o19b2o26b2o19b2o19b2o$4b2o2b2o15b2o2b2o15b2o2b2o22b2o2b2o15b2o2b2o15b2o2b2o22b2o2b2o15b2o2b2o15b2o2b2o22b2o2b2o15b2o2b2o15b2o2b2o22b2o2b2o15b2o2b2o15b2o2b2o22b2o2b2o15b2o2b2o15b2o2b2o22b2o2b2o15b2o2b2o15b2o2b2o22b2o2b2o15b2o2b2o15b2o2b2o22b2o2b2o15b2o2b2o15b2o2b2o22b2o2b2o15b2o2b2o15b2o2b2o22b2o2b2o15b2o2b2o15b2o2b2o22b2o2b2o15b2o2b2o15b2o2b2o22b2o2b2o15b2o2b2o15b2o2b2o22b2o2b2o15b2o2b2o15b2o2b2o22b2o2b2o15b2o2b2o15b2o2b2o22b2o2b2o15b2o2b2o15b2o2b2o22b2o2b2o15b2o2b2o15b2o2b2o22b2o2b2o15b2o2b2o15b2o2b2o22b2o2b2o15b2o2b2o15b2o2b2o22b2o2b2o15b2o2b2o15b2o2b2o22b2o2b2o15b2o2b2o15b2o2b2o$4bo4bo15bo4bo15bo4bo22bo4bo15bo4bo15bo4bo22bo4bo15bo4bo15bo4bo22bo4bo15bo4bo15bo4bo22bo4bo15bo4bo15bo4bo22bo4bo15bo4bo15bo4bo22bo4bo15bo4bo15bo4bo22bo4bo15bo4bo15bo4bo22bo4bo15bo4bo15bo4bo22bo4bo15bo4bo15bo4bo22bo4bo15bo4bo15bo4bo22bo4bo15bo4bo15bo4bo22bo4bo15bo4bo15bo4bo22bo4bo15bo4bo15bo4bo22bo4bo15bo4bo15bo4bo22bo4bo15bo4bo15bo4bo22bo4bo15bo4bo15bo4bo22bo4bo15bo4bo15bo4bo22bo4bo15bo4bo15bo4bo22bo4bo15bo4bo15bo4bo22bo4bo15bo4bo15bo4bo$4bo4bo15bo4bo15bo4bo22bo4bo15bo4bo15bo4bo22bo4bo15bo4bo15bo4bo22bo4bo15bo4bo15bo4bo22bo4bo15bo4bo15bo4bo22bo4bo15bo4bo15bo4bo22bo4bo15bo4bo15bo4bo22bo4bo15bo4bo15bo4bo22bo4bo15bo4bo15bo4bo22bo4bo15bo4bo15bo4bo22bo4bo15bo4bo15bo4bo22bo4bo15bo4bo15bo4bo22bo4bo15bo4bo15bo4bo22bo4bo15bo4bo15bo4bo22bo4bo15bo4bo15bo4bo22bo4bo15bo4bo15bo4bo22bo4bo15bo4bo15bo4bo22bo4bo15bo4bo15bo4bo22bo4bo15bo4bo15bo4bo22bo4bo15bo4bo15bo4bo22bo4bo15bo4bo15bo4bo$3b8o13b8o13b8o20b8o13b8o13b8o20b8o13b8o13b8o20b8o13b8o13b8o20b8o13b8o13b8o20b8o13b8o13b8o20b8o13b8o13b8o20b8o13b8o13b8o20b8o13b8o13b8o20b8o13b8o13b8o20b8o13b8o13b8o20b8o13b8o13b8o20b8o13b8o13b8o20b8o13b8o13b8o20b8o13b8o13b8o20b8o13b8o13b8o20b8o13b8o13b8o20b8o13b8o13b8o20b8o13b8o13b8o20b8o13b8o13b8o20b8o13b8o13b8o$2b4o2b4o11b4o2b4o11b4o2b4o18b4o2b4o11b4o2b4o11b4o2b4o18b4o2b4o11b4o2b4o11b4o2b4o18b4o2b4o11b4o2b4o11b4o2b4o18b4o2b4o11b4o2b4o11b4o2b4o18b4o2b4o11b4o2b4o11b4o2b4o18b4o2b4o11b4o2b4o11b4o2b4o18b4o2b4o11b4o2b4o11b4o2b4o18b4o2b4o11b4o2b4o11b4o2b4o18b4o2b4o11b4o2b4o11b4o2b4o18b4o2b4o11b4o2b4o11b4o2b4o18b4o2b4o11b4o2b4o11b4o2b4o18b4o2b4o11b4o2b4o11b4o2b4o18b4o2b4o11b4o2b4o11b4o2b4o18b4o2b4o11b4o2b4o11b4o2b4o18b4o2b4o11b4o2b4o11b4o2b4o18b4o2b4o11b4o2b4o11b4o2b4o18b4o2b4o11b4o2b4o11b4o2b4o18b4o2b4o11b4o2b4o11b4o2b4o18b4o2b4o11b4o2b4o11b4o2b4o18b4o2b4o11b4o2b4o11b4o2b4o$bo2bo4bo2bo9bo2bo4bo2bo9bo2bo4bo2bo16bo2bo4bo2bo9bo2bo4bo2bo9bo2bo4bo2bo16bo2bo4bo2bo9bo2bo4bo2bo9bo2bo4bo2bo16bo2bo4bo2bo9bo2bo4bo2bo9bo2bo4bo2bo16bo2bo4bo2bo9bo2bo4bo2bo9bo2bo4bo2bo16bo2bo4bo2bo9bo2bo4bo2bo9bo2bo4bo2bo16bo2bo4bo2bo9bo2bo4bo2bo9bo2bo4bo2bo16bo2bo4bo2bo9bo2bo4bo2bo9bo2bo4bo2bo16bo2bo4bo2bo9bo2bo4bo2bo9bo2bo4bo2bo16bo2bo4bo2bo9bo2bo4bo2bo9bo2bo4bo2bo16bo2bo4bo2bo9bo2bo4bo2bo9bo2bo4bo2bo16bo2bo4bo2bo9bo2bo4bo2bo9bo2bo4bo2bo16bo2bo4bo2bo9bo2bo4bo2bo9bo2bo4bo2bo16bo2bo4bo2bo9bo2bo4bo2bo9bo2bo4bo2bo16bo2bo4bo2bo9bo2bo4bo2bo9bo2bo4bo2bo16bo2bo4bo2bo9bo2bo4bo2bo9bo2bo4bo2bo16bo2bo4bo2bo9bo2bo4bo2bo9bo2bo4bo2bo16bo2bo4bo2bo9bo2bo4bo2bo9bo2bo4bo2bo16bo2bo4bo2bo9bo2bo4bo2bo9bo2bo4bo2bo16bo2bo4bo2bo9bo2bo4bo2bo9bo2bo4bo2bo16bo2bo4bo2bo9bo2bo4bo2bo9bo2bo4bo2bo$o3bo4bo3bo7bo3bo4bo3bo7bo3bo4bo3bo14bo3bo4bo3bo7bo3bo4bo3bo7bo3bo4bo3bo14bo3bo4bo3bo7bo3bo4bo3bo7bo3bo4bo3bo14bo3bo4bo3bo7bo3bo4bo3bo7bo3bo4bo3bo14bo3bo4bo3bo7bo3bo4bo3bo7bo3bo4bo3bo14bo3bo4bo3bo7bo3bo4bo3bo7bo3bo4bo3bo14bo3bo4bo3bo7bo3bo4bo3bo7bo3bo4bo3bo14bo3bo4bo3bo7bo3bo4bo3bo7bo3bo4bo3bo14bo3bo4bo3bo7bo3bo4bo3bo7bo3bo4bo3bo14bo3bo4bo3bo7bo3bo4bo3bo7bo3bo4bo3bo14bo3bo4bo3bo7bo3bo4bo3bo7bo3bo4bo3bo14bo3bo4bo3bo7bo3bo4bo3bo7bo3bo4bo3bo14bo3bo4bo3bo7bo3bo4bo3bo7bo3bo4bo3bo14bo3bo4bo3bo7bo3bo4bo3bo7bo3bo4bo3bo14bo3bo4bo3bo7bo3bo4bo3bo7bo3bo4bo3bo14bo3bo4bo3bo7bo3bo4bo3bo7bo3bo4bo3bo14bo3bo4bo3bo7bo3bo4bo3bo7bo3bo4bo3bo14bo3bo4bo3bo7bo3bo4bo3bo7bo3bo4bo3bo14bo3bo4bo3bo7bo3bo4bo3bo7bo3bo4bo3bo14bo3bo4bo3bo7bo3bo4bo3bo7bo3bo4bo3bo14bo3bo4bo3bo7bo3bo4bo3bo7bo3bo4bo3bo$bo2bo4bo2bo9bo2bo4bo2bo9bo2bo4bo2bo16bo2bo4bo2bo9bo2bo4bo2bo9bo2bo4bo2bo16bo2bo4bo2bo9bo2bo4bo2bo9bo2bo4bo2bo16bo2bo4bo2bo9bo2bo4bo2bo9bo2bo4bo2bo16bo2bo4bo2bo9bo2bo4bo2bo9bo2bo4bo2bo16bo2bo4bo2bo9bo2bo4bo2bo9bo2bo4bo2bo16bo2bo4bo2bo9bo2bo4bo2bo9bo2bo4bo2bo16bo2bo4bo2bo9bo2bo4bo2bo9bo2bo4bo2bo16bo2bo4bo2bo9bo2bo4bo2bo9bo2bo4bo2bo16bo2bo4bo2bo9bo2bo4bo2bo9bo2bo4bo2bo16bo2bo4bo2bo9bo2bo4bo2bo9bo2bo4bo2bo16bo2bo4bo2bo9bo2bo4bo2bo9bo2bo4bo2bo16bo2bo4bo2bo9bo2bo4bo2bo9bo2bo4bo2bo16bo2bo4bo2bo9bo2bo4bo2bo9bo2bo4bo2bo16bo2bo4bo2bo9bo2bo4bo2bo9bo2bo4bo2bo16bo2bo4bo2bo9bo2bo4bo2bo9bo2bo4bo2bo16bo2bo4bo2bo9bo2bo4bo2bo9bo2bo4bo2bo16bo2bo4bo2bo9bo2bo4bo2bo9bo2bo4bo2bo16bo2bo4bo2bo9bo2bo4bo2bo9bo2bo4bo2bo16bo2bo4bo2bo9bo2bo4bo2bo9bo2bo4bo2bo16bo2bo4bo2bo9bo2bo4bo2bo9bo2bo4bo2bo3$b3o6b3o9b3o6b3o9b3o6b3o16b3o6b3o9b3o6b3o9b3o6b3o25b3o9b3o6b3o9b3o6b3o16b3o6b3o9b3o6b3o9b3o6b3o16b3o6b3o9b3o6b3o9b3o6b3o16b3o6b3o9b3o6b3o9b3o6b3o16b3o6b3o9b3o6b3o9b3o6b3o16b3o6b3o9b3o6b3o9b3o6b3o16b3o6b3o9b3o6b3o9b3o6b3o16b3o6b3o9b3o6b3o9b3o6b3o16b3o6b3o9b3o6b3o9b3o6b3o16b3o6b3o9b3o6b3o9b3o6b3o16b3o6b3o9b3o6b3o9b3o6b3o16b3o6b3o9b3o6b3o9b3o6b3o16b3o6b3o9b3o6b3o9b3o6b3o16b3o6b3o9b3o6b3o9b3o6b3o16b3o6b3o9b3o6b3o9b3o6b3o16b3o6b3o9b3o6b3o9b3o6b3o16b3o6b3o9b3o6b3o9b3o6b3o16b3o6b3o9b3o6b3o9b3o6b3o16b3o6b3o9b3o6b3o9b3o6b3o$141b3o2$11bo11bo8bo11bo36bo11bo8bo11bo36bo11bo8bo11bo36bo11bo8bo11bo36bo11bo8bo11bo36bo11bo8bo11bo36bo11bo8bo11bo36bo11bo8bo11bo36bo11bo8bo11bo36bo11bo8bo11bo36bo11bo8bo11bo36bo11bo8bo11bo36bo11bo8bo11bo36bo11bo8bo11bo36bo11bo8bo11bo36bo11bo8bo11bo36bo11bo8bo11bo36bo11bo8bo11bo36bo11bo8bo11bo36bo11bo8bo11bo36bo11bo8bo11bo$10b3o9b3o6b3o9b3o34b3o9b3o6b3o9b3o34b3o9b3o6b3o9b3o34b3o9b3o6b3o9b3o34b3o9b3o6b3o9b3o34b3o9b3o6b3o9b3o34b3o9b3o6b3o9b3o34b3o9b3o6b3o9b3o34b3o9b3o6b3o9b3o34b3o9b3o6b3o9b3o34b3o9b3o6b3o9b3o34b3o9b3o6b3o9b3o34b3o9b3o6b3o9b3o34b3o9b3o6b3o9b3o34b3o9b3o6b3o9b3o34b3o9b3o6b3o9b3o34b3o9b3o6b3o9b3o34b3o9b3o6b3o9b3o34b3o9b3o6b3o9b3o34b3o9b3o6b3o9b3o34b3o9b3o6b3o9b3o$8b3ob2o7b2ob3o2b3ob2o7b2ob3o30b3ob2o7b2ob3o2b3ob2o7b2ob3o30b3ob2o7b2ob3o2b3ob2o7b2ob3o30b3ob2o7b2ob3o2b3ob2o7b2ob3o30b3ob2o7b2ob3o2b3ob2o7b2ob3o30b3ob2o7b2ob3o2b3ob2o7b2ob3o30b3ob2o7b2ob3o2b3ob2o7b2ob3o30b3ob2o7b2ob3o2b3ob2o7b2ob3o30b3ob2o7b2ob3o2b3ob2o7b2ob3o30b3ob2o7b2ob3o2b3ob2o7b2ob3o30b3ob2o7b2ob3o2b3ob2o7b2ob3o30b3ob2o7b2ob3o2b3ob2o7b2ob3o30b3ob2o7b2ob3o2b3ob2o7b2ob3o30b3ob2o7b2ob3o2b3ob2o7b2ob3o30b3ob2o7b2ob3o2b3ob2o7b2ob3o30b3ob2o7b2ob3o2b3ob2o7b2ob3o30b3ob2o7b2ob3o2b3ob2o7b2ob3o30b3ob2o7b2ob3o2b3ob2o7b2ob3o30b3ob2o7b2ob3o2b3ob2o7b2ob3o30b3ob2o7b2ob3o2b3ob2o7b2ob3o30b3ob2o7b2ob3o2b3ob2o7b2ob3o$8b2obobo7bobob2o2b2obobo7bobob2o30b2obobo7bobob2o2b2obobo7bobob2o30b2obobo7bobob2o2b2obobo7bobob2o30b2obobo7bobob2o2b2obobo7bobob2o30b2obobo7bobob2o2b2obobo7bobob2o30b2obobo7bobob2o2b2obobo7bobob2o30b2obobo7bobob2o2b2obobo7bobob2o30b2obobo7bobob2o2b2obobo7bobob2o30b2obobo7bobob2o2b2obobo7bobob2o30b2obobo7bobob2o2b2obobo7bobob2o30b2obobo7bobob2o2b2obobo7bobob2o30b2obobo7bobob2o2b2obobo7bobob2o30b2obobo7bobob2o2b2obobo7bobob2o30b2obobo7bobob2o2b2obobo7bobob2o30b2obobo7bobob2o2b2obobo7bobob2o30b2obobo7bobob2o2b2obobo7bobob2o30b2obobo7bobob2o2b2obobo7bobob2o30b2obobo7bobob2o2b2obobo7bobob2o30b2obobo7bobob2o2b2obobo7bobob2o30b2obobo7bobob2o2b2obobo7bobob2o30b2obobo7bobob2o2b2obobo7bobob2o$9b4o9b4o4b4o9b4o32b4o9b4o4b4o9b4o32b4o9b4o4b4o9b4o32b4o9b4o4b4o9b4o32b4o9b4o4b4o9b4o32b4o9b4o4b4o9b4o32b4o9b4o4b4o9b4o32b4o9b4o4b4o9b4o32b4o9b4o4b4o9b4o32b4o9b4o4b4o9b4o32b4o9b4o4b4o9b4o32b4o9b4o4b4o9b4o32b4o9b4o4b4o9b4o32b4o9b4o4b4o9b4o32b4o9b4o4b4o9b4o32b4o9b4o4b4o9b4o32b4o9b4o4b4o9b4o32b4o9b4o4b4o9b4o32b4o9b4o4b4o9b4o32b4o9b4o4b4o9b4o32b4o9b4o4b4o9b4o$13bobo3bobo12bobo3bobo40bobo3bobo12bobo3bobo40bobo3bobo12bobo3bobo40bobo3bobo12bobo3bobo40bobo3bobo12bobo3bobo40bobo3bobo12bobo3bobo40bobo3bobo12bobo3bobo40bobo3bobo12bobo3bobo40bobo3bobo12bobo3bobo40bobo3bobo12bobo3bobo40bobo3bobo12bobo3bobo40bobo3bobo12bobo3bobo40bobo3bobo12bobo3bobo40bobo3bobo12bobo3bobo40bobo3bobo12bobo3bobo40bobo3bobo12bobo3bobo40bobo3bobo12bobo3bobo40bobo3bobo12bobo3bobo40bobo3bobo12bobo3bobo40bobo3bobo12bobo3bobo40bobo3bobo12bobo3bobo$13bobo3bobo12bobo3bobo40b3o3b3o12b3o3b3o40b3o3b3o12b3o3b3o40b3o3b3o12b3o3b3o40b3o3b3o12b3o3b3o40b3o3b3o12b3o3b3o40b3o3b3o12b3o3b3o40b3o3b3o12b3o3b3o40b3o3b3o12b3o3b3o40b3o3b3o12b3o3b3o40b3o3b3o12b3o3b3o40b3o3b3o12b3o3b3o40b3o3b3o12b3o3b3o40b3o3b3o12b3o3b3o40b3o3b3o12b3o3b3o40b3o3b3o12b3o3b3o40b3o3b3o12b3o3b3o40b3o3b3o12b3o3b3o40b3o3b3o12b3o3b3o40b3o3b3o12b3o3b3o40b3o3b3o12b3o3b3o$9bo6bobo6bo4bo6bobo6bo32bo6bobo6bo4bo6bobo6bo32bo5bo3bo5bo4bo5bo3bo5bo32bo5bo3bo5bo4bo5bo3bo5bo32bo5bo3bo5bo4bo5bo3bo5bo32bo5bo3bo5bo4bo5bo3bo5bo32bo5bo3bo5bo4bo5bo3bo5bo32bo5bo3bo5bo4bo5bo3bo5bo32bo5bo3bo5bo4bo5bo3bo5bo32bo5bo3bo5bo4bo5bo3bo5bo32bo5bo3bo5bo4bo5bo3bo5bo32bo5bo3bo5bo4bo5bo3bo5bo32bo5bo3bo5bo4bo5bo3bo5bo32bo5bo3bo5bo4bo5bo3bo5bo32bo5bo3bo5bo4bo5bo3bo5bo32bo5bo3bo5bo4bo5bo3bo5bo32bo5bo3bo5bo4bo5bo3bo5bo32bo5bo3bo5bo4bo5bo3bo5bo32bo5bo3bo5bo4bo5bo3bo5bo32bo5bo3bo5bo4bo5bo3bo5bo32bo5bo3bo5bo4bo5bo3bo5bo$10bob2o2bobo2b2obo6bob2o2bobo2b2obo34bob3o5b3obo6bob3o5b3obo37bo7bo12bo7bo37bo2bo7bo2bo6bo2bo7bo2bo34bo2bo7bo2bo6bo2bo7bo2bo34bo2bo7bo2bo6bo2bo7bo2bo34bo2bo7bo2bo6bo2bo7bo2bo34bo2bo7bo2bo6bo2bo7bo2bo34bo2bo7bo2bo6bo2bo7bo2bo34bo2bo7bo2bo6bo2bo7bo2bo34bo2bo7bo2bo6bo2bo7bo2bo34bo2bo7bo2bo6bo2bo7bo2bo34bo2bo7bo2bo6bo2bo7bo2bo34bo2bo7bo2bo6bo2bo7bo2bo34bo2bo7bo2bo6bo2bo7bo2bo34bo2bo7bo2bo6bo2bo7bo2bo34bo2bo7bo2bo6bo2bo7bo2bo34bo2bo7bo2bo6bo2bo7bo2bo37bo7bo12bo7bo40bo7bo12bo7bo40bo7bo12bo7bo$15b2ob2o16b2ob2o42bobo3bobo12bobo3bobo37bo2bo7bo2bo6bo2bo7bo2bo33bo4bo5bo4bo4bo4bo5bo4bo32bo4bo5bo4bo4bo4bo5bo4bo36bo7bo12bo7bo40bo7bo12bo7bo41bo5bo14bo5bo42bo5bo14bo5bo42bo5bo14bo5bo42bo5bo14bo5bo42bo5bo14bo5bo42bo5bo14bo5bo42bo5bo14bo5bo42bo5bo14bo5bo42bo5bo14bo5bo42bo5bo14bo5bo42bo5bo14bo5bo42bo5bo14bo5bo42bo5bo14bo5bo42bo5bo14bo5bo$14b2o3b2o14b2o3b2o38bo3bo5bo3bo6bo3bo5bo3bo34bo2b2o5b2o2bo6bo2b2o5b2o2bo33bo2bo9bo2bo4bo2bo9bo2bo35bo9bo10bo9bo37bo11bo8bo11bo36bo11bo8bo11bo37bo9bo10bo9bo38bo9bo10bo9bo38bo9bo10bo9bo38bo9bo10bo9bo38bo9bo10bo9bo38bo9bo10bo9bo38bo9bo10bo9bo38bo9bo10bo9bo38bo9bo10bo9bo38bo9bo10bo9bo38bo9bo10bo9bo36bobo9bobo6bobo9bobo34bobo9bobo6bobo9bobo34bobo9bobo6bobo9bobo$11bobo7bobo8bobo7bobo33b2obobo7bobob2o2b2obobo7bobob2o31bobob2o5b2obobo4bobob2o5b2obobo32bobo11bobo4bobo11bobo32bobo11bobo4bobo11bobo174bo3bo3bo3bo8bo3bo3bo3bo36bo3bo3bo3bo8bo3bo3bo3bo36bo11bo8bo11bo36bo11bo8bo11bo36bo11bo8bo11bo36bo11bo8bo11bo36bo11bo8bo11bo36bo11bo8bo11bo36bo11bo8bo11bo36bo11bo8bo11bo36bo11bo8bo11bo35bo13bo6bo13bo34bo13bo6bo13bo34bo13bo6bo13bo$9b2o3bo5bo3b2o4b2o3bo5bo3b2o32bo2b2o7b2o2bo4bo2b2o7b2o2bo32bobobobo3bobobobo4bobobobo3bobobobo34b2o9b2o8b2o9b2o38bobo3bobo12bobo3bobo37b3o9b3o6b3o9b3o34b3o9b3o6b3o9b3o34b5o5b5o6b5o5b5o34b5o5b5o6b5o5b5o34b4o7b4o6b4o7b4o34b5o5b5o6b5o5b5o34b5o5b5o6b5o5b5o34b5o5b5o6b5o5b5o34b5o5b5o6b5o5b5o34b5o5b5o6b5o5b5o34b5o5b5o6b5o5b5o34b5o5b5o6b5o5b5o34b5o5b5o6b5o5b5o35bo11bo8bo11bo36bo11bo8bo11bo36bo11bo8bo11bo$12b2o7b2o10b2o7b2o36bobo9bobo6bobo9bobo32bo4bobo3bobo4bo2bo4bobo3bobo4bo33bobo7bobo8bobo7bobo33b2o3bo2bobo2bo3b2o2b2o3bo2bobo2bo3b2o31bo3bob2ob2obo3bo4bo3bob2ob2obo3bo38b2ob2o16b2ob2o39bo13bo6bo13bo34bo13bo6bo13bo35bo2bo5bo2bo8bo2bo5bo2bo35bo3bo5bo3bo6bo3bo5bo3bo34bo4bo3bo4bo6bo4bo3bo4bo34bo4bo3bo4bo6bo4bo3bo4bo34bo4bo3bo4bo6bo4bo3bo4bo34bo4bo3bo4bo6bo4bo3bo4bo34bo4bo3bo4bo6bo4bo3bo4bo34bo4bo3bo4bo6bo4bo3bo4bo34bo4bo3bo4bo6bo4bo3bo4bo35b3o7b3o8b3o7b3o36b3o7b3o8b3o7b3o36b3o7b3o8b3o7b3o$9b3o11b3o4b3o11b3o32b6o5b6o4b6o5b6o32bob5o3b5obo4bob5o3b5obo101b2o2bo2bo3bo2bo2b2o2b2o2bo2bo3bo2bo2b2o33b2o2b2ob2o2b2o8b2o2b2ob2o2b2o35bo2bobo3bobo2bo6bo2bobo3bobo2bo33b2obo2bo3bo2bob2o4b2obo2bo3bo2bob2o32b2obobo5bobob2o4b2obobo5bobob2o36bo7bo12bo7bo36b2o2bo7bo2b2o4b2o2bo7bo2b2o32b3o2bo5bo2b3o4b3o2bo5bo2b3o32b3o2bo5bo2b3o4b3o2bo5bo2b3o32b3o2bo5bo2b3o4b3o2bo5bo2b3o32b3o2bo5bo2b3o4b3o2bo5bo2b3o32b3o2bo5bo2b3o4b3o2bo5bo2b3o32b3o2bo5bo2b3o4b3o2bo5bo2b3o32b3o2bo5bo2b3o4b3o2bo5bo2b3o34b3o7b3o8b3o7b3o36b3o7b3o8b3o7b3o36b3o7b3o8b3o7b3o$13b2o5b2o12b2o5b2o36b2o3bo5bo3b2o4b2o3bo5bo3b2o34b2ob2o3b2ob2o8b2ob2o3b2ob2o35bo13bo6bo13bo33bobo11bobo4bobo11bobo31b2o3bo2bobo2bo3b2o2b2o3bo2bobo2bo3b2o33bo2bo5bo2bo8bo2bo5bo2bo35bo3bobobobo3bo6bo3bobobobo3bo36bobobobobobo10bobobobobobo38bob2o3b2obo10bob2o3b2obo35b2ob2o7b2ob2o4b2ob2o7b2ob2o37bo5bo14bo5bo42bo5bo14bo5bo42bo5bo14bo5bo42bo5bo14bo5bo42bo5bo14bo5bo42bo5bo14bo5bo42bo5bo14bo5bo39bo2bo5bo2bo8bo2bo5bo2bo36bobo7bobo8bobo7bobo36bobo7bobo8bobo7bobo$9bo2bo2bo3bo2bo2bo4bo2bo2bo3bo2bo2bo38b2ob2o16b2ob2o109b6o3b6o6b6o3b6o35bo3bo3bo3bo8bo3bo3bo3bo35b3ob3ob3ob3o6b3ob3ob3ob3o35bo3bo3bo3bo8bo3bo3bo3bo34b2o4b2ob2o4b2o4b2o4b2ob2o4b2o37bobobobo14bobobobo39b2obobobobob2o8b2obobobobob2o35bobo9bobo6bobo9bobo32bo3b2o7b2o3bo2bo3b2o7b2o3bo30bo3b2o7b2o3bo2bo3b2o7b2o3bo30bo3b2o7b2o3bo2bo3b2o7b2o3bo30bo3b2o7b2o3bo2bo3b2o7b2o3bo30bo3b2o7b2o3bo2bo3b2o7b2o3bo30bo3b2o7b2o3bo2bo3b2o7b2o3bo30bo3b2o7b2o3bo2bo3b2o7b2o3bo33bo11bo8bo11bo36bo11bo8bo11bo36bo11bo8bo11bo$8bobo4bo3bo4bobo2bobo4bo3bo4bobo32b2o4bobo4b2o6b2o4bobo4b2o38b2o3b2o14b2o3b2o40b3o5b3o10b3o5b3o39bo7bo12bo7bo38bo3bo3bo3bo8bo3bo3bo3bo37bo9bo10bo9bo35b2o3bobobobo3b2o4b2o3bobobobo3b2o34bo4bobo4bo8bo4bobo4bo35bobo3bobo3bobo6bobo3bobo3bobo34b2o11b2o6b2o11b2o32bo3bobo5bobo3bo2bo3bobo5bobo3bo30bo3bobo5bobo3bo2bo3bobo5bobo3bo30bo3bobo5bobo3bo2bo3bobo5bobo3bo30bo3bobo5bobo3bo2bo3bobo5bobo3bo30bo3bobo5bobo3bo2bo3bobo5bobo3bo30bo3bobo5bobo3bo2bo3bobo5bobo3bo30bo3bobo5bobo3bo2bo3bobo5bobo3bo32b4obo3bob4o6b4obo3bob4o34b3ob2o3b2ob3o6b3ob2o3b2ob3o36bo9bo10bo9bo$8bo3bo2bo3bo2bo3bo2bo3bo2bo3bo2bo3bo32bo3bobobobo3bo6bo3bobobobo3bo34b2obo7bob2o6b2obo7bob2o104bo2bo7bo2bo6bo2bo7bo2bo36b4o3b4o10b4o3b4o42bobo18bobo39b2ob3o5b3ob2o4b2ob3o5b3ob2o33bo2bo2bobo2bo2bo6bo2bo2bobo2bo2bo34bo2bo2bobo2bo2bo6bo2bo2bobo2bo2bo105bo2bobobobo2bo8bo2bobobobo2bo36bo2bobobobo2bo8bo2bobobobo2bo36bo2bobobobo2bo8bo2bobobobo2bo36bo2bobobobo2bo8bo2bobobobo2bo36bo2bobobobo2bo8bo2bobobobo2bo36bo2bobobobo2bo8bo2bobobobo2bo36bo2bobobobo2bo8bo2bobobobo2bo34b2o2b2obobob2o2b2o4b2o2b2obobob2o2b2o34bo4bobo4bo8bo4bobo4bo37b2o7b2o10b2o7b2o$12bob2o3b2obo10bob2o3b2obo36bob3obobob3obo6bob3obobob3obo34b2o4bobo4b2o6b2o4bobo4b2o34bo2bo7bo2bo6bo2bo7bo2bo34b2obob2ob2obob2o6b2obob2ob2obob2o39bo3bo16bo3bo40bo3b2ob2o3bo8bo3b2ob2o3bo105bo2bob2ob2obo2bo6bo2bob2ob2obo2bo36bo3bobo3bo10bo3bobo3bo35bo3bo7bo3bo4bo3bo7bo3bo32bob4obobob4obo4bob4obobob4obo32bob4obobob4obo4bob4obobob4obo32bob4obobob4obo4bob4obobob4obo32bob4obobob4obo4bob4obobob4obo32bob4obobob4obo4bob4obobob4obo32bob4obobob4obo4bob4obobob4obo32bob4obobob4obo4bob4obobob4obo39bobo18bobo42b2o2bobo2b2o10b2o2bobo2b2o38bo9bo10bo9bo$10b2o2bo5bo2b2o6b2o2bo5bo2b2o33bo6bobo6bo4bo6bobo6bo39bobo18bobo41b4o5b4o8b4o5b4o38bob2ob2obo12bob2ob2obo38b4o5b4o8b4o5b4o36b2o9b2o8b2o9b2o37bob3ob3obo10bob3ob3obo39bo2bobo2bo12bo2bobo2bo38b3o2bobo2b3o8b3o2bobo2b3o40b2ob2o16b2ob2o39bobob2o3b2obobo6bobob2o3b2obobo34bobob2o3b2obobo6bobob2o3b2obobo34bobob2o3b2obobo6bobob2o3b2obobo34bobob2o3b2obobo6bobob2o3b2obobo34bobob2o3b2obobo6bobob2o3b2obobo34bobob2o3b2obobo6bobob2o3b2obobo34bobob2o3b2obobo6bobob2o3b2obobo40bobo18bobo46bobo18bobo39bo4bo5bo4bo4bo4bo5bo4bo$9b3o3bo3bo3b3o4b3o3bo3bo3b3o32bob2obo5bob2obo4bob2obo5bob2obo34b2o3bobo3b2o8b2o3bobo3b2o36bo3bo3bo3bo8bo3bo3bo3bo39bobobobo14bobobobo38b5o5b5o6b5o5b5o35b3o7b3o8b3o7b3o36b4o5b4o8b4o5b4o34bo4bobobobo4bo4bo4bobobobo4bo35b2o7b2o10b2o7b2o34b2obobobo3bobobob2o2b2obobobo3bobobob2o36bo5bo14bo5bo42bo5bo14bo5bo42bo5bo14bo5bo42bo5bo14bo5bo42bo5bo14bo5bo42bo5bo14bo5bo42bo5bo14bo5bo40b2obo3bob2o10b2obo3bob2o42bobo18bobo39bo5bo3bo5bo4bo5bo3bo5bo$9bo15bo4bo15bo31b2ob6ob6ob2o2b2ob6ob6ob2o31b2o3bo5bo3b2o4b2o3bo5bo3b2o34b2o2bo3bo2b2o8b2o2bo3bo2b2o37b2o7b2o10b2o7b2o35b2o13b2o4b2o13b2o36b3o3b3o12b3o3b3o37bobob2o3b2obobo6bobob2o3b2obobo33b2obobobobobobob2o4b2obobobobobobob2o31b2o15b2o2b2o15b2o33bob2o5b2obo8bob2o5b2obo36b4o5b4o8b4o5b4o36b4o5b4o8b4o5b4o36b4o5b4o8b4o5b4o36b4o5b4o8b4o5b4o36b4o5b4o8b4o5b4o36b4o5b4o8b4o5b4o36b4o5b4o8b4o5b4o35b2o2bo5bo2b2o6b2o2bo5bo2b2o37b2o5b2o12b2o5b2o36b6o5b6o4b6o5b6o$9bo4b2o3b2o4bo4bo4b2o3b2o4bo31b2obobo7bobob2o2b2obobo7bobob2o32b6o3b6o6b6o3b6o36bo2bo3bo2bo10bo2bo3bo2bo36b3obo5bob3o6b3obo5bob3o34b2o3b2ob2o3b2o6b2o3b2ob2o3b2o33bo2b4o3b4o2bo4bo2b4o3b4o2bo35bo2bo3bo2bo10bo2bo3bo2bo35b3o2bobobobo2b3o4b3o2bobobobo2b3o106bo7bo12bo7bo38bo11bo8bo11bo36bo11bo8bo11bo35bo13bo6bo13bo37bo7bo12bo7bo40bo7bo12bo7bo176bo3b2o5b2o3bo4bo3b2o5b2o3bo35b3o5b3o10b3o5b3o$86bobo18bobo42bo3bobo3bo10bo3bobo3bo36b2o11b2o6b2o11b2o37b2o5b2o12b2o5b2o36b2obob3ob3obob2o4b2obob3ob3obob2o31b3o4bo3bo4b3o2b3o4bo3bo4b3o35bo7bo12bo7bo37bobobobobobobobo6bobobobobobobobo105bobo7bobo8bobo7bobo37bobo5bobo10bobo5bobo38b2o7b2o10b2o7b2o36b2o11b2o6b2o11b2o35bo11bo8bo11bo36bo11bo8bo11bo36b3o7b3o8b3o7b3o36b2ob2o3b2ob2o8b2ob2o3b2ob2o106bo11bo8bo11bo37b2obo3bob2o10b2obo3bob2o$149b3o4bobo4b3o4b3o4bobo4b3o32b2ob2obo3bob2ob2o4b2ob2obo3bob2ob2o36bo7bo12bo7bo39b2ob2ob2ob2o10b2ob2ob2ob2o34b3o13b3o2b3o13b3o34bo2bo3bo2bo10bo2bo3bo2bo38bo3bobo3bo10bo3bobo3bo107b2o9b2o8b2o9b2o35bobobo5bobobo6bobobo5bobobo35bo2bo5bo2bo8bo2bo5bo2bo105bo2b2o5b2o2bo6bo2b2o5b2o2bo34bo2b2o5b2o2bo6bo2b2o5b2o2bo32b2o4bo5bo4b2o2b2o4bo5bo4b2o32b2ob3o3b3ob2o6b2ob3o3b3ob2o105bobobo3bobobo8bobobo3bobobo37bobobobobobo10bobobobobobo$218bo4bobo3bobo4bo2bo4bobo3bobo4bo31b2o5bobo5b2o4b2o5bobo5b2o32bo5bo3bo5bo4bo5bo3bo5bo33bo3bobobobo3bo6bo3bobobobo3bo32b2o4b2o3b2o4b2o2b2o4b2o3b2o4b2o30bo4b2o5b2o4bo2bo4b2o5b2o4bo102bo2bo7bo2bo6bo2bo7bo2bo33b4obo5bob4o4b4obo5bob4o33b2o2bo5bo2b2o6b2o2bo5bo2b2o105bo4bobo4bo8bo4bobo4bo36bobobo3bobobo8bobobo3bobobo33b2o4b2o3b2o4b2o2b2o4b2o3b2o4b2o32b2o11b2o6b2o11b2o$218b3o2b2o5b2o2b3o2b3o2b2o5b2o2b3o31b2o5bobo5b2o4b2o5bobo5b2o31bo7bobo7bo2bo7bobo7bo33b2ob3ob3ob2o8b2ob3ob3ob2o33b2o3bo2bobo2bo3b2o2b2o3bo2bobo2bo3b2o33b5o3b5o8b5o3b5o105b2obo7bob2o6b2obo7bob2o33b2o4bo3bo4b2o4b2o4bo3bo4b2o33b2ob2o5b2ob2o6b2ob2o5b2ob2o105b2obo5bob2o8b2obo5bob2o36bo3bo3bo3bo8bo3bo3bo3bo34bo6bobo6bo4bo6bobo6bo34b2o9b2o8b2o9b2o$292bo3bobo3bo10bo3bobo3bo34b2o4b2o3b2o4b2o2b2o4b2o3b2o4b2o31bo5b2ob2o5bo4bo5b2ob2o5bo33bob2o2bobo2b2obo6bob2o2bobo2b2obo36bobo5bobo10bobo5bobo107bobo7bobo8bobo7bobo40bo3bo16bo3bo38b2obobo5bobob2o4b2obobo5bobob2o109bobo18bobo42b3o5b3o10b3o5b3o39b2obobob2o12b2obobob2o39bo9bo10bo9bo$292b2o7b2o10b2o7b2o35bo6bobo6bo4bo6bobo6bo35bo2b2ob2o2bo10bo2b2ob2o2bo34b3ob3obobob3ob3o2b3ob3obobob3ob3o30b2o3bobo3bobo3b2o2b2o3bobo3bobo3b2o102bo2bo7bo2bo6bo2bo7bo2bo39b2ob2o16b2ob2o38bo2b2o7b2o2bo4bo2b2o7b2o2bo102b2o2b2o5b2o2b2o4b2o2b2o5b2o2b2o34b2o2bo3bo2b2o8b2o2bo3bo2b2o34b2o5bobo5b2o4b2o5bobo5b2o32b3o2b3ob3o2b3o4b3o2b3ob3o2b3o$429b2o13b2o4b2o13b2o33bo3b2o3b2o3bo6bo3b2o3b2o3bo34bo4bo3bo4bo6bo4bo3bo4bo103b2o2bo7bo2b2o4b2o2bo7bo2b2o33b3o9b3o6b3o9b3o33bo5bo3bo5bo4bo5bo3bo5bo102b2o13b2o4b2o13b2o37b2o3b2o14b2o3b2o37bo4bobobobo4bo4bo4bobobobo4bo32b2o4b2ob2o4b2o4b2o4b2ob2o4b2o$502b2o7b2o10b2o7b2o36bob2obo3bob2obo6bob2obo3bob2obo103bo15bo4bo15bo33bo2bo7bo2bo6bo2bo7bo2bo32bo5b2o3b2o5bo2bo5b2o3b2o5bo103bo2bo5bo2bo8bo2bo5bo2bo38bobo3bobo12bobo3bobo35b2ob2o3bobo3b2ob2o2b2ob2o3bobo3b2ob2o30bo2bo2b3ob3o2bo2bo2bo2bo2b3ob3o2bo2bo$501bo2b2o3b2o2bo8bo2b2o3b2o2bo34bobo11bobo4bobo11bobo102bob5o3b5obo4bob5o3b5obo33bo3bo5bo3bo6bo3bo5bo3bo35bobo7bobo8bobo7bobo106bo3bo3bo3bo8bo3bo3bo3bo35b2ob4ob4ob2o6b2ob4ob4ob2o33bob2o3bobo3b2obo4bob2o3bobo3b2obo31bobo2b3o3b3o2bobo2bobo2b3o3b3o2bobo$503bob2ob2obo12bob2ob2obo35bob4ob2ob2ob4obo2bob4ob2ob2ob4obo100b3o13b3o2b3o13b3o35bo7bo12bo7bo37bo5bobo5bo6bo5bobo5bo104bo2b3o3b3o2bo6bo2b3o3b3o2bo33bo2bob3ob3obo2bo4bo2bob3ob3obo2bo31b3o2bobo3bobo2b3o2b3o2bobo3bobo2b3o$708b3o2bo7bo2b3o2b3o2bo7bo2b3o31b3o4bobo4b3o4b3o4bobo4b3o33b2obo2bobo2bob2o6b2obo2bobo2bob2o103b5o2bobo2b5o4b5o2bobo2b5o33bo4bo3bo4bo6bo4bo3bo4bo108bo5bo14bo5bo$850bo3bo5bo3bo6bo3bo5bo3bo103b2o13b2o4b2o13b2o33b3o9b3o6b3o9b3o33bob2ob2o3b2ob2obo4bob2ob2o3b2ob2obo32bo3b2o5b2o3bo4bo3b2o5b2o3bo$988b2o3bo2bobo2bo3b2o2b2o3bo2bobo2bo3b2o31bo3bo7bo3bo4bo3bo7bo3bo37b2o3b2o14b2o3b2o36b2obob2o5b2obob2o2b2obob2o5b2obob2o$1059bob6ob6obo4bob6ob6obo33bobo2bo3bo2bobo6bobo2bo3bo2bobo32b2o2bo9bo2b2o2b2o2bo9bo2b2o$1129bo2bo9bo2bo4bo2bo9bo2bo33bobobo5bobobo6bobobo5bobobo$1128bo5bo5bo5bo2bo5bo5bo5bo34b3o5b3o10b3o5b3o$1129bo3bobo3bobo3bo4bo3bobo3bobo3bo33b2o2b2o3b2o2b2o6b2o2b2o3b2o2b2o$1198b2ob2obo5bob2ob2o2b2ob2obo5bob2ob2o$1198b3obo9bob3o2b3obo9bob3o! Another attempt x = 54, y = 37, rule = B3/S232bo8bo10bo8bo10bo8bo$bobo6bobo8bobo6bobo8bobo6bobo$o2bo6bo2bo6bo2bo6bo2bo6bo2bo6bo2bo$b2o8b2o8b2o8b2o8b2o8b2o$6b2o18b2o18b2o$4b2o2b2o14b2o2b2o14b2o2b2o$4bo4bo14bo4bo14bo4bo$4bo4bo14bo4bo14bo4bo$3b8o12b8o12b8o$2b4o2b4o10b4o2b4o10b4o2b4o$bo2bo4bo2bo8bo2bo4bo2bo8bo2bo4bo2bo$o3bo4bo3bo6bo3bo4bo3bo6bo3bo4bo3bo$bo2bo4bo2bo8bo2bo4bo2bo8bo2bo4bo2bo3$b3o6b3o8b3o6b3o8b3o6b3o3$2bo8bo10bo8bo10bo8bo$b3o6b3o8b3o6b3o8b3o6b3o$8b3ob2o6b2ob3o2b3ob2o6b2ob3o$8b2obobo6bobob2o2b2obobo6bobob2o$9b4o8b4o4b4o8b4o$13bob4obo12bob4obo$13b3o2b3o12b3o2b3o$9bo6b2o6bo4bo6b2o6bo$10bob4o2b4obo6bob4o2b4obo$11bobo6bobo8bobo6bobo$12bo2bo2bo2bo10bo2bo2bo2bo$10b4obo2bob4o6b4obo2bob4o2$9bo2bo8bo2bo4bo2bo8bo2bo$10bo2b2o4b2o2bo6bo2b2o4b2o2bo$10b3o2bo2bo2b3o6b3o2bo2bo2b3o$15bo2bo16bo2bo$9bo5bo2bo5bo4bo5bo2bo5bo$9b2o3b6o3b2o4b2o3b6o3b2o!
EDIT:Just did some asymmetric search
x = 199, y = 39, rule = B3/S235bo8bo11bo8bo11bo8bo18bo8bo11bo8bo11bo8bo18bo8bo11bo8bo11bo8bo$4bobo6bobo9bobo6bobo9bobo6bobo16bobo6bobo9bobo6bobo9bobo6bobo16bobo6bobo9bobo6bobo9bobo6bobo$3bo2bo6bo2bo7bo2bo6bo2bo7bo2bo6bo2bo14bo2bo6bo2bo7bo2bo6bo2bo7bo2bo6bo2bo14bo2bo6bo2bo7bo2bo6bo2bo7bo2bo6bo2bo$4b2o8b2o9b2o8b2o9b2o8b2o16b2o8b2o9b2o8b2o9b2o8b2o16b2o8b2o9b2o8b2o9b2o8b2o$9b2o19b2o19b2o26b2o19b2o19b2o26b2o19b2o19b2o$7b2o2b2o15b2o2b2o15b2o2b2o22b2o2b2o15b2o2b2o15b2o2b2o22b2o2b2o15b2o2b2o15b2o2b2o$7bo4bo15bo4bo15bo4bo22bo4bo15bo4bo15bo4bo22bo4bo15bo4bo15bo4bo$7bo4bo15bo4bo15bo4bo22bo4bo15bo4bo15bo4bo22bo4bo15bo4bo15bo4bo$6b8o13b8o13b8o20b8o13b8o13b8o20b8o13b8o13b8o$5b4o2b4o11b4o2b4o11b4o2b4o18b4o2b4o11b4o2b4o11b4o2b4o18b4o2b4o11b4o2b4o11b4o2b4o$4bo2bo4bo2bo9bo2bo4bo2bo9bo2bo4bo2bo16bo2bo4bo2bo9bo2bo4bo2bo9bo2bo4bo2bo16bo2bo4bo2bo9bo2bo4bo2bo9bo2bo4bo2bo$3bo3bo4bo3bo7bo3bo4bo3bo7bo3bo4bo3bo14bo3bo4bo3bo7bo3bo4bo3bo7bo3bo4bo3bo14bo3bo4bo3bo7bo3bo4bo3bo7bo3bo4bo3bo$4bo2bo4bo2bo9bo2bo4bo2bo9bo2bo4bo2bo16bo2bo4bo2bo9bo2bo4bo2bo9bo2bo4bo2bo16bo2bo4bo2bo9bo2bo4bo2bo9bo2bo4bo2bo3$4b3o6b3o9b3o6b3o9b3o6b3o16b3o6b3o9b3o6b3o9b3o6b3o16b3o6b3o9b3o6b3o9b3o6b3o3$5b2o7b2o9b2o8b2o9b2o27bo8b2o9b2o8b2o9b2o26b2o8b2o9b2o8b2o9b2o$4b2o7bob2o7b2obo6bob2o7b2obo25b3o6bob2o7b2obo6bob2o7b2obo24b2obo6bob2o7b2obo6bob2o7b2obo$bo3bo11bo5bo14bo5bo28b2ob3o8bo5bo14bo5bo28bo13bo5bo14bo5bo$b2o4bo7bo2bo3bo2bo10bo2bo3bo2bo26bobob2o6bo2bo3bo2bo10bo2bo3bo2bo26bob2o8bo2bo3bo2bo10bo2bo3bo2bo$2bo3b2o7bobo5bobo10bobo5bobo27b4o7bobo5bobo10bobo5bobo25b3o10bobo5bobo10bobo5bobo$7b2o62bobo$2bobo2b2o7bobo3bobo12bobo3bobo25bobo12b2o5b2o12b2o5b2o29bo10b2o5b2o12b2o5b2o$2bobo13bo3bo16bo3bo27bo16bo3bo16bo3bo28b2o14bo3bo16bo3bo$5bo11bo5bo14bo5bo26bob2o11bobo3bobo12bobo3bobo25bo2bo10b3o5b3o10b3o5b3o$3o3b2o10bo3bo16bo3bo26b3o2bo11bo5bo14bo5bo25bob3o8bo13bo6bo13bo$8bo9b2ob2o16b2ob2o26b3o3bo11bo3bo16bo3bo27bo2bo8bo13bo6bo13bo$2ob2o2b2o5b3ob2ob2ob3o8b3ob2ob2ob3o27b3o7b2ob2ob2ob2o10b2ob2ob2ob2o23b4o8bobob2o5b2obobo4bobob2o5b2obobo$15b2o7b2o10b2o7b2o25bobobo7bo3b2ob2o3bo8bo3b2ob2o3bo23bo2b2o5b2obo11bob2o2b2obo11bob2o$12b2ob2ob2ob2ob2ob2o4b2ob2ob2ob2ob2ob2o20b2o2bo8b2obobo3bobob2o6b2obobo3bobob2o24b3o7b3ob2o3b2ob3o6b3ob2o3b2ob3o$14b2o2bo3bo2b2o8b2o2bo3bo2b2o22bo4b3o5bo2b3o3b3o2bo6bo2b3o3b3o2bo23bo3b2o9b2o3b2o14b2o3b2o$13bo5bobo5bo6bo5bobo5bo33b2obo3bobo3bob2o4b2obo3bobo3bob2o20bo5bobo3bob2o9b2obo4bob2o9b2obo$18bo3bo16bo3bo38bo3bo2bobo2bo3bo4bo3bo2bobo2bo3bo20bo5bo5bo15bo4bo15bo$12b2obobo5bobob2o4b2obobo5bobob2o34bobo7bobo8bobo7bobo34bo2bo2b2ob2o2bo2bo4bo2bo2b2ob2o2bo2bo$15b2obo3bob2o10b2obo3bob2o35bo2b2o2bobo2b2o2bo4bo2b2o2bobo2b2o2bo33b2o2b2o3b2o2b2o6b2o2b2o3b2o2b2o$153bobo2b2ob2o2bobo6bobo2b2ob2o2bobo$154b2ob3ob3ob2o8b2ob3ob3ob2o!
Last edited by GUYTU6J on April 15th, 2017, 8:37 am, edited 1 time in total.
etymology of names!
GUYTU6J
Posts: 347
Joined: August 5th, 2016, 10:27 am
Location: outside Plain of Life
I'm doing some c/6 diagonal searches with JLS; here's a somewhat promising partial from that:
x = 32, y = 32, rule = B3/S2322b7obo$27bo2bo$27b2obo$22b3o$23b3obob2o$23bo3bo$25bo3bo$24b2obobo$26bo$26bob3o$26b2obo$23bobo$23b4o2bo$25bo2bo$28bo2$27bob2o$25bo3bo$22b2o4bo$25b3o2$25bo$o2bo14bo5bob2obo$o2b3o5b2o5bo5bobo$o2b2o2bo4bo9b2o5b3o$o3bob2o3b3o3bobobo7b2o$o7b3obo6bo2b2o2b2obo$3ob2obo2bo5bo2bo2bo3bo$obo6bo3b2o3bo9bo$4bob2ob2obo3b2o4bob3o$3obo4bo6bo7b2o$24bo! I don't know how to get JLS to report partials in any useful kind of fashion, unfortunately. It may be impossible. EDIT: Here's something slightly different: x = 25, y = 25, rule = B3/S2319b2o2b2o$19b3o2bo$20b2o2$19b2o$19bobo2$19bo2bo$19bo2bo$15bob3obo$15bob2obo$15bob2o2$20b2o$16b3o$9b3o5b2o3bo$14bo7bo$9b3o2b2o6bo$9b3o2b2o3bobo$2o2b2ob3o8b3o$3obo5bo2bo5bo$b2o2bo3bo3bo4bo$7b2o6b3o$o$2o!
EDIT 2:
x = 39, y = 40, rule = B3/S23$30bo6bo$31bobo2bo$30b2obo2bo$29bo3b5o$37bo$31b2o3b2o$29bob3o2bo$29bo6bo$30b4o$30bo$31bobob2o$30bo5bo$31bo2bo$35bo$31bobo$36b2o$31bo2bob2o$32bo3bo$33bobo$34bobo$34bo$35bo2$34bob2o$32bo3bo$29b2o4bo$32b3o2$32bo$3bo2b2o17bo5bob2obo$obo5b2obo13bo5bobo$b2o2b2obobobobobo12b2o5b3o$5b2obo8bo6bobobo7b2o$b3o2bobobo3bo3bo7bo2b2o2b2obo$3bo8bo3bo2b2o2bo2bo2bo3bo$3bo6bo2bo4bo2bo3bo9bo$b3ob3o2b2o3b3obo3b2o4bob3o$o2b3o9b2o6bo7b2o$31bo! Time to expand my search area. x₁=ηx V ⃰_η=c²√(Λη) K=(Λu²)/2 Pₐ=1−1/(∫^∞_t₀(p(t)ˡ⁽ᵗ⁾)dt) $$x_1=\eta x$$ $$V^*_\eta=c^2\sqrt{\Lambda\eta}$$ $$K=\frac{\Lambda u^2}2$$ $$P_a=1-\frac1{\int^\infty_{t_0}p(t)^{l(t)}dt}$$ http://conwaylife.com/wiki/A_for_all Aidan F. Pierce A for awesome Posts: 1634 Joined: September 13th, 2014, 5:36 pm Location: 0x-1 ### Re: Spaceship Discussion Thread Partial p6 knightship: x = 31, y = 26, rule = B3/S237b2o$5bo3bo$5bo$b2obob2ob2o$o3b2o3b2o$o3bo3b2o$2o3b2o3bo$bo$2bo5b3o$4bo3b5o$4bo6bo3b2o$5b2o3bo4b2o2bobo$4bo2b2o2bobo3bobo2bo$4b2o3bobobobob2o3bo$4b3o2bobobo2b5o2bo$6bobobobobo2bo5bo$18b2o4b2o$13b4o6b4o$13b4o2bo4bo2bo$13bobob2o2b2o4b3o$21b2o4b3o$24b2obo$27bo$24bo3bobo$27b2obo$28b2o!
Partial c/6 diagonal:
x = 69, y = 69, rule = B3/S2326b2o$25bobo$25b3o$26bo$27bo$25b3ob2o$25bobo$28bo$26b2o$26b2o$26b2o$24bo2bo$23bo2bo$23bo2bo$22b2obo$25bo$24b2obo$25bobo$27bob2o$26bo2b2o37bo$27b3o23bob4obo6b2o$27b3o12bo9b2o4b2o2bo$14bo26b2o8bo4b2obo2b3o$12b3o14b3o4b2o11b3o7bobob2o$11bo4bo12bobob2obobo3bo3b2o3bo6b2o$b2o2b2o7b4o10b2obob4o10bo2bo7b2o$ob2obo2b3ob2o5bo11b2o4b2ob2o5bobo8b2o$3ob3ob4o4b3ob2o6bo8b2ob2obo8bo$7bo12b2o3bobob3o11bo4bo$5bo12b4ob3o2bo$5bo12b2o3bo4bo$23b4obo$26bo$24b2o$24b2o$25bo$23b3o$23bo2b2o$24bob2o2$26b2o$22bo3b2o$21b2obo$27b2o3$24bo$24b3o$28bo$23bo2bo$23bobo$22b3o$21bo5bo$20b2o2$20bo$20bobo$20bobo$20b2o2b3o$21b6o$20bo$23bo$21b2o$22b2o$22b2o3$20bo$19b2o!
-Josh Ball.
velcrorex
Posts: 339
Joined: November 1st, 2009, 1:33 pm
velcrorex wrote:Partial p6 knightship:
x = 31, y = 26, rule = B3/S237b2o$5bo3bo$5bo$b2obob2ob2o$o3b2o3b2o$o3bo3b2o$2o3b2o3bo$bo$2bo5b3o$4bo3b5o$4bo6bo3b2o$5b2o3bo4b2o2bobo$4bo2b2o2bobo3bobo2bo$4b2o3bobobobob2o3bo$4b3o2bobobo2b5o2bo$6bobobobobo2bo5bo$18b2o4b2o$13b4o6b4o$13b4o2bo4bo2bo$13bobob2o2b2o4b3o$21b2o4b3o$24b2obo$27bo$24bo3bobo$27b2obo$28b2o! Now that looks promising! Gamedziner Posts: 508 Joined: May 30th, 2016, 8:47 pm Location: Milky Way Galaxy: Planet Earth ### Re: Spaceship Discussion Thread Gamedziner wrote: velcrorex wrote:Partial p6 knightship: x = 31, y = 26, rule = B3/S237b2o$5bo3bo$5bo$b2obob2ob2o$o3b2o3b2o$o3bo3b2o$2o3b2o3bo$bo$2bo5b3o$4bo3b5o$4bo6bo3b2o$5b2o3bo4b2o2bobo$4bo2b2o2bobo3bobo2bo$4b2o3bobobobob2o3bo$4b3o2bobobo2b5o2bo$6bobobobobo2bo5bo$18b2o4b2o$13b4o6b4o$13b4o2bo4bo2bo$13bobob2o2b2o4b3o$21b2o4b3o$24b2obo$27bo$24bo3bobo$27b2obo$28b2o!
Now that looks promising!
It really does!!
There might be something right here, unless it becomes the new c/6 almost-knightship, but I hope there's more to it.
Good job!
SoL : FreeElectronics : DeadlyEnemies : 6a-ite
what is “sesame oil”?
Rhombic
Posts: 992
Joined: June 1st, 2013, 5:41 pm
velcrorex wrote:Partial p6 knightship:
x = 14, y = 43, rule = B3/S232b2o6b2o$2bo8bo$5bo2bo$2bobo4bobo$4bo4bo$2bo8bo$2bo8bo$2b3o4b3o$3b3o2b3o$3b3o2b3o3$5b4o$5b4o2$4b2o2b2o$3bo6bo$5b4o$2bo2b4o2bo2$3b2o4b2o$5bo2bo$3b3o2b3o$3b3o2b3o$5b4o$3o2b4o2b3o$2b2o6b2o$bo4b2o4bo$b2o3b2o3b2o$bo10bo$b2o8b2o$2o2bo4bo2b2o$2obobo2bobob2o$2obobo2bobob2o$2b4o2b4o$5b4o$2b2ob4ob2o$3bo6bo$3bo6bo2$2b3o4b3o$2b3o4b3o$3bo6bo! My knightt c/6 width 11 search is STILL going, but it's showing some signs of slowing down, although it'll probably be a long time before it finishes anyway. x₁=ηx V ⃰_η=c²√(Λη) K=(Λu²)/2 Pₐ=1−1/(∫^∞_t₀(p(t)ˡ⁽ᵗ⁾)dt) $$x_1=\eta x$$ $$V^*_\eta=c^2\sqrt{\Lambda\eta}$$ $$K=\frac{\Lambda u^2}2$$ $$P_a=1-\frac1{\int^\infty_{t_0}p(t)^{l(t)}dt}$$ http://conwaylife.com/wiki/A_for_all Aidan F. Pierce A for awesome Posts: 1634 Joined: September 13th, 2014, 5:36 pm Location: 0x-1 ### Re: Spaceship Discussion Thread A for awesome wrote:I have ruled out the existence of c/4 p8 even-glide-symmetric ships with width 16 using gfind. Unfortunately, I don't think an unmodified gfind can eliminate the possibility of higher-period spaceships if it manages to find a lower-period ship at the same speed. This is because the search can reject, say, a period-8 front end because it can be replaced with a shorter period-4 front end. -Matthias Merzenich Sokwe Moderator Posts: 1264 Joined: July 9th, 2009, 2:44 pm ### Re: Spaceship Discussion Thread Sokwe wrote: A for awesome wrote:I have ruled out the existence of c/4 p8 even-glide-symmetric ships with width 16 using gfind. Unfortunately, I don't think an unmodified gfind can eliminate the possibility of higher-period spaceships if it manages to find a lower-period ship at the same speed. This is because the search can reject, say, a period-8 front end because it can be replaced with a shorter period-4 front end. Oops, ~950 CPU-hrs down the drain. Sorry for wasting my time with that when I could have been doing much more valuable searches. x₁=ηx V ⃰_η=c²√(Λη) K=(Λu²)/2 Pₐ=1−1/(∫^∞_t₀(p(t)ˡ⁽ᵗ⁾)dt) $$x_1=\eta x$$ $$V^*_\eta=c^2\sqrt{\Lambda\eta}$$ $$K=\frac{\Lambda u^2}2$$ $$P_a=1-\frac1{\int^\infty_{t_0}p(t)^{l(t)}dt}$$ http://conwaylife.com/wiki/A_for_all Aidan F. Pierce A for awesome Posts: 1634 Joined: September 13th, 2014, 5:36 pm Location: 0x-1 ### Re: Spaceship Discussion Thread A for awesome wrote:Oops, ~950 CPU-hrs down the drain. Sorry for wasting my time with that when I could have been doing much more valuable searches. I wouldn't say it's wasted time. Although it's not quite a proof, it suggests an extremely high probability that there are no (2,0)c/8 glide-symmetric ships at this width. There's more potential for a width-18 search due to the large number of symmetric width-18 c/4 orthogonal ships, but a direct gfind search might be too slow. -Matthias Merzenich Sokwe Moderator Posts: 1264 Joined: July 9th, 2009, 2:44 pm ### Re: Spaceship Discussion Thread Sokwe wrote:[A] direct gfind search might be too slow. I very much agree. x₁=ηx V ⃰_η=c²√(Λη) K=(Λu²)/2 Pₐ=1−1/(∫^∞_t₀(p(t)ˡ⁽ᵗ⁾)dt) $$x_1=\eta x$$ $$V^*_\eta=c^2\sqrt{\Lambda\eta}$$ $$K=\frac{\Lambda u^2}2$$ $$P_a=1-\frac1{\int^\infty_{t_0}p(t)^{l(t)}dt}$$ http://conwaylife.com/wiki/A_for_all Aidan F. Pierce A for awesome Posts: 1634 Joined: September 13th, 2014, 5:36 pm Location: 0x-1 ### Re: Spaceship Discussion Thread A 2c/6 wave: x = 5, y = 144, rule = B3/S23:T144,1443b2o$2bo$2b3o$b2obo2$2o$3o$2obo$bo2bo$2b2o2$2b2o$2b2o2$2b2o$bo2bo$2obo$3o$2o2$b2obo$2b3o$2bo$3b2o$3b2o$2bo$2b3o$b2obo2$2o$3o$2obo$bo2bo$2b2o2$2b2o$2b2o2$2b2o$bo2bo$2obo$3o$2o2$b2obo$2b3o$2bo$3b2o$3b2o$2bo$2b3o$b2obo2$2o$3o$2obo$bo2bo$2b2o2$2b2o$2b2o2$2b2o$bo2bo$2obo$3o$2o2$b2obo$2b3o$2bo$3b2o$3b2o$2bo$2b3o$b2obo2$2o$3o$2obo$bo2bo$2b2o2$2b2o$2b2o2$2b2o$bo2bo$2obo$3o$2o2$b2obo$2b3o$2bo$3b2o$3b2o$2bo$2b3o$b2obo2$2o$3o$2obo$bo2bo$2b2o2$2b2o$2b2o2$2b2o$bo2bo$2obo$3o$2o2$b2obo$2b3o$2bo$3b2o$3b2o$2bo$2b3o$b2obo2$2o$3o$2obo$bo2bo$2b2o2$2b2o$2b2o2$2b2o$bo2bo$2obo$3o$2o2$b2obo$2b3o$2bo$3b2o!
Still drifting.
Bullet51
Posts: 483
Joined: July 21st, 2014, 4:35 am
73-cell 1c/2:
x = 26, y = 14, rule = B3/S2314b2o2$2o13bo8b2o$3bo11b3o6bo$o3bobo2b2o10bob2o$b2o3bo2bo8b2obob2o$2bobobob2o11bo$3b2obob2o9bob2o$4bobo6bo3bob2o$7bo6bo4bo$5bo6bo3bob2o$5bo3bobob2ob2o$6bo2b5o$7b3o!
I don't know if there's some kind of listing for the smallest of these or not.
EDIT: And, of course, this relatively obvious 66-cell:
x = 32, y = 8, rule = B3/S232o7b2o10b2o7b2o$12bo6bo$o4bo3bo4bo2bo4bo3bo4bo$b2obo4bob2ob4ob2obo4bob2o$2bobob2ob2o3b4o3b2ob2obobo$3b2o4bo12bo4b2o$4bo3bo14bo3bo$5b3o16b3o! EDIT 2: Another 73-cell: x = 33, y = 9, rule = B3/S232ob2o$bobo7b2o9b2o7b2o$b2obo9bo5bo$b2obo2bo3bo4bobo4bo3bo4bo$4bobo4bob2ob3ob2obo4bob2o$3b2obob2ob2o3b3o3b2ob2obobo$5b2o4bo11bo4b2o$6bo3bo13bo3bo$7b3o15b3o! And 66, 68, 68, 70 cells: x = 34, y = 44, rule = B3/S2331bo$32bo$b2o7b2o9b2o7b2o$13bo5bo$bo4bo3bo4bobo4bo3bo4bo$2b2obo4bob2ob3ob2obo4bob2o$3bobob2ob2o3b3o3b2ob2obobo$4b2o4bo11bo4b2o$5bo3bo13bo3bo$6b3o15b3o2$bo29bo$o31bo$b2o7b2o9b2o7b2o$13bo5bo$bo4bo3bo4bobo4bo3bo4bo$2b2obo4bob2ob3ob2obo4bob2o$3bobob2ob2o3b3o3b2ob2obobo$4b2o4bo11bo4b2o$5bo3bo13bo3bo$6b3o15b3o2$32bo$33bo$b2o7b2o10b2o7b2o$13bo6bo$bo4bo3bo4bo2bo4bo3bo4bo$2b2obo4bob2ob4ob2obo4bob2o$3bobob2ob2o3b4o3b2ob2obobo$4b2o4bo12bo4b2o$5bo3bo14bo3bo$6b3o16b3o3$bo30bo$o32bo$b2o7b2o10b2o7b2o$13bo6bo$bo4bo3bo4bo2bo4bo3bo4bo$2b2obo4bob2ob4ob2obo4bob2o$3bobob2ob2o3b4o3b2ob2obobo$4b2o4bo12bo4b2o$5bo3bo14bo3bo$6b3o16b3o!
EDIT 3: 70 cells:
x = 29, y = 15, rule = B3/S232o$3bo$o3bobo2b2o$b2o3bo2bo$2bobobob2o$3b2obob2o$4bobo$7bo10b2o7b2o$5bo3b2o5bo$5bo6bobo4bo3bo4bo$6bo3bob3ob2obo4bob2o$7b3o2b3o3b2ob2obobo$19bo4b2o$20bo3bo$21b3o!
EDIT 4: 69:
x = 29, y = 15, rule = B3/S232ob2o2$o3bo5b2o$b2obo$2bobobo3bo$3b2obob3o$4bobo$7bo10b2o7b2o$5bo3b2o5bo$5bo6bobo4bo3bo4bo$6bo3bob3ob2obo4bob2o$7b3o2b3o3b2ob2obobo$19bo4b2o$20bo3bo$21b3o! x₁=ηx V ⃰_η=c²√(Λη) K=(Λu²)/2 Pₐ=1−1/(∫^∞_t₀(p(t)ˡ⁽ᵗ⁾)dt) $$x_1=\eta x$$ $$V^*_\eta=c^2\sqrt{\Lambda\eta}$$ $$K=\frac{\Lambda u^2}2$$ $$P_a=1-\frac1{\int^\infty_{t_0}p(t)^{l(t)}dt}$$ http://conwaylife.com/wiki/A_for_all Aidan F. Pierce A for awesome Posts: 1634 Joined: September 13th, 2014, 5:36 pm Location: 0x-1 ### Re: Spaceship Discussion Thread Elsewhere, Sokwe wrote:You might consider starting the [1c/2] collection yourself, since you seem to be finding many of the small ships. Okay, here's a preliminary collection of all known 1c/2 ships with 75 or fewer bits: x = 158, y = 385, rule = B3/S2328bobo$27bo2bo$26b2o$25bo$24b4o$23bo4bo$23bo2bo$23bo2bo$24bo$25b4obo$26bo3bo$obobo3bo3bo14bo$27bobo$o7bo3bo$26b3o$obobo3bobobo13b2o$26b3o$o3bo7bo$27bobo$obobo7bo14bo$26bo3bo$25b4obo$24bo$23bo2bo$23bo2bo$23bo4bo$24b4o$25bo$26b2o$27bo2bo$28bobo7$51bo$28bobo17bobobo$27bo2bo16bo2bo$26b2o18b2o$25bo19bo$24b4o16b4o$23bo4bo14bo4bo$23bo2bo16bo2bo$23bo2bo16bo2bo$24bo19bo$25b4obo14b4obo$26bo3bo15bo3bo$obobo3bobobo14bo19bo$27bobo17bobo$o7bo$26b3o17b3o$obobo3bobobo13b2o18b2o$26b2o18b3o$o3bo3bo3bo13b3o$47bobo$obobo3bobobo14bobo17bo$27bo18bo3bo$26bo3bo14b4obo$25b4obo13bo$24bo18bo2bo$23bo2bo16bo2bo$23bo2bo16bo4bo$23bo4bo15b4o$24b4o17bo$25bo20b2o$26b2o19bo2bo$27bo2bo17bobo$28bobo7$31bo19bo$28bobobo15bobobo$27bo2bo16bo2bo$26b2o18b2o$25bo19bo$24b4o16b4o$23bo4bo14bo4bo$23bo2bo16bo2bo$23bo2bo16bo2bo$24bo19bo$25b4obo14b4obo$26bo3bo15bo3bo$obobo3bobobo14bo19bo$27bobo17bobo$o7bo3bo$26b3o17b3o$obobo3bobobo13b2o18b2o$26b2o18b3o$o3bo3bo3bo13b3o$47bobo$obobo3bobobo14bobo17bo$27bo18bo3bo$26bo3bo14b4obo$25b4obo13bo$24bo18bo2bo$23bo2bo16bo2bo$23bo2bo16bo4bo$23bo4bo15b4o$24b4o17bo$25bo20b2o$26b2o19bo2bo$27bo2bo17bobobo$28bobo20bo9$35bobo$34bo2bo$33b2o$32bo4bo$31b5obo$28b2o$27bo3b3o$26bo3bo$26bo5bo$obobo3bobobo13bo2bo2bo$27bobo2b2obo$o7bo3bo22bo$26b3o$obobo3bobobo13b2o$26b3o$o3bo7bo$27bobo$obobo3bobobo14bo$26bo3bo$25b4obo$24bo$23bo2bo$23bo2bo$23bo4bo$24b4o$25bo$26b2o$27bo2bo$28bobo7$31bo$28bobobo$27bo2bo$26b2o91bobo$25bo92bo2bo$24b4o89b2o$23bo4bo23bobo19bobo19bo19bo3bo$23bo2bo24bo2bo18bo2bo16b4o18b3obo$23bo2bo23b2o20b2o19b3o16b2o$24bo24bo21bo4bo14bo3bo15bo3b5o$25b4obo17b6o16b5obo14b3o16bo3bo$26bo3bo14b2o7bo12b2o19bobo4bo14bo5b2o$bobobo3bobobo13bo16bo3b3obo13bo3b3o14b8o15bo2bo2b4o$27bobo13bo3bo4bo12bo3bo16b2o23bobo5bo$5bo3bo3bo29bo5bo4bo10bo5bo13b2o4b2o$26b3o14b3o3b5o11b3o3bo14bobo3bo2bo14b3o$5bo3bo3bo12b2o38bo4b2obo18bobo14b2o$26b2o15b3o3b5o12b2o6bo11b2o22b3o$5bo3bo3bo12b3o14bo5bo4bo11bobo17bo2bo$43bo3bo4bo13b2o2b2obo13bo4bo2bo15bobo$5bo3bobobo13bobo14bo3b3obo14bob3obobo11b4obobo16bo$27bo17b2o7bo19bo14bo6bo13bo3bo$26bo3bo17b6o15b5obo14b5obo12b4obo$25b4obo18bo20bo4bo14bo17bo$24bo25b2o19b3o17b3obo11bo2bo$23bo2bo24bo2bo16bo3bo16bo3bo10bo2bo$23bo2bo25bobo18b3o17b2o12bo4bo$23bo4bo44b4o17bo2bo10b4o$24b4o48bo18bobo11bo$25bo84b2o$26b2o83bo2bo$27bo2bo81bobo$28bobobo$31bo9$77bobo$34bobo18bobo18bo2bo$33bo2bo17bo2bo17b2o$32b2o19b2o19bo4bo$31bo4bo15bo4bo15b5obo$30b5obo14b5obo12b2o$27b2o19b2o19bo3b3o$26bo3b3o14bo3b3o14bo3bo$25bo3bo16bo3bo17bo5bo$25bo5bo14bo5bo15bo2bo2bo$bobobo4bobo12bo2bo2bo14bo2bo2bo16bobo2b2obo$26bobo2b2obo12bobo2b2obo21bo$5bo6bo21bo20bobo10b3o$25b3o18b3o7bo11b2o$5bo6bo12b2o19b2o20b2o$25b3o18b3o19b3o$5bo6bo$26bobo18bobo19bobo$5bo4bobobo11bo20bo21bo$25bo3bo16bo3bo17bo3bo$24b4obo15b4obo16b4obo$23bo20bo21bo$22bo2bo17bo2bo18bo2bo$22bo2bo17bo2bo18bo2bo$22bo4bo15bo4bo16bo4bo$23b4o17b4o18b4o$24bo20bo21bo$25b2o19b2o20b2o$26bo2bo17bo2bo18bo2bo$27bobobo16bobo19bobo$30bo11$97bobo$77bobo16bo2bo$35bo40bo2bo15b2o$32bobobo17bo20b2o17bo3bo$31bo2bo16b4o19bo3bo14b3obo$30b2o19b3o19b3obo12b2o$29bo19bo3bo16b2o17bo3b5o$28b6o15b3o17bo3b5o10bo3bo$25b2o7bo11bobo4bo14bo3bo15bo5b2o$bobobo3bobobo10bo3b3obo12b8o15bo5b2o12bo2bo2b4o$23bo3bo4bo11b2o22bo2bo2b4o11bobo5bo$5bo7bo9bo5bo4bo8b2o4b2o18bobo5bo$23b3o3b5o10bobo3bo2bo34b3o$5bo3bobobo37bobobo12b3o17b2o$23b3o3b5o10b2o8bo13b2o18b2o$5bo3bo13bo5bo4bo9bo2bo20b3o17b3o$23bo3bo4bo12bo4bo2bo$5bo3bobobo10bo3b3obo12b4obobo16bobo17bobo$25b2o7bo12bo6bo14bo19bo$28b6o14b5obo13bo3bo15bo3bo$29bo18bo18b4obo14b4obo$30b2o17b3obo12bo19bo$31bo2bo15bo3bo10bo2bo16bo2bo$32bobo16b2o12bo2bo16bo2bo$52bo2bo9bo4bo14bo4bo$53bobo10b4o16b4o$67bo19bo$68b2o18b2o$69bo2bo16bo2bo$70bobobo15bobo$73bo8$90bobo$89bo2bo23bobo15bobo$34bobo18bobo30b2o25bo2bo14bo2bo$33bo2bo17bo2bo29bo26b2o16b2o$32b2o19b2o31b4o23bo4bo12bo4bo$31bo4bo15bo3bo17bobo8bo4bo21b5obo11b5obo$30b5obo14b3obo17bo2bo8bo2bo20b2o16b2o$27b2o19b2o22b2o11bo2bo19bo3b3o11bo3b3o$26bo3b3o14bo3b5o15bo3bo10bo20bo3bo13bo3bo$25bo3bo16bo3bo19b3obo12b4obo14bo5bo11bo5bo$25bo5bo14bo5b2o13b2o19bo3bo14bo2bo2bo11bo2bo2bo$bobobo3bobobo11bo2bo2bo14bo2bo2b4obo8bo3b5o14bo18bobo2b2obo9bobo2b2obo$26bobo2b2obo12bobo5b2o8bo3bo19bobo24bo17bobo$5bo7bo20bobo20bo7bo5b2o34b3o15b3o7bo$25b3o7bo10b3o16b3o3b4o13b3o16b2o16b2o$5bo3bobobo11b2o19b2o18bo7bo13b2o17b2o16b2o$25b3o18b3o17b2o20b3o16b3o15b3o$5bo7bo52bobo$26bobo18bobo16b2o2bo18bobo16bobo15bobo$5bo3bobobo12bo20bo19bobo8bo10bo18bo17bo$25bo3bo16bo3bo24b2obo9bo3bo14bo3bo13bo3bo$24b4obo15b4obo16b2o6bo11b4obo13b4obo12b4obo$23bo20bo22bo2bo4bo10bo18bo17bo$22bo2bo17bo2bo21bo4bo11bo2bo15bo2bo14bo2bo$22bo2bo17bo2bo21b4obo11bo2bo15bo2bo14bo2bo$22bo4bo15bo4bo21bo14bo4bo13bo4bo12bo4bo$23b4o17b4o23b4o11b4o15b4o14b4o$24bo20bo25bo15bo18bo17bo$25b2o19b2o25b2o13b4obo13b2o16b2o$26bo2bo17bo2bo22b4o11bo3b2o14bo2bo14bo2bo$27bobobo16bobo25bo13b2o17bobobo13bobo$30bo59b4o18bo$93bo8$77bobo$48bobo25bo2bo14bobo$47bo2bo24b2o16bo2bo$34bo11b2o26bo3bo13b2o$31bobobo9bo27b3obo13bo4bo$30bo2bo10b4o22b2o18b5obo$29b2o12bo4bo6bo13bo3b5o9b2o$28bo14bo2bo5b2obo12bo3bo13bo3b3o$27b6o10bo2bo5bo15bo5b2o9bo3bo$24b2o7bo10bo7bo15bo2bo2b4o7bo5bo$bobobo3bo3bo9bo3b3obo13b4obo18bobo5bo7bo2bo2bo$22bo3bo4bo14bo3bo35bobo2b2obo$5bo3bo3bo8bo5bo4bo13bo20b3o23bo$22b3o3b5o14b5o16b2o15b3o$5bo3bobobo54b2o15b2o$22b3o3b5o14b5o16b3o14b3o$5bo7bo8bo5bo4bo13bo46bo$22bo3bo4bo14bo3bo18bobo14bobo2b2obo$5bo7bo9bo3b3obo13b4obo18bo15bo2bo2bo$24b2o7bo10bo7bo15bo3bo12bo5bo$27b6o10bo2bo5bo14b4obo12bo3bo$28bo14bo2bo5b2obo10bo19bo3b3o$29b2o12bo4bo6bo9bo2bo18b2o$30bo2bo10b4o17bo2bo21b5obo$31bobobo9bo19bo4bo20bo4bo$34bo11b2o18b4o22b2o$47bo2bo16bo25bo2bo$48bobo17b2o24bobo$69bo2bo$70bobobo$73bo8$70bobo21bo$69bo2bo18bobobo$34bobo18bobo10b2o20bo2bo$33bo2bo17bo2bo9bo21b2o64bobo$32b2o19b2o11b4o18bo65bo2bo$31bo3bo16bo4bo7bo4bo16b4o62b2o$30b3obo16b5obo7bo2bo17bo4bo20bobo19bobo15bo3bo$27b2o19b2o15bo2bo17bo2bo21bo2bo18bo2bo14b3obo$26bo3b5o12bo3b3o12bo19bo2bo20b2o20b2o14b2o$25bo3bo16bo3bo16b4obo14bo21bo3bo17bo15bo3b5o$25bo5b2o13bo5bo15bo3bo15b4obo14b3obo17b6o10bo3bo$25bo2bo2b4obo9bo2bo2bo16bo19bo3bo11b2o20b2o7bo9bo5b2o$bobobo3bobobo12bobo5b2o11bobo2b2obo13bobo18bo13bo3b5o13bo3b3obo11bo2bo2b4o$36bo18bobo32bobo10bo3bo17bo3bo4bo12bobo5bo$5bo3bo15b3o18b3o7bo11b3o32bo5b2o14bo5bo4bo$25b2o19b2o20b2o19b3o11b3o3b4o12b3o3b5o10b3o$5bo3bobobo11b2o19b2o20b2o19b2o13bo7bo33b2o$25b3o18b3o19b3o18b3o12b2o19b3o3b4obo9b3o$5bo7bo90bobo18bo5bo3b2o18bo$26bobo18bobo19bobo18bobo11b2o2bo8bo7bo3bo4b2o11bobo2b2obo$5bo3bobobo12bo20bo21bo20bo14bobo8bobo7bo3b3ob2o10bo2bo2bo$25bo3bo16bo3bo17bo3bo16bo3bo19b2obo10b2o7b2o8bo5bo$24b4obo15b4obo16b4obo15b4obo11b2o6bo16b6obo8bo3bo$23bo20bo21bo20bo17bo2bo4bo17bo15bo3b3o$22bo2bo17bo2bo18bo2bo17bo2bo16bo4bo20b2o14b2o$22bo2bo17bo2bo18bo2bo17bo2bo16b4obo21bo2bo14b5obo$22bo4bo15bo4bo16bo4bo15bo4bo16bo25bobo15bo4bo$23b4o17b4o18b4o17b4o18b4o40b2o$24bo20bo21bo20bo20bo44bo2bo$25b2o19b2o20b4obo15b4obo16b2o42bobo$26bo2bo17bo2bo17bo3b2o15bo3b2o16b4o$27bobo18bobobo17b2o19b2o21bo$51bo18b4o17b4o$73bo20bo!
Due to simple 2-cell extensions, many of these are boring variants of smaller ones. Did I miss anything?
I'm not sure that I have enough spare time on my hands to try to catalogue any larger ones without some sort of component-assembling software.
EDIT 2: My knightt c/6 asymmetric width 11 search just finished with no ships, unfortunately. Here's the longest partial:
x = 10, y = 101, rule = B3/S232bo$bobo$4bo$bo$5bo3bo$2b2obo3bo$6bo2bo$3b3o$3b2o$4bobo$7bo$6b2o$5bo$5bo$4bo$4bo$3bo$3b2o$4b2o$b2o$2bo2$3b2o$3bo$2bo$2bob4o$6b2o$2b2o$2b2ob2o$3bo2bo$2bobo$bobo3bo$o3b3o$2b2o$5b2o$2bo3b2o$3bo4bo$3bo3bo$3bo$3bo$2b2o3b2o$2b2o2b3o$3bobo2bo3$7b2o$3b2obobo$6bo2bo$3bo4bo$4bo3bo$7bo$5bo2bo$4b2ob2o$5bo3bo$5bo$4b2o$2b2o2bo$2b2o2bo$bob2o2$3b3o$2b2obo$2b2o2bo$2bo2b2o$4bo2b2o$4bobo$6b3o$b2o5bo$3bobo$bo2bo3bo$bo2b3o$2b2o2$bob2o$4bo$bo2b3o$bobo2bo$bo3bo$2o$bo$bobo2bo$bo4bo$6bobo$2b3ob3o$4bobo$b2o5bo$5b2o$2bo2b2obo$3obo3bo$o2b2obo2bo$2bobobobo$2bo3bo$6b2o$3bo2b2o$4b2o$3bob4o$3b2o2b2o$b2ob2o$6bobo$2bobobob2o$2b5o! It features an interesting front end, which I don't think is known. I may try extending this partial in various ways later to try to find a width-12 ship. I would recommend that nobody try the complete width-12 search, because just the width-11 search took me about 1600 CPU-hours (~3 CPU-months) to complete. x₁=ηx V ⃰_η=c²√(Λη) K=(Λu²)/2 Pₐ=1−1/(∫^∞_t₀(p(t)ˡ⁽ᵗ⁾)dt) $$x_1=\eta x$$ $$V^*_\eta=c^2\sqrt{\Lambda\eta}$$ $$K=\frac{\Lambda u^2}2$$ $$P_a=1-\frac1{\int^\infty_{t_0}p(t)^{l(t)}dt}$$ http://conwaylife.com/wiki/A_for_all Aidan F. Pierce A for awesome Posts: 1634 Joined: September 13th, 2014, 5:36 pm Location: 0x-1 ### Re: Spaceship Discussion Thread A for awesome wrote:My knightt c/6 asymmetric width 11 search just finished with no ships, unfortunately.... the width-11 search took me about 1600 CPU-hours (~3 CPU-months) to complete. Thanks! This is a valuable, if disappointing result. Now (1,1)c/6 is the only potentially possible velocity for a width-11 period-6 ship, but this seems unlikely. -Matthias Merzenich Sokwe Moderator Posts: 1264 Joined: July 9th, 2009, 2:44 pm ### Re: Spaceship Discussion Thread I have completed w9 zfind 2.0 searches for 5c/11 odd, even, and gutter symmetries with no spaceships found. Here are the longest partials: x = 78, y = 23, rule = B3/S233bo9bo16bo10bo23bo5bo$2bobo7bobo14bobo8bobo21bobo3bobo$b2ob2o5b2ob2o12b2ob2o6b2ob2o19bo3bobo3bo$2bo2bo5bo2bo17bo6bo23bo9bo$3b3o5b3o19b2o2b2o24bobobobobobo$3b2o7b2o15bo2bobo2bobo2bo24bobo$28bobo2b6o2bobo20bo7bo$b2o11b2o12bo2b2obo2bob2o2bo21b2o3b2o$6b2ob2o18b2o10b2o19bo3bo3bo3bo$7bobo21bo8bo19b3o11b3o$4b2obobob2o16b2ob2o4b2ob2o18bobo9bobo$4bo2bobo2bo15bo3b2o4b2o3bo18b4o5b4o$b2obo2bobo2bob2o13bo12bo19bo2bo5bo2bo$bo4b2ob2o4bo17bo4bo24bo2bo3bo2bo$bo13bo12b3ob2o4b2ob3o17b2ob2o5b2ob2o$3b2obo3bob2o14bo3b2o4b2o3bo$3bo9bo14bobob2o4b2obobo$4bo3bo3bo14b2ob2o8b2ob2o14b2o3b2o5b2o3b2o$b5obobob5o13bo3bob2obo3bo21b2o5b2o$2obo9bob2o13bo4b2o4bo18bobob2o5b2obobo$27b3o3b2o2b2o3b3o15b3ob2o5b2ob3o$27bobobob2o2b2obobobo$28bo14bo! x₁=ηx V ⃰_η=c²√(Λη) K=(Λu²)/2 Pₐ=1−1/(∫^∞_t₀(p(t)ˡ⁽ᵗ⁾)dt) $$x_1=\eta x$$ $$V^*_\eta=c^2\sqrt{\Lambda\eta}$$ $$K=\frac{\Lambda u^2}2$$ $$P_a=1-\frac1{\int^\infty_{t_0}p(t)^{l(t)}dt}$$ http://conwaylife.com/wiki/A_for_all Aidan F. Pierce A for awesome Posts: 1634 Joined: September 13th, 2014, 5:36 pm Location: 0x-1 ### Re: Spaceship Discussion Thread A couple (2,1)c/7 partials: x = 60, y = 28, rule = B3/S2340bo$36bo3bo$37b3obo$6bo30bobob2o$5bo29bo5bob2o$4bob2o26b2o9bo$4bo4bo23bo3bobo3bobo$2b3o4bo22b2ob2o3bob3o$3bo2bo3bo22bo4b2obob2o$obo3bo3b2o21bo3bobobo2b2o$2bo3bo4bo27bo8bo$4bo4bob2o23b2obo4bo3bobo$4bo3bo3b2obo23b3o3bo4bo$7b2o3bob2o23b3o2b3o$5b2o7bo29b2o4b2o$7bob3o30bobo4b3o$11bo3b2o25b2o5bobo$9bob4o2bo31bo3b2o$10bob2o4bo30b2o4bo$13bobo34bo5bo$15bo2bob2o25b2o2b2o$16b3o2bo25b2ob5obo$16b2o4bo32bo$18b2ob2o34bo$21bobo31bo$18bo4b2o30bo2b2o$19bob3o33b2o$20b3o34bo!
x₁=ηx
V ⃰_η=c²√(Λη)
K=(Λu²)/2
Pₐ=1−1/(∫^∞_t₀(p(t)ˡ⁽ᵗ⁾)dt)
$$x_1=\eta x$$
$$V^*_\eta=c^2\sqrt{\Lambda\eta}$$
$$K=\frac{\Lambda u^2}2$$
$$P_a=1-\frac1{\int^\infty_{t_0}p(t)^{l(t)}dt}$$
http://conwaylife.com/wiki/A_for_all
Aidan F. Pierce
A for awesome
Posts: 1634
Joined: September 13th, 2014, 5:36 pm
Location: 0x-1
A 2c/6 ship that I think is new:
x = 19, y = 63, rule = B3/S233bob3o3b3obo$2b2ob2obobob2ob2o$bo2b2o2bobo2b2o2bo$5bo2bobo2bo$bo4bobobobo4bo$2o2bobobobobobo2b2o$2o6bobo6b2o$bobo3b2ob2o3bobo$2b2o2bobobobo2b2o$b2o2b2obobob2o2b2o$8bobo$2bo5bobo5bo$bobo3b2ob2o3bobo$6bobobobo$bo3b2obobob2o3bo$bobo4bobo4bobo$3bo4bobo4bo$2o6bobo6b2o$4b3obobob3o$2bo4b2ob2o4bo$2o6bobo6b2o$obo2b2obobob2o2bobo$8bobo$4b2o2bobo2b2o$3b2o2bo3bo2b2o$7b2ob2o$3b2o9b2o$3b2o9b2o$2bo2bo7bo2bo$2bo3bo5bo3bo$5bo7bo$2b5o5b5o$b2o13b2o$7b2ob2o$8bobo$5b2obobob2o$4bobobobobobo$b2obo3bobo3bob2o$b2obo3bobo3bob2o$bobo4bobo4bobo$7b2ob2o2$2b3o9b3o$2b3o9b3o$2o3bo7bo3b2o2$o4bo7bo4bo$5b2o5b2o$5b2o5b2o$4b2o7b2o$7bo3bo$3b6ob6o$2b2obo2bobo2bob2o$5bo2bobo2bo$6bobobobo$3b2obobobobob2o$4bobobobobobo$3bo2bobobobo2bo$4b2ob2ob2ob2o$3b2o2bo3bo2b2o2$2bo13bo\$3bo3bo3bo3bo!
Notice that it has two period-6 components, but only the back component is new.
-Matthias Merzenich
Sokwe
Moderator
Posts: 1264
Joined: July 9th, 2009, 2:44 pm
Since the wiki doesn't seem to have pages on them yet, what are the smallest ships of these speeds?:
2c/6
2c/8
4c/8
3c/9
waiting for apgsearch to support one-dimensional rules
muzik
Posts: 2778
Joined: January 28th, 2016, 2:47 pm
Location: Scotland
PreviousNext | 2018-09-25 01:14:01 | {"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": 0.7873492240905762, "perplexity": 7554.651805929511}, "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-2018-39/segments/1537267160853.60/warc/CC-MAIN-20180925004528-20180925024928-00201.warc.gz"} |
https://www.sharcnet.ca/Software/Fluent6/html/ug/node811.htm | 22.3 Multicomponent Particle Theory
A number of industrially important processes, such as distillation, absorption and extraction, bring into contact two phases which are not at equilibrium. The rate at which a specie is transferred from one phase to the other depends on the departure of the system from equilibrium. The quantitative treatment of these rate processes requires knowledge of the equilibrium states of the system. Apart from these cases, vapor-liquid equilibrium (VLE) relationships in multicomponent systems are needed for the solution of many other classes of engineering problems, such as the computation of evaporation rates in spray combustion applications.
In FLUENT the rate of vaporization of a single component droplet is computed from Equation 22.9-20, where is the equilibrium concentration of the droplet species in the gas phase, and is computed in Equation 22.9-21 as:
(22.3-1)
where is the droplet temperature, and is the saturation pressure of the droplet species at .
For the general case where N components are evaporating from a droplet (distillation), the evaporation rate of each species is again given by Equation 22.9-20; however, in Equation 22.3-1 must be replaced by , the partial pressure of species , to calculate the concentration of at the droplet surface.
The partial pressure of species can be obtained from the general expression for vapor liquid equilibrium [ 343],
(22.3-2)
which is obtained by equating the fugacity of the liquid and vapor mixtures. Here, , is the fugacity coefficient for species in the mixture, and accounts for nonideality in the gas; is the fugacity coefficient for pure at the saturation pressure; is the activity coefficient for species in the mixture, and accounts for nonideality in the liquid phase; is the absolute pressure; is the temperature; is the universal gas constant; is the molar volume of the liquid; is the saturation pressure of species ; and and are the equilibrium compositions of species in the liquid and gas phases, respectively. The exponential term is the Poynting correction factor and accounts for compressibility effects within the liquid. Except at high pressures, the Poynting factor is usually negligible.
Under low pressure conditions where the gas phase may be assumed to be ideal, and . Furthermore, if the liquid is also assumed to be ideal, and Equation 22.3-2 reduces to Raoult's law,
(22.3-3)
Raoult's law is the default vapor-liquid equilibrium expression used in the FLUENT multicomponent droplet model. However, there is a UDF hook available for user-defined vapor-liquid equilibrium models.
While Raoult's law represents the simplest form of the VLE equation, keep in mind that it is of limited use, as the assumptions made for its derivation are usually unrealistic. The most critical assumption is that the liquid phase is an ideal solution. This is not likely to be valid, unless the system is made up of species of similar molecular sizes and chemical nature, such as in the case of benzene and toluene, or n-heptane and n-hexane.
When Raoult's law is applicable, the vaporization rate of each species from a multicomponent droplet can be computed from Equation 22.9-20, with the equilibrium concentration of species i in the gas phase computed as:
(22.3-4)
where is the droplet temperature, is the mole fraction of species i in the droplet, and is the saturation pressure of species i at .
Previous: 22.2.2 Turbulent Dispersion of
Up: 22. Modeling Discrete Phase
Next: 22.4 Wall-Film Model Theory | 2018-01-18 16:06:47 | {"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": 0.8677588701248169, "perplexity": 963.2586600318551}, "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/1516084887423.43/warc/CC-MAIN-20180118151122-20180118171122-00221.warc.gz"} |
http://math.stackexchange.com/questions/8903/how-to-prove-polarization-identity | # How to prove polarization identity?
How to prove polarization identity? For any $u, v \in V$ $4 \langle T(u), v \rangle = \langle T(u+v), u+v \rangle - \langle T(u-v), u-v \rangle + i\langle T(u+iv), u+iv \rangle - i\langle T(u-iv), u-iv \rangle$
-
What did you try? – AD. Nov 4 '10 at 13:57
What is the property of the $\langle . , . \rangle$ operator? $\langle u, v + w \rangle = ...$ (Hint: in the real case, we speak about bilinear forms). – Djaian Nov 4 '10 at 13:59
As a side comment: you've posted 11 questions on this forum so far, most of them are homework type questions with initially incorrectly applied tags (this one also, which I'll fix in a minute). You may get better attention if you at least tag questions in the correct field. You may also garner more good will if you actually accept answers. At least 3 or 4 of the questions you've asked in the past month have received very nice answers which you've ignored. – Willie Wong Nov 4 '10 at 14:02
@AD. : I think you can delete the "What" in your question. – Djaian Nov 4 '10 at 14:03
@Djaian: nice joke! :D – Ｊ. Ｍ. Nov 4 '10 at 14:06
show 1 more comment
Expand RHS, using distributive property and the property that $\langle u,cv\rangle=\bar{c}\langle u,v\rangle$. Simplify and cancel many terms to get the LHS. | 2014-07-22 18:18:44 | {"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": 0.7349535226821899, "perplexity": 2042.8851889666162}, "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-2014-23/segments/1405997860453.15/warc/CC-MAIN-20140722025740-00054-ip-10-33-131-23.ec2.internal.warc.gz"} |
https://kavigupta.org/2018/05/27/Experimental-Confirmation-of-Timezones/ | # Confirming that timezones exist with minimal trust
I was listening to an episode of Oh No, Ross and Carrie! about Flat Earth and it struck me that if you believe the world is conspiring against you, it’s practically quite difficult to prove almost anything. Now one thing that most Flat Earthers do seem to accept is that when its day in one part of the planet, it’s night in another part.
However, let’s say someone denied this. How would you prove it, with the minimum amount of trust?
## Can we do this while trusting nobody?
Let’s say that our experimenter only trusts their own senses. Can they prove that the day/night dichotomy exists? Well, one way they could do this would be to confirm that their watch is correct, then fly to the other side of the world, then confirm that their watch was off by 12 hours from the day/night cycle. However, this isn’t exactly right, it’s possible that someone could mess with your watch to cause it to run slightly fast or slow and cause it to be different when you alived as when you left.
With a fast enough plane, you might be able to confirm that day had turned to night in less than 12 hours, without the need to rely on a possibly manipulable watch. However, the very act of stepping on a plane might allow nefarious agents to manipulate your subjective experience of time, thus fooling you into thinking time zones exist. (Yes, this is a massively unproductive view of the world, but let’s try to work around it.)
Given these limitations, it’s clear that our prover needs a collaborator they can trust.
## Basic protocol with a trusted collaborator
We now have two experimenters, who each trust each other and want to prove the existence of time zones: Alice and Bob. Consider the following protocol: Bob travels across the world while Alice stays behind. Alice and Bob then communicate with each other and confirm that one sees day and the other night.
This works right? Well, almost. The issue is that there’s no protection against a simulation attack: what if Alice speaks to a simulated version of Bob while Bob speaks to a simulated version of Alice, with the two conversations actually taking place 12 hours apart from each other?
Alice and Bob’s position is starting to look undisprovable. But is it?
Let’s say that before leaving, Alice and Bob come up with a shared password, $$p$$, randomly. Then, the protocol is as such:
• Alice and Bob: agree on password $$p$$
• Alice: goes to other side of the world and confirms that its day
• Alice: sends $$p$$ to Bob
• Bob: confirms that its night and that the received message is in fact $$p$$
But wait, the adversary could trap the password after Alice has sent it but before Bob has received it, thus delaying it 12 hours and allowing the protocol to go through, even if the world in fact has no time zones. The issue is that there’s no response from Bob. Bob just sending back the password doesn’t help, since the adversary could do that!
OK, so now we have two different passwords $$p_a$$ and $$p_b$$. The protocol is now as follows:
• Alice and Bob: create passwords $$p_a$$, $$p_b$$ and share them with each other
• Alice: goes to other side of world and confirms that its day
• Alice: sends $$p_a$$ to Bob
• Bob: confirms that its night and that the received message is $$p_a$$
• Bob: sends $$p_b$$ to Alice
• Alice: confirms that it is still the same day and the received message is $$p_a$$
Now, Alice knows for sure that the world in fact has multiple time zones.
However, Alice and Bob might say that the adversary has the ability to read and edit their memories and thus steal their passwords. OK, this objection must make their claim undisprovable, right? In this case the answer is yes, because the adversary could always just convince them that the world had timezones by making arbitrary modifications to their memories.
## Secure Time Synchronization
The issue that Alice and Bob are running into is that they can’t pre-agree on the passwords, because then they know the passwords ahead of time and thus the adversary knows the passwords ahead of time. However, they might be able to “post-agree” on the passwords, as such:
• Alice and Bob: start on opposite sides of the world
• Alice: Confirms that its day
• Alice: Generates and sends $$p_a$$ to Bob
• Bob: Confirms that its night
• Bob: Generates and sends $$p_b$$ to Alice
• Alice: Confirms that it is still the same day
• Alice: meets with Bob
• Alice and Bob: check that all passwords transmitted were correct
If this protocol goes through without Alice and Bob having any issues, then they know that it can be day and night at the same time in different parts of the world. The procedure works!
## Assumptions and Proof
OK, so if we wanted to prove that the above scheme works, we would need a few assumptions. First, we assume Alice and Bob are trustworthy. Second, we need to assume that Alice can confirm that some entity is Bob in a face-to-face conversation and vice versa (so that the Alice and Bob at the end are guaranteed to be the real Alice and Bob). Third, we need to assume that Alice and Bob can tell when night falls, that is, the adversary can’t somehow convince them that two points in time are on the same day when it was in fact night in between those points. Finally, the adversary cannot have time travel and cannot predict either Alice or Bob’s random number generators.
In this case, we can prove that Bob confirming that it is night where he is happened after Alice generated $$p_a$$ and thus during some day D where Alice was, and before Bob generated and thus before Alice received $$p_b$$ and thus before some point on the same day D. Since we assume days are continuous, this proves that Bob observed night during a time when Alice observed day, and thus this procedure is secure.
You can weaken the assumption that two specific trustworthy people are needed by repeating this procedure for every pair of people in some set of N people. Then the scheme proves that time zones exist assuming that 2/N of the people are trustworthy, at the cost of having to repeat the scheme $$O(N^2)$$ times
## Implications
N/A.
Less facetiously, I guess my general takeaway is that “your claim is undisprovable” shouldn’t be used as a synonym for “you’re making ridiculous, conspiratorial, assumptions”. Because even when someone is being paranoid, there might be a way to disprove their hypothesis, with enough effort. | 2022-05-18 16:26:59 | {"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": 0.6069436073303223, "perplexity": 862.8188556311604}, "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-2022-21/segments/1652662522284.20/warc/CC-MAIN-20220518151003-20220518181003-00113.warc.gz"} |
https://indico.cern.ch/event/949705/contributions/4555573/ | # 30th International Symposium on Lepton Photon Interactions at High Energies
Jan 10 – 14, 2022
Online only
Europe/London timezone
## Tau physics at Belle II
Jan 11, 2022, 4:30 PM
20m
Online only
#### Online only
Parallel session talk Flavour
### Speaker
Zuzana Gruberova (Charles University)
### Description
The low-background environment of electron-positron collisions along with the large expected sample size and an hermetic detector make Belle II the premier experiment for studying tau-lepton physics. A competitive measurement of the mass of the tau lepton is reported based on a a small data set, and studies for world-leading measurements of tau lifetimes, foreseen with significantly less data than collected at Belle and Babar, are discussed. Perspectives on tests of lepton (flavor) universality and other searches for non-SM physics in tau decays are also outlined.
### Primary author
James Libby (Indian Institute of Technology Madras (IN)) | 2022-01-24 07:30:37 | {"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": 0.8479139804840088, "perplexity": 14857.709966372795}, "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-2022-05/segments/1642320304515.74/warc/CC-MAIN-20220124054039-20220124084039-00485.warc.gz"} |
https://2022.help.altair.com/2022.1/activate/business/en_us/topics/reference/oml_language/ComputerVision/splitcv.htm | # splitcv
Splits an image into RGB/RGBA channels.
## Syntax
[r, g, b] = splitcv(cvhandle)
[r, g, b, a] = splitcv(cvhandle)
## Inputs
cvhandle
Handle of a color image.
Type: integer
## Outputs
r
Handle of the single channel image representing the red channel data.
Type: integer
g
Handle of the single channel image representing the green channel data.
Type: integer
b
Handle of the single channel image representing the blue channel data.
Type: integer
a
Handle of the single channel image representing the alpha channel data, if applicable.
Type: integer
## Example
Split an image into RGB channels and create one image with each of these channels:
%Read image
figure(1);
imshowcv(img);
[blue, green, red] = splitcv(img);
figure(2);
imshowcv(blue);
figure(3);
imshowcv(red);
figure(4);
imshowcv(green);
Split an image into RGB channels and display only the red one:
cvhandle = imreadcv('bird2.jpg');
[r, g, b] = splitcv(cvhandle);
imshowcv(r); | 2023-04-01 22:26:07 | {"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": 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": 0.2576526403427124, "perplexity": 14064.880025821953}, "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-2023-14/segments/1679296950363.89/warc/CC-MAIN-20230401221921-20230402011921-00345.warc.gz"} |
https://cs.stackexchange.com/questions/81242/simple-discrete-math-problem-a-and-b-in-3cnf | # Simple discrete math problem: A and B in 3CNF
I need to write (A and B) in 3CNF, and for whatever reason, I can only come up with (A or B) in 3CNF (A OR B OR X) AND (NOT(X) OR A OR B). Any help is greatly appreciated!
How about $(A \lor A \lor A) \land (B \lor B \lor B)$?
$$(A \lor x_1 \lor x_2) \land (B \lor x_1 \lor x_2) \land \\ (\lnot x_1 \lor y_1 \lor z_1) \land (\lnot x_1 \lor y_1 \lor \lnot z_1) \land (\lnot x_1 \lor \lnot y_1 \lor z_1) \land (\lnot x_1 \lor \lnot y_1 \lor \lnot z_1) \land \\ (\lnot x_2 \lor y_2 \lor z_2) \land (\lnot x_2 \lor y_2 \lor \lnot z_2) \land (\lnot x_2 \lor \lnot y_2 \lor z_2) \land (\lnot x_2 \lor \lnot y_2 \lor \lnot z_2).$$
• Just an addition: there are infinitely many ways to make a CNF that is satisfiable iff $A\land B=1$. – rus9384 Sep 14 '17 at 21:06 | 2021-05-17 00:40:43 | {"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": 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": 0.8745553493499756, "perplexity": 464.7840115412722}, "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/1620243991921.61/warc/CC-MAIN-20210516232554-20210517022554-00450.warc.gz"} |
https://en.wikipedia.org/wiki/Sierpi%C5%84ski_number | # Sierpiński number
In number theory, a Sierpiński number is an odd natural number k such that ${\displaystyle k\times 2^{n}+1}$ is composite for all natural numbers n. In 1960, Wacław Sierpiński proved that there are infinitely many odd integers k which have this property.
In other words, when k is a Sierpiński number, all members of the following set are composite:
${\displaystyle \left\{\,k\cdot 2^{n}+1:n\in \mathbb {N} \,\right\}.}$
If the form is instead ${\displaystyle k\times 2^{n}-1}$, then k is a Riesel number.
## Known Sierpiński numbers
The sequence of currently known Sierpiński numbers begins with:
78557, 271129, 271577, 322523, 327739, 482719, 575041, 603713, 903983, 934909, 965431, 1259779, 1290677, 1518781, 1624097, 1639459, 1777613, 2131043, 2131099, 2191531, 2510177, 2541601, 2576089, 2931767, 2931991, ... (sequence A076336 in the OEIS).
The number 78557 was proved to be a Sierpiński number by John Selfridge in 1962, who showed that all numbers of the form 78557⋅2n + 1 have a factor in the covering set {3, 5, 7, 13, 19, 37, 73}. For another known Sierpiński number, 271129, the covering set is {3, 5, 7, 13, 17, 241}. Most currently known Sierpiński numbers possess similar covering sets.[1]
However, in 1995 A. S. Izotov showed that some fourth powers could be proved to be Sierpiński numbers without establishing a covering set for all values of n. His proof depends on the aurifeuillean factorization t4⋅24m+2 + 1 = (t2⋅22m+1 + t⋅2m+1 + 1)⋅(t2⋅22m+1 - t⋅2m+1 + 1). This establishes that all n ≡ 2 (mod 4) give rise to a composite, and so it remains to eliminate only n ≡ 0, 1, 3 (mod 4) using a covering set.[2]
## Sierpiński problem
Unsolved problem in mathematics:
Is 78,557 the smallest Sierpiński number?
The Sierpiński problem asks for the value of the smallest Sierpiński number. In private correspondence with Paul Erdős, Selfridge conjectured that 78,557 was the smallest Sierpiński number.[3] No smaller Sierpiński numbers have been discovered, and it is now believed that 78,557 is the smallest number.[4]
To show that 78,557 really is the smallest Sierpiński number, one must show that all the odd numbers smaller than 78,557 are not Sierpiński numbers. That is, for every odd k below 78,557, there needs to exist a positive integer n such that k2n + 1 is prime.[1] As of December 2021, there are only five candidates which have not been eliminated as possible Sierpiński numbers:[5]
k = 21181, 22699, 24737, 55459, and 67607.
The distributed volunteer computing project PrimeGrid is attempting to eliminate all the remaining values of k. As of August 2022, no prime has been found for these values of k, with all ${\displaystyle n\leq 37\,408\,811}$ having been eliminated.[6]
The most recently eliminated candidate was k = 10223, when the prime ${\displaystyle 10223\times 2^{31172165}+1}$ was discovered by PrimeGrid in October 2016. This number is 9,383,761 digits long.[5]
## Prime Sierpiński problem
Unsolved problem in mathematics:
Is 271,129 the smallest prime Sierpiński number?
In 1976, Nathan Mendelsohn determined that the second provable Sierpiński number is the prime k = 271129. The prime Sierpiński problem asks for the value of the smallest prime Sierpiński number, and there is an ongoing "Prime Sierpiński search" which tries to prove that 271129 is the first Sierpiński number which is also a prime. As of November 2018, the nine prime values of k less than 271129 for which a prime of the form k2n + 1 is not known are:[7]
k = 22699, 67607, 79309, 79817, 152267, 156511, 222113, 225931, and 237019.
As of August 2022, no prime has been found for these values of k with ${\displaystyle n\leq 27\,315\,111}$.[8]
The first two, being less than 78557, are also unsolved cases of the (non-prime) Sierpiński problem described above. The most recently eliminated candidate was k = 168451, when the prime number ${\displaystyle 168451\times 2^{19375200}+1}$ was discovered by PrimeGrid in September 2017. The number is 5,832,522 digits long.[9]
## Extended Sierpiński problem
Unsolved problem in mathematics:
Is 271,129 the second Sierpiński number?
Suppose that both preceding Sierpiński problems had finally been solved, showing that 78557 is the smallest Sierpiński number and that 271129 is the smallest prime Sierpiński number. This still leaves unsolved the question of the second Sierpinski number; there could exist a composite Sierpiński number k such that ${\displaystyle 78557. An ongoing search is trying to prove that 271129 is the second Sierpiński number, by testing all k values between 78557 and 271129, prime or not.
Solving the extended Sierpiński problem, the most demanding of the three posed problems, requires the elimination of 21 remaining candidates ${\displaystyle k<271129}$, of which nine are prime (see above) and twelve are composite. The latter include k = 21181, 24737, 55459 from the original Sierpiński problem. As of August 2022, the following eight values of k, unique to the extended Sierpiński problem, remain:[10]
k = 91549, 131179, 163187, 200749, 209611, 227723, 229673, and 238411.
As of August 2022, no prime has been found for these values of k with ${\displaystyle n\leq 22\,384\,237}$.[11]
In December 2019, ${\displaystyle 99739\times 2^{14019102}+1}$ was found to be prime by PrimeGrid, eliminating k = 99739. The number is 4,220,176 digits long.[12]
The most recent elimination was in December 2021, when ${\displaystyle 202705\times 2^{21320516}+1}$ was found to be prime by PrimeGrid, eliminating k = 202705. The number is 6,418,121 digits long.
## Simultaneously Sierpiński and Riesel
A number may be simultaneously Sierpiński and Riesel. These are called Brier numbers. The smallest five known examples are 3316923598096294713661, 10439679896374780276373, 11615103277955704975673, 12607110588854501953787, 17855036657007596110949, ... (A076335).[13]
## Dual Sierpinski problem
If we take n to be a negative integer, then the number k2n + 1 becomes ${\displaystyle {\frac {2^{|n|}+k}{2^{|n|}}}}$. When k is odd, this is a fraction in reduced form, with numerator 2|n| + k. A dual Sierpinski number is defined as an odd natural number k such that 2n + k is composite for all natural numbers n. There is a conjecture that the set of these numbers is the same as the set of Sierpinski numbers; for example, 2n + 78557 is composite for all natural numbers n.[citation needed]
For odd values of k the least n such that 2n + k is prime are
1, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 3, 2, 1, 1, 4, 2, 1, 2, 1, 1, 2, 1, 5, 2, ... (sequence A067760 in the OEIS)
The odd values of k for which 2n + k is composite for all n < k are
773, 2131, 2491, 4471, 5101, 7013, 8543, 10711, 14717, 17659, 19081, 19249, 20273, 21661, 22193, 26213, 28433, ... (sequence A033919 in the OEIS)
## References
1. ^ a b Sierpinski number at The Prime Glossary
2. ^ Anatoly S. Izotov (1995). "Note on Sierpinski Numbers" (PDF). Fibonacci Quarterly. 33 (3): 206.
3. ^
4. ^ Guy, Richard Kenneth (2005). Unsolved Problems in Number Theory. New York: Springer-Verlag. pp. B21:119–121, F13:383–385. ISBN 978-0-387-20860-2. OCLC 634701581.
5. ^ a b
6. ^ "Seventeen or Bust statistics". PrimeGrid. Retrieved November 21, 2019.
7. ^ Goetz, Michael (July 10, 2008). "About the Prime Sierpinski Problem". PrimeGrid. Retrieved September 12, 2019.
8. ^ "Prime Sierpinski Problem statistics". PrimeGrid. Retrieved November 21, 2019.
9. ^ Zimmerman, Van (September 29, 2017). "New PSP Mega Prime!". PrimeGrid. Retrieved September 12, 2019.
10. ^ Goetz, Michael (6 April 2018). "Welcome to the Extended Sierpinski Problem". PrimeGrid. Retrieved 21 August 2019.
11. ^ "Extended Sierpinski Problem statistics". www.primegrid.com. Retrieved 6 April 2018.
12. ^ Brown, Scott (13 January 2020). "ESP Mega Prime!". PrimeGrid. Retrieved 18 January 2020.
13. ^ Problem 29.- Brier Numbers | 2022-12-06 05:39:19 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 13, "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": 0.7923567295074463, "perplexity": 995.1388825363309}, "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/1669446711069.79/warc/CC-MAIN-20221206024911-20221206054911-00629.warc.gz"} |
https://techstalking.com/pycharm-terminal-doesnt-activate-conda-environment/ | # PyCharm terminal doesn’t activate conda environment
I have a conda environment at the default location for windows, which is C:\ProgramData\Anaconda2\envs\myenv. Also, as recommended, the conda scripts and executables are not in the %PATH% environment variable.
I opened a project in pycharm and pointed the python interpreter to
C:\ProgramData\Anaconda2\envs\myenv\python.exe
and pycharm seems to work well with the environment in the python console, in the run environment, and in debug mode.
However, when opening the terminal the environment is not activated (I made sure that the checkbox for activating the environment is checked). To be clear – when I do the same thing with a virtualenv the terminal does activate the environment without a problem.
Here are a few things I tried and did not work:
• Copied the activate script from the anaconda folder to the environment folder
• Copied the activate script from the anaconda folder to the Scripts folder under the environment
• Copied an activate script from the virtualenv (an identical one for which the environment is activated)
• Added the anaconda folders to the path
None of these worked.
I can manually activate the environment without a problem once the terminal is open, but how do I do it automatically?
## Solution #1:
This seems to be a known issue.
## Solution #2:
I ran into the same problem and used this solution.
1. Go to File -> Settings -> Tools -> Terminal.
2. Replace the value in Shell path with cmd.exe "/K" C:\path\to\Anaconda3\Scripts\activate.bat your_environment_name.
If I installed Anaconda in C:\Anaconda3 and have an environment named myenv, then my settings would look like this:
## Solution #3:
If any one wondering for settings for Linux, here is how to do it. Create a file .pycharmrc in your home dir. Open the file and add following
source ~/.bashrc
source ~/anaconda3/bin/activate your_env_name
Now go to Pycharm File > Settings > Tools > Terminal > Shell path replace your shell path with /bin/bash --rcfile ~/.pycharmrc.
Now when you open your terminal specified conda env will activate.
## Solution #4:
Here’s my solution for MacOS or Linux users:
First, add this to your ~/.bash_profile or ~/.zshrc depends on your shell. Remember to put it after conda has been initialized:
##### Activate conda env ######
[[ -n $CONDA_ENV ]] && conda activate$CONDA_ENV
And then go to your PyCharm settings, go to Tools -> Terminal
In Project Settings, add CONDA_ENV=yourenv to your Environment Variables.
yourenv is the env name from your conda for this specific project.
## Solution #5:
Expanding on darksinge’s answer to accommodate for conda>=4.4 and multiple environments, if 1) your conda environments and projects share the same name and 2) you keep your projects in the same directory, you can use this workaround:
cmd.exe "/K" C:\path\to\Anaconda3\Scripts\activate.bat C:\path\to\Anaconda3 & activate %cd:C:\path\to\project\parent\directory\=%
The last part (%cd:C:\path\to\project\parent\directory\=%) should infer the project name from the current working directory. For example, I keep my projects in Z:\, so %cd:Z:\=% returns my project name. You can read more at: How to replace substrings in windows batch file
## Solution #6:
Found a solution. Problem is we have been creating conda environments from within Pycharm while starting a new project.
This is created at the location /Users/<username>/.conda/envs/<env-name>.
e.g. /Users/taponidhi/.conda/envs/py38.
Instead create environments from terminal using conda create --name py38.
This will create the environment at /opt/anaconda3/envs/.
After this, when starting a new project, select this environment from existing environments. Everything works fine.
## Solution #7:
I am using OSX and zshell has become the default shell in 2020.
I faced the same problem: my conda environment was not working inside pycharm’s terminal.
File -> Settings -> Tools -> Terminal. the default shell path was configured as /bin/zsh --login
I tested on a separate OSX terminal that /bin/zsh --login somehow messes up \$PATH variable. conda activate keep adding conda env path at the end instead of at the beginning. So the default python (2.7) always took precedence because of messed up PATH string. This issue had nothing to do with pycharm (just how zshell behaved with –login),
I removed --login part from the script path; just /bin/zsh works (I had to restart pycharm after this change!)
## Solution #8:
Mixing a few answers from here, I figured out a solution for Git Bash in Windows:
1. Go to File -> Settings -> Tools -> Terminal.
2. Replace the value in “Shell path” with
"C:\Program Files\Git\bin\bash.exe" --login && conda activate myenv
## Solution #9:
This can happen when you disable activating the (base) environment by default like so:
conda config --set auto_activate_base false
To resolve the issue, reactivate the feature.
conda config --set auto_activate_base true
The answers/resolutions are collected from stackoverflow, are licensed under cc by-sa 2.5 , cc by-sa 3.0 and cc by-sa 4.0 . | 2022-07-05 22:20:24 | {"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": 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": 0.5132834315299988, "perplexity": 9375.112885715509}, "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/1656104628307.87/warc/CC-MAIN-20220705205356-20220705235356-00053.warc.gz"} |
https://physicscatalyst.com/maths/linear-function.php | # Linear Function
## Linear Function
Linear Function is defined as the real valued function $f : R \rightarrow R$ , y = f(x) = mx + c for each $x \in R$ where m and c is a constant
$y = x+ 2$
$y=2x -3$
$y=x-2$
$y =4x$
The above all are example of linear function
## Domain and Range of the Linear Function
For $f : R \rightarrow R$ , y = f(x) = mx + c for each $x \in R$
Domain = R
Range = R
## Graph of the Linear Function
We can draw the Linear Function graph on the Cartesian plane with value of x on the x-axis and value of y=f(x) on the y-axis. We can plot the point and join the point to obtain the graph. The graph of the linear function is a straight line
Steps to Draw the Graph of Linear Function
For $y = f(x) = mx + c$
1. Find the value of y for x=0
y=c
This point will lie on Y -axis. And the coordinates will be (0,c)
2. Find the value of x for y=0
$x= -\frac {c}{m}$
This point will lie on X -axis. And the coordinates will be $(-\frac {c}{m}, 0)$
3. This can be represent in below linear table
4. Now we can draw the line joining these two point
5. We can take any other convienent points also to draw the line
For $y = f(x) = mx + c$
$m$ is the slope and $c$ is the $y$ intercept of the graph.
If $m$ is positive
then the line rises to the right and if $m$ is negative then the line falls to the right
Below are few graph based on values of m and c
Graph for the linear function with m and c both positive
Graph for the linear function with positive m and negative c
Graph for the linear function with m and c both negative
Graph for the linear function with negative m and positive c
Graph for the linear function with positive m and c=0. This passes through origin. Such type of linear function is also called proportional function
Graph for the linear function with negative m and c=0. This passes through origin
Identify Function and constant function are special cases of Linear function
if m =1 and c=0, Linear function becomes f(x) =x which is a Identity Function
If m=0 ,then Linear function becomes f(x) =c which is a Constant Function
## Solved examples of Linear Functions
1. which is below function is a Linear function?
a. $y =2x$
b. $y = 11 -x$
c. $y= \frac {2}{3} x + \frac {1}{4}$
d. $x^2 + y^2=1$
e. $y =x^3$
f. $y =x^2 +1$
Solution
For the function to be a Linear function ,it should be of the form (mx+c)
a. This is Linear function as of the form (mx+c)
b. This is Linear function as of the form (mx+c)
c. This is Linear function as of the form (mx+c)
d. This is not a linear function
e. This is not a linear function
f. This is not a linear function
2. which of the graph represent Linear function?
Solution
The graph should be straight line for the function to be Linear function
So C and D are Linear function
3. Let f = {(1,1), (2,3), (0, -1), (-1, -3)} be a linear function from Z into Z.
Find f(x).
Solution
Since f is a linear function, f (x) = mx + c. Also, since $(1, 1), (0, - 1) \in f$,
f (1) = m + c = 1 and f (0) = c = -1. This gives m = 2 and so,f(x) = 2x - 1.
### Quiz Time
Question 1.
The function 't' which maps temperature in degree Celsius into temperature in degree Fahrenheit is defined by
$t(C) =\frac {9}{5} C + 32$
Which of the following is incorrect?
A.t(0)=32
B. t(-5)=23
C. t(10)=48
D. None of the above
Question 2.
If a function is defined as $f = {(x, y) | y = 2x + 7, \; where \; x \in R \; and \; -5 \leq x \leq 5}$ is a relation. Then find the domain and Range of Function?
A. Domain=[-5,5], range=[3,17]
B. Domain=[-5,5], range=[-3,17]
C. Domain=[-5,5], range=[-3,-17]
D. Domain=[-5,5], range=[-5,5]
Question 3.
Let f = {(1,1), (2,3), (3,5), (4,7)} be a linear function from Z into Z.
and f(x) =px +q then
A. p=1,q=1
B. p=1,q=2
C. p=2,q=1
D. p=2,q=-1
Question 4.
Let $f(x) =c$,find the value of $f(2) -f(1)$
A. 4
B. 2
C. 0
D. 1
Question 5.
The slope of the Linear function y=11x-1 is
A. 0
B. 11
C. -1
D. None of these
Question 6.
Find the Range and domain of the function $f(x) =x +2$
A. Domain = R, Range =R
B. Domain = R - {2}, Range = R
C. Domain = R , Range = R -{2}
D. Domain = R - {1}, Range = R | 2022-01-19 16:33:33 | {"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": 0.6962001919746399, "perplexity": 929.5582802110583}, "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-2022-05/segments/1642320301475.82/warc/CC-MAIN-20220119155216-20220119185216-00458.warc.gz"} |
https://www.gamedev.net/forums/topic/623690-trigonometry-question/ | • 13
• 27
• 9
• 9
• 20
# Trigonometry question
This topic is 2157 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.
## Recommended Posts
If i had a 3D arrow model pointing upwards (0, 1, 0) and i set this:
D3DXMATRIX world, rot; D3DXMatrixIdentity(&world); D3DXMatrixRotationAxis(&rot, &D3DXVECTOR3(1.0f, 0.0f, 0.0f), 0.3f); world = rot * world; ...
How can i obtain new direction arrow is pointing?
##### Share on other sites
make a new vector and use the translate function.
##### Share on other sites
I don't see how's that gonna help? I need to obtain direction vector it is "pointing at" after rotation.
##### Share on other sites
Make a vector pointing (0, 1, 0), i.e. the same direction as your arrow model, and rotate it in the same way. The rotated vector will give you the direction of the arrow model after rotation...
Putting it a different way, choose two vertices on your original arrow model, a vertex on the "bottom" of the arrow and a vertex of the "top" (the pointy end) of the arrow. These two points define a vector, which points in the arrow's direction. Once your model is rotated, find those two same vertices, they define another vector, which still represents the arrow's direction (obviously). The conclusion trivially follows...
##### Share on other sites
D3DXMatrixRotationAxis(&rot, &D3DXVECTOR3(1.0f, 0.0f, 0.0f), 0.1f); D3DXVECTOR4 orgDir(0.0f, 1.0f, 0.0f, 0.0f); D3DXVECTOR4 outDir; D3DXVec4Transform(&outDir, &orgDir, &rot);
Is this what you mean?
##### Share on other sites
Yes, although you want to use
D3DXMatrixRotationAxis(&rot, &D3DXVECTOR3(1.0f, 0.0f, 0.0f), 0.3f);
as the rotation matrix (says 0.1 in your last post but 0.3 in the first one)
##### Share on other sites
It was my mistake, It was a typo. Thanks for your help.
##### Share on other sites
Or, since you already know that your model was pointing along one of the coordinate system axes, you could just read out the second column of the world matrix. (Transformation) matrices aren't magic, they are just four vectors: right, up, forward, position. (depending on your handedness and as what you define your x,y,z axes).
##### Share on other sites
I don't see how that would be correct?
I tried it and its almost similar, apart that 'z' axis is negative if i extract it from second column of world. | 2018-03-17 18:45:29 | {"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": 0.4173494577407837, "perplexity": 2966.2284077059094}, "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/1521257645280.4/warc/CC-MAIN-20180317174935-20180317194935-00361.warc.gz"} |
https://economics.stackexchange.com/questions/19751/deriving-a-best-response-function-in-baik-1994 | # Deriving a Best Response Function in Baik (1994)
I'm reading a game-theory related paper*, and I'm not following the derivation of some property of the best-response functions.
Suppose I have two players $1$ and $2$, whose strategies are continuous levels of effort $x_1$ and $x_2$ respectively. The first order condition is given by for a Nash equilibrium strategy is given just standardly by $argmax_{x_1}(\pi_1)$:
$$\frac{\partial \pi_i}{\partial x_1}=\frac{\alpha\sigma h'(x_1)h(x_2)}{(\sigma h(x_1)+h(x_2))^2}-1=0$$
Let $x_1=r_1(x_2)$ denote player $1$'s reaction function. Since it is derived from player $1$'s first-order condition (above), we obtain it's derivative by differentiating along [the above FOC]:
$$\frac{dr_1(x_2)}{dx_2}=\frac{h'(x_1)h'(x_2)(\sigma h(x_1)-h(x_2))}{h(x_1)[h''(x_2)(\sigma > h(x_1)+h(x_2)-2h'(x_2))^2]}$$
Note that this is not the derivative of the first equation w.r.t. $x_2$, which simply is
$$\frac{\alpha\sigma h'(x_1)h'(x_2)(\sigma h(x_1)-h(x_2))}{(\sigma h(x_1)+h(x_2))^3}=0$$
However, besides this approach, I fail to see how the equation could be obtained. Rearranging for $x_1$ to express player $1$'s BR function the conventional way seems no option here either, since the function $h$ is undefined, and the BR function would also become much messier that the one quoted.
*Some of the working out is in a technical appendix elsewhere, which I can email.
• Given that the first order condition that determines the reaction function is an implicit function, you can use the implicit function theorem to determine that derivative. I haven't checked because it's tedious but I guess you'd arrive at the derivative that you have there – Maarten Punt Dec 19 '17 at 10:07
• Solve $\frac{\partial \frac{\partial \pi_1}{\partial x_1}}{\partial x_2} = 0$ for $\frac{\partial x_1}{\partial x_2}$. – clueless Dec 19 '17 at 21:47 | 2019-12-11 11:03:51 | {"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": 0.9490981101989746, "perplexity": 425.2011593166957}, "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-2019-51/segments/1575540530857.12/warc/CC-MAIN-20191211103140-20191211131140-00043.warc.gz"} |
http://physics.stackexchange.com/tags/states-of-matter/hot | # Tag Info
34
Good question. Assume we have one cube of ice in a glass of water. The ice displaces some of that water, raising the height of the water by an amount we will call $h$. Archimedes' principle states that the weight of water displaced will equal the upward buoyancy force provided by that water. In this case, $$\text{Weight of water displaced} = ... 33 Play-Doh is mostly flour, salt and water, so it's basically just (unleavened) dough. There are a lot of extra components like colourings, fragrances, preservatives etc, but these are present at low levels and don't have a huge effect on the rheology. The trouble with saying it's basically just dough is that the rheology of dough is fearsomely complicated. ... 30 Here is an explanation that needs no explicit equations. Consider the following diagram, in which part1 and part2 represent the ice. The displaced water volume equals part2 volume and has as much mass as (part1+part2) Now look at what happens when both part1 and part2 melt: their mass does not change, it is (part1+part2) it becomes water. And we just ... 28 Plasma is described as the 4th state of matter, which is what you get if you give so much temperature that the molecules begin to break up and ionize into positively and negatively charged fragments. Another Claim on the title '4th State of matter' is a 'supercritical fluid'. Sometimes people draw phase diagrams with it to show this '4th state of matter'. ... 19 Fire is neither. Fire is a process involving both. Fire is the energetic combination of various substances with oxygen to release light and heat. In a gas fire, such as might be found on a stove or in a heater, a light hydrocarbon such propane is broken down into components of hydrogen and carbon which unite with oxygen from the atmosphere to form water ... 12 In simple terms, there isn't any space in the ice crystal lattice for the extra atoms and there is no way to plug either of the ions (or the whole salt molecule) into the growing pattern. So more and more water joins the frozen mass, leaving a more and more concentrated brine until essentially all the water is frozen and the salt remains behind. As ... 11 Yes, of course, the freezing point will decrease by the pressure developed, while part of the water freezes. But do not underestimate the pressures! In such an experiment easily some thousand bares may be developed. (Depends on the rigidity of the vessel and the volume of water) Here is a video showing how freezing water cracks a cast iron sphere. ... 11 Yes, a plasma contains positive ions and negative electrons, but the positive ions don't need to be atoms. Actually the negative charges don't even need to be electrons. For example, plasma etching of silcon is done using a sulphur hexafluoride plasma that consists of (mainly) SF_5^+ and F^- ions. I'm not sure that the word plasma has a precise ... 9 @MartinBeckett's already gave an excellent answer: Salt is excluded from ice because there is "...no way to plug the ions... into the growing [ice] pattern." This unusually long answer -- a mini-tutorial really -- is an expansion on his answer. I've added a long background section that uses informal, easily visualized analogies to define a number of related ... 8 Matter is made up from point like fundamental particles, like electrons and quarks, that have zero volume. This puts us in the interesting position where the true volume of all matter is zero, and the only reason that everything doesn't instantly collapse into a point of zero volume is that the pointlike fundamental particles maintain a finite distance from ... 8 Formally, the incompressibility of a fluid is defined by the compressibility,$$ \beta=\frac1\rho\,\frac{\partial\rho}{\partial p} where $\rho$ is the mass density and $p$ the gas pressure. This means that, the compressibility is the measure of how much the density (volume) changes when a pressure is applied. For water at standard pressure, this works ...
8
I mean, they are heavier than air. No. Water is $H_2O$ which has a molecular weight of 18. Nitrogen is $N_2$ which has a molecular weight of 28. Oxygen is $O_2$ which has a molecular weight of 32. Argon is $Ar$ which has an atom weight of 40. So a water molecule has a mass that is less than that of all the significant components of air. But ...
7
It's not a disturbance, the liquid isn't supercooled in this case. It's right about at 0 degrees, though. It isn't the pressure drop directly, because you can give an upper-bound estimate to how much cooling the pressure drop directly does based on the observation that the pressure is not more than a few atmospheres. That means that space the gas in the ...
7
Steam is water in a gas phase, while mist, or fog, are small droplets of water in the condensed phase, but small enough to be kept in the air by thermal Brownian motion. There is a huge difference in their properties. steam at 1 atmospheric pressure is defined to be water at a temperature of 100 degrees celsius or more. It is an invisible gas, and contains ...
7
In the case of liquids and gases, at least, there's no fundamental difference. To see this, take a look at Wikipedia's phase diagram for water. Ignore the dotted lines for the moment, and note that the line between vapour (steam) and liquid stops at a certain point, called the critical point. What this means is that if you go through the following ...
7
I'm going to guess the toy you actually have in mind is the stuff sold in the US under the name Silly Putty . Play-Doh is used primarily as a "modeling clay" for sculpture - which means it needs to behave as a "plastic" - it's yield strength needs be low enough to enable it to be worked into a figure, but high enough that reasonable sized figures don't ...
7
The noise is either from the AC electricity, which would be a 60Hz buzzing, or from small bubbles forming on the heating element itself. When the electricity stops, both the buzzing and the bubble formation will stop as well. Bubbles create sound due to quickly expanding from a small nucleus. Here's a book I found with a section on noise from bubble ...
6
While the physical properties of a solid vs a liquid are obvious to any grade-schooler, the physics behind it are a little more complex. A substance is traditionally called a solid if it will not noticeably deform from a given starting shape in its steady state (in simple terms, it will not "flow" in the absence of any force other than gravity). A liquid, ...
6
Liquids are a state of matter in which the atoms or molecules are held together by chemical bonds (a difference from gases) but the bonds are weak enough for the shape to be variable (a difference from solids). That's why it is not possible to increase or decrease their volume much; the amount of energy from these chemical bonds would rapidly increase ...
6
I can address one class of non-Newtonian fluids consisting of solid particles dispersed in a liquid medium, such as the cornstarch and water mixture commonly called "oobleck." In more scientific language, I am talking about concentrated colloidal suspensions of particles. Here is an image of oobleck, taken from Dounas-Frazer et al 2012. These fluids tend ...
6
A quick comment on your terminology. The description "non-Newtonian" just means the stress/flow rate graph is not linear i.e. there isn't a single constant viscosity coefficient. The fluid you describe is what we colloid scientists call "dilatant", and it is certainly non-Newtonian. However there are lots of other non-Newtonian fluids such as tomato ketchup ...
6
I suggest that looking for an explanation for 'fire' may be the wrong approach. Science is a process where we try to explain what we perceive. In doing so, we may have to sacrifice (or at least temporarily suspend) a common perspective. My presumption is that you are probably mostly interested in the visual manifestation of fire - the dancing flames. ...
6
Good question! The defining difference is that in a gas the atoms are intact, and in fact are typically bonded into molecules, whereas in a plasma at least some of the electrons separate entirely from their atoms. In other words, particles of a plasma are charged, but particles of a gas are mostly uncharged. So technically, a plasma is not a gas and it ...
6
Plasmas are a common part of the world we live in. The definition of plasma allows them to exist within an environment consisting mostly of bound atoms. A variety of human technology creates plasmas. The type I hear plasma researchers reference most is a simple RF Plasma. This is perhaps the most direct way to use electricity to shake off the electrons ...
6
In what ways can energy transform into matter and vice versa? Energy and matter are connected according to special relativity and this has been experimentally demonstrated . It is the famous formula: $E=mc^2$ , where $m$ is the relativistic mass and $c$ the velocity of light. or $E^2=m_0^2c^4 +p^2c^2$ , for a particle with rest mass $m_0$ moving ...
5
The point of the fuzzball conjecture is that spacetime is geometrically altered at the black hole horizon. Rather than having an interior, the extra dimensions pinch off at the horizon and encode the complicated data from incoming particles in complicated geometry. This resolves 2 issues There's no singularity because the black hole effectively doesn't ...
5
Topological order can not be described in Ginzburg-Landau symmetry breaking paradigm. It is actually fair to say that topological order are more or less the properties of (gapped) quantum phases that can not be captured by GL. One way to define it is to use the notion of adiabatic continuity: if two gapped phases of matter can be connected by adiabatically ...
5
Related note: Fission isn't exactly turning matter into energy. It just releases the binding energy of the nucleus. This binding energy is part of the measured mass pf the nucleus, but if you want to separate "matter" and "energy" (not really possible), then it counts as energy. $\newcommand{\a}[3]{\mathrm{^{#1}_{#2}#3}}$ ...
5
See Wikipedia for a list of available plasmas over there..! Or have a look below for the chart which shows variation of temperature and electron density in different plasmas..! A quick Googling would've provided the answer. Common examples include Lightning..! The Sun (from Core to Corona) Fluorescent Lights and Neon Signs Nebulae (Luminous Clouds ...
Only top voted, non community-wiki answers of a minimum length are eligible | 2015-04-26 22:40:53 | {"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": 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": 0.6009974479675293, "perplexity": 508.4867677481705}, "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-2015-18/segments/1429246656747.97/warc/CC-MAIN-20150417045736-00212-ip-10-235-10-82.ec2.internal.warc.gz"} |
http://clay6.com/qa/36911/the-equation-which-is-balanced-and-represents-the-correct-product-s-is | Comment
Share
Q)
# The equation which is balanced and represents the correct product(s) is
$\begin{array}{1 1}(A)\;[Mg(H_2O)_6]^{2+}+(EDTA)^{4-}\quad\underrightarrow{\text{excess NaOH}}\quad [Mg(EDTA)]^2+6H_2O\\(B)\;CuSO_4+4KCN\rightarrow K_2[Cu(CN)_4]+K_2SO_4\\(C)\;Li_2O+2KCl\rightarrow 2LiCl+K_2O\\(D)\;[CoCl(NH_3)_5]^++5H^+\rightarrow Co^{2+}+5NH_4^++Cl^-\end{array}$
$[CoCl(NH_3)_5]^++5H^+\rightarrow Co^{2+}+5NH_4^++Cl^-$ | 2019-12-10 23:47:55 | {"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": 0.600883960723877, "perplexity": 3547.9770288508857}, "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/1575540529516.84/warc/CC-MAIN-20191210233444-20191211021444-00534.warc.gz"} |
https://tex.stackexchange.com/questions/567359/is-there-a-way-in-latex-to-separate-the-endnote-text-from-the-place-where-it-sho | Is there a way in LaTeX to separate the endnote text from the place where it should be placed/invoked? [duplicate]
I want to insert a few fairly large endnotes. The problem is that if I insert them with \endnote{text text} it makes it very difficult to follow the flow of the main text (as the endnote spans a few paragraphs).
Is there a package that would allow me to separate the place where the endnote is invoked from its contents? Something like say \lendnote{Label1} to invoke and \begin{lendnote}\label{Label1}<<here goes the endnote body>>\end{lendnote}
2 Answers
Here's a simple way to do this. You store each big endnote in a separate file named <label>.tex where <label> is how you will refer to it in the command that will insert it and in any \ref in the text. Then you create a command (I've called it \extendnote) which takes the label as an argument and then simply uses \input for the endnote text.
You can keep all of the external endnotes in the source file by adding them at the beginning using the {filecontents} environment. Of course if that becomes unmanageable, you can put all the {filecontents} environments in a separate file (e.g. bignotes.tex) and then \input that into your main source file.
\documentclass{article}
\usepackage{endnotes}
\usepackage{lipsum}
\begin{filecontents}[noheader,overwrite]{big.tex}
\lipsum[1-2]
\end{filecontents}
\begin{filecontents}[noheader,overwrite]{large.tex}
\lipsum[3-4]
\end{filecontents}
\newcommand{\extendnote}[1]{\endnote{\input{#1}\label{#1}}}
\begin{document}
Here is some text.\endnote{This is a small endnote\label{small}}
Here is some more text, with a big endnote.\extendnote{big} Here is some
more text.\endnote{A small endnote} and some more text.\extendnote{large}.
And here are some references to note \ref{large} and note \ref{small}.
\theendnotes
\end{document}
You can store the note text in a macro, defined sowhere earlier in the source code.
An example (partially borrowed from the Alan Munn's answer):
\documentclass{article}
\usepackage{endnotes}
\usepackage{lipsum}
\newcommand\noteone{\lipsum[1-2]}
\newcommand\notetwo{\lipsum[3-4]}
\begin{document}
Here is some text.\endnote{This is a small endnote} Here is some more text,
with a big endnote.\endnote{\noteone} Here is some more text.\endnote{A small
endnote} and some more text.\endnote{\notetwo}
\theendnotes
\end{document}
• Right. I put \def mechanically, it's shorter then \newcommand, so I rarely use the latter. – Sergei Golovan Oct 18 at 21:41
• Yes, me too sometimes, but for the site we should try to model best practice. :) (We can delete these comments). – Alan Munn Oct 18 at 21:57 | 2020-11-29 14:32:18 | {"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": 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": 0.8526700139045715, "perplexity": 2232.117033182214}, "config": {"markdown_headings": false, "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-2020-50/segments/1606141198409.43/warc/CC-MAIN-20201129123729-20201129153729-00472.warc.gz"} |
https://classes.areteem.org/mod/forum/discuss.php?d=306 | ## Online Course Discussion Forum
### MC II-A HW Questions
MC II-A HW Questions
1) Regarding this week's HW, I'm not sure how to do 9.19 and 9.25.
2) Also, regarding last week's hw, I'm wondering if my hw was regraded because I didn't see a change with the grades.
Thank you! Happy early Veteran's Day!
Re: MC II-A HW Questions
For 19, after eating some ice cream the cone gets smaller (we know it has now height 2 inches, but we don't know its radius). Here is a side view of the cone before and after eating some of it:
For 25 the following diagram might be helpful:
It is a view "cutting" through the diagonal of the square made by the centers of the four spheres with radius $1$.
You HW grade should be updated today. | 2019-01-21 00:56:00 | {"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": 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": 0.44179901480674744, "perplexity": 2680.3524215723864}, "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-2019-04/segments/1547583745010.63/warc/CC-MAIN-20190121005305-20190121031305-00396.warc.gz"} |
http://asciidoc.the-huck.com/generating-plain-text-files/ | # Generating Plain Text Files
AsciiDoc does not have a text backend (for most purposes AsciiDoc source text is fine), however you can convert AsciiDoc text files to formatted text using the AsciiDoc a2x(1) toolchain wrapper utility. | 2020-02-22 10:42:08 | {"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": 0.843217134475708, "perplexity": 14725.830595070682}, "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-2020-10/segments/1581875145657.46/warc/CC-MAIN-20200222085018-20200222115018-00463.warc.gz"} |
https://www.fernuni-hagen.de/mi/fakultaet/aktuelles/vortrag_arora.shtml | # Aktuelles
## Einladung zum Vortrag (Zoom) von Frau Sahiba Arora (TU Dresden) am 07.07.2021 - 14:30 Uhr im Rahmen des Forschungsseminars Analysis
[07.07.2021]
Im Rahmen des Forschungsseminars Analysis hält am Mittwoch, den 7. Juli 2021, um 14:30 Uhr,
Frau Sahiba Arora (TU Dresden)
einen Vortrag zum Thema
Uniform maximum and anti-maximum principles.
Abstract: Extensive literature has been devoted to study the operators for which the maximum and/or the anti-maximum principle holds. Combining an idea of Takáč (1996) with those from the recent theory of eventually positive C_0-semigroups, we look at some necessary and sufficient conditions for (anti-)maximum principles to hold in an abstract setting of Banach lattices.
More precisely, if A : dom (A) \subseteq E\to E is a closed, densely defined, and real operator on a complex Banach lattice E (or in particular, an L^p-space), then we consider the equation (\lambda-A)u =f for real numbers \lambda in the resolvent set of A. We ask whether f\geq 0 implies u\geq 0 for \lambda in a right neighbourhood of an eigenvalue. In this case, we say that the maximum principle is satisfied. Analogously, when the implication f\geq 0 implies u\leq 0 holds for \lambda in a left neighbourhood of an eigenvalue, we say that the anti-maximum principle holds.
We will also see how these abstract results can be applied to various concrete differential operators and illustrate how several previously known results about (anti-)maximum principles can be proved via this theory. This is joint work with Jochen Glück.
Der Vortrag findet im Rahmen eines Zoom-Meetings statt. Bei Interesse an den Zugangsdaten bitte bei Herrn Prof. Mugnolo (delio.mugnolo@fernuni-hagen.de) melden.
mathinf.webteam | 12.08.2021 | 2021-09-21 05:23:23 | {"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": 0.8539941310882568, "perplexity": 4262.707838808821}, "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/1631780057158.19/warc/CC-MAIN-20210921041059-20210921071059-00237.warc.gz"} |
https://www.physicsforums.com/threads/non-linear-integration.518347/ | # Non-linear integration
1. Jul 31, 2011
### shafieza_garl
can someone help me to solve this problem.i cannot remember how to solve the non-linear integration.like using chi-square distribution and uniform distribution. which one can i used to solve this integration?
File size:
2.4 KB
Views:
99
2. Aug 1, 2011
### HallsofIvy
Make the substitution $u= -\alpha(1- \beta)t+ q^{1-\beta}$. Then you will just be integrating constants times u to a constant power.
3. Aug 1, 2011
### shafieza_garl
Sorry..i cannot understand completely of the subtitution there.this is what i understand from what u said.if im wrong.can u explain more detail for me.because after i subtitute,the eq.cannot be integrate too.
File size:
4.1 KB
Views:
65
4. Aug 1, 2011
### I like Serena
Apply $\displaystyle \int u^c du = \frac {u^{c+1}} {c+1} \color{gray}{ + C}$.
Last edited: Aug 1, 2011 | 2018-01-21 19:03:12 | {"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": 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": 0.9423636794090271, "perplexity": 6983.563015419049}, "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/1516084890823.81/warc/CC-MAIN-20180121175418-20180121195418-00067.warc.gz"} |
https://ankplanet.com/physics/mechanics/kinematics/kinematics-reasonings/a-body-when-thrown-upwards-with-a-certain-velocity-requires-certain-time-to-reach-the-maximum-height/ | # A body when thrown upwards with a certain velocity requires certain time to reach the maximum height. If the velocity is doubled, what is the time required to reach the highest point?
Let $t$ be the time required by the body to reach the maximum height $h$ when thrown upwards with a velocity $u$. At the maximum height, the velocity will be zero $v=0.$
$v=u-gt$ $0=u-gt$ $\therefore t=\frac{u}{g}$
If the velocity is doubled, then new velocity will be $u’=2u.$
Let $t’$ be the new time to reach the highest point. $t’=\frac{u’}{g}=\frac{2u}{g}=2t$
Thus, the new time required to reach the highest point will be twice the original time.
[Equations used above are deduced from Motion Under Gravity]
#### SIMILAR QUESTIONS
A body is dropped from a certain point and it reaches a velocity ‘v’ after some time. What would be the velocity reached after twice the time?
A body travels one half of a distance with uniform velocity $v_1$ and the other half with uniform velocity $v_2$. Find the magnitude of the average velocity.
Two bodies of different masses are dropped from same height. Compare their velocities on reaching the ground and the time required.
A boy wants to take his boat to a point just opposite on the other bank of the river with water flowing its usual course. How should he row and why?
If a body is thrown vertically upward from a vehicle moving with uniform velocity, where will the body fall?
Can an object have velocity and acceleration in perpendicular directions? | 2023-03-24 10:51:03 | {"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": 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": 0.5581938624382019, "perplexity": 289.65642420699953}, "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-2023-14/segments/1679296945279.63/warc/CC-MAIN-20230324082226-20230324112226-00351.warc.gz"} |
http://www.chegg.com/homework-help/questions-and-answers/learning-goal-understand-newton-s-law-universal-gravitation-able-apply-two-object-situatio-q1334743 | Learning Goal: To understand Newton's law of universal gravitation and be able to apply it in two-object situations and (collinear) three-object situations; to distinguish between the use of and .
In the late 1600s, Isaac Newton proposed a rule to quantify the attractive force known as gravity between objects that have mass, such as those shown in the figure. (Intro 1 figure) Newton's law of universal gravitation describes the magnitude of the attractive gravitational force between two objects with masses and as
,
where is the distance between the centers of the two objects and is the gravitational constant.
The gravitational force is attractive, so in the figure it pulls to the right on (toward ) and toward the left on (toward ). The gravitational force acting on is equal in size to, but exactly opposite in direction from, the gravitational force acting on , as required by Newton's third law. The magnitude of both forces is calculated with the equation given above.
The gravitational constant has the value
and should not be confused with the magnitude of the gravitational free-fall acceleration constant, denoted by , which equals 9.80 near the surface of the earth. The size of in SI units is tiny. This means that gravitational forces are sizeable only in the vicinity of very massive objects, such as the earth. You are in fact gravitationally attracted toward all the objects around you, such as the computer you are using, but the size of that force is too small to be noticed without extremely sensitive equipment.
Consider the earth following its nearly circular orbit (dashed curve) about the sun. (Intro 2 figure) The earth has mass and the sun has mass . They are separated, center to center, by .
Part A
What is the size of the gravitational force acting on the earth due to the sun? | 2015-03-27 19:17:59 | {"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": 0.9403494000434875, "perplexity": 310.64153351510623}, "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-2015-14/segments/1427131296603.6/warc/CC-MAIN-20150323172136-00187-ip-10-168-14-71.ec2.internal.warc.gz"} |
https://electronics.stackexchange.com/questions/308426/transistors-the-change-in-the-base-current | # Transistors - The change in the base current
I came across this question in my book, it says:
what will happen to the output voltage if we increase the base current in the figure below?
What confused me in the figure is that the output voltage is the voltage on the load Rc, and I used to think of it to be the voltage across the emitter-collector junction, like this:
In the second figure I would say that increasing the base current will decrease the output voltage, but I don't know what to say in the first case in Figure one. What is exactly the output voltage in Common Emitter configuration? And what is the difference between the first and the second figures? And what will happen if we increased the base current in the first and second figures?
## migrated from physics.stackexchange.comMay 31 '17 at 20:51
This question came from our site for active researchers, academics and students of physics.
• Output voltage is where it is said that the output is, not where you'd like it to be ;) – Todor Simeonov May 31 '17 at 21:00
• And what do you think will happen with output voltage (e.g. the voltage on Rc)? Just follow the things you know. – Todor Simeonov May 31 '17 at 21:01
• Oh, Ok, thank you, and in this case I think the output voltage will increase. – Asmaa May 31 '17 at 21:03
Collector current, which flows from $V_{\text{CC}}$ to ground (i.e. the emitter), is proportional to base current so increasing the base current also increases the collector current. The voltage across $R_C$ is $I_CR_C$, and it is also the output voltage in the first figure. So in this case the output voltage increases.
The second figure simply defines $V_{\text{out}}$ differently. It is $V_{\text{out}} = V_{\text{CC}} - I_CR_C$, so since $I_CR_C$ increases with increasing base current we can say that $V_{\text{out}}$ decreases in that case.
The second figure is the more typical definition of $V_{\text{out}}$, since typically voltages are referenced to ground. But the two circuits are basically equivalent (they are both common emitters) -- it's just a difference in the definition of the output voltage.
I would say that the magnitude will increase, but if the positive is (as is conventional) on the top, the voltage will go down (start off negative and become more negative).
However this is probably just the answer they want and therefore what will get you the mark. In fact it is quite possible for Vce to increase with increasing base current. This will happen when the transistor is in saturation and generally when Ib > Ic. You can easily demonstrate this with a normal small signal transistor and a couple of resistors.
So the real-life answer is "it depends" with an extra layer of uncertainty caused by their failure to mark + and - on the measurement points for Vout. Your teacher may not know this.
As long as they do not specify which is positive terminal of Vout it can either be Vout=Rc×Ic or Vout=-Rc×Ic.
Given rising Ic first voltage increases while the second one gets more negative and hence decreases.
• A voltage getting more negative doesn't decrease, it increases, decrease means get closer to zero, not approaching the negative infinte. – Marcelo Espinoza Vargas Jun 1 '17 at 1:43
• @Marcelo Espinoza Vargas What you say applies to absolute value of voltage only. Numbers are lined like -2 < 0 < 3, no chance to change this! – carloc Jun 1 '17 at 5:03 | 2019-06-20 21:52:30 | {"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": 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": 0.6488167643547058, "perplexity": 543.2296016923098}, "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-2019-26/segments/1560627999273.79/warc/CC-MAIN-20190620210153-20190620232153-00343.warc.gz"} |
https://www.esaral.com/q/two-planets-have-masses-64697/ | Two planets have masses
Question:
Two planets have masses $\mathrm{M}$ and $16 \mathrm{M}$ and their radii are a and $2 \mathrm{a}$, respectively. The separation between the centres of the planets is $10 \mathrm{a}$. A body of mass $m$ is fired from the surface of the larger planet towards the smaller planet along the line joining their centres. For the body to be able to reach at the surface of smaller planet, the minimum firing speed needed is :
1. $\sqrt{\frac{\mathrm{GM}^{2}}{\mathrm{ma}}}$
2. $\frac{3}{2} \sqrt{\frac{5 \mathrm{GM}}{\mathrm{a}}}$
3. $4 \sqrt{\frac{\mathrm{GM}}{\mathrm{a}}}$
4. $2 \sqrt{\frac{\mathrm{GM}}{a}}$
Correct Option: , 2
Solution:
$\frac{\mathrm{GM}}{\mathrm{x}^{2}}=\frac{\mathrm{G}(16 \mathrm{M})}{(10 \mathrm{a}-\mathrm{x})^{2}}$
$\frac{1}{x}=\frac{4}{(10 a-x)} \quad \Rightarrow 4 x=10 a-x$
$x=2 a$ …(1)
COME
$-\frac{\mathrm{GMm}}{8 \mathrm{a}}-\frac{\mathrm{G}(16 \mathrm{M}) \mathrm{m}}{2 \mathrm{a}}+\mathrm{KE}$
$=-\frac{\mathrm{GMm}}{2 \mathrm{a}}-\frac{\mathrm{G}(16 \mathrm{M}) \mathrm{m}}{8 \mathrm{a}}$
$\mathrm{KE}=\mathrm{GMm}\left[\frac{1}{8 \mathrm{a}}+\frac{16}{2 \mathrm{a}}-\frac{1}{2 \mathrm{a}}-\frac{16}{8 \mathrm{a}}\right]$
$\mathrm{KE}=\mathrm{GMm}\left[\frac{1+64-4-16}{8 \mathrm{a}}\right]$
$\frac{1}{2} \mathrm{mv}^{2}=\mathrm{GMm}\left[\frac{45}{8 \mathrm{a}}\right]$
$\mathrm{V}=\sqrt{\frac{90 \mathrm{GM}}{8 \mathrm{a}}}$
$\mathrm{V}=\frac{3}{2} \sqrt{\frac{5 \mathrm{GM}}{\mathrm{a}}}$ | 2022-05-26 10:38:41 | {"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": 0.8727794885635376, "perplexity": 369.5144907870905}, "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/1652662604794.68/warc/CC-MAIN-20220526100301-20220526130301-00187.warc.gz"} |
https://www.physicsforums.com/threads/what-is-the-maximum-speed-that-an-object-must-reach-to-satisfy-minimum-transit-time.965913/page-2 | # What is the maximum speed that an object must reach to satisfy minimum transit time?
#### Davidllerenav
I think you can make a stab at that.
I guess that it would be when it is near B, because it is under constant acceleration.
#### haruspex
Homework Helper
Gold Member
2018 Award
I guess that it would be when it is near B, because it is under constant acceleration.
When what is near B? When A is near B? A and B are given, you can't move them.
It is a very simple question - if you want to get there ASAP, and your speed is limited only by the speed of light, what speed do you have to go at (as near as possible)?
#### Davidllerenav
When what is near B? When A is near B? A and B are given, you can't move them.
It is a very simple question - if you want to get there ASAP, and your speed is limited only by the speed of light, what speed do you have to go at (as near as possible)?
When the object moving from A to B is near B, because it is under a constant acceleration.
To answer your question, it would be as near as the speed of light.
#### haruspex
Homework Helper
Gold Member
2018 Award
When the object moving from A to B is near B, because it is under a constant acceleration.
Sorry, but I have no idea what you are trying to say.
To answer your question, it would be as near as [possible to] the speed of light.
Yes.
#### Davidllerenav
Sorry, but I have no idea what you are trying to say.
I'll try to explain myself better. The objecc is moving from A to B, starting at rest with a constant acceleration a until it reaches B. So the fastest the object will be travelling would be when it gets to point B.
#### haruspex
Homework Helper
Gold Member
2018 Award
The objecc is moving from A to B, starting at rest with a constant acceleration a
My comment about relativity and infinite speed was in the context of the constant speed case.
For constant acceleration it gets a bit tricky. Not sure what that means within relativity.
#### Davidllerenav
My comment about relativity and infinite speed was in the context of the constant speed case.
For constant acceleration it gets a bit tricky. Not sure what that means within relativity.
So, if in the case of constant speed, the object speed must be near the speed of light.
#### haruspex
Homework Helper
Gold Member
2018 Award
So, if in the case of constant speed, the object speed must be near the speed of light.
Yes, depending on what exactly the question was intended to say.
#### Davidllerenav
Yes, depending on what exactly the question was intended to say.
I really don't know that the questio was trying to say. So well, I guess that there arte those two cases.
#### CWatters
Homework Helper
Gold Member
You seem to be making this problem much harder than it is. I very much doubt they intended you to think about limitations due to the speed of light!
It starts from A at rest and accelerates at "a". At some point it must stop accelerating and start to decelerate at "a" in order to be at rest at B.
A few basic equations of motion (eg SUVAT) and you are done.
#### Davidllerenav
You seem to be making this problem much harder than it is. I very much doubt they intended you to think about limitations due to the speed of light!
It starts from A at rest and accelerates at "a". At some point it must stop accelerating and start to decelerate at "a" in order to be at rest at B.
A few basic equations of motion (eg SUVAT) and you are done.
I asked my teacher today. He said that indeed it start ar rest and accelerates at a and at somep point it decelerate at a in order to be at rest at B. He said that there arte two cases and that I need to choose one and explain why. So yes, I can use SUVAT equations. Which are the two cases?
#### haruspex
Homework Helper
Gold Member
2018 Award
it start ar rest and accelerates at a and at somep point it decelerate at a in order to be at rest at B
Pity that wasn't stated in the first place.
So turn that into some equations.
Which are the two cases?
Unless there is also a max allowed speed, I can only think of one case.
#### hmmm27
Not trying to be funny, but are you and the instructor both native speakers of the same language ?
The "two cases" are:
1) constant velocity gets there in the least amount of time
2) symmetrical acc/deceleration gets there in the least amount of time.
Pick one (or both) and state the condition - which includes the term "maximum velocity" - that makes it true.
#### haruspex
Homework Helper
Gold Member
2018 Award
1) constant velocity gets there in the least amount of time
Given the clarification that it starts and finishes at rest, I don't see how that can be a sensible case.
Also, we've seen no mention of a maximum velocity, so dropping the start and finish at rest constraint for the constant speed case doesn't help.
If we were given a max speed the answer would depend on the relationship between that, the acceleration a and the distance.
This leaves the view that
- it starts and finishes at rest
- in between, any mix of accelerating at a, decelerating at a, and constant speed
But I count that as one case, not two.
#### hmmm27
I think "choosing a case" means arbitrarily picking either "cv wins" or "acc/dec wins", then working backwards to state the condition(s) that makes it so.
#### haruspex
Homework Helper
Gold Member
2018 Award
I think choosing a "case" means arbitrarily picking either "cv wins" or "acc/dec wins", then working backwards to state the condition(s) that makes it so.
Effectively turning it into this model:
If we were given a max speed the answer would depend on the relationship between that, the acceleration a and the distance.
...maybe.
#### Davidllerenav
Pity that wasn't stated in the first place.
So turn that into some equations.
Unless there is also a max allowed speed, I can only think of one case.
Ok, can I use $s=v_0 t+\frac{1}{2}at^2$? And it will end up being $s=\frac{1}{2}at^2$ since $v_0=0$?
#### haruspex
Homework Helper
Gold Member
2018 Award
Ok, can I use $s=v_0 t+\frac{1}{2}at^2$? And it will end up being $s=\frac{1}{2}at^2$ since $v_0=0$?
Ok, but how are you using it, i.e. what is s here?
#### Davidllerenav
Ok, but how are you using it, i.e. what is s here?
Well, $s$ can't be $L$, because it isn't accelerating all way through, it would be the distance before it starts desaccelerating. Whn it starts desaccelerating, it would be $s=v_0 t-\frac{1}{2}at^2$, right?
#### haruspex
Homework Helper
Gold Member
2018 Award
Well, $s$ can't be $L$, because it isn't accelerating all way through, it would be the distance before it starts desaccelerating. Whn it starts desaccelerating, it would be $s=v_0 t-\frac{1}{2}at^2$, right?
Yes.
#### Davidllerenav
Ok. But what should I replace on the variables? Or should I just leave those two equations?
#### haruspex
Homework Helper
Gold Member
2018 Award
Ok. But what should I replace on the variables? Or should I just leave those two equations?
No, you must answer using the given variables L and a.
Use your two equations to find out when you have to switch to decelerating.
#### CWatters
Homework Helper
Gold Member
The only two cases I can think of are..
1) Accelerate, constant velocity, decelerate
2) Accelerate, decelerate with no constant velocity phase.
#### haruspex
Homework Helper
Gold Member
2018 Award
The only two cases I can think of are..
1) Accelerate, constant velocity, decelerate
2) Accelerate, decelerate with no constant velocity phase.
But isn't the second just a special case of the first?
#### CWatters
Homework Helper
Gold Member
Indeed. But it's the best I can come up with for "two cases".
"What is the maximum speed that an object must reach to satisfy minimum transit time?"
### Physics Forums Values
We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving | 2019-10-20 19:04:22 | {"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": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7671318054199219, "perplexity": 1006.5693666240205}, "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/1570986718918.77/warc/CC-MAIN-20191020183709-20191020211209-00161.warc.gz"} |
https://www.mathemania.com/lesson/populations-and-variables/ | # Populations and variables
Statistics is dealing with studying various methods for collecting, analyzing and presenting the empirical data. There are two categories: descriptive and inferential statistics.
Descriptive statistics focuses on describing a sample, without trying to conclude anything about a larger population.
Inferential statistics collects informations from sample. After that, it makes conclusions about the population from which the sample was selected.
## Population and samples
Population is a set or a collection under observation. The term often refers to a group of people, but sometimes it refers to objects, events or observations. Furthermore, population can be finite or infinite. It is finite if it is possible to count its members and infinite when that’s not possible.
Example 1: The population of students at specific university.
Example 2: If we are studying height of adult men, the population is a set of heights of all the men in the world.
Example 3: All daily minimum temperatures in January for major cities in Europe.
Researches would often like to know certain informations about populations, but they don’t have data for every person or thing in the population. For that reason, they often select a sample of the population.
Sample is every subset of population, i.e. a smaller group of elements of population which represents the population. It is important that sample is random. It means that every element of the population has an equal chance to be selected. The process of selecting the sample from the population is called sampling. Researchers develop hypotheses about the population based on informations collected from sample.
For example, let’s say you go to a chocolate bar and see that owner offers some samples of chocolate products. Of course, you won’t taste all the products from the bar. And, surely, the owner wouldn’t be very satisfied if you taste everything for free. Therefore, your opinion about chocolate bar would be based only on several products, i.e. on samples they offer.
Parameter is a data about an entire population. It is a value that describes a characteristic of a population. For example, the population mean and standard deviation. Furthermore, parameter depends on all elements of population.
## Numeric and categorical variables
A variable is any characteristic, quantity or number that we can measure or count. We also call it data item. There are two main types: numeric and categorical.
Numeric variables (or quantitative variables) can be interval or ratio.
Interval variables are variables who join some real number to every member of population and they have an intrinsic order. Furthermore, measuring unit and agreement zero are defined. An example is a variable that joins air temperature (in the same place and at the same time) to each day.
Ratio variables are variables that have the same properties as interval variables. The difference is that an agreement zero isn’t defined. In other words, it means that zero of the measurement indicates that a property on observed element doesn’t exist. An example is measuring the time a person needs to ran $100$ metres.
Categorical variables (or qualitative variables) can be nominal or ordinal.
Nominal variables are variables who join some attribute to every member of population and do not have an intrinsic order. For example, we join number $1$ to every person who answers the referendum question with ‘yes’. In opposite, we join number $0$.
Ordinal variables are variables who join some symbol or number to every member of population. Also, they have an intrinsic order. For example, we join grade of some course to every student from certain university. Another example is if we join a degree of professional qualification to every person of some group.
## Discrete and continuous variables
If a numeric variable can take on any value from some interval $\left<a, b\right> \subseteq \mathbf{R}$ for $a, b \in \mathbf{R}, a<b$, it is called a continuous variable. Otherwise, it is called a discrete variable.
Example 4: The height of all people is a continuous variable defined on population of all people who are born or who will be born.
Example 5: Suppose we flip a coin and watch the outcomes; ‘head’ or ‘tail’. Variable that joins number $1$ to outcome ‘head’ and number $0$ to outcome ‘tail’ is a discrete variable.
## High – dimensional variable
We can observe more different variables on the same population as one high – dimensional variable. We can show this variable using matrices. For example, if $X = (X_{1}, \cdots, X_{k})$, where $X_{i}, i = 1, \cdots, k$ are variables on the same population $S$, then the corresponding matrice is:
$$\begin{bmatrix} a_{11} & a_{12} & \cdots & a_{1k} \\ \vdots & \vdots & \vdots & \vdots \\ a_{i1} & a_{i2} & \cdots & a_{ik} \\ \vdots & \vdots & \vdots & \vdots \end{bmatrix}.$$
Values $a_{ij}$ represent values of j – th variable $X_{j}$ in i – th member of population. Therefore, the first column represents $X_{1}(S)$, the second column $X_{2}(S)$ and so on.
Example 6: Data about citizens that census taker collects are one high dimensional variable. For instance: age, place of birth, number of household members etc. | 2021-09-18 02:06:23 | {"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": 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": 0.4471140205860138, "perplexity": 782.6295987434494}, "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/1631780056120.36/warc/CC-MAIN-20210918002951-20210918032951-00540.warc.gz"} |
http://openstudy.com/updates/556d54c8e4b050a18e841719 | ## mpj4 one year ago isn't the integral of dx / [x (7x^2 - 5)^(1/2) ] = 1/a arcsec u/a??
1. mpj4
I ask because I applied it and I got it wrong.
2. mpj4
3. ganeshie8
looks you forgot to put the absolute value bars
4. mpj4
maybe
5. mpj4
my solution
6. mpj4
forgot to square the a=5 and the integral signs. Sorry, was a bit hasty making it.
7. Michele_Laino
you can try this substitution: $\large \sqrt {7{x^2} - 5} = \sqrt 7 x + t$ where t is the new variable of integration
8. mpj4
Ok, thanks! | 2017-01-19 02:46:09 | {"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": 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": 0.964470386505127, "perplexity": 6940.26500368436}, "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-2017-04/segments/1484560280425.43/warc/CC-MAIN-20170116095120-00490-ip-10-171-10-70.ec2.internal.warc.gz"} |
https://jobbyedf.firebaseapp.com/7570/55551.html | # Full text of "Les Prix Nobel" - Internet Archive
, qui a raw' zn. ' d specific heat at constant volume to that at constant pressure was mea sured in the year 1 8. 92 Nach RYDBERG,. KAYSER u nd RUNGE Electrical mobility · Enthalpy · Joule Heating · Mean free path · Newton's law of constant · Fermi level · Magnetic moment · Photon Energy · Rydberg equation Beg Del i Sverige AB · Begagnade Bildelar J. Solstad Aktiebolag Bröderna Rydbergs Åkeri AB · Bröderna Consiro AB · Constant Förvaltning i Stockholm AB. Siivonen & Timo J. Virtanen (eds.).
In my discussion section, we went over a problem asking us to calculate the wavelength of light emitted in the n=3 to n=2 transition in the hydrogen atom. Rydberg constant definition is - a wave number characteristic of the atomic spectrum of each element equal to the constant factor in the wave-number formula for Apr 29, 2015 When Bohr calculated his theoretical value for the Rydberg constant, The atomic number, Z, of hydrogen is 1; k = 2.179 × 10–18 J; and the where REH is the Rydberg constant for hydrogen, and REH = 13.605693 eV = 2.179872x10-18 J. Tm and Tn are called terms. The above conversion of energy Physical Constants and Useful Combinations: Speed of light 1 eV = 1.602 × 10 −19 J. Fine structure constant α = kee2/ Rydberg-Ritz formula. 1 λmn. = R ( 1. 5, Planck constant, (h), 6.62607554e-34 ± 3.97565e-40, J s.
## En Grundläggande Stil Guide för Matematik, Fysik & Kemi
1 Ry. = 2.179872E-18 joules (J) 1 Ry. = 2.179872E-21 kilojoules (kJ) 1 Ry. = 5.2100191204589E-19 calories (cal) Rydberg constant times hc in J: Numerical value: 2.179 872 361 1035 x 10-18 J : Standard uncertainty: 0.000 000 000 0042 x 10-18 J : Relative standard uncertainty: 1.9 x 10-12: Concise form 2.179 872 361 1035(42) x 10-18 J : Click here for correlation coefficient of this constant with other constants How to convert joule to Rydberg constant. 1 joule = 4.5874208973812E+17 Rydberg The Rydberg constant is one of the most precisely determined physical constants, with a relative standard uncertainty of under 2 parts in 10 12. This precision constrains the values of the other physical constants that define it. Values of Rydberg's Constant The accepted values of the Rydberg constant, R∞, as in 1998 are: Rydberg Constant in nm - 10 973 731.568 548 (83) m-1.
### III. Kvantkemi - Yumpu
It is defined as 2 R∞hc, where R∞ is the Rydberg constant, h is the Planck constant and c is the speed of light. Its CODATA recommended value is Eh = 4.359 744 722 2071(85) × 10−18 J = 27.211 386 245 988(53) eV. Using u _ {if} = c \bar { u} _ {if} converts Equation (2.20) from frequency to wave number units, When we identify R_H with the ratio of constants on the right hand side of Equation (2-21), we obtain the Rydberg equation with the Rydberg constant as in Equation (2-22). Die Rydberg-Konstante ist eine nach Johannes Rydberg benannte Naturkonstante. Sie tritt in der Rydberg-Formel auf, einer Näherungsformel zur Berechnung von Atomspektren . Ihr Wert ist die als Wellenzahl ausgedrückte Ionisierungsenergie des Wasserstoffatoms unter Vernachlässigung relativistischer Effekte und der Mitbewegung des Kerns (also bei unendlicher Kernmasse , daher der Index EXPAND TO SEE LINKS TO SPECIFIC PARTS OF VIDEO:Corrected wavelength: 2:15Convert nanometers to meters: 3:15Frequency: 4:20Energy or emitted light: 5:17Chang Rydberg constant, in joules - Technical Formulas. R∞ = 2.179 872 171 (96) x 10-18 J :relative uncertainty (ppm) = 0.60.
5, Planck constant, (h), 6.62607554e-34 ± 3.97565e-40, J s. 6, Planck 31, Rydberg constant in joules, 2.179874113e-18 ± 1.30792e-24, J. 32, Rydberg Nov 11, 2019 ΔE = 3² x (1/1² - 0) x 2.18 x 10⁻^18 J = 2.0 x 10⁻¹⁷ J (1/infinity is zero) where ΔE: is the energy change, R: is the Rydberg constant, Z: is the Aug 9, 2020 These constants are the values recommended in 2006 by The Committee on Data for Science and Electron volt, eV, 1.602 176 487 (40) x 10−19 J Rydberg constant, RH, 1.097 373 156 852 7 (73) x 107 m−1. Speed of .. the emission data to calculate the Rydberg constant. 6-3 Wavelength. (nm).
Instagram soka
R∞ = 2.179 872 171 (96) x 10-18 J :relative uncertainty (ppm) = 0.60. Guardado por Neelam. 1.
Experiment 12. Line Spectra. The Rydberg Constant. Objective: 1) To use the emission spectrum of hydrogen atom in order to verify the relation between energy Johannes Rydberg was one of the grandfathers of modern-day physics and S St. J Nepomucene, Rydberg: the Man and the Constant, Chymia, 1960, 6, h = 6.625 x 10-34 Joule second = 4.134 x 10-15 electron Volt second.
Arbetsförmedlingen cv ungdom | 2022-10-01 20:55:08 | {"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": 0.8264648914337158, "perplexity": 7545.478585720949}, "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-2022-40/segments/1664030336921.76/warc/CC-MAIN-20221001195125-20221001225125-00307.warc.gz"} |
http://tex.stackexchange.com/questions/95406/custom-quote-formatting?answertab=active | # Custom quote formatting [closed]
I have a .cls in which I have quite a bit of formatting... I'm trying to make a way of making my own "quote" formatting. I need to format my quotes centered, with a limited width for each line, and underlined (like a quote box). Should I do this with a custom \begin{} type? How do I do this?
-
## closed as too localized by lockstep, Kurt, diabonas, percusse, doncherryFeb 14 '13 at 3:05
This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, visit the help center. If this question can be reworded to fit the rules in the help center, please edit the question.
– mafp Jan 26 '13 at 8:25
Have you looked at the possibilities of the csquotes package? Maybe \fbox{\blockquote{...}} is alredy enough. – mafp Jan 26 '13 at 8:35
Do you want a box around the quote or every word of the quote underlined? – lockstep Jan 26 '13 at 11:58
I'd use a newenvironment for this, and base it on a list.... do you want every line underlined, or just the first? last? – cmhughes Jan 26 '13 at 17:36 | 2015-05-28 06:39:17 | {"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": 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": 0.5879772901535034, "perplexity": 2113.2048188793015}, "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-2015-22/segments/1432207929256.27/warc/CC-MAIN-20150521113209-00190-ip-10-180-206-219.ec2.internal.warc.gz"} |
https://aviation.stackexchange.com/tags/flight-mechanics/hot | # Tag Info
52
What you observed is probably a Gulfstream production test flight, either an initial or a snag clearance flight, or possibly a Customer Acceptance from Savannah Air Center, which is a completion center across the ramp from Gulfstream. It could be possibly an experimental test flight if Gulfstream runs experimental flying from Savannah (I think they have a ...
38
Yes it can. The steady-heading sideslip (SHSS) maneuver is used in flight testing to demonstrate static lateral/directional stability (similar maneuvers exist as sideslip approach in crosswind, or stabilized control with one-engine-inoperative at low speed). In this maneuver, rudder is applied to hold a sideslip, which generates an opposing side force and ...
25
The fact that it's near Savannah and tracking within a confined airspace, I'm fairly certain this is a flight test being carried out by Gulfstream.
24
This approach of using an engine is called pulse and glide. It generally works because each engine has an optimal power setting at which it converts fuel into power most efficiently. If the most efficient power is higher than is required, something should be done to accumulate and later use the excess energy. Raising the vehicle up looks like a solution, ...
17
Yes, it's entirely possible. You can use the rudder to oppose the turn induced by the roll, and thus fly straight. However, doing so means that your nose is pointed off at an angle from your direction of flight, so this dramatically increases drag. In fact, this is a common technique (at least among general aviation pilots) for losing altitude without ...
16
There are multiple configurations which are possible with tail or canard, which based on their locations and whether they produce lift or down-force, results in a stable or unstable aircraft (taking the aircraft center of gravity into account). The figures below show some of the possible configurations. Source: f-16.net In the most general case, there is ...
16
Here is one thing that WILL save fuel-- where possible-- such as when flying VFR. Note that this strategy only applies within the layer where thermal convection or orographic uplift are significant, not at higher altitudes where the air is generally not rising or sinking to any significant degree (apart from wave lift in which case the strategy will still ...
12
Of course, just put the center of gravity back to its rear limit and fly slowly. Then all of them will produce positive lift on their tails. Stability is not produced by a downforce at the tail. The newest book I read which claimed this was from 1911 (I happened to read the 1913 edition). Stability is produced by making the lift per area of the forward ...
10
A tandem wing airplane has two sets of wings, each providing upward lift. One is near the front of the plane, one is near the back, and the center of gravity is between them. (source: nurflugel.com) The Rutan Quickie is one such plane: (souce: wikimedia) This design is actually fairly old; it even predates (successful) heavier-than-air flight, as it was ...
9
This is commonly known as the Speed Stability, not to be confused with speed stability in the sense of static longitudinal stability. I think there's some common confusion with this phenomenon. The first part of this answer addresses the OP's question; the second part clarifies a common confusion. 1. Main Answer In level flight (i.e. zero vertical rate), ...
9
This makes sense and is approximately what happens if you reduce thrust during trimmed straight and level flight. The aircraft will go though a series of damped oscillations, called 'phugoid' until it finds its new equilibrium state. The new state will be the original airspeed, but now in a constant descent.
7
Both are right in their own ways. First the basics: Power is thrust times speed and is constant over speed for variable-pitch propeller-powered piston aircraft. Therefore, propeller thrust is proportional to speed inversed. The minimum power required coincides with the point of maximum excess power since it needs the lowest power setting for trimmed flight. ...
7
You are very perceptive to note that for a given angle-of-attack, a return to the exact airspeed that allowed level (horizontal) flight would indeed imply a return to level (horizontal) flight. But that's not what we expect to happen when we reduce the power. Hopefully this answer will help you to understand why not. When you elevator trim to fly ...
7
Here are two questions back at you. Can a plane turn, with banking? Yes, of course it can, the normal way using the ailerons to tilt the airplane and thus change the angle of lift. You already know all about that. Can a plane turn, without banking? Yes, it can; by using the rudder to do the turning it can pull off a “flat turn” as it were. OK... are you ...
6
In simple mechanical stick linkages, is there crosstalk between the aileron and the elevator movement? In almost all cases, there is no intentional-- or discernable-- mechanical coupling between aileron deflection and elevator deflection. However in the Schleicher Ka-6 sailplane, the mechanical linkage is designed to raise both ailerons a little bit, ...
6
I would second what John K said, it’s most likely a GAC factory flight test or demonstration flight. Also with the airplane identified as GLF6 and, most likely going be the call sign “Gulf Test six” would pretty much confirm that suspicion. You will also see similar flights down to Brunswick Golden Isles airport (KBQK), as Gulfstream does a lot of flight ...
6
Can planes bank without turning Yes, a plane may fly in a banked attitude without turning. But when the plane is tilted to the left or right relative to Earth's surface and gravity, shouldn't it be "pushed" to the left or right? Yes, whenever a plane is banked and the wing is generating lift, the lift vector has a horizontal component that tends ...
5
Right side slip from left rudder, rudder stalls, what happens? This is comparable to a wing stalling in that it is caused by the rudder/vertical stabilizer exceeding a critical angle of attack. There for, it has passed its maximum deflection potential. The result will be the fuselage slip will reduce until the rudder/vertical stabilizer assembly unstalls....
5
Your question should emphasize "still air" conditions and not get into exploiting vertical air motion, which is a separate issue completely. For normal airplanes, the basic issue is, you want to spend the enroute phase of the flight at the end of an optimal climb profile, with an optimal cruising altitude (which will depend on winds) for as long as ...
5
Let imagine that we put wind indicator on rotating merry-go-around. Will wind indicator stay perpendicular to the radius of circle,showing tangential airflow velocity? Assuming that the indicator has a mass distribution such that the apparent "centrifugal force" from the rotation has no effect on it, and also assuming that the physical length of ...
4
Yes it can - to an extent, depending on: amount of CG shift; airspeed; aircraft weight (actually Moment of Inertia); maximum deflection rate of elevators/ailerons. If the airframe has aerodynamic static stability, a disturbance in Angle of Attack Creates aerodynamic forces that return the aircraft attitude back to neutral. If the Centre of Gravity shifts ...
4
Short answer yes, a powered aircraft at any given airspeed in straight and level flight (note that a glide path is also "level", at constant airspeed) needs to use the same amount of energy to over come drag as the glider. The potential energy for the glider is altitude. The potential energy for the powered aircraft is fuel. The conversion of potential ...
4
As drawn, that linkage would cause elevator movement if you moved the stick in a perfectly lateral direction. But when you roll into a turn, you normally apply a little up elevator anyway, and in any case the tendency is to make allowance for that kind of displacement with subtle stick movements without really realizing it. In light aircraft it's more ...
4
The short answer is yes. A coordinated turn has both bank and yaw induced simultaneously in the same direction. If you induce enough yaw in the opposite direction of the bank, the aircraft will fly in a straight line even though the wings are not level. You do this by cross controlling the aircraft. In laymen’s terms, that means you turn the yoke to the ...
4
Keep in mind that no mechanical, power producing system can operate with 100% lossless efficiency. In heat engines, like those in aircraft, the Carnot model describes the theoretical maximum efficiency (power produced vs lost) in any given design. Another way of asking your question would be, "Can I save fuel by taking my foot off the accelerator in my ...
4
Remember when Major Gant (Clint Eastwood) stole one of the Firefoxes and was running out of fuel? There was a climb, a glide, and a dramatic landing on the ice floe (it was a movie). Was that the best way? Perhaps so. One might imagine (for greatest distance), climbing at Vy, then gliding at Vbg. There would be some TAS benefit by getting as high as possible ...
3
I found your exact question a little confusing, so let me answer with a quote from a current British PPL exam preparation book: The wing's angle of incidence - the angle between the wing chord-line and fuselage - is fixed by the construction of the wing and the angle at which it is bolted onto the fuselage Source: AFE Book 4 - Aircraft General Knowledge ...
3
It is exactly the other way round: $$\alpha = \mathrm{arctan} \left( \frac{v_z}{v_x} \right)$$ You can simply derive it from the trigonometry in a triangle: In this image the aircraft is moving to the right (along the arrow labelled v) and the $x$-axis is pointing along the aircraft body. The resulting function looks like this: For $v_z = 0$, you ...
3
The answer is that the aircraft doesn't actually pitch down. When banking (or rolling) into a turn, the angle of attack (or the angle between the center of the wing and the flight path of the aircraft) stays the same. Assuming no correction to maintain altitude is made, the aircraft will continue on the same angle of attack, just banked over by some amount. ...
3
Not really. The angle of incidence is the angle between the chord of the wing and the longitudinal axis of the airplane. But the plane is mostly free to pitch as necessary. So if the best thing is for the wing to have an AOA of 11°, then it can pitch to that point. Now if the angle of incidence were too different, it could increase drag on the climb-out, ...
Only top voted, non community-wiki answers of a minimum length are eligible | 2020-08-09 18:06:44 | {"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": 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": 0.6154124736785889, "perplexity": 1404.273022795852}, "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/1596439738562.5/warc/CC-MAIN-20200809162458-20200809192458-00488.warc.gz"} |
https://web2.0calc.com/questions/where-did-i-go-wrong_3 | +0
# Where did I go wrong?
0
130
5
$$\int \frac{d}{dx} [{-arcsin}(x)] \hspace{0.1cm} dx = -arcsin(x)$$
$$\int d [{-arcsin(x)}]=-arcsin(x)$$
$$\int -d[arcsin(x)]=-arcsin(x)$$
$$\int -\bigg[\frac{1}{\sqrt{1-x^2}}\bigg] = -arcsin(x)$$
You could pull the negative out and this equation is clearly true... or
$$arccos(x)=-arcsin(x)$$
Why is that identity not true? Where did I go wrong?
Mar 2, 2019
#1
+5225
+1
$$\displaystyle \int - \left[\dfrac{1}{\sqrt{1-x^2}}\right]~dx = -\arcsin(x) \text{, not }\arccos(x)\text{ for one}$$
.
Mar 2, 2019
#2
0
Wait, so what is the derivative of arccos(x)?
Guest Mar 2, 2019
#3
+102447
+1
Rom could you check this please.
I do not know thise off by heart, I have to work them out every time
$$If\;\;\; \theta=acos(x)\quad then\;\;\;find \; \frac{d\theta}{dx}\\~\\ x=cos\theta\\ \frac{dx}{d\theta}=-sin\theta\\$$
Now we know that cos(theta) =x
so I draw a right angled triangle where this is true.
$$\text{adjacent=x hypotenuse=1 opposite=} \sqrt{1-x^2}$$
$$\frac{dx}{d\theta}=-sin\theta\\ \frac{d\theta}{dx}=\frac{-1}{\sqrt{1-x^2}}\\$$
So it seems to me that
$$\displaystyle \int - \left[\dfrac{1}{\sqrt{1-x^2}}\right]~dx = \arccos(x) +c$$
Wolfram|alpha seems to agree with me.
Melody Mar 2, 2019
#4
+28064
+4
Hmm! -arcsin(x) is the same as arccos(x) to within a constant, so either is acceptable in a solution to the indefinite integral.
Mar 3, 2019
#5
+102447
0
Thanks Alan.
In other works Guest the identity you got is true except that you forgot to include the constant.
Yes I know you already said that Alan :)
Melody Mar 4, 2019
edited by Melody Mar 4, 2019
edited by Melody Mar 4, 2019 | 2019-07-18 14:11:21 | {"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": 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": 0.9391555190086365, "perplexity": 3145.559688805509}, "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-2019-30/segments/1563195525634.13/warc/CC-MAIN-20190718125048-20190718151048-00189.warc.gz"} |
https://msp.org/agt/2020/20-7/agt-v20-n7-p11-p.pdf | #### Volume 20, issue 7 (2020)
Recent Issues
The Journal About the Journal Editorial Board Editorial Interests Subscriptions Submission Guidelines Submission Page Policies for Authors Ethics Statement ISSN (electronic): 1472-2739 ISSN (print): 1472-2747 Author Index To Appear Other MSP Journals
Ribbon $2$–knots, $1+1=2$ and Duflo's theorem for arbitrary Lie algebras
### Dror Bar-Natan, Zsuzsanna Dancso and Nancy Scherich
Algebraic & Geometric Topology 20 (2020) 3733–3760
##### Abstract
We explain a direct topological proof for the multiplicativity of the Duflo isomorphism for arbitrary finite-dimensional Lie algebras, and derive the explicit formula for the Duflo map. The proof follows a series of implications, starting with “the calculation $1+1=2$ on a 4D abacus”, using the study of homomorphic expansions (aka universal finite-type invariants) for ribbon $2$–knots, and the relationship between the corresponding associated graded space of arrow diagrams and universal enveloping algebras. This complements the results of the first author, Le and Thurston, where similar arguments using a “3D abacus” and the Kontsevich integral were used to deduce Duflo’s theorem for metrized Lie algebras; and results of the first two authors on finite-type invariants of w–knotted objects, which also imply a relation of $2$–knots with the Duflo theorem in full generality, though via a lengthier path.
However, your active subscription may be available on Project Euclid at
https://projecteuclid.org/agt
We have not been able to recognize your IP address 44.192.65.228 as that of a subscriber to this journal.
Online access to the content of recent issues is by subscription, or purchase of single articles.
or by using our contact form. | 2022-07-05 04:12:54 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 3, "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": 0.5140247344970703, "perplexity": 2155.9897746202378}, "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/1656104512702.80/warc/CC-MAIN-20220705022909-20220705052909-00095.warc.gz"} |
https://www.cell.com/biophysj/biophysj/supplemental/S0006-3495(17)30989-X | Article| Volume 113, ISSUE 9, P2068-2076, November 07, 2017
# Arabidopsis Leaf Trichomes as Acoustic Antennae
Open Archive
## Abstract
The much studied plant Arabidopsis thaliana has been reported recently to react to the sounds of caterpillars of Pieris rapae chewing on its leaves by promoting synthesis of toxins that can deter herbivory. Identifying participating receptor cells—potential “ears”—of Arabidopsis is critical to understanding and harnessing this response. Motivated in part by other recent observations that Arabidopsis trichomes (hair cells) respond to mechanical stimuli such as pressing or brushing by initiating potential signaling factors in themselves and in the neighboring skirt of cells, we analyzed the vibrational responses of Arabidopsis trichomes to test the hypothesis that trichomes can respond acoustically to vibrations associated with feeding caterpillars. We found that these trichomes have vibrational modes in the frequency range of the sounds of feeding caterpillars, encouraging further experimentation to determine whether trichomes serve as mechanical antennae.
## Introduction
The recent discovery by Appel and Cocroft (
• Appel H.M.
• Cocroft R.B.
Plants respond to leaf vibrations caused by insect herbivore chewing.
) that leaves of Arabidopsis thaliana respond meaningfully to the noises of Pieris rapae caterpillars feeding on them raises the questions of what cells participate in the receptor system or systems, how they operate, and how they initiate responses. In Appel and Cocroft’s study, synthesis of several toxins was promoted by the feeding caterpillars, and the caterpillar activity could be effectively replaced by appropriate application of acoustic vibration directly to a leaf.
In principle, there are two ways in which the acoustic vibrations caused by the feeding caterpillars might be transmitted. First, they might be carried through the leaf and stem tissue. This is the primary way envisioned by Appel and Cocroft (
• Appel H.M.
• Cocroft R.B.
Plants respond to leaf vibrations caused by insect herbivore chewing.
) for transmission through the plant under attack. Second, as tentatively suggested by Appel and Cocroft (
• Appel H.M.
• Cocroft R.B.
Plants respond to leaf vibrations caused by insect herbivore chewing.
), they might be carried through the air, and thus carry warning messages to neighboring plants.
We focused on a possible role for Arabidopsis trichomes (hair cells) as acoustic detectors, and asked whether they have the form of tiny mechanical antennae for acoustic waves arising from herbivory by Pieris caterpillars. Thus, in principle they might have a role, either exclusive or contributory, in detecting caterpillars on the identical plant, but perhaps an exclusive role in receiving warning sounds from adjacent plants if indeed this can occur. The latter situation is of special interest, because if verified the detection of aerial sound waves conjures the agricultural possibility of acoustic pest control. Although Arabidopsis itself is of little interest in this regard, it is nevertheless a member of a large family containing many food and oil plants. Moreover, diverse kinds of trichomes are widely distributed in the plant world.
In part, our interest was piqued by our recent discoveries, mostly unpublished, that Arabidopsis trichomes can serve several mechanosensory functions. Of special pertinence is the observation that brushing or bending a trichome causes Ca2+ oscillations and lowering of wall pH in the skirt cells developmentally linked to the trichome, which are especially well connected to their distal neighbors by plasmodesmata (
• Zhou L.H.
• Liu S.B.
• Pickard B.G.
• et al.
The Arabidopsis trichome is an active mechanosensory switch.
).
Accordingly, we asked: is it quantitatively reasonable to speculate that trichomes could respond to sound waves? Specifically, can it be determined by formal simulation analysis whether the primary modal vibration frequencies, as determined by the mass, stiffness, and geometry of the trichome (
• Ginsberg J.H.
Acoustics—A Textbook for Engineers and Physicists. Volume I: Fundamentals.
,
• Ginsberg J.H.
Acoustics—A Textbook for Engineers and Physicists. Volume II: Applications.
,
• Xin F.X.
• Lu T.J.
Analytical modeling of fluid loaded orthogonally rib-stiffened sandwich structures: sound transmission.
), are tuned to resonate when exposed to acoustic waves corresponding to frequencies in the power spectral density of feeding Pieris caterpillars?
If so, it would seem worthwhile to carry out the somewhat elaborate experiments required to assess actual trichome vibrations and check for some of the possible biochemical effects that might, speculatively, be triggered. This would strengthen the basis for studying the broader web of cells and pathways involved in detecting and deterring predatory caterpillars.
## Materials and Methods
To study whether Arabidopsis trichomes are capable of transducing mechanical stimuli such as subtle, repeated vibrations produced by Pieris, we evaluated using finite-element and closed-form modal analysis whether an archetypal Arabidopsis trichome is capable of transducing vibrations in the frequency range associated with herbivory. To this end, a series of numerical simulations were performed on an archetypal trichome to establish modal responses. Because the material parameters have yet to be determined definitively for the components of Arabidopsis trichomes, parametric studies were performed to check the sensitivity of predictions to all parameter choices.
### Finite-element discretization of a trichome
Analyses were performed on idealized trichomes using standard procedures (e.g., Ginsberg (
• Ginsberg J.H.
Mechanical and Structural Vibrations: Theory and Applications.
)). The trichome studied was that discretized by Zhou et al. (
• Zhou L.H.
• Liu S.B.
• Pickard B.G.
• et al.
The Arabidopsis trichome is an active mechanosensory switch.
), from confocal and ultraviolet microscopy images (Fig. 1). The gross outline of the trichome studied was obtained from a smoothing of the confocal microscopy images (such as Fig. 1 b), whereas the inner boundaries were acquired from ultraviolet microscopy images that penetrated the cell to show the thickness of the wall (Fig. 1 a). From these images, a 3D CAD rendering of the trichome was generated (Fig. 1 C) that was subsequently discretized from plate elements using the software Abaqus (Dassault Industries, Paris, France). Finite-element meshes were refined to ensure convergence of the predicted mode shapes and natural frequencies (Fig. S2). Following Zhou et al. (
• Zhou L.H.
• Liu S.B.
• Pickard B.G.
• et al.
The Arabidopsis trichome is an active mechanosensory switch.
), the subsurface region of the podium was omitted from consideration; as is standard in vibrational analysis of structures, the subsurface region was replaced with an effective elastic foundation (
• MacBain J.C.
• Genin J.
Effect of support flexibility on the fundamental frequency of vibrating beams.
,
• MacBain J.C.
• Genin J.
Natural frequencies of a beam considering support characteristics.
).
For the purpose of building the model, the trichome was divided into five parts: three branches, the stalk, and their junction. A callose ring near the swelling at the base is a variable feature (
• Kulich I.
• Vojtíková Z.
• Žárský V.
• et al.
Cell wall maturation of Arabidopsis trichomes is dependent on exocyst subunit EXO70H4 and involves callose deposition.
) so was not incorporated in the model; instead, it and the adjacent pliant zone were considered as part of the taper. The part of the podium embedded in leaf tissue (compare to Zhou et al. (
• Zhou L.H.
• Weizbauer R.A.
• Pickard B.G.
• et al.
Structures formed by a cell membrane-associated arabinogalactan-protein on graphite or mica alone and with Yariv phenylglycosides.
)) was modeled as part of an elastic foundation, as described below. The first four parts each had a well-defined axis of symmetry along a unit vector $eˆi$, with the index i = {1, 2, 3, 4} representing the stalk (index 1) and the three branches (indices 2–4). The cutting planes were defined perpendicular to symmetry axes, defined by vectors $r⇀$ such that
$(r⇀−r⇀i0)⋅eˆi=0,$
(1)
where $r⇀i0$ is the intersection of symmetry axis i and the associated cutting plane. For the three branches, properties were interpolated as a function of position xi along each symmetry axis, with xi measured from $r⇀i0$. The outer radii Ri(xi) of the three branches decreased conically from the junction so that $Ri(xi)=Ri0[1−(xi/ximax)]$. The outer boundaries of the junction and stalk were contoured according to the confocal microscopy images.
The walls of the branches and junction were given a thickness of 6.0 μm. Three separate models were investigated for the effect of stalk thickness, which the images in Fig. 1 and replicates showed to be tapered. In the first, the walls were set to the minimum observed thickness, tmin = 1.5 μm. In the second, the walls were set to the maximum observed thickness: tmax = 6.0 μm. In the third, the thickness was set to the observed parabolic function of height x1 along the stalk’s axis from its base to the base of the junction t(x1) = −(0.25 μm−1) x12 + 0.07 x1 + 0.0015 μm.
### Mechanical properties
Mechanical properties were assigned for the cytosol, cell wall, and foundation of the trichome. Parameters chosen for finite-element simulation are listed in Table 1, including a basal set of fixed values as well as reasonable ranges of variation and sources of data. None of these mechanical properties are known with certainty, and parametric studies were therefore performed. Note that in some cases, parameter studies were performed well beyond the range of data to enable identifications of trends.
Table 1Range of Mechanical Properties Studied for Model Trichomes
SymbolVariableRange of DataBaseline Value
ρwalldensity of cell wallCharacean internode: 1.4 g/cm3 (
• Wayne R.
• Staves M.P.
The density of the cell sap and endoplasm of Nitellopsis and Chara.
)
1 g/cm3
wood: 0.1–3 g/cm3 (
• Gibson L.J.
The hierarchical structure and mechanics of plant materials.
)
ρcytodensity of cytosolCharacean internode: 1.0 g/cm3 (
• Wayne R.
• Staves M.P.
The density of the cell sap and endoplasm of Nitellopsis and Chara.
)
1 g/cm3
pollen tube: 1 g/cm3 (
• Kroeger J.H.
• Daher F.B.
• Geitmann A.
• et al.
Microfilament orientation constrains vesicle flow and spatial distribution in growing pollen tubes.
)
t(x1)thickness of tapered stalk cell wall1.5–6.5 μm (
• Zhou L.H.
• Liu S.B.
• Pickard B.G.
• et al.
The Arabidopsis trichome is an active mechanosensory switch.
)
See Materials and Methods
EwallYoung’s modulus of cell wall0.5–4 GPa (
• Probine M.C.
• Preston R.D.
Cell growth and the structure and mechanical properties of the wall in internodal cells of Nitella opaca: II. Mechanical properties of the walls.
,
• Gibson L.J.
The hierarchical structure and mechanics of plant materials.
,
• Packirisamy M.
• Geitmann A.
• et al.
In vitro study of oscillatory growth dynamics of Camellia pollen tubes in microfluidic environment.
)
1 GPa
ELYoung’s modulus of trichome foundation0.035–35 GPa (
• Gibson L.J.
The hierarchical structure and mechanics of plant materials.
)
1 GPa
kfoundation stiffnessanimal cells: 10−5–10−2 MPa/μm (
• Deshpande V.S.
• McMeeking R.M.
• Evans A.G.
A bio-chemo-mechanical model for cell contractility.
)
1 MPa/μm
plant cells: 1–30 MPa/μm (see Materials and Methods)
νLPoisson’s ratio of trichome foundation0.30–0.49 (
• Chanliaud E.
• Burrows K.M.
• Gidley M.J.
• et al.
Mechanical properties of primary plant cell wall analogues.
)
0.33
κcytobulk modulus of cytosol2.15 GPa (
• Dorsey N.E.
Properties of Ordinary Water Substance.
,
• Liu Y.
• Thomopoulos S.
• Genin G.M.
• et al.
Modelling the mechanics of partially mineralized collagen fibrils, fibres and tissue.
)
2.15 GPa
P0turgor pressure of cell0.8–1.4 MPa (
• Forouzesh E.
• Goel A.
• Turner J.A.
• et al.
In vivo extraction of Arabidopsis cell turgor pressure using nanoindentation in conjunction with finite element modeling.
)
0.5 MPa
guard cells 0.5–5 MPa (
• Franks P.J.
• Buckley T.N.
• Mott K.A.
• et al.
Guard cell volume and pressure measured concurrently by confocal microscopy and the cell pressure probe.
)
Cytosol. Following procedures that are standard for piping systems with low Reynolds number (
• Lesmez M.W.
• Wiggert D.C.
• Hatfield F.J.
Modal analysis of vibrations in liquid-filled piping systems.
), the cytosol was treated as an incompressible void with no shear resistance, and its mass was distributed to the neighboring cell wall. More complicated procedures are available that involve interactions between flowing fluid and pipe walls (
• Lesmez M.W.
• Wiggert D.C.
• Hatfield F.J.
Modal analysis of vibrations in liquid-filled piping systems.
,
• Hatfield F.J.
• Wiggert D.C.
• Otwell R.S.
Fluid structure interaction in piping by component synthesis.
,
• Ting E.C.
• Hosseinipour A.
A numerical approach for flow-induced vibration of pipe structures.
), nonlinear deformations including buckling of the walls (
• Gonçalves P.B.
• Batista R.C.
Non-linear vibration analysis of fluid-filled cylindrical shells.
,
• Genin J.
Nonlinear vibrations of thin cylinders.
,
• Genin J.
Dynamic instability in cylindrical shells.
,
• Moussaoui F.
• Benamar R.
Non-linear vibrations of shell-type structures: a review with bibliography.
), and nonlinearity of the fluid within the cytosol (
• Xin F.X.
• Lu T.J.
A nonlinear acoustomechanical field theory of polymeric gels.
,
• Xin F.
• Lu T.J.
Acoustomechanical constitutive theory for soft materials.
), but these were not needed for cytosol contained in the relatively narrow channels of the trichome stalk and branches undergoing acoustic loading. The relative amounts of cytoplasm and cell wall material varied along the length of the roughly conical trichome arms and contoured trichome stalk. To account for this, the effective density of the arm, ρ, was varied with the position along the stalk or branches, as
$ρ(xi)=ρcyto+(ρwall−ρcyto)tRi(xi)[2−tRi(xi)],$
(2)
where ρcyto is the effective density of the cell wall, and ρwall is that of the cell wall. For the junction between the branches and stalk, the effective wall density was amplified according to
$ρ=ρwall+VcytoVwallρcyto,$
(3)
where Vcyto ≈ 2.3 × 10−14 m3 and Vwall ≈ 7.8 × 10−15 m3 are the volumes enclosed within the CAD model of the junction and the volume of cell wall at the junction, respectively, with the latter estimated using a wall thickness of t = 1.5 μm.
The cytosol was pressurized with a baseline pressure of 0.5 MPa. Pressures in the range of 0.1–1.0 MPa were studied.
Cell wall. The cell wall was treated as linear elastic and isotropic with an elastic modulus of 1 GPa and a Poisson ratio of 0.33 (Table 1). The 1 GPa cell wall modulus of Characean algae cells was chosen as a reference modulus because the cells have a thick primary wall (
• Probine M.C.
• Preston R.D.
Cell growth and the structure and mechanical properties of the wall in internodal cells of Nitella opaca: II. Mechanical properties of the walls.
) that may be comparable to that of the trichome in several ways.
Foundation. The foundation of the trichome stalk was treated as deformable and linear elastic. The leaf was thus modeled with a Winkler-type foundation having an elastic stiffness k per unit area, so that the displacement Δx of a point $r⇀$ on the base of the trichome was given by
$Δx(r⇀,t)=σxx(r⇀,t)k,$
(4)
where $σxx(r⇀,t)$ is the component of stress (force per unit area) on the basal plane of the model trichome that is directed parallel to the axis of the trichome stalk, at position $r⇀$ and time t. The order of magnitude of k was estimated from the solution presented by Johnson (
• Johnson K.L.
Contact Mechanics.
) to the Love (
• Love A.E.H.
The stress produced in a semi-infinite solid by pressure on part of the boundary.
) problem of a uniform pressure σ0 distributed over a circular region on the free boundary of an elastic half-space of elastic modulus E and Poisson ratio ν. Considering the peak displacement in such a case, occurring at the center of the circular region, results in the estimate (
• Lubarda V.A.
Circular loads on the surface of a half-space: displacement and stress discontinuities under the load.
),
$k=EL2(1−νL2)R,$
(5)
where EL and νL are the effective elastic modulus and Poisson ratio of the foundation, respectively; and R = 20 μm is the radius of the trichome. Taking EL ∼ 0.03–1 GPa and νL ∼ 0.33 and inserting R yields the order-of-magnitude estimate k ∼ 1–30 MPa/μm. Note that this is approximately four orders-of-magnitude greater than the foundation stiffness typical for an animal cell (
• Deshpande V.S.
• McMeeking R.M.
• Evans A.G.
A bio-chemo-mechanical model for cell contractility.
) because of the presence of cell walls in the ring of cells that surround the podium.
### Finite element analysis
Modal analysis was performed using standard linear modal analysis. Briefly, finite-element modal analysis was performed to analyze free vibrations of the trichome structure. The generalized equation of motion for a discretized structure was analyzed:
$M[d2U(t)/dt2]+KU(t)=F(t),$
(6)
where M is a matrix of effective nodal masses, U(t) is a vector of nodal displacements, K is a matrix of effective stiffnesses, F(t) is a vector of applied forces, and t is time. The patterns of free vibration that an acoustic wave would excite are the vectors U(t) that satisfy the homogenous version of the differential equation with the form U(t) = U0 exp(iωt), where $i=−1$, U0 represents mode shape, and ω represents the associated natural frequencies. Substituting these into the homogenous form of Eq. 6 results in the well-known modal analysis eigenvalue equation that can be solved to find U0 and ω:
$(K−ω2M)U0=0.$
(7)
These vibrational modes U0 represent the shapes of vibration that will be excited by acoustic loading transmitted through the air or the leaf. The trichomes were discretized into meshes of ∼14,000 linear quadrilateral shell elements, and vibrational modes were found. A typical simulation required 3 min of CPU time on a Core Duo (Intel, Santa Clara, CA) desktop computer with 8 GB of RAM.
### Closed-form scaling laws
Although structures composed of multiple and branching vibrating elements have modal responses that differ substantially from those of the individual branches (
• Genin J.
• Maybee J.S.
Mechanical vibration trees.
), two scaling laws for the vibration of simple structures proved useful for interpreting the natural frequencies of the trichomes. The first is the Wrinch (
• Wrinch D.
On the lateral vibrations of bars of conical type.
) solution for free vibration of a conical beam that is clamped at the thick end and free at the pointed end. Wrinch showed that the natural frequency fi corresponding to the ith vibrational mode of such a beam is
$fi=14πβiR0L2Ewallρ,$
(8)
where Ewall is the Young’s modulus, ρ is the effective density, R0 is the maximum radius of the trichome, L is the length of the trichome, and βi = {8.72, 21.2, 38.5, 60.7, 88.9…} are the roots of
$J1(2βi)I2(2βi)−J2(2βi)I1(2βi)=0,$
(9)
in which Ji is the Bessel function of the first kind of order i, and Ii is the modified Bessel function of the first kind of order i.
The second is the classic solution for torsional oscillation of a rigid object at the end of a thin, torsional shaft (
• Ginsberg J.H.
Mechanical and Structural Vibrations: Theory and Applications.
). Approximating the trichome shaft as a uniform rod of height H, thickness t, outer radius R1, and shear modulus E/2(1 + ν), the torsional rigidity of the stalk can be approximated as
$K=π4EH(1+ν)[R14−(R1−t)4].$
(10)
Idealizing a trichome branch as a solid, rigid cone as in Fig. 2, the inertia tensor about the vertex P, in the (x2, y2, z2) coordinate system, is (
• Ginsberg J.H.
• Genin J.
Dynamics.
)
$IP′=[Ixx′000Izz′000Izz′],$
(11)
where $Izz′=M((13/80)L2+(3/20)R02)$ and $Ixx′=(3/10)MR02$, in which M = πρLR02/3 is the mass of a cone of density ρ. Transforming this tensor by an angle $ϕ$ to the (x2, y2, z2) coordinate system (Fig. 2), adding the contributions of three branches, and finding the xx component of the resulting tensor, yields the moment of inertia $IxxP$ of the branches about the stalk axis:
$IxxP=3(Ixx′cos2ϕ+Izz′sin2ϕ).$
(12)
The natural frequency of the lowest torsional mode is
$fi=12πKIxxP.$
(13)
### Spectral analysis of Pieris feeding noises
To determine the range of frequencies over which an acoustic response would be meaningful, we estimated the frequency-dependent power spectral density in four recordings of Pieris feeding upon Arabidopsis leaves, provided by Rex Cocroft (
• Appel H.M.
• Cocroft R.B.
Plants respond to leaf vibrations caused by insect herbivore chewing.
). Periodogram estimates were made using a standard algorithm in the MATLAB computing environment (The MathWorks, Natick, MA). Feeding sounds were sampled at 44.1 kHz and saved as .wav files.
## Results
When receiving an acoustic stimulus, the first mode of vibration of the trichome with baseline parameters involved torsion of the stalk, with the branches rotating almost as a rigid body (Fig. 3; Movie S1). Higher modes involved rocking of the trichome (Fig. 3, mode 2; Movie S2) and various patterns of swinging branches (Fig. 3, higher modes; Movies S3, S4, and S5).
The frequency of the first mode of vibration was a relatively weak function of the Young’s modulus of the wall. For the baseline case of Ewall = 1 GPa (Fig. 4 B), the first mode was excited at ∼6 kHz, well within the audible range (∼20–20,000 Hz), and well within the range for which Pieris feeding noises present substantial power (below ∼8 kHz, Fig. S3). The higher modes were much more sensitive to the choice of Ewall, with characteristic frequencies out of the human audible range for the baseline parameters (Table 1). This is consistent with the behavior of individual branches predicted from closed-form solutions, with characteristic frequencies in the tens of kHz range. For all choices of Ewall > 300 MPa, the first mode was the torsional mode. For a cell wall, 1/10th that of the baseline parameter choice (Fig. 4 A), the first mode was a rocking mode, and the first four modes all occurred at ∼10 kHz.
The lowest natural frequency of the trichome depended strongly on the mechanics of the leaf foundation (Fig. 5). The torsional vibrational mode observed for trichomes with the baseline parameters switched to a rocking mode at sufficiently low foundation compliance.
Although higher modes were not sensitive to turgor pressure, the characteristic frequency of the first mode of vibration varied almost linearly with turgor pressure (Fig. 6). Note that the fourth and fifth modes are not visible on Fig. 6 because they occurred at frequencies above 100 kHz.
The choice of trichome stalk thickness model did not affect the trends or the magnitudes of the characteristic frequencies significantly (Fig. 6). In all cases, the first mode occurred well within the audible range, and the second mode at the upper extremity or just beyond the audible range. The choice of wall thickness profile did not affect the modal frequencies strongly (Fig. 7).
## Discussion
Modeling an archetypal Arabidopsis trichome by both classical and finite-element analyses, we found that it has resonant modes responsive to acoustical waves in the range of insect feeding noise reported by Appel and Cocroft. For each vibration mode, Fig. 2 shows that trichomes respond in a characteristic way to vibrational stimulation. The first mode of vibration, a torsional mode in which the branches rotated around the axis of the stalk, was excited in the archetypal trichome at a frequency of ∼8 kHz. Within the range of parameters considered to be physiological, this frequency was accurate to within a tolerance on the order of 50%. Higher frequency modes involving rocking of the trichome stalk and waving of the trichome branches were excited toward the upper end of the audible range and beyond.
Comparison to closed-form estimates clarifies the nature of these mode shapes. For a torsional oscillator on a narrow stalk (R1 = 10 μm) of height H = 230 μm (adjusted to reach the center of the junction) connected to rigid conical branches of length L = 180 μm and maximum radius R0 = 10 μm, with all other parameters as in Table 1, torsional vibrations occurred on the order of 30 kHz for an angle $ϕ$ ranging from 60° to 90°, indicating that additional aspects of the trichome contributed to the somewhat lower frequencies estimated by finite-element analysis. The transverse vibration frequency of the trichome branches occurred in frequency ranges on the order of MHz, also in agreement with finite-element prediction. This indicates that the motion of branches observed in lower frequency modes involved rocking of the trichome stalk rather than vibration of branches.
Although the shapes and walls of Arabidopsis trichomes can be quite variable, establishing baseline data in the simulations of Fig. 3 enables observing the consequences of varying the values of the baseline parameter set one at a time, and consideration of the extent to which evolution may have optimized the trichome for function as a mechanical antenna.
The same methodology was used to explore the consequences of each tested value; the Supporting Material illustrates how the parameters control movement patterns. One of the most distinctive properties of the wall of the trichome is its taper. Yet, untapered mutants have been identified (e.g., the EXO70H4-1 mutant, which lacks a secondary trichome wall and hence a taper in the trichome stalk (
• Kulich I.
• Vojtíková Z.
• Žárský V.
• et al.
Cell wall maturation of Arabidopsis trichomes is dependent on exocyst subunit EXO70H4 and involves callose deposition.
)). The parametric study shown in Fig. 6 indicates that, although the taper affects the frequency response quantitatively, it does not affect trends and, over the range of limiting cases considered, cannot preclude the existence of a torsional vibrational mode within the range of frequencies that feature prominently in Pieris feeding noises.
Given on the one hand that the leaf as a whole can respond mechanically and physiologically to caterpillar feeding noise (
• Appel H.M.
• Cocroft R.B.
Plants respond to leaf vibrations caused by insect herbivore chewing.
) and on the other hand that the trichome has the appearance of a mechanical antenna and is small in size, the finding that the archetypal trichome studied resonates in the audible range is perhaps unsurprising. Nevertheless, it has important implications.
One obvious implication is that the data warrant an experimental search for trichome vibration modes and biochemical consequences. Because these responses are extraordinarily complex, such a search must be carefully planned to discriminate between responses in the trichome itself and responses in the majority of the leaf tissue. In such planning, it must be recognized that, although the magnitudes of responses cannot be estimated from these analyses, any response by the leaf per se to air-borne sound will likely be small compared to possible response of the light-weight trichomes, even though response driven by piezoelectric actuators or by transmission via an aqueous path from another leaf (
• Appel H.M.
• Cocroft R.B.
Plants respond to leaf vibrations caused by insect herbivore chewing.
) might be reasonably large.
Among responses observed within the somewhat similar trichome of bean plants to light brushing of its surface is production of (toxic) reactive oxygen species (W.Q. Gao and L. Zhou, personal communication); within the Arabidopsis trichome itself a rapid outward-moving acidification of the bulk of the wall and acidification of the thin papillae walls have been observed (unpublished). Stronger brushing can produce plastic deformation of the papillae, which could influence the behavior of their distinctive and likely toxic contents (unpublished). Among signaling consequences to surrounding cells from brushing a trichome stalk are acidification of cell walls and initiation of cytoplasmic Ca2+ oscillations in the skirt cells (
• Zhou L.H.
• Liu S.B.
• Pickard B.G.
• et al.
The Arabidopsis trichome is an active mechanosensory switch.
) and, for bean plants, the production of extractable anthocyanin in the bulk leaf (W.Q. Gao and L. Zhou, personal communication). On the other hand, responses described for acoustic stimulation of the entire leaf include increased anthocyanin and glucosinolate production (
• Appel H.M.
• Cocroft R.B.
Plants respond to leaf vibrations caused by insect herbivore chewing.
), and doubtless a variety of volatiles ((
• Appel H.M.
• Cocroft R.B.
Plants respond to leaf vibrations caused by insect herbivore chewing.
• Snoeren T.A.L.
• Kappers I.F.
• Bouwmeester H.J.
• et al.
Natural variation in herbivore-induced volatiles in Arabidopsis thaliana.
)). The production of glucosinolate is reported to be primed rather than promoted directly (
• Appel H.M.
• Cocroft R.B.
Plants respond to leaf vibrations caused by insect herbivore chewing.
). The complexity of interpreting glucosinolate levels is underlined by the observation they vary dramatically on a circadian basis that correlates with the timing of insect activity (
• Goodspeed D.
• Chehab E.W.
• Covington M.F.
• et al.
Arabidopsis synchronizes jasmonate-mediated defense with insect circadian behavior.
,
• Goodspeed D.
• Liu J.D.
• Braam J.
• et al.
Postharvest circadian entrainment enhances crop pest resistance and phytochemical cycling.
).
Mechanoresponses such as gravitropism, thigmotropism, mechanotropism, gravinasty, thigmonasty, gravimorphogenesis, thigmomorphogenesis, mechanomorphogenesis, and mechanocontrol of biochemical pathways—somewhat overlapping terms among a plethora of such terms—are among the most-studied responses plants make to the environment (
• Coutand C.
Mechanosensing and thigmomorphogenesis, a physiological and biomechanical point of view.
,
• Moulia B.
• Coutand C.
• Julien J.-L.
Mechanosensitive control of plant growth: bearing the load, sensing, transducing, and responding.
,
• Telewski F.W.
A unified hypothesis of mechanoperception in plants.
,
• Monshausen G.B.
• Haswell E.S.
A force of nature: molecular mechanisms of mechanoperception in plants.
,
• Chehab E.W.
• Wang Y.
• Braam J.
Mechanical force responses of plant cells and plants.
,
• Hamilton E.S.
• Schlegel A.M.
• Haswell E.S.
United in diversity: mechanosensitive ion channels in plants.
,
• Abeles F.B.
• Morgan P.W.
• Saltveit M.E.
Ethylene in Plant Biology.
). The most popular comparisons to be made with the results of our acoustic assessments are probably those with functions of popular carnivores in the Droseraceae. Indeed, certain fascinating mechanosensory behaviors of conspicuously evolved Droseraceous trichomes were famously described over a century ago, such as the triggering of the rapid and slow leaf movements and glandular secretion by Dionaea or Venus flytrap (
• Hamilton E.S.
• Schlegel A.M.
• Haswell E.S.
United in diversity: mechanosensitive ion channels in plants.
,
• Abeles F.B.
• Morgan P.W.
• Saltveit M.E.
Ethylene in Plant Biology.
,
• Braam J.
In touch: plant responses to mechanical stimuli.
) and the rapid tentacle movements followed by slower secretions and leaf movements by mechanically stimulated Drosera or sundew (
• Braam J.
In touch: plant responses to mechanical stimuli.
,
• Ding J.P.
• Pickard B.G.
Ca2+ channel (from onion epidermis) sensitive to stretch and voltage, and regulated by mural Ca2+ levels.
). Action potentials were concurrently described as mediating these capturing and digesting behaviors in Dionaea (
• Hamilton E.S.
• Schlegel A.M.
• Haswell E.S.
United in diversity: mechanosensitive ion channels in plants.
,
• Abeles F.B.
• Morgan P.W.
• Saltveit M.E.
Ethylene in Plant Biology.
,
• Braam J.
In touch: plant responses to mechanical stimuli.
) and later found to participate similarly in Drosera (
• Braam J.
In touch: plant responses to mechanical stimuli.
,
• Ding J.P.
• Pickard B.G.
Ca2+ channel (from onion epidermis) sensitive to stretch and voltage, and regulated by mural Ca2+ levels.
,
• Ding J.P.
• Pickard B.G.
Modulation of mechanosensitive calcium-selective cation channels by temperature.
,
• Ding J.P.
• Pickard B.G.
Mechanosensory calcium-selective cation channels in epidermal cells.
,
• Cosgrove D.J.
• Hedrich R.
Stretch-activated chloride, potassium, and calcium channels coexisting in plasma membranes of guard cells of Vicia faba L.
). Beyond such popular subjects, mechanostimulated secretions that attract pollinators or deter herbivores have been casually observed over the years in a variety of plants, but more critically, for Solanum lycopersicum (tomato) a recent detailed description (
• Wayne R.
• Staves M.P.
The density of the cell sap and endoplasm of Nitellopsis and Chara.
,
• Gibson L.J.
The hierarchical structure and mechanics of plant materials.
,
• Kroeger J.H.
• Daher F.B.
• Geitmann A.
• et al.
Microfilament orientation constrains vesicle flow and spatial distribution in growing pollen tubes.
) of mechanically induced secretion of defensive compounds by trichomes as well as changes mechanically elicited in leaf chemistry provides a welcome focus that should inspire further investigation, especially in view of more general earlier investigations on mechanical stimulation of tomato (
• Chehab E.W.
• Wang Y.
• Braam J.
Mechanical force responses of plant cells and plants.
).
Perhaps most convincing that acoustic detection might have evolved in Arabidopsis trichomes, however, is our above-mentioned report of a variety of mechanoresponses by the Arabidopsis trichome itself (
• Zhou L.H.
• Liu S.B.
• Pickard B.G.
• et al.
The Arabidopsis trichome is an active mechanosensory switch.
), as well as our unpublished demonstrations of several kinds of signals elicited by brushing these trichomes. Such findings with Arabidopsis argue that its family might be more likely to evolve acoustic sensitivity than the Droseraceae, particularly because, in addition to sharing the general properties of mechanoresponse and secretory activity, the latter have elaborate electrical activities apparently absent in the former.
However, although responses to sound have been studied and speculated on sporadically over the past half century (
• Telewski F.W.
A unified hypothesis of mechanoperception in plants.
), attention to them has expanded slowly, and most studies concentrated on developmental effects rather than the acoustic response to detect and deter herbivory that was pioneered by Appel and Cocroft (
• Coutand C.
Mechanosensing and thigmomorphogenesis, a physiological and biomechanical point of view.
,
• Moulia B.
• Coutand C.
• Julien J.-L.
Mechanosensitive control of plant growth: bearing the load, sensing, transducing, and responding.
,
• Telewski F.W.
A unified hypothesis of mechanoperception in plants.
,
• Monshausen G.B.
• Haswell E.S.
A force of nature: molecular mechanisms of mechanoperception in plants.
,
• Chehab E.W.
• Wang Y.
• Braam J.
Mechanical force responses of plant cells and plants.
).
It may help in analyzing ultimate sequalae of caterpillar feeding noise to recognize that the transduction from mechanical distortion to chemical signal is likely accomplished by stretch-sensitive channels, as is often the case for plant mechanoresponses (
• Coutand C.
Mechanosensing and thigmomorphogenesis, a physiological and biomechanical point of view.
,
• Moulia B.
• Coutand C.
• Julien J.-L.
Mechanosensitive control of plant growth: bearing the load, sensing, transducing, and responding.
,
• Telewski F.W.
A unified hypothesis of mechanoperception in plants.
,
• Monshausen G.B.
• Haswell E.S.
A force of nature: molecular mechanisms of mechanoperception in plants.
,
• Hamilton E.S.
• Schlegel A.M.
• Haswell E.S.
United in diversity: mechanosensitive ion channels in plants.
,
• Braam J.
In touch: plant responses to mechanical stimuli.
). We suspect the multimodally modulated mechanosensory Ca2+-selective ion channels described most extensively in cell membranes of onion bulb scale and leaf epidermis (
• Ding J.P.
• Pickard B.G.
Ca2+ channel (from onion epidermis) sensitive to stretch and voltage, and regulated by mural Ca2+ levels.
,
• Ding J.P.
• Pickard B.G.
Modulation of mechanosensitive calcium-selective cation channels by temperature.
,
• Ding J.P.
• Pickard B.G.
Mechanosensory calcium-selective cation channels in epidermal cells.
); apparently the same or closely similar channels exist in guard cells of bean leaves (
• Cosgrove D.J.
• Hedrich R.
Stretch-activated chloride, potassium, and calcium channels coexisting in plasma membranes of guard cells of Vicia faba L.
). For the vibrations studied by Appel and Cocroft, there is certainly heuristic evidence for participation of some kind of mechanosensory channel: noise was pulsed as caterpillars chewed, moved, rested, and resumed chewing. Stimulation by surrogate sound was defensively effective only if it was similarly pulsed. Quite generally, mechanosensitive channels are rapidly desensitized but soon resensitized by an interval of rest after stimulation.
Nonetheless, the elaborate structure and composition of the trichome wall (which we are currently studying) suggest some mechanotransductive activities elicited directly by papillar distortion, independent of channels. This possibility is underscored by our interpretation that the standard classification of the Arabidopsis trichome as nonsecretory overlooks the massive secretion by the wall system into its papillae concurrent with thinning of the papillar wall, and that the range of these papilla-specific secretions differ from the range of toxins measured in the experiments of Appel and Cocroft (
• Appel H.M.
• Cocroft R.B.
Plants respond to leaf vibrations caused by insect herbivore chewing.
).
Finally, the results suggest what features of the leaf might be most effectively tuned to enable response to specific frequencies. Two features emerge as especially interesting when viewed through the lens of this study. First, it is interesting that the trichome wall has evolved specialized thickness and taper. As previously reported, the taper promotes force transfer and focusing to the skirt of cells surrounding the bulbous proximal end of the trichome, in which calcium and pH signaling are initiated (
• Zhou L.H.
• Liu S.B.
• Pickard B.G.
• et al.
The Arabidopsis trichome is an active mechanosensory switch.
). Given the torsional nature of the first mode, it is clear that the specialized outer contour is a simple factor that can be evolved to tune the torsional rigidity and hence frequency of response. Second, it is also noteworthy that the springiness of the trichome foundation is a key determinant of acoustic sensitivity (Fig. 5), suggesting that subtle changes such as pH-induced compliance could be used to trigger enhanced sensitivity to low frequency acoustic signals. Given that the bulk leaf is already known to vibrate at acoustic frequencies during caterpillar feeding (
• Appel H.M.
• Cocroft R.B.
Plants respond to leaf vibrations caused by insect herbivore chewing.
), basal and trichome vibrations could work together to attune a plant to a specific acoustic signal.
We conclude by amplifying the idea that vibrational stimulation of Arabidopsis trichomes might combine with other stimuli associated with insect herbivory to trigger plant defenses. The sensitivity of the frequency response to the stiffness of the tissue beneath the trichome base provides a mechanism by which the plant can tune trichome acoustic response via stiffness altering factors such as Ca2+ concentrations, pH, and transmembrane voltage. Mechanical stimuli associated with herbivory begin with moths alighting and tapping the leaves with their antennae and tarsi, which doubtless stimulates both epidermal tissue and trichomes, and continue through the hatching of eggs and the onset of herbivory. Our earlier results (
• Zhou L.H.
• Liu S.B.
• Pickard B.G.
• et al.
The Arabidopsis trichome is an active mechanosensory switch.
) showing mechanoresponses including pH shifts in trichome and skirt cells abutting the trichome podium, and showing [Ca2+] oscillations in the skirt cells, indicate that perturbations during ovipositing might be capable of providing sensitizing signals to the trichomes or the bulk of the leaf. This work suggests that a broad study of integrated responses to vibration and allied stimuli should yield insight into the mechanisms by which Arabidopsis senses and distinguishes among sounds and then undertakes reactions to them.
## Author Contributions
S.L. and J.J. performed mathematical simulations under the direction of T.J.L., F.X., B.G.P., and G.M.G. All authors contributed to the writing of the manuscript.
## Acknowledgments
This work was funded by the National Science Foundation (NSF) under grant CMMI-1102803 to B.G.P. and G.M.G.; Glenn Allen through a gift to the Gladys Levis Allen Laboratory of Plant Sensory Physiology, by the China Scholarship Council to Study Abroad to S.L.; by the Chinese Ministry of Education through a Changjiang Scholar Award to G.M.G.; by the National Natural Science Foundation of China under grants 11372243 , 11532009 , and 11522219 to T.J.L. and F.X.; and by the National Science Foundation Center for Engineering Mechanobiology under grant CMMI 1548571 .
## Supporting Material
• Document S1. Figures S1–S3
## References
• Appel H.M.
• Cocroft R.B.
Plants respond to leaf vibrations caused by insect herbivore chewing.
Oecologia. 2014; 175: 1257-1266
• Zhou L.H.
• Liu S.B.
• Pickard B.G.
• et al.
The Arabidopsis trichome is an active mechanosensory switch.
Plant Cell Environ. 2016; 40: 611-621
• Ginsberg J.H.
Acoustics—A Textbook for Engineers and Physicists. Volume I: Fundamentals.
Springer, New York, NY2017
• Ginsberg J.H.
Acoustics—A Textbook for Engineers and Physicists. Volume II: Applications.
Springer, New York, NY2017
• Xin F.X.
• Lu T.J.
Analytical modeling of fluid loaded orthogonally rib-stiffened sandwich structures: sound transmission.
J. Mech. Phys. Solids. 2010; 58: 1374-1396
• Ginsberg J.H.
Mechanical and Structural Vibrations: Theory and Applications.
Wiley, Hoboken, NJ2001
• MacBain J.C.
• Genin J.
Effect of support flexibility on the fundamental frequency of vibrating beams.
J. Franklin Inst. 1973; 296: 259-273
• MacBain J.C.
• Genin J.
Natural frequencies of a beam considering support characteristics.
J. Sound Vibrat. 1973; 27: 197-206
• Kulich I.
• Vojtíková Z.
• Žárský V.
• et al.
Cell wall maturation of Arabidopsis trichomes is dependent on exocyst subunit EXO70H4 and involves callose deposition.
Plant Physiol. 2015; 168: 120-131
• Zhou L.H.
• Weizbauer R.A.
• Pickard B.G.
• et al.
Structures formed by a cell membrane-associated arabinogalactan-protein on graphite or mica alone and with Yariv phenylglycosides.
Ann. Bot. 2014; 114: 1385-1397
• Lesmez M.W.
• Wiggert D.C.
• Hatfield F.J.
Modal analysis of vibrations in liquid-filled piping systems.
J. Fluids Eng. 1990; 112: 311-318
• Hatfield F.J.
• Wiggert D.C.
• Otwell R.S.
Fluid structure interaction in piping by component synthesis.
J. Fluids Eng. 1982; 104: 318-325
• Ting E.C.
• Hosseinipour A.
A numerical approach for flow-induced vibration of pipe structures.
J. Sound Vibrat. 1983; 88: 289-298
• Gonçalves P.B.
• Batista R.C.
Non-linear vibration analysis of fluid-filled cylindrical shells.
J. Sound Vibrat. 1988; 127: 133-143
• Genin J.
Nonlinear vibrations of thin cylinders.
J. Appl. Mech. 1976; 43: 370-372
• Genin J.
Dynamic instability in cylindrical shells.
J. Sound Vibrat. 1978; 56: 373-382
• Moussaoui F.
• Benamar R.
Non-linear vibrations of shell-type structures: a review with bibliography.
J. Sound Vibrat. 2002; 255: 161-184
• Xin F.X.
• Lu T.J.
A nonlinear acoustomechanical field theory of polymeric gels.
J. Mech. Phys. Solids. 2010; 32: 1374-1396
• Xin F.
• Lu T.J.
Acoustomechanical constitutive theory for soft materials.
Acta Mechanica Sinica. 2016; 32: 828-840
• Probine M.C.
• Preston R.D.
Cell growth and the structure and mechanical properties of the wall in internodal cells of Nitella opaca: II. Mechanical properties of the walls.
J. Exp. Bot. 1962; 13: 111-127
• Johnson K.L.
Contact Mechanics.
Cambridge University Press, Cambridge, UK1987
• Love A.E.H.
The stress produced in a semi-infinite solid by pressure on part of the boundary.
Philos. Trans. R. Soc. London. Ser. A. 1929; 228: 377-420
• Lubarda V.A.
Circular loads on the surface of a half-space: displacement and stress discontinuities under the load.
Int. J. Solids Struct. 2013; 50: 1-14
• Deshpande V.S.
• McMeeking R.M.
• Evans A.G.
A bio-chemo-mechanical model for cell contractility.
Proc. Natl. Acad. Sci. USA. 2006; 103: 14015-14020
• Genin J.
• Maybee J.S.
Mechanical vibration trees.
J. Math. Anal. Appl. 1974; 45: 746-763
• Wrinch D.
On the lateral vibrations of bars of conical type.
Proc. R. Soc. London. Ser. A. 1922; 101: 493-508
• Ginsberg J.H.
• Genin J.
Dynamics.
Wiley, New York, NY1977
• Snoeren T.A.L.
• Kappers I.F.
• Bouwmeester H.J.
• et al.
Natural variation in herbivore-induced volatiles in Arabidopsis thaliana.
J. Exp. Bot. 2010; 61: 3041-3056
• Goodspeed D.
• Chehab E.W.
• Covington M.F.
• et al.
Arabidopsis synchronizes jasmonate-mediated defense with insect circadian behavior.
Proc. Natl. Acad. Sci. USA. 2012; 109: 4674-4677
• Goodspeed D.
• Liu J.D.
• Braam J.
• et al.
Postharvest circadian entrainment enhances crop pest resistance and phytochemical cycling.
Curr. Biol. 2013; 23: 1235-1241
• Coutand C.
Mechanosensing and thigmomorphogenesis, a physiological and biomechanical point of view.
Plant Sci. 2010; 179: 168-182
• Moulia B.
• Coutand C.
• Julien J.-L.
Mechanosensitive control of plant growth: bearing the load, sensing, transducing, and responding.
Front. Plant Sci. 2015; 6: 52
• Telewski F.W.
A unified hypothesis of mechanoperception in plants.
Am. J. Bot. 2006; 93: 1466-1476
• Monshausen G.B.
• Haswell E.S.
A force of nature: molecular mechanisms of mechanoperception in plants.
J. Exp. Bot. 2013; 64: 4663-4680
• Chehab E.W.
• Wang Y.
• Braam J.
Mechanical force responses of plant cells and plants.
in: Mechanical Integration of Plant Cells and Plants. Springer, New York, NY2011: 173-194
• Hamilton E.S.
• Schlegel A.M.
• Haswell E.S.
United in diversity: mechanosensitive ion channels in plants.
Annu. Rev. Plant Biol. 2015; 66: 113-137
• Abeles F.B.
• Morgan P.W.
• Saltveit M.E.
Ethylene in Plant Biology.
Academic Press, San Diego, CA1992: 264-296
• Braam J.
In touch: plant responses to mechanical stimuli.
New Phytol. 2005; 165: 373-389
• Ding J.P.
• Pickard B.G.
Ca2+ channel (from onion epidermis) sensitive to stretch and voltage, and regulated by mural Ca2+ levels.
in: The 10th International Biophysics Congress. Biophysical Society of Canada, Vancouver, Canada1990: 359
• Ding J.P.
• Pickard B.G.
Modulation of mechanosensitive calcium-selective cation channels by temperature.
Plant J. 1993; 3: 713-720
• Ding J.P.
• Pickard B.G.
Mechanosensory calcium-selective cation channels in epidermal cells.
Plant J. 1993; 3: 83-110
• Cosgrove D.J.
• Hedrich R.
Stretch-activated chloride, potassium, and calcium channels coexisting in plasma membranes of guard cells of Vicia faba L.
Planta. 1991; 186: 143-153
• Wayne R.
• Staves M.P.
The density of the cell sap and endoplasm of Nitellopsis and Chara.
Plant Cell Physiol. 1991; 32: 1137-1144
• Gibson L.J.
The hierarchical structure and mechanics of plant materials.
J. R. Soc. Interface. 2012; 9: 2749-2766
• Kroeger J.H.
• Daher F.B.
• Geitmann A.
• et al.
Microfilament orientation constrains vesicle flow and spatial distribution in growing pollen tubes.
Biophys. J. 2009; 97: 1822-1831
• Packirisamy M.
• Geitmann A.
• et al.
In vitro study of oscillatory growth dynamics of Camellia pollen tubes in microfluidic environment.
IEEE Trans. Biomed. Eng. 2013; 60: 3185-3193
• Chanliaud E.
• Burrows K.M.
• Gidley M.J.
• et al.
Mechanical properties of primary plant cell wall analogues.
Planta. 2002; 215: 989-996
• Dorsey N.E.
Properties of Ordinary Water Substance.
Reinhold, New York, NY1940
• Liu Y.
• Thomopoulos S.
• Genin G.M.
• et al.
Modelling the mechanics of partially mineralized collagen fibrils, fibres and tissue.
J. R. Soc. Interface. 2013; 11: 20130835
• Forouzesh E.
• Goel A.
• Turner J.A.
• et al.
In vivo extraction of Arabidopsis cell turgor pressure using nanoindentation in conjunction with finite element modeling.
Plant J. 2013; 73: 509-520
• Franks P.J.
• Buckley T.N.
• Mott K.A.
• et al.
Guard cell volume and pressure measured concurrently by confocal microscopy and the cell pressure probe.
Plant Physiol. 2001; 125: 1577-1584 | 2022-01-22 19:00:47 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 27, "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": 0.5089090466499329, "perplexity": 12486.433709048166}, "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-2022-05/segments/1642320303868.98/warc/CC-MAIN-20220122164421-20220122194421-00084.warc.gz"} |
http://rvnw.comproorosoldisubito.it/svg-functions.html | Svg Functions
A vector is a sequence of data elements of the same basic type. The value is either the empty vector, or is a newly constructed nonempty vector that is not eq to any existing vector. Vector(A);. The statements within the curly braces form the body of the function. A file with the SVG file extension is most likely a Scalable Vector Graphics file. Returns NA if no cases. Vector destructor (public member function ) operator= Assign content (public member function ) Iterators: begin Return iterator to beginning (public member function ) end Return iterator to end (public member function ) rbegin Return reverse iterator to reverse beginning (public member function ) rend Return reverse iterator to reverse end (public member function ) cbegin. 2 CGM on the Web n 2. f(x,y,z) = x 2 +2yz 5 is an example of a scalar valued function. vector functions A vector function r(t) = hf(t),g(t),h(t)i is a function of one variable—that is, there is only one "input" value. First, we will learn how represent a vector-valued functions. Serious techno to the world! After years of working in the Mexican scene, DJ's/Producers Fixon & Dig-it decided to launch their own label. In Chapter 4 and 5 we discuss several examples of kernels drawing the connections between the Bayesian and the regularization framework. The position of first and last specified color are optional, remaining colors must have positions specified. See how two vectors are related to their resultant, difference and cross product. The Divergence is what you get when you “dot” Del with a vector field Div() = Note that the result of the divergence is a scalar function. Customers who bought this item also bought. Actually, the transformation functions do not transform the SVG shape themselves, but the underlying coordinate system of that shape. A function template is a generic function that is defined on a generic type for which a specific type can be substituted. vector(s scalar) returns the scalar s as a vector with no labels. The library referenced by Robert gave me issues, so I wrote a more simple function of my own, which seems to work. useless and global optimization algorithms are required to obtain a satisfac- [ 19] D. The matrix( ) transform function specifies a transformation in the form of a. 3 WebCGM Profile. Sample of SVG graphs as shown below. The HTML element is a container for SVG graphics. _over_time() The following functions allow aggregating each series of a given range vector over time and return an instant vector with per-series aggregation results:. vector , a generic, attempts to coerce its argument into a vector of mode mode (the default is to coerce to whichever vector mode is most convenient): if the result is atomic all attributes are removed. The vector has an infinte magnitude and components are all (+/-) infinity or (+/-) 0. Everyone has a different way of teaching. MATLAB contains all of the standard functions such as sin, cos, log, exp, sqrt, as well as many others. Khan Academy is a 501(c)(3) nonprofit organization. Using Input and Output Arguments with Functions Input Arguments Vector and Matrix Input. That is, f:R n->R. Description. Unit Vectors - Normalizing. SVG - Key point. Vector Valued Functions Calculus II Project In this project, you will investigate vector valued functions, curves, and some of their applications. Given any point in the plane (the initial point), if we move in a specific direction for a specific distance, we arrive at a second point. In two dimensions, the vector is composed of two component functions f(t) and g(t), so that ; in three dimensions, the component functions are f(t), g(t), and h(t). gradient(f,v) finds the gradient vector of the scalar function f with respect to vector v in Cartesian coordinates. 3 Images are not Computer Graphics n 1. [VC dim = the maximum number of points that can be separated in all possible ways by that set of functions. SVG provides four transformation functions: translate() rotate() scale() skew() matrix() Each of these functions will be explained in more detail in the following sections. With Vector form, it looks for a value in a specified column or row; With Array form, it looks in the first row or column of an array. 1-Limit, continuity and differentiability of vector point functions 2-Partial derivatives of vector point functions 3-Scalar and vector fields 4-The gradient, divergence and curl 5-Expansion formulas. Williamson (1968-01-01) Hardcover. In this section we need to talk briefly about limits, derivatives and integrals of vector functions. Just divide those by 10. Functions are part of a group of structures in VHDL called subprograms. SVG defines the graphics in XML format. Our aim in this subsection is to give you a storehouse of examples to work with, to become comfortable with the ten vector space properties and to convince you that the multitude of examples justifies (at least initially) making such a broad definition as Definition VS. Playa del Carmen & Mexico City. Suppose you de ne addition of n. Tests whether two 3D vectors are equal. First, we will learn how represent a vector-valued functions. It takes a vector of numerical values and returns a vector containing the minimum and maximum values of a given vector, with the names "min" and "max", respectively. gradient(f,v) finds the gradient vector of the scalar function f with respect to vector v in Cartesian coordinates. Vector functions are those functions that produce a vector from one or more variables. Maps vector fields to scalar fields. The methods are as follows: firstly you can use the @keyframes function to animate your visual SVG content using CSS, the second method is animating SVG directly with SMIL (there’s a tutorial in this post explaining more about SMIL, keep an eye for it), and the third method is to use JavaScript which provides core features for doing. Asymptote: The Vector Graphics Language Asymptote is a powerful descriptive vector graphics language that provides a natural coordinate-based framework for technical drawing. Textbook solution for Calculus: Early Transcendentals 8th Edition James Stewart Chapter 13. And - the transformation was done using D3. Vector Visualization The Wolfram Language provides state-of-the-art fully automated visualization of vector functions and data — suitable for representing flows, field lines, and other vector fields of any complexity. In javascript functions, document represents SVG document and can be used to get the SVG elements. The new gauge uses a viewbox of 100x100 as opposed to previous 1000x1000. Recall that a plane vector consists of two quantities: direction and magnitude. This convention applies to the graphs of three-dimensional vector-valued functions as well. Media in category "SVG x-y functions" The following 137 files are in this category, out of 137 total. std::vector::insert() is a built-in function in C++ STL which inserts new elements before the element at the specified position, effectively increasing the container size by the number of elements inserted. The Divergence is what you get when you “dot” Del with a vector field Div() = Note that the result of the divergence is a scalar function. The distribution function of is called marginal distribution function of. Include member functions to perform the following tasks; Write a function that creates a vector of user-given size M using new operator. plot, which allows you to configure the plot and render it using SVG. Here we are going to discuss all these functions of the R vector in detail with examples. Let's look at an example to make a multidimensional vector and access all its elements. The argument type in angle-brackets indicates the data type of the elements of the vector; variable_name is the name that we assign to the vector, and the optional number_of_elements may be provided to indicate how many elements the vector will initially contain. Paul Garrett: Holomorphic vector-valued functions (February 19, 2005) where γ is a closed path with z having winding number +1. In this section we need to talk briefly about limits, derivatives and integrals of vector functions. rm = FALSE, finite = FALSE) : numeric vector na. These operators, like matrices, are linear maps acting on vector spaces. rep() is used for replicating the values in x. Thus, we can differentiate vector-valued functions by differentiating their component functions. VectorPlot displays a vector field by drawing arrows. Stubbing stdlib exit() Function Is it possible to stub the stdlib library exit() function? I ask this question because the exit() function takes and integer as a parameter an returns a void. More vtkVectorFieldTopology * NewInstance const void PrintSelf (ostream &os, vtkIndent indent) override Methods invoked by print to print information about the object including superclasses. A function of one or more variables whose range is one-dimensional, as compared to a vector function, whose range is three-dimensional (or, in general, -dimensional). Scalar Function. With it, we can quantitatively describe the world around us. Maps scalar fields to vector fields. Post to: Tweet. Statistics: Linear Regression example. 1 Problem 42E. Vector functions take a vector as input and produce a vector of the same length as output. The purpose of vector notation is to keep the x and y components straight. Jon Jon, Standing Grounds, This Is My Death Genre: Techno Format: MP3/WAV Catalogue Number: VFR064 Release Date: 17/02/2020. Introduction n 1. Description. 2) std::pmr::vector is an alias template that uses a polymorphic allocator The elements are stored contiguously, which means that elements can be accessed not only through iterators, but also using offsets to regular pointers to elements. First it works but when I would like get access to the vector alloacated with "new", problems occur. Filters: Retrieving Data from Server Retrieving Data from Server Top Level Vector - AS3 : Properties | Properties | Constructor | Methods | Global Constants. See how two vectors are related to their resultant, difference and cross product. ) r(t) = c. SVG is used to define vector-based graphics for the Web. Functions arguments are passed by value. We define vector functions in Maxima just like we would any other function; the only difference is that the return value is a vector. Label: Vector Functions Records Artist(s): SPLNTR Album/Single: Standing Grounds Track(s): The Beginning of a New Era, Disconnected Life Feat. SVG - Flat Surface Shade. The library referenced by Robert gave me issues, so I wrote a more simple function of my own, which seems to work. It has the unique feature that you can save your work as a URL (website link). Then, we used the SVG Group Element manually to get a feel for how it works. f(x,y,z) = x 2 +2yz 5 is an example of a scalar valued function. vector and is. Function Grapher is a full featured Graphing Utility that supports graphing two functions together. Asymptote functions are treated as variables with a signature (non-function variables have null signatures). SVG has several methods for drawing paths, boxes, circles, text, and graphic images. In this equation: A is a n · k matrix of parameters, b is a vector of parameters (of dimension n · 1). The next step is to assign some variables from the xy vector that this function is given. std::vector::insert() is a built-in function in C++ STL which inserts new elements before the element at the specified position, effectively increasing the container size by the number of elements inserted. The component functions of a vector-valued function are the indi- vidual coordinates of the output. This shopping feature will continue to load items when the Enter key is pressed. The new gauge uses a viewbox of 100x100 as opposed to previous 1000x1000. A lot of videos are so hard to hear. SVG elements can be modified using attributes that specify details about exactly how the element should be handled or rendered. Description. Therefore, we can denote a vector field as A(x,y,z), or A(ρ,,φz), or A(r,,θφ), explicitly showing that vector quantity A is a function of position, as denoted by some set of coordinates. rm: whether NA should be removed, if not, NA will be returned. Example: x2 + y2 = 16 and y2 x2 = z 1, set x = 4cost and y = 4sint, and solve for z in the second equation 13. You can also repeat a pattern using the rep() function. It even works as one would hope when test is a vector. Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Sequence containers store elements strictly in linear s. This API contains predefined functions that can be called from outside the SVG map, typically from the HTML document in which the SVG map is embedded. A vector-valued function in the plane is a function that associates a vector in the plane with each value of t in its domain. And, thanks to the Internet, it's easier than ever to follow in their footsteps (or just finish your homework or study for that next big test). The HTML element is a container for SVG graphics. The length function returns the length of R objects such as vectors, lists, or strings (find a little trick in Example 3). In mechanics, we will use two types of quantities to represent concepts like force, mass and time numerically. Add to Favourites. Scalar Valued Functions Definition: A scalar valued function is a function that takes one or more values but returns a single value. Caller CallerRows =. Similarly the integral of a vector is also a vector. This plugin not only provides SVG Support like the name says, it also allows you to easily embed your full SVG file's code using a simple IMG tag. SVG Tutorial David Duce *, Ivan Herman +, Bob Hopgood * * Oxford Brookes University, + World Wide Web Consortium Contents ¡ 1. id, that the renderer failed to draw: reportError. For example: rep(), seq(), using all() and any(), more on c() etc. Function Spaces. The library referenced by Robert gave me issues, so I wrote a more simple function of my own, which seems to work. If our function has three inputs, the math works out the same. Inject life into your SVG With a rich animation library and easy event handing, Snap. Visit Stack Exchange. You'll find many others in R packages. The function has three input parameters of the type floating scalar or float vector: N, the vector to orient, I, the incident vector, and Nref, the reference vector. The class returns the chart data in SVG format. svg is designed for modern browsers and therefore supports the newest SVG features like masking, clipping, patterns, full gradients, groups, and more. Yes, it is called Vector3. Usage of this Vector Graphics library should be easy even if you don't have JavaScript experience. share | cite. This function returns the last element of vector. The HTML element is a container for SVG graphics. The output is one-dimensional. Vector-Valued Functions - Velocity, Acceleration, and Speed. SEE ALSO: Complex Function, Real Function, Vector Function. The calculus of scalar valued functions of scalars is just the ordinary calculus. share | cite. test – A logical expression, which may be a vector. Mathematically you cannot normalise a vector with a magnitude of 0, however, for convenience to the programmer, many vector implementations will return a zero vector (0,0,0). So summary: scalar-function and vector-function refer to the *codomain* of a function, and general, the domain is assumed to be R^2 or R^3. That vector space is called Cn. 332 Chapter 13 Vector Functions. Use the LOOKUP function to look up a value in a one-column or one-row range, and retrieve a value from the same position in another one-column or one-row range. We can use pointy brackets. The article is necessarily a bit technical. We will examine both 2- and 3-dimensional vectors. SVG to HTML5 Canvas Converter This tool converts SVG into an HTML5 Canvas JavaScript function. A time vector can be created to compute the step response over this range. An empirical study of bit vector function optimization. The vector derivative admits the following physical interpretation: if r(t) represents the position of a particle, then the derivative is the velocity of the particle. VEX Functions See VEX contexts to learn about the different contexts (such as surface shaders or Rotates a vector by a rotation that would bring the x-axis to a given direction. 1 Is there any difference between 'vector field' and 'vector function'? 'vector function' is also called 'vector-valued function' (Thomas calculus). vector::erase() is a library function of "vector" header, it is used to erase/delete elements from the vector, it either removes one element from a specified iterator position or removes a range of elements. So wanting to learn how to make SVG files, I did what anyone would do. First, we will learn how represent a vector-valued functions. rm = FALSE, finite = FALSE) : numeric vector na. Update: Edge supports CSS transforms on SVG elements starting with EdgeHTML 17 released on the 30th of April 2018. C++ Vector Example | Vector in C++ Tutorial is today's topic. extend('line polyline polygon', { // add methods here which should be added to the draw-object // e. In Chapter 4 and 5 we discuss several examples of kernels drawing the connections between the Bayesian and the regularization framework. Actually, the transformation functions do not transform the SVG shape themselves, but the underlying coordinate system of that shape. Figure 3: Vector plot of the user field function tauWS showing that this wall shear stress function does yield vectors that are parallel to the wall. The definition of the derivative of a vector-valued function is nearly identical to the definition of a real-valued function of one variable. 3 WebCGM Profile. In the following R tutorial, I’m going to show you eight examples for the application of max and min in the R programming language. The reflect function returns a vector that points in the direction of reflection. This environment enables density mapping, distance analysis, surface analysis, advanced analysis with map algebra, grid statistics, spatial modeling, and surface creation. vector::swap ( right). SVG elements support mouse events, keyboard events. O(1) Indexing in a monad. To remove a graph, leave its text box blank. 2) std::pmr::vector is an alias template that uses a polymorphic allocator The elements are stored contiguously, which means that elements can be accessed not only through iterators, but also using offsets to regular pointers to elements. Scalar Function. In this case the maximum difference in magnitude between the build in WallShearStress function that has a component normal to the wall and the user defined function tauWS that is parallel to the. Since a position vector cannot be translated, it is technically not a vector, so it instead should be considered as a means of using. This is uncommon in C++, where operations are typically done in-place for performance. In this case, our x-component doesn't add much to the value of the function: the partial derivative is always 1. Because the output is a vector, physicists often call this a vector field or a vector function. The VC dimension of a set of functions is a measure of their capacity or complexity. That vector space is called Cn. able function in the direction of a vector ~ v. A vector is a fixed-length array with constant-time access and update of the vector slots, which are numbered from 0 to one less than the number of slots in the vector. Calling Javascript functions in HTML documents from events in SVG DOM. dest= argument accepts either the BLACS grid position or the MPI rank if the user desires a single process to own the matrix. Functions that we use in R vectors are known as the vector functions. Implementing Template Functions. Scalar point function. Both for HTML and SVG elements, when using CSS transforms, we have three translation functions available for 2D: translateX (tx), translateY (ty) and translate (tx [, ty]). The Vector Field function is used to composite and convert two rasters into a two-band raster that is of either data type Magnitude-Direction or data type U-V. In this section we need to talk briefly about limits, derivatives and integrals of vector functions. Use the LOOKUP function to look up a value in a one-column or one-row range, and retrieve a value from the same position in another one-column or one-row range. Stack Exchange network consists of 177 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. empty () – Returns whether the container is empty. The notation represents the norm, or magnitude, of vector v. Mathematics is the language of physics. Now, the word vector can mean a lot of different things. capacity () – Returns the size of the storage space currently allocated to the vector expressed as number of elements. SVG - amChart. Label each point with an appropriate t= value. Vector Valued Functions Calculus II Project In this project, you will investigate vector valued functions, curves, and some of their applications. SVG provides four transformation functions: translate() rotate() scale() skew() matrix() Each of these functions will be explained in more detail in the following sections. First, we will learn how represent a vector-valued functions. By default the direction of the vector is indicated by the direction of the arrow, and the magnitude is indicated by its color. In order to allow fine-grain features checks, we support three builtin function-like macros. In two dimensions, the vector is composed of two component functions f(t) and g(t), so that ; in three dimensions, the component functions are f(t), g(t), and h(t). Customers who bought this item also bought. JavaScript SVG parser and renderer on Canvas. Vectors in The Racket Guide introduces vectors. The inner vector (std::vector) is a vector of integers and the outer vector is a vector of such inner vectors (std::vector). SVG - Dialog. Vector, in physics, a quantity that has both magnitude and direction. This function returns the last element of vector. test – A logical expression, which may be a vector. Description. Using Symbolic Math Toolbox™, you can differentiate and integrate symbolic expressions, perform series expansions, find transforms of symbolic expressions, and perform vector calculus operations by using the listed functions. It is a great advantage to understand its basic principles when learning to write scripts for it. Developed in C++ and Java, it supports also multi-class classification, weighted SVM for unbalanced data, cross-validation and automatic model selection. org/wiki/File:mplwp_hyperbolic functions. The space curve generated by this vector function is called a circular helix. In Chapter 4 and 5 we discuss several examples of kernels drawing the connections between the Bayesian and the regularization framework. : Author: Original concepts by Katharine Cook Briggs (1875-1968) and Isabel Briggs Myers (1897-1980) Author of the referenced image is JakeBeech. To change a row vector into a column vector, use the transpose operator [Ctrl]l. vector::swap ( right). Vector calculus, or vector analysis, is concerned with differentiation and integration of vector fields, primarily in 3-dimensional Euclidean space. Here we are going to discuss all these functions of the R vector in detail with examples. The function changes the position or dimensions of a shape. extend('line polyline polygon', { // add methods here which should be added to the draw-object // e. Because the output is a vector, physicists often call this a vector field or a vector function. Call member function for each element in vector using std::mem_fun; // call member function print() for each referred object for_each (coll. This will avoid potential problems with trailing space removal or character set conversion that would change data values, such as may occur if you use a nonbinary string data type. By design, MATLAB ® software can efficiently perform repeated operations on collections of data stored in vectors and matrices. O(1) Indexing in a monad. Vector functions are those functions that produce a vector from one or more variables. As an example, consider the derivative of a particular scalar, d/da (a. Use the LOOKUP function to look up a value in a one-column or one-row range, and retrieve a value from the same position in another one-column or one-row range. Most vector art packages (Illustrator, Inkscape etc) can export as SVG. You must also decide whether the vector is a row or column vector. Each component function in a vector valued function represents the location of the value in a different dimension. foo: function(){ // can access this } // or even variables bar:5 } Method calc is always needed which updates the point of the shape. SVG - Drag. Many vector-valued functions, like scalar-valued functions, can be differentiated by simply differentiating the components in the Cartesian coordinate system. The quantity is known as the magnetic vector potential. Simple SSE and SSE2 (and now NEON) optimized sin, cos, log and exp The story. From Ramanujan to calculus co-creator Gottfried Leibniz, many of the world's best and brightest mathematical minds have belonged to autodidacts. svg Plot created with mplwp, the Matplotlib extension for Wikipedia plots. The second is the number of columns. Vector Functions and their Graphs. If our function has three inputs, the math works out the same. It uses a modified version of the excellent Canvg library. Then you establish x, y (and z if applicable) according to the equations, then plot using the plot(x,y) for 2D or the plot3(x,y,z) for 3D command. a scalar) to each point of some region of space. Transformations: Inverse of a Function example. Here is a sample with a function template. year() year(v=vector(time()) instant-vector) returns the year for each of the given times in UTC. Similarly the integral of a vector is also a vector. We can try to survive. Let's look at an example to make a multidimensional vector and access all its elements. The SVG Group Element that was transformed contains the 2 SVG Circles. These braces are optional if the body contains only a single expression. Function templates are implemented like regular functions, except they are prefixed with the keyword template. The lookup function has two forms, vector and array. To change a row vector into a column vector, use the transpose operator [Ctrl]l. Buying a poster from posters. 2 The set of real-valued functions of a real variable, de ned on the domain [a x b]. Vector Visualization The Wolfram Language provides state-of-the-art fully automated visualization of vector functions and data — suitable for representing flows, field lines, and other vector fields of any complexity. The space curve generated by this vector function is called a circular helix. $$n$$-dimensional space). The next step is to assign some variables from the xy vector that this function is given. vector::erase() is a library function of "vector" header, it is used to erase/delete elements from the vector, it either removes one element from a specified iterator position or removes a range of elements. In javascript functions, document represents SVG document and can be used to get the SVG elements. t the vector a. To see an interactive demonstration showing several ways to visualize x- and y-components of a vector see the Components of a Vector page. Fleet operators benefit above all from the functions during test drives. For example: rep(), seq(), using all() and any(), more on c() etc. As you will see, these behave in a fairly predictable manner. As a presentation attribute, transform can be used by any element (in SVG 1. a function named c. You must also decide whether the vector is a row or column vector. a log-likelihood) However, an. foo: function(){ // can access this } // or even variables bar:5 } Method calc is always needed which updates the point of the shape. The matrix( ) transform function specifies a transformation in the form of a. We've used onClick event to call a javascript functions. Container is a objects that hold data of same type. ) for a better understanding of pointers. In two dimensions, the vector is composed of two component functions f(t) and g(t), so that ; in three dimensions, the component functions are f(t), g(t), and h(t). Vector(A);. For example, Hadamard's 'gamma' function (Hadamard 1894) which, unlike the gamma function, is an entire function. These are scalar-valued functions in the sense that the result of applying such a function is a real number, which is a. SVG - Maps. Paul Garrett: Holomorphic vector-valued functions (February 19, 2005) where γ is a closed path with z having winding number +1. In this equation: A is a n · k matrix of parameters, b is a vector of parameters (of dimension n · 1). The data types can be logical, integer, double, character, complex or raw. If we do not know the step, as well as how many numbers we want, we can use along. The demo above allows you to enter up to three vectors in the form (x,y,z). Label focused on exploring the most serious, dark and abstract aspects of techno. • v is a vector. Vector is the name of a new wave rock band formed in Sacramento, CA in the early 1980s. {\displaystyle \mathbb {R} ^ {3}}. Actually, the transformation functions do not transform the SVG shape themselves, but the underlying coordinate system of that shape. Here are some significant features of Asymptote functions:. The new feature is that the elements of the vector spaces are functions, and the spaces are in nite dimensional. resize (N, 0xDEADC0DE); return v;} Here the vector instance is being returned by value, which means potential deep copying of the. Here is a sample with a function template. This theory is one of the fundamental tools employed in modern physics, the spectral theory of operators, approximation of analytic operators, analytic mappings between vectors, and vector-valued functions of several variables. MATLAB code that is written to operate simultaneously on different arrays is said to be vectorized. From Ramanujan to calculus co-creator Gottfried Leibniz, many of the world's best and brightest mathematical minds have belonged to autodidacts. 1 Vector-valued functions In the previous chapters we have considered real functions of several (usually two) variables f: D → R, where D is a subset of Rn, where n is the number of variables. Thus z is really a function of the single variable x. Vector destructor (public member function ) operator= Assign content (public member function ) Iterators: begin Return iterator to beginning (public member function ) end Return iterator to end (public member function ) rbegin Return reverse iterator to reverse beginning (public member function ) rend Return reverse iterator to reverse end (public member function ) cbegin. 2) std::pmr::vector is an alias template that uses a polymorphic allocator The elements are stored contiguously, which means that elements can be accessed not only through iterators, but also using offsets to regular pointers to elements. Stack Exchange network consists of 177 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Python Classes Consulting (312) 285-6886 Big Data Science. Types defined include: bit bit_vector typical signals integer natural positive typical variables boolean string character typical variables real time delay_length typical variables Click on standard to see the functions defined Note: This package must be provided with compiler, do not use this one. The examples show the function as it appears in a Derivation field in the Transformer stage. SVG - Flat Surface Shade. The intersection is an ellipse, with each of the two vector-valued functions describing half of it. The lookup function has two forms, vector and array. Here's a selection of statistical functions that come with the standard R installation. A smooth vector function is one where the derivative is continuous and where the derivative is not equal to zero. Show that all 8 properties are again satis ed. We calculate the components of the vector by subtracting the coordinates of the initial point. empty () – Returns whether the container is empty. test – A logical expression, which may be a vector. js was created by Lauren McCarthy and is developed by a community of collaborators, with support from the Processing Foundation and NYU ITP. Many vector-valued functions, like scalar-valued functions, can be differentiated by simply differentiating the components in the Cartesian coordinate system. It is typically represented by an arrow whose direction is the same as that of the quantity and whose length is proportional to the quantity’s magnitude. Use the LOOKUP function to look up a value in a one-column or one-row range, and retrieve a value from the same position in another one-column or one-row range. Serious techno to the world! After years of working in the Mexican scene, DJ's/Producers Fixon & Dig-it decided to launch their own label. Functions are part of a group of structures in VHDL called subprograms. Coming Soon: The much anticipated return of our milled receiver AK-47 line! Stay tuned and we will update you here when they will be available to order/buy from our authorized dealers. The vector derivative admits the following physical interpretation: if r(t) represents the position of a particle, then the derivative is the velocity of the particle. A vector functions of a single variable in R2 R 2 and R3 R 3 have the form, respectively, where f (t) f ( t), g(t) g ( t) and h(t) h ( t) are called the component functions. Smith, Robert Geva, Milind Girkar and Serguei V. The function should return a vector of correlations for the monitors that meet the threshold requirement. Since a position vector cannot be translated, it is technically not a vector, so it instead should be considered as a means of using. 1 The R function plot() The plot() function is one of the most frequently used plotting functions in R. 1-Limit, continuity and differentiability of vector point functions 2-Partial derivatives of vector point functions 3-Scalar and vector fields 4-The gradient, divergence and curl 5-Expansion formulas. This environment enables density mapping, distance analysis, surface analysis, advanced analysis with map algebra, grid statistics, spatial modeling, and surface creation. This serves to cleanup code as well as allow for reusability. We can use pointy brackets. A vector-valued function, also referred to as a vector function, is a mathematical function of one or more variables whose range is a set of multidimensional vectors or infinite-dimensional vectors. Free vector calculator - solve vector operations and functions step-by-step This website uses cookies to ensure you get the best experience. The problem is to find an input vector which produces a result equal to. Construct and insert element (public member function ) emplace_back Construct and insert element at the end (public member function ) Allocator: get_allocator Get allocator (public member function ) Non-member function overloads relational operators Relational operators for vector (function template ) swap Exchange contents of vectors (function. )To avoid copying, accept a std::vector& instead. But, it is widely used in classification objectives. The monad allows operations to be strict in the vector when necessary. Both sapply () and lapply () consider every value in the vector to be an element on which they can apply a function. The class returns the chart data in SVG format. The next step is to assign some variables from the xy vector that this function is given. Asymptote functions are treated as variables with a signature (non-function variables have null signatures). Usage To plot a function just type it into the function box. The basic R code for the max and min functions is shown above. Coming Soon: The much anticipated return of our milled receiver AK-47 line! Stay tuned and we will update you here when they will be available to order/buy from our authorized dealers. Exponential growth of large scale data industrial internet of things is evident due to the enormous deployment of IoT data acquisition devices. Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Mathematically you cannot normalise a vector with a magnitude of 0, however, for convenience to the programmer, many vector implementations will return a zero vector (0,0,0). If is discrete, then is a discrete random variable and its probability mass function is called marginal probability mass function of. This is actually quite simple. Otherwise -N is returned. When template functions are overloaded in such a way that the match of the template with the function call is not unique. 1-Limit, continuity and differentiability of vector point functions 2-Partial derivatives of vector point functions 3-Scalar and vector fields 4-The gradient, divergence and curl 5-Expansion formulas. InverseTransformPoint(target. C++ compiles C programs but adds object oriented (OO) features (classes, inheritance, polymorphism), templates (generic functions and classes), function and operator overloading, namespaces (packages), exception handling, a library of standard data structures (string, vector, map. All these functions have the same syntax: is a dot and then the name of the type. If you can describe a lot of different phenomena with a set of functions then the value of his large. So wanting to learn how to make SVG files, I did what anyone would do. Stack Exchange network consists of 177 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. The operation + (vector addition) must satisfy the following conditions: Closure: If u and v are any vectors in V, then the sum u + v belongs to V. Definition of Vector-Valued Functions A vector-valued function of one variable in Cartesian 3D space has the form r(t) = f (t)i +g(t)j +h(t)k or r(t) = ⟨f (t),g(t),h(t)⟩, where f (t), g(t), h(t) are called the component functions. We have step-by-step solutions for your textbooks written by Bartleby experts!. style() operator!. This is comparable to what you already know from basic continuity where a graph is continuous and does not contain any sharp. SVG is used to define vector-based graphics for the Web. Khan Academy is a 501(c)(3) nonprofit organization. We will examine both 2- and 3-dimensional vectors. A vector function is a mathematical function that outputs a vector. Graphing Vector-Valued Functions. The methods are as follows: firstly you can use the @keyframes function to animate your visual SVG content using CSS, the second method is animating SVG directly with SMIL (there’s a tutorial in this post explaining more about SMIL, keep an eye for it), and the third method is to use JavaScript which provides core features for doing. Arithmetic Package Functions- VII • Conversion to/ from std_logic_vector – Used to convert signed and unsigned values to and from std_logic_vector » Really just copies each bit Function Description Argument Returns std_logic_vector unsigned signed unsigned Convert to unsigned signed Convert to signed std_logic_vector std_logic_vector std. Note that as. How do you find the derivative and the integral of a vector function?. Try typing in (3cos(t), 3sin(t), t) in the input bar. It will work with any host, and I hope it helps people to experiment with Canvas. After all, there is a reason for the word Vector in Vector Graphics. To refer to a JavaScript function named "f( )" that resides in the HTML document (in which the SVG document is embedded), let the SVG document call the function through "top. e (in 2D) where f and g are real-valued functions of t. We calculate the components of the vector by subtracting the coordinates of the initial point. Vectors in Two and Three Dimensions Part 3: Vector-Valued Functions and their Derivatives. 2) std::pmr::vector is an alias template that uses a polymorphic allocator The elements are stored contiguously, which means that elements can be accessed not only through iterators, but also using offsets to regular pointers to elements. The gradient vector can be computed by finding the partial derivatives of a function: Find the gradient vector of the function : Visualize the direction of the gradient vector using a unit vector representation:. Inject life into your SVG With a rich animation library and easy event handing, Snap. For example, the vector u= −(1, 2,4) is defined by u:= < 1, -2, 4 > ; Note that if you use square parenthesis A:= [ 4, 5, 6 ] ; then Maple will see A as a point, and not a vector. However, if we use the transform attribute approach, all parameters for transform functions are numbers, meaning we cannot control and combine units anymore. Vector Functions 3. Transform functions. From Wikimedia Commons, the free media repository English: A diagram depicting the cognitive functions of each Myers-Briggs. However, there exist complex functions that are probably simpler in the sense of analytic function theory and which interpolate the factorial values. It can also render a chart in SVG format from an array of coordinates passed to the class. Edit your functions and then click the "Graph it" button below. To pass an argument by reference, simply enclose it in a structure (see Structures). It is typically represented by an arrow whose direction is the same as that of the quantity and whose length is proportional to the quantity’s magnitude. The possible SVG transformations are: rotation, scaling, translation, and skewing. Vector functions are used in a number of differential operations, such as gradient (measures the rate and direction of change in a scalar field), curl (measures the tendency of the vector function to rotate about a point in a vector field), and divergence (measures the magnitude of a source at a given point in a vector field). In mathematics, a function space is a set of functions between two fixed sets. Vector Form. It takes a vector xy (which has all three of the variables we said we had to give matlab to define our differential equation in it) and a time vector, and gives us the derivative of the three things in the the xy vector. a function named c. SVG - Clock. Public Member Functions: virtual vtkTypeBool IsA (const char *type) Return 1 if this class is the same type of (or a subclass of) the named class. I was originally going to write two functions, but noticed a clever way to achieve both functionalities using only the function shown above. The excerpt below shows my vector structure and two functions: vec2_magnitude calculates the magnitude (length) of a vector Stack Exchange Network Stack Exchange network consists of 175 Q&A communities including Stack Overflow , the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Vector destructor (public member function ) operator= Assign content (public member function ) Iterators: begin Return iterator to beginning (public member function ) end Return iterator to end (public member function ) rbegin Return reverse iterator to reverse beginning (public member function ) rend Return reverse iterator to reverse end (public member function ) cbegin. Tests whether two 3D vectors are equal. js file and include at the bottom of your SVG, inside the svg tag. All these functions have the same syntax: is a dot and then the name of the type. Transformations: Scaling a Function example. Container is a objects that hold data of same type. More vtkVectorFieldTopology * NewInstance const void PrintSelf (ostream &os, vtkIndent indent) override Methods invoked by print to print information about the object including superclasses. A vector's type can be checked with the typeof() function. SVG supports JavaScript/ECMAScript functions. This is the second article in a series I'm writing about SVG coordinate systems and transformations. According to their definitions, they are all the same things to me. Types defined include: bit bit_vector typical signals integer natural positive typical variables boolean string character typical variables real time delay_length typical variables Click on standard to see the functions defined Note: This package must be provided with compiler, do not use this one. Often, the domain and/or codomain will have additional structure which is inherited by the function space. Continuous objects) or model initial states (param. Note that the ^ and \$ surrounding alpha are there to ensure that the entire string matches. Try typing in (3cos(t), 3sin(t), t) in the input bar. Vector functions make working with vectors easy. 1-Limit, continuity and differentiability of vector point functions 2-Partial derivatives of vector point functions 3-Scalar and vector fields 4-The gradient, divergence and curl 5-Expansion formulas. resize (N, 0xDEADC0DE); return v;} Here the vector instance is being returned by value, which means potential deep copying of the. The domain of a vector function is all values for which its components are defined. To make it a vector, you can use the Vector command. ResetVariantArrayToDefaults. C++ is a compiled language, an upward compatible superset of C and an (incompatible) predecessor to Java. (cue the dramatic music) Don’t get me wrong, Youtube can be great. vector insert() function in C++ STL std::vector::insert() is a built-in function in C++ STL which inserts new elements before the element at the specified position, effectively increasing the container size by the number of elements inserted. Vector-Valued Functions - Velocity, Acceleration, and Speed. In vector analysis we compute derivatives of vector functions of a real variable; that is we compute derivatives of functions of the type F(t) = f1(t)i + f2(t)j+ f3(t) k or, in different notation,. Starting in R2019b, you can also specify dim as a vector of positive integers to query multiple dimension lengths at a time. C++ vector is a data structure, helps you to perform many operations on elements. A vector's type can be checked with the typeof() function. Scoping Rules require you to pass a function whose argument is a vector of parameters (e. The distribution function of is called marginal distribution function of. The order of variables in this vector is defined by symvar. Basically, vector functions work exactly as matrix functions. Below is a list of all of the attributes available in SVG along with links to reference documentation to help you learn which elements support them and how they work. It's the name of a breakfast cereal manufactured by Kellogg's Canada. Obvious applications of the gradient are finding the max/min of multivariable functions. The site was built and is maintained by Alexis Deveria, with occasional updates provided by the web development community. foo: function(){ // can access this } // or even variables bar:5 } Method calc is always needed which updates the point of the shape. I was originally going to write two functions, but noticed a clever way to achieve both functionalities using only the function shown above. Angle(transform. The questions here ask you about identifying functions from example graphs and the first step in graphing a vector function, among other aspects of this mathematical subject. Note that functions which expect vectors always expect column vectors rather than row vectors. Physical Interpretation If $$\mathbf{r}\left( t \right)$$ represents the position of a particle, then the derivative is the velocity of the particle:. From Wikimedia Commons, the free media repository English: A diagram depicting the cognitive functions of each Myers-Briggs. The interface follows a language that is designed to look a lot like the notation use in linear algebra. Create a function to compute the cross product of two vectors. Parameters first, last Forward iterators to the initial and final positions in a sequence. Definition of Vector-Valued Functions A vector-valued function of one variable in Cartesian 3D space has the form r(t) = f (t)i +g(t)j +h(t)k or r(t) = ⟨f (t),g(t),h(t)⟩, where f (t), g(t), h(t) are called the component functions. Stubbing stdlib exit() Function Is it possible to stub the stdlib library exit() function? I ask this question because the exit() function takes and integer as a parameter an returns a void. The statements within the curly braces form the body of the function. Introduction n 1. From Wikimedia Commons, the free media repository. If you're not familiar with the syntax for the SVG transform attribute functions, and for the sake of brevity of this article, and because the syntax details and how it works is outside the scope of this article, I recommend you read the article I've written about this a while back: Understanding SVG Coordinate Systems and Transformations. The lookup function has two forms, vector and array. Vector functions make working with vectors easy. 2 CGM on the Web n 2. Wait! Have you checked - R Matrix Functions. Download over 794 icons of function in SVG, PSD, PNG, EPS format or as webfonts. SVG integrates with other W3C standards such as the DOM and XSL. In the case where U is used, the renderer will convert to magnitude on the fly. All these functions have the same syntax: is a dot and then the name of the type. The LOOKUP function vector form syntax has the following arguments: lookup_value Required. If you want to store these results, use a column with a VARBINARY or BLOB binary string data type. R range Function. If you do not specify v, then gradient(f) finds the gradient vector of the scalar function f with respect to a vector constructed from all symbolic variables found in f. We can try to survive. R rep() Function. sk\s*Jeeves#i','#HP\s*Web\s*PrintSmart#i','#HTTrack#i','#IDBot#i','#Indy\s*Library#','#ListChecker#i','#MSIECrawler#i','#NetCache#i','#Nutch#i','#RPT-HTTPClient#i','#. Vector functions make working with vectors easy. For example: rep(), seq(), using all() and any(), more on c() etc. SVG - Drag. Actually, the transformation functions do not transform the SVG shape themselves, but the underlying coordinate system of that shape. Matlab is a software package that makes it easier for you to enter matrices and vectors, and manipulate them. Vector Functions is a techno label based in Mexico. Functions are small sections of code that perform an operation that is reused throughout your code. To change a row vector into a column vector, use the transpose operator [Ctrl]l. Let me say I have a vector valued function r, and I'll put a little vector arrow on top of it. If you do not specify v, then gradient(f) finds the gradient vector of the scalar function f with respect to a vector constructed from all symbolic variables found in f. svg Plot created with mplwp, the Matplotlib extension for Wikipedia plots. 1 MultivariableVector-Valued Functions InCalculus I,westudiedfunctionsoftheform y f (x),forexample f (x) x2. If you're interested, take a look. Online Graphing Calculator: Plot your own SVG Math Graphs. The LOOKUP function vector form syntax has the following arguments: lookup_value Required. 1 The R function plot() The plot() function is one of the most frequently used plotting functions in R. And they are all some kind of mapping, which assigns a vector to every point in a space or manifold or whatever. If the dot product of I and Nref is smaller than zero the return value is N. The lookup function has two forms, vector and array. In javascript functions, document represents SVG document and can be used to get the SVG elements. In this equation: A is a n · k matrix of parameters, b is a vector of parameters (of dimension n · 1). vector returns TRUE if x is a vector of the specified mode having no. rep() is used for replicating the values in x. vector are quite distinct from the meaning of the formal class "vector" in the methods package, and hence as(x, "vector") and is(x, "vector"). KanjiVG is a description of the sinographs (or kanji) used by the Japanese language. Types defined include: bit bit_vector typical signals integer natural positive typical variables boolean string character typical variables real time delay_length typical variables Click on standard to see the functions defined Note: This package must be provided with compiler, do not use this one. Many encryption and compression functions return strings for which the result might contain arbitrary byte values. Advanced calculus: Of real-valued functions of a real variable and vector-valued functions of a vector variable. In the case where V is used, the renderer will convert to direction on the fly. We need the draw package to make this. This will avoid potential problems with trailing space removal or character set conversion that would change data values, such as may occur if you use a nonbinary string data type. http://commons. vector insert() function in C++ STL std::vector::insert() is a built-in function in C++ STL which inserts new elements before the element at the specified position, effectively increasing the container size by the number of elements inserted. It can also render a chart in SVG format from an array of coordinates passed to the class. Usage To plot a function just type it into the function box. Maths Geometry Graph plot vector. A vector function is simply a function whose domain is a set of real numbers and whose range is a set of vectors. The vector has an infinte magnitude and components are all (+/-) infinity or (+/-) 0. Scalar Valued Functions Definition: A scalar valued function is a function that takes one or more values but returns a single value. In this video we will learn how to find Derivatives and Integrals of Vector Functions. a log-likelihood) However, an. Often, the domain and/or codomain will have additional structure which is inherited by the function space. size() exceeds v. This results in a coupled set of equations, called the Navier-Stokes equations , which describe how fluids behave when subjected to external forces. Without them, if there were a value named alphabet, it would also match, and the replacement would be onebet. So, if I define a vector valued function, and if you don't remember what those are we'll have a little bit of review here. vector functions A vector function r(t) = hf(t),g(t),h(t)i is a function of one variable—that is, there is only one "input" value. You can plot 2 functions, function 1 (in dark green) and function 2 (in magenta). szdim = size(A,dim) returns the length of dimension dim when dim is a positive integer scalar. Members in a vector are officially called components. Transformations: Scaling a Function example. A function of one or more variables whose range is one-dimensional, as compared to a vector function, whose range is three-dimensional (or, in general, -dimensional). Vector valued function derivative example Our mission is to provide a free, world-class education to anyone, anywhere. Vector)} filter {OpenLayers. The intersection is an ellipse, with each of the two vector-valued functions describing half of it. Count End With Test = CallerRows * CallerCols End Function In this code, the variables CallerRows and CallerCols get the number of rows and columns in the range from which the function was called. : rowSums, rowMeans, table, tabulate) or one can design custom functions that avoid expensive R loops by using vector- or matrix-based approaches. Here is an example of a three-dimensional vector function: which is plotted below for 0<=t<=7*pi. org helps support GraphSketch and gets you a neat, high-quality, mathematically-generated poster. Let’s dive. js file and include at the bottom of your SVG, inside the svg tag. To create a vector with one of these functions you must (atleast initially) decide how long do you want the vector to be. Exponential growth of large scale data industrial internet of things is evident due to the enormous deployment of IoT data acquisition devices. What we have just walked through is the explanation of the gradient theorem. Vector functions take a vector as input and produce a vector of the same length as output. empty () – Returns whether the container is empty. t the vector a. This API contains predefined functions that can be called from outside the SVG map, typically from the HTML document in which the SVG map is embedded. js was created by Lauren McCarthy and is developed by a community of collaborators, with support from the Processing Foundation and NYU ITP. That is, the derivative of the "dot" product (a. Thus if a vector is a function of, say time, then its derivative with respect to time is also a vector. The function has three input parameters of the type floating scalar or float vector: N, the vector to orient, I, the incident vector, and Nref, the reference vector. Subsection EVS Examples of Vector Spaces. share | cite. De ne addition in the obvious way: f(x) + g(x) h(x) another real function, and scalar multiplication: f(x) = F(x) yet another real function. Labels and equations are typeset with LaTeX, for high-quality PostScript output. jQuery SVG helps manage the diagram on the Web page. Chapter 2 Function Spaces Many di erential equations of physics are relations involving linear di er-ential operators. The Vector Field function is used to composite and convert two rasters into a two-band raster that is of either data type Magnitude-Direction or data type U-V. Vector Visualization The Wolfram Language provides state-of-the-art fully automated visualization of vector functions and data — suitable for representing flows, field lines, and other vector fields of any complexity. Bitmap and vector are two types of images found online or as a supported image type in graphics software. And so this is where we begin. An empirical study of bit vector function optimization. Functions arguments are passed by value. org helps support GraphSketch and gets you a neat, high-quality, mathematically-generated poster. A vector-valued function, also referred to as a vector function, is a mathematical function of one or more variables whose range is a set of multidimensional vectors or infinite-dimensional vectors. See how two vectors are related to their resultant, difference and cross product. The input of a vector-valued function could be a scalar or a vector (that is, the dimension of the domain could be 1 or greater than 1); the dimension of the domain is not defined by the dimension of the range. We define vector functions in Maxima just like we would any other function; the only difference is that the return value is a vector. 3D Vector Plotter. ) r(t) = c. Linear Vector Function. However, as we have emphasized before, the physical reality. Arithmetic Package Functions- VII • Conversion to/ from std_logic_vector – Used to convert signed and unsigned values to and from std_logic_vector » Really just copies each bit Function Description Argument Returns std_logic_vector unsigned signed unsigned Convert to unsigned signed Convert to signed std_logic_vector std_logic_vector std. The R Language. Just divide those by 10. Variables with the same name are allowed, so long as they have distinct signatures. Another important property of a vector is its length. It allows the same natural syntax that is used with plain arrays but offers a series of services that free the C++ programmer from taking care of the allocated memory and help to operate consistently on the contained objects. In other scenarios, the function space might. It may use cartesian or polar coordinates. Vectors implement IFn, for invoke() of one argument, which they presume is an index and look up in themselves as if by nth, i. Everyone has a different way of teaching. A lot of videos are so hard to hear. Many encryption and compression functions return strings for which the result might contain arbitrary byte values. This function returns the last element of vector. An empirical study of bit vector function optimization. org/wiki/File:mplwp_hyperbolic functions. 1 The R function plot() The plot() function is one of the most frequently used plotting functions in R. _over_time() The following functions allow aggregating each series of a given range vector over time and return an instant vector with per-series aggregation results:. Plot the terminal point of each position vector.
venx0dz57mnxa 6h90b6dd431y9xh ff8wxy9km8c saw457b9rd0 v85xox4d5b 3saw1yojymdzsri b2j1i3i61e71 i52pxha4wlyq qs2bu7f391fg4 f3ghxto1pyn ng4hc7sb3qz5gl nf9atac3coc4 r6t5l9b7dk lzyg1lsbz6 d659p37hdxy apzij37ivf8d0g zo58bnwkamm7v 0u9mjf92uagar1b qgba3dqxoh6 12bknej9bf309 41r356hgegw yqqyoop6gx2kgk wcl0j0qojcc8nu 7lu20bpewg vwxj6xz2yn9d g3ibspye6520mu5 | 2020-10-27 15:56:33 | {"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": 0.39277416467666626, "perplexity": 1228.4842773161427}, "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-2020-45/segments/1603107894203.73/warc/CC-MAIN-20201027140911-20201027170911-00294.warc.gz"} |
https://makondo.ugr.es/event/0/session/95/contribution/373 | # The 35th International Symposium on Lattice Field Theory
18-24 June 2017
Palacio de Congresos
Home > Timetable > Session details > Contribution details
Andalucía III
# Light-by-light forward scattering amplitudes in Lattice QCD
## Speakers
• Dr. Antoine GERARDIN
## Content
We present our preliminary results on the calculation of hadronic light-by-light forward scattering amplitudes using vector four-point correlation functions computed on the lattice. Using a dispersive approach, these forward scattering amplitudes can be described by $\gamma^ \gamma^ \to$ hadrons fusion cross sections and then compared with phenomenology. We show that only a few states are needed to reproduce our data. In particular, the sum rules considered in this study imply intricate relations between mesons and provide valuable information about individual form factors which are often used to estimate the meson pole contributions to the hadronic light-by-light contribution to the $(g-2)$ of the muon. | 2020-01-24 23:58:34 | {"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": 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": 0.7378182411193848, "perplexity": 1664.9341382541454}, "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/1579250626449.79/warc/CC-MAIN-20200124221147-20200125010147-00064.warc.gz"} |
http://en.wikipedia.org/wiki/Talk:Azimuthal_quantum_number | # Talk:Azimuthal quantum number
WikiProject Physics (Rated C-class, High-importance)
This article is within the scope of WikiProject Physics, a collaborative effort to improve the coverage of Physics on Wikipedia. If you would like to participate, please visit the project page, where you can join the discussion and see a list of open tasks.
C This article has been rated as C-Class on the project's quality scale.
High This article has been rated as High-importance on the project's importance scale.
WikiProject Chemistry (Rated C-class, Mid-importance)
This article is within the scope of WikiProject Chemistry, a collaborative effort to improve the coverage of chemistry on Wikipedia. If you would like to participate, please visit the project page, where you can join the discussion and see a list of open tasks.
C This article has been rated as C-Class on the project's quality scale.
Mid This article has been rated as Mid-importance on the project's importance scale.
I've replaced the term state with the more appropriate orbital, as state refers to the combined condition of all the electrons in the atom, rather than the one (or two) in the orbital.--Ian 08:21, 30 Jan 2005 (UTC)
## Contents
the azimuthal quantum number maybe could be merged with the orbital angular momentum quantum number as they are the same thing! agreed67.70.129.22 19:13, 12 September 2006 (UTC)
## Contradictory sentences in history section.
In the history section, is the following:
"Bohr argued that the angular momentum in any orbit n was nKh, where h is Planck's constant and K is some multiplying factor, the same for all the orbits, which was later determined to be 1/2π. The lowest quantum level therefore had an angular momentum of zero."
Since the lowest value for n is 1, we get L = h/2π, not zero.
The last sentence makes no sense, since according to the first sentence the angular momentum was not zero.
This should be sorted out, as the two sentences totally contradict each other.
This history section is very poor. There is the contradiction mentioned above, and most of the paragraph is concerned with things somewhat unrelated to the orbital quantum number. It desperately needs fixing, and for now I'll remove some of the worst chunks. --Latch.r 05:04, 22 May 2007 (UTC)
## Proof needed
Can anybody put the proof of $\mathbf{L^2\boldsymbol{\psi}} = \hbar^2{l(l+1)}\boldsymbol{\psi}$ into the article? --83.131.82.102 (talk) 13:50, 9 March 2008 (UTC)
## Derivation
Interesting. A derivation section in which nothing is derived. I agree w/ the above comment. —Preceding unsigned comment added by 160.94.156.170 (talk) 20:33, 16 July 2008 (UTC)
"The letters after the f sub-shell just follow f in alphabetical order except those already used. One mnemonic to remember the sequence S. P. D. F. G. H. ... is "Sober Physicists Don't Find Giraffes Hiding In Kitchens Like My Nephew" "
Surely this is wrong? J comes after I, not K. — Preceding unsigned comment added by 94.173.197.85 (talk) 19:43, 29 November 2013 (UTC) | 2014-08-27 13:05:21 | {"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": 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": 0.7440307140350342, "perplexity": 1620.4302883860694}, "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-2014-35/segments/1408500829210.91/warc/CC-MAIN-20140820021349-00338-ip-10-180-136-8.ec2.internal.warc.gz"} |
https://blog.canpan.info/asae-e/index-5.html | 2019N1105
11̑SZW
@{́A11̑SZW܂B́A3ɊJÂsyɊwZ\ďoꂷlN̔\܂Bt炵ASZ傫Ȕ肪܂B
Z搶̂b
SN̔\
ySZ̍ŐVLz
posted by ac at 20:27 | SZ
2019N1102
CwsL8
yЂƎƂԂɉ߂܂BԂ̌o߂Ƃ́Â܂B
́Â݂₰b̉Ԃ炩Ă鍠ł傤B
͂xŁÅwтɔĂˁB
CwsLAɂĂ܂B
{Ă݂ȂA肪Ƃ܂B
posted by ac at 00:53 | @UN
2019N1101
CwsL7
yЂƎAƂԂɉ߂Ă܂B
wZ܂ŁA炭܂B
́A݂₰bɉԂ炭Ƃł傤I
qǂ̋AA炭҂B
posted by ac at 13:28 | @UN
2019N1101
CwsL6
O[hŗVԎqǂ́AȂƖCȂƁB
݂ȎvVł܂B
posted by ac at 09:55 | @UN
2019N1101
ԒdÂ
@6ŃACwsɏoĂ܂B́Aō̎OO[hałB
@wZł́AHtɌẲԒdÂ肪sĂ܂BZ搶Ǘ@gĉԒdkĂĂ܂BsXی쏗܂Ǘ@劈Ă܂B
ŁAԂJ̐]wZɂȂ܂B
posted by ac at 09:16 | ]wZ
2019N1101
CwsL5
͂悤܂I
̋o߂ʂ܂܁ACɒ}Ă܂B
C߂Ƃ͈߂āA璩H܂B
posted by ac at 07:15 | @UN
2019N1031
CwsL4
ɃzeɓAוقǂĂH܂B
yH|Xʼn€Ay[HƂȂ܂B
̌́AeŃerςQ[肵āAk̂ЂƎłB
posted by ac at 19:25 | @UN
2019N1031
CwsL3
BقցB
̓Wi͈ɎIf炵WiłBBe֎~̂ŏЉłȂ̂cOI
݂ȑ喞̗lqłB
posted by ac at 15:39 | @UN
2019N1031
CwsL2
HAɕ{ւwɍs܂B
ꂼɂ낢Ȏv悤łB
ɂ́Auꂳ邩AlɍiFBl̂߂ɁAMĂꐶĂĂBĺAAABvĂė܂oĂ܂B
Ȏvl܂CwsłB
݂ȁACłB
posted by ac at 14:08 | @UN
2019N1031
CwsL1@̂̂є
̂̂єقŊwł܂B̂̐Aj̐AĉȊw̐Ɉ|Ă݂ȋÁXłB
̏o͋CƈЗ͂͐łˁB
posted by ac at 10:35 | @UN | 2019-12-10 15:03:11 | {"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": 0.9352450966835022, "perplexity": 12642.534762839228}, "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-2019-51/segments/1575540527620.19/warc/CC-MAIN-20191210123634-20191210151634-00276.warc.gz"} |
http://mathoverflow.net/revisions/27942/list | The least positive integer for which the equality $$\left\lceil \frac{2}{2^{1/n}-1}\right\rceil = \left\lfloor \frac{2n}{\log 2} \right\rfloor.$$ fails is $n=777,451,915,729,368$. See http://research.att.com/~njas/sequences/A129935http://oeis.org/A129935.
Another example that I like is the number $f(n)$ of inequivalent differentiable structures on $\mathbb{R}^n$. We have $f(n)=1$ if $n\neq 4$, while $f(4)=c$, the cardinality of the continuum.
The least positive integer for which the equality $$\left\lceil \frac{2}{2^{1/n}-1}\right\rceil = \left\lfloor \frac{2n}{\log 2} \right\rfloor.$$ fails is $n=777,451,915,729,368$. See http://research.att.com/~njas/sequences/A129935.
Another example that I like is the number $f(n)$ of inequivalent differentiable structures on $\mathbb{R}^n$. We have $f(n)=1$ if $n\neq 4$, while $f(4)=c$, the cardinality of the continuum. | 2013-05-23 08:09:03 | {"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": 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": 0.9667249321937561, "perplexity": 95.34151334205373}, "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/1368703035278/warc/CC-MAIN-20130516111715-00039-ip-10-60-113-184.ec2.internal.warc.gz"} |
http://mathhelpforum.com/pre-calculus/134922-dot-product-help.html | Math Help - dot product help.
1. dot product help.
v = <5,6> & p = <1,10>
what is v*p?
2. for $a = (x_1,y_1)$and $b = (x_2,y_2)$
$a . b = x_1\times x_2+y_1\times y_2$ | 2015-08-29 14:18:46 | {"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": 3, "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": 0.5506922602653503, "perplexity": 8660.228941631167}, "config": {"markdown_headings": false, "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-2015-35/segments/1440644064445.47/warc/CC-MAIN-20150827025424-00162-ip-10-171-96-226.ec2.internal.warc.gz"} |
http://www.reference.com/browse/Quantum+Cryptography | Definitions
# Quantum cryptography
Quantum cryptography, or quantum key distribution (QKD), uses quantum mechanics to guarantee secure communication. It enables two parties to produce a shared random bit string known only to them, which can be used as a key to encrypt and decrypt messages.
An important and unique property of quantum cryptography is the ability of the two communicating users to detect the presence of any third party trying to gain knowledge of the key. This results from a fundamental part of quantum mechanics: the process of measuring a quantum system in general disturbs the system. A third party trying to eavesdrop on the key must in some way measure it, thus introducing detectable anomalies. By using quantum superpositions or quantum entanglement and transmitting information in quantum states, a communication system can be implemented which detects eavesdropping. If the level of eavesdropping is below a certain threshold a key can be produced which is guaranteed as secure (i.e. the eavesdropper has no information about), otherwise no secure key is possible and communication is aborted.
The security of quantum cryptography relies on the foundations of quantum mechanics, in contrast to traditional public key cryptography which relies on the computational difficulty of certain mathematical functions, and cannot provide any indication of eavesdropping or guarantee of key security.
Quantum cryptography is only used to produce and distribute a key, not to transmit any message data. This key can then be used with any chosen encryption algorithm to encrypt (and decrypt) a message, which can then be transmitted over a standard communication channel. The algorithm most commonly associated with QKD is the one-time pad, as it is provably secure when used with a secret, random key.
## Quantum key exchange
Quantum communication involves encoding information in quantum states, or qubits, as opposed to classical communications use of bits. Usually, photons are used for these quantum states. Quantum cryptography exploits certain properties of these quantum states to ensure its security. There are several different approaches to quantum key distribution, but they can be divided into two main categories depending of which property they exploit. Prepare and measure protocols : Unlike in classical physics, the act of measurement is an integral part of quantum mechanics. In general, measuring an unknown quantum state will change that state in some way. This is known as quantum indeterminacy, and underlies results such as the Heisenberg uncertainty principle, information-disturbance theorem and no cloning theorem. This can be exploited in order to detect any eavesdropping on communication (which necessarily will involve measurement), and more importantly calculate the amount of information which has been intercepted. Entanglement based protocols : The quantum states of two (or more) separate objects can become linked together in such a way that they must be described by a combined quantum state, not as individual objects. This is known as Entanglement and means, for example, performing a measurement on one object will affect the other. If an entangled pair of objects is shared between two parties, anyone intercepting either particle will alter the overall system, allowing their presence (and the amount of information they have gained) to be determined.
These two approaches can both further be divided into three families of protocols; discrete variable, continuous variable and distributed phase reference coding. Discrete variable protocols were the first to be invented, and they remain the most widely implemented. The other two families are mainly concerned with overcoming practical limitations of experiments. The two protocols described below both use discrete variable coding.
### BB84 protocol - Charles H. Bennett and Gilles Brassard (1984)
This protocol, known as BB84 after its inventors and year of publication, was originally described using photon polarization states to transmit the information. However, any two pairs of conjugate states can be used for the protocol, and many optical fibre based implementations described as BB84 use phase encoded states. The sender (traditionally referred to as Alice) and the receiver (Bob) are connected by a quantum communication channel which allows quantum states to be transmitted. In the case of photons this channel is generally either an optical fibre or simply free space. In addition they communicate via a public classical channel, for example using broadcast radio or the internet. Neither of these channels need to be secure; the protocol is designed with the assumption that an eavesdropper (referred to as Eve) can interfere in any way with both.
The security of the protocol comes from encoding the information in non-orthogonal states. Quantum indeterminacy means that these states cannot in general be measured without disturbing the original state (see No cloning theorem). BB84 uses two pairs of states, with each pair conjugate to the other pair, and the two states within a pair orthogonal to each other. Pairs of orthogonal states are referred to as a basis. The usual polarization state pairs used are either the rectilinear basis of vertical (0°) and horizontal (90°), the diagonal basis of 45° and 135° or the circular basis of left- and right-handedness. Any two of these bases are conjugate to each other, and so any two can be used in the protocol. Below the rectilinear and diagonal bases are used.
Basis 0 1
X
The first step in BB84 is quantum transmission. Alice creates a random bit (0 or 1) and then randomly selects one of her two bases (rectilinear or diagonal in this case) to transmit it in. She then prepares a photon polarization state depending both on the bit value and basis, as shown in the table to the left. So for example a 0 is encoded in the rectilinear basis (+) as a vertical polarization state, and a 1 is encoded in the diagonal basis (x) as a 135° state. Alice then transmits a single photon in the state specified to Bob, using the quantum channel. This process is then repeated from the random bit stage, with Alice recording the state, basis and time of each photon sent.
Quantum mechanics (particularly quantum indeterminacy) says there is no possible measurement that will distinguish between the 4 different polarization states, as they are not all orthogonal. The only measurement possible is between any two orthogonal states (a basis), so for example measuring in the rectilinear basis will give a result of horizontal or vertical. If the photon was created as horizontal or vertical (as a rectilinear eigenstate) then this will measure the correct state, but if it was created as 45° or 135° (diagonal eigenstates) then the rectilinear measurement will instead return either horizontal or vertical at random. Furthermore, after this measurement the photon will be polarized in the state it was measured in (horizontal or vertical), with all information about its initial polarization lost.
As Bob does not know the basis the photons were encoded in, all he can do is select a basis at random to measure in, either rectilinear or diagonal. He does this for each photon he receives, recording the time, measurement basis used and measurement result. After Bob has measured all the photons, he communicates with Alice over the public classical channel. Alice broadcasts the basis each photon was sent in, and Bob the basis each was measured in. They both discard photon measurements (bits) where Bob used a different basis, which will be half on average, leaving half the bits as a shared key.
Alice's random bit 0 1 1 0 1 0 0 1
Alice's random sending basis X X X X
Photon polarization Alice sends
Bob's random measuring basis X X X X
Photon polarization Bob measures
PUBLIC DISCUSSION OF BASIS
Shared secret key 0 1 0 1
To check for the presence of eavesdropping Alice and Bob now compare a certain subset of their remaining bit strings. If a third party (usually referred to as Eve, for 'eavesdropper') has gained any information about the photons polarization, this will have introduced errors in Bobs' measurements. If more than $p$ bits differ they abort the key and try again, possibly with a different quantum channel, as the security of the key cannot be guaranteed. $p$ is chosen so that if the number of bits known to Eve is less than this, privacy amplification can be used to reduce Eve's knowledge of the key to an arbitrarily small amount, by reducing the length of the key.
### E91 protocol - Artur Ekert (1991)
The Ekert scheme uses entangled pairs of photons. These can be created by Alice, by Bob, or by some source separate from both of them, including eavesdropper Eve. The photons are distributed so that Alice and Bob each end up with one photon from each pair.
The scheme relies on two properties of entanglement. First, the entangled states are perfectly correlated in the sense that if Alice and Bob both measure whether their particles have vertical or horizontal polarizations, they will always get the same answer with 100% probability. The same is true if they both measure any other pair of complementary (orthogonal) polarizations. However, the particular results are completely random; it is impossible for Alice to predict if she (and thus Bob) will get vertical polarization or horizontal polarization.
Second, any attempt at eavesdropping by Eve will destroy these correlations in a way that Alice and Bob can detect.
## Privacy amplification and information reconciliation
The quantum cryptography protocols described above will provide Alice and Bob with nearly identical shared keys, and also with an estimate of the discrepancy between the keys. These differences can be caused by eavesdropping, but will also be caused by imperfections in the transmission line and detectors. As it is impossible to distinguish between these two types of errors, it is assumed all errors are due to eavesdropping in order to guarantee security. Provided the error rate between the keys is lower than a certain threshold (20% as of April 2007), two steps can be performed to first remove the erroneous bits and then reduce Eve's knowledge of the key to an arbitrary small value. These two steps are known as information reconciliation and privacy amplification respectively, and were first described in 1992.
Information reconciliation is a form of error correction carried out between Alice and Bob's keys, in order to ensure both keys are identical. It is conducted over the public channel and as such it is vital to minimise the information sent about each key, as this can be read by Eve. A common protocol used for information reconciliation is the cascade protocol, proposed in 1994. This operates in several rounds, with both keys divided into blocks in each round and the parity of those blocks compared. If a difference in parity is found then a binary search is performed to find and correct the error. If an error is found in a block from a previous round that had correct parity then another error must be contained in that block; this error is found and corrected as before. This process is repeated recursively, which is the source of the cascade name. After all blocks have been compared, Alice and Bob both reorder their keys in the same random way, and a new round begins. At the end of multiple rounds Alice and Bob will have identical keys with high probability, however Eve will have gained additional information about the key from the parity information exchanged.
Privacy Amplification is a method for reducing (and effectively eliminating) Eve's partial information about Alice and Bob's key. This partial information could have been gained both by eavesdropping on the quantum channel during key transmission (thus introducing detectable errors), and on the public channel during information reconciliation (where it is assumed Eve gains all possible parity information). Privacy amplification uses Alice and Bob's key to produce a new, shorter key, in such a way that Eve has only negligible information about the new key. This can be done using a universal hash function, chosen at random from a publicly known set of such functions, which takes as its input a binary string of length equal to the key and outputs a binary string of a chosen shorter length. The amount by which this new key is shortened is calculated, based on how much information Eve could have gained about the old key (which is known due to the errors this would introduce), in order to reduce the probability of Eve having any knowledge of the new key to a very low value.
## Implementations
As of March 2007 the longest distance over which quantum key distribution has been demonstrated using optic fibre is 148.7 km, achieved by Los Alamos/NIST using the BB84 protocol. Significantly, this distance is long enough for almost all the spans found in today's fibre networks. The distance record for free space QKD is 144km between two of the Canary Islands, achieved by a European collaboration using entangled photons (the Ekert scheme) in 2006, and using BB84 enhanced with decoy states in 2007 . The experiments suggest transmission to satellites is possible, due to the lower atmospheric density at higher altitudes. For example although the minimum distance from the International Space Station to the ESA Space Debris Telescope is about 400 km, the atmospheric thickness is about an order of magnitude less than in the European experiment, thus yielding less attenuation compared to this experiment.
The DARPA Quantum Network, a 10-node quantum cryptography network, has been running since 2004 in Massachusetts, USA. It is being developed by BBN Technologies, Harvard University, Boston University and QinetiQ.
There are currently three companies offering commercial quantum cryptography systems; id Quantique (Geneva), MagiQ Technologies (New York) and SmartQuantum (France). Several other companies also have active research programmes, including Toshiba, HP, IBM, Mitsubishi, NEC and NTT (See External links for direct research links).
Quantum encryption technology provided by the Swiss company Id Quantique was used in the Swiss canton (state) of Geneva to transmit ballot results to the capitol in the national election occurring on Oct. 21, 2007.
In 2004, the world's first bank transfer using quantum cryptography was carried in Vienna, Austria. An important cheque, which needed absolute security, was transmitted from the Mayor of the city to an Austrian bank.
The world's first computer network protected by quantum cryptography was implemented in October 2008, at a scientific conference in Vienna. The network used 200 km of standard fibre optic cable to interconnect six locations across Vienna and the town of St Poelten located 69 km to the west. The event was witnessed by Gilles Brassard and Anton Zeilinger.
## Attacks
### Example: Intercept and resend
The simplest type of possible attack is the intercept-resend attack, where Eve measures the quantum states (photons) sent by Alice and then sends replacement states to Bob, prepared in the state she measures. In the BB84 protocol this will produce errors in the key shared between Alice and Bob. As Eve has no knowledge of the basis a state sent by Alice is encoded in, she can only guess which basis to measure in, in the same way as Bob. If she chooses correctly then she will measure the correct photon polarization state as sent by Alice, and will resend the correct state to Bob. However if she chooses incorrectly then the state she measures will be random, and the state sent to Bob will not be the same as the state sent by Alice. If Bob then measures this state in the same basis Alice sent he will get a random result, as Eve has sent him a state in the opposite basis, instead of the correct result he would get without the presence of Eve. An example of this type of attack is shown in the table below.
Alice's random bit 0 1 1 0 1 0 0 1
Alice's random sending basis X X X X
Photon polarization Alice sends
Eve's random measuring basis X X X
Polarization Eve measures and sends
Bob's random measuring basis X X X X
Photon polarization Bob measures
PUBLIC DISCUSSION OF BASIS
Shared secret key 0 0 0 1
Errors in key
The probability Eve chooses the incorrect basis is 50% (assuming Alice chooses her basis randomly), and if Bob measures this intercepted photon in the basis Alice sent he will get a random result, i.e. an incorrect result with probability of 50%. The probability an intercepted photon generates an error in the key string is then 50% x 50% = 25%. If Alice and Bob publicly compare $n$ of their key bits (thus discarding them as key bits, as they are no longer secret) the probability they find disagreement and identify the presence of Eve is
$P_d = 1 - left\left(frac\left\{3\right\}\left\{4\right\}right\right)^n$
So to detect an eavesdropper with probability $P_d = 0.999999999$ Alice and Bob need to compare $n = 72$ key bits.
### Security Proofs
The above is just a simple example of an attack. If Eve is assumed to have unlimited resources, for example classical and quantum computing power, there are many more attacks possible. BB84 has been proven secure against any attacks allowed by quantum mechanics, both for sending information using an ideal photon source which only ever emits a single photon at a time, and also using practical photon sources which sometimes emit multiphoton pulses. These proofs are unconditionally secure in the sense that no conditions are imposed on the resources available to the Eavesdropper, however there are other conditions required:
1. Eve cannot access Alice and Bob's encoding and decoding devices.
2. The random number generators used by Alice and Bob must be trusted and truly random (for example a Quantum random number generator).
3. The classical communication channel must be authenticated using an unconditionally secure authentication scheme.
### Man in the middle attack
Quantum cryptography is vulnerable to a man-in-the-middle attack when used without authentication to the same extent as any classical protocol, since no principle of quantum mechanics can distinguish friend from foe. As in the classical case, Alice and Bob cannot authenticate each other and establish a secure connection without some means of verifying each other's identities (such as an initial shared secret). If Alice and Bob have an initial shared secret then they can use an unconditionally secure authentication scheme (such as Carter-Wegman,) along with quantum key distribution to exponentially expand this key, using a small amount of the new key to authenticate the next session. Several methods to create this initial shared secret have been proposed, for example using a 3rd party or chaos theory.
### Photon number splitting attack
In the BB84 protocol Alice sends quantum states to Bob using single photons. In practice many implementations use laser pulses attenuated to a very low level to send the quantum states. These laser pulses contain a very small number of photons, for example 0.2 photons per pulse, which are distributed according to a Poissonian distribution. This means most pulses actually contain no photons (no pulse is sent), some pulses contain 1 photon (which is desired) and a few pulses contain 2 or more photons. If the pulse contains more than one photon, then Eve can split off the extra photons and transmit the remaining single photon to Bob. This is the basis of the photon number splitting attack, where Eve stores these extra photons in a quantum memory until Bob detects the remaining single photon and Alice reveals the encoding basis. Eve can then measure her photons in the correct basis and obtain information on the key without introducing detectable errors.
Even with the possibility of a PNS attack a secure key can still be generated, as shown in the GLLP security proof, however a much higher amount of privacy amplification is needed reducing the secure key rate significantly (with PNS the rate scales as $t^2$ as compared to $t$ for a single photon sources, where $t$ is the transmittance of the quantum channel).
There are several solutions to this problem. The most obvious is to use a true single photon source instead of an attenuated laser. While such sources are still at a developmental stage QKD has been carried out successfully with them. However as current sources operate at a low efficiency and frequency key rates and transmission distances are limited. Another solution is to modify the BB84 protocol, as is done for example in the SARG04 protocol, in which the secure key rate scales as $t^\left\{3/2\right\}$. The most promising solution is the decoy state idea, in which Alice randomly sends some of her laser pulses with a lower average photon number. These decoy states can be used to detect a PNS attack, as Eve has no way to tell which pulses are signal and which decoy. Using this idea the secure key rate scales as $t$, the same as for a single photon source. This idea has been implemented successfully in several QKD experiments, allowing for high key rates secure against all known attacks.
### Hacking attacks
Hacking attacks target imperfections in the implementation of the protocol instead of the protocol directly. If the equipment used in quantum cryptography can be tampered with, it could be made to generate keys that were not secure using a random number generator attack. Another common class of attacks is the Trojan horse attack which does not require physical access to the endpoints: rather than attempt to read Alice and Bob's single photons, Mallory sends a large pulse of light back to Alice in between transmitted photons. Alice's equipment reflects some of Mallory's light, revealing the state of Alice's polarizer. This attack is easy to avoid, for example using an optical isolator to prevent light from entering Alice's system, and all other hacking attacks can similarly be defeated by modifying the implementation. Apart from Trojan horse there are several other known attacks including faked state attacks , phase remapping attacks and time-shift attacks . The time-shift attack has even been successfully demonstrated on a commercial quantum crypto-system . This demonstration is the first successful demonstration of quantum hacking against a non-homemade quantum key distribution system.
### Denial of service
Because currently a dedicated fibre optic line (or line of sight in free space) is required between the two points linked by quantum cryptography, a denial of service attack can be mounted by simply cutting or blocking the line or, perhaps more surreptitiously, by attempting to tap it.
## History
Quantum cryptography was proposed first by Stephen Wiesner, then at Columbia University in New York, who, in the early 1970s, introduced the concept of quantum conjugate coding. His seminal paper titled "Conjugate Coding" was rejected by IEEE Information Theory but was eventually published in 1983 in SIGACT News (15:1 pp. 78-88, 1983). In this paper he showed how to store or transmit two messages by encoding them in two “conjugate observables”, such as linear and circular polarization of light, so that either, but not both, of which may be received and decoded. He illustrated his idea with a design of unforgeable bank notes. A decade later, building upon this work, Charles H. Bennett, of the IBM Thomas J. Watson Research Center, and Gilles Brassard, of the Université de Montréal, proposed a method for secure communication based on Wiesner’s “conjugate observables”. In 1990, independently and initially unaware of the earlier work, Artur Ekert, then a Ph.D. student at Wolfson College, University of Oxford, developed a different approach to quantum cryptography based on peculiar quantum correlations known as quantum entanglement.
## Prospects
The current commercial systems are aimed mainly at governments and corporations with high security requirements. Key distribution by courier is typically used in such cases, where traditional key distribution schemes are not believed to offer enough guarantee. This has the advantage of not being intrinsically distance limited, and despite long travel times the transfer rate can be high due to the availability of large capacity portable storage devices. The major difference of quantum cryptography is the ability to detect any interception of the key, whereas with courier the key security cannot be proven or tested. QKD systems also have the advantage of being automatic, with greater reliability and lower operating costs than a secure human courier network.
Factors preventing wide adoption of quantum cryptography outside high security areas include the cost of equipment, and the lack of a demonstrated threat to existing key exchange protocols. However, with optic fibre networks already present in many countries the infrastructure is in place for a more widespread use.
## References
Search another word or see Quantum Cryptographyon Dictionary | Thesaurus |Spanish
Copyright © 2014 Dictionary.com, LLC. All rights reserved.
• Please Login or Sign Up to use the Recent Searches feature
FAVORITES
RECENT
; | 2014-04-25 06:36:10 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 11, "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": 0.4863176643848419, "perplexity": 815.538895214636}, "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-2014-15/segments/1398223210034.18/warc/CC-MAIN-20140423032010-00248-ip-10-147-4-33.ec2.internal.warc.gz"} |
http://sites.psu.edu/johnroe/tag/kadison-singer-problem/ | The Kadison-Singer Problem is an old question from von Neumann algebra theory, whose roots lie in Dirac’s writings on quantum mechanics. Its classical expression is as follows: Consider the algebra $$B$$ of all bounded operators on a separable Hilbert space, and the maximal abelian subalgebra $$D$$ of all diagonal operators (relative to some fixed orthonormal basis). Does every pure state of $$D$$ extend uniquely to a pure state of $$B$$?
Some remarks of Dirac can be read as asserting that the answer should be “yes”, not merely for $$D$$ but for every masa. For other masa’s this is definitely false, but the original question remained tantalizingly open. Penn State’s Joel Anderson showed in the seventies or early eighties how to relate this question to a number of other “paving” problems, including some which are purely finite-dimensional. More recently these have gotten connected to engineering questions in signal processing and time-frequency analysis.
This summer, the problem has finally been solved (affirmatively) by finite-dimensional methods. As I understand it, these methods have to do with “discrepancy theory”: you have a large number of individually not-too-large objects (here vectors in a Euclidean space) and you want to split them into two subsets in such a way that some notion of “total mass” is evenly distributed between the two subsets. The catch here is that the “total mass” is measured by a quadratic form
$\sum_{v\in S} \langle x,v\rangle^2$
defined on the whole unit sphere in $${\mathbb R}^n$$. The method does not just prove Kadison-Singer but also gives an procedure for splitting any reasonably well-connected graph into two sparser but still well-connected graphs; iterating this constructs expanders.
#### References
Anderson, J., 1979. Extensions, restrictions, and representations of states on C*-algebras. Trans. Amer. Math. Soc. 249, 303–329.
Casazza, P.G., Tremain, J.C., 2006. The Kadison–Singer Problem in mathematics and engineering. PNAS 103, 2032–2039.
Harvey, N., 2013. An introduction to the Kadison-Singer problem and the paving conjecture.
Marcus, A., Spielman, D.A., Srivastava, N., 2013a. Interlacing Families I: Bipartite Ramanujan Graphs of All Degrees (arXiv e-print No. 1304.4132).
Marcus, A., Spielman, D.A., Srivastava, N., 2013b. Interlacing Families II: Mixed Characteristic Polynomials and the Kadison-Singer Problem (arXiv e-print No. 1306.3969).
Srivastava, N., 2013. Discrepancy, Graphs, and the Kadison-Singer Problem [WWW Document]. Windows On Theory. URL http://windowsontheory.org/2013/07/11/discrepancy-graphs-and-the-kadison-singer-conjecture-2/ (accessed 9.9.13). | 2016-04-29 10:05:32 | {"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": 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": 0.716852068901062, "perplexity": 1328.9169138619702}, "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-2016-18/segments/1461860111313.83/warc/CC-MAIN-20160428161511-00192-ip-10-239-7-51.ec2.internal.warc.gz"} |
https://physicstasks.eu/1995/annies-ride | ## Annie's Ride
Annie went for a bike ride. At first, she was going up a hill with approximately constant speed of 10 km·h−1. Like that she drove one sixth of the overall route. The next part of her ride equaled one third of the overall route and was straight, but because she rode through a beautiful countryside which she wanted to enjoy, she accelerated but a little and was now moving with a speed of 20 km·h−1. The rest of the trip she was hurrying home, so she drove with a constant speed of 30 km·h−1.
a) What was her average speed on the overall route?
b) How great would her average speed be in case she moved for a third of the way with a speed of 10 km·h−1, one third with a speed of 20 km·h−1 and the rest of the way with speed of 30 km·h−1?
• #### Analysis
v1 = 10 km·h−1 Annie’s speed in the first part v2 = 20 km·h−1 Annie’s speed in the second part v3 = 30 km·h−1 Annie’s speed in the third part vp = ? (km·h−1) Annie’s average speed
• #### Hint 1 for a): Average speed on the overall route
Express each of the times Annie needed to make the parts of the route. How long did it take her to travel the whole route? How do you find average velocity with overall time and overall distance?
• #### Hint 2 for b): verage speed in case b)
Again, express the lengths of each of the parts of the route using velocity and time. Then express the overall distance traveled. Now you can easily find the average speed as you did in the previous part.
• #### COMPLETE SOLUTION
We mark the overall route s, its parts s1, s2, s3.
We know that:
$s_1\,=\,\frac{s}{6} \,,\hspace{30px} s_2\,=\,\frac{s}{3} \,,\hspace{30px} s_3\,=\,\frac{s}{2}\,.$
It also holds that:
$s_1\,=\,v_1t_1 \,,\hspace{30px} s_2\,=\,v_2t_2 \,,\hspace{30px} s_3\,=\,v_3t_3\,.$
So:
$t_1 \,=\, \frac{s_1}{v_1} \,=\, \frac{s}{6v_1} \,,\hspace{30px} t_2\,=\,\frac{s_2}{v_2} \,=\, \frac{s}{3v_2} \,,\hspace{30px} t_3\,=\,\frac{s_3}{v_3} \,=\, \frac{s}{2v_3}\,.$
We find the average velocity as distance traveled over time of travel:
$v_p\,=\,\frac{s}{t_1+t_2+t_3}\,=\,\frac{s}{\frac{s}{6v_1}+\frac{s}{3v_2}+\frac{s}{2v_3}}\,=\,\frac{6v_1v_2v_3}{3v_1v_2+2v_1v_3+v_2v_3}\,.$
Numerically:
$v_p\,=\,\frac{(6{\cdot} 10\cdot 20{\cdot} 30)\,\mathrm{km}}{(3{\cdot} 10\cdot 20+2{\cdot} 10\cdot 30+20{\cdot} 30)\,\mathrm{h}}\,=\,20\,\mathrm{km\cdot h^{-1}}\,.$
Note:
We can solve this problem graphically, too. We draw a graph of speed dependence on time (t1 = t2 = t3) and a graph of average speed dependence on time.
Area under the curve in both graphs represents the distance traveled and must equal. From graphs it is obvious that vp = v2.
We mark the parts of Annie’s route s1, s2, s3 the speeds with which she traveled v1, v2, v3 and times t1, t2, t3, the overall time tc.
We know that:
$t_1\,=\,\frac{t_c}{6}\,,\hspace{30px}t_2\,=\,\frac{t_c}{3}\,,\hspace{30px}t_3\,=\,\frac{t_c}{2}\,.$
For the parts of the route it holds that:
$s_1\,=\,\frac{v_1t_c}{6}\,,\hspace{30px}s_2\,=\,\frac{v_2t_c}{3}\,,\hspace{30px}s_3\,=\,\frac{v_3t_c}{2}\,.$
This is how we calculate the average speed vp :
$v_p\,=\,\frac{s_1+s_2+s_3}{t_c}\,=\,\frac{\frac{v_1t_c}{6}+\frac{v_2t_c}{3}+\frac{v_3t_c}{2}}{t_c}\,=\,\frac{v_1+2v_2+3v_3}{6}\,.$
Numerically:
$v_p\,=\,\frac{(10+2{\cdot} 20+3{\cdot} 30)}{6}\,\mathrm{km\cdot h^{-1}}\dot=23{,}3\,\mathrm{km\cdot h^{-1}}\,.$
a) Annie’s average speed on the overall route is:
$v_p=\frac{6v_1v_2v_3}{3v_1v_2+2v_1v_3+v_2v_3}\,.$
Numerically:
$v_p\,=\,20\,\mathrm{km\cdot h^{-1}}\,.$
b) In case Annie was moving with a speed v1 for a sixth of time she traveled, third of the time with a speed v2 and the rest of overall time with a speed of v3, her average speed is:
$v_p = \frac{v_1+2v_2+3v_3}{6}\,.$
Numerically:
$v_p\,\dot=\,23{,}3\,\mathrm{km\cdot h^{-1}}\,.$
×Original source: http://fo.cuni.cz - upraveno
Zpracováno v diplomové práci Jany Moltašové (2011). | 2021-06-15 13:50:41 | {"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": 0.5727325677871704, "perplexity": 1253.9930730031958}, "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-2021-25/segments/1623487621273.31/warc/CC-MAIN-20210615114909-20210615144909-00332.warc.gz"} |
https://www.txcorp.com/images/docs/vsim/latest/html/VSimExamples/VSimMD/RadiationGeneration/helixTwt2ImpedAtten/helixTwt2ImpedAtten.html | # Helix Traveling Wave Tube 2: Impedance and Attenuation (helixTwt2ImpedAtten.sdf)
Keywords:
Helix TWT Impedance and Attenuation Run
## Problem description
This VSimMD example is one of a set of simulations showing different calculations to aid the design of a helix traveling wave tube (TWT) in three dimensions. The 100-turn helix with end feeds is imported from a CAD file, but all other geometrical parts are created with the Constructive Solid Geometry (CSG) capabilities within VSimComposer.
An input signal is sent into a short section of coaxial input waveguide and a similar section of coaxial waveguide at the opposite end of the tube provides an output power port. The geometry includes three dielectric support rods, each clad by a section of resistive tubing for attenuation. The interaction impedance run enables the user to calculate the transverse impedance and Pierce interaction impedance of the helix TWT. The transverse impedance is relevant for impedance matching at the input and output coaxial ports, and the Pierce interaction impedance is related to the interaction of the particles with the signal, and thus the signal gain.
The user may wish to run this simulation type multiple times, varying parameters such as the coax radius and dielectric permittivity, in order to result in a design with the correct impedance parameters.
Related simulations:
This simulation can be performed with a VSimMD or VSimPD license.
## Opening the Simulation
The Helix TWT Impedance and Attenuation example is accessed from within VSimComposer by the following actions:
• In the resulting Examples window expand the VSim for Microwave Devices option.
• Expand the Radiation Generation option.
• Select “Helix Traveling Wave Tube 2: Impedance and Attenuation” and press the Choose button.
• In the resulting dialog, create a New Folder if desired, and press the Save button to create a copy of this example.
All of the properties and values that create the simulation are now available in the Setup Window as shown in Fig. 364, with the Elements Tree in the upper center, and the Property Editor in the lower center. The right pane shows a 3D view of the geometry as well as the grid, if its visibility has been activated (which it was not when this image was captured).
Fig. 364 Setup Window for the Helix TWT impedance and attenuation example.
## Geometry details
The various geometrical objects can all be seen in the tree by pulling down the bar separating the Elements Tree from the Property Editor and then expanding Geometries, CSG, and Grids. Make sure tube and Grid are unclicked, helixWithFeedsGeom is clicked, and that the Toggle Axes button is set to remove the axes from the view. This allows one to see the interior geometrical objects, including the incoming feed, the dielectric support rods, the resistive tubes in the center, the particle emission disk at the left, and various planes where measurements are taken. The geometry is shown in Fig. 365.
Fig. 365 Interior geometry for the Helix TWT impedance and attenuation example.
## Constants and Parameters
Pulling the separator bar between the Elements Tree and the Property Editor and opening the Constants part of the tree gives the view shown in Fig. 366.
Fig. 366 Constants for the Helix TWT impedance and attenuation example.
There are three types of constants. The first set of constants, from PI through ELECMASSEV are not changeable by the user. These are the various mathematical and physical constants that the simulation will use. The second set of constants are those with HELIX in the name. These must correspond to the helix geometry, the beginning, mid-radius, wire radius, pitch, and number of turns of the helix. These cannot be set arbitrarily, as the helix was imported as an STL file. Instead they must be set to match the imported helix parameters. The remaining constants define fundamental geometry quantities, such as where the tube begins and its radius, other physical simulation parameters, such as the wave frequency, and numerical parameters, such as the number of cells in each direction.
Moving the scroll bar and opening the Parameters part of the tree shows the Parameters, values that come from arithmetically combining constants and other parameters. This is shown in Fig. 367.
Fig. 367 Parameters for the Helix TWT impedance and attenuation example.
As an example, LENGTH_HELIX is shown. The expression shows that this is the number of helix turns times the helix pitch. It also shows the value. Of course, the expression is editable, while the value is not.
Both constants and parameters have a description field that allows the user to document the quantity.
## Materials
To bring materials in the simulation, in the right pane, select the Database tab, select one or more materials (with ctrl-click) and hit the button Add To Simulation. The materials will then appear under Materials in the tree view. At this point one can change the properties of the materials including the name. In this example we imported resistive damper, changed its name to LossyMaterial, and changed its value of conductivity to 0.55. This is shown in Fig. 368.
Fig. 368 Materials for the Helix TWT impedance and attenuation example.
Once one has materials in the simulation, one can set the materials of any of the geometries. Click on the geometry, then in the Property Editor pane, double click on the material value. A context menu will allow you to set the material of the geometry to any material in the simulation.
## Running the simulation
After performing the above actions, continue as follows:
• Proceed to the run window by pressing the Run button in the left column of buttons.
• To run the file, click on the Run button in the upper left corner of the right panel. You will see the output of the run in the right pane. The run has completed when you see the output, “Engine completed successfully.” This is shown in Fig. 369.
Fig. 369 The Run Window at the end of execution.
The simulation to determine the impedance should run long for any mismatch at the outgoing boundary to stabilize. That is, the simulation must be run long enough for the electromagnetic wave to reach the far end of the tube and for any reflections to return some distance to the last history in the tube. This will take about 100,000 steps. On a 4-core machine, we have observed 0.23s/step, so this simulation will take about 7 hours. This simulation parallelizes well up to 16 cores, so with a sufficient license and workstation or cluster, one can bring this simulation time down to about 2 hours.
## Visualizing the results
After performing the above actions, continue as follows:
• Proceed to the Visualize Window by pressing the Visualize button in the left column of buttons.
To see the fields inside the tube as shown in Fig. 370, continue as follows:
• Select Data View: Data Overview
• Expand Scalar Data then D, then select field D_y.
• Expand Geometries then select poly (rod1).
• Check Clip All Plots, which cuts through the data at the $$z=0$$ plane.
• Select the Colors option from the upper menu bar in the Visualuization Results pane.
• In the Color Options window: check Fix Minimum and set it to -400, then check Fix Maximum, set it to 400, and click OK.
• Move the dump slider forward in time to see the evolution.
Fig. 370 The transverse displacement field, Dy, on the central x-y plane at dump 31.
This plot shows the transverse displacement field. Once can see that it is confined inside the tube (sanity check), and that it is most intense inside the dielectric rod at the bottom. The field is larger at the left, as it is just entering and propagating down the tube.
At any time one can leave this visualization pane to move back to the run pane to see how the simulation is progressing.
The longitudinal field inside the tube is shown in Fig. 371, which can be obtained by the steps:
• Select Data View: Field Analysis
• Select Field E_x.
• Select Horizontal under Lineout Settings, set the Intercept to 0, and click Perform Lineout.
• Under Layout select Stacked 2d/1d
• Move the dump slider forward in time to see the evolution.
Fig. 371 The longitudinal field, Ex, on axis at dump 27, which is time step 54,000.
As seen in Fig. 371, the longitudinal field has dropped from about 240 V/m to about 68 V/m in the center of the simulation where the resistive tubes are. This corresponds to about 14 dB of attenuation. The purpose of this attenuation is to have sufficient damping so that reflections coming back from the end to the beginning and then reflection again do not grow, as that would change the device into an oscillator, with energy growth that could destroy the system. If the Helix Traveling Wave Tube 3: Power Run (helixTwt3PowerRun.sdf) shows that this is happening, one can return to this run and increase the conductivity of the LossyMaterial or the length of the resistive tube (through BGNTURN_ATTENUATOR and ENDTURN_ATTENUATOR) to provide more attenuation.
Histories contain the time evolution of quantities defined in the input file. These can be seen by selecting the Data View, History. To determine various impedances we want particular histories obtained by the process:
• Select Data View: History
• Under Graph 1 select poyntingA
• Under Graph 2 select transverseVoltageA
• Under Graph 3 select EonAxisA_0
• Under Graph 4 select <None>
The result is shown in Fig. 372. The power through the plane was defined such that incoming is negative.
Fig. 372 Poynting power (W), transverse voltage (V), and electric field (V/m), at Plane A, along the helix TWT as a function of time (s).
Impedance parameters of interest are the transverse impedance
$Z_t = \frac{V_t^2}{2 P}$
and the Pierce interaction impedance
$Z_p = \frac{E_x^2 \lambda^2}{8 \pi^2 P}$
where $$P$$ is the poynting power (recorded in the history poyntingA), $$V_t$$ is the transverse Voltage amplitude (recorded in the histories transverseVoltage), $$E_x$$ is the electric field amplitude (recorded in the history EonAxis), and $$\lambda$$ is the wavelength of the field along the helix TWT axis.
The histories show the graphs of these quantities. To get precise values for any of these, one can press the Limits button, which will pop up a window withe the precise values. First, the X-Axis limits show that the units are $$ns$$. Secondly, one needs to choose a consistent time period, where all amplitudes are roughly constants. The period $$28ns < t < 32ns$$ is chosen. One can now adjust the limits until the peaks line up with the limits. We want average poynting power. We find $$P_{min} = 18. \ mW$$ and $$P_{max} = 44. \ mW$$. Hence, $$P_{av} = 31. \ mW$$. During that same time interval we find $$V_t = 2.4 \ V$$ and $$E_x = 240 \ V/m$$.
Fig. 371 can be used to obtain the wavelength. One can see four wavelengths between $$0.20m$$ and $$0.357m$$. Therefore the wavelength is $$(.357m-.20m)/4 = 0.039m$$
We now compute
$Z_t = \frac{2.4*2.4}{2*.031} = 92.9 \ \Omega$
and the Pierce interaction impedance is
$Z_p = \frac{240^{2}\cdot 0.039^{2}}{8\pi 0.031} = 35.8 \ \Omega .$
## Further Experiments
As noted above, one can change the attenuation by varying the conductivity of the resistive tubs or their length. For any given length, there is a maximum attainable attenuation, as there is no attenuation at zero conductivity (infinite resistance, i.e., vacuum) and none as well at infinite conductivity (metallic shielding). So if more than 14 dB attenuation is needed one can vary the conductivity, but a maximum will be observed, and if that is insufficient one will have to vary the rod length.
With additional computing resources, one could increase the grid resolution so that the resistive tube could be made thinner. As it is made thinner, one can go to greater conductivity without having the skin depth less than the resistive tube thickness.
The coupling is determined by the transverse impedance of the structure, which in turn depends on the capacitance provided by the rods. Varying the relative permittivity changes the transverse impedance. | 2019-01-20 22:15:48 | {"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": 0.3607419431209564, "perplexity": 1365.4426961000859}, "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/1547583739170.35/warc/CC-MAIN-20190120204649-20190120230649-00538.warc.gz"} |
http://web.stanford.edu/class/stats366/Mixt.html | # Modelling Mixtures: the Dirichlet distribution¶
## A measure on discrete measures.¶
You are given a set $$\mathcal{X}$$ (here taken as finite) and a probability density $$p(x), (p(x)\geq 0, \sum p(x)=1)$$. Also given is a set $$A$$ in $$\mathcal{X}$$. The problem is to compute or approximate $$p(A)$$. We consider the Birthday Problem from a Bayesian standpoint.
In order to go further we need to extend what we did before for the binomial and its Conjugate Prior to the multinomial and the the Dirichlet Prior. This is a probability distribution on the $$n$$ simplex
$\Delta_n=\{{\tilde{p} }=(p_1,\cdots,p_n),\; p_1+\cdots +p_n=1,\; p_i\geq 0\;\}$
It is a $$n$$-dimensional version of the beta density. Wilks (1962) is a standard reference for Dirichlet c omputations.
The Dirichlet has a parameter vector: $$\tilde{a} =(a_1,\ldots,a_n)$$. Throughout we write $$A=a_1+\cdots+a_n$$.
With respect to Lebesgue measure on $$\Delta_n$$ normalised to have total mass 1 the Dirichlet has density:
$D_{\tilde{a}}(\tilde{x})=\frac{\Gamma(A)}{\prod \Gamma (a_i)} x_1^{a_1-1} x_2^{a_2-1} \cdots x_n^{a_n-1}$
The uniform distribution on $$\Delta_n$$ results from choosing all $$a_i=1$$. The multinomial distribution corresponding to $$k$$ balls dropped into $$n$$ boxes with fixed probability $$(p_1,\cdots,p_n)$$ (with the ith box containing $$k_i$$ balls) is
${k \choose {k_1 \ldots k_n}} p_1^{k_1} \cdots p_n^{k_n}$
If this is averaged with respect to $$D_{\tilde{a}}$$ one gets the marginal (or Dirichlet/ Multinomial):
$P (k_1,\ldots,k_n) = P(\tilde{k})= \frac{(a_1)_{(k_1)} (a_2)_{(k_2)} \ldots (a_n)_{(k_n)}} {A_{(k)}} \mbox{ where }\;\; m_{(j)} \stackrel{\mbox{def}}{=} m(m+1)\cdots (m+(j-1))$
From a more practical point of view there are two simple procedures worth recalling here:
• To pick $$\tilde{p}$$ from a Dirichlet prior; just pick $$X_1, X_2, \ldots ,X_n$$ independant from gamma densities
$\frac{e^{-x}x^{a_i-1}}{\Gamma(a_i)} \mbox{ and set } p_i=\frac{X_i}{X_1+\cdots X_n}, 1\leq i\leq N$
• To generate sequential samples from the marginal distribution use Polya’s Urn: Consider an urn containing $$a_i$$ balls of color $$i$$ (actually fractions are allowed).
Each time, choose a color $$i$$ with probability proportional to the number of balls of that color in the urn. If $$i$$ is drawn, replace it along with another ball of the same color.
The Dirichlet is a convenient prior because the posterior for $$\tilde{p}$$ having observed $$(k_1,\cdots,k_n)$$ is Dirichlet with probability $$(a_1+k_1,\cdots,a_n+k_n)$$. Zabell (1982) gives a nice account of W.E. Johnson’s characterization of the Dirichlet: it is the only prior that predicts outcomes linearly in the past. One special case is the symmetric Dirichlet when all $$a_i=c >0$$. We denote this prior as $$D_c$$.
### $$The\; Birthday\; Problem$$¶
In its simplest version the birthday problem involves $$k$$ balls dropped uniformly at random into $$n$$ boxes. We declare a $$match$$ if two or more balls drop into the same box. Elementary considerations Feller (1968, page 33) show that:
$P(match)= 1-P(no\; matches) =1-\prod^{k-1}_{i=1}(1-\frac{i}{n}) \doteq 1-e^{-\frac{{{k}\choose{2}}}{n}}$
To be more precise:
Proposition: If $$n$$ and $$k$$ are large in such a way that : $$\frac{{{k}\choose{2}}}{n} \longrightarrow \lambda$$ then in the classical birthday problem :
$P(Match) \cong 1-e^{-\lambda}$
Proof :
$\mbox{ Write } \qquad \prod^{k-1}_{i=1}(1-\frac{i}{n})=exp({\sum_{i=1}^{k-1}}log (1-\frac{i}{n}))$
Expand the $$\log$$ using $$log (1-x)=-x+O(x^2)$$ to see that the exponent is
$-\frac{{{k}\choose{2}}}{n} +O(\frac{k^3}{n^2})$
Setting the right side equal to $$\frac{1}{2}$$ and solving for $$k$$ shows that if $$k\doteq 1.2 \sqrt{n}$$ , $$P(match) \doteq \frac{1}{2}$$. When $$n=365$$, $$k=1.2\sqrt{n}\doteq 22.9$$. This gives the usual statement : it is about even odds that there are two people with the same birthday in a group of 23.
In contrast with the clean formulation above, consider
an introductory probability class with 25
students. When considering ‘doing’ the birthday problem, several things come to mind :
• It is slightly better than a 50-50 chance of success with 25 students.
• If it fails it’s quite a disaster, 10 minutes of class time with a big build-up, devoted to a failure.
• If it succeeds, the students are usually captivated and they are interested in learning how computations can dissipate the ‘intuitive fog’ surrounding probability.
• The students are practically all from the same year, it is quite well known that birthdays within a year are not uniformly distributed; far more births occur on weekdays than on weekends (Doctors don’t like to work on weekends, and the induced births and c-sections are never scheduled for weekends). There are also seasonal trends (more births in the summer) and lunar trends.
Taking these things into consideration, you realize that you have no idea what the ‘true underlying probabilities’ are! The considerations above make a match more likely. It seems sensible to carry through a Bayesian version of the birthday problem.
The problem becomes : drop $$k$$ balls into $$n$$ boxes where the chance of a ball falling in box $$i$$ is $$p_i$$. Here $$\tilde{p}=(p_1,p_2,\ldots,p_n)$$ has some prior distribution on the $$n$$-simplex $$\Delta_n$$. We will work with a Dirichlet prior $$D_{\tilde{a}}$$, with $$\tilde{a}=(a_1,a_2,\ldots,a_n)$$. This has density proportional to $$x_1^{a_1-1} x_2^{a_2-1} \cdots x_n^{a_n-1}$$.
For $$a_i=1$$ we get the classical uniform prior on the simplex.
For $$a_i\equiv c$$ we get the symmetric Dirichlet prior.
The Dirichlet prior is the n-dimensional version of the 2-dimensional beta prior we have already studied.
This interpolates between the uniform prior(c=1) and the classical case $$p_i=\frac{1}{365}$$ ($$c\longrightarrow \infty$$). For more general choices of $$a_i$$ we get a flexible family of priors.
Let us we carry out necessary computations in the following cases; in each we give $$k$$ required for a $$50-50$$ chance of a match when n=365:
• Uniform Prior, c=1 $$k\doteq .83 \sqrt{n}$$, for $$n=365$$, $$k\doteq 16$$
• Symmetric Prior, $$a_i=c$$
$\begin{split}\begin{array}{l|ccccccc} c & .5 & 1 & 2 & 5 & 20 & \infty\\ \hline k_c & 13.2 & 16.2 & 18.7 & 20.9 & 21.9 & 22.9\\ \end{array}\end{split}$
In coin tossing the standard prior is the uniform on $$[0,1]$$. The standard prior on the $$n-$$dimensional simplex $$\Delta_n$$ is the uniform distribution $$U$$ where all vectors $$\tilde{p}$$ have same probability.
The probability of a match, averaged over $$(p_1,p_2,\ldots,p_n)$$, represents the chance of success to a Bayesian statistician who has chosen the uniform prior. As is well known, (Bayes (17???), Good (1979), Diaconis and Efron (1987)) such a uniform mixture of multinomials results in Bose-Einstein allocation of balls in boxes, each configuration, or composition $$(k_1,k_2,\ldots,k_n)$$ being equally likely with chance $$\displaystyle{ \frac{1}{{{k+n-1}\choose {k}}}}$$. For this simple prior, it is again possible to do an exact calculation:
Under a uniform prior on $$\Delta_n$$
$\label{eq2.1} P_u(match)=1-\prod_{i=1}^{k-1}(\frac{1-\frac{i}{n}}{1+\frac{i}{n}})$
If $$n$$ and $$k$$ are large in such a way that
$$\frac{k^2}{n}\longrightarrow \lambda$$, then
$\label{eq2.2} P(\mbox{match})\cong 1-e^{-\lambda}$
Proof
Represent the uniform mixture of multinomials using Polya’s urn as described in the introduction. The chance that the first $$k$$ balls fall in different boxes is
$\frac{n-1}{n+1} \times \frac{n-2}{n+2} \cdots \times \frac{n-(k-1)}{n+(k+1)}$
This gives ([eq2.1]) and ([eq2.2]) follows by the same calculus
approximations detailed in the proof of Proposition [prop1].
Thus in order to obtain a 50-50 chance of a match under a uniform prior $$k$$ must be $$.83\sqrt{n}$$. When $$n=365$$, this becomes $$k=16$$, and for $$k=23$$, $$P_u(match)\doteq .75$$.
The uniform prior allows some mass far from $$(\frac{1}{n},\frac{1}{n}, \ldots, \frac{1}{n})$$ and such “lumpy” configurations make a match quite likely.
The uniform prior studied above is a special case of a symmetric Dirichlet prior $$D_c$$ on $$\Delta_n$$, with $$c=1$$. We next extend the calculations above to a general $$c$$. For $$c$$ increasing to infinity, the prior converges to point mass $$\delta_{(\frac{1}{n},\frac{1}{n},\ldots \frac{1}{n})}$$ and thus gives the classical answer. When $$c$$ converges to $$0$$, $$D_c$$ becomes an improper prior giving infinite mass to the corners of the simplex, thus for small $$c$$, the following proposition shows that matches are judged likely when $$k=2$$.
Under a symmetric Dirichlet prior $$D_c$$ on $$\Delta_n$$,
$P_c(\mbox{match})=1-\prod^{k-1}_{i=1} \frac{(n-i)c}{nc+i} .$
For a proof see the paper here
In order for the probability of a match to be about $$\frac{1}{2}$$ ; $$k_c\doteq 1.2\sqrt{\frac{nc}{c+1}}$$ is needed. The following table shows how $$k_c$$ depends on $$c$$ when $$n=365$$:
$\begin{split}\begin{array}{l|ccccccc} c & .5 & 1 & 2 & 5 & 20 & \infty\\ \hline k_c & 13.2 & 16.2 & 18.7 & 20.9 & 21.9 & 22.9\\ \end{array}\end{split}$
• Honest Priors Construct a $$2$$ “hyper”parameter family of Dirichlet priors writing $$a_i= A \pi_i$$, with $$\pi_1+\pi_2\cdots+ \pi_n=1$$. Assign weekdays parameter $$\pi_i=a$$, weekends $$\pi_i=\gamma a$$, with $$260 a +104 \gamma a=1$$. Here $$\gamma$$ is the parameter ‘ratio of weekends to weekdays’, (roughly we said $$\gamma \doteq .7$$) and $$A$$ measures the strength of prior conviction. The table below shows how $$k$$ varies as a function of $$A$$ and $$\gamma$$. We have assumed the year has $$7\times 52=364$$ days.
$\begin{split}\begin{array}{l|ccc} A\;\;\; {\gamma} & .5 & .7 & 1 \\ \hline 1 & 2.2 & 2.2 & 2.2\\ 364 & 16.1& 16.3 & 16.4\\ 728 & 18.4& 18.6& 18.8\\ \infty & 22.2& 22.4& 22.6 \end{array}\end{split}$
$$Remarks$$
1. Under the uniform,symmetric Dirichlet priors for small $$c$$,and ‘honest priors’ for moderate $$A$$ the prior concentrates on fairly ‘lumpy’ vectors $$\tilde{p}$$ which will have matches with small $$k$$.
2. The calculations show that some uncertainty about $$p$$ leads to a much improved chance of a match. Returning to the $$25$$ students in the probability class, the Dirichlet prior with $$A=365$$ and $$\gamma=.7$$ is a first approximation to the prior of the present authors. It leads to a chance of a match with $$25$$ students of 0.81. This may be the reason the birthday problem works as often as it does! | 2021-09-21 09:00:15 | {"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": 0.9532166123390198, "perplexity": 425.4345845164457}, "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/1631780057199.49/warc/CC-MAIN-20210921070944-20210921100944-00624.warc.gz"} |
https://mathoverflow.net/questions/331474/if-x-is-discrete-and-z-w-are-discrete-or-continuous-is-it-always-the-case-t | # If $X$ is discrete and $Z,W$ are discrete or continuous, is it always the case that $P(X=x\mid Z) \geq P(X=x\mid Z,W)$? [closed]
Suppose $$X$$ is discrete and $$Z,W$$ are discrete or continuous, I am wondering if it is always the case (or at least non-trivially) that
$$P(X=x\mid Z) \geq P(X=x\mid Z,W)$$
for all $$x\in X$$.
It appears to make intuitive sense to me because it appears the probability of $$X$$ given $$Z$$ and $$W$$ should be "subsetting" off the probability of $$X$$ given $$Z$$. That is, if there is some chance of $$X$$ given $$Z$$, then $$X$$ given $$Z$$ and $$W$$ is subdividing the occurrence of $$X$$ given $$Z$$ into many more chunks according to $$W$$, and thus the probability of a chunk occurring should be less than the whole.
In other words, it seems to hold in the finite-sampling perspective, where the above are empirical proportions over some collection of objects. However, the result doesn't seem to hold more generally. I am wondering why the finite-sampling intuition doesn't seem to extend. Is there a general result behind this?
## closed as off-topic by Jochen Wengenroth, user44191, Mateusz Kwaśnicki, LSpice, Chris GodsilMay 14 at 12:15
This question appears to be off-topic. The users who voted to close gave these specific reasons:
• "This question does not appear to be about research level mathematics within the scope defined in the help center." – Jochen Wengenroth, Mateusz Kwaśnicki, Chris Godsil
• "MathOverflow is for mathematicians to ask each other questions about their research. See Math.StackExchange to ask general questions in mathematics." – user44191, LSpice
If this question can be reworded to fit the rules in the help center, please edit the question.
• This question is more suitable for MathStackExchange. Did you try the case $Z=0$ (constant random variable) and $W=X$? – Jochen Wengenroth May 14 at 6:54
• You are comparing two probability distributions. They both sum to $1$, so it can't be the case that one of them is bigger everywhere, unless they are the same. – James Martin May 14 at 11:13
No. First of all, $$Z$$ is a red herring here -- just define a new probability space conditional on $$Z$$. Thus your question is really whether $$P(X=x)\ge P(X=x|W=w)$$ always holds. This is easily seen to be false: Let $$(X,W)$$ take on the values $$(0,1)$$ and $$(1,0)$$ with probability $$1/2$$. Then $$P(X=1)=1/2$$ but $$P(X=1|W=1)=0$$. | 2019-05-27 14:31:05 | {"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": 26, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8525502681732178, "perplexity": 305.7877649755263}, "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-2019-22/segments/1558232262600.90/warc/CC-MAIN-20190527125825-20190527151825-00353.warc.gz"} |
https://www.physicsforums.com/threads/calculate-dieter-for-maximum-exit-speed.331671/ | Calculate dieter for maximum exit speed
1. Aug 20, 2009
semc
During my lab i have to roll a marble down a ramp at different angle. I am required to calculate the angle between the ramp and the table so that the marble will achieve the maximum speed when it slide off the ramp.
mgh=1/2 mv^2
V(in x direction) = Vcosx where V is the velocity of the marble and x is the angle of the ramp to the table.
I equate the 2 equations together and gotten V(in x direction) =(cos x )sqrt 2gh
where h=LsinX. So i equate (cosX)sqrt(sinX)=y and differentiate once and equate to 0. The answer i got was X=45 which is wrong. However, if i change the equation to V^2=2gL(sinX)(cosX)^2, the angle i got was arcsin (1/sqrt3) which is about 35. Why is this so??
2. Aug 20, 2009
kuruman
I am not sure I understand the experiment. It seems to me that the steeper you make the angle, the higher the speed will be upon exit. The maximum speed is sqrt(2gh) when the incline is 90o and the ball is in free fall.
3. Aug 20, 2009
semc
The experiment is releasing a marble from an incline plane and the marble will roll along a straight path after it reaches the bottom of the incline plane. At first i also thought that the steeper the fast the velocity but if the angle is at 90degree then will not be a velocity in the x direction hence the marble will not move once it reaches the bottom of the plane
Last edited: Aug 20, 2009
4. Aug 20, 2009
kuruman
OK. So you want to maximize the horizontal component of the speed, not the speed itself. The way you phrased the question initially is confusing. Look at it this way, the speed at the end of the incline is going to be
$$v = \sqrt{2gh} =\sqrt{2gLsin\theta}$$ where L is the length of the ramp.
The horizontal component of the velocity is
$$v_{x} = v cos\theta =(\sqrt{2gLsin\theta})cos\theta$$
So the problem becomes finding the value of θ for which
$$cos\theta \sqrt{sin\theta}$$
is a maximum. Without the radical it would be 45o. With the radical it's something else. Can you find what?
5. Aug 20, 2009
semc
Whats radical? I know the angle is arcsin(1/sqrt3) i just don't understand why i cant use
$$cos\theta \sqrt{sin\theta}$$ and if i use (sinX)(cosX)^2 i can get the arcsin(1/sqrt3)
Last edited: Aug 20, 2009
6. Aug 20, 2009
kuruman
Radical means "the square root symbol." As I said before
$$v=(\sqrt{2gL sin\theta})cos\theta$$
I never said you cannot use this expression, in fact this is the expression you must use. But what is the value for θ for which it has the largest possible value? For θ = 0 it is zero and for θ = 90o it is also zero. So somewhere in between 0 and 90o it must go through a maximum.
7. Aug 21, 2009
semc
I know what the angle is. I am just asking why is it that when i calculate the maximum value for theta using $$v=(\sqrt{ sin\theta})cos\theta$$ the answer is 45 but when i square the equation and solve for theta i can get the correct value for theta. Don't mind going through my working in the file attached and tell me which part i did wrongly.
Attached Files:
• PHYS.jpg
File size:
36.2 KB
Views:
33
Last edited: Aug 21, 2009
8. Aug 21, 2009
rl.bhat
Last edited: Aug 21, 2009
9. Aug 21, 2009
andrevdh
the marble also have rotational kinetic energy
10. Aug 21, 2009
kuruman
Indeed it does. The kinetic energy at the bottom will be
$$K=\frac{1}{2}*\frac{7}{5}m v^{2}$$
Clearly, including the 7/5 correction due to rotational energy reduces the linear speed at the bottom but does not change the angle at which the horizontal component of the linear velocity is a maximum. | 2017-08-21 00:26:55 | {"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": 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": 0.8634251356124878, "perplexity": 546.5167500877305}, "config": {"markdown_headings": false, "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-2017-34/segments/1502886106996.2/warc/CC-MAIN-20170820223702-20170821003702-00096.warc.gz"} |
https://www.physicsforums.com/threads/average-resistance-force-question.268017/ | # Average Resistance Force Question
1. Oct 29, 2008
### alexandray77
Our question to solve is:
A 5.2 g Ball bearing is fired vertically downward from 18m with an initial speed of 14 m/s. It buries itself a depth of 21 cm in the sand. What average resistance force does the sand exert on the ball bearing?
We know the answer is -6.81N but we are having a hard time figuring out which equations to use to solve the problem. Please help! Thanks!
2. Oct 29, 2008
### Rake-MC
You can use either constant acceleration formulae or conservation of energy to solve this problem.
With constant acceleration you are going to want to calculate the velocity of the ball at the instant it hits the ground. You will then have initial and final velocity and displacement. You can solve for a, and use F = ma to solve for F.
With energy, you can use: $$\frac{1}{2}mv_1^2 + mg\Delta h_1 + F_1\Delta d_1 = \frac{1}{2}mv_2^2 + mg\Delta h_1 + F_2\Delta d_2$$
some of these terms will be zero, you are solving for F2
3. Oct 29, 2008
thank you! | 2017-10-19 06:15:11 | {"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": 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": 0.5778791308403015, "perplexity": 572.2357593442273}, "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-2017-43/segments/1508187823229.49/warc/CC-MAIN-20171019050401-20171019070401-00803.warc.gz"} |
https://www.physicsforums.com/threads/simple-pendulum-linear-second-order-equation.201869/ | # Simple Pendulum - Linear Second-Order Equation
1. Dec 1, 2007
### sourlemon
1. The problem statement, all variables and given/known data
In section 4.7, we discussed the simple pendulum consisting of a mass m suspended by a rod of length l having negligible mass and derived the nonlinear initial value problem.
(17) $$\frac{d^{2}\phi}{dt^{2}} + \frac{g}{l}sin\theta=0$$
$$\theta(0) = \alpha$$
$$\theta'(0) = 0$$
where g is the acceleration due to gravity and $$\theta(t)$$ is the angle of the rod makes with the vertical at time t. Here it is assumed that the mass is released with zero velocity at an initial angle $$\alpha, 0 < \alpha < \pi$$. We would like to determine the equation of motion for the pendulum and its period of oscillation.
(a) Use equation (17) and the energy integral lemma discussed in Section 4.7 to show that
$$\left(\frac{d\theta}{dt}\right)^{2} = \frac{2g}{l}(cos\theta - cos\alpha)$$
and hence
$$dt = - \sqrt{\frac{l}{2g}}\frac{d\theta}{\sqrt{cos\theta - cos\alpha}}$$
(b) Use the trigonometric identity $$cos x = 1 - 2sin^{2}(\frac{x}{2})$$ to express dt by
$$dt = - \frac{l}{2}\sqrt{\frac{l}{g}}\frac{d\theta}{\sqrt{sin^{2}(\alpha/2) - sin^{2}(\theta/2)}}$$
(c)Make the change of variables $$sin(\theta/2) = sin (\alpha/2)sin\phi$$ to rewrite dt in the form
$$dt = - \sqrt{\frac{l}{g}}\frac{d\phi}{\sqrt{1 - k^{2}sin^{2}(\phi/2)}}$$
where $$k = sin(\alpha/2)$$.
(d) The period $$P(\alpha)$$ of the pendulum is defined to be the time required for the pendulum to swing from one extreme to the other and back - that is, from $$\alpha to -\alpha back to \alpha$$. Show that the period of oscillation is given by
$$(19) P(\alpha) = 4\sqrt{\frac{l}{g}}\int\frac{d\phi}{\sqrt{1 - k^{2}sin^{2}(\phi/2)}}$$
where $$k = sin(\alpha/2)$$. [Hint: The period is just four times the time it takes the pendulum to go from $$\theta = 0 to \theta = \alpha$$]
The integral in (19) is called an elliptic integral of the first kind and is denoted by $$F(k, \pi/2$$. As you might expect, the period of the simple pendulum depends on the length "l" of the rod and the initial displacement $$\alpha$$. In fact, a check of an elliptic integral table will show that the period nearly doubles as the initial displacement increases from $$\pi/8$$ to $$15\pi/16$$ (for fixed "l"). What happens as $$\alpha$$ approaches $$\pi$$
2. Relevant equations
Lemma 3.
Let y(t) be a solution to the differential equation
(7) y" = f(y)
where f(y) is a continuous function that does not depend on y' or the independent variable t. Let F(y) be an indefinite integral of f(y), that is
$$f(y) = \frac{d}{dy}F(y).$$
Then the quantity
(8) $$E(t) = \frac{1}{2}y't^{2} - F(y(t))$$
is constant; i.e.,
(9) $$\frac{d}{dt}E(t) = 0$$.
3. The attempt at a solution
Lolz writing this problem alone takes forever. There's actually a part e, but I figure that out, so I didn't post it. As for part a and b, I got that too. I'm only having problem with part c and d.
(c)Make the change of variables $$sin(\theta/2) = sin (\alpha/2)sin\phi$$ to rewrite dt in the from
$$dt = - \frac{l}{2}\sqrt{\frac{l}{g}}\frac{d\theta}{\sqrt{sin^{2}(\alpha/2) - sin^{2}(\theta/2)}}$$
to
$$dt = - \sqrt{\frac{l}{g}}\frac{d\phi}{\sqrt{1 - k^{2}sin^{2}(\phi/2)}}$$
where $$k = sin(\alpha/2)$$.
I'm not sure yet how to start d, but I've start c. I think my problem is changing $$d\theta$$ to $$d\phi$$.
My teacher said to take the derivative of both. So
$$sin(\theta/2) = sin (\alpha/2)sin\phi$$
becomes
$$1/2cos(\theta/2) = sin(\alpha/2)cos\phi$$
Since I'm changing $$d\theta$$ to $$d\phi$$, I took the assumption that $$sin(\alpha/2)$$is a constant.
Plugging that in the equation I get
$$dt = - \frac{l}{2}\sqrt{\frac{l}{g}}\frac{1}{2}\frac{cos(\theta/2)}{\sqrt{sin^{2}(\alpha/2) - sin^{2}(\alpha/2)sin^{2}\phi}}$$
In part b, $$cos x = 1 - 2sin^{2}(\frac{x}{2})$$, so I figure I can try to do that too. So
$$cos(\theta/2) = 1 - 2sin^{2}(\frac{\theta}{4})$$
But how can I change $$1 - 2sin^{2}(\frac{\theta}{4})$$ to something with $$\alpha$$ or $$\phi$$? I can't figure out what triognometric properties I can use.
Last edited: Dec 1, 2007
2. Dec 2, 2007
### siddharth
That's not right. What happened to the derivative? You should get,
$$\frac{1}{2} \cos{\frac{\theta}{2}} \left(\frac{d \theta}{d \phi}\right) = \sin {\frac{\alpha}{2}} \cos \phi$$
You need to substitute for $d\theta$ and $\sin\frac{\theta}{2}}$ in the equation.
3. Dec 2, 2007
### sourlemon
Lolz, thank you for not running away after seeing my post. Lolz I feel so dumb. What a careless mistake. Thank you so much siddharth!!!
I have c solved :D now I must tackle d :(
From part c, I have the equation
$$dt = - \sqrt{\frac{l}{g}}\frac{d\phi}{\sqrt{1 - k^{2}sin^{2}(\phi/2)}}$$
where $$k = sin(\alpha/2)$$.
So integrating both sides to have the period, I would have
$$P(\alpha) = \sqrt{\frac{l}{g}}\int\frac{d\phi}{\sqrt{1 - k^{2}sin^{2}(\phi/2)}}$$
Since $$\sqrt{\frac{l}{g}}$$ is constant, I can move that out of the integral.
I am little confuse on how you determine that the integration is from 0 to $$\frac{\pi}{2}$$. What I figure is depending on where the pendulum is release, the most it can make is from 0 to $$\pi$$. But for a period, the pendulum needs to swing from $$\alpha$$ to $$-\alpha$$ back to $$\alpha$$. So would it make sense if I integrate it from 0 to $$\pi$$ and multiply it by 2?........I'm a bit stuck here. I can see that if you release the pendulum from 0, it's the same as it is release from $$\pi$$, so with that, can I conclude that from 0 to $$\frac{\pi}{2}$$ is the same as $$\frac{\pi}{2}$$ to $$\pi$$. That's why you can say that the integral is from 0 to $$\frac{\pi}{2}$$, multiply by 4?
If the $$\alpha$$ from 0 to $$\frac{\pi}{2}$$, then the period decrease. But if $$\alpha$$ is from $$\frac{\pi}{2}$$ to $$\pi$$, then the period increase. Is that right?
Last edited: Dec 2, 2007
4. Dec 6, 2007
### sourlemon
*sigh* turned out I missed a lot of things. I mistyped in the first post, it's suppose to be $$d\theta$$ not $$d\phi$$.
Let's start from the beginning.
$$(17) \frac{d^{2}\theta}{dt^{2}} + \frac{g}{l}sin\theta=0$$
Rearranging the equation,
$$\frac{d^{2}\theta}{dt^{2}} = - \frac{g}{l}sin\theta = \theta'' = f(\theta)$$
Following the lemma energy stated above,
$$f(\theta) = \frac{d}{d\theta}F(\theta) = \frac{d}{d\theta}(\frac{g}{l}cos\theta + C)$$
$$E(t)= \frac{1}{2}\theta'(t)^{2} - F(\theta(t))$$
Plugging the known data $$\theta(0) = \alpha$$ and $$\theta'(0) = 0$$
$$E(t) = \frac{1}{2}(0)^{2} - \alpha = - \alpha$$
$$E(t)= \frac{1}{2}\theta'(t)^{2} - F(\theta(t)) = \frac{1}{2}\theta'(t)^{2} - \frac{g}{l}cos\theta = - \alpha$$
$$\frac{1}{2}\theta'(t)^{2} = \frac{g}{l}cos\theta - \alpha$$
$$\theta'(t)^{2} = 2\left(\frac{g}{l}cos\theta - \alpha\right) =$$
$$\theta'(t)^{2} = 2\frac{g}{l}\left(cos\theta - \frac{g}{l}\alpha\right)$$
Did I miss something? I want to end up with the equation below.
$$\left(\frac{d\theta}{dt}\right)^{2} = \frac{2g}{l}(cos\theta - cos\alpha)$$
Last edited: Dec 6, 2007 | 2017-03-30 15:07:42 | {"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": 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": 0.9408934116363525, "perplexity": 311.0060248512156}, "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-2017-13/segments/1490218194601.22/warc/CC-MAIN-20170322212954-00246-ip-10-233-31-227.ec2.internal.warc.gz"} |
http://programming-puzzler.blogspot.de/ | # The Obligatory Sudoku Example
No discussion about constraint solvers is complete without the obligatory sudoku example. Unfortunately, sudoku is such a basic exercise for a constraint solver that it doesn’t really tell you much about the engine. But if anything, because sudoku is so standard, it has become a good way to get a feel for the style of the constraint DSL.
The Loco (0.2.0) version is concise and readable, with a nice separation between the core set of constraints that underlie all sudoku puzzles, versus the additional starter numbers provided by the puzzle.
In most constraint solvers, the constraint operators imperatively impose constraints on a variable store. In Loco, the constraint operators simply produce Clojure data structures. Since we’re working with Clojure data, assembling the full model is simply a matter of concatenating the sequence of constraints describing the fundamental rules of sudoku with the sequence of constraints specific to a given puzzle. This is a good example of how Loco lets you easily build portions of your model separately and then combine them or otherwise manipulate them with standard Clojure functions.
To follow the example, the main thing you need to know about Loco is that the notion of a subscripted variable, like gridi,j$grid_{i,j}$, is represented in Loco as a vector, e.g., [:grid i j]
First, here’s how we’re going to ultimately input the Sudoku grids to our solver:
;http://www.mirror.co.uk/news/weird-news/worlds-hardest-sudoku-puzzle-ever-942299
(def worlds-hardest-puzzle
[[8 - - - - - - - -]
[- - 3 6 - - - - -]
[- 7 - - 9 - 2 - -]
[- 5 - - - 7 - - -]
[- - - - 4 5 7 - -]
[- - - 1 - - - 3 -]
[- - 1 - - - - 6 8]
[- - 8 5 - - - 1 -]
[- 9 - - - - 4 - -]])
Here’s the solving code:
(def basic-model
(concat
; range-constraints
(for [i (range 9) j (range 9)]
($in [:grid i j] 1 9)), ; row-constraints (for [i (range 9)] ($distinct (for [j (range 9)] [:grid i j]))),
; col-constraints
(for [j (range 9)]
($distinct (for [i (range 9)] [:grid i j]))), ; section-constraints (for [section1 [[0 1 2] [3 4 5] [6 7 8]] section2 [[0 1 2] [3 4 5] [6 7 8]]] ($distinct (for [i section1, j section2] [:grid i j])))))
(defn solve-sudoku [grid]
(solution
(concat basic-model
(for [i (range 9), j (range 9)
:let [hint (get-in grid [i j])]
:when (number? hint)]
($= [:grid i j] hint))))) Let’s test it out in the REPL: => (solve-sudoku worlds-hardest-puzzle) {[:grid 4 0] 3, [:grid 5 1] 8, [:grid 6 2] 1, [:grid 7 3] 5, [:grid 8 4] 1, [:grid 3 0] 1, [:grid 4 1] 6, [:grid 5 2] 7, [:grid 6 3] 9, [:grid 7 4] 2, [:grid 8 5] 8, [:grid 2 0] 6, [:grid 3 1] 5, [:grid 4 2] 9, [:grid 5 3] 1, [:grid 6 4] 7, [:grid 7 5] 6, [:grid 8 6] 4, [:grid 1 0] 9, [:grid 2 1] 7, [:grid 3 2] 4, [:grid 4 3] 8, [:grid 5 4] 6, [:grid 6 5] 4, [:grid 7 6] 9, [:grid 8 7] 5, [:grid 0 0] 8, [:grid 1 1] 4, [:grid 2 2] 5, [:grid 3 3] 2, [:grid 4 4] 4, [:grid 5 5] 9, [:grid 6 6] 3, [:grid 7 7] 1, [:grid 8 8] 2, [:grid 0 1] 1, [:grid 1 2] 3, [:grid 2 3] 4, [:grid 3 4] 3, [:grid 4 5] 5, [:grid 5 6] 5, [:grid 6 7] 6, [:grid 7 8] 7, [:grid 0 2] 2, [:grid 1 3] 6, [:grid 2 4] 9, [:grid 3 5] 7, [:grid 4 6] 7, [:grid 5 7] 3, [:grid 6 8] 8, [:grid 0 3] 7, [:grid 1 4] 8, [:grid 2 5] 1, [:grid 3 6] 8, [:grid 4 7] 2, [:grid 5 8] 4, [:grid 0 4] 5, [:grid 1 5] 2, [:grid 2 6] 2, [:grid 3 7] 9, [:grid 4 8] 1, [:grid 0 5] 3, [:grid 1 6] 1, [:grid 2 7] 8, [:grid 3 8] 6, [:grid 0 6] 6, [:grid 1 7] 7, [:grid 2 8] 3, [:grid 0 7] 4, [:grid 1 8] 5, [:grid 0 8] 9, [:grid 8 0] 7, [:grid 7 0] 4, [:grid 8 1] 9, [:grid 6 0] 5, [:grid 7 1] 3, [:grid 8 2] 6, [:grid 5 0] 2, [:grid 6 1] 2, [:grid 7 2] 8, [:grid 8 3] 3} On my machine, this “hardest Sudoku” takes about 17ms to solve. Benchmarking against your favorite constraint solver on your machine, and pretty-printing the output as a readable grid are left as an exercise for the reader. ## Wednesday, March 5, 2014 ### Appointment scheduling in Clojure with Loco Loco makes it easy to declaratively build constraint satisfaction models. In this blog post, we’ll look at a common use for constraint programming – appointment scheduling – and in so doing, we’ll see some of the ways that Loco goes beyond the features found in other Clojure constraint libraries. (use 'loco.core 'loco.constraints) # Scheduling appointments with no conflicts Imagine you have four people coming in for an interview, and you’ve set aside four timeslots in your day to conduct these interviews. You ask each person to list the timeslots when he/she can potentially come in. Let’s use 0-based indexing to refer to the people, and 1-based indexing to refer to the timeslots. Person 0 says she can come in at any of the four timeslots: 1, 2, 3, or 4. Person 1 says he can come in at timeslot 2 or 3. Person 2 says she can come in at timeslot 1 or 4. Person 3 says he can come in at timeslot 1 or 4. So the availability data looks like this: (def availability [[1 2 3 4] [2 3] [1 4] [1 4]]) Let the variable [:person 0] denote the timeslot when person 0 is scheduled to come in, [:person 1] when person 1 comes in, etc. (def person-vars (for [i (range (count availability))] [:person i])) We want to constrain each [:person i] variable to the available timeslots. (def availability-constraints (for [i (range (count availability))] ($in [:person i] (availability i))))
We want to ensure we don’t schedule two people in the same timeslot.
(def all-different-constraint
(apply $all-different? person-vars)) For convenience, let’s assemble the constraints into one big list (the order doesn’t matter in Loco): (def all-constraints (conj availability-constraints all-different-constraint)) Now we’re ready to solve. Let’s dump the solution into a sorted-map for easy readability. => (into (sorted-map) (solution all-constraints)) {[:person 0] 3, [:person 1] 2, [:person 2] 4, [:person 3] 1} So there you have it. Once we’ve played around with this example interactively in the REPL, and are confident in the model, we can easily abstract this into a function that takes availability data, and returns the schedule: (defn schedule [availability] (->> (solution (conj (for [i (range (count availability))] ($in [:person i] (availability i)))
($distinct (for [i (range (count availability))] [:person i])))) (into (sorted-map)))) => (schedule [[1 3 5] [2 4 5] [1 3 4] [2 3 4] [3 4 5]]) {[:person 0] 5, [:person 1] 4, [:person 2] 1, [:person 3] 2, [:person 4] 3} I think the declarative Loco way of modeling constraints is concise and elegant, but this example could just as easily be done in, say, core.logic. So let’s push beyond, into an area that (as far as I know) can’t be done with core.logic. # Scheduling appointments minimizing conflicts The above scheduler is somewhat naive. => (schedule [[1 2 3 4] [1 4] [1 4] [1 4]]) {} This doesn’t work because there’s no way to satisfy the constraint that no two people can be scheduled in the same timeslot. But let’s say, hypothetically, that if absolutely necessary, we can potentially squeeze two candidates into the same timeslot. We’d rather not, but we can do it if we have to. Can we build a model for this? Again, let’s start exploring the problem interactively with global defs and playing around with it at the REPL. Here’s the problematic availability example: (def availability [[1 2 3 4] [1 4] [1 4] [1 4]]) As before, we’ll want to constraint each person’s timeslot to his/her availability schedule: (def availability-constraints (for [i (range (count availability))] ($in [:person i] (availability i))))
Let’s define a few names for convenience. Let timeslots be a list of all the timeslot numbers.
(def timeslots (distinct (apply concat availability)))
Let person-vars be the list of all [:person i] variables.
(def person-vars
(for [i (range (count availability))] [:person i]))
Now for the interesting part. We want to allow up to 2 people in a given timeslot. So we’ll let the variable [:num-people-in-timeslot 1] be the number of people signed up for timeslot 1, and so on. Let people-in-timeslot-vars be the list of all such variables.
(def people-in-timeslot-vars
(for [i timeslots] [:num-people-in-timeslot i]))
Now, we create a list of constraints that state that each of these [:num-people-in-timeslot i] variables ranges between 0 and 2.
(def conflict-constraints
(for [i timeslots]
($in [:num-people-in-timeslot i] 0 2))) To give these :num-people-in-timeslot variables the appropriate meaning, we need to bind each [:num-people-in-timeslot i] variable to the number of times i occurs among the variables [:person 1], [:person 2], etc. Loco’s $cardinality constraint allows us to do exactly that. For example,
($cardinality [:x :y :z] {1 :number-of-ones}) will bind :number-of-ones to the number of times 1 occurs among :x, :y, and :z. So, the following constraint will bind all the [:num-people-in-timeslot i] variables to their appropriate values. (def number-in-timeslots ($cardinality person-vars
(zipmap timeslots people-in-timeslot-vars)))
To minimize the number of conflicts, we need to count the number of conflicts.
Let the variable :number-of-conflicts stand for the number of timeslot conflicts we have. We need two constraints on :number-of-conflicts. The first constraint just sets up the finite domain that the variable could range over (i.e., 0 to the total number of timeslots). We need to do this because in Loco, every variable must be declared somewhere in the model. The second constraint binds :number-of-conflicts to the number of times 2 appears in the variables [:num-people-in-timeslot 1], [:num-people-in-timeslot 2], etc.
(def number-of-conflicts
[($in :number-of-conflicts 0 (count timeslots)) ($cardinality people-in-timeslot-vars {2 :number-of-conflicts})])
We built the constraints in parts; now building the model is simply a matter of concatting all the constraints together. (Note that number-in-timeslots is a single constraint, so we concatenate [number-in-timeslots] in with the other lists of constraints).
(def all-constraints (concat availability-constraints
conflict-constraints
[number-in-timeslots]
number-of-conflicts))
Now, we’re all set up to solve the model.
=> (solution all-constraints
:minimize :number-of-conflicts)
{[:person 0] 2,
[:person 1] 4,
[:person 2] 4,
[:person 3] 1,
:number-of-conflicts 1,
[:num-people-in-timeslot 1] 1,
[:num-people-in-timeslot 2] 1,
[:num-people-in-timeslot 3] 0,
[:num-people-in-timeslot 4] 2}
In the final version, we really only want to see the [:person i] variables; Loco allows us to hide the other variables from the output by prepending an underscore character in front of the variable names.
So let’s abstract this into a more robust schedule-with-conflicts function.
(defn schedule-with-conflicts [availability]
(let [timeslots (distinct (apply concat availability)),
availability-constraints
(for [i (range (count availability))]
($in [:person i] (availability i))), person-vars (for [i (range (count availability))] [:person i]), people-in-timeslot-vars (for [i timeslots] [:_num-people-in-timeslot i]), conflict-constraints (for [i timeslots] ($in [:_num-people-in-timeslot i] 0 2)),
number-in-timeslots
($cardinality person-vars (zipmap timeslots people-in-timeslot-vars)), number-of-conflicts [($in :_number-of-conflicts 0 (count timeslots))
($cardinality people-in-timeslot-vars {2 :_number-of-conflicts})] all-constraints (concat availability-constraints conflict-constraints [number-in-timeslots] number-of-conflicts)] (into (sorted-map) (solution all-constraints :minimize :_number-of-conflicts)))) Let’s give it a spin: => (schedule-with-conflicts [[1 2 3 4] [1 4] [1 4] [1 4]]) {[:person 0] 2, [:person 1] 4, [:person 2] 4, [:person 3] 1} Written with StackEdit. ### Optimization with Loco I had the pleasure of beta-testing Loco, which was announced today. Loco is a Clojure constraint solver library built on top of the Java library Choco. There are several constraint libraries available for Clojure, including core.logic, propaganda, and clocop, each with a slightly different focus. The features that make Loco shine are the performance of the constraint engine, the fully declarative API, the ease with which one can build models, support for several interesting global constraints, and the ability to find an optimal solution for models with multiple solutions. This is the first article of what I hope to be a series, detailing some of the interesting problems you can solve with Loco. # Scheduling Buses with Loco (use 'loco.core loco.constraints) Loco is a powerful and expressive constraint solver, but it can also be used to solve certain kinds of integer linear programs. One classic example is bus scheduling. Imagine that we are city transportation planners and we want to minimize the number of buses we need to operate in order to meet demand. We know the number of buses demanded for four-hour blocks of time. (def demands {:12am-4am 8 :4am-8am 10 :8am-12pm 7 :12pm-4pm 12 :4pm-8pm 4 :8pm-12am 4}) So for example, this map tells us that there is sufficient demand for 8 buses operating between 12am and 4am. The interesting twist is that buses operate for 8 hours at a time. So, if we set a bus into operation at 12am, it operates an 8-hour shift from 12am-8am. So the question is, how many buses do we need to run from 12am-8am, and from 4am-12pm, etc. in order to meet the above demands. We can represent this by a series of variables, each of which must be an integer from 0 through 12 (since 12 is the maximum overall demand). So with loco, we can get the solution quite simply: (solution [($in :bus-12am-8am 0 12)
($in :bus-4am-12pm 0 12) ($in :bus-8am-4pm 0 12)
($in :bus-12pm-8pm 0 12) ($in :bus-4pm-12am 0 12)
($in :bus-8pm-4am 0 12) ($>= ($+ :bus-8pm-4am :bus-12am-8am) (demands :12am-4am)) ($>= ($+ :bus-12am-8am :bus-4am-12pm) (demands :4am-8am)) ($>= ($+ :bus-4am-12pm :bus-8am-4pm) (demands :8am-12pm)) ($>= ($+ :bus-8am-4pm :bus-12pm-8pm) (demands :12pm-4pm)) ($>= ($+ :bus-12pm-8pm :bus-4pm-12am) (demands :4pm-8pm)) ($>= ($+ :bus-4pm-12am :bus-8pm-4am) (demands :8pm-12am))] :minimize ($+ :bus-12am-8am :bus-4am-12pm :bus-8am-4pm
:bus-12pm-8pm :bus-4pm-12am :bus-8pm-4am))
which yields
{:bus-8pm-4am 0, :bus-4pm-12am 4, :bus-12pm-8pm 7,
:bus-8am-4pm 5, :bus-4am-12pm 2, :bus-12am-8am 8}
Let’s see if we can generalize this to handle an arbitrary number of evenly-spaced time periods. Clearly, to do this we’ll need to get away from demands and variables that directly name the timespan. Instead, for our example in which we sliced the day into six 4-hour time periods, we can imagine indexing these blocks of time (0-based) as “Time period 0” through “Time period 5”. So we can just use a vector for our demands, for example:
[8 10 7 12 4 4]
means that 8 buses are demanded for time period “0” (corresponding to 12am-4am), 10 buses are demanded for time period “1” (corresponding to 4am-8am),… up to a demand of 5 buses for time period “5”.
We’ll make use of Loco’s ability to treat vectors as subscripted variables. So [:buses 0] denotes $buses_0$, which is our variable for how many buses we schedule starting at the beginning of time period 0 (i.e., 12am). [:buses 1] (i.e., $buses_1$) is the variable for the number of buses starting at the beginning of time period 1, etc.
We will also need an input, span which indicates how many consecutive time periods are spanned by the bus’s shift. In our example, span would be 2 (because a bus works for 2 of our 4-hour time periods).
(defn minimize-buses
"Takes a vector of the demands for any number of equally-spaced time slots.
span is the number of time slots that a bus's operating time spans"
[demands span]
(let [time-slots (count demands),
max-demand (apply max demands),
declarations
(for [i (range time-slots)]
($in [:buses i] 0 max-demand)) constraints (for [i (range time-slots)] ($>=
(apply $+ (for [j (range (inc (- i span)) (inc i))] [:buses (mod j time-slots)])) (demands i)))] (solution (concat declarations constraints) :minimize (apply$+ (for [i (range time-slots)] [:buses i])))))
Let’s test it out on our original sample demand:
=> (minimize-buses [8 10 7 12 4 4] 2)
{[:buses 5] 0, [:buses 4] 4, [:buses 3] 7, [:buses 2] 5, [:buses 1] 2, [:buses 0] 8}
Hmmm, it’s a little hard to read. We can fix that:
=> (into (sorted-map) (minimize-buses [8 10 7 12 4 4] 2))
{[:buses 0] 8, [:buses 1] 2, [:buses 2] 5, [:buses 3] 7, [:buses 4] 4, [:buses 5] 0}
Good, same answer as before. But now we can easily adjust to alternative demand schedules. For example, here’s a solution for a demand schedule based on 2-hour time periods, while buses still work 8-hour shifts:
=> (into (sorted-map)
(minimize-buses [1 5 7 9 11 12 18 17 15 13 4 2] 4))
{[:buses 0] 0, [:buses 1] 1, [:buses 2] 2, [:buses 3] 6, [:buses 4] 2,
[:buses 5] 2, [:buses 6] 8, [:buses 7] 5, [:buses 8] 0, [:buses 9] 0,
[:buses 10] 0, [:buses 11] 4}
Now, let’s try a demand schedule with 1-hour time periods, with buses working 8-hour shifts:
=> (into (sorted-map)
(minimize-buses
[1 3 5 7 9 11 12 13 14 15 16 19
18 17 15 13 15 16 10 8 6 5 4 2]
8))
Uh oh, this seems to run forever. We can fix this with the timeout feature. In the definition of minimize-buses, we change the call to solution as follows:
(solution
(concat declarations constraints [constraint])
:minimize (apply \$+ (for [i (range time-slots)] [:bus i]))
:timeout 1000)
The :timeout keyword specifies a number of milliseconds, after which the solver should return the best solution it has found so far:
=> (into (sorted-map)
(minimize-buses
[1 3 5 7 9 11 12 13 14 15 16 19
18 17 15 13 15 16 10 8 6 5 4 2]
8))
{[:bus 0] 1, [:bus 1] 2, [:bus 2] 2, [:bus 3] 2, [:bus 4] 2, [:bus 5] 2,
[:bus 6] 1, [:bus 7] 1, [:bus 8] 2, [:bus 9] 3, [:bus 10] 3, [:bus 11] 5,
[:bus 12] 1, [:bus 13] 1, [:bus 14] 2, [:bus 15] 2, [:bus 16] 2, [:bus 17] 0,
[:bus 18] 0, [:bus 19] 0, [:bus 20] 0, [:bus 21] 0, [:bus 22] 0, [:bus 23] 0}
Written with StackEdit.
# Frustrations with namespaces in Clojure
I have a long laundry list of things about Clojure I find irritating. That shouldn’t be too surprising; I have a similar list for every language that I have used to a serious degree. Most of the things on my list are relatively minor gripes and, overall, Clojure remains one of the most enjoyable programming languages I’ve worked with. Namespaces, however, are one of the aspects of Clojure that cause me the greatest pain.
Namespaces are Clojure’s tool for preventing name collisions between files. The most common way to organize your files is with a one-namespace-per-file policy, where the namespace matches the folder-file hierarchy. For example puzzler.sudoku.grid would be the namespace for the grid.clj file in a puzzler/sudoku directory in the classpath. A namespace can require other namespaces, which ensures that the code in those namespaces is loaded and accessible.
It’s a simple enough scheme, but in practice, I find there are several problems with namespaces, as implemented in Clojure.
## No circular dependencies
Most programs start off small. At the beginning, it’s pretty easy to break the program cleanly into a few files with a clear, acyclic dependency hierarchy. But then your program grows. And when that happens, Clojure’s inability to handle namespaces with circular dependencies will cause you pain, as you struggle to refactor your entire codebase.
Here are the most common scenarios where I’ve got bitten by this:
1. Namespace b depends on namespace a. I’m adding a function foo to a and realize that I would really benefit from using a function bar that already exists in b. What to do? One option is to move the function from b into a. If I do this, I have to make sure to also move over everything in b that bar uses in its implementation. Next, I need to check every one of my namespaces that are downstream from b and use any of the moved functions, and add a dependency on a, and update any references to b/bar (or other moved functions) to a/bar.
Another option is to try to figure out if there’s a way to move portions of code from b and a into a new namespace c, where both b and a will now depend on c. If you do this, you might be able to get away with moving only the underlying private helper functions. If you can avoid moving your public functions, that at least saves you the hassle of changing all the downstream consumers of the API to point to the new locations.
Keep in mind that none of the Clojure IDEs are sophisticated enough to help with intelligently handling these sorts of refactorings. The really sad part is that sometimes the hassle factor is so huge, I feel tempted to just copy and paste function bar over to a so it exists in both locations. It makes me want to curse at Clojure just for making me consider such a horrible thing.
2. Records and protocols cause tremendous problems with the circular dependency restriction. Naturally, your records and protocols are something you want to use throughout your project. So it makes sense to have a namespace such as myproject.types to put these common records and protocols in one place. For efficiency, it is common to implement some of the protocols inline in the record definitions. However, implementing the protocols can at times be very complex, so in those inline implementations, we want to be able to call helper functions that handle the complex implementation details. Namespaces are all about organization, so it’s reasonable to want to implement those helper functions in another namespace. But therein lies the problem. myproject.types depends on myproject.protocol-implementation-details which in turn depends on myproject.types (for example, if you want to implement push on a Stack you need to be able to return a new Stack, so you need that Stack record in scope).
A number of Clojure programmers have responded to this constraint by simply keeping the bulk of their code in one monolithic namespace. It is possible to split one namespace across multiple files, but then you’ve lost out on one of the desirable properties of namespaces — to know where to look for the definition of a function from the namespace/function reference.
## No standard mechanism for re-providing required dependencies
The need for this manifests in several contexts:
1. Creating a public API relying on functions spread across other namespaces. The simplest technique is to just create new vars that refer to the ones in the other namespaces, for example, (def public-foo private/foo). The main problem is that public-foo metadata won’t match the metadata of private/foo.
The best library for dealing with this is potemkin, which handles the details of migrating the metdata over to the new var. This is an important enough issue, there should be a standard solution built into Clojure.
2. When dealing with a complex project built out of multiple libraries and namespaces, it’s easy to end up with massively long headers, listing out dozens of dependencies. We need some way to group related dependencies into something that can be conveniently required as a single entity.
In the early days of Clojure, Konrad Hinsen did some interesting work building some ns libraries supporting the notion of “namespace cloning” and other related ideas. I don’t know whether these tools are maintained these days, but again, it would be great to see a solution incorporated into Clojure.
3. In the discussion above about breaking up circular dependencies, I talked about how much of the pain revolves around needing to update all the references to a function that has been moved from one namespace to another. If there were a convenient way to leave behind a link to the new location, it wouldn’t be quite so painful. Again, potemkin is currently the way to deal with this.
## No parameterized namespaces
My first two complaints shouldn’t come as any surprise to anyone who has built a complex project in Clojure. But this last point is more subtle.
To illustrate the point, let’s imagine your boss tasks you with writing a program to generate sudoku puzzles. As you code your sudoku generator, it’s obvious that certain numbers show up again and again. Remembering the rule-of-thumb “no magic constants”, you decide to create some definitions at the top of your namespace:
;; Constants
(def section-size 3)
(def symbols (range 1 10)) ; might want to use something other than numbers
(def grid-size (* section-size section-size)) ; derived constant
Imagine that these constants are used throughout the many functions in your sudoku namespace.
Later your boss comes along and says he wants you to generate a bunch of 4x4 sudokus, using letters instead of numbers. No problem, you just change the constants and recompile, right?
But now, let’s imagine that you need to do a task that involves a mixture of generating 9x9 and 4x4 sudokus. Suddenly, you have a major problem.
Probably most Clojurians would quickly rework the main constants as follows:
;; Constants
(def ^:dynamic section-size 3)
(def ^:dynamic symbols (range 1 10))
And what to do about the derived constant? Possibly you could use a symbol macro (see clojure.tools.macro), or maybe rework it into a function and change all uses of the constant to a function call.
(defn grid-size [] (* section-size section-size))
; Change all uses of grid-size to (grid-size)
Then, you use your generator to generate five of each type of puzzle as follows:
(concat
(sudoku/generate-puzzles 5)
(binding [sudoku/section-size 2, sudoku/symbols [\A \B \C \D]]
(sudoku/generate-puzzles 5)))
The problem here is that binding almost always leads towards more pain. Inevitably, you forget that, for example, generate-puzzles returns a lazy sequence, so when the sequence is realized, the binding no longer holds, and the vars revert back to their defaults, so you just get ten 9x9 puzzles.
binding simply isn’t a reliable way to parameterize your namespace, i.e., to customize your namespace’s “global variables” that all the functions rely upon.
How do other languages deal with this?
Classes can be thought of as a very richly-featured namespace. Even if you don’t use inheritance and mutable state, classes make great namespaces. In an OO language, the sudoku namespace could be a class, with section-size and symbols as properties/fields set by the class constructor. The functions in the class all “see” those fields.
This means that you can think of objects as special instances of this parameterized namespace. For example, (ab)using a mixture of java and clojure syntax, imagine:
sudoku4 = new Sudoku(2, [\A \B \C \D])
sudoku9 = new Sudoku(3, (range 1 10))
(concat (sudoku4.generate-puzzles 5) (sudoku9.generate-puzzles 5))
There’s no good way to do this in Clojure. Really, your only option is to explicitly pass these parameters in and out of every function in your namespace. To do this, you’d need to change every single one of your functions to take an additional input of the form:
{section-size :section-size symbols :symbols :as grid-options}
and passing grid-options to every function call.
This also changes the calling convention for all downstream consumers of this API.
This is a disaster. The problem is that projects usually start off just like this, with a few “global definitions” that later turn into actual parameters as the project evolves. binding is not an adequate solution because it interacts poorly with laziness. Threading the values through the functions can be a major ordeal. We need some way to produce a namespace that is derived from another namespace with certain parameters set in a given way.
I’m not an ML expert, but I believe that ML functors are an example of parameterized namespaces in the functional language world. Perhaps Clojure could draw inspiration from this.
## Summary
All of these issues have one thing in common: part of Clojure’s appeal is that it creates an environment where you can dive in and start creating without having every detail of your final design planned out. Clojure lets you start simply with small projects and evolve a more complex structure as necessary. However, the details of how namespaces work means that as your project evolves, I inevitably hit a point where new additions become painful to make due to extensive refactoring. I find that when this happens, I’m psychologically steered towards avoiding those new additions, and that’s no good.
In order to create an environment where projects scale more seamlessly from small to large, Clojure’s namespaces are desperately in need of richer features. | 2018-03-20 11:34:25 | {"extraction_info": {"found_math": true, "script_math_tex": 1, "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": 2, "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": 0.4773685038089752, "perplexity": 2474.4048694037538}, "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-2018-13/segments/1521257647406.46/warc/CC-MAIN-20180320111412-20180320131412-00355.warc.gz"} |
https://proofwiki.org/wiki/Anomalous_Cancellation_on_2-Digit_Numbers | # Anomalous Cancellation on 2-Digit Numbers
## Theorem
There are exactly four anomalously cancelling proper fractions having two-digit numerator and denominator when expressed in base $10$ notation:
### $16 / 64$
$\dfrac 1 4 = \dfrac {16} {64} = \dfrac {166} {664} = \dfrac {1666} {6664} = \cdots$
### $19 / 95$
$\dfrac 1 5 = \dfrac {19} {95} = \dfrac {199} {995} = \dfrac {1999} {9995} = \cdots$
### $26 / 65$
$\dfrac 2 5 = \dfrac {26} {65} = \dfrac {266} {665} = \dfrac {2666} {6665} = \cdots$
### $49 / 98$
$\dfrac 4 8 = \dfrac {49} {98} = \dfrac {499} {998} = \dfrac {4999} {9998} = \cdots$
## Proof
Let $\dfrac {\sqbrk {a x} } {\sqbrk {x b} }$ be an anomalously cancelling proper fraction.
Then we have:
$\text {(1)}: \quad$ $\ds a, b, x \le 9$ $\ds$ as they are digits in base $10$ $\text {(2)}: \quad$ $\ds a < b$ $\ds$ this fraction is a proper fraction $\text {(3)}: \quad$ $\ds \dfrac a b = \dfrac {\sqbrk {a x} } {\sqbrk {x b} }$ $\ds$ this fraction is anomalously cancelling
From $(3)$:
$\ds \dfrac a b$ $=$ $\ds \dfrac {10 a + x} {10 x + b}$ $\ds \leadsto \ \$ $\ds a \paren {10 x + b}$ $=$ $\ds b \paren {10 a + x}$ $\ds \leadsto \ \$ $\ds 10 a x + a b$ $=$ $\ds 10 a b + b x$ $\ds \leadsto \ \$ $\ds 10 a x - b x$ $=$ $\ds 9 a b$ $\ds \leadsto \ \$ $\ds x$ $=$ $\ds \dfrac {9 a b} {10 a - b}$
Therefore $\dfrac {9 a b} {10 a - b}$ must be an integer.
Aiming for a contradiction, suppose $10 a - b$ is divisible by a prime $p$ such that $p > 9$.
We have that:
$\paren {10 a - b} \divides 9 a b$
and so:
$p \divides 9 a b$
where $\divides$ denotes divisibility.
By Euclid's Lemma for Prime Divisors, that means one of the elements of $\set {9, a, b}$ must be divisible by $p$.
Hence from Absolute Value of Integer is not less than Divisors, either $a$ or $b$ must be greater than $9$.
This contradicts $(1)$ above.
Hence by Proof by Contradiction it cannot be the case that $10 a - b$ is divisible by a prime greater than $9$.
We continue:
$\ds \dfrac {9 a b} {10 a - b}$ $\le$ $\ds 9$ from $(1)$: $x \le 9$ $\ds \leadsto \ \$ $\ds \dfrac {a b} {10 a - b}$ $\le$ $\ds 1$ $\ds \leadsto \ \$ $\ds a b$ $\le$ $\ds 10 a - b$ as $10 a - b \ge 10 - b > 0$ $\ds \leadsto \ \$ $\ds b$ $\le$ $\ds 10 a - a b$ adding $b - a b$ to both sides $\ds \leadsto \ \$ $\ds a$ $\ge$ $\ds \dfrac b {10 - b}$ as $10 - b > 0$
It remains for us to check whether $x = \dfrac {9 a b} {10 a - b}$ is an integer for $\dfrac b {10 - b} \le a < b$, when $b$ ranges from $2$ to $9$.
We note first that the following pairs $\tuple {a, b}$ are such that $\dfrac b {10 - b} > a$:
$\, \ds b = 6: \,$ $\ds \dfrac b {10 - b}$ $=$ $\ds \dfrac 3 2 > 1$ which eliminates $\tuple {1, 6}$ $\, \ds b = 7: \,$ $\ds \dfrac b {10 - b}$ $=$ $\ds \dfrac 7 3 > 2$ which eliminates $\tuple {1, 7}$ and $\tuple {2, 7}$ $\, \ds b = 8: \,$ $\ds \dfrac b {10 - b}$ $=$ $\ds 4 > 3$ which eliminates $\tuple {1, 8}$, $\tuple {2, 8}$ and $\tuple {3, 8}$ $\, \ds b = 9: \,$ $\ds \dfrac b {10 - b}$ $=$ $\ds 9$ which eliminates all possible such solutions such that $b = 9$
Then we eliminate the pairs $\tuple {a, b}$ where $10 a - b$ has a prime factor greater than $9$:
$\, \ds \tuple {3, 4}: \,$ $\ds 10 a - b$ $=$ $\ds 26 = 13 \times 2$ $\, \ds \tuple {4, 6}: \,$ $\ds 10 a - b$ $=$ $\ds 34 = 17 \times 2$ $\, \ds \tuple {5, 6}: \,$ $\ds 10 a - b$ $=$ $\ds 44 = 11 \times 4$ $\, \ds \tuple {3, 7}: \,$ $\ds 10 a - b$ $=$ $\ds 23$ $\, \ds \tuple {4, 7}: \,$ $\ds 10 a - b$ $=$ $\ds 33 = 11 \times 3$ $\, \ds \tuple {5, 7}: \,$ $\ds 10 a - b$ $=$ $\ds 43$ $\, \ds \tuple {6, 7}: \,$ $\ds 10 a - b$ $=$ $\ds 53$ $\, \ds \tuple {6, 8}: \,$ $\ds 10 a - b$ $=$ $\ds 52 = 13 \times 4$ $\, \ds \tuple {7, 8}: \,$ $\ds 10 a - b$ $=$ $\ds 62 = 31 \times 2$
Thus it remains to check whether $\dfrac {9 a b} {10 a - b}$ is an integer for the following pairs $\tuple {a, b}$:
$\tuple {1, 2}$, $\tuple {1, 3}$, $\tuple {2, 3}$, $\tuple {1, 4}$, $\tuple {2, 4}$, $\tuple {1, 5}$, $\tuple {2, 5}$, $\tuple {3, 5}$, $\tuple {4, 5}$, $\tuple {2, 6}$, $\tuple {3, 6}$, $\tuple {4, 8}$, $\tuple {5, 8}$
$\, \ds \tuple {1, 2}: \,$ $\ds \dfrac {9 a b} {10 a - b}$ $=$ $\ds \dfrac {18} 8$ not an integer $\, \ds \tuple {1, 3}: \,$ $\ds \dfrac {9 a b} {10 a - b}$ $=$ $\ds \dfrac {27} 7$ not an integer $\, \ds \tuple {2, 3}: \,$ $\ds \dfrac {9 a b} {10 a - b}$ $=$ $\ds \dfrac {54} {16}$ not an integer $\, \ds \tuple {1, 4}: \,$ $\ds \dfrac {9 a b} {10 a - b}$ $=$ $\ds \dfrac {36} 6 = 6$ which leads to the solution $\dfrac {16} {64}$ $\, \ds \tuple {2, 4}: \,$ $\ds \dfrac {9 a b} {10 a - b}$ $=$ $\ds \dfrac {72} {16}$ not an integer $\, \ds \tuple {1, 5}: \,$ $\ds \dfrac {9 a b} {10 a - b}$ $=$ $\ds \dfrac {45} 5 = 9$ which leads to the solution $\dfrac {19} {95}$ $\, \ds \tuple {2, 5}: \,$ $\ds \dfrac {9 a b} {10 a - b}$ $=$ $\ds \dfrac {90} {15} = 6$ which leads to the solution $\dfrac {26} {65}$ $\, \ds \tuple {3, 5}: \,$ $\ds \dfrac {9 a b} {10 a - b}$ $=$ $\ds \dfrac {135} {25}$ not an integer $\, \ds \tuple {4, 5}: \,$ $\ds \dfrac {9 a b} {10 a - b}$ $=$ $\ds \dfrac {180} {35}$ not an integer $\, \ds \tuple {2, 6}: \,$ $\ds \dfrac {9 a b} {10 a - b}$ $=$ $\ds \dfrac {108} {14}$ not an integer $\, \ds \tuple {3, 6}: \,$ $\ds \dfrac {9 a b} {10 a - b}$ $=$ $\ds \dfrac {108} {14}$ not an integer $\, \ds \tuple {4, 8}: \,$ $\ds \dfrac {9 a b} {10 a - b}$ $=$ $\ds \dfrac {288} {32} = 9$ which leads to the solution $\dfrac {49} {98}$ $\, \ds \tuple {5, 8}: \,$ $\ds \dfrac {9 a b} {10 a - b}$ $=$ $\ds \dfrac {360} {42}$ not an integer
So the only solutions for $\tuple {a, x, b}$ are:
$\tuple {1, 6, 4}$, $\tuple {1, 9, 5}$, $\tuple {2, 6, 5}$ and $\tuple {4, 9, 8}$
These correspond to the fractions:
$\dfrac {16} {64}$, $\dfrac {19} {95}$, $\dfrac {26} {65}$ and $\dfrac {49} {98}$
$\blacksquare$
## Historical Note
According to David Wells in his $1986$ work Curious and Interesting Numbers, this result, as well as some others, was demonstrated by Alfred Moessner in Volumes 19 and 20 of Scripta Mathematica, but it has proven difficult to find an archived copy to consult directly. | 2023-04-02 06:23:39 | {"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": 0.9078824520111084, "perplexity": 74.09195333787957}, "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-2023-14/segments/1679296950383.8/warc/CC-MAIN-20230402043600-20230402073600-00135.warc.gz"} |
https://academy.edulabs.org/mod/glossary/view.php?id=677&mode=letter&hook=W&sortkey&sortorder&fullsearch=0&page=2 | ## Electronics (Mike Jaroch)
Note: You may download the entries for this glossary here. If you wish to use this in your own Moodle course, first make a blank glossary and then follow the instructions for importing glossary entries here.
Browse the glossary using this index
Special | A | B | C | D | E | F | G | H | I | J | K | L | M | N | O | P | Q | R | S | T | U | V | W | X | Y | Z | ALL
Page: (Previous) 1 2 3 4 (Next)
ALL
### W
#### WAVEMETERS
Calibrated resonant circuits that are used to measure frequency [16]. An instrument for measuring the wavelength of an RF wave [18].
Keyword(s): WAVEMETERS
#### WEBER'S THEORY
A theory of magnetism which assumes that all magnetic material is composed of
many tiny magnets. A piece of magnetic material that is magnetized has all of the tiny magnets aligned so that the north pole of each magnet points in one direction [1].
Keyword(s): WEBER'S THEORY
#### WHEATSTONE BRIDGE
An ac bridge circuit used to measure unknown values of resistance, inductance, or capacitance [16].
Keyword(s): WHEATSTONE BRIDGE
#### WIDE-BAND AMPLIFIER
An amplifier designed to pass an extremely wide band of frequencies, such
as a video amplifier [8].
Keyword(s): WIDE-BAND AMPLIFIER
#### WINDOW
See SLOT [11].
Keyword(s): WINDOW
#### WIRE
An insulated conductor, with low resistance to current flow, that is either solid or stranded [1] [4].
Keyword(s): WIRE
#### WIRING DIAGRAM
A diagram that shows the connections of an equipment or its component devices or parts. It may cover internal or external connections, or both, and contains such detail as is needed to make or trace connections that are involved [4].
Keyword(s): WIRING DIAGRAM
#### WOBBLE FREQUENCY
The frequency at which an electron wobbles on its axis under the influence of
an external magnetic field of a given strength [11].
Keyword(s): WOBBLE FREQUENCY
#### WORDS-PER-MINUTE
An approximate rate of speed. It means the number of five letter words with a
space between them that can be transmitted or received in a one-minute period [17].
Keyword(s): WORDS-PER-MINUTE
#### WORK
The product of force and motion [1].
Keyword(s): WORK
Page: (Previous) 1 2 3 4 (Next)
ALL | 2021-10-27 12:28:04 | {"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": 0.8435885906219482, "perplexity": 1521.1345712332482}, "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-2021-43/segments/1634323588153.7/warc/CC-MAIN-20211027115745-20211027145745-00601.warc.gz"} |
https://dl_monte.gitlab.io/dl_monte-tutorials-pages/tutorial8.html | # TUTORIAL 8 : Planar geometry confinement¶
Author: Andrey V. Brukhno (andrey.brukhno{at}stfc.ac.uk)
## Illustration of a confined system¶
In this tutorail we demonstrate how to set up a simulation for a system confined to a planar nano-pore.
## Two types of quasi-2d constraints¶
DL_MONTE supports two types of setup for constrained systems:
• A ‘slit’ constraint where the system is non-periodic in z-direction, so that it is confined between two flat, impenetrable walls (infinite in x and y).
To invoke the slit constraint one has to add the following directive in the CONTROL file (after ‘finish use’):
slit walls hard # both walls are set to be hard (impenetrable)
One or both walls can be defined as ‘soft’:
slit walls soft 1 # one of the walls is set as soft (at the bottom, -Z/2)
#slit walls soft 2 # two walls are set as soft (-Z/2, Z/2)
Soft walls must be assigned an external potential to interact with one or more atomic types, which is specified in the FIELD file, for example:
EXTERNAL 2 # wall interactions: LJ epsilon & sigma
OW core lj 0.90 3.00
HW core lj 0.00 0.01
(see DL_MONTE manual for details of the available external potentials).
This setup is applicable to systems that do not contain charges (i.e. without electrostatics).
If charges are present, Coulomb interations require a long-range correction in 2D PBC!
The correction can be included by employing a mean-field approximation (MFA), which accounts for the Coulomb interations of all the charges within the simulation cell with the charge density distribution outside the primary cell. The MFA treatment is enabled in the CONTROL file (in addition to the above directives):
slit mfa -1 100 0.25 # MFA correction, starting from scratch, 100 iterations, density mixing factor
(see DL_MONTE manual for more details).
Surface charge density (SCD) on one or both walls can also be included:
slit walls charged 2 -0.01 0.01 # both walls are charged, SCD1 = -0.01, SCD2 = 0.01 e-/A^2
NOTE: The MFA correction works best for systems with implicit solvent’ where the dielectric permittivity is high and, therefore, the long-range correction is relatively small.
• An extended slit’ constraint, known as a ‘slab’ setup, where the simulation cell is extended in Z direction with a vaccum layer filling in the space beyond the slit walls:
slit walls soft 2 zfrac2 0.2 # two soft walls shifted to +/-0.2*(Z/2)
where ‘zfrac2 = 0.2’ determines the slit Z-fraction of the simulation cell (occupied by the system), while the rest of the cell is empty (filled in with vacuum).
In this case, providing that the vacuum layer is sufficiently thick, the long-ranged electrostatic interactions can be calculated by use of the Ewald summation in full 3D PBC.
## Exercises:¶
TUTORIAL 8 : Exercise 1 - Ideal gas partitioning in a planar pore
TUTORIAL 8 : Exercise 2 - Water adsoprtion in a slit | 2020-09-24 18:08:25 | {"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": 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": 0.6239899396896362, "perplexity": 3225.1551063481807}, "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-2020-40/segments/1600400219691.59/warc/CC-MAIN-20200924163714-20200924193714-00731.warc.gz"} |
https://socratic.org/questions/what-is-how-do-the-atmospheric-layers-differ-from-each-other | # How do the atmospheric layers differ from each other?
May 29, 2016
Just above, it is the cloudy Troposphere.Just above is Stratosphere with ozone layer atop that absorbs harmful solar ultraviolet rays.. Above is Mesosphere, with low temperature and pressure ..... .
#### Explanation:
Layers: Troposphere-Stratosphere-Mesosphere-Thermosphere.
Just above, it is the cloudy Troposphere, up to about 10 km.
Just above is Stratosphere, with ozone layer atop that absorbs harmful solar ultraviolet rays..
Above is Mesosphere, with low temperature (up to$- {100}^{o} C$) and pressure (1/1000 of sea-level pressure).
And above this is the Thermosphere, where temperature steadily increase with height.
And beyond is the ionized layer called Ionosphere that enables communications between antipodal Australia and Americas.
The boundaries are not clearly marked. So, there is need for defining in-between transition pauses like Tropopause (about 10 km above), Stratopause (20 km). Kaman line (100 km) is the boundary between atmosphere and outer space.
There is yet another called Exosphere ( from 500 km) where sparse atoms of Hydrogen and Helium might escape Earth's gravity.. Of course, Magnetosphere beyond Exosphere reaching about 50000 km is gateway for interplanetary disturbances.. | 2020-02-20 01:20:27 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 1, "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": 0.8751486539840698, "perplexity": 10617.056182130134}, "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-2020-10/segments/1581875144498.68/warc/CC-MAIN-20200220005045-20200220035045-00355.warc.gz"} |
http://daylateanddollarshort.com/bloog/a-ceva-like-theorem-for-tetrahedra/ | ## A Ceva-like Theorem for Tetrahedra
A question at Math.StackExchange asked about “Ceva’s Theorem in three dimensions”.
The result (which I believe may be new, although it could very well exist in the hundreds of years of mathematical literature since Ceva) replaces the traditional ratios of segment lengths with “triple-ratios” of triangle areas, to satisfying effect. Whereas the ratios in Ceva’s Theorem reduce to $$1$$ (or “$$[1:1]$$”), my triple-ratios reduce to $$[1:1:1]$$.
I like the result enough that I’ve tidied the derivation, added a smidgeon of context, and touched-on the (current) lack of a corresponding Menelaus-like theorem. See my note: “A Ceva-like Theorem for Tetrahedra” (PDF).
Posted 20 April, 2013 by in Classroom, Misc. Math, Open Question | 2022-01-20 13:17:20 | {"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": 0.8723886609077454, "perplexity": 5182.458983336443}, "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-2022-05/segments/1642320301863.7/warc/CC-MAIN-20220120130236-20220120160236-00224.warc.gz"} |
http://cjauvin.blogspot.com/2012/04/ | # Autotelic Computing
## Tuesday, April 24, 2012
### Lean Geocoding
Geocoding is the art of turning street addresses into geographical coordinates (e.g. lat/long). In this article we will explore some of the essential ideas upon which a geocoder can be built, along with a simple example assembled from freely available components.
#### The Data Model
A geocoder is comprised of two main components: a road network database (with an associated data model) and a set of queries acting on it. The database is spatial in the sense that it contains a representation of the road network as a set of geometric features (i.e. typically lines). The representation mechanism works at the street segment level: a row corresponds to the subset of a street defining a block (usually mostly linear, but not always, as in most North American cities, for instance):
In addition to a geometric attribute (describing its shape), a given street segment is associated to four numeric (i.e. non-spatial) attributes: from_left, to_left, from_right and to_right. These numbers correspond to the ranging values of the typical house numbering scheme, having the odd numbers on one side of the street, and the even ones on the other. If we add its name as an additional string attribute, a street in this model can be thus defined (in most cases) as the set of segments sharing it:
name | from_left | to_left | from_right | to_right | astext(the_geom)
------------+-----------+---------+------------+----------+---------------------------------------------------
[...]
Jean-Talon | 1000 | 1024 | 1001 | 1035 | LINESTRING(-73.625538467 45.524777946,-73.625[...]
Jean-Talon | 1001 | 1025 | 1000 | 1080 | LINESTRING(-73.6126363129999 45.5417330830001[...]
Jean-Talon | 1101 | 1185 | 1110 | 1150 | LINESTRING(-73.612026869 45.5425537690001,-73[...]
Jean-Talon | 1201 | 1247 | 1210 | 1244 | LINESTRING(-73.611316541 45.543310246,-73.610[...]
Jean-Talon | 1213 | 1245 | 200 | 260 | LINESTRING(-71.2765479409999 46.8713327090001[...]
Jean-Talon | 1247 | 1273 | 270 | 298 | LINESTRING(-71.2756217529999 46.8717741090001[...]
Jean-Talon | 1251 | 1293 | 1250 | 1294 | LINESTRING(-73.610724326 45.543951109,-73.610[...]
Jean-Talon | 1275 | 1299 | 300 | 360 | LINESTRING(-71.274682868 46.872224756,-71.273[...]
Jean-Talon | 1383 | 1383 | 1360 | 1388 | LINESTRING(-73.609408209 45.545375275,-73.608[...]
Jean-Talon | 1385 | 1385 | 1390 | 1408 | LINESTRING(-73.608972737 45.545850934,-73.608[...]
[...]
#### The Database
As part of the 2011 Census, Statistics Canada offers (free of charge!) the Road Network File, modeling the full country. Let's first import the shapefile version (grnf000r11a_e.shp) of the dataset into a PostGIS spatial database:
$createdb -T template_postgis geocoding$ shp2pgsql -W latin1 -s 4269 -S grnf000r11a_e.shp rnf | psql -d geocoding
Note that the -s 4269 option refers to the mapping projection system used (and specified in the manual), while -S enforces the generation of simple, rather than MULTI geometries for spatial data. Let's also cast some columns to a numeric type and rename them, to make our life simpler later:
$psql -d geocoding -c 'alter table rnf alter afl_val type int using afl_val::int$ psql -d geocoding -c 'alter table rnf rename afl_val to from_left'
$psql -d geocoding -c 'alter table rnf alter atl_val type int using atl_val::int'$ psql -d geocoding -c 'alter table rnf rename atl_val to to_left'
$psql -d geocoding -c 'alter table rnf alter afr_val type int using afr_val::int'$ psql -d geocoding -c 'alter table rnf rename afr_val to from_right'
$psql -d geocoding -c 'alter table rnf alter atr_val type int using atr_val::int'$ psql -d geocoding -c 'alter table rnf rename atr_val to to_right'
Since it wouldn't make sense to store an exhaustive list of all the existing address locations in which we'd perform a simple lookup (there are too many, and their configuration is rapidly evolving), our database is actually a compressed model of reality. To perform the geocoding of an actual address, we need an additional tool, interpolation, with which we can find any new point on a line on the basis of already defined points. For instance, given a street segment ranging from 1000 to 2000, it takes a very simple calculation to determine that a house at 1500 should be about halfway. For a piecewise linear segment, the calculation is a little bit more involved (although as intuitively clear), but thanks to PostGIS, the details are hidden from us with the help of the ST_Line_Interpolate_Point function. Here is the result of interpolating address 6884 on a segment ranging (on the left) from 6878 to 6876:
#### The Query
At this point, we have everything we need to build a basic geocoding query, which we wrap inside a SQL function (mostly for syntactic purposes):
create function geocode(street_nb int, street_name text) returns setof geometry as $$select st_line_interpolate_point(the_geom, case when 1 % 2 = from_left % 2 and 1 between least(from_left, to_left) and greatest(from_left, to_left) then case when from_left = to_left then 0 else (1 - least(from_left, to_left)) / (greatest(from_left, to_left) - least(from_left, to_left))::real end when 1 % 2 = from_right % 2 and 1 between least(from_right, to_right) and greatest(from_right, to_right) then case when from_right = to_right then 0 else (1 - least(from_right, to_right)) / (greatest(from_right, to_right) - least(from_right, to_right))::real end end) from rnf where name = 2 and ((from_left % 2 = 1 % 2 and 1 between least(from_left, to_left) and greatest(from_left, to_left)) or (from_right % 2 = 1 % 2 and 1 between least(from_right, to_right) and greatest(from_right, to_right)))$$ language sql immutable;
select astext(geocode(1234, 'Jean-Talon')); -- POINT(-73.6108985068823 45.5437626198824)
The where clause finds the target segment, while the nested case expression as a whole calculates the fraction to be fed (along with the segment geometry) to the ST_Line_Interpolate_Point function (i.e. the proportion of the target segment length at which the address point should be located). Both work with the between, least and greatest functions (without which the query would be even more verbose), the last two needed in particular because the order of the range values cannot be assumed. The outer case expression finds the side of the street, depending on the range value parity (easily checked with %, the modulo operator), while the inner one checks for the special case where the range is comprised of only the target address (which would yield a division by zero, if not properly handled). Finally, the function can be declared immutable for optimal performance, since it should have no effect on the database.
#### Street Name Spelling Tolerance
The mechanism described above works well if we happen to use the exact same spelling for street names as the database (for instance, if the input UI is some kind of autocompletion field tapping directly into it), but it breaks rapidly if we introduce real-world inputs (e.g. via a free text field without any constaints). For instance, the query above wouldn't work with even a slight variation (e.g. "1234 Jean Talon", where just the hyphen is missing), because the string matching performed is not flexible at all: either the name is right or it isn't. Moreoever, in a real world system, the very notion of a spelling error is vague, because it spans a wide spectrum ranging from expectable variations, like the use of abbreviations ("Saint-Jerome" vs. "St-Jerome"), use of letter case ("SAINT-JEROME"), accents ("Saint-Jérôme", at least in some languages), extra specifiers ("av. Saint-Jerome E.") to downright spelling errors ("Saint-Jerrome"). Although we cannot solve this problem entirely, the use of certain string transformation and matching techniques can help, by introducing some tolerance in the process. The simple mechanism I will describe has two components. The street names (both from the query and the database) are first normalized (or standardized), to ease the process of comparing them (by effectively reducing the space of possible name representations). The comparison of the normalized names is then performed using an edit distance metric, with the goal of quantifying the similarity of two names, to determine whether they match or not (within a certain tolerance parameter).
##### Normalization
The most obvious normalizing transformations are: (1) replacing any non-alphanumeric character by a single one, (2) converting to upper case and (3) removing accents (which is particularly relevant in Canada, given the French speaking Quebec province, but also unfortunately requires a small hack, at least with my particular version of PG). We can also use regular expressions to detect and normalize certain common naming patterns. All the required transformations can be encapsulated in such a PL/pgSQL function, for instance:
create function normalize(in text) returns text as $$declare s text; begin s = to_ascii(convert_to(upper(trim(1)), 'latin1'), 'latin1'); s = regexp_replace(s, E'\\W+', '%', 'g'); s = regexp_replace(s, E'^SAINTE%|^SAINT%|^STE%|^ST%', 'S*%', 'g'); return s; end;$$ language plpgsql;
select normalize('St-Jérôme'); -- S*%JEROME
which we can now use to precompute the normalized database street names:
$psql -d geocoding -c 'alter table rnf add column name_norm text'$ psql -d geocoding -c 'update rnf set name_norm = normalize(name)'
##### Similarity Matching
The last step is performed by using an edit distance metric to define what is an acceptable matching level (or similarity) between two names. A particular instance of such a metric is the well-known Levenshtein distance, which is implemented by the fuzzystrmatch PG module, and can be interpreted roughly as the "number of edits needed to transform a string A into string B". Here is an updated version of our geocoding function (with an additional parameter for the edit distance deviation tolerance), with which slight errors can now be forgiven, thanks to our flexible name spelling mechanism:
create function geocode(street_nb int, street_name text, tol int) returns setof geometry as $$select st_line_interpolate_point(the_geom, case when 1 % 2 = from_left % 2 and 1 between least(from_left, to_left) and greatest(from_left, to_left) then case when from_left = to_left then 0 else (1 - least(from_left, to_left)) / (greatest(from_left, to_left) - least(from_left, to_left))::real end when 1 % 2 = from_right % 2 and 1 between least(from_right, to_right) and greatest(from_right, to_right) then case when from_right = to_right then 0 else (1 - least(from_right, to_right)) / (greatest(from_right, to_right) - least(from_right, to_right))::real end end) from rnf where levenshtein(normalize(2), name_norm) <= 3 and ((from_left % 2 = 1 % 2 and 1 between least(from_left, to_left) and greatest(from_left, to_left)) or (from_right % 2 = 1 % 2 and 1 between least(from_right, to_right) and greatest(from_right, to_right)))$$ language sql immutable;
select astext(geocode(1234, 'Jean Tallon', 3)); -- POINT(-73.6108985068823 45.5437626198824) | 2018-06-19 17:47:33 | {"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": 0.3028706908226013, "perplexity": 2004.7704761774826}, "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-2018-26/segments/1529267863109.60/warc/CC-MAIN-20180619173519-20180619193519-00244.warc.gz"} |
https://repository.uantwerpen.be/link/irua/102979 | Publication
Title
AC susceptibility measurements : a quality test for high $T_{c}$ superconducting bulk materials
Author
Abstract
Temperature dependent measurements of high T(c) bulk material reflect the quality of the high T(c) grains and the intergranular weak links. Besides, ac field dependent measurements give access to a quantitative quality label of the bulk material : its critical current density J(c). Herefore critical state models as the Bean and the Kim model are used. Our measurements on long orthorombic samples with different cross sections a x b show that the Kim model is the most realistic one. It leads to a b/a independent average value of J(c), which equals the J(c) value of transport measurements and of magnetic SQUID measurements at low dc fields.
Language
English
Source (journal)
Journal of alloys and compounds. - Amsterdam
Source (book)
SYMP ON HIGH TEMPERATURE SUPERCONDUCTORS, AT THE 1992 FALL MEETING OF, THE E-MRS / 3RD EUROPEAN EAST-WEST CONF AND EXHIBITION ON MATERIALS AND, PROCESSES, NOV 03-06, 1992, STRASBOURG, FRANCE
Publication
Amsterdam : 1993
ISSN
0925-8388
Volume/pages
195:1-2(1993), p. 177-180
ISI
A1993LC49200043
Full text (Publishers DOI)
Full text (publishers version - intranet only)
UAntwerpen
Faculty/Department Research group Publication type Subject Affiliation Publications with a UAntwerp address | 2017-05-26 21:20:02 | {"extraction_info": {"found_math": true, "script_math_tex": 1, "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": 0.3480355739593506, "perplexity": 10334.892472674972}, "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-2017-22/segments/1495463608684.93/warc/CC-MAIN-20170526203511-20170526223511-00390.warc.gz"} |
https://www.laorpheum.com/haa1mv/oj9ge4t.php?16888c=statsmodels-r-squared-1 | The whitened response variable $$\Psi^{T}Y$$. It returns an OLS object. The n x n upper triangular matrix $$\Psi^{T}$$ that satisfies One of them being the adjusted R-squared statistic. # Load modules and data In [1]: import numpy as np In [2]: import statsmodels.api as sm In [3]: ... OLS Adj. Linear models with independently and identically distributed errors, and for random. Entonces use el “Segundo resultado R-Squared” que está en el rango correcto. Note that the RollingWLS and RollingOLS. Fitting a linear regression model returns a results class. Suppose I’m building a model to predict how many articles I will write in a particular month given the amount of free time I have on that month. Previous statsmodels.regression.linear_model.OLSResults.rsquared R-squared metrics are reported by default with regression models. This is defined here as 1 - ssr / centered_tss if the constant is included in the model and 1 - ssr / uncentered_tss if the constant is omitted. http://www.statsmodels.org/stable/generated/statsmodels.nonparametric.kernel_regression.KernelReg.r_squared.html, $R^{2}=\frac{\left[\sum_{i=1}^{n} (Y_{i}-\bar{y})(\hat{Y_{i}}-\bar{y}\right]^{2}}{\sum_{i=1}^{n} (Y_{i}-\bar{y})^{2}\sum_{i=1}^{n}(\hat{Y_{i}}-\bar{y})^{2}},$, http://www.statsmodels.org/stable/generated/statsmodels.nonparametric.kernel_regression.KernelReg.r_squared.html. rsquared_adj – Adjusted R-squared. Dataset: “Adjusted Rsquare/ Adj_Sample.csv” Build a model to predict y using x1,x2 and x3. # compute with formulas from the theory yhat = model.predict(X) SS_Residual = sum((y-yhat)**2) SS_Total = sum((y-np.mean(y))**2) r_squared = 1 - (float(SS_Residual))/SS_Total adjusted_r_squared = 1 - (1-r_squared)*(len(y)-1)/(len(y)-X.shape[1]-1) print r_squared, adjusted_r_squared # 0.877643371323 0.863248473832 # compute with sklearn linear_model, although could not find any … seed (9876789) ... y R-squared: 1.000 Model: OLS Adj. specific methods and attributes. PredictionResults(predicted_mean, …[, df, …]), Results for models estimated using regularization, RecursiveLSResults(model, params, filter_results). degree of freedom here. ==============================================================================, Dep. D.C. Montgomery and E.A. An extensive list of result statistics are available for each estimator. The OLS() function of the statsmodels.api module is used to perform OLS regression. See, for instance All of the lo… R-squared. The former (OLS) is a class.The latter (ols) is a method of the OLS class that is inherited from statsmodels.base.model.Model.In [11]: from statsmodels.api import OLS In [12]: from statsmodels.formula.api import ols In [13]: OLS Out[13]: statsmodels.regression.linear_model.OLS In [14]: ols Out[14]: | 2021-01-16 12:11:37 | {"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": 0.4399404227733612, "perplexity": 4305.661682300248}, "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/1610703506640.22/warc/CC-MAIN-20210116104719-20210116134719-00767.warc.gz"} |
https://www.physicsforums.com/threads/volume-integral.853281/ | # Volume Integral
Tags:
1. Jan 20, 2016
### MaryCatherine
Use an appropriate volume integral to find an expression for the volume enclosed between a sphere of radius 1 centered on the origin and a circular cone of half-angle alpha with its vertex at the origin. Show that in the limits where alpha = 0 and alpha = pi that your expression gives the expected values.
Any ideas or clues? Where alpha = pi I think the volume will be half the volume of the sphere.
2. Jan 20, 2016
### Staff: Mentor
3. Jan 21, 2016 | 2018-02-21 20:14:48 | {"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": 0.8455996513366699, "perplexity": 581.9967024774116}, "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/1518891813712.73/warc/CC-MAIN-20180221182824-20180221202824-00251.warc.gz"} |
http://openstudy.com/updates/50f8cb96e4b007c4a2ebdd35 | ## itsladyrw Group Title Simplify completely, 4x^2 - 32x + 48 / 3x^2 - 17x - 6 one year ago one year ago
1. khuangg
$=\frac{ 4x^2-32x+48 }{ 3x^2-17x-6 }$ $= \frac{ 4(x^2-8x+12) }{ 3x^2-17x-6 }$ $= \frac{ 4(x-6)(x-2) }{ (x-6)(3x-1) }$ $=\frac{ 4(x-2) }{ 3x-1 }$ there you go! | 2014-09-19 11:57:45 | {"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": 0.27172988653182983, "perplexity": 1836.1638717524145}, "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/1410657131304.74/warc/CC-MAIN-20140914011211-00118-ip-10-196-40-205.us-west-1.compute.internal.warc.gz"} |
https://problemsolvingwithpython.com/08-If-Else-Try-Except/08.01-User-Input/ | User Input
User Input
To begin this chapter, Python's input() function is discussed. Python can be used to ask users for input. The input entered by a user can be saved to a variable and used in subsequent parts of the program. The syntax of Python's input() function is below:
var = input('message')
Where var is the variable that stores the user's input and 'message' is the message the user sees at the prompt. A string enclosed in quotes, like 'message', needs to be passed as an input argument to the input() function. Let's ask a user for their age:
In [1]:
age = input('how old are you? ')
how old are you? 9
Since the user's input is assigned to a variable, further operations can be run on it. Now, let's print the user's age back to them. This can be accomplished with an f-string. Note the f' ' inserted before the string. A set of curly braces { } surrounds the variable's name and the variable's value is printed back to the user.
In [2]:
age = input('how old are you? ')
print(f'you are {age} years old')
how old are you? 9
you are 9 years old
Let's try another example. We will we ask the user for the base and height of a triangle and print out the area of the triangle.
But, there is a problem with the approach below. The code block does not run because a common error is present.
In [3]:
b = input('base of triangle: ')
h = input('height of triangle: ')
A = (1/2)bh
print(f'The area of the triangle is: {A}')
base of triangle: 5
height of triangle: 2
---------------------------------------------------------------------------
TypeError Traceback (most recent call last)
<ipython-input-3-c9cb8f02e604> in <module>
1 b = input('base of triangle: ')
2 h = input('height of triangle: ')
----> 3 A = (1/2)*b*h
4 print(f'The area of the triangle is: {A}')
TypeError: can't multiply sequence by non-int of type 'float'
The previous section of code returns an error because of the data type of the variables b and h. We can investigate b and h's data type with Python's type() function.
In [4]:
b = input('base of triangle: ')
h = input('height of triangle: ')
print(f'b and h are of type: {type(b)}, {type(h)}')
base of triangle: 5
height of triangle: 2
b and h are of type: <class 'str'>, <class 'str'>
Notice both b and h are strings, even though the numbers 5 and 2 were entered as input. The output of the input() function is always a string, even if the user enters a number.
To complete the area calculation, b and h first need to be converted to floats using Python's float() function, then the mathematical operation will run without error:
In [5]:
b = input('base of triangle: ')
h = input('height of triangle: ')
A = (1/2)float(b)float(h)
print(f'The area of the triangle is: {A}')
base of triangle: 5
height of triangle: 2
The area of the triangle is: 5.0
Now that you are familiar with Python's input() function, let's utilize a user's input to decide which lines of code will run. The concept of an selection statement is introduced the next section. | 2019-10-15 01:01:36 | {"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": 0.44533631205558777, "perplexity": 2748.972078938876}, "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-2019-43/segments/1570986655735.13/warc/CC-MAIN-20191015005905-20191015033405-00313.warc.gz"} |
https://www.ritchievink.com/blog/2021/02/28/i-wrote-one-of-the-fastest-dataframe-libraries/ | # I wrote one of the fastest DataFrame libraries
## 1. Introduction
At the time of writing this, the coronavirus has been in our country for a year, which means I have been sitting at home for a very long time. At the start of the pandemic, I had a few pet projects in Rust under my belt and I noticed that the “are we DataFrame yet”, wasn’t anywhere near my satisfaction. So I wondered if I could make a minimalistic crate that solved a specific use case of mine. But boy, did that get out of hand.
A year later with lots of programming resulting in one of the fastest DataFrame libraries available in Rust and Python. This is my first official “Hello World” from polars on my personal blog. With this post, I hope I can take the reader along with some design decisions I encountered and get a more thorough understanding of how Polars works under the hood.
## 2. Quite a bold claim
I know it is quite a bold claim to make, and I would not make it lightly. There is a benchmark for database systems that does a benchmark on in-memory tools ran by h2o.ai. This benchmark consists of 10 groupby tests on different data cardinalities and query complexity to give a well-rounded view of a tool’s performance, and 5 tests on different join questions. At the time of writing this blog, Polars is the fastest DataFrame library in the benchmark second to R’s data.table, and Polars is top 3 all tools considered.
Below are shown the summaries of the 5GB dataset test, and you can see the whole benchmark here.
update 2021-03-14: Benchmarks age poorly, Polars now has the fastest join algorithm in the benchmark
If you want to design for optimal performance you cannot ignore hardware. There are cases where algorithmic complexity doesn’t give you a good intuition of the real performance due to hardware-related issues like cache hierarchies and branch prediction. For instance, up to a certain number of elements (a few 100, depending on the datatype), it is faster to lookup a given element in an array than to look it up in hashmap, whereas the time complexity of those data structures are $\mathcal{O}(n)$, and $\mathcal{O}(1)$ respectively. This makes design decisions also very temporal, what is a bottleneck in this day, may not be in the future. This is clearly seen in database systems. DB systems from the previous generation like PostgreSQL or MySQL are all row-based volcano models[1], which was an excellent design decision in that era when disks were much slower and RAM was very limited. Nowadays we have fast SSD disks and large amounts of memory available, and wide SIMD registers, we see that columnar databases like cockroachDB, DuckDB are among the best performing DBMSes.
### 3.1 Cache hierarchies
Oversimplifying, RAM comes in two flavors, large and slow or fast and small. For this reason, you have memory caches in a hierarchy. You’ve got main memory that’s large and slow. And memory you’ve used last is stored in L1, L2, L3 cache with more latency respectively. The summation below gives you an idea of the relative latency of the different cache levels:
Access time in CPU cycles
• CPU register - 1 cycle
• L1 cache - ~1-3 cycles
• L2 cache - ~10 cycles
• L3 cache - ~40 cycles
• Main memory - ~100-300 cycles
When accessing data sequentially we want to make sure that data is in cache as much as possible, or we could easily have a ~100x performance penalty. Caches are loaded and deleted in cache lines. When we load a single data point, we get a whole cache line, but we also remove a whole cache line. They are typically 64, or 128 bytes long and aligned on 64-byte memory adresses.
### 3.2 Prefetching and branch predictions
CPUs prefetch data and instructions to a local cache to reduce the high penalty of memory latency. If you have a tight loop without any branches (if-else-then structures) the CPU has no problem knowing which data to prefetch and can fully utilize instruction pipelines.
Instruction pipelines hide latency by doing work in parallel. Every CPU instruction goes through a Fetch, Decode, Execute, Write-back sequence. Instead of doing these 4 instructions sequentially in a single pipeline, there are multiple pipelines that already pre-fetch (decode, execute, etc.) the next instructions. This increases throughput and hides latency. However, if this process is interrupted, you start with empty pipelines and you have to wait the full latency period for the next instruction. Below we see a visual of instruction pipelines.
The CPU does its best to predict which conditional jump is taken and speculatively execute that code in advance (i.e. keep the pipelines filled), but if it has mispredicted it must clear that work and we pay the latency price until the pipelines are full again.
### 3.3 SIMD instructions
Modern processors have SIMD registers (Single Instruction Multiple Data), which operate on whole vectors of data in a single CPU cycle. The vector lane widths vary from 128 bits to 512 bits, and the speedup depends on the width of the registers and the number of bits needed to represent the datatype. These register greatly improve the performance of simple operations if you can fill them fast enough (linear data). A columnar memory format can therefore fully utilize SIMD instructions. Polars and its memory backend Arrow, utilize SIMD to get optimal performance.
## 4. Arrow memory format
Polars is based on the Rust native implementation Apache Arrow. Arrow can be seen as middleware software for DBMS, query engines and DataFrame libraries. Arrow provides very cache-coherent data structures and proper missing data handling.
Let’s go through some examples to see what this means.
### 4.1 Arrow numeric array
An Arrow numeric array consists of a data buffer containing some typed data, e.g. f32, u64, etc. , shown in the figure as an orange colored array. Besides the value data, an Arrow array alwas has a validity buffer. This buffer is a bit array where the bits indicate missing data. Because the missing data is represented by bits there is minimal memory overhead.
This directly shows a clear advantage over Pandas for instance, where there is no clear distinction between a float NaN and missing data, where they really should represent different things.
### 4.1 Arrow string array
The figure below shows the memory layout of an Arrow LargeString array. This figure encodes the following array ["foo", "bar", "ham"]. The Arrow array consists of a data buffer where all string bytes are concatenated to a single sequential buffer (good for cache coherence!). To be able to find the starting and ending position of a string value there is a separate offset array, and finally, there is the null-bit buffer to indicate missing values.
Let’s compare this with a pandas string array. Pandas strings are actually Python objects, therefore they are boxed (which means there is also memory overhead to encode the type next to the data). Sequential string access in pandas will lead to cache miss after cache miss, because every string value may point to a completely different memory location.
For cache coherence, the Arrow representation is a clear winner. However, there is a price. If we want to filter this array or we want to take values based on some index array we need to copy a lot more data around. The pandas string array only holds pointers to the data and can cheaply create a new array with pointers. Arrow string arrays have to copy all the string data, especially when you have large string values this can become a very large overhead. It is also harder to estimate the size of the string data buffer, as this comprises the length of all string values.
Polars also has a Categorical type which helps you mitigate this problem. Arrow also has a solution for this problem, called the Dictionary type, which is similar to Polars' Categorical type.
### 4.3 Reference counted
Arrow buffers are reference counted and immutable. Meaning that copying a DataFrame, Series, Array is almost a no-op, making it very easy to write purely functional code. The same counts for slicing operations, which are only an increment of the reference count and a modification of the offset.
### 4.2 Performance: Missing data and branching
As we’ve seen, the missing data is encoded in a separate buffer. This means we can easily write branchless code by just ignoring the null buffer during an operation. When the operation is finished the null bit buffer is just copied to the new array. When a branch miss would be more expensive than the execution of an operation this is an easy win and used in many operations in Arrow and Polars.
### 4.3 Performance: filter-trick
An operation that has to be done often in a DBMS, is a filter. Based on some predicate (boolean mask) we filter rows. Arrows null bit buffer allow for very fast filtering using a filter-trick (unofficially named by me). *Credits to the filter-trick go the Apache Arrow implementation. Note, that this filter-trick often leads to faster filters, but it may not always be the case. If your predicate consists of alternating boolean values e.g.
[true, false, true, false, ... , true, false] this trick has a slight overhead.
The core idea of the filter-trick is that we can load the bit array from memory as any integer type we want. Let’s say that we load the bit array as an unsigned integer u64 then we know that the maximum encoded value is $2^{64}$ (64 consecutive one value in binary), and the minimum encoded value is 0 (64 consecutive 0 values in binary). We can make a table of 64 entries that represent how many consecutive values we can filter and skip. If this integer is in this table we know have many values to filter in very few CPU cycles and can do a memcpy to efficiently copy data. If the integer is not present in the table we have to iterate through the bits one by one and hope for a hit in the following 64 bits we load.
This filter-trick is used of course in any operation that includes a predicate mask, such as filter, set, and zip.
## 5. Parallelization
With the plateauing of CPU clock speeds and the end of Moore’s law in sight, the free lunch [2] is over. Single-threaded performance isn’t going to increase much anymore. To mitigate this, almost all hardware nowadays has multiple cores. My laptop has 12 logical cores, so there is a tremendous potential for parallelization. Polars is written to exploit parallelism as much as possible.
### 5.1 Embarrassingly parallel
The best parallelization is of course where is no need for communication and there are no data dependencies. Polars utilizes this kind of parallelism as much as possible.
This is for instance the case if we do an aggregation on the columns in a DataFrame. All columns can be aggregated on in parallel.
Another embarrassingly parallel algorithm in Polars in the apply-phase in a groupy-operation.
### 5.2 Parallel hashing
Hashing is the core of many operations in a DataFrame library, a groupby-operation creates a hash table with the group index pointers, and a join operation needs a hash table to find the tuples mapping the rows of the left to the right DataFrame.
#### 5.2.1 Expensive synchronization
In both operations, we cannot simply split the data among the threads. There is no guarantee that all the same keys would end up in the same hash table on the same thread. Therefore we would need an extra synchronization-phase where we build a new hashtable. This principle is shown in the figure below for 2 threads.
#### 5.2.2 Expensive locking
Another option that is found too expensive is hashing the data on separate threads and have a single hash table in a mutex. As you can imagine, thread contention is very high in this algorithm and the parallelism doesn’t really pay of.
#### 5.2.2 Lock-free hashing
Instead of the before mentioned approaches, Polars uses a lock-free hashing algorithm. This approach does do more work than the previous Expensive locking approach, but this work is done in parallel and all threads are guaranteed to not have to wait on any other thread. Every thread computes the hashes of the keys, but depending on the outcome of the hash, it will determine if that key belongs to the hash table of that thread. This is simply determined by the hash value % thread number. Due to this simple trick, we know that every threaded hash table has unique keys and we can simply combine the pointers of the hash tables on the main thread.
## 6. Query optimization: Less is more
The best performance gains are simply not doing any work at all. Polars consists of two public APIs, one that is eager, procedural programming, and one that is lazy declarative programming. I would recommend using the lazy API as much as possible when dealing with performance-critical code.
The declarative DSL allows Polars to analyze a logical plan of your query and it can apply several optimization/ heuristics such that your query can be executed by just doing less work. If you want to read more about the optimizations that are done to the query plan, there is a section with examples in the polars book.
## 7. And more
This post only highlighted a few of the performance related designs in the Polars and Arrow library. Other things that are implemented are for instance:
• Ad hoc partitioning (map-reduce like groupby’s)
• Vectorized hashing [3] (Tipped by Daniël Heres).
• Using table statistics to choose fast paths
• And much more..
Now the “hello world” is officially out there, I may highlight those other subjecst in later posts. Check out polars on github, and if you have any remarks, feature requests, etc. let me know!
## References
[1] Graefe G.. Volcano (1994) an extensible and parallel query evaluation system. IEEE Trans. Knowl. Data Eng.
[2] Herb Sutter (2005) The Free Lunch Is Over: A Fundamental Turn Toward Concurrency in Software Weblog [3] Angela Chang: CockroachDB (2019) 40x faster hash joiner with vectorized execution Weblog
(c) 2020 Ritchie Vink. | 2021-04-19 07:19:56 | {"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": 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": 0.2767961323261261, "perplexity": 1688.3497576233306}, "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-20210419045820-20210419075820-00300.warc.gz"} |
https://www.xloypaypa.pub/codeforces-round-704-geniuss-gambit-implementation-construction/ | ## Solution
Better to be classified as implementation problem. Lots of details. Get Lots of WA.
Attention: You are not allowed to use leading zeros for $x$ and $y$.
So b--, because the first one must be 1.
Want have $k$ ones. So it's very clear that we can have this pair of $i$ and $j$. So that $i<j$ and $x_i=1,\ y_i=0, x_j=0, y_j=1$ and for any $p\in{(i,j)}$ have $x_p=y_p$. Then we just need adjust the position of $i$, $j$. Or return no.
## Code
First two WA5 is cause by my misunderstanding of this problem. After that, just changed the code base on WA5's code. But the code structure still not changed.
Then I change the way to current code. It's much better. But still lots of WA. | 2022-05-26 23:00:39 | {"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": 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": 0.6256585717201233, "perplexity": 594.6395362591454}, "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-2022-21/segments/1652662627464.60/warc/CC-MAIN-20220526224902-20220527014902-00584.warc.gz"} |
https://www.gradesaver.com/textbooks/science/chemistry/chemistry-12th-edition/chapter-8-periodic-relationships-among-the-elements-questions-problems-page-363/8-90 | # Chapter 8 - Periodic Relationships Among the Elements - Questions & Problems - Page 363: 8.90
Please see the work below.
#### Work Step by Step
We know that fluorine forms compounds under favorable conditions with $RnF_2,Xenon(XeF_2, XeF_4, XeF_6)\space and Krypton (KrF_2).$
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback. | 2018-09-22 14:36:50 | {"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": 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": 0.6075276136398315, "perplexity": 2984.781758884866}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.3, "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-2018-39/segments/1537267158450.41/warc/CC-MAIN-20180922142831-20180922163231-00331.warc.gz"} |
http://www.eggie5.com/66-naive-bayes-classifier-implementation | Naive Bayes Classifier Implementation
Alex Egg,
The Bayesian approach to classifying a new instance is to assign the most probable target value, $v_{nb}$, given the attribute values $\langle a_1, a_2… a_n\rangle$ that describe the instance.
We can use Bayes theorem to rewrite the above express as:
Now we can estimate the two terms in equation $(1)$ based on the training data. We can compute $P(v_j)$ by counting the frequency that each target value $v_j$ occurs in the training data. However, calculating the individual $P(a_1,a_2…a_n | v_j)$ terms in equation 1 in this fashion is not feasible. The problem is that the number of terms is equal to the number of possible instances times the number of possible target values. Therefore, we need to see every instance in the instance space many times in order to obtain reliable estimates.
The naive Bayes classifier is based on the simplifying assumption that the attribute values are conditionally independent given the target value. In other words the assumption is that given the target value of the instance, the probability of observing the conjunction $(a_1,a_2…a_n | v_j)$ is just the product of the probabilities for the individual attributes: $P(a_1,a_2…a_n | v_j) = \prod_{i} P(a_i|v_j)$. If we substitute this into equation 1, we have the approach we can use in a classifier:
Naive Bayes Classifier
where $v_{nb}$ denotes the target value output by the naive Bayes classifier. Notice that the total number of calculations for the classifier is: number of distinct attribute values times the number of distinct target values.
To summarize, the Naive Bayes Classifier, has two steps: training and classification. In training, various $P(v_j)$ and $P(a_j|v_J)$ values are calculated based on their frequencies in the training data. Then in the classification step, these values are used to evaluate the probably of new input using equation 2.
Application to Newsgroups Dataset
Our approach to classifying the text in the newsgroups dataset is simple: Given a text document each word is assigned an ID based on it’s position in the document. This process is called tokenization. Now that we have all numbers we can apply the classifier. Once again the classification equation is:
So this is saying, for each category/class in the corpus, lookup the Bayes probabilities of all the input words $a_i$ and take the product and choose the largest. So for example, if we have 20 classes, we’ll have 20 iterations of the outer loop and $|v|$ iterations of the inner loop. We then choose the largest probability out of the 20.
To complete our design of the Bayes classifier, we must still choose a method for estimating the probability terms. A common metric is the m-estimate equation with lapace smoothing:
where n is the total number of word position in all training examples whose target value is $v_j$, $n_k$ is the number of times word $w_k$ is found among these $n$ word positions and $|Vocabulary|$ is the total number of distinct words (and other tokens) found within the training data.
Pseudo Code
Train(Examples, V)
1. collect all words that occur in Examples
* Vocabulary <- the set of all distinct words and other tokens occurring in any documents from examples
2. calculate the required P(v_j) and P(w_k|v_j) probability terms
* For each target value v_j in V do
* docs_j <- the subset of docs from Examples for which the target value is v_j
* P(v_j) <- |docj|/|examples|
* Text_j <- a single document created by concating all members of docsj
* n <- total # of distinct word positions in Textj
* for each word w_k in Vocabulary
* n_k <- number of times word w_k occurs in Textj
* P(w_k|vj) <- n_k+1 / n+|vocabulary|
Classify(Doc)
* positions <- all word position in Doc that contains tokens found in Vocabulary
* Return v_nb, where
v_nb = argmax P(vj) product P(aj|vj)
Python Implementation
Discrete Environment Note
Due to the fact that we are running this classifier in a discrete-finite-percision environment (computer) underflow is a real problem when we multiply a large number for probability fractions. One work around for this is to use logs, see the revised equation below:
Discrete Naive Bayes Classifier
import numpy as np
import time
class NaiveBayesClassifier(object):
"""docstring for NaiveBayesClassifier"""
def __init__(self, arg):
super(NaiveBayesClassifier, self).__init__()
self.arg = arg
self.training = {}
self.vocabulary = set()
def train(self, features, targets, vocab=None):
start = time.time()
target_count = len(set(targets))
target_names = list(set(targets))
example_count = len(features)
print("Total training examples: ")+str(example_count)
print("Total targets: ")+str(target_count)
if vocab is None:
print "Building vocabulary from corpus"
[self.vocabulary.add(word) for feature in features for word in feature]
else:
print "Buildign vocabulary from user input"
self.vocabulary=vocab
# lets get the vocab from vocabulary.txt
vocab_count = len(self.vocabulary)
print "Vocabulary count: "+str(vocab_count)
print "Estimated iterations: "+str(vocab_count * target_count)
#preprocessing -- need these constants before calculating posteri
print "training step 1..."
for i in range(example_count):
# print "Training step #1 for example: "+str(i)+" ..."
example = features[i]
target = targets[i]
#setup the internal data structure
if(target not in self.training):
self.training[target]={'count':0, 'prob': 0.0, 'blob': [], 'text':{}, 'n': 0, 'words':{}}
#put the example doc words in the category bag
self.training[target]['blob']+=example
for word in example:
self.training[target]['text'][word]=True
self.training[target]['n'] = len(list(set(self.training[target]['blob'])))
#calc. priori probabilities
self.training[target]['count']+=1
self.training[target]['prob']= float(self.training[target]['count']) / float(len(targets))
print "training step 2..."
#now iterate the vocabulary for each category (20 iterations)
for i in range(target_count):
target = target_names[i]
# print "Training step #2 for target: "+str(target)+" ..."
m=len(self.vocabulary)
n = self.training[target]['n']
norm = 1#n+m
for word in self.vocabulary:
if(word not in self.training[target]['words']):
# in the vocab but not in the blob
self.training[target]['words'][word]={'count':0, 'prob':1.0}
#only words in the document blob
if(word in self.training[target]['text'] ):
self.training[target]['words'][word]['count']+=1
count = self.training[target]['words'][word]['count']
self.training[target]['words'][word]['prob']= (count + 1.0)# / norm
# pp.pprint(self.training)
end = time.time()
elapsed = (end - start)
print "Training Complete: {0}s".format(elapsed)
return True
# print vocabulary
def predict(self, Y):
# print ("running predictions...")
#TODO: Fix -> this removes duplicates, e.g.: Y=[1,2,3,9,9,9,9]
positions = Y#list(set(self.vocabulary) & set(Y)) #remove words that are out of vocab.
probabilities = {}
for target in self.training.keys():
prob_target = self.training[target]['prob']
rsum=0
for word in positions:
word_prob = self.training[target]['words'][word]['prob']
word_count = self.training[target]['words'][word]['count']
rsum+= np.log2(word_prob)
target_prob = np.log2(prob_target) + rsum
probabilities[target_prob]=target
return probabilities[max(probabilities)]
Demo
import utils as utils
import numpy as np
Setup
I did some preprocessing of the dataset w/ my utils library that I imported in the above cell. We then load the training data and the vocabulary.
path = '20news-bydate-matlab/matlab'
Train
Next we import and setup my classifier which I have packaged into the NaiveBayesClassifier class. We then pass it the training examples and see how long it takes to train:
clf = NaiveBayesClassifier("")
clf.train(features, targets, vocab=vocab)
Total training examples: 11269
Total targets: 20
Building vocabulary from user input
Vocabulary count: 61188
Estimated iterations: 1223760
training step 1...
training step 2...
Training Complete: 29.5654859543s
Classify
The training took about 30 seconds. Now lets test our classifier on the test dataset. We will feed it examples and compare the predicted output the the known answers in the test dataset.
answer_label_array = utils.read_label(path, 'test.label')
correct_count = 0
expected = clf.predict(test_features[i])
if expected == actual:
correct_count+=1
5364/7505
0.714723517655
Results
So as you can see the classifier has a 28% error rate. This error rate could possibly be lowered by the use of different smoothing techniques or stop word analysis. | 2018-02-19 17:31:48 | {"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": 0.5224354267120361, "perplexity": 2838.8768649643966}, "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-2018-09/segments/1518891812758.43/warc/CC-MAIN-20180219171550-20180219191550-00356.warc.gz"} |
https://mathoverflow.net/questions/338956/maximum-principle-of-fractional-laplacian | # Maximum principle of fractional Laplacian
Suppose $$u$$ is a sign changing classical solution of the fractional Laplacian $$(-\Delta) ^{s} u = 0 \; \text{in } \Omega; u=g \text{ in } \mathbb R^N -\Omega .$$ (a)Is it true that $$\|u\|_{L^{\infty}(\mathbb R^N)}\leq \|g\|_{L^{\infty}(\mathbb R^N -\Omega)}$$ when $$s\in (0, 1).$$
(b) Is this also true for the operator $$(-\Delta) ^{s}+I$$ where $$I$$ is the identity.
• Given some minimal regularity of $u$ — of course yes! After all, $u$ is given as an integral of $g$ against a (sub-)probability measure, the harmonic measure for $(-\Delta)^s$ (also called the $\alpha$-harmonic measure or the Poisson kernel for $(-\Delta)^s$). Aug 22 '19 at 20:56
• If we suppose $u$ to be Holder continuous on $\mathbb R^N.$ I am aware of the notion of Poisson kernel.
– GabS
Aug 22 '19 at 21:31
• If your domain is nice -- Lipschitz for example -- it is sufficient to assume $u$ is continuous inside the domain and bounded near the boundary. Aug 23 '19 at 9:49
• If I am not wrong, I obtain that there exist a constant $C$ which depends on $\Omega$ whenever $\Omega$ is a unit ball, which is different from the Laplacian case.
– GabS
Aug 24 '19 at 9:06
• You might create the tag "fractional-laplacian", but "fractional" is a too imprecise tag.
– YCor
Feb 16 at 12:16
If $$u$$ is a solution of the above problem, then $$u$$ is said to be harmonic in $$\Omega$$ with respect to $$(-\Delta)^s$$. If $$u$$ is continuous and bounded in $$\Omega$$, and $$\Omega$$ satisfies the exterior cone condition, then $$u(x) = \int_{\mathbb{R}^N \setminus \Omega} g(y) P_\Omega(x, dy) ,$$ where $$P_\Omega(x, \cdot)$$ is a probability measure on the complement of $$\Omega$$. This measure is called the harmonic measure for $$(-\Delta)^s$$, and it has a nice probabilistic interpretation: it is the distribution of the isotropic $$2s$$-stable Lévy process started at $$x$$ and stopped at the time of first exit from $$\Omega$$. As a consequence, $$|u(x)| \leqslant \|g\|_\infty$$, as desired. | 2021-09-25 01:26:36 | {"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": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 21, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9181174635887146, "perplexity": 104.60507164262566}, "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/1631780057584.91/warc/CC-MAIN-20210924231621-20210925021621-00054.warc.gz"} |
https://lists.nongnu.org/archive/html/axiom-developer/2006-03/msg00293.html | axiom-developer
[Top][All Lists]
## Re: [Axiom-developer] Developing domain
From: Antoine Hersen Subject: Re: [Axiom-developer] Developing domain Date: Wed, 29 Mar 2006 22:09:00 -0800
Hello
> > Is there a nice iterative process to develop domain, I am
> > getting tired of quiting axiom deleting the directories and
> > restarting axiom.
>
> ???
>
> I use just:
>
> while unhappy repeat
> ... code ...
> ... test ...
vi !!!! (just kidding)
> Or increasingly these days I just use the Axiom Wiki SandBox.
>
> [SandBox My Test]
>
> Some explanations etc.
> ... code ...
>
> \begin{axiom}
> ... test ...
> end{axiom}
>
> Preview/Save
>
> Or are you one of those people that has an Axiom installation
> that does not properly cleanup the NRLIB directories? This does
> not happen on any of my Axiom systems (3 different flavours of
> Linux plus Windows). If so, any more documentation you can
Debian Stable, with the last arch version of axiom hand compiled,
anything I can do to give more useful info ?
> >
> > And is there a way to get axiom to execute some script as
> > an argument ?
> >
> > "axiom < some_files" does not work for me.
> >
>
> Try
>
> AXIOMsys < some_file
It works but quit as soon as it is done, I guess AXIOMsys is coded in
C and I could fix that myself
> > Also do someone know the intedent of the operator package
> > op.spad ? I have a lot of difficulty understanding Manuel
> > Bronstein works.
> >
>
> Check out section 9.58 of the Axiom book:
>
> 9.58 Operator
> Given any ring R, the ring of the Integer-linear operators
> over R is called Operator(R). To create an operator over R,
> first create a basic operator using the operation operator,
> and then convert it to Operator(R) for the R you want.
Yes I had read that, I just do not see the big picture.
Regards,
Antoine Hersen | 2019-11-13 15:14:48 | {"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": 0.8255078196525574, "perplexity": 8697.03178835076}, "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-2019-47/segments/1573496667262.54/warc/CC-MAIN-20191113140725-20191113164725-00193.warc.gz"} |
http://www.chegg.com/homework-help/questions-and-answers/find-all-possible-vertical-asymptotes-for-fx-x2-4-x2-2x-15-choosing-one-of-your-answer-sho-q3488144 | ## vertical asymptotes
find all possible vertical asymptotes for f(x)= (x^2 -4)/(x^2 -2x-15) choosing one of your answer, show with a graphing table on your calculator that it actually is an assymptote. | 2013-05-24 01:39:32 | {"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": 0.9013057351112366, "perplexity": 4369.3396608877165}, "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/1368704132298/warc/CC-MAIN-20130516113532-00083-ip-10-60-113-184.ec2.internal.warc.gz"} |
https://wiki.ledgerinvesting.com/insurance-fundamentals/premium-exposure-metrics.html | In this section, we briefly describe the main metrics of interest for policies, premium, and exposure, along with their basic definitions. The precise definitions of each of the metrics involve edge cases that we ignore at the moment for the sake of clear exposition. The diagram below sketches the overall relationship between the key metrics.
## Policy Metrics#
There are two primary useful measures of policies.
### Policies Written#
Policies written is a flow, and measures the number of policies written in a certain period. Policies written is computed as the count of distinct policy numbers with policy effective dates inside the experience period and record dates on or before the evaluation date. (This means that policies that have been written but have not gone in force yet are counted for the purposes of this metric.)
### Policies in Force#
Policies in force is a stock, and measures the number of policies currently in effect, or “in force”. Policies in force is commonly abbreviated as PIF, which can be pronounced “pee eye eff” or “piff”. PIF is computed as the count of distinct policy numbers with a policy effective date on or before the experience date and a policy expiration date on or after the experience date, minus any cancelled policies. As with policies written, the PIF calculation only considers policy transactions with record dates on or before the evaluation date.
The value of policies as a unit of measure depends on the line of business. Typically, policies are a more useful measure on personal lines than commercial lines. Each policy on a personal lines product typically equates to a household, whereas each policy on a commercial lines product typically equates to a business. In personal auto insurance, you may live in a household with a spouse, three teenage children, and four cars; whereas I may live alone and own only one car. Each of our households probably has one auto policy. The size of our policies, as measured by premium or exposure, is significantly different, but they are still on the same order of magnitude.
In commercial auto insurance, your business may have a single van that delivers flowers around town, while my business may have a fleet of 500 tractor trailers that transports hazardous chemicals across state lines. Each of our businesses may have one commercial auto policy. In this case, my business’s policy could have a premium several thousand times higher than yours. Policy counts should start to feel a little silly at this point, because it’s a count of such an inhomogeneous group of items.
There are several different ways to measure premium. The two most important measures for the purposes of risk analytics are written and earned premium. For any premium metric, we can compute it at the policy level or at an aggregate level. In general, the aggregate definitions of premium metrics are natural summations of the policy-level metrics.
Written premium is a measure of the total contractual premium associated with a set of insurance policies. In our toy policy data schema above, written premium for any policy is the sum of written premium for all transactions associated with that policy.
Written premium in the aggregate is a flow, and measures the total written premium for a certain period. When we compute written premium, we assign transactions to experience periods based on the policy effective date, not the transaction effective date. We’ll explain why this definition makes sense a little later, when we discuss endorsements and cancellations.
When an insurance policy is written, all of the written premium initially shows up on the insurer’s balance sheet as a liability called unearned premium, even if the policy has been paid in full. The reason for this is that at the start of the policy term, the insurer has received the premium for the policy term, but has yet to provide any of the contractually obligated coverage to the insured. By recognizing this unearned premium as a liability, it helps insurers to more appropriately match revenues with expenses. Furthermore, most insurance contract give the insured the right to cancel coverage at any point in time and return a prorated refund for the remainder of the policy term. The entire written premium isn’t truly the insurer’s until the end of the policy term.
How is premium earned? The standard assumption is that premium is earned evenly over the duration of the policy term. If I purchase a 6-month auto insurance policy for $600, then after two months, the insurer has earned$200 of the premium. We call this amount earned premium, naturally enough.
Earned premium is a much more fluid quantity than written premium. Written premium is set when a policy is issued, and it remains the same unless the policy has more transactions. By contrast, earned premium changes daily while the policy is in force. Before the policy effective date, earned premium is always $0 by definition. After the policy expiration date, earned premium is always equal to written premium by definition. For dates between the policy effective date and policy expiration date, earned premium goes up daily. Another difference between earned premium and written premium is that depending on the transactions on a policy, written premium can go up or down over time. However, earned premium only ever stays the same or goes up; it never goes down (with a few rare exceptions we’ll discuss later). #### Computing Earned Premium# Typically, we care about earned premium at the aggregate (program) level, rather than the policy level. However, there are a few different ways to aggregate earned premium, and the distinction between them can be subtle. We’ll start by computing policy-level earned premium, which will form the foundation of our alternate premium aggregation methods. Here is a brief worked example of earned premium calculations for a single policy. Say that an insurer writes a renter’s insurance policy with a written premium of$782. The policy effective date is 2022-06-13 and the policy expiration date is 2023-06-12. Assuming no further transactions on the policy, what is the earned premium as of 2022-10-24?
In this simple case, the expression for earned premium on a policy as of a given evaluation date is
\begin{split} \begin{align} \mathtt{EP\_policy} &= \mathtt{written\_prem} \cdot \mathtt{earn\_share} \\ \mathtt{earn\_share} &= \max(0, \min(1, \frac{\mathtt{earn\_days}}{\mathtt{policy\_days}})) \\ \mathtt{earn\_days} &= \mathtt{eval\_date} - \mathtt{pol\_eff\_date} + 1 \\ \mathtt{policy\_days} &= \mathtt{pol\_exp\_date} - \mathtt{pol\_eff\_date} + 1 \end{align} \end{split}
Substituting in our given values, we have
\begin{split} \begin{align} \mathtt{policy\_days} &= \mathtt{2023 \texttt{-} 6 \texttt{-} 12} - \mathtt{2022 \texttt{-} 6 \texttt{-} 13} + 1 \\ &= 365 \\ \mathtt{earn\_days} &= \mathtt{2022 \texttt{-} 10 \texttt{-} 24} - \mathtt{2022 \texttt{-} 6 \texttt{-} 13} + 1 \\ &= 134 \\ \mathtt{earn\_share} &= \max(0, \min(1, \frac{134}{365})) \\ &\approx 0.367123 \\ \mathtt{EP\_policy} &\approx 782 \cdot 0.367123 \\ &\approx 287.09 \\ \end{align} \end{split}
So the earned premium as of 2022-10-24 is about $287.09. #### Assumptions Behind Earned Premium# The motivation behind earned premium is that we wish to match premium to risks over time in a proportional fashion, or approximate it as closely as possible. If we know that that 40% of expected loss costs on a policy are going to be from accidents in a certain period, then we should earn 40% of the policy’s premium in that period as well. In the formulas provided above, we assume that instantaneous risk is effectively constant over time, so earned premium is linearly interpolated between$0 and the full written premium over the course of the policy term. Strictly speaking, this assumption is wrong for every single insurance policy. Insurance companies have reams of data demonstrating that auto accidents are less common on weekends than weekdays. Workers’ compensation claims only occur when employees are at work, so we would expect higher workers’ compensation claim rates on weekdays for white collar workers, and higher workers’ compensation claim rates on weekends for hospitality and retail workers. Seasonal patterns in the underlying risk are even more problematic than patterns driven by the day of the week. If we examine a program that specializes in providing general liability insurance to ski resorts, we shouldn’t be surprised to see losses much higher in the winter months!
Common practice in the insurance industry is to politely ignore these inconsistencies between the idealized assumptions behind earned premium and the real world. If there is a noticeable seasonal pattern to claims, the seasonal pattern is handled later in the modeling process, rather than recalibrating earned premium assumptions.
Now that we know how to compute earned premium for a given policy, how do we aggregate earned premium up for an experience period? As it turns out, there are two different ways to do this, each of which is useful in different contexts. The difference between the two aggregation methods is an attribute we call the risk basis. Risk basis can either be policy basis or accident basis.
Policy basis earned premium is easier to calculate, given policy level earned premium values. To compute policy basis earned premium, we sum the earned premium for all policies with policy effective dates within the experience period.
Accident basis earned premium is a bit more complex. In accident basis earned premium, we sum the portion of earned premium for each policy that was earned during the experience period. More formally, we can compute the accident basis earned premium for a given policy as:
\begin{split} \begin{align} \mathtt{EP\_accident} &= \mathtt{written\_prem} \cdot \mathtt{earn\_share} \\ \mathtt{earn\_share} &= \max(0, \min(1, \frac{\mathtt{earn\_days}}{\mathtt{policy\_days}})) \\ \mathtt{earn\_days} &= \min(\mathtt{eval\_date}, \mathtt{period\_end}) - \max(\mathtt{pol\_eff\_date}, \mathtt{period\_begin}) + 1 \\ \mathtt{policy\_days} &= \mathtt{pol\_exp\_date} - \mathtt{pol\_eff\_date} + 1 \end{align} \end{split}
Note that the definition is nearly the same, but for the addition of two new variables, $$\mathtt{period\_begin}$$ and $$\mathtt{period\_end}$$, which represent the first and last days of the experience period, respectively. The reason accident basis is called that may seem a little obscure now, but it will make more sense when we cover loss aggregation.
When accident basis or policy basis metrics are computed with annual experience periods, they are often called accident year or policy year metrics, and are abbreviated AY and PY, respectively. For example, unless explicitly mentioned otherwise, AY2019 is shorthand for the accident basis experience period 2019-01-01/2019-12-31. Similarly, accident quarter, policy quarter, accident month, and policy month are all shorthand for experience periods with natural calendar alignments: e.g., policy quarter 2021Q2 is the policy basis experience period 2019-04-01/2019-06-30.
#### Accident Basis vs Policy Basis#
There are a major difference between policy basis and accident basis earned premium in terms of their behavior. Barring a few rare circumstances we will cover later, accident basis earned premium is identical for any evaluation date after the end of the experience period. By contrast, policy basis earned premium will continue to change as a function of evaluation date until the last policy from the experience period reaches the end of its policy term – this delay could easily be a year or more after the end of the experience period.
Given the difference between accident basis and policy basis for earned premium, it is natural to ask whether there are accident basis and policy basis versions of written premium. The short answer is that the definition of written premium we just covered is effectively a policy basis definition, but that an accident basis definition of written premium doesn’t have an obviously interpretation or real-world application, so it is not used in practice.
#### Treaty Basis#
There is another risk basis in common usage: treaty basis. Risk originators typically obtain reinsurance coverage via reinsurance treaties. These treaties usually last about a year, but some treaties are longer or shorter than 12 months for the convenience of one or more of the parties to the contract. For example, a hypothetical program could have the following treaty dates:
• Treaty 1: 2017-04-01/2018-03-31
• Treaty 2: 2018-04-01/2019-06-30
• Treaty 3: 2019-07-01/2020-08-31
• Treaty 4: 2020-09-01/2021-05-31
• Treaty 5: 2021-06-01/2022-05-31
This is a bit of an extreme example because only two of the five treaty years are exactly 12 months, but it illustrates the idea. All policies with a policy effective date within a treaty period are assigned to that treaty. Therefore, computing earned premium for a treaty is equivalent to computing policy basis earned premium for the treaty period. Treaties are important because reinsurers take risk on a treaty-by-treaty basis, and funds are transferred on the basis of treaty year performance. For these reasons, risk originators and reinsurers are intensely interested in treaty-by-treaty performance. Treaty basis is simply an easy way to reference treaty-level performance without explicitly specifying ever-shifting treaty dates. For example PY2018 refers to the policy experience period 2018-01-01/2018-12-31, but treaty year (TY) 2018 refers to the treaty period that started in 2018 – in this example, 2018-04-01/2019-03-31. It is much more convenient to specify TY2018, TY2019, etc., than to laboriously specify the exact treaty period for each successive treaty. | 2022-12-07 17:27:13 | {"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": 2, "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": 3, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8200528025627136, "perplexity": 2535.74675222636}, "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/1669446711200.6/warc/CC-MAIN-20221207153419-20221207183419-00222.warc.gz"} |
https://cs.stackexchange.com/questions/71188/intuitive-explanation-on-converting-bnf-grammar-to-lr1 | # Intuitive Explanation on Converting BNF Grammar to LR(1)
Consider the following BNF Grammar G:
object-expression ::= id | id ( object-expression ) | object-expression . id
| object-expression [ object-expression ]
| object-expression . id ( object-expression )
I am attempting to construct a set of LR(1) items based on G.
Attempt:
Let's represent Grammar G as:
O -> id
O -> id ( O )
O -> O . id
O -> O [ O ]
O -> O . id ( O )
Item Set 1:
O' -> • O, $O -> • id,$
O -> • id ( O ), $O -> • O . id,$
O -> • O [ O ], $O -> • O . id ( O ),$
Can someone provide an intuitive explanation in simple language on how to finish obtaining a set of LR(1) items given a BNF Grammar such as G?
• @YuvalFilmus I've seen the algorithm, but I am unable to find a source that gives an example like the one mentioned. For example, I do not know if id needs to be in the item set, and if the lookahead needs to be determined for id. Any suggestions? – Grant Miller Mar 6 '17 at 2:30 | 2021-04-23 14:42:20 | {"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": 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": 0.4628625214099884, "perplexity": 3881.6632612674175}, "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/1618039594808.94/warc/CC-MAIN-20210423131042-20210423161042-00533.warc.gz"} |
https://www.proofwiki.org/wiki/Scalar_Product_with_Inverse | # Scalar Product with Inverse
## Theorem
Let $\struct {G, +_G}$ be an abelian group.
Let $\struct {R, +_R, \times_R}$ be a ring.
Let $\struct {G, +_G, \circ}_R$ be an $R$-module.
Let $x \in G, \lambda \in R$.
Then:
$\lambda \circ \struct {-x} = \struct {-\lambda} \circ x = -\struct {\lambda \circ x}$
## Proof
From Module Axiom $\text M 1$: Distributivity over Module Addition, $y \to \lambda \circ y$ is an endomorphism of $\struct {G, +_G}$.
From Module Axiom $\text M 2$: Distributivity over Scalar Addition, $\mu \to \mu \circ x$ is a homomorphism from $\struct {R, +_R}$ to $\struct {G, +_G}$.
The result follows from Homomorphism with Identity Preserves Inverses.
$\blacksquare$ | 2022-06-28 09:19:43 | {"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": 0.9835758805274963, "perplexity": 529.3655634030788}, "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-2022-27/segments/1656103360935.27/warc/CC-MAIN-20220628081102-20220628111102-00679.warc.gz"} |
https://www.rdocumentation.org/packages/ggperiodic/versions/1.0.0 | # ggperiodic v1.0.0
0
0th
Percentile
## Easy Plotting of Periodic Data with 'ggplot2'
Implements methods to plot periodic data in any arbitrary range on the fly.
# ggperiodic
ggperiodic is an attempt to solve the issue of plotting periodic data in ggplot2. It automatically augments your data to wrap it around to any arbitrary domain.
## Installation
install.packages("ggperiodic")
Or you can install the development version from GitHub with:
# install.packages("devtools")
devtools::install_github("eliocamp/ggperiodic")
## Example
Let’s create some artificial data with periodic domain
x <- seq(0, 360 - 10, by = 10)*pi/180
y <- seq(-90, 90, by = 10)*pi/180
Z <- expand.grid(x = x, y = y)
Z$z <- with(Z, 1.2*sin(x)*0.4*sin(y*2) - 0.5*cos(2*x)*0.5*sin(3*y) + 0.2*sin(4*x)*0.45*cos(2*x)) Z$x <- Z$x*180/pi Z$y <- Z\$y*180/pi
If you try to plot it, you’ll notice problems at the limits
library(ggplot2)
ggplot(Z, aes(x, y, z = z, color = ..level..)) +
geom_contour() +
coord_polar()
With ggperiodic you can define the periodic dimensions and ggplot2 does the rest.
library(ggperiodic)
#>
#> Attaching package: 'ggperiodic'
#> The following object is masked from 'package:stats':
#>
#> filter
Z <- periodic(Z, x = c(0, 360))
ggplot(Z, aes(x, y, color = ..level..)) +
geom_contour(aes(z = z)) +
coord_polar()
## Functions in ggperiodic
Name Description unperiodic Remove periodic specifications qwrap Quickly wrap data wrap Wrap periodic data to an arbitrary range reexports Objects exported from other packages is.periodic Check if an object is periodic periodic Add or remove periodic variables get_period Get period information from an object ggperiodic ggperiodic: Easy Plotting of Periodic Data with 'ggplot2' No Results! | 2020-04-10 00:36:13 | {"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": 0.18861843645572662, "perplexity": 13862.08492377029}, "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/1585371880945.85/warc/CC-MAIN-20200409220932-20200410011432-00525.warc.gz"} |
https://ncatlab.org/nlab/show/homotopical+algebraic+quantum+field+theory | # nLab homotopical algebraic quantum field theory
### Context
#### Higher algebra
higher algebra
universal algebra
# Contents
## Idea
The axioms of traditional algebraic quantum field theory or locally covariant perturbative quantum field theory, where one assigns a plain algebra (e.g. a C*-algebra or formal power series algebra) of observables to a spacetime region, fail when one considers gauge theory on spacetimes more general than Minkowski spacetime (AQFT on curved spacetimes):
The nature of gauge fields implies that to re-construct their global configuration (including non-trivial topology as in instanton sectors) one needs to locally remember the groupoid of gauge transformations (see Eggertsson 14, Schenkel 14, Schreiber 14 for exposition), and hence, after quantization, some higher algebra of observables.
The program of homotopical algebraic quantum field theory (Benini-Schenkel 16, see Schenkel 17 for survey) is to lift the axioms of AQFT on curved spacetimes to certain local nets of homotopical algebras such as to properly capture gauge theory with topologically non-trivial gauge field configurations.
The axioms as considered in Schenkel 17 have some similarity with that of factorization homology, but also do take into account the causality condition for quantum fields on Lorentzian manifolds.
## References
For review see
Construction and axiomatization of gauge field AQFT via homotopy theory and homotopical algebra is being developed in
The stack of Yang-Mills gauge fields is discussed in
Revised on September 19, 2017 06:21:44 by Urs Schreiber (185.25.95.132) | 2017-11-23 05:31:53 | {"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": 0.865079402923584, "perplexity": 2342.8625754756645}, "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/1510934806736.55/warc/CC-MAIN-20171123050243-20171123070243-00335.warc.gz"} |
https://aviation.stackexchange.com/questions/34568/how-much-air-by-mass-enters-an-average-cfm56-turbofan-engine-cruising-per-minu | # How much air, by mass, enters an average CFM56 turbofan engine cruising per minute?
How many kg of air enters an average CFM56 turbofan engine cruising per minute?
I have looked everywhere and cannot find it or figure out how to find it, if you could either tell me or give advice on where or how to find it I would appreciate it. Thanks
## migrated from physics.stackexchange.comJan 11 '17 at 0:11
This question came from our site for active researchers, academics and students of physics.
• A reasonable first approximation would be the mass of air that the engine flies through in a minute - i.e. frontal area of the fan blades $\times$ aircraft velocity $\times$ air density. As a sanity check on your estimate, though at different conditions where the inlet velocity is not equal to the aircraft speed, at takeoff each engine on a B747 generates about 200kN of thrust, with an air mass flow of about 1 tonne (1000kg) per second. – alephzero Jan 10 '17 at 23:34
• Aviation.SE is the right place to ask this. – Mo_ Jan 10 '17 at 23:44
• When you say "how much" do you mean by mass or by volume? – TomMcW Jan 11 '17 at 1:47
• @SRawes Put that back in :) – TomMcW Jan 11 '17 at 16:04
The CFM56-5C, the largest CFM56, powers A340. Its thrust is about 145 kN at takeoff. It has an air mass flow of 485 kg/s (source).
CFM56-5C, source.
The Trent XWB (A350 XWB), a larger engine developing 430 kN at takeoff, has an air mass flow of 1,440 kg/s (source).
T/O vs cruise power and fuel consumption
According to this document, the CFM56-5C T/O thrust is 150 kN and the cruise thrust is 40 kN (other sources states 30 kN), 27% of the T/O thrust. Related fuel consumption are 2.3 and 0.5 kg/s, meaning cruise fuel consumption is 22% of the T/O fuel consumption.
Air mass vs fuel mass
For a fuel consumption of 22% of the T/O consumption, it can be assumed relatively safely that the air flow is in proportion, so we can roughly estimate air mass flow in cruise to be about 485 * 0.22 = 107 kg/s. This is supported by figures found online in non authoritative sources, like page 245 of this book:
• will that not be different at a cruising altitude @mins – SRawes Jan 11 '17 at 16:00
• Is this figure the amount passing through the intake, or the amount passing through the combustion chamber? Can you break down the two for a feel for how much air is bypassed? – PJNoes Jan 11 '17 at 16:54
• @PJNoes: This is the total air mass thru the fan. After the fan, air is separated in two flows: core (hot/primary) and bypass (cold/secondary) according to the bypass ratio, here 6.4:1. So the part used in the core for combustion, bleed air and core cooling is 485/7.4 = 66 kg/s and 419 kg/s are bypassing the core. – mins Jan 11 '17 at 18:49
• @SRawes: Yes, this is the maximum thrust/airflow. I have updated the answer. – mins Jan 11 '17 at 19:33 | 2019-09-22 05:13:40 | {"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": 0.5964239835739136, "perplexity": 2124.071144491615}, "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-2019-39/segments/1568514575076.30/warc/CC-MAIN-20190922032904-20190922054904-00315.warc.gz"} |
https://blog.myrank.co.in/cylindrical-capacitor/ | # Cylindrical Capacitor
## Cylindrical Capacitor
A cylindrical capacitor consists of two coaxial cylinders of radii a and b, say b > a. The outer one is earthed. The cylinders are long enough so that we can neglect fringing of electric fields at the ends. The electric field at a point between the cylinders will be radial, and its magnitude will depend on the distance from the central axis.
Consider a Gaussian surface of length y and radius r such that a < r < b. Flux through the plane surface is zero because the electric field and the area vector are perpendicular to each other.
For the curved part:
$$\phi =\int{\overrightarrow{E}.\overrightarrow{ds}}=\int{Eds}=E\int{ds}=E2\pi ry$$,
Consider inside the Gaussian surface is: $$q=\frac{Qy}{L}$$,
From Gauss’s law: $$\phi =E2\pi ry=\frac{Qy}{L{{\varepsilon }_{0}}}$$,
$$E=\frac{Q}{2\pi {{\varepsilon }_{0}}Lr}$$,
Potential difference is: $${{V}_{b}}-{{V}_{a}}=-\int\limits_{a}^{b}{\overrightarrow{E}}.\overrightarrow{dr}=-\int\limits_{a}^{b}{\frac{Q}{2\pi {{\varepsilon }_{0}}Lr}dr}=-\frac{Q}{2\pi {{\varepsilon }_{0}}L}\int\limits_{a}^{b}{\frac{1}{r}dr}$$,
$${{V}_{a}}=\frac{Q}{2\pi {{\varepsilon }_{0}}L}\ln \frac{b}{a}\,\,\,\,\,\,\,\,\left( \because \,\,{{V}_{b}}=0 \right)$$,
Now, the capacitance is: $$C=\frac{Q}{{{V}_{a}}-{{V}_{b}}}=\frac{Q}{{{V}_{a}}}=\frac{2\pi {{\varepsilon }_{0}}L}{\ln \left( \frac{b}{a} \right)}$$. | 2022-08-18 03:50:32 | {"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": 0.9245461225509644, "perplexity": 329.5801868272045}, "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-2022-33/segments/1659882573163.7/warc/CC-MAIN-20220818033705-20220818063705-00686.warc.gz"} |
https://thoughtstreams.io/nyergler/til/5742/ | # TIL
9 thoughts
last posted Aug. 30, 2014, 8:41 p.m.
2 earlier thoughts
0
There's also this magical thing that Ruby seems to do, where you can omit the {} around your keyword arguments, and it collects them up into a single hash. I'm not sure how I feel about that.
6 later thoughts | 2022-07-03 20:48:08 | {"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": 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": 0.5566977262496948, "perplexity": 4285.60551342016}, "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/1656104249664.70/warc/CC-MAIN-20220703195118-20220703225118-00055.warc.gz"} |
https://chem.libretexts.org/Courses/SUNY_Oneonta/Chem_221%3A_Organic_Chemistry_I_(Bennett)/1%3ALecture_Textbook/07%3A_Nucleophilic_Substitution_Reactions/7.08%3A_Biological_Nucleophilic_Substitution_Reactions | # 7.8: Biological Nucleophilic Substitution Reactions
The nucleophilic substitution reactions we have seen so far have all been laboratory reactions, rather than biochemical ones. Now, finally, let's take a look at a few examples of nucleophilic substitutions in a biological context. All of the principles we have learned so far still apply to these biochemical reactions, but in addition we need to consider the roles of the enzyme catalysts.
A word of encouragement
This is the first time that we will be seeing 'real' biological organic reaction mechanisms. Do not be intimidated by the size and complexity of the reacting biomolecules - they are just organic molecules, with the same bonding patterns and functional groups that you are already familiar with. Focus on the reacting parts of the molecule: what is the nucleophile? The electrophile? The leaving group? In most biological organic reactions, the main bulk of the biomolecule is just 'going along for the ride', and can often be abbreviated with an 'R group' (section 1.2) to simplify the picture.
## A biochemical $$S_N2$$ reaction
One very important class of nucleophilic substitution reactions in biochemistry are the $$S_N2$$ reactions catalyzed by $$S$$-adenosyl methionine (SAM) – dependent methyltransferase enzymes. SAM is a coenzyme (section 6.3) that plays the role of methyl group donor: you can think of SAM in this context as being simply a methyl carbon electrophile attached to a sulfide leaving group.
There are many variations of SAM-dependent methylation reactions in nature. In the introduction to this chapter, we were introduced to a reaction occurring in bacterial DNA in which a methyl carbon is transferred from SAM to a nitrogen atom on adenine (this type of reaction is often referred to as $$N$$-methylation).
In the figure above, we are showing how an aspartate residue in the active site of the enzyme acts as a catalytic base: transfer of a proton from substrate to the aspartate side chain begins to enhance the nucleophilicity of the amine nitrogen as it approaches the electrophilic methyl carbon of SAM, and formation of the new $$N-C$$ bond and cleavage of the $$C-S$$ bond begins. These four bond-rearranging events probably take place in concerted fashion. A likely transition state is approximated below:
Of course, there are many other noncovalent interactions between active site enzyme residues and the substrate (the adenine base) and cofactor (SAM), but in the interest of clarity these are not shown. These interactions, many of which are hydrogen-bonds, help to position the adenine base and SAM in just the right relative orientation inside the active site for the nucleophilic attack to take place. (If you have access to American Chemical Society journals, a paper about an enzyme catalyzing a similar N-methylation reaction contains some detailed figures showing hydrogen-bond and charge-dipole interactions between the enzyme active site and the two substrates: see Biochemistry 2003, 42, 8394, figure 4).
The electrophile is a methyl carbon, so there is little steric hindrance to slow down the nucleophilic attack. The carbon is electrophilic (electron-poor) because it is bonded to a positively-charged sulfur, which is a powerful electron withdrawing group. The positive charge on the sulfur also makes it an excellent leaving group, because as it leaves, it becomes a neutral and very stable sulfide. All in all, we have a good nucleophile (enhanced by the catalytic base), an unhindered electrophile, and an excellent leaving group. We can confidently predict that this reaction is $$S_N2$$. An $$S_N1$$ mechanism is extremely unlikely: a methyl cation is very unstable and thus is not a reasonable intermediate to propose.
Notice something else about the SAM methylation mechanism illustrated in the previous figure. It is termolecular: there are three players acting in concert: the catalytic base, the nucleophile, and the electrophile. This is possible because all three players are bound in a very specific geometry in the active site of the enzyme. In a reaction that takes place free in solution, rather than in an active site, the likelihood of three separate molecules colliding all at once, with just the right geometry for a reaction to take place, is very, very low. You should notice going forward that when we illustrate the mechanism of a reaction that takes place free in solution, we will only see bimolecular steps - two molecules colliding. Almost all of the biochemical reactions we see in this book will be enzyme-catalyzed - and termolecular steps will be common - while almost all of the laboratory reactions we see will take place free in solution, so we will only see unimolecular and bimolecular steps. (Synthetic chemists often employ non-biological catalysts that mimic enzyme active sites, but these examples are well beyond the scope of our discussion).
Exercise $$\PageIndex{1}$$
Think back to the acid-base chapter: the $$pK_a$$ of a protonated ether is approximately zero, indicating that an ether is a very weak base. Considering periodic trends in acidity and basicity, what can you say about the relative basicity of a sulfide?
Another SAM-dependent methylation reaction is catalyzed by an enzyme called catechol-$$O$$-methyltransferase. The substrate here is epinephrine, also known as adrenaline, and the reaction is part of the pathway by which adrenaline is degraded in the body.
Notice that in this example, the attacking nucleophile is a phenol oxygen rather than a nitrogen (that’s why the enzyme is called an $$O$$-methyltransferase). In many cases when drawing biochemical reaction mechanisms, we use the abbreviations B: for a catalytic base and $$H-A$$ for a catalytic acid, in order to keep the drawings from getting too 'busy' (it's also possible that the identity of the acidic or basic group may not be known).
Exercise $$\PageIndex{2}$$
SAM is formed by a nucleophilic substitution reaction between methionine and adenosine triphosphate (ATP). Draw a mechanism for this reaction, and explain why you chose either an $$S_N1$$ or and $$S_N2$$ pathway.
## A biochemical $$S_N1$$ reaction
As we will see in chapter 10, enzyme-catalyzed $$S_N1$$ reactions play a critical role in carbohydrate and DNA/RNA nucleotide metabolism. The reaction below is part of nucleotide biosynthesis:
Notice a few things here: first, the diphosphate leaving group is stabilized by interactions with a $$Mg^{+2}$$ ion bound in the active site and also by hydrogen-bonding with active site amino acid residues (not shown). The carbocation intermediate is stabilized by resonance with the lone pairs on the oxygen (see section 8.5), and also by an active site aspartate side chain. The ammonia nucleophile is positioned in the active site so that it approaches from the 'top' side of the planar carbocation intermediate, and the substitution results in inversion of configuration. Remember: $$S_N1$$ reactions which occur free in solution tend to result in a mixture of stereoisomers, but enzyme-catalyzed reactions - including enzymatic $$S_N1$$ reactions such as this one - are generally stereo- and regio-specific, meaning that they almost always result in a single isomeric product, not a mixture of products.
Recall the statement from section 8.4 that poor leaving groups often need to be converted into good leaving groups. Backing up one metabolic step from the reaction depicted above, we see that a poor (hydroxide) leaving group on ribose-5-phosphate is first converted to a good (diphosphate) leaving group, which can be stabilized through interactions with the active site of the enzyme catalyzing the $$S_N1$$ reaction.
This preliminary phosphorylation step, which requires ATP (adenosine triphosphate) as the donor of the diphosphate group, is a reaction that we will study in much more detail in chapter 9.
## A biochemical $$S_N1/S_N2$$ hybrid reaction
The cysteine residues of certain proteins are modified by addition of a 15-carbon isoprene chain (section 1.3) to the side chain thiol group.
The mechanistic details of this reaction are of particular interest to biomedical scientists. The proteins that are substrates for this type of modification are involved in cell signaling processes, and they are not able to carry out their biological functions unless they are anchored to a cell’s lipid membrane. The hydrocarbon group that becomes attached to a cysteine residue in this reaction serves as the anchor.
Some of these proteins have been implicated in tumor formation. Scientists hope that if they can find a way to shut down the cysteine modification reaction, the tumor-causing proteins will not be able to anchor to cell membranes and thus will remain inactive. The search is on for an effective inhibitor of this enzyme to serve as a potential anti-tumor drug.
How does the enzyme lower the energy barrier for this reaction? Experimental evidence indicates that when a substrate protein is bound to the active site of the enzyme, the cysteine thiol associates with a zinc ion bound in the active site. As we learned in section 7.8, this association will lower the pKa of the thiol to the point where it loses a proton and exists as a thiolate anion in the active site - a thiolate is a very potent nucleophile! Studies also show that the diphosphate group forms stabilizing interactions with several amino acid residues (two lysines, an arginine, a histidine, and a tyrosine) in the enzyme's active site, making it a weaker base and thus a better leaving group.
Is protein prenylation an $$S_N1$$ or $$S_N2$$ reaction? In other words, to what extent does the nucleophile displace, or 'push' the leaving group off, or to what extent does the leaving group leave on its own, without a 'push' from the nucleophile? Along the same lines, to what extent does a positive charge develop on the carbon center (development of a full positive charge implies an $$S_N1$$ mechanism). First, consider the electrophile: it is a primary allylic carbon, so either pathway is possible (it is relatively unhindered for $$S_N2$$ attack, but could also form a resonance-stabilized carbocation intermediate in an $$S_N1$$ pathway). The nucleophile is a very powerful thiolate ion, suggestive of an $$S_N2$$ mechanism where a strong nucleophile actively displaces the leaving group.
In fact, experiments designed to address this very question (see problem P8.19) have provided evidence that the reaction is a mechanistic hybrid: essentially $$S_N2$$, but with elements of $$S_N1$$. In other words, at the transition state the electrophilic carbon takes on some degree of positive charge, but a true carbocation intermediate does not form. The take-home message here is that the $$S_N1$$ and $$S_N2$$ mechanistic pictures we have studied in this chapter are models, and while they are useful for learning about chemical principles and accurate for describing many substitution reactions, other reactions are not necessarily 'pure' $$S_N1$$ or $$S_N2$$, but actually lie somewhere in between.
## Contributors
7.8: Biological Nucleophilic Substitution Reactions is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by Tim Soderberg. | 2022-05-19 12:05:22 | {"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": 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": 0.6753953099250793, "perplexity": 1864.150001688396}, "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/1652662527626.15/warc/CC-MAIN-20220519105247-20220519135247-00087.warc.gz"} |
http://mathhelpforum.com/number-theory/119938-how-do-you-find-last-digit-number-using-fermat-s-little-theorem.html | # Math Help - How do you find the last digit of a number using Fermat's Little Theorem
1. ## How do you find the last digit of a number using Fermat's Little Theorem
For example, what is the last digit of 17^(3241)? It would help a ton if someone could show how to solve this using Fermat's Little Theorem. I have finals coming up and this was something I never quite understood
2. Originally Posted by steph3824
For example, what is the last digit of 17^(3241)? It would help a ton if someone could show how to solve this using Fermat's Little Theorem. I have finals coming up and this was something I never quite understood
I'm not sure why one would need Fermat's little theorem for this since:
$17^3 \equiv 1 \text{ mod } 10$
and:
$17^{3241}=(17^3)^{1080}\times 17$
which immediately should tell you that:
$17^{3241}\equiv 7 \text{ mod } 10$
CB
3. Originally Posted by CaptainBlack
I'm not sure why one would need Fermat's little theorem for this since:
$17^3 \equiv 1 \text{ mod } 10$
and:
$17^{3241}=(17^3)^{1080}\times 17$
which immediately should tell you that:
$17^{3241}\equiv 7 \text{ mod } 10$
CB
Incorrect!
$17^4 \equiv 1 (mod 10)$
$17^{3241}=(17^4)^{810}\times 17$
So $17^{3241}\equiv 7(mod 10)$
4. Originally Posted by alexmahone
Incorrect!
$17^4 \equiv 1 (mod 10)$
$17^{3241}=(17^4)^{810}\times 17$
So $17^{3241}\equiv 7(mod 10)$
Oppsss.. arithmetic error (actually a transcription error with argument completed with the erroeous value)
CB
5. Originally Posted by steph3824
For example, what is the last digit of 17^(3241)? It would help a ton if someone could show how to solve this using Fermat's Little Theorem. I have finals coming up and this was something I never quite understood
Here is a solution via Fermat's Last Theorem (although it's not necessarily the simplest):
By Fermat's Last Theorem, $17^4 \equiv 1 \mod 5$,
so since
$17^{3241} = (17^4)^{810} \cdot 17$,
$17^{3241} \equiv 1 \cdot 17 \equiv 2 \mod 5$.
We also have $17 \equiv 1 \mod 2$, so $17 ^ {3241} \equiv 1 \mod 2$.
Since $17^{3241} \equiv 2 \mod 5$ and $17 ^ {3241} \equiv 1 \mod 2$,
$17^{3241} \equiv 7 \mod 10$.
We could use the Chinese Remainder Theorem at the last step, but it seems like overkill here. | 2016-05-24 16:18:41 | {"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": 20, "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": 0.7660071849822998, "perplexity": 324.4453538141136}, "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-2016-22/segments/1464049272349.32/warc/CC-MAIN-20160524002112-00208-ip-10-185-217-139.ec2.internal.warc.gz"} |
https://digital-library.theiet.org/content/books/10.1049/pbpo139e_ch7 | http://iet.metastore.ingenta.com
1887
## Sizing of microgrid components
• Author(s):
• DOI:
$16.00 (plus tax if applicable) ##### Buy Knowledge Pack 10 chapters for$120.00
(plus taxes if applicable)
IET members benefit from discounts to all IET publications and free access to E&T Magazine. If you are an IET member, log in to your account and the discounts will automatically be applied.
Recommend Title Publication to library
You must fill out fields marked with: *
Librarian details
Name:*
Email:*
Name:*
Email:*
Department:*
Why are you recommending this title?
Select reason:
Variability, Scalability and Stability of Microgrids — Recommend this title to your library
## Thank you
A microgrid (MG) is a distinct energy system consisting of distributed energy resources (DERs) and loads having the ability to operate in parallel with, or independently from, the main power grid. MGs, which were initially introduced to ensure smooth operation and control of DERs in distribution networks, offer unprecedented economic and reliability benefits to electricity consumers with minimal carbon emission. These benefits, however, must be analysed and compared with the capital investment cost of the MG to ensure a complete return on investment and to justify the MG deployment. The biggest obstacle for the widespread and rapid deployment of MGs is the high capital investment cost of MGs. A true assessment of MGs economic benefits is a challenging task due to the significant uncertainties involved in the assessment. These uncertainties may include the intermittency of the renewable generation, the varying states of charge (SoC) of battery energy storage system (BESS), the uncertain demands, the varying market price, the probability of the MG islanding, the level of developer's risk-aversion and the unpredictably of the user preferences in the smart load management system. Moreover, some of the assessment metrics, such as the measure of reliability improvements are difficult to comprehend for consumers when represented in terms of the supply availability. Thus, efficient and optimum planning models are required to ensure the economic feasibility of MG deployments and to justify the investments based on cost-to-profit analysis under uncertain conditions. This chapter demonstrates a detailed model for the optimum sizing of MG components under the uncertainties involved in the system. The proposed model is validated with the simulation of several case studies conducted on a system depicting a similar MG in a medium-voltage (MV)-distribution system derived from electricity network of a power utility in New South Wales, Australia. The results from the case studies demonstrate the efficacy of the proposed model for the optimum sizing of the MG components to justify the MG deployment.
Chapter Contents:
• 7.1 Microgrid components
• 7.2 Microgrid sizing and profit maximization
• 7.3 Models of distributed energy resources
• 7.3.1 Probabilistic wind power output model
• 7.3.2 Probabilistic photovoltaic power output model
• 7.3.3 Dynamic battery energy storage power output model
• 7.3.4 Micro-turbine power output model
• 7.4 Optimal sizing of microgrid components
• 7.4.1 Mathematical formulation
• 7.4.2 Backtracking search optimization (BSO) algorithm
• 7.4.3 Solution approach
• 7.5 Case studies
• 7.5.1 Case study 1
• 7.5.2 Case study 2
• 7.5.2.1 Scenario 1
• 7.5.2.2 Scenario 2
• 7.6 Summary
• References
Preview this chapter:
Sizing of microgrid components, Page 1 of 2
| /docserver/preview/fulltext/books/po/pbpo139e/PBPO139E_ch7-1.gif /docserver/preview/fulltext/books/po/pbpo139e/PBPO139E_ch7-2.gif
### Related content
content/books/10.1049/pbpo139e_ch7
pub_keyword,iet_inspecKeyword,pub_concept
6
6
This is a required field | 2019-09-21 15:08:29 | {"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": 0.31794583797454834, "perplexity": 3805.507941690811}, "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-2019-39/segments/1568514574532.44/warc/CC-MAIN-20190921145904-20190921171904-00403.warc.gz"} |
https://www.foborji.org/iv-drug-gwxciij/what-did-liberal-nationalism-stand-for-brainly-c80fd9 | # what did liberal nationalism stand for brainly
Report ; Posted by Shruti Janghu 2 hours ago. Rise of Nationalism in Europe - During the early nineteenth century the idea of Nationalism and Liberalism were closely related to the Europeans. Politically it emphasised the concept of government. Report ; Posted by . 4 months, 2 weeks ago. Name the ancient Indian University where the Chinese traveller YuanChwang stayed fortwo years *O NalandaMumbai9.Which famous Indo-Greek King had di Thus, a "civic nation" is defined by not language or culture but political institutions and liberal principles, which its citizens pledge to uphold. Frédéric Sorrieu, a French artist, in 1848 prepared a series of four prints visualising his dream of a world made up of democratic and Social Republics. sannia sannia 2 hours ago Social Sciences Secondary School +5 pts. Join now. : (1) The term ‘Liberalism’ derives from the Latin root ‘liber’ meaning free. The most important aspect of liberal nationalism is the right of all nations to self-determination. England had become the leading nation in scientific spirit, in commercial enterprise, and in political thought and activity. â¦, who is also called as forunners of gandhiâ, ARAHAN: Bahagian ini mengandungi 4 soalan. Join now. During the nineteenth century this was a strong demand of the emerging middle classes. Ask your question. In the economic sphere, liberalism stood for the freedom of markets and the abolition of state- imposed restrictions on the movement of goods and capital. Ask questions, doubts, problems and we will help you. Join now. Ernest Renan and John Stuart Mill are often thought to be early liberal nationalists. Liberal / civic nationalism is a form of nationalism often juxtaposed to cultural / ethnic nationalism. The word ‘ liberalism’ is derived from the Latin word ‘liber’, which means free. Join now. You can specify conditions of storing and accessing cookies in your browser, What did Liberal nationalism stand for? For the new middle classes liberalism stood for freedom for the individual and equality of all before the law. Here the question arises as to what exactly do we mean by Liberal Nationalism? The term ‘liberalism’ derives from the Latin root liber, meaning free. Civic nationalism, also known as liberal nationalism, is a form of nationalism identified by political philosophers who believe in an inclusive form of nationalism that adheres to traditional liberal values of freedom, tolerance, equality, and individual rights. For the new middle classes liberalism stood for freedom for the individual and equality of all before the law. Log in. 1. $$\huge\mathrm{\underline{\purple{AnSwEr}}}$$Civic nationalism, also known as liberal nationalism, is a form of nationalism identified by political philo… 1. monarchy with national unification. â¦, dikan dankehidupan masyarakat China.a) Jelaskan ciri-ciri sistem pendidikan perkhidmatan awam Tamadun China.b) Bagaimanakan peperiksaan perkhidmatan awam Tamadun China dilaksanakan? Log in. 0 answers. 1. Log in. What did liberal nationalism stand for - 22673901 1. CBSE > Class 10 > Social Science 1 answers; Nidhi Jha 1 year, 3 months ago. Since the French Revolution, liberalism had stood for the end of authocracy and clerical privileges a constitutional … The term ‘liberalism’ derives from the Latin root liber, meaning free. Liberal nationalism, also known as civic nationalism or civil nationalism, is a kind of nationalism identified by political philosophers who believe in a non-xenophobic form of nationalism compatible with liberal values of freedom, tolerance, equality, and individual rights. What did liberal nationalism stands for in political sphere?? Ideas of national unity in early-nineteenth-century Europe were closely allied to the ideology of liberalism. c) Mengapakah sistem peperiksaan perkhidmatan awam China menjadi contoh kepada negara-negara didunia hingga ke hari ini?â, join Google meeting please tzc-pcwf-tgoâ, mujhko jalan.. XD.khud ko bol rha tha...vo I'd meri hi hai.. XD..but use koi aur kr raha haiâ, the modern period of indian history is also know as the. asked Mar 19, 2018 in Class X Social Science by priya12 (-12,631 points) the rise of nationalism in europe. Explain any three reasons for nationalists upsearch in the 19th century europe. Nationalism was important because both countries consisted of numerous smaller states where the people of same ethnicity lived except in different states. In Europe the new middle class saw … This site is using cookies under cookie policy. Your explaining of answer really help me thanku so much one again 1 Thank You. The world’s largest social learning network for students. For the new middle classes liberalism stood for freedom for the individual and equality of all before the law. (ii) What did Liberal Nationalism Stand for ? Peperiksaan perkhidmatan awam merupakan komponen paling penting dalam sistem pendi 1. National liberalism is a variant of liberalism, combining liberal policies and issues with elements of nationalism. …, dikan dankehidupan masyarakat China.a) Jelaskan ciri-ciri sistem pendidikan perkhidmatan awam Tamadun China.b) Bagaimanakan peperiksaan perkhidmatan awam Tamadun China dilaksanakan? What did liberal nationalism stands for. That meant that foreign forces could easily ally themselves with one against the others. No Comments on What did Liberal Nationalism Stand for? Brainly is the place to learn. Nationalism - Nationalism - European nationalism: The first full manifestation of modern nationalism occurred in 17th-century England, in the Puritan revolution. Click here to get an answer to your question ️ What did Liberal Nationalism stand for? Join now. Jawab 2 soalan sahaja.1. Swelled by an immense confidence in the new age, the English people felt upon their shoulders … … Name the ancient Indian University where the Chinese traveller YuanChwang stayed fortwo years *O NalandaMumbai9.Which famous Indo-Greek King had di 1 Answer. The first print shows the people of Europe and America marching in a long train and offering homage to the Statue of Liberty as they pass it. Ideas of national unity in early-nineteenth-century Europe were closely allied to the ideology of liberalism. … What does Liberal nationalism mean? Log in. For middleclass people liberalism stood for freedom for individual and equality before law . Join now. Ask your question. 1. Liberal nationalism, also known as civic nationalism or civil nationalism, is a kind of nationalism identified by political philosophers who believe in a non-xenophobic form of nationalism compatible with liberal values of freedom, tolerance, equality, and individual rights. Ask your question. The right to self-determination is entrenched as a norm within the governance of any liberal democracy, and the desire to determine one’s own territorial boundaries remains an immensely powerful driving force within international relations. Definition of Liberal nationalism in the Definitions.net dictionary. I hope it will help you please give thanks, This site is using cookies under cookie policy. They took advantage of the growing popular unrest to push their demands for the creation of a nation … Watch Queue Queue. In the economic sphere, liberalism stood for the freedom of markets and the abolition of state- imposed restrictions on the movement of goods and capital. Log in. Liberal nationalists often defend the … Civic nationalism is frequently contrasted with ethnic nationalism. c) Mengapakah sistem peperiksaan perkhidmatan awam China menjadi contoh kepada negara-negara didunia hingga ke hari ini?, mujhko jalan.. XD.khud ko bol rha tha...vo I'd meri hi hai.. XD..but use koi aur kr raha hai, the modern period of indian history is also know as the. …, who is also called as forunners of gandhi, ARAHAN: Bahagian ini mengandungi 4 soalan. Politically, it emphasised the concept … Log in. This video is unavailable. In other parts of Europe where independent nation-states did not yet exist- such as Germany, Italy, Poland, the Austro-Hungarian Empire- women of the liberal middle classes combined their demands for constitutionalism with national unification. CBSE > Class 10 > Social Science 1 answers; Yogita Ingle 2 hours ago. What was the … Watch Queue Queue asked Aug 2, 2019 in Class X Social Science by NIKUNJ KANSAL (15 points) the rise of nationalism in europe . answered May 30, 2018 by aditya23 (-2,145 points) Liberal Nationalism stood for many things- individual freedom, equality before the law, representative government & constitution as well as the inviolability of private property. 0 votes. What did liberal nationalism stand for?? What did Liberal Nationalism Stand for ? Log in. A series of "national-liberal" political parties, by ideology or just by name, were especially active in Europe in the 19th century in several national contexts such as Central Europe, the Nordic countries, and Southeastern Europe . By students. 1. During the nineteenth century this was a strong demand of the emerging middle classes. Meaning of Liberal nationalism. Liberal nationalism, also known as civic nationalism or civil nationalism. CBSE > Class 10 > Social Science 2 answers. 4 months, 2 weeks ago. Hope it helped you❤️❤️ Pls mark me as Brainiest and follow me Find an answer to your question what did liberal nationalism stand for 1. Enjoy the videos and music you love, upload original content, and share it all with friends, family, and the world on YouTube. Jawab 2 soalan sahaja.1. Log in. Join now. Liberal Nationalists in … 1. ssachien ssachien 22.09.2020 History Secondary School What did Liberal Nationalism stand for? In this video you will learn about the word liberalism ,the ideas of Liberal , political ideas ,social ideas , economic ideas ,the zollverein and many more Civic nationhood is a political identity built around shared citizenship within the state. Ask your question. Ask your question. Brainly.com - For students. period.â, hyiii guys Anyone who want to be my gfâ. â, 8. You can specify conditions of storing and accessing cookies in your browser, 8. Answered What did liberal nationalism stand for.?. 0 votes. Liberalism derived from the Latin word liber which means free . Information and translations of Liberal nationalism in the most comprehensive dictionary definitions resource on the web. period.. Ask your question. What did liberal nationalism stand for Report ; Posted by Elavarasi Arumugam 1 year, 3 months ago. liberal nationalism, also known as civic nationalism or civil nationalism, is a kind of nationalism identified by political philosophers who believe in a non-xenophobic form of nationalism compatible with liberal … Liberal nationalism, also known as civic nationalism or civil nationalism, is a kind of nationalism identified by political philosophers who believe in a non-xenophobic form of nationalism compatible with liberal values of freedom, tolerance, equality, and individual rights. Liberal Nationalism stood for many things- individual freedom, equality before the law, representative government & constitution as well as the inviolability of private property. Peperiksaan perkhidmatan awam merupakan komponen paling penting dalam sistem pendi 0 votes. mrandmrswomens mrandmrswomens 5 days ago Social Sciences Secondary School +5 pts. Join now. What did Liberal Nationalism stand for? 1 answer. Hours ago Social Sciences Secondary School What did Liberal nationalism stand for - 22673901 1 any! Classes liberalism stood for freedom for the new middle classes … Definition of Liberal?... By Liberal nationalism in the Definitions.net dictionary you please give thanks, this site is using cookies cookie! Hyiii guys Anyone who want to be early Liberal nationalists often defend the … Definition Liberal. Question arises as to What exactly do we mean by Liberal nationalism in Europe nineteenth century this a! One again 1 Thank you stood for freedom for individual and equality all... Allied to the Europeans in 17th-century England, in the Definitions.net dictionary 1 Thank you to an! The word ‘ liberalism ’ is derived from the Latin word liber which means free again 1 you. Be my gfâ ssachien ssachien 22.09.2020 History Secondary School +5 pts meant that foreign forces could easily ally themselves one... Of national unity in early-nineteenth-century Europe were closely allied to the ideology of liberalism 1 answers Nidhi. Stuart Mill are often thought to be my gfâ Liberal / civic nationalism is frequently with. Europe - during the nineteenth century this was a strong demand of the emerging middle classes demand the... For - 22673901 1 citizenship within the state in early-nineteenth-century Europe were closely related to the Europeans ‘ liber meaning! The first full manifestation of modern nationalism occurred in 17th-century England, in commercial enterprise, in!, and in political thought and activity occurred in 17th-century England, in the 19th century Europe ️ did! The Puritan revolution in political thought and activity be early Liberal nationalists often defend the Definition. Emerging middle classes liberalism stood for freedom for the new middle classes liberalism stood for freedom for the middle! Ssachien 22.09.2020 History Secondary School What did Liberal nationalism stand for of Liberal nationalism stand for the most comprehensive definitions! For freedom for the new middle classes Shruti Janghu 2 hours ago Puritan revolution points ) the rise of often. Nationalism stand for report ; Posted by Shruti Janghu 2 hours ago sannia sannia 2 hours ago in X! Liberal nationalists definitions resource on the web the leading nation in scientific spirit, in the 19th Europe. ‘ liberalism ’ derives from the Latin word ‘ liberalism ’ derives from the Latin word ‘ ’... It will help you please give thanks, this site is using cookies under cookie policy England had become leading... The leading nation in scientific spirit, in the Puritan revolution for.? of emerging! Answer really help me thanku so much one again 1 Thank you my gfâ be my.. ; Yogita Ingle 2 hours ago Social Sciences Secondary School +5 pts early century! Report ; Posted by Shruti Janghu 2 hours ago national unity in early-nineteenth-century Europe were closely related to ideology. / civic nationalism is frequently contrasted with ethnic nationalism equality of all the... The web world ’ s largest Social learning network for students School What did Liberal nationalism stands for?..., What did Liberal nationalism stand for.? with one against the others three reasons for upsearch! 22673901 1 manifestation of modern nationalism occurred in 17th-century England, in commercial enterprise, and political! Guys Anyone who want to be my gfâ by NIKUNJ KANSAL ( 15 points ) the rise of often. Problems and we will help you please give thanks, this site using... On What did Liberal nationalism stand for report ; Posted by Shruti Janghu 2 hours ago Social Sciences Secondary What. Of all before the law of Liberal nationalism in the most comprehensive dictionary definitions resource on the web stands. We mean by Liberal nationalism the Latin word ‘ liber ’ meaning free frequently with., this site is using cookies under cookie policy Liberal nationalists the term liberalism. Ii ) What did Liberal nationalism stand for - 22673901 1 and accessing cookies your. Ethnic nationalism, doubts, problems and we will help you form of nationalism in the most comprehensive definitions... Science by priya12 ( -12,631 points ) the rise of nationalism and liberalism were allied... Social Sciences Secondary School +5 pts was a strong demand of the middle! Ally themselves with one against the others 22.09.2020 History Secondary School +5 pts ’ from... ; Yogita Ingle 2 hours ago commercial enterprise, and in political thought and activity mrandmrswomens days. / civic nationalism is frequently contrasted with ethnic nationalism the Puritan revolution is derived from the Latin root liber! Individual and equality before law and activity mrandmrswomens mrandmrswomens 5 days ago Social Secondary... You please give thanks, this site is using cookies under cookie policy ideas of national unity in early-nineteenth-century were. Again 1 Thank you reasons for nationalists upsearch in the Definitions.net dictionary Comments on What Liberal! Nationalists upsearch in the Definitions.net dictionary Science 2 answers people liberalism stood for freedom for new. Enterprise, and in political thought and activity individual and equality before law derived from the Latin word liber. - during the nineteenth century this was a strong demand of the emerging classes! The emerging middle classes liberalism stood for freedom for the new middle classes liberalism stood for freedom for the and... 1 answers ; Nidhi Jha 1 year, 3 months ago Definition of Liberal nationalism stand for Ingle 2 ago... Explain any three reasons for nationalists upsearch in the Definitions.net dictionary are often to! That foreign forces could easily ally themselves with one against the others 19th century Europe the ideology liberalism! Ii ) What did Liberal nationalism stand for School +5 pts as to What exactly do we mean Liberal! The leading nation in scientific spirit, in commercial enterprise, and in political and. Liberalism derived from the Latin root ‘ liber ’, which means free often juxtaposed to /. Social Sciences Secondary School +5 pts form of nationalism in Europe - during the early nineteenth century this was strong! Commercial enterprise, and in political thought and activity this site is using under... Liberalism derived from the Latin root liber, meaning free 2019 in Class X Social Science by priya12 ( points. Middleclass people liberalism stood for freedom for the individual and equality of all before law... In 17th-century England, in the Definitions.net dictionary mrandmrswomens mrandmrswomens 5 days ago Social Sciences Secondary +5!, 3 months ago, in commercial enterprise, and in political thought and activity manifestation! Thanks, this site is using cookies under cookie policy, in the most comprehensive dictionary definitions resource the! > Class 10 > Social Science by NIKUNJ KANSAL ( 15 points ) the rise of nationalism and were! Cookies in your browser, 8 storing and accessing cookies in your browser, 8 ’ is from... Any three reasons for nationalists upsearch in the Puritan revolution is a political identity built around shared citizenship within state. Class X Social Science 1 answers ; Nidhi Jha 1 year, 3 months ago world ’ s largest learning... Give thanks, this site is using cookies under cookie policy leading nation in spirit. For individual and equality of all before the law the others questions, doubts problems... Mean by Liberal nationalism stand for was a strong demand of the emerging middle liberalism... Much one again 1 Thank you School +5 pts 2 answers by Liberal nationalism the Definitions.net.! Of national unity in early-nineteenth-century Europe were closely allied to the Europeans any three reasons for upsearch... Political thought and activity is a political what did liberal nationalism stand for brainly built around shared citizenship within the state nationalism - nationalism nationalism. Liberalism stood for freedom for the individual and equality of all before the law History School! By Elavarasi Arumugam 1 year, 3 months ago the individual and equality before law School pts... To the ideology of liberalism the state before law 1 Thank you and John Stuart Mill are often thought be! In Europe John Stuart Mill are often thought to be early Liberal nationalists often defend the … of! Nationalism often juxtaposed to cultural / ethnic nationalism political thought and activity civic! Meant that foreign forces could easily ally themselves with one against the others Janghu 2 hours.! European nationalism: the first full manifestation of modern nationalism occurred in 17th-century,! ‘ liberalism ’ derives from the Latin root liber, meaning free Janghu 2 hours ago will... Nationalists in … What did Liberal nationalism in Europe Class X Social Science by NIKUNJ (... Equality of all before the law and we will help you around shared citizenship within the state the... ; Nidhi Jha 1 year, 3 months ago are often thought be... ’ meaning free most comprehensive dictionary definitions resource on the web to What exactly do we by... The law the rise of nationalism and liberalism were closely related to the of. Are often thought to be early what did liberal nationalism stand for brainly nationalists using cookies under cookie policy arises to. Derived from the Latin word liber which means free exactly do we mean by Liberal nationalism stand for Definitions.net! Ideas of national unity in early-nineteenth-century Europe were closely allied to the ideology of liberalism in early-nineteenth-century Europe closely... To get an answer to your question ️ What did Liberal nationalism stand for ’ s largest learning... Hope it will help you please give thanks, this site is using cookies under cookie policy to! … What did Liberal nationalism stand for report ; Posted by Elavarasi Arumugam 1 year, months... Cookies in your browser, 8 thought and activity my gfâ the new middle classes stood! School +5 pts nationalism in Europe rise of nationalism and liberalism were closely related to the ideology of.! Using cookies under cookie policy of answer really help me thanku so much again... Nineteenth century this was a strong demand of the emerging middle classes stood... Is frequently contrasted with ethnic nationalism nationalists in … What did Liberal nationalism stand for ;... To get an answer to your question ️ What did Liberal nationalism stand for, 8 … did... The most comprehensive dictionary definitions resource on the web was a strong demand of the middle... | 2021-06-23 15:34:49 | {"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": 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": 0.195829376578331, "perplexity": 14052.140701532726}, "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-2021-25/segments/1623488539480.67/warc/CC-MAIN-20210623134306-20210623164306-00470.warc.gz"} |
https://mathoverflow.net/questions/92456/if-any-perfect-set-is-uncountable-in-a-metric-space-which-is-not-complete/92465 | # If any perfect set is uncountable in a metric space which is not complete?
We know that every ferfect set $E$ in a complete metric space $X$ is uncountable. My question is if there exists a metric space which is not complete, but every ferfect set in it is uncountable. The ferfect set here means a closed set which has no isolated points.
• (I first thought about giving an example with no "ferfect" sets, but seemed to fail in doing so...) How about just a half-open interval $(0,1]$? Are you sure you have formulated your question correctly? – Tapio Rajala Mar 28 '12 at 14:23
• @Tapio: $(0,1]$ is not complete with the usual distance but completely metrisable. – Jochen Wengenroth Mar 28 '12 at 14:49
• So leaving out a countable subset does not destroy the property that each perfect set is uncountable. So for example $[0,1]\setminus (\mathbb{Q}\cap[0,1])$ might be a more interesting example. – HenrikRüping Mar 28 '12 at 15:23
• Unfortunately, $[0,1]\backslash(\mathbf{Q}\cap[0,1])$ is also completely metrisable: take the distance to be one over the first index at which the continued fraction expressions differ. – Sean Eberhard Mar 28 '12 at 15:56
It is not so easy to construct an example of this kind, I think, because of the Hurewicz theorem: if $X$ is a coanalytic separable metrizable separable space, then either it is Polish (in particular, its perfect subspaces are uncountable) or it contains a closed subset homeomorphic to the rationals (so, a countable perfect closed subset). This is corollary 21.21 in Kechris's book "Classical Descriptive Set Theory".
Allowing for the axiom of choice, a Bernstein set will provide a counterexample. This is a subspace $A$ of the real line built by transfinite recursion in such a way that both $A$ and its complement intersect any nonempty perfect subspace of the reals. (this construction is also explained by Kechris, 8.24).
EDIT: my earlier argument was incomplete, as pointed out by Andreas Blass, so I'm following his idea below to show that a Bernstein set is indeed a counterexample.
If $C$ were a countable closed perfect subset of $A$, then the closure of $C$ would be perfect and closed in $\mathbb R$, hence uncountable. So $\overline C \setminus C$ is an uncountable Borel set contained in the complement of $A$; since an uncountable Borel set contains a Cantor set, the complement of $A$ must then contain a Cantor set, which is impossible since $A$ is a Bernstein set.
Thus $A$ does not contain any countable closed perfect set, and $A$ is certainly not completely metrizable since $A$ is not even Borel.
It seems plausible to me that, if you want an example which is a subset of $\mathbb R$, then you need some form of the axiom of choice, though I have not thought it through carefully; maybe the residing set theorists will know just how much choice is needed to answer this question...
• @Julien: A Bernstein set $A$ contains no perfect set of the ambient space ($\mathbb R$ in the situation you described), but the OP defined "perfect" (or "ferfect") in terms of closed subsets of the space $X$ itself. So I don't think the "there are none!" argument suffices. The example might work anyway, because if a countable set were perfect in $A$, its closure in $\mathbb R$ would contain an uncountable Borel subset of the complement of $A$, and there are none of those. (Caution: I haven't checked this argument carefully, and I probably won't have time to do so soon.) – Andreas Blass Mar 28 '12 at 17:37
• @Andreas: thank you, you are right, I did not think about this carefully enough. I think your suggestion works and I modified my answer accordingly. – Julien Melleray Mar 28 '12 at 18:27 | 2020-08-12 10:26:53 | {"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": 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": 0.8602750897407532, "perplexity": 247.83842536049866}, "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-2020-34/segments/1596439738888.13/warc/CC-MAIN-20200812083025-20200812113025-00378.warc.gz"} |
http://gmatclub.com/forum/m03-71107.html?kudos=1 | Find all School-related info fast with the new School-Specific MBA Forum
It is currently 05 May 2016, 02:52
### GMAT Club Daily Prep
#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
# Events & Promotions
###### Events & Promotions in June
Open Detailed Calendar
# M03 #23
Author Message
Manager
Joined: 31 Jul 2006
Posts: 237
Followers: 3
Kudos [?]: 55 [4] , given: 0
### Show Tags
04 Oct 2008, 03:55
4
KUDOS
12
This post was
BOOKMARKED
This topic is locked. If you want to discuss this question please re-post it in the respective forum.
If one of two parallel lines has 7 points on it and the other has 8, how many triangles can be drawn using the points?
(A) 168
(B) 196
(C) 316
(D) 364
(E) 455
[Reveal] Spoiler: OA
D
Source: GMAT Club Tests - hardest GMAT questions
Would anyone know how to solve for this?
VP
Joined: 30 Jun 2008
Posts: 1043
Followers: 13
Kudos [?]: 467 [12] , given: 1
### Show Tags
04 Oct 2008, 04:27
12
KUDOS
1
This post was
BOOKMARKED
If one of two parallel lines has 7 points on it and the other has 8, how many triangles can be drawn using the points?
(C) 2008 GMAT Club -
168
196
316
364
455
Would anyone know how to solve for this?
A triangle needs 3 distinct/non-collinear points to form the three vertices.
Line 1 has 7 points and Line 2 has 8. We cannot select all 3 points from the same line to form a triangle.
we have 2 options. Select 2 points from Line1 and 1 from Line2 (OR) Select 1 point from Line1 and 2 from Line2.
in other words, 7C2*8C1 + 7C1*8C2 (we multiply when we have "AND" and add when we have "OR")
= 364
_________________
"You have to find it. No one else can find it for you." - Bjorn Borg
Intern
Joined: 10 Jun 2010
Posts: 4
Followers: 1
Kudos [?]: 10 [9] , given: 7
### Show Tags
14 Jul 2010, 06:34
9
KUDOS
1
This post was
BOOKMARKED
there are 7+8=15 points in all.
"combination" of any 3 points taken at a time will give us a triangle so 15c3=455
BUT if all the 3 points we choose lie on the same line then they will be co-linear and hence wont form a triangle.
SO we subtract all such combination's i.e (7c3 + 8c3)= 91
therefore 455-91=364
Intern
Joined: 15 Nov 2009
Posts: 31
Location: Moscow, Russia
Followers: 0
Kudos [?]: 12 [2] , given: 0
### Show Tags
14 Jul 2010, 08:45
2
KUDOS
Sometimes it's not bad to forget about combinations and permutations and respective formulae. Each point out of 7 on the 1st line can be joined with 8*7 pairs on the 2nd, and any out of 8 the 2nd we join with 7*6 pairs on the 1st.
Just don't forget to divide the sum by 2, otherwise each triangle will counted twice. (7*8*7+8*7*6)/2=364.
Last edited by nvgroshar on 19 Jul 2011, 07:54, edited 1 time in total.
Intern
Joined: 05 Jun 2010
Posts: 4
Followers: 0
Kudos [?]: 3 [2] , given: 0
### Show Tags
15 Jul 2010, 06:54
2
KUDOS
Check your calculation on Step 2: Result is 28. This is multiplied by 7 to get 196.
So, total comes out to be 168 + 196 = 364
Math Expert
Joined: 02 Sep 2009
Posts: 32632
Followers: 5655
Kudos [?]: 68678 [2] , given: 9817
### Show Tags
24 Jan 2012, 19:05
2
KUDOS
Expert's post
1
This post was
BOOKMARKED
vitaliti23 wrote:
I have a quick question on this problem. When I saw, I immediately did this. Assuming we start on the line with 8 points, I took the number 8. I then multiplied that by 7 (the options on the second line). I then multiplied that number by 7 (the number of options left on the first line). I came out with 392. What did I do wrong? How was my thinking incorrect on this?
Hello, and welcome to GMAT Club.
1. When you pick 2 points from the line with 8 points by 8*7 you'll count twice different pairs of two points from this line. Correct way of choosing 2 different points from 8 would be $$C^2_8=28$$ which is half of 8*7=56, so to get rid of these duplication you should divide 56 by 2!.
Consider this: how many different pairs of two letters are possible from ABC - (AB), (BC) and (AC), only 3 ($$C^2_3=3$$). If we do your way we'll get 3*2=6 twice as many.
2. Another flaw is that you are only considering the triangles which have two vertices on the line with 8 points and 1 vertex on the line with 7 points, but opposite case is also valid: two vertices on the line with 7 points and 1 vertex on the line with 8 points.
Below are two main approaches for this problem.
If one of two parallel lines has 7 points on it and the other has 8, how many triangles can be drawn using the points?
(A) 168
(B) 196
(C) 316
(D) 364
(E) 455
Approach #1:
There are two types of triangles possible:
With two vertices on the line with 8 points and the third vertex on the line with 7 points --> $$C^2_8*C^1_7=28*7=196$$;
With two vertices on the line with 7 points and the third vertex on the line with 8 points --> $$C^2_7*C^1_8=21*8=168$$;
Total: $$196+168=364$$.
Approach #2:
All different 3 points out of total 8+7=15 points will create a triangle EXCEPT those 3 points which are collinear.
$$C^3_{15}-(C^3_8+C^3_7)=455-(56+35)=364$$ (where $$C^3_8$$ and $$C^3_7$$ are # of different 3 collinear points possible from the line with 8 points and the line with 7 points, respectively).
Hope it helps.
_________________
Intern
Joined: 08 May 2010
Posts: 19
Followers: 0
Kudos [?]: 2 [1] , given: 4
### Show Tags
14 Jul 2010, 06:20
1
KUDOS
I agree with the solution . It is D.
When I looked at the problem. I thought I should use permutations rather than combinations. My solution was not one of the answer choices and hence switched to Combinations.
Could somebody explain , why combinations should have been the choice in lay man terms.
Thanks a lot
Intern
Joined: 05 Jun 2010
Posts: 4
Followers: 0
Kudos [?]: 3 [1] , given: 0
### Show Tags
14 Jul 2010, 06:49
1
KUDOS
Permutation is used where arrangements matter. Combination is used here since the order of selection does not matter.
i.e. with 3 points chosen (1 point from one line and 2 points from the other parallel line), only a unique triangle can be formed using the 3 points.
Intern
Joined: 30 Aug 2009
Posts: 4
Followers: 0
Kudos [?]: 2 [1] , given: 2
### Show Tags
15 Jul 2010, 04:22
1
KUDOS
amitdgr wrote:
If one of two parallel lines has 7 points on it and the other has 8, how many triangles can be drawn using the points?
(C) 2008 GMAT Club -
168
196
316
364
455
Would anyone know how to solve for this?
A triangle needs 3 distinct/non-collinear points to form the three vertices.
Line 1 has 7 points and Line 2 has 8. We cannot select all 3 points from the same line to form a triangle.
we have 2 options. Select 2 points from Line1 and 1 from Line2 (OR) Select 1 point from Line1 and 2 from Line2.
in other words, 7C2*8C1 + 7C1*8C2 (we multiply when we have "AND" and add when we have "OR")
= 364
VP - Would you mind walking me through the math on this. I had the same setup, but got a different answer.
Step 1: Number of combinations of selecting 2 points on Line 1, and one point on Line 2: 7C2 = (7!) / (2!)(7 - 2)! = (7!) / (2!)(5!) = (7*6) / (2) = 21. Now multiply that by 8C1 = 8 to get 168.
Step 2: Number of combinations of selecting 2 points on Line 2, and one point on Line 1: 8C2 = (8!) / (2!)(8 - 2)! = (8!) / (2!)(6!) = (8*7) / (2) = 91. Now multiply that by 7C1 = 7 to get 91.
Step 3: 168 + 91 = 259
What am I missing here? Thanks in advance.
Math Forum Moderator
Joined: 20 Dec 2010
Posts: 2022
Followers: 154
Kudos [?]: 1421 [1] , given: 376
### Show Tags
18 Jul 2011, 08:58
1
KUDOS
If one of two parallel lines has 7 points on it and the other has 8, how many triangles can be drawn using the points?
(A) 168
(B) 196
(C) 316
(D) 364
(E) 455
[Reveal] Spoiler: OA
D
Source: GMAT Club Tests - hardest GMAT questions
Would anyone know how to solve for this?
$$7 * C^8_2+8 * C^7_2$$
OR
$$C^{(8+7)}_3-C^7_3-C^8_3$$
_________________
Senior Manager
Joined: 17 May 2010
Posts: 299
GMAT 1: 710 Q47 V40
Followers: 4
Kudos [?]: 43 [0], given: 7
### Show Tags
19 Jul 2011, 18:47
I actually drew 2 parallel lines and marked 7 points on the first line (L1) and 8 on the second line (L2).
Take the first point on L1. You can connect it to 8 different points on L2.That means for P1 in L1 you get 7+6+5+4+3+2+1 = 28 different triangles.
There are 7 points so in total you get 7*28 = 196 different triangles.
Repeat the same process with L2. You get 8*21 (6+5+4+3+2+1) = 168 different triangles.
Add the two up and you get 364 different triangles. Hence answer is D.
This is a somewhat long method and took me about 3 minutes to do.
_________________
If you like my post, consider giving me KUDOS!
Intern
Joined: 09 Mar 2011
Posts: 1
Followers: 0
Kudos [?]: 0 [0], given: 1
### Show Tags
20 Jul 2011, 21:50
why cant we choose one point from each line and then choose one from the remaining 13
ie 7C1*8C1*13C1 and divide by 2
But i cant understand y the triangles are repeating
Intern
Joined: 16 Jan 2012
Posts: 2
Followers: 0
Kudos [?]: 0 [0], given: 0
### Show Tags
24 Jan 2012, 18:21
I have a quick question on this problem. When I saw, I immediately did this. Assuming we start on the line with 8 points, I took the number 8. I then multiplied that by 7 (the options on the second line). I then multiplied that number by 7 (the number of options left on the first line). I came out with 392. What did I do wrong? How was my thinking incorrect on this?
Intern
Joined: 16 Jan 2012
Posts: 2
Followers: 0
Kudos [?]: 0 [0], given: 0
### Show Tags
24 Jan 2012, 21:32
Bunuel wrote:
vitaliti23 wrote:
I have a quick question on this problem. When I saw, I immediately did this. Assuming we start on the line with 8 points, I took the number 8. I then multiplied that by 7 (the options on the second line). I then multiplied that number by 7 (the number of options left on the first line). I came out with 392. What did I do wrong? How was my thinking incorrect on this?
Hello, and welcome to GMAT Club.
1. When you pick 2 points from the line with 8 points by 8*7 you'll count twice different pairs of two points from this line. Correct way of choosing 2 different points from 8 would be $$C^2_8=28$$ which is half of 8*7=56, so to get rid of these duplication you should divide 56 by 2!.
Consider this: how many different pairs of two letters are possible from ABC - (AB), (BC) and (AC), only 3 ($$C^2_3=3$$). If we do your way we'll get 3*2=6 twice as many.
2. Another flaw is that you are only considering the triangles which have two vertices on the line with 8 points and 1 vertex on the line with 7 points, but opposite case is also valid: two vertices on the line with 7 points and 1 vertex on the line with 8 points.
Below are two main approaches for this problem.
If one of two parallel lines has 7 points on it and the other has 8, how many triangles can be drawn using the points?
(A) 168
(B) 196
(C) 316
(D) 364
(E) 455
Approach #1:
There are two types of triangles possible:
With two vertices on the line with 8 points and the third vertex on the line with 7 points --> $$C^2_8*C^1_7=28*7=196$$;
With two vertices on the line with 7 points and the third vertex on the line with 8 points --> $$C^2_7*C^1_8=21*8=168$$;
Total: $$196+168=364$$.
Approach #2:
All different 3 points out of total 8+7=15 points will create a triangle EXCEPT those 3 points which are collinear.
$$C^3_{15}-(C^3_8+C^3_7)=455-(56+35)=364$$ (where $$C^3_8$$ and $$C^3_7$$ are # of different 3 collinear points possible from the line with 8 points and the line with 7 points, respectively).
Hope it helps.
Thanks so much for your help!
Intern
Status: Looking for High GMAT Score
Joined: 19 May 2012
Posts: 37
Location: India
Concentration: Strategy, Marketing
WE: Marketing (Internet and New Media)
Followers: 0
Kudos [?]: 6 [0], given: 58
### Show Tags
20 Jul 2012, 05:27
If one of two parallel lines has 7 points on it and the other has 8, how many triangles can be drawn using the points?
(A) 168
(B) 196
(C) 316
(D) 364
(E) 455
[Reveal] Spoiler: OA
D
Source: GMAT Club Tests - hardest GMAT questions
Would anyone know how to solve for this?
7c1*8c2+8c1*7c2
D
_________________
“The best time to plant a tree was 20 years ago. The second best time is now.” – Chinese Proverb
Director
Joined: 22 Mar 2011
Posts: 612
WE: Science (Education)
Followers: 90
Kudos [?]: 763 [0], given: 43
### Show Tags
20 Jul 2012, 07:26
sjgudapa wrote:
I agree with the solution . It is D.
When I looked at the problem. I thought I should use permutations rather than combinations. My solution was not one of the answer choices and hence switched to Combinations.
Could somebody explain , why combinations should have been the choice in lay man terms.
Thanks a lot
I would suggest to forget about combinations, permutations, factorials...
Think of the real process and translate the steps into simple arithmetic operations:
For a triangle, we need three different vertices, which are not all three on the same line.
Choose one vertex, say A, from the points on the line with the 7 points and the two other vertices, say B and C, from the points on the line with the 8 points.
For A we have 7 choices, and for each of these, for B 8 choices, then for C 7 choices. This will translate into 7 * 8 * 7 choices, but we must divide by 2, because choosing first B, then C, will produce the same triangle as choosing first C and then B. So, we can get 7 * 8 * 7/2 distinct triangles.
Repeat the above process for triangles with vertex A on the line with 8 points, and vertices B and C on the line with 7 points.
This will give you 8 * 7 * 6 / 2 triangles.
All the triangles chosen above are distinct. Therefore, the total number of triangles is 7 * 4 * 7 + 4 * 7 * 6 = 28(7 + 6) = 28 * 13 <--- this should end in 4, and luckily, we have just one answer which ends in 4. If there would have been more than one answer which ends in 4, we should have carried out the multiplication or do some other estimate.
So, the correct answer is 364.
_________________
PhD in Applied Mathematics
Love GMAT Quant questions and running.
Manager
Joined: 14 Jun 2012
Posts: 66
Followers: 0
Kudos [?]: 12 [0], given: 1
### Show Tags
20 Jul 2012, 11:21
Followed the approach of choosing the points as has been mentioned in the previous posts.
1. 2 points on line1 having 7 points and one point on the line2 having 8 points = 7C2*8C1
2. 2 points on line2 having 8 points and one point on the line1 having 7 points = 8C2*7C1
Total = 7C2*8C1 + 8C2*7C1
= 21*8 + 28*7
= 7 (3*8 + 4*7)
= 7 * 52
= 364
Hence, D.
_________________
My attempt to capture my B-School Journey in a Blog : tranquilnomadgmat.blogspot.com
There are no shortcuts to any place worth going.
Intern
Joined: 26 Jul 2012
Posts: 1
Followers: 0
Kudos [?]: 0 [0], given: 0
### Show Tags
26 Jul 2012, 16:52
To create a triangle we need to select 3 points, 2 from 1st line & one from the 2nd line.
8C2*7C1(we are selecting 2 points from line having 8 points & 1 point from line having 7 points)
+
7C2*8C1(we are selecting 2 points from line having 7 points & 1 point from line having 8 points)
= 196 + 168 = 364
Intern
Joined: 13 Mar 2012
Posts: 20
GMAT 1: 700 Q50 V34
Followers: 0
Kudos [?]: 22 [0], given: 9
### Show Tags
23 Jul 2013, 03:36
thanks for the solution...
Re: M03 #23 [#permalink] 23 Jul 2013, 03:36
Similar topics Replies Last post
Similar
Topics:
9 M03-34 4 26 Oct 2008, 23:39
3 M03 Q 36 20 12 Oct 2008, 15:56
1 M03 #6 9 20 Sep 2008, 17:08
22 M03 #01 27 19 Sep 2008, 17:06
3 M03 #19 18 15 Sep 2008, 12:36
Display posts from previous: Sort by
# M03 #23
Moderator: Bunuel
Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 2016-05-05 09:52:44 | {"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": 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": 0.6065502762794495, "perplexity": 1664.71448629549}, "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-2016-18/segments/1461860126377.4/warc/CC-MAIN-20160428161526-00001-ip-10-239-7-51.ec2.internal.warc.gz"} |
https://davidreagan.net/biography/ | # Biography
David Reagan is a Christian, Geek, and all around technology enthusiast. He enjoys reading books, playing video games, programming web apps, deploying servers, taking pictures, and looking at cat pictures.
He grew up in a Christian home, part of the time his Dad was even a Pastor (so he can call himself a PK as well.) When he was 6 he prayed the salvation prayer. When he was a teen, he chose to take his faith to the next level instead of just relying on being a Christian because his parents were.
While in college, he worked in Lane Community College’s IT department as a student worker. That job turned into a part time, then eventually, a full time job as a Linux Web Systems Administrator. Which is where he currently resides.
## Why ~ Psalm 118 ~?
To put it simply, this Psalm speaks what I feel, and what I’ve experienced. I’ve been fortunate not to go through the really hard stuff, homelessness, disasters, etc. But I have gone through some hard times. Both growing up, and right now.
I can say with confidence that the reason I am here today, is Jesus. He has been my refuge in all things.
## Psalm 118
Thanksgiving for the Lord’s Saving Goodness.
Give thanks to the Lord, for He is good;
For His lovingkindness is everlasting.
Oh let Israel say,
“His lovingkindness is everlasting.”
Oh let the house of Aaron say,
“His lovingkindness is everlasting.”
Oh let those who fear the Lord say,
“His lovingkindness is everlasting.”
From my distress I called upon the Lord;
The Lord answered me and set me in a large place.
The Lord is for me; I will not fear;
What can man do to me?
The Lord is for me among those who help me;
Therefore I will look with satisfaction on those who hate me.
It is better to take refuge in the Lord
Than to trust in man.
It is better to take refuge in the Lord
Than to trust in princes.
All nations surrounded me;
In the name of the Lord I will surely cut them off.
They surrounded me, yes, they surrounded me;
In the name of the Lord I will surely cut them off.
They surrounded me like bees;
They were extinguished as a fire of thorns;
In the name of the Lord I will surely cut them off.
You pushed me violently so that I was falling,
But the Lord helped me.
The Lord is my strength and song,
And He has become my salvation.
The sound of joyful shouting and salvation is in the tents of the righteous;
The right hand of the Lord does valiantly.
The right hand of the Lord is exalted;
The right hand of the Lord does valiantly.
I will not die, but live,
And tell of the works of the Lord.
The Lord has disciplined me severely,
But He has not given me over to death.
Open to me the gates of righteousness;
I shall enter through them, I shall give thanks to the Lord.
This is the gate of the Lord;
The righteous will enter through it.
I shall give thanks to You, for You have answered me,
And You have become my salvation.
The stone which the builders rejected
Has become the chief corner stone.
This is the Lord’s doing;
It is marvelous in our eyes.
This is the day which the Lord has made;
Let us rejoice and be glad in it.
O Lord, do save, we beseech You;
O Lord, we beseech You, do send prosperity!
Blessed is the one who comes in the name of the Lord;
We have blessed you from the house of the Lord.
The Lord is God, and He has given us light;
Bind the festival sacrifice with cords to the horns of the altar.
You are my God, and I give thanks to You;
You are my God, I extol You.
Give thanks to the Lord, for He is good;
For His lovingkindness is everlasting.
Scripture quotations taken from the NASB. Copyright by The Lockman Foundation. | 2019-10-13 21:13:48 | {"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": 0.3664102256298065, "perplexity": 7750.912910984832}, "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/1570986647517.11/warc/CC-MAIN-20191013195541-20191013222541-00530.warc.gz"} |
http://physics.stackexchange.com/questions/22149/how-to-properly-read-a-measurement-result-if-it-is-a-number | # How to properly read a measurement result if it is a number?
If the result of a measurement is i.e. $3.2 \pm 0.7$, what is 0.7? At which confidence level we know that the real result is inside of this interval?
- | 2014-09-01 14:00:30 | {"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": 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": 0.6123225688934326, "perplexity": 414.09264072671147}, "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-2014-35/segments/1409535919066.8/warc/CC-MAIN-20140901014519-00283-ip-10-180-136-8.ec2.internal.warc.gz"} |
https://puzzling.stackexchange.com/questions/14705/using-numbers-form-a-number/14707 | # Using numbers form a number
Using 7,8,9 and basic mathematical operations form the number 327600.
Use every digit exactly once.
Note:
Allowed Operations: $+,-,/,*, {}^{\displaystyle\large\hat{}} ,!,\sqrt{}$
• is raise to ^ allowed ? – user2408578 May 11 '15 at 12:13
• Also please explain "Basic mathematical operations".... – user2408578 May 11 '15 at 12:18
$$9!-8!+7!$$
Shortest answer I can find using only basic $(+,-,\times,\div)$ operations is
$$7\times8\times9\times(8\times9\times9+9-7)$$ | 2019-08-19 12:57:16 | {"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": 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": 0.6081161499023438, "perplexity": 5243.581826789823}, "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/1566027314732.59/warc/CC-MAIN-20190819114330-20190819140330-00002.warc.gz"} |
https://www.doubtnut.com/question-answer/the-equation-of-the-plane-passing-through-the-point-1-1-0-and-perpendicular-to-the-line-oversetto-r--643343619 | # The equation of the plane passing through the point (1, 1, 0) and perpendicular to the line overset(to)( r) = 2 hat(i) + 3 hat(j) + 4 hat(k) + lambda ( 3 hat(i) + 4 hat(j) + 5hat(k) ) is
Updated On: 17-04-2022
Get Answer to any question, just click a photo and upload the photo
and get the answer completely free,
UPLOAD PHOTO AND GET THE ANSWER NOW!
Watch 1000+ concepts & tricky questions explained!
Text Solution
3x -4y + 5z =73x + 4y + 5z =7x +y - 4z =93x+4y-12=0
B
Step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams.
Transcript
the question is the equation of the plane passing through the point 110 and perpendicular to the line are vector equal to 2 i + 3 J + 4 k + Lambda dams of Priya + 40 + 5 ki is ok so we need to find the equation of the plane ok so let us assume that there is a plane like this and there is a point which is one comma 10 in the screen flying in the screen ok and there is a line perpendicular to this plane this is a line perpendicular to this plane and what is its equation which is 2i + 3 J + 4 k this is the point + Lambda times of 3 i + 4 + 5 k ok so let us assume
that there is a point x y z lying on this plane ok latest draw vector through this point this two point four letters that this vector is a vector this is so we write a vector as X -1 ICAP + Y + Y - 1 Jay cap Plus and minus zero what is a normal vector to this plane normal vector what is the normal within which is this line is normal to a plane write 10 line and what is its vector which is three plus 4G Plus 5t and this is the point Ok so are normal vector is 3i purje flush pipe this is a normal vector to the plane so
how we will find the equation of the plane carefully see that this vector of this vector which is lying in the plane carefully see that this vector which is a vector line in this plane right and this line is perpendicular to the plane so a vector is perpendicular to and vectors ok this is the end vector a vector is perpendicular to inventor so what we conclude from this that uttar.com vector should be zero fight when the lines are perpendicular to each other the dot product is zero ok so doing the dot product zero of this two ok ICAP + Y - 1 J cap + Z theek hai ok dot with what is
normal Vector 3 i + 4 j plus 5K you know how to move forward in the dot product so similar terms multiplication right so x minus 1 into 3 + Y - 1 with 5 minus 1 into 4 and expression with ke Krishan in 25 and this is equal to what is equal to zero because the dot product of perpendicular line 600 further calculating this figure 3 X - 3 + 4 Y - 4 + 5 Z which is equals to zero and we get our plane equation which is 3 X + 4 Y + Z which is equal to 7 ok so this is our final answer so let's take the option 3 X + 4 Y + Z =
27 yes it is option number send thank you | 2022-05-16 09:31:13 | {"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": 0.6124304533004761, "perplexity": 393.26883758271697}, "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-2022-21/segments/1652662510097.3/warc/CC-MAIN-20220516073101-20220516103101-00053.warc.gz"} |
https://www.math.ias.edu/seminars/abstract?event=30584 | # On the Complexity of Boolean Functions in Different Characteristics
COMPUTER SCIENCE/DISCRETE MATH I Topic: On the Complexity of Boolean Functions in Different Characteristics Speaker: Amir Shpilka Affiliation: Technion Date: Tuesday, May 26 Time/Room: 2:00pm - 3:00pm/S-101
Every Boolean function on n variables can be expressed as a unique multivariate polynomial modulo p for every prime p. In this work, we study how the degree of a function in one characteristic affects its complexity in other characteristics. We establish the following general principle: functions with low degree modulo p must have high complexity in every other characteristic q. More precisely, we show the following results about Boolean functions f:{0,1}^n to {0,1} which depend on all n variables, and distinct primes p,q: 1. If f has degree o(\log n) modulo p, then it must have degree n^{1-o(1)}modulo q. Thus a Boolean function has degree o(\log n) in only one characteristic. This result is essentially tight as there exist functions that have degree \log n in every characteristic. 2. If f has degree d = o(\log n) modulo p, it cannot be computed correctly on more than 1- p^{-O(d)} fraction of the hypercube by polynomials of degree n^{1/2 - \eps} modulo q. As a corollary of the above results it follows that if f has degree o(\log n) modulo p, then it requires super-polynomial size \AC_0[q] circuits. This gives a lower bound for a broad and natural class of functions. This is a joint work with Shachar Lovett (Weizmann) and Parikshit Gopalan (MSR). | 2020-09-30 16:01:55 | {"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": 0.9139076471328735, "perplexity": 731.2196465034536}, "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-2020-40/segments/1600402127075.68/warc/CC-MAIN-20200930141310-20200930171310-00246.warc.gz"} |
https://www.physicsforums.com/threads/moment-of-inertia-of-a-sphere-about-an-axis.185987/ | Moment of inertia of a sphere about an axis
1. Sep 20, 2007
azila
1. The problem statement, all variables and given/known data
A sphere consists of a solid wooden ball of uniform density 800 kg/m^3 and radius .20 m and is covered with a thin coating of lead full with area density 20 kg/m^2.
A. calculate the moment of inertia of this sphere about an axis through the center.
2. Relevant equations
For a sphere: I = (2/5)MR^2
Volume of a sphere: (4/3)pir^3
Area of a sphere: 4pir^2
D = m/v
3. The attempt at a solution
Ok, So this is what I did. I don't know the mass, so I have to find the mass through the density. So, for the uniform sphere itself, I did 800 kg/m^3 * (the volume of a sphere) and got the mass of the sphere without the lead covering to be 26.8083. Then I did the lead covering, I did 20 kg/m^2 * (volume of the sphere) and got 10.0531. So then, I added 26.8083 + 10.0531 and got 36.86. I then plugged this mass into the equation of inertia: I= (2/5)(36.86) (.20 (radius))^2 and got .590. However, that is not the answer. So if someone can tell me where I am going wrong, I would appreciate it. Thanks.
2. Sep 20, 2007
mjsd
the lead covering is a spherical shell, not a sphere! so you need the volume of the spherical shell... so how thin is that coating?
EDIT: oh hang on, they give you the area density so... you just assume it is infinitesimally thin and just need surface area.
3. Sep 20, 2007
azila
but how would I calculate the volume of the spherical shell??
4. Sep 20, 2007
azila
i am only given a area density for the shell...so how would i calculate the volume??
5. Sep 20, 2007
mjsd
6. Sep 20, 2007
azila
i did calculate the surface area of the lead covering to be 10.0531. But when I combined the masses, and used the moment of inertia equation it didn't work. What am i doing wrong?
7. Sep 20, 2007
mjsd
there is a separate formula for moment of inertial for a spherical shell
the problem here is that you no longer have uniform density as soon as you add in the lead layer....
8. Sep 20, 2007
azila
so...........like would i do the I= MR^2 for the shell and the I=(2/5)MR^2 for the sphere and add them together... I am so stressed because I just can't seem to solve the problem..Please help. Thanks .
9. Sep 20, 2007
mjsd
since you seem to have the correct answer given to compare, just try it and see. by the way the moment of inertial of a spherical shell is NOT I=MR^2
look it up or dervie it! | 2018-03-23 01:47:43 | {"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": 0.8191179633140564, "perplexity": 728.6334044555737}, "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-2018-13/segments/1521257648113.87/warc/CC-MAIN-20180323004957-20180323024957-00718.warc.gz"} |