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https://crypto.stackexchange.com/questions/31177/chosen-ciphertext-attack-on-rabin-factorize-n	# Chosen-Ciphertext Attack on Rabin: Factorize n At the end of this Wikipedia section, you can see that $$\gcd(\|r-s\|,n)=q$$ or $$\gcd(\|r-s\|,n)=p$$ holds. I don't understand why. • Take the equations for $r$ and $s$ given above the part you reference and write out $r-s$. – yyyyyyy Dec 10 '15 at 11:56	2019-11-12 05:18:08	{"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": 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": 0.9315359592437744, "perplexity": 395.8265378300256}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-47/segments/1573496664752.70/warc/CC-MAIN-20191112051214-20191112075214-00258.warc.gz"}
https://janmr.com/blog/2015/01/typesetting-math-with-html-and-css-fractions/	janmr blog Typesetting Math Using HTML and CSS: Fractions January 24, 2015 Currently, there is no best way of showing math on the web. An HTML5 standard exists, MathML, but unfortunately it doesn't have broad browser support. Instead, many alternatives exist, all with varying quality and speed. I would like to explore how far you can get by using just HTML and CSS (including web fonts). My findings should be considered experimental and in no way authoritative. This post will deal with one way of typesetting fractions, inspired by the approach taken by Kahn Academy's KaTeX project. Consider first the following layout: before 8 1234 after which is obtained by the following HTML and CSS: before<span class="math-box"> <span class="vstack"> <div style="top: 0.686em;">8</div> <div style="top: -0.677em;">1234</div> <span class="baseline-fix"></span> </span></span>after .math-box { display: inline-block;}.math-box .vstack { display: inline-block; position: relative;}.math-box .vstack > div { position: relative; text-align: center; height: 0;}.math-box .baseline-fix { display: inline-table; table-layout: fixed;} (Live demo.) Worth noting is: • The two outermost elements span.math-box and span.vstack are inline-block elements, meaning that they are positioned as inline elements (aligning baselines, for one, which is very important here), but otherwise behave as block elements (able to contain other block elements). • The divs will be stacked on top of each other because of their block style. Therefore the width of the enclosing span.vstack element will be the widest of the divs. • The divs have height zero which means that, by themselves, they will be displayed on top of each other, all following the same baseline. • The divs are positioned relatively, so the top property can position the elements correctly in the vertical direction. • The .baseline-fix is necessary in Internet Explorer, since otherwise the elements outside .math-box will not be aligned correctly (in the vertical direction). I don't know exactly why this fix works. But we would also like to display a horizontal line between the numerator and denominator, like so: before 8 1234 after We aim for this markup before<span class="math-box"> <span class="vstack"> <div style="top: 0.686em;">8</div> <div style="top: -0.677em;">1234</div> <div style="top: -0.23em;"><span class="frac-line"></span></div> <span class="baseline-fix"></span> </span></span>after but what should the style be for span.frac-line? Using something like display: inline-block; width: 100%; border-bottom: 1px solid black; will work, but how thick should the line be? Using something like 0.04em makes the line scale with the font size, but using a small font size can result in a line thickness less than 1 pixel (leading to the line disappearing or having a modified color). Here, KaTeX has a nice trick up their sleeve: .math-box .frac-line { width: 100%; display: inline-block;}.math-box .frac-line:before { display: block; border-bottom-style: solid; border-bottom-width: 1px; content: "";}.math-box .frac-line:after { display: block; margin-top: -1px; border-bottom-style: solid; border-bottom-width: 0.04em; content: "";} (Live demo.) Using the pseudo-elements :before and :after, two lines will be drawn on top of each other. This way the line will be 0.04em, but never less than 1px. Now we have a complete fraction, but there is one remaining issue. We would like the outer-most box span.math-box to exactly enclose the inner elements, like this: before 8 1234 after (If the fit is not perfect then see the comment in the final paragraph.) This will not be the case for the HTML/CSS presented above because of the relative positioning of the elements. The outer box will fit perfectly in the horizontal direction, but not in the vertical direction. A fix is to use a so-called strut, widely used in TeX: before<span class="math-box" style="border: 1px solid red;"> <span class="strut" style="height: 2.008em; vertical-align: -0.686em;"></span> <span class="vstack"> <div style="top: 0.686em;">8</div> <div style="top: -0.677em;">1234</div> <div style="top: -0.23em;"><span class="frac-line"></span></div> <span class="baseline-fix"></span> </span></span>after .math-box .strut { display: inline-block;} (Live demo.) A strut is just a zero-width element, which can be made to control the vertical extent both below and above the baseline. This concludes the demonstration of how to typeset a fraction using HTML and CSS. Note one important thing here: The browser's layout engine will take care of all spacing and alignment in the horizontal direction, but you have to position everything yourself in the vertical direction. And to do that, you need precise information of how tall your font's characters are. Commenting is not possible for this post, but feel free to leave a question, correction or any comment by using the contact page	2022-07-05 04:31:04	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.594314694404602, "perplexity": 5318.132909582842}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656104512702.80/warc/CC-MAIN-20220705022909-20220705052909-00457.warc.gz"}
http://openstudy.com/updates/4dd90edad95c8b0bfdda63c4	## anonymous 5 years ago solve for the approximate value of x in the following expression lnx=3 1. anonymous You'll have to take e of both sides :$\ln e x= e^3$ and ln e , cancels out so you'll get : $x = e^3 \approx20.09$ Correct me if I'm wrong ^_^ 2. anonymous i still not get it	2016-10-25 01:42:12	{"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": 0.9215225577354431, "perplexity": 933.5896617440291}, "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-2016-44/segments/1476988719843.44/warc/CC-MAIN-20161020183839-00544-ip-10-171-6-4.ec2.internal.warc.gz"}
http://www.water-campus.de/hydro:open-lake-energy	# WEM \| Water Engineering Applied Ecohydrology ### Sidebar hydro:open-lake-energy # Energy budget for open lake evaporation The energy transferred from the water by the energy for evaporation $Q_ {ve}$ equals: $Q_{ve}= Q_ec \frac {\left ( T_s-T_b\right )} {L}$ where $c$ is the specific heat capacity of water (cal/gm/°C) and $T_b$ is an arbitrarily chosen base temperature, in general 0 degrees Celsius, while $L$ is the latent heat of vaporization (590 cal/gm). Re-combining the first two equations, we obtain: $Q_{e}=\frac {Q_s-Q_{rs}-Q_{lw}+Q_v-Q_{\theta} } {1+R+c(T_s-T_b)/L }$ with $Q_s$ incoming solar radiation and $Q_{rs}$ reflected solar radiation and $Q_{lw}$ net long wave radiation from the water body to the atmosphere, $Q_v$ net energy advected into the lake by flows of water, $Q_{\theta}$ change of energy storage in the lake. R is the Bowen Ratio. As the total amount of energy used for evaporation is: $E_o = \frac {Q_e} {L\rho}$ where $\rho$ is the density $g/cm^3$, evaporation from an open water surface can be expressed in terms of the energy balance components and conditions at the lake surface: $E_{o} = \frac {Q_s-Q_{rs}-Q_{lw}+Q_v-Q_{\theta} } { \rho * \left [ L(1+R)+c(T_s-T_b) \right ] }$ References Dunne, T. & Leopold, L. B. (1978). Water in Environmental Planning. New York: Freeman and Company.	2023-03-31 06:30:54	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6384283900260925, "perplexity": 968.063647063627}, "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/1679296949573.84/warc/CC-MAIN-20230331051439-20230331081439-00426.warc.gz"}
https://math.stackexchange.com/questions/3197346/the-paradox-of-random-occurrence-the-family-problem	The paradox of random occurrence the family problem we have 100 families: 10 families have no children, 40 families have 1 child for each one, 30 families have 2 children for each one, 10 families have 3 children for each one and 10 families have 4 children for each one A- suppose we selected a family randomly, what is the expected number of children in that family? B-suppose we selected a child randomly, what is the expected number of children in his family? is it the same? I've tried the expectation to solve it but only for "A" , for "B" I didn't know how to solve it the answer for A is 1.7 children but it is wrong because prof said the correct one is 3. • How many children are there all in all? How many of them have no siblings? How many have one sibling? Two? Three? – lulu Apr 22 at 17:54 • A) How many children in all? How many families? You're done! – David G. Stork Apr 22 at 17:56 • What do you mean by "the answer for A is 1.7 (2) children"? Is this a multiple-choice question, but you forgot to tell us what the options are? – TonyK Apr 22 at 18:53 • I meant my answer is 1.7 so Approximately it's 2 kid ( what is the 0.7 of a child is it his arm or leg? so Approximately it is 2 kids) anyways my prof said the correct answer is NOT 2 – Nidal Apr 22 at 18:57 • There is no reason that an average number of children has to be a whole number. – Ross Millikan Apr 22 at 19:39 For (B), the mean is $$S/T$$ where $$S = 10(0^2) + 40(1^2) + 30(2^2) + 10(3^2) + 10(4^2) = 410$$ $$T = 10(0) + 40(1) + 30(2) + 10(3) + 10(4) = 170$$ so the mean is $$410/170 = 2.412$$ The idea is, for example, a kid from a $$3$$ child family contributes a count of $$3$$ to the sum of family sizes, and that is the case for all $$103$$ such children. So those kids contribute a total of $$10(3^2)$$ to the total. Etc. Denominator is the total number of kids. • I solved the problem a different way and got the same final answer as you for (B)(look at the answer above yours), and your solution looks alot more efficient. Although, I still can't seem to understand why you squared the number of children in a group then divided the sum of all that by the total number of children. – Allan Henriques Apr 22 at 20:22 • @AllanHenriques imagine asking every one of the children "how many children in your family?" There are, e.g., 103=30 kids who would answer "3" so those 30 3's contribute 330 =90 = 10(33) to the "sum of answers". Etc. The mean in B is the "average answer", i.e. the sum of all those answers divided by the total number of answers, i.e. divided by the total number of children. I think your method is must amount to the same thing except that your final value is exactly 10 times too big, so I'm not sure what you did. – Ned Apr 23 at 0:16 • I found the average number of families in the grouping that will be selected, The average number is 24, which is closer to the group with 30 families. The group with 30 families has 2 children per family. But your solution works better because it is more intuitive and understandable – Allan Henriques Apr 23 at 0:37 Intuitively, the expected number of children must be higher when you select a child because you are more likely to select a family with lots of children. In particular, if you select a child, you will never select the families with no children. For each question, you just add up the product of the number of children and the probability of selecting a family of that size. For A, you have $$0.1$$ chance of selecting a family with no children, $$0.4$$ chance of selecting a family with $$1$$ child and so on. The expected number is then $$0.1\cdot 0 + 0.4 \cdot 1 +0.3 \cdot 2 + 0.1 \cdot 3 + 0.1 \cdot 4=1.7$$ For B, what fraction of the children are in families of $$4$$ children? That is the probability you select a family of $$4$$. The approach is the same as part A, but the probabilities are different. Over to you. • I think you have a mistake with the numbers. it should be 0.41 (not 0.44) – adhg Apr 22 at 18:19 • @adhg Correct. Fixed. Thanks. – Ross Millikan Apr 22 at 18:23 • I did the same, but my prof said it's wrong, he gave me the answer for A and it is 3 children, not 1.7 – Nidal Apr 22 at 18:28 • The average for A is clearly less than $3$, so your professor is wrong. I get $\frac {410}{170}\approx 2.412$ for B. – Ross Millikan Apr 22 at 19:18 • The mean for B must be bigger than A, since the distribution in B, relative to that in A, is biased toward larger families. In general the distribution in B is called the "size biasing" of A, and it's mean is always bigger (unless A is a constant distribution in which case the means are the same). – Ned Apr 22 at 19:32 I got the same for part A for part B I made chart and pooled the number of children for each family grouping, then I calculated what percentage of the total child pool a certain family made up, then I multiplied the number of families in the grouping by the percentage, then I added up all the numbers. *family grouping refers to the number of families with a certain number of children format: family grouping, # of children in grouping, percentage of total child pool, family grouping multiplied by percentage 10, 0, 0%, 0 40, 40, 23.53%, 9.412 30, 60, 35.29%, 10.587 10, 30, 17.65%, 1.765 10, 40, 23.53%, 2.353 otal 170, 100% 24.117 10<< 24.117-family grouping <30 30-family group= 2 kids from part a) 1<< 1.7 kids <2 2 kids so the number of kids is the same in A) and B) even though the probabilities change slightly • prof said the correct one is 3 . – Nidal Apr 22 at 19:01 • @NidalAlSaqqa Since he is a professor, he is probably more reliable than a single random guy or a highschool student having a different answer. But the fact that 3 people got the same answer of 1.7 while having bias against the answer 1.7 (since the professor said otherwise), our probability of being right is increased drastically. So bring your solution to classmates and then your teacher. If we are wrong after all this, come back and show us why. – Allan Henriques Apr 22 at 19:06 • 10<< 24.117-family grouping <30 30-family group= 2 kids _____________ I didn't get what you do ? the total was 24 , then 2 ?? – Nidal Apr 22 at 21:02 • @NidalAlSaqqa I found the average number of families in the grouping that will be selected, the average number of families in the selected group is 24.117. A family group with 24.117 families does not exist, and rightly so, because there is never a 100% chance of selecting a certain family. So we need to round the number. This number is closer to 30 families. Thirty corresponds to the group that has 2 children per family, which is the same answer we got in question (A). btw I'm using the << to mean much greater than. – Allan Henriques Apr 22 at 21:18 • thanks alot that was helpful , yet I'm waiting for ned to explain what he did , you both got the same result – Nidal Apr 22 at 21:28	2019-05-27 07:19:39	{"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": 14, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5736070871353149, "perplexity": 553.7316204302002}, "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-2019-22/segments/1558232262029.97/warc/CC-MAIN-20190527065651-20190527091651-00255.warc.gz"}
https://www.semanticscholar.org/paper/Mapping-color-fluctuations-in-the-photon-in-heavy-Alvioli-Frankfurt/7788c6b20341e4d3dc98c3d989f928efeae39296	# Mapping color fluctuations in the photon in ultraperipheral heavy ion collisions at the Large Hadron Collider @article{Alvioli2017MappingCF, title={Mapping color fluctuations in the photon in ultraperipheral heavy ion collisions at the Large Hadron Collider}, author={Massimiliano Alvioli and Leonid L. Frankfurt and Vadim Guzey and Mark Strikman and Mikhail Zhalov}, journal={Physics Letters B}, year={2017}, volume={767}, pages={450-457} } • Published 21 May 2016 • Physics • Physics Letters B 8 Citations ## Figures from this paper Global analysis of color fluctuation effects in proton– and deuteron–nucleus collisions at RHIC and the LHC • Physics Physical Review D • 2018 We test the hypothesis that configurations of a proton with a large-$x$ parton, $x_p \gtrsim 0.1$, have a smaller than average size. The QCD $Q^2$ evolution equations suggest that these small Neutron skin effect in W$^+$ and W$^-$ production in high-energy proton-lead collisions • Physics • 2018 We extend our Monte Carlo generator of global configurations in nuclei to include different spatial distributions of protons and neutrons in heavy nuclei taking into account the difference of spatial Two-particle azimuthal correlations in photonuclear ultraperipheral Pb+Pb collisions at 5.02 TeV with ATLAS Two-particle long-range azimuthal correlations are measured in photonuclear collisions using 1.7 nb−1 of 5.02 TeV Pb+Pb collision data collected by the ATLAS experiment at the LHC. Candidate events Slow-neutron production as a probe of hadron formation in high-energy γ*A reactions • Physics Physical Review C • 2020 Studies of the hadron formation in hard processes on nuclei were focusing on hadrons in the current fragmentation region. Little is known about the space-time picture of formation of hadrons in the New opportunities at the photon energy frontier • Physics • 2020 Ultra-peripheral collisions (UPCs) involving heavy ions and protons are the energy frontier for photon-mediated interactions. UPC photons can be used for many purposes, including probing low-$x$ UvA-DARE (Digital Academic Repository) Two-particle azimuthal correlations in photonuclear ultraperipheral Pb-Pb collisions at 5.02 TeV with ATLAS Two-particle long-range azimuthal correlations are measured in photonuclear collisions using 1 . 7 nb − 1 of 5.02 TeV Pb + Pb collision data collected by the ATLAS experiment at the CERN Large Hadron Two-particle azimuthal correlations in photonuclear ultraperipheral Pb + Pb collisions at 5 . 02 TeV with ATLAS • Physics • 2021 Citation for published version (APA): Aad, G., Abbott, B., Abbott, DC., Abud, AA., Abeling, K., Abhayasinghe, D. K., Abidi, S. H., AbouZeid, O. S. A., Abraham, NL., Abramowicz, H., Dam, M., Camplani, Snowmass 2021-Letter of Interest New opportunities at the photon energy frontier Coordinators of this LoI : • 2020 ## References SHOWING 1-10 OF 86 REFERENCES Revealing “flickering” of the interaction strength in p A collisions at the CERN LHC • Physics • 2014 We analyze $pA$ interactions at ultra-high energies within the semiclassical approximation for high energy processes accounting for the diffractive processes and a rapid increase with the incident Evidence for x-dependent proton color fluctuations in pA collisions at the LHC • Physics • 2014 We argue that the pattern of the deviation from the Glauber approximation prediction for the centrality dependence of the rate of forward jet production observed in pA collisions at the LHC provides AbstractWe present the perturbative QCD analysis of nuclear shadowing in the deep inelastic scattering at smallx in terms of the spatial wave function ofq $$\bar q$$ fluctuations of virtual photons.	2022-08-12 16:16:10	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.9118890762329102, "perplexity": 8629.703006417843}, "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-2022-33/segments/1659882571719.48/warc/CC-MAIN-20220812140019-20220812170019-00733.warc.gz"}
https://tex.stackexchange.com/questions/276720/lyx-cant-import-bioinformatics-oxford-journal-latex-file	# Lyx can't import Bioinformatics (Oxford journal) Latex file Hi I am trying to import a latex file given hereto lyx. http://www.oxfordjournals.org/our_journals/bioinformatics/for_authors/submission_online.html Lyx is throwing an error. Can anyone help? • Welcome to TeX.SX! Please help us help you and add a minimal working example (MWE) that illustrates your problem. Reproducing the problem and finding out what the issue is will be much easier when we see compilable code, starting with \documentclass{...} and ending with \end{document}. Nov 5 '15 at 8:56 If you put the bioinfo.cls in the proper directory, such as (on Ubuntu): /usr/share/texlive/texmf-dist/tex/latex/bioinfo/ Then you can make a Lyx layout as -- #% Do not delete the line below; configure depends on this # \DeclareLaTeXClass[bioinfo]{article (Bioinformatics)} # Read the definitions from article.layout Input article.layout -- And then put that in the proper directory, namely: /usr/share/lyx/layouts/ Then make sure you do sudo texhash on the command line, and tools-> reconfigure in lyx, and you should have it set up. There is a probably with the .eps image (OUP_First_SBk_Bot_8401.eps) in Lyx that I can't figure out, don't know if that will cause you problems. • Welcome to TeX.SX! You can format text passages as code by using the {} button on the top of the editor. Small in line passages you can format as code with the ` accent. Nov 5 '15 at 12:30 • Thanks for the answer, but I am a Mac user and I could not find any folder of the kind you suggested. Do you know what to do over Mac? – Mr K Nov 6 '15 at 0:10	2022-01-27 13:59:02	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.7114408612251282, "perplexity": 3001.068405488993}, "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/1642320305266.34/warc/CC-MAIN-20220127133107-20220127163107-00288.warc.gz"}
https://tex.stackexchange.com/questions/318562/font-size-of-table-much-bigger-than-rest-of-text	# Font size of table much bigger than rest of text I have created a table (two rows, four columns) in a seperate .tex file. The table is smaller than the text width, so I stretch it to the text width. Unfortunately, the font size is much bigger than the font size of the rest of my text. This also appears when I don't stretch the table. How can I get the same font size of my table as with the rest of my text or is this not usual? Here is my table (table.tex): \renewcommand{\arraystretch}{1.5} \begin{tabular}{cccc}\toprule ~ & \textbf{Col1} & \textbf{Col2} & \textbf{Col3} \\ \midrule $\beta = 1$ & 1234 (0.01234) & 4.524 (0.2345) & -0.54 (0.0423) \\ \midrule $\beta = 2$ & 0.0345 (0.023) & 0.0012(0.0125) & -0.0161 (0.0174) \\ \bottomrule \end{tabular} And I include the table in the following way: \begin{table}[!htbp] \caption[Caption]{This is my caption.} \centering \resizebox{\textwidth}{!}{\input{table}} \end{table} • \resizebox stretches anything, so what do you expect? ;-) If you want to have a \textwidth wide table, use tabularx for example. And please don't post such fragments only. – user31729 Jul 8 '16 at 15:28 • never use \resizebox on tables, otherwise you get what you describe. – David Carlisle Jul 8 '16 at 15:58 The following example will fix the problem: \documentclass{article} \usepackage{booktabs} \usepackage{tabularx} \usepackage{lipsum} \newcolumntype{C}{>{\small\centering\arraybackslash}X} \begin{document} \parindent=0pt \renewcommand{\arraystretch}{1.5} \lipsum[1-2] \begin{table}[!htbp] \caption[Caption]{This is my caption.} \begin{tabularx}{\textwidth}{CCCC}\toprule ~ & \textbf{Col1} & \textbf{Col2} & \textbf{Col3} \\ \midrule $\beta = 1$ & 1234 (0.01234) & 4.524 (0.2345) & -0.54 (0.0423) \\ \midrule $\beta = 2$ & 0.0345 (0.023) & 0.0012(0.0125) & -0.0161 (0.0174) \\ \bottomrule \end{tabularx} \end{table} \end{document} The \resizebox will resize and will make the fonts bigger. Hence, use tabularx package with C column specifier. The C column specifier is defined using \newcolumntype command, which sets the content in center. • Thank you very much. This worked very well. I have another bigger table with 7 columns. Some of the columns are again centered but some are left aligned and for one column I'm using p{7cm}. This table is wider than the text width (and also pretty long so landscape does also not work) . If I'm using resizebox to resize it to text width the fonts get smaller but I think there is no way without making the fonts smaller (otherwise it does not fit into the page). How can I use your approach in this case (also for the left aligned and p{7cm} columns) ? – machinery Jul 8 '16 at 18:18	2019-09-15 16:10:25	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 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": 0.9904218316078186, "perplexity": 1646.9337578280242}, "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/1568514571651.9/warc/CC-MAIN-20190915155225-20190915181225-00036.warc.gz"}
https://accesscardiology.mhmedical.com/content.aspx?bookid=2432&sectionid=190801452	Skip to Main Content ## CHAPTER 1 • 1–1. The lungs always receive more blood flow than any other organ because 100% of the cardiac output always passes through the lungs. • 1–2. False. Flow through any vascular bed depends on its resistance to flow and the arterial pressure. As long as this pressure is maintained constant (a critical point that is dependent on adjustments in cardiac output), alterations in flow through any individual bed will have no influence on flow through other beds in parallel with it. • 1–3. A leaky aortic valve will cause a diastolic murmur. Normally, the aortic valve is closed during diastole when there is a large reverse pressure difference between the aorta and left ventricle. • 1–4. False. Slowing conduction through the AV node will have no effect on the heart rate but will increase the interval between atrial and ventricular excitation. The heart rate is normally slowed by decreases in the rate of action potential initiation by pacemaker cells in the SA node. • 1–5. 1. Saying that the diameter of a vessel increases by 10% is equivalent to saying that its radius increases by 10%; that is, the radius after the change is 1.1 times the radius before the change. The Poiseuille equation says that, other factors equal, resistance is proportional to 1 over radius to the fourth power: $R ∝ 1 / r 4$ Thus, $R before ∝ 1 / ( r before ) 4$ $R after ∝ 1 / ( 1.1 r before ) 4 = [ 1 / ( 1.1 ) 4 ] [ 1 / ( r before ) 4 ] = ( 1 / 1.46 ) R before = 0.68 R before$ Therefore, a 10% increase in vessel radius will reduce its resistance by 32%. 2. Because $Q · = Δ P / R$ , and R after = 0.68 R before $Q · after = Δ P / 0.68 R before$ … and for a given ΔP $Q · after = 1 / 0.68 R before = 1.46 Q · before$ Therefore, for a given ΔP, a 10% increase in vessel diameter will increase flow by 46%. [Note: One must always use change factors NOT percentage changes in these equations.] • 1–6. 1. Wrong! The pulmonary and systemic circuits are arranged in series and therefore must have the same output flow in the steady state. 2. Wrong! The right and left hearts are served by a common electrical excitation system and therefore beat at the same rate. 3. Wrong! Although it is true that the right ventricle is less muscular than the left, that fact does not explain why pulmonary pressure is so low. Actually, it is the other way around because pulmonary arterial pressure is low, the right ventricle does not have to be very muscular to pump blood into the lungs. 4. Right! $ΔP= Q ⋅ ×R$ . Because $Q ⋅$ is the same (= CO) through the lungs and the systemic circulation, the only way ... ### Pop-up div Successfully Displayed This div only appears when the trigger link is hovered over. Otherwise it is hidden from view.	2019-08-18 14:54:08	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 8, "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": 0.597104012966156, "perplexity": 1472.455467889372}, "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-2019-35/segments/1566027313936.42/warc/CC-MAIN-20190818145013-20190818171013-00444.warc.gz"}
https://www.clutchprep.com/chemistry/practice-problems/73491/the-decomposition-of-n2o5-can-be-described-by-the-equation2n2o5-soln-4no2-soln-o	Chemistry Practice Problems Average Rate of Reaction Practice Problems Solution: The decomposition of N2O5 can be described by the ... 🤓 Based on our data, we think this question is relevant for Professor Miao's class at CSUN. # Solution: The decomposition of N2O5 can be described by the equation2N2O5 (soln) → 4NO2 (soln) + O2 (g)Given these data for the reaction at 45° C in carbon tetrachloride solution, calculate the average rate of reaction for each successive time interval. ###### Problem The decomposition of N2O5 can be described by the equation 2N2O5 (soln) → 4NO2 (soln) + O2 (g) Given these data for the reaction at 45° C in carbon tetrachloride solution, calculate the average rate of reaction for each successive time interval. View Complete Written Solution Average Rate of Reaction Average Rate of Reaction #### Q. The rate of the following reaction is 0.720 M/s. What is the relative rate of each species in the reaction? Solved • Sun May 27 2018 08:30:31 GMT-0400 (EDT) Average Rate of Reaction #### Q. Consider the reaction: 8H2S(g) + 4O2(g) → 8H2O(g) + S8(g) Δ[H 2S]/Δt = -0.031 M/s a. Find Δ[O2]/Δt.b. Find Δ[H2O]/Δt.c. Find Δ[S8]/Δt.d. Fi... Solved • Mon Apr 02 2018 10:43:19 GMT-0400 (EDT) Average Rate of Reaction #### Q. Consider the reaction: 8H2S(g) + 4O2(g) → 8H2O(g) + S8(g) Δ[H2S]/Δt = -0.071 M/s a. Find Δ[O2]/Δt. Express your answer to two significant figur... Solved • Wed Mar 28 2018 13:58:26 GMT-0400 (EDT) Average Rate of Reaction #### Q. Consider the following reaction and data:H2 + 2ICl → 2HCl + I 2Time (s) I 2 concentration (M) 5 1.11 ... Solved • Tue Mar 27 2018 11:11:12 GMT-0400 (EDT)	2020-03-30 04:17:34	{"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": 0.8377814888954163, "perplexity": 11052.998870276326}, "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/1585370496523.5/warc/CC-MAIN-20200330023050-20200330053050-00538.warc.gz"}
http://mathhelpforum.com/pre-calculus/22290-generating-linear-programming-problem.html	# Thread: Generating a Linear Programming problem 1. ## Generating a Linear Programming problem Thanks. It was an error in the question. Problem solved. 2. Hello, DooBeeDoo! I can't tell what the right-hand side values of the constraints are. They weren't given . . . And without them, we can't write the constraints (exactly). A company manufactures two products from three scarce resources. The net profits are 4 and 3, respectively. Each unit of Product 1 uses 2 units of Resource1, 1 unit of Resource 2, and 3 units of Resource 3. Each unit of Product 2 uses 3 units of Resource 1, 2 units of Resource 2, and 2 units of Resource 3. Let $x$ = number of Product 1 to be produced: . $x \:\geq\:0$ Let $y$ = number of Product 2 to be produced: . $y \:\geq \:0$ I assume they want us to maximize the Profit Function: . $P \:=\:4x + 3y$ We have the following information: $\begin{array}{cccccccccc}& & \text{Resource 1} &\|& \text{Resource 2} &\|& \text{Resource 3} &\|& \text{Profit} &\| \\ \hline \text{Product 1 }(x) &\|& 2x &\|& x &\|& 3x &\|& 4x & \| \\ \text{Product 2 }(y) &\|& 3y &\|& 2y &\|& 2y &\|& 3y &\| \\ \hline \text{Available} &\|& T_1 &\|& T_2 &\|& T_3 &\|& &\| \end{array}$ If we were given $T_1,\:T_2,\:T_3$, we could write the inequalities: . . $\begin{array}{ccc}x & \geq & 0 \\ y & \geq & 0 \\ 2x + 3y & \leq & T_1 \\ x + 2y & \leq & T_2 \\ 3x +2y & \leq & T_3 \end{array}$ and solve the problem . . .	2016-08-24 03:11:34	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 8, "wp_latex": 0, "mimetex.cgi": 0, "/images/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": 0.7766358256340027, "perplexity": 961.7221308116812}, "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-2016-36/segments/1471982290765.41/warc/CC-MAIN-20160823195810-00221-ip-10-153-172-175.ec2.internal.warc.gz"}
https://stats.stackexchange.com/questions/430933/relation-between-location-of-a-sample-mean-in-sampling-distribution-and-the-stan	# relation between location of a sample mean in sampling distribution and the standard error Can someone explain the following statement with an example We'll describe the location of the sample mean by calculating how many standard errors it is away from the center of the sampling distribution. That will give us a z-score for our sample mean. Z-scores for test results. Suppose exam scores on a statewide math test are normally distributed with mean $$\mu = 300, \sigma =25.$$ A particular student scores $$X = 330,$$ what is her z-score? $$Z = \frac{X - \mu}{\sigma} = \frac{330 - 300}{25} = \frac{30}{25} = 1.2.$$ She scores 30 points above the mean $$\mu,$$ which is $$30/25 = 1.2$$ standard deviations above the mean. Suppose that last year's scores on an equivalent test were normally distributed with $$\mu = 305$$ and $$\sigma=27,$$ and that her older brother scored $$Y = 335.$$ Is it fair for him to brag that he's better at math than his sister? Aside from mentioning that bragging is futile and maybe nasty, we could point out that his z-score was only $$(335 - 305)/27 = 30/27 = 1.11.$$ One purpose of z-scores is to try to make results from slightly different situations more comparable. If his sister knows about z-scores, then he might decide to turn the conversation to one of his non-mathematical accomplishments. Z-scores for standard errors. In your question you mention 'standard error'. That term is used in statistical inference. Suppose you have a sample of size $$n = 100$$ from a normal population with mean $$\mu = 50$$ and standard deviation $$\sigma = 5$$ (variance $$\sigma^2 = 25).$$ Also suppose the sample mean is $$\bar X = 50.8$$ Then the 'standard error of the mean' is $$\sigma_{\bar X} = SD(\bar X) = \frac{\sigma}{\sqrt{n}} = \frac{5}{10} = 0.5.$$ If you want to know how many standard errors the sample mean $$\bar X = 50.8$$ is above the population mean $$\mu = 50,$$ then you use the same formula as for test scores above: $$Z =\frac{\bar X - \mu}{SD(\bar X)} = \frac{\bar X - \mu}{\sigma/\sqrt{n}} = \frac{50.8-50}{0.5} = \frac{0.8}{0.5} = 1.6.$$ So the sample mean is 1.6 standard errors above the population mean. In many applications, such a z-score often lies in the interval $$(-2, 2)$$ when nothing remarkable has occurred. But I should leave the specific applications to your text and lectures. Extra. More on z-scores for test results: Another purpose of z-scores is to find various kinds of probabilities and proportions. We could use a printed table of the standard normal distribution to find the percentage of students who scored lower than the sister this year and the percentage who scored lower than her brother last year. These are called percentiles. Maybe you can use the printed standard normal cumulative distribution table in your textbook to find the percentiles shown below. (I'm using R statistical software, but you should get similar percentiles, to about four significant digits, from the printed table.) pnorm(1.2) [1] 0.8849303 # sister's percentile is about 88.5 pnorm(1.11) [1] 0.8665005 # brother's percentile is about 86.7 • Thanks a lot man!! This has been the best explanation to this question. I knew what is Z score and all this but now I know how it is used in real world. – Deshwal Oct 12 '19 at 13:11	2021-07-27 15:52:09	{"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": 19, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.683064341545105, "perplexity": 362.1060541489085}, "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-2021-31/segments/1627046153392.43/warc/CC-MAIN-20210727135323-20210727165323-00571.warc.gz"}
https://kb.osu.edu/handle/1811/19393?show=full	dc.creator Lee, P. C. en_US dc.creator Nee, J. B. en_US dc.date.accessioned 2006-06-15T19:18:37Z dc.date.available 2006-06-15T19:18:37Z dc.date.issued 1999 en_US dc.identifier 1999-RI-04 en_US dc.identifier.uri http://hdl.handle.net/1811/19393 dc.description Author Institution: Department of Physics and Chemistry, National Central Univ. en_US dc.description.abstract The metastable $O(^{1}D)$ produced in the photodissociation of $O_{2}$ in the Rydberg states lying in the wavelength region 105-130 nm were investigated. The detection of $O(^{1}D)$ was made by measuring the infrared emission at 762 nm from the transition $O_{2}(b^{1}\Sigma^{+}_{g}- X {^{3}}\Sigma^{-}_{g})$ produced by $O({^{1}}D) + O_{2}$. The excited states of $O_{2}$ can be classified as either the ${^{3}}\Sigma^{-}_{u}$ states, which are correlated with $O(^{1}D) + O(^{3}P)$, or the ${^{3}}\Pi_{u}$ states, which are correlated with $O(^{3}P) + O(^{3}P)$. Our studies resulted in the determination of the quantum yields for producing $O(^{1}D)$ for many bands of the $E {^{3}}\Sigma^{+}_{u}$ states. We assigned several series of Rydberg states, and found the mixing of the ${^{3}}\Sigma^{+}_{u}$ and ${^{3}}\Pi_{u}$ at some wavelengths. Furthermore, a band at 116.3 nm was observed to emit weakly in the visible region. en_US dc.format.extent 90537 bytes dc.format.mimetype image/jpeg dc.language.iso English en_US dc.publisher Ohio State University en_US dc.title PRODUCTION OF $O(^{1}D)$ IN THE RYDBERG STATES OF $O_{2}$ BY PHOTODISSOCIATION IN THE WAVELENGTH REGION 105-130 NM en_US dc.type article en_US ### Files in this item Items in Knowledge Bank are protected by copyright, with all rights reserved, unless otherwise indicated.	2021-09-19 21:16:11	{"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": 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": 0.334276407957077, "perplexity": 3090.0139009184154}, "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-2021-39/segments/1631780056900.32/warc/CC-MAIN-20210919190128-20210919220128-00519.warc.gz"}
https://readpaper.com/paper/4705295618352889857	This website requires JavaScript. # Improved Kernel Alignment Regret Bound for Online Kernel Learning Dec 2022 In this paper, we improve the kernel alignment regret bound for online kernellearning in the regime of the Hinge loss function. Previous algorithm achievesa regret of $O((\mathcal{A}_TT\ln{T})^{\frac{1}{4}})$ at a computationalcomplexity (space and per-round time) of $O(\sqrt{\mathcal{A}_TT\ln{T}})$,where $\mathcal{A}_T$ is called \textit{kernel alignment}. We propose analgorithm whose regret bound and computational complexity are better thanprevious results. Our results depend on the decay rate of eigenvalues of thekernel matrix. If the eigenvalues of the kernel matrix decay exponentially,then our algorithm enjoys a regret of $O(\sqrt{\mathcal{A}_T})$ at acomputational complexity of $O(\ln^2{T})$. Otherwise, our algorithm enjoys aregret of $O((\mathcal{A}_TT)^{\frac{1}{4}})$ at a computational complexity of$O(\sqrt{\mathcal{A}_TT})$. We extend our algorithm to batch learning andobtain a $O(\frac{1}{T}\sqrt{\mathbb{E}[\mathcal{A}_T]})$ excess risk boundwhich improves the previous $O(1/\sqrt{T})$ bound. Q1论文试图解决什么问题？ Q2这是否是一个新的问题？ Q3这篇文章要验证一个什么科学假设？ 0	2023-01-29 09:05:44	{"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": 0.9431009292602539, "perplexity": 1631.8449054256819}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764499710.49/warc/CC-MAIN-20230129080341-20230129110341-00385.warc.gz"}
http://mathhelpforum.com/math-topics/27516-pka-pkb-calculation-2.html	# Math Help - pKa and pKb calculation 1. rightyo I tried to put in values for the equation but I have no idea what goes where ... Don't worry I'll just make up the marks on the organic chemistry bit lol I give up on this stuff 2. Originally Posted by Dan167 rightyo I tried to put in values for the equation but I have no idea what goes where ... Don't worry I'll just make up the marks on the organic chemistry bit lol I give up on this stuff i don't see your hang up, Dan. you have $x^2 + K_ax - 0.1K_a = 0$ there are not many choices of where to put anything. you know that $K_a = 3.3 \times 10^{-2}$, so just replace $K_a$ with that wherever you see it, so you get $x^2 + {\color{red}(3.3 \times 10^{-2})}x - 0.1{\color{red}(3.3 \times 10^{-2})} = 0$ so you get $x^2 + 0.033x - 0.0033 = 0$ so, $x = \frac {-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac {-0.033 \pm \sqrt{(0.033)^2 - 4(1)(0.0033)}}{2}$ and just plug that in your calculator when you get the answer, take the logarithm of it and then take it's negative and you have the pH 3. If your having difficulty with the qudratic formula just assume that the change in conceration of the acid can be nelgated and use the following formula instead. $Ka = \frac{x^2}{[Acid]}$ your answer will be a bit off, but your chemistry teachers may not mind, you should ask them is that is allowed on your course (because it is on mine). Yeah we have the same general feeling in my chemistry class as well most people make up all the marks on the organic chemistry, I however make up the marks with analytical chemistry . Page 2 of 2 First 12	2015-10-04 12:10:08	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 7, "wp_latex": 0, "mimetex.cgi": 0, "/images/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": 0.848652184009552, "perplexity": 500.3761351560889}, "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-2015-40/segments/1443736673439.5/warc/CC-MAIN-20151001215753-00247-ip-10-137-6-227.ec2.internal.warc.gz"}
http://mathhelpforum.com/geometry/223021-finding-midpoint-line-connecting-two-points.html	# Thread: Finding the Midpoint of the line connecting two points which are... 1. ## Finding the Midpoint of the line connecting two points which are... Hi everyone. I can't fully describe things in the title, so here is the full question: $P(ct, \frac {c}{t})$ is any point on the rectangular hyperbola $xy=c^{2}$. The tangent at P cuts the y-axis at T and the normal at P cuts the x-axis at N. Find the Cartesian equation of the locus of the midpoint of TN. My attempted solution: First, I define T to be the point $T(0,q)$ and N to be the point $N(p,0)$. To find the Gradient of the tangent, I implicitly differentiate the hyperbola. Therefore: $\frac{\mathrm{d} y}{\mathrm{d} x} (xy=c^{2})$ $\frac{\mathrm{d} y}{\mathrm{d} x} = - \frac {y}{x}$ Here, I substitute the $x, y$ values of P into the gradient, and try to solve for PT and PN. So using this to solve for PT: The gradient of the tangent using the values of P would be $- \frac {1}{t^{2}}$ This makes the equation PT to be: $(y-0)= - \frac {1}{t^{2}} (x -q)$ Therefore: $q= t^{2}y+x$ The gradient for the normal would be $t^{2}$, and the equation would be $(y-p)= t^{2} (x -0)$ $p=y-t^{2}x$ Using the found values of p and q, let M be the midpoint of TN. Therefore, $M(X,Y)=(\frac {1}{2} (y-t^{2}x), \frac {1}{2} (t^{2}y+x))$ Here I look like this. How do I find the gradient to the midpoint so that I can find the Cartesian equation. After being like for some time, I turned over to the answer and it was given as: $y^{4}=c^{2}(c^{2}-2xy)$ What am I doing wrong? Or can c be substituted into the equation later? Can I have some pointers please? 2. ## Re: Finding the Midpoint of the line connecting two points which are... I suspect that my differentiation is incorrect, but I can't find the correct differentiation. 3. ## Re: Finding the Midpoint of the line connecting two points which are... I am having a little trouble understanding your notation. The notation $\dfrac{\mathrm{d}y}{\mathrm{d}x}$ means the derivative of $y$ with respect to $x$. The notation $\dfrac{\mathrm{d}y}{\mathrm{d}x}(xy=c^2)$ doesn't make sense. Am I correct in assuming you mean implicitly differentiate both sides of the equation $xy=c^2$ with respect to $x$? If that is the case, the correct notation would be $\dfrac{\mathrm{d}}{\mathrm{d}x}(xy) = \dfrac{\mathrm{d}}{\mathrm{d}x}(c^2)$ which reads the derivative of $xy$ with respect to $x$ is equal to the derivative of $c^2$ with respect to $x$. Anyway, assuming my understanding of your notation is correct, then the equation you are using for the tangent and normal lines are wrong. The tangent line at $\left(ct,ct^{-1}\right)$ is given by the equation $y-ct^{-1} = -t^{-2}(x-ct)$. If $x=0$, then you will get a point on the $y$-axis. So, let $x=0$ and solve for $y$: $y = ct^{-1} + ct^{-1} = \dfrac{2c}{t}$ So $T = \left(0,\dfrac{2c}{t} \right)$. Next, the equation of the normal line is given by $y-ct^{-1} = t^2(x-ct)$. To find the $x$-intercept, let $y=0$ and solve for $x$: $x = -ct^{-3}+ct$ So $N = \left(c\dfrac{t^4-1}{t^3}, 0 \right)$. Now, the midpoint is $\left( c\dfrac{t^4-1}{2t^3}, \dfrac{c}{t} \right)$. 4. ## Re: Finding the Midpoint of the line connecting two points which are... Thanks a lot. Like I thought, my derivative was wrong. Yes, I was speaking of implicit differentiation. I think I pressed the wrong button in the editor. (I use a LaTeX editor). From your workings, it looks like I don't need to use differentiation in my question at all after all. Am I correct? 5. ## Re: Finding the Midpoint of the line connecting two points which are... Actually, you calculated the derivative correctly. I did not rewrite it, but in order to get the slope of the tangent and normal lines, I used the derivative you obtained. $\left. \dfrac{\mathrm{d}y}{\mathrm{d}x}\right]_{\left(ct,ct^{-1}\right)} = -\dfrac{y}{x} = -\dfrac{ct^{-1}}{ct} = -\dfrac{1}{t^2}$ where the notation $\left. \dfrac{\mathrm{d}y}{\mathrm{d}x}\right]_{\left(ct,ct^{-1}\right)}$ means the evaluation of the derivative at the point $(ct,ct^{-1})$. So, the slope of the tangent line is $-\dfrac{1}{t^2}$ and the slope of the normal line is $t^2$ (the slope of the normal line is negative the reciprocal of the slope of the tangent line). So, everything else was your work. The only difference between our two results is that rather than writing the equations of the lines using the unknown points $T,N$, I just wrote the equations using the same known point $(ct,ct^{-1})$. Everything else was your work. Do you see how to get from the point $\left(c\dfrac{t^4-1}{2t^3},\dfrac{c}{t}\right)$ to the equation $y^4 = c^2(c^2-2xy)$? 6. ## Re: Finding the Midpoint of the line connecting two points which are... The Gradient of M is $\frac {y_T-y_N}{x_T-x_N}= \frac {(\frac {2c}{t})}{-\frac {c(t^{4}-1)}{t^{3}}}$ Simplifying, this would be $-\frac {2t^{2}}{t^{4}-1}$ Inputting then $y-\frac {c}{t}= (-\frac {2t^{2}}{t^{4}-1})(x-\frac {c(t^{4}-1)}{t^{3}})$ Simplifying, I get $t^{5}(y)-t^{4}(3c)+t^{3}(2x)+t(c-y)+2=0$ I don't know how to get rid of the variable t. Help on that would be appreciated! 7. ## Re: Finding the Midpoint of the line connecting two points which are... Originally Posted by LimpSpider The Gradient of M is $\frac {y_T-y_N}{x_T-x_N}= \frac {(\frac {2c}{t})}{-\frac {c(t^{4}-1)}{t^{3}}}$ Simplifying, this would be $-\frac {2t^{2}}{t^{4}-1}$ Inputting then $y-\frac {c}{t}= (-\frac {2t^{2}}{t^{4}-1})(x-\frac {c(t^{4}-1)}{t^{3}})$ Simplifying, I get $t^{5}(y)-t^{4}(3c)+t^{3}(2x)+t(c-y)+2=0$ I don't know how to get rid of the variable t. Help on that would be appreciated! You should not be taking any gradients. You are using the midpoint $(x,y) = \left(\dfrac{c(t^4-1)}{2t^3}, \dfrac{c}{t}\right)$, so $x = \dfrac{c(t^4-1)}{2t^3}, y = \dfrac{c}{t}$. Solve for $t$ in the second equation and plug it into the first: \begin{align}t & = \dfrac{c}{y} \\ x & = \dfrac{c\left[\left(\tfrac{c}{y}\right)^4-1\right]}{2\left(\tfrac{c}{y}\right)^3}\end{align} Multiply top and bottom of the RHS by $c\left(\dfrac{y}{c}\right)^4$: $x = \dfrac{c^2\left(1 - \tfrac{y^4}{c^4}\right)}{2y}$ Multiply both sides by 2y: \begin{align}2xy & = c^2 - \dfrac{y^4}{c^2} \\ \dfrac{y^4}{c^2} & = c^2 - 2xy \\ y^4 & = c^2(c^2-2xy)\end{align} 8. ## Re: Finding the Midpoint of the line connecting two points which are... Thanks! That helps!	2016-12-05 08:56:33	{"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": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 65, "wp_latex": 0, "mimetex.cgi": 0, "/images/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": 0.9624226689338684, "perplexity": 158.67044400334706}, "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-2016-50/segments/1480698541556.70/warc/CC-MAIN-20161202170901-00289-ip-10-31-129-80.ec2.internal.warc.gz"}
https://www.law.cornell.edu/cfr/text/40/63.11646	40 CFR § 63.11646 - What are my compliance requirements? § 63.11646 What are my compliance requirements? (a) Except as provided in paragraph (b) of this section, you must conduct a mercury compliance emission test within 180 days of the compliance date for all process units at new and existing affected sources according to the requirements in paragraphs (a)(1) through (a)(13) of this section. This compliance testing must be repeated annually thereafter, with no two consecutive annual compliance tests occurring less than 3 months apart or more than 15 months apart. (1) You must determine the concentration of mercury and the volumetric flow rate of the stack gas according to the following test methods and procedures: (i) Method 1 or 1A (40 CFR part 60, appendix A-1) to select sampling port locations and the number of traverse points in each stack or duct. Sampling sites must be located at the outlet of the control device (or at the outlet of the emissions source if no control device is present) and prior to any releases to the atmosphere. (ii) Method 2, 2A, 2C, 2D, 2F (40 CFR part 60, appendix A-1), or Method 2G (40 CFR part 60, appendix A-2) to determine the volumetric flow rate of the stack gas. (iii) Method 3, 3A, or 3B (40 CFR part 60, appendix A-2) to determine the dry molecular weight of the stack gas. You may use ANSI/ASME PTC 19.10, “Flue and Exhaust Gas Analyses” (incorporated by reference-see § 63.14) as an alternative to EPA Method 3B. (iv) Method 4 (40 CFR part 60, appendix A-3) to determine the moisture content of the stack gas. (v) Method 29 (40 CFR part 60, appendix A-8) to determine the concentration of mercury, except as provided in paragraphs (a)(1)(vi) and (vii) of this section. (vi) Upon approval by the permitting authority, ASTM D6784; “Standard Test Method for Elemental, Oxidized, Particle-Bound and Total Mercury in Flue Gas Generated from Coal-Fired Stationary Sources (Ontario Hydro Method)” (incorporated by reference - see § 63.14) may be used as an alternative to Method 29 to determine the concentration of mercury. (vii) Upon approval by the permitting authority, Method 30B (40 CFR part 60, appendix A-8) may be used as an alternative to Method 29 to determine the concentration of mercury for those process units with relatively low particulate-bound mercury as specified in Section 1.2 of Method 30B. (2) A minimum of three test runs must be conducted for each performance test of each process unit. Each test run conducted with Method 29 must collect a minimum sample volume of 0.85 dry standard cubic meters (30 dry standard cubic feet). If conducted with Method 30B or ASTM D6784, determine sample time and volume according to the testing criteria set forth in the relevant method. If the emission testing results for any of the emission points yields a non-detect value, then the minimum detection limit (MDL) must be used to calculate the mass emissions rate (lb/hr) used to calculate the emissions factor (lb/ton) for that emission point and, in turn, for calculating the sum of the emissions (in units of pounds of mercury per ton of concentrate, or pounds of mercury per million tons of ore) for all emission points subject to the emission standard for determining compliance. If the resulting mercury emissions are greater than the MACT emission standard, the owner or operator may use procedures that produce lower MDL results and repeat the mercury emissions testing one additional time for any emission point for which the measured result was below the MDL. If this additional testing is performed, the results from that testing must be used to determine compliance (i.e., there are no additional opportunities allowed to lower the MDL). (3) Performance tests shall be conducted under such conditions as the Administrator specifies to the owner or operator based on representative performance of the affected source for the period being tested. Upon request, the owner or operator shall make available to the Administrator such records as may be necessary to determine the conditions of performance tests. Performance tests must be conducted under operating conditions (including process or production throughputs) that are based on representative performance. Record and report to the permit authority the process throughput for each test run. For sources with multiple emission units (e.g., two roasters, or a furnace, electrowinning circuit and a mercury retort) ducted to a common control device and stack, compliance testing must be performed either by conducting a single compliance test with all affected emissions units in operation or by conducting a separate compliance test on each emissions unit. Alternatively, the owner or operator may request approval from the permit authority for an alternative testing approach. If the units are tested separately, any emissions unit that is not tested initially must be tested as soon as is practicable. If the performance test is conducted when all affected units are operating, then the number of hours of operation used for calculating emissions pursuant to paragraphs (a)(6) and (7) of this section must be the total number of hours for the unit that has the greatest total operating hours for that period of time, or based on an appropriate alternative method approved by the permit authority to account for the hours of operation for each separate unit in these calculations. (4) Calculate the mercury emission rate (lb/hr), based on the average of 3 test run values, for each process unit (or combination of units that are ducted to a common stack and are tested when all affected sources are operating pursuant to paragraph (a)(3) of this section) using Equation (1) of this section: $E=\mathrm{Cs}\mathrm{Qs}K\phantom{\rule{0ex}{0ex}}\text{(Eq. 1)}$ Where: E = mercury emissions in lb/hr; Cs = concentration of mercury in the stack gas, in grains per dry standard cubic foot (gr/dscf); Qs = volumetric flow rate of the stack gas, in dry standard cubic feet per hour; and K = conversion factor for grains (gr) to pounds (lb), 1.43 × 10−4. (5) Monitor and record the number of one-hour periods each process unit operates during each month. (6) For the initial compliance determination for both new and existing sources, determine the total mercury emissions for all the full calendar months between the compliance date and the date of the initial compliance test by multiplying the emission rate in lb/hr for each process unit (or combination of units ducted to a common stack that are tested together) by the number of one-hour periods each process unit (or the unit that had the greatest total operating hours among the combination of multiple units with one stack that are tested together, or an alternative method approved by the permit authority, pursuant to paragraph (a)(3) of this section) operated during those full calendar months prior to the initial compliance test. This initial period must include at least 1 full month of operations. After the initial compliance test, for subsequent compliance tests, determine the mercury mass emissions for the 12 full calendar months prior to the compliance test in accordance with the procedures in paragraph (a)(7) of this section. Existing sources may use a previous emission test for their initial compliance determination in lieu of conducting a new test if the test was conducted within one year of the compliance date using the methods specified in paragraphs (a)(1) through (a)(4) of this section, and the tests were representative of current operating processes and conditions. If a previous test is used for their initial compliance determination, 3 to 12 full months of data on hours of operation and production (i.e., million tons of ore or tons of concentrate), including the month the test was conducted, must be used to calculate the emissions rate (in units of pounds of mercury per million tons of ore for the ore pretreatment affected sources, or in units of pounds of mercury per tons of concentrate for the other affected sources). (7) For compliance determinations following the initial compliance test for new and existing sources, determine the total mercury mass emissions for each process unit for the 12 full calendar months preceding the performance test by multiplying the emission rate in lb/hr for each process unit (or combination of units ducted to a common stack that are tested together) by the number of one-hour periods each process unit (or the unit that had the greatest total operating hours among the combination of multiple units with one stack that are tested together, or an alternative method approved by the permit authority, pursuant to paragraph (a)(3) of this section) operated during the 12 full calendar months preceding the completion of the performance tests. (8) You must install, calibrate, maintain and operate an appropriate weight measurement device, mass flow meter, or densitometer and volumetric flow meter to measure ore throughput for each roasting operation and autoclave and calculate hourly, daily and monthly totals in tons of ore according to paragraphs (a)(8)(i) and (a)(8)(ii) of this section. (i) Measure the weight or the density and volumetric flow rate of the oxidized ore slurry as it exits the roaster oxidation circuit(s) and before the carbon-in-leach tanks. Alternatively, the weight of the ore can be measured “as fed” if approved by the permit authority as an acceptable equivalent method to measure amount of ore processed. (ii) Measure the weight or the density and volumetric flow rate of the ore slurry as it is fed to the autoclave(s). Alternatively, the weight or the density and volumetric flow rate of the oxidized ore slurry can be measured as it exits the autoclave and before the carbon-in-leach tanks if approved by the permit authority as an acceptable equivalent method to measure amount of ore processed. (9) Measure the weight of concentrate (produced by electrowinning, Merrill Crowe process, gravity feed, or other methods) using weigh scales for each batch prior to processing in mercury retorts or melt furnaces. For facilities with mercury retorts, the concentrate must be weighed in the same state and condition as it is when fed to the mercury retort. For facilities without mercury retorts, the concentrate must be weighed prior to being fed to the melt furnace before drying in any ovens. For facilities that ship concentrate offsite, measure the weight of concentrate as shipped offsite. You must keep accurate records of the weights of each batch of concentrate processed and calculate, and record the total weight of concentrate processed each month. (10) You must maintain the systems for measuring density, volumetric flow rate, and weight within ±5 percent accuracy. You must describe the specific equipment used to make measurements at your facility and how that equipment is periodically calibrated. You must also explain, document, and maintain written procedures for determining the accuracy of the measurements and make these written procedures available to your permitting authority upon request. You must determine, record, and maintain a record of the accuracy of the measuring systems before the beginning of your initial compliance test and during each subsequent quarter of affected source operation. (11) Record the weight in tons of ore for ore pretreatment processes and concentrate for carbon processes with mercury retorts, carbon processes without mercury retorts, and for non-carbon concentrate processes on a daily and monthly basis. (12) Calculate the emissions from each new and existing affected source for the sum of all full months between the compliance date and the date of the initial compliance test in pounds of mercury per ton of process input using the procedures in paragraphs (a)(12)(i) through (a)(12)(iv) of this section to determine initial compliance with the emission standards in § 63.11645. This must include at least 1 full month of data. Or, if a previous test is used pursuant to paragraph (a)(6) of this section for the initial compliance test, use a period of time pursuant to paragraph (a)(6) of this section to calculate the emissions for the affected source. After this initial compliance test period, determine annual compliance using the procedures in paragraph (a)(13) of this section for existing sources. (i) For ore pretreatment processes, divide the sum of mercury mass emissions (in pounds) from all roasting operations and autoclaves during the number of full months between the compliance date and the initial compliance test by the sum of the total amount of gold mine ore processed (in million tons) in these process units during those same full months following the compliance date. Or, if a previous test is used to determine initial compliance, pursuant to paragraph (a)(6) of this section, then the same 3 to 12 full months of production data (i.e., million tons of ore) and hours of operation referred to in paragraph (a)(6) of this section, must be used to determine the emissions in pounds of mercury per million tons of ore. (ii) For carbon processes with mercury retorts, divide the sum of mercury mass emissions (in pounds) from all carbon kilns, preg tanks, electrowinning, mercury retorts, and melt furnaces during the initial number of full months between the compliance date and the initial compliance tests by the total amount of concentrate (in tons) processed in these process units during those same full months following the compliance date. If a previous test is used to determine initial compliance, pursuant to paragraph (a)(6) of this section, then the same 3 to 12 full months of production data (i.e., tons of concentrate) and hours of operation referred to in paragraph (a)(6) of this section, must be used to determine the emissions in pounds of mercury per tons of concentrate. (iii) For carbon processes without mercury retorts, divide the sum of mercury mass emissions (in pounds) from all carbon kilns, preg tanks, electrowinning, and melt furnaces during the initial number of full months between the compliance date and the initial compliance tests by the total amount of concentrate (in tons) processed in these process units during those same full months following the compliance date. If a previous test is used to determine initial compliance, pursuant to paragraph (a)(6) of this section, then the same 3 to 12 full months of production data (i.e., tons of concentrate) and hours of operation referred to in paragraph (a)(6) of this section, must be used to determine the emissions in pounds of mercury per tons of concentrate. (iv) For non-carbon concentrate processes, divide the sum of mercury mass emissions (in pounds) from mercury retorts and melt furnaces during the initial number of full months between the compliance date and the initial compliance tests by the total amount of concentrate (in tons) processed in these process units during those same full months following the compliance date. If a previous test is used to determine initial compliance, pursuant to paragraph (a)(6) of this section, then the same 3 to 12 full months of production data (i.e., tons of concentrate) and hours of operation referred to in paragraph (a)(6) of this section, must be used to determine the emissions in pounds of mercury per tons of concentrate. (13) After the initial compliance test, calculate the emissions from each new and existing affected source for each 12-month period preceding each subsequent compliance test in pounds of mercury per ton of process input using the procedures in paragraphs (a)(13)(i) through (iv) of this section to determine compliance with the emission standards in § 63.11645. (i) For ore pretreatment processes, divide the sum of mercury mass emissions (in pounds) from all roasting operations and autoclaves in the 12-month period preceding a compliance test by the sum of the total amount of gold mine ore processed (in million tons) in that 12-month period. (ii) For carbon processes with mercury retorts, divide the sum of mercury mass emissions (in pounds) from all carbon kilns, preg tanks, electrowinning, mercury retorts, and melt furnaces in the 12-month period preceding a compliance test by the total amount of concentrate (in tons) processed in these process units in that 12-month period. (iii) For carbon processes without mercury retorts, divide the sum of mercury mass emissions (in pounds) from all carbon kilns, preg tanks, electrowinning, and melt furnaces in the 12-month period preceding a compliance test by the total amount of concentrate (in tons) processed in these process units in that 12-month period. (iv) For non-carbon concentrate processes, divide the sum of mercury mass emissions (in pounds) from mercury retorts and melt furnaces in the 12-month period preceding a compliance test by the total amount of concentrate (in tons) processed in these process units in that 12-month period. (b) At all times, you must operate and maintain any affected source, including associated air pollution control equipment and monitoring equipment, in a manner consistent with safety and good air pollution control practices for minimizing emissions. Determination of whether such operation and maintenance procedures are being used will be based on information available to the Administrator which may include, but is not limited to, monitoring results, review of operation and maintenance procedures, review of operation and maintenance records, and inspection of the source.	2022-05-16 06:32:22	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 1, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5825657248497009, "perplexity": 2974.9074048976113}, "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-2022-21/segments/1652662509990.19/warc/CC-MAIN-20220516041337-20220516071337-00597.warc.gz"}
https://labs.tib.eu/arxiv/?author=M.%20Mai	• Sigma resonance parameters from a $N_f=2$ lattice QCD simulation(1804.10225) April 26, 2018 hep-lat In this work we present the analysis of the energy spectrum from a recent two-flavor ($N_f=2$) lattice QCD calculation for pion-pion scattering in the scalar, isoscalar channel (the $\sigma$-meson). The lattice simulation was performed for two quark masses corresponding to a pion mass of 315 MeV and 227 MeV. The $\sigma$-meson parameters are extracted using various parametrizations of the scattering amplitude. The results obtained from a chiral unitary parametrization are extrapolated to the physical point and read $M_\sigma =( 440^{+10}_{-16}(50) - i\,240(20)(25))$ MeV, where the uncertainties in the parentheses denote the stochastic and systematic ones. The behavior of the $\sigma$-meson parameters with increasing pion mass is discussed as well. • Workshop on Pion-Kaon Interactions (PKI2018) Mini-Proceedings. Editors: M. Amaryan, Ulf-G. Mei{\ss}ner, C. Meyer, J. Ritman, and I. Strakovsky(1804.06528) April 18, 2018 hep-ph, hep-ex, nucl-ex, nucl-th This volume is a short summary of talks given at the PKI2018 Workshop organized to discuss current status and future prospects of pi-K interactions. The precise data on pi-K interaction will have a strong impact on strange meson spectroscopy and form factors that are important ingredients in the Dalitz plot analysis of a decays of heavy mesons as well as precision measurement of Vus matrix element and therefore on a test of unitarity in the first raw of the CKM matrix. The workshop has combined the efforts of experimentalists, Lattice QCD, and phenomenology communities. Experimental data relevant to the topic of the workshop were presented from the broad range of different collaborations like CLAS, GlueX, COMPASS, BaBar, BELLE, BESIII, VEPP-2000, and LHCb. One of the main goals of this workshop was to outline a need for a new high intensity and high precision secondary KL beam facility at JLab produced with the 12 GeV electron beam of CEBAF accelerator. This workshop is a successor of the workshops Physics with Neutral Kaon Beam at JLab [1] held at JLab, February, 2016; Excited Hyperons in QCD Thermodynamics at Freeze-Out [2] held at JLab, November, 2016; New Opportunities with High-Intensity Photon Sources [3] held at CUA, February, 2017. Further details about the PKI2018 Workshop can be found on the web page of the conference: http://www.jlab.org/conferences/pki2018/ . • Three-body spectrum in a finite volume: the role of cubic symmetry(1802.03362) Feb. 9, 2018 nucl-th, hep-lat The three-particle quantization condition is partially diagonalized in the center-of-mass frame by using cubic symmetry on the lattice. To this end, instead of spherical harmonics, the kernel of the Bethe-Salpeter equation for particle-dimer scattering is expanded in the basis functions of different irreducible representations of the octahedral group. Such a projection is of particular importance for the three-body problem in the finite volume due to the occurrence of three-body singularities above breakup. Additionally, we study the numerical solution and properties of such a projected quantization condition in a simple model. It is shown that, for large volumes, these solutions allow for an instructive interpretation of the energy eigenvalues in terms of bound and scattering states. • Three-body Unitarity in the Finite Volume(1709.08222) Nov. 19, 2017 nucl-th, hep-lat The physical interpretation of lattice QCD simulations, performed in a small volume, requires an extrapolation to the infinite volume. A method is proposed to perform such an extrapolation for three interacting particles at energies above threshold. For this, a recently formulated relativistic $3\to 3$ amplitude based on the isobar formulation is adapted to the finite volume. The guiding principle is two- and three-body unitarity that imposes the imaginary parts of the amplitude in the infinite volume. In turn, these imaginary parts dictate the leading power-law finite-volume effects. It is demonstrated that finite-volume poles arising from the singular interaction, from the external two-body sub-amplitudes, and from the disconnected topology cancel exactly leaving only the genuine three-body eigenvalues. The corresponding quantization condition is derived for the case of three identical scalar-isoscalar particles and its numerical implementation is demonstrated. • Three-body Unitarity with Isobars Revisited(1706.06118) Aug. 17, 2017 hep-ph, nucl-th The particle exchange model of hadron interactions can be used to describe three-body scattering under the isobar assumption. In this study we start from the 3->3 scattering amplitude for spinless particles, which contains an isobar-spectator scattering amplitude. Using a Bethe-Salpeter Ansatz for the latter, we derive a relativistic three-dimensional scattering equation that manifestly fulfills three-body unitarity and two-body unitarity for the sub-amplitudes. This property holds for energies above breakup and also in the presence of resonances in the sub-amplitudes. • Chiral Extrapolations of the $\boldsymbol{\rho(770)}$ Meson in $\mathbf{N_f=2+1}$ Lattice QCD Simulations(1704.06248) July 25, 2017 nucl-th, hep-lat Recent $N_f=2+1$ lattice data for meson-meson scattering in $p$-wave and isospin $I=1$ are analyzed using a unitarized model inspired by Chiral Perturbation Theory in the inverse-amplitude formulation for two and three flavors. Chiral extrapolations are performed that postdict phase shifts extracted from experiment quite well. In addition, the low-energy constants are compared to the ones from a recent analysis of $N_f=2$ lattice QCD simulations to check for the consistency of the hadronic model used here. Some inconsistencies are detected in the fits to $N_f=2+1$ data, in contrast to the previous analysis of $N_f=2$ data. • Workshop on High-Intensity Photon Sources (HIPS2017) Mini-Proceedings(1704.00816) April 3, 2017 nucl-ex This workshop aimed at producing an optimized photon source concept with potential increase of scientific output at Jefferson Lab, and at refining the science for hadron physics experiments benefitting from such a high-intensity photon source. The workshop brought together the communities directly using such sources for photo-production experiments, or for conversion into $K_L$ beams. The combination of high precision calorimetry and high intensity photon sources greatly enhances scientific benefit to (deep) exclusive processes like wide-angle and time-like Compton scattering. Potential prospects of such a high-intensity source with modern polarized targets were also discussed. The availability of $K_L$ beams would open new avenues for hadron spectroscopy, for example for the investigation of "missing" hyperon resonances, with potential impact on QCD thermodynamics and on freeze-out both in heavy ion collisions and in the early universe. • Workshop on Excited Hyperons in QCD Thermodynamics at Freeze-Out (YSTAR2016) Mini-Proceedings(1701.07346) Feb. 2, 2017 hep-ph This Workshop brought top experts, researchers, postdocs, and students from high-energy heavy ion interactions, lattice QCD and hadronic physics communities together. YSTAR2016 discussed the impact of "missing" hyperon resonances on QCD thermodynamics, on freeze-out in heavy ion collisions, on the evolution of early universe, and on the spectroscopy of strange particles. Recent studies that compared lattice QCD predictions of thermodynamic properties of quark-gluon plasma at freeze-out with calculations based on statistical hadron resonance gas models as well as experimentally measured ratios between yields of different hadron species in heavy ion collisions provide indirect evidence for the presence of "missing" resonances in all of these contexts. The aim of the YSTAR2016 Workshop was to sharpen these comparisons and advance our understanding of the formation of strange hadrons from quarks and gluons microseconds after the Big Bang and in todays experiments at LHC and RHIC as well as at future facilities like FAIR, J-PARC and KL at JLab. It was concluded that the new initiative to create a secondary beam of neutral kaons at JLab will make a bridge between the hardron spectroscopy, heavy-ion experiments and lattice QCD studies addressing some major issues related to thermodynamics of the early universe and cosmology in general. • On the pole content of coupled channels chiral approaches used for the $\bar{K}N$ system(1603.02531) April 22, 2016 hep-ph, nucl-th Several theoretical groups describe the antikaon-nucleon interaction at low energies within approaches based on the chiral SU(3) dynamics and including next-to-leading order contributions. We present a comparative analysis of the pertinent models and discuss in detail their pole contents. It is demonstrated that the approaches lead to very different predictions for the $K^{-}p$ amplitude extrapolated to subthreshold energies as well as for the $K^{-}n$ amplitude. The origin of the poles generated by the models is traced to the so-called zero coupling limit, in which the inter-channel couplings are switched off. This provides new insights into the pole contents of the various approaches. In particular, different concepts of forming the $\Lambda(1405)$ resonance are revealed and constraints related to the appearance of such poles in a given approach are discussed. • Workshop on Physics with Neutral Kaon Beam at JLab (KL2016) Mini-Proceedings(1604.02141) April 6, 2016 hep-ph, hep-ex, nucl-ex, nucl-th The KL2016 Workshop is following the Letter of Intent LoI12-15-001 "Physics Opportunities with Secondary KL beam at JLab" submitted to PAC43 with the main focus on the physics of excited hyperons produced by the Kaon beam on unpolarized and polarized targets with GlueX setup in Hall D. Such studies will broaden a physics program of hadron spectroscopy extending it to the strange sector. The Workshop was organized to get a feedback from the community to strengthen physics motivation of the LoI and prepare a full proposal. Further details about the Workshop can be found on the web page of the conference: http://www.jlab.org/conferences/kl2016/index.html . • Solution of two-center time-dependent Dirac equation in spherical coordinates: Application of the multipole expansion of the electron-nuclei interaction(1208.4731) Oct. 24, 2012 physics.atom-ph A non-perturbative approach to the solution of the time-dependent, two-center Dirac equation is presented with a special emphasis on the proper treatment of the potential of the nuclei. In order to account for the full multipole expansion of this potential, we express eigenfunctions of the two-center Hamiltonian in terms of well-known solutions of the "monopole" problem that employs solely the spherically-symmetric part of the interaction. When combined with the coupled-channel method, such a wavefunction-expansion technique allows for an accurate description of the electron dynamics in the field of moving ions for a wide range of internuclear distances. To illustrate the applicability of the proposed approach, the probabilities of the K- as well as L- shell ionization of hydrogen-like ions in the course of nuclear alpha-decay and slow ion-ion collisions have been calculated.	2020-05-27 03:31:53	{"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": 0.6499380469322205, "perplexity": 1641.6373958319805}, "config": {"markdown_headings": false, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-24/segments/1590347392057.6/warc/CC-MAIN-20200527013445-20200527043445-00439.warc.gz"}
https://api-project-1022638073839.appspot.com/questions/how-do-you-solve-1-5-x-10	# How do you solve (1/5)^x=10? Sep 10, 2015 $x = \log \frac{10}{\log 1 - \log 5}$ #### Explanation: By taking logarithms in both sides we get $\log {\left(\frac{1}{5}\right)}^{x} = \log 10 \implies x \left(\log 1 - \log 5\right) = \log 10 \implies x = \log \frac{10}{\log 1 - \log 5}$ Sep 10, 2015 Use properties of exponents and logs to reformulate and solve, finding: $x = - \frac{1}{\log} \left(5\right)$ #### Explanation: ${\left(\frac{1}{5}\right)}^{x} = {5}^{- x}$ So $\log \left({\left(\frac{1}{5}\right)}^{x}\right) = \log \left({5}^{- x}\right) = - x \cdot \log \left(5\right)$ $\log \left(10\right) = 1$ So taking logs of both sides, the original equation becomes: $- x \cdot \log \left(5\right) = 1$ Divide both sides by $- \log \left(5\right)$ to get: $x = - \frac{1}{\log} \left(5\right)$	2021-10-26 12:51:16	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 9, "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": 0.9655874967575073, "perplexity": 2437.86708263839}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323587877.85/warc/CC-MAIN-20211026103840-20211026133840-00487.warc.gz"}
https://zbmath.org/?q=an%3A1099.90013	# zbMATH — the first resource for mathematics Analysis of a $$GI/M/1$$ queue with multiple working vacations. (English) Zbl 1099.90013 Summary: Consider a $$GI/M/1$$ queue with vacations such that the server works with different rates rather than completely stops during a vacation period. We derive the steady-state distributions for the number of customers in the system both at arrival and arbitrary epochs, and for the sojourn time for an arbitrary customer. ##### MSC: 90B22 Queues and service in operations research 60K25 Queueing theory (aspects of probability theory) Full Text: ##### References: [1] Doshi, B.T., Queueing systems with vacations—a survey, Queueing systems, 1, 29-66, (1986) · Zbl 0655.60089 [2] Doshi, B.T., Single server queues with vacations, (), 217-264 [3] Gross, D.; Harris, C.M., Fundamentals of queueing theory, (1998), Wiley New York · Zbl 0949.60002 [4] Neuts, M.F., Matrix-geometric solutions in stochastic models, (1981), Johns Hopkins University Press Baltimore · Zbl 0469.60002 [5] Servi, L.D.; Finn, S.G., M/M/1 queues with working vacations (M/M/1/WV), Performance evaluation, 50, 41-52, (2002) [6] H. Takagi, Queueing Analysis: A Foundation of Performance Evaluation, Vol 1: Vacation and Priority Systems, Part 1, Elsevier, Amsterdam, 1991. · Zbl 0744.60114 [7] Tian, N.; Zhang, D.; Cao, C., The GI/M/1 queue with exponential vacations, Queueing systems, 5, 331-344, (1989) · Zbl 0684.60072 [8] D.A. Wu, H. Takagi, M/G/1 queues with multiple working vacations, Proceedings of the Queueing Symposium, Stochastic Models and their Applications, Kakegawa, 2003, pp. 51-60. This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.	2021-07-31 06:03:54	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.5733450055122375, "perplexity": 5991.523624929843}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-31/segments/1627046154053.17/warc/CC-MAIN-20210731043043-20210731073043-00048.warc.gz"}
https://www.projecteuclid.org/euclid.aoap/1535443243	## The Annals of Applied Probability ### Dictator functions maximize mutual information #### Abstract Let $(\boldsymbol{\mathsf{X}},\boldsymbol{\mathsf{Y}})$ denote $n$ independent, identically distributed copies of two arbitrarily correlated Rademacher random variables $(\mathsf{X},\mathsf{Y})$. We prove that the inequality $\mathrm{I}(f(\boldsymbol{\mathsf{X}});g(\boldsymbol{\mathsf{Y}}))\le \mathrm{I}(\mathsf{X};\mathsf{Y})$ holds for any two Boolean functions: $f,g\colon \{-1,1\}^{n}\to \{-1,1\}$ [$\mathrm{I}(\cdot ;\cdot)$ denotes mutual information]. We further show that equality in general is achieved only by the dictator functions $f(\boldsymbol{x})=\pm g(\boldsymbol{x})=\pm x_{i}$, $i\in \{1,2,\ldots,n\}$. #### Article information Source Ann. Appl. Probab., Volume 28, Number 5 (2018), 3094-3101. Dates Revised: January 2018 First available in Project Euclid: 28 August 2018 https://projecteuclid.org/euclid.aoap/1535443243 Digital Object Identifier doi:10.1214/18-AAP1384 Mathematical Reviews number (MathSciNet) MR3847982 Zentralblatt MATH identifier 06974774 #### Citation Pichler, Georg; Piantanida, Pablo; Matz, Gerald. Dictator functions maximize mutual information. Ann. Appl. Probab. 28 (2018), no. 5, 3094--3101. doi:10.1214/18-AAP1384. https://projecteuclid.org/euclid.aoap/1535443243 #### References • [1] Anantharam, V., Gohari, A. A., Kamath, S. and Nair, C. (2013). On hypercontractivity and the mutual information between Boolean functions. In Proc. 51st Annual Allerton Conference on Communication, Control, and Computing 13–19. • [2] Courtade, T. A. and Kumar, G. R. (2014). Which Boolean functions maximize mutual information on noisy inputs? IEEE Trans. Inform. Theory 60 4515–4525. • [3] Cover, T. M. and Thomas, J. A. (2006). Elements of Information Theory, 2nd ed. Wiley, Hoboken, NJ. • [4] Klotz, J. G., Kracht, D., Bossert, M. and Schober, S. (2014). Canalizing Boolean functions maximize mutual information. IEEE Trans. Inform. Theory 60 2139–2147. • [5] Kumar, G. R. and Courtade, T. A. (2013). Which Boolean functions are most informative? In Proc. IEEE Int. Symp. on Inform. Theory 226–230. DOI:10.1109/ISIT.2013.6620221. • [6] O’Donnell, R. (2014). Analysis of Boolean Functions. Cambridge Univ. Press, New York. • [7] Pichler, G. (2017). Clustering by mutual information. Ph.D. thesis, Vienna Univ. Technology. • [8] Roberts, A. W. and Varberg, D. E. (1973). Convex Functions. Pure and Applied Mathematics 57. Academic Press, New York.	2019-10-21 16:29:36	{"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": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6877211332321167, "perplexity": 5583.268637744501}, "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/1570987779528.82/warc/CC-MAIN-20191021143945-20191021171445-00111.warc.gz"}
https://www.nature.com/articles/s43246-020-0031-4?utm_campaign=MultipleJournals_USG_DEVICE&utm_source=Nature_community&utm_medium=Community_sites&utm_content=BenJoh-Nature-MultipleJournals-Engineering-Global&error=cookies_not_supported&code=941a4ef0-f6fb-4344-bc7e-3f3bca001435	# Simple and fast fabrication of single crystal VO2 microtube arrays ## Abstract Single crystal VO2 is a strongly correlated electron material that has shown great potential for a wide range of high-performance modern device applications, such as microbolometers, lithium ion batteries, microactuators and strain sensors. However, the present fabrication methods for single crystal VO2 almost always require complicated procedures, strict conditions and long reaction times of up to one week. Here, we report a simple, fast, low-cost and green method for fabricating single crystal VO2 using a thermal oxidation route based on resistive heating of a vanadium foil in air. Our method not only reduces the complete fabrication time from hours to tens of seconds but also naturally forms single crystal VO2 microtube arrays that are nearly vertically aligned on the surface of a V2O5 substrate. Microstructure characteristics and the reversible phase transition between the monoclinic VO2 and rutile VO2 phases demonstrate that the obtained single crystal VO2 is the same as that achieved by other fabrication methods. ## Introduction Vanadium dioxide (VO2) is an archetypal metal-insulator transition (MIT) material, which has received increasing attention in the exploration of the MIT mechanism1,2,3 and has applications in various modern devices, such as electronic inks4, supercapacitors5, thermal diodes6, thermal emitters7, smart windows8, temperature sensors9, and radiative thermal memristors10. The MIT of VO2 features a dramatic change in the electrical conductivity, optical reflectivity, magnetic susceptibility, and dielectric function because of the accompanying reversible structural phase transition between the monoclinic insulating phase (M1) and the tetragonal metallic phase (rutile structure, R) at approximately 67 °C11. The phase transition characteristics and the physical properties of VO2 are strongly related to its stoichiometry, crystallinity, and morphology, which are generally determined by the fabrication methods12. Representative fabrication methods include reactive sputtering1, pulsed laser deposition2,3,6, hydrothermal methods4, electrodeposition5, molecular beam epitaxy8, chemical vapor deposition13, thermolysis14, and magnetron sputtering15 for polycrystalline VO2 films or nanoparticles. Despite the potential large-scale applications of polycrystalline VO2, the poor crystal quality and weak MIT properties12 make it difficult for polycrystalline VO2 to meet the requirements of high-performance modern devices. In contrast, single crystal VO2 has excellent MIT properties11,12, which not only has been utilized for exploring intrinsic MIT physics16,17 but also has demonstrated great potential for manufacturing high-performance microbolometers18, lithium-ion batteries19, microactuators20, and highly sensitive strain sensors21. So far, various methods have been utilized for fabricating single crystal VO211,12. Several representative fabrication methods such as self-flux method, solution growth technique, vapor transport method, and hydrothermal method are briefly introduced as follows. The self-flux method16,22 can fabricate high-quality single crystal VO2 microrods with lengths of a few millimeters and widths of ~100 µm by heating V2O5 powder to 800–1100 °C in a vacuum tube furnace. During heating, the V2O5 is partially reduced and converts to single crystal VO2 in the molten V2O5. However, the fabrication time is very long and is approximately 1 week23. The solution growth technique24,25 can fabricate bulk single crystal VO2 with lengths of up to greater than 10 mm and widths of up to 1 mm by heating a mixture of VO2 powder and V2O5 powder, which are sealed in a silica tube up to 1050 °C and slowly cooled up to 100 h to 775 °C25. The rod-like bulk single crystal VO2 grows along the high-temperature rutile c axis during heating and cooling process24. The vapor transport method17,18,20 can fabricate rectangular cross-sectional single crystal VO2 nanowires with lengths of tens of micrometers and widths of tens of nanometers to a few micrometers by evaporating V2O517,18,20 powder or VO226 powder at approximately 950 °C and transporting the vapor by Ar carrier gas to a substrate in a quartz tube furnace. Although the fabrication time of the vapor transport method is much shorter than that of the self-flux method or solution growth technique, the fabrication time is still too long, approximately 5 h26. Recently, it was reported that the hydrothermal method can also be utilized to fabricate nearly round cross-sectional single crystal VO2 nanowires with lengths of tens of micrometers and widths of hundreds of nanometers by heating an aqueous solution of V2O5, H2C2O4·2H2O, and H2SO4 at 100 °C for 10 h, followed by heating the as-obtained precursor solution at 260 °C for 24 h27. This method realized massively production of single crystal VO2 nanowires. However, the fabrication time is also very long. To date, high-quality single crystal VO2 with different sizes from nanometers to millimeters can be fabricated as reviewed above; however, fabricating single crystal VO2 via these present methods not only requires complicated procedures17,18,19,21,24,25,27, strict conditions17,18,19,21,24,25,26,27, or special equipment16,17,18,21,22,24,25,26, but also consumes a great deal of energy because of the very long heating times; thus, these methods are costly. In this work, we report a simple method for fabricating single crystal VO2 via a thermal oxidation route28 based on the resistive heating29 of V foils in air using an ordinary laboratory power supply. The complete fabrication time is only tens of seconds, which is the fastest method for fabricating single crystal VO2 to date. Interestingly, the as-fabricated single crystal VO2 with a length of up to 430 µm and an average width of approximately 6.8 µm naturally forms high-density microtube arrays that are nearly vertically aligned on the surface of a V2O5 substrate without additional treatments. Moreover, the only required raw materials are commercially pure V foil and air, the as-fabricated products are only VO2 and V2O5, and there is no waste; thus, our fabrication method has a low cost and is environmentally friendly. ## Results ### Fabrication of single crystal VO2 microtube arrays The fabrication of single crystal VO2 microtube arrays contains two successive processes, i.e., heating a V foil in air with a direct current to approximately 1700 °C in tens of seconds to synthesize vanadium oxide on the surface of the V foil and then cooling the as-synthesized vanadium oxide to below 250 °C in a few seconds (Supplementary Movie 1). Figure 1 depicts a typical temperature vs. time curve for the growth process of single crystal VO2 microtube arrays. Here, each temperature recorded by a pyrometer (temperature range of 250–2000 °C) refers to the temperature of the central region of the V foil, i.e., the highest temperature throughout the V foil. At the beginning of the heating process, the surface of the V foil is oxidized gradually in air (photograph no. 1, inset in Fig. 1), and the as-synthesized product is a mixture of solid vanadium oxide (VOx, where x is the O–V atomic ratio), mainly V2O5 and VO229, when the temperature of the V foil is below 678 °C (the melting point of bulk V2O530). With continuous heating, the temperature of the V foil and as-synthesized solid VOx continually increases to over 678 °C. The solid VOx melts into a liquid and gradually forms two droplets31 that cover both sides of the V foil (photograph no. 2, inset in Fig. 1). Obviously, the liquid droplets increase in thickness as the temperature continually increases, which means that additional VOx is synthesized. When the temperature of the V foil as well as the VOx liquid droplets further increases to approximately 1700 °C by continuous heating, the V foil reaches an incandescent state (photograph no. 3, inset in Fig. 1). At this point, the V atoms can largely dissociate from the V foil and diffuse into the VOx liquid droplets, which makes the V content and the thickness of the liquid droplets increase largely. Next, after the heating is stopped by turning off the direct current manually, the temperature of the VOx liquid droplets decreases from ~1700 °C to below 250 °C at a very rapid rate of approximately 370 °C/s because of the natural heat dissipation. During cooling, it is found that the liquid droplets experience two crystallization processes in sequence that correspond to the platform and the trough in the temperature vs. time curve (marked with two rectangles in Fig. 1). When the temperature drops to approximately 833 °C, a large number of single crystal VO2 microtubes grow in the liquid droplets and form arrays that stand in the residual liquid VOx in less than 1 s (photograph no. 4, inset in Fig. 1). This crystallization process is exothermic, which maintains the crystallization starting temperature of VO2 for a short time. As a result, a platform at ~833 °C appears in the temperature vs. time curve, as shown in the navy box inset in Fig. 1. After the VO2 crystallization finishes, the whole product resumes rapid cooling to approximately 382 °C. The residual liquid VOx crystallizes and finally forms a solid V2O5 substrate with the as-crystallized single crystal VO2 microtube arrays nearly vertically aligned on the surface of it (photograph no. 5, inset in Fig. 1). This crystallization process is also exothermic, which can even increase the temperature by up to 27 °C compared with the crystallization equilibrium temperature of V2O5 (382 °C) in more than 1 s. As a result, a trough at ~382 °C appears in the temperature vs. time curve, as shown in the magenta box inset in Fig. 1. After the V2O5 crystallization finishes, the whole product resumes rapid cooling again until it reaches room temperature. To find the key fabrication conditions for single crystal VO2 microtube arrays, we carried out many fabrication runs with different fabrication parameters. Supplementary Fig. 1 depicts the temperature vs. time curves of 14 successful fabrication runs for single crystal VO2 microtube arrays using 14 V foils with different heating rates, cooling rates, peak temperatures, fabrication times, and crystallization temperatures, as summarized in Supplementary Table 1. We found that these fabrication parameters have wide ranges. The heating rate can occur over a range from 18 to 46 °C/s with an average of 28 °C/s, the peak temperature can occur over a range from 1613 to 1955 °C with an average of 1782 °C, and the fabrication time can occur over a range from 44 to 99 s with an average of 73 s; all can be easily controlled by adjusting the direct current manually. Moreover, the cooling rates range from 312 to 677 °C/s, with an average of 442 °C/s, without being controlled. The crystallization temperatures of VO2 and V2O5 can occur over a range from 813 to 834 °C with an average of 824 °C and from 317 to 398 °C with an average of 353 °C, respectively. Additionally, we also carried out two unsuccessful fabrication runs with the following fabrication parameters. (1) The as-synthesized VOx liquid droplets are rapidly cooled to below 250 °C (with a cooling rate of ~345 °C/s, shown by the red curve in Supplementary Fig. 2a) after gradually heating the V foil to a relatively low temperature of 1005 °C. It is found that irregular bumps are produced on the surface of the V foil, and there are no microtube arrays, as shown in Supplementary Fig. 2b. X-ray diffraction (XRD) examination demonstrated that the obtained product is mainly V2O5 with small amounts of VO2 (the red curve in Supplementary Fig. 2d). An explanation is presented as Supplementary Note 1. (2) The as-synthesized VOx liquid droplets are slowly cooled to below 250 °C (with a cooling rate of ~29 °C/s, as shown by the blue curve in Supplementary Fig. 2a) after gradually heating the V foil to a high temperature of 1785 °C. Only plate-shaped solids are produced on the surface of V foil, and there are no microtube arrays, as shown in Supplementary Fig. 2c. XRD examination demonstrated that the obtained product is a pure phase of V2O5 (blue curve in Supplementary Fig. 2d). An explanation is presented as Supplementary Note 2. Consequently, we propose that the V foil must be gradually heated to incandescent temperature and then rapidly cooled to fabricate single crystal VO2 microtube arrays. ### Characterization of single crystal VO2 microtube arrays The morphology, composition, and microstructure of the as-fabricated product were examined using scanning electron microscopy (SEM), X-ray energy dispersive spectroscopy (EDS), X-ray photoelectron spectroscopy (XPS), XRD, and transmission electron microscopy (TEM). As shown in the representative SEM images in Fig. 2a (side view), Fig. 2b (top view), and in the lower-left inset in Fig. 2a, single crystal VO2 are straight, rod-like and nearly vertically aligned on the surface of a layered V2O5 substrate (Supplementary Fig. 3). The upper-right inset in Fig. 2a and the upper-right inset in Fig. 2b show that these single crystal VO2 microtubes are all hollow and have rectangular cross-sections and steeple-shaped tops. It is also noticed that there are small amounts of solidified V2O5 flux on the side surface of the VO2 microtubes25, which are further observed by high-resolution TEM (HRTEM) and analyzed below. The EDS mapping of an individual microtube confirms that only V and O are present and uniformly distributed over the whole microtube, as shown in the lower-right inset of Fig. 2b. Additional EDS spectrums (Supplementary Fig. 4) demonstrate that the O–V atomic ratio are ~1.98 and ~2.49 for the VO2 microtube and the V2O5 substrate, respectively. The widths of different microtubes vary from 1 to 12 µm (Fig. 2c) with an average width of ~6.8 µm, while each microtube has a uniform width along its entire length. The length of those microtubes can reach up to 430 µm. Obviously, the density (~450 mm−2) of the VO2 microtube arrays is very high and reaches that of ultra-dense (400–2000 mm−2) VO2 nanowires fabricated by the vapor transport method32. To the best of our knowledge, this special morphology of our single crystal VO2 microtube arrays has never been reported before for single crystal or polycrystalline VO211,12. Moreover, the as-fabricated product was characterized by XPS. As shown in Supplementary Fig. 5, the XPS spectra show a mixture of two vanadium oxidation states (i.e., V5+ and V4+), which indicates that the vanadium in as-fabricated product features a mixed 4–5 oxidation state33. The crystal structure of the as-fabricated product was first examined using XRD at room temperature, and the diffraction pattern is shown in Fig. 2d. Clear and narrow diffraction peaks reveal that the as-fabricated product consists of high-quality crystals. The XRD peaks can be indexed unambiguously to the M1 phase of VO2 (JCPDS card no. 72-0514) and orthorhombic phase of V2O5 (JCPDS card no. 41-1426), which confirms that the as-fabricated product only contains VO2 and V2O5. Additional powder XRD pattern (Supplementary Fig. 6) also demonstrates that no other vanadium oxide can be identified except for VO2 and V2O5. Moreover, very strong diffraction peaks that belong to the {2kl} family of the M1 phase of VO2 strongly suggest that the VO2 microtube arrays have a preferred growth direction. Next, the microstructures of the VO2 microtube and the V2O5 substrate were observed using TEM. Two cross-sectional TEM samples were prepared using focused ion beam (FIB) milling from different position of an individual microtube (Supplementary Fig. 7a, b). To avoid contamination of the sample surface by the FIB bombardment, a protective Pt layer was coated on the surface of the microtube during TEM sample preparation. Figure 2e depicts a TEM bright-field image of a corner of the microtube with a VO2 wall thickness of ~2.1 µm (the VO2 is outlined with the navy dash lines). The corresponding HRTEM image (inset in Fig. 2e) displays the typical single crystal morphology of the M1 phase of VO2 along the [100] zone axis (ZA), while the interplanar distance of 0.452 and 0.453 nm can be indexed as the (010) crystal plane and (001) crystal plane of the M1 phase of VO2, respectively. Furthermore, a thin layer of V2O5 with a thickness of less than 500 nm (marked in Fig. 2e) that is located on the side surface of the VO2 microtube can be observed, which is the solidified V2O5 flux25. The microstructure of the solidified V2O5 flux and the crystallographic relation between the VO2 and V2O5 were analyzed with an HRTEM image (Fig. 2f), which is a magnified image of the region in the cyan box located at the boundary between the VO2 and V2O5 in Fig. 2e. The lower-left inset shows an electron diffraction image of single crystal VO2 located in the lower-left region of Fig. 2f, which can be indexed to the [100] ZA of the M1 phase of VO2. Thus, the side surface of the VO2 microtube (marked with navy dashed lines in Fig. 2f) can be indexed to the (01$$\bar{1}$$) crystal plane with an interplanar distance of 0.320 nm, while another side surface of the VO2 microtube shown in Fig. 2e can be indexed to the (011) crystal plane. The angle of these two crystal planes is 89.86°; thus, the VO2 microtube appears to have a rectangular cross-section. The upper-right inset in Fig. 2f shows the corresponding electron diffraction image of the upper-right region in Fig. 2f, which can be indexed to the [010] ZA of V2O5. It is obvious that the boundary between the VO2 and V2O5 is noncoherent. Electron diffraction pattern obtained from another TEM sample also demonstrates a single crystal VO2 in [100]M1 crystalline orientation (Supplementary Fig. 7c). Hence, TEM observations for two TEM samples clearly indicate that the microtube is single crystal VO2 growing in [100]M1 direction and confirm that the crystalline orientation remains the same across the entire VO2 microtube. The growth direction of our single crystal VO2 microtubes is the same as that of single crystal VO2 fabricated by the vapor transport method20,26, the self-flux method22,23, and the hydrothermal method27. It is noted that the solidified V2O5 flux on the side surface of VO2 microtubes can be easily removed (Supplementary Figs. 4a and 8b) by immersing the as-fabricated microtubes in a dilute aqueous solution of sodium carbonate with a concentration of 30 g/L for 30 min at room temperature based on the following reactions: V2O5 + Na2CO3 → 2NaVO3 + CO234. ### Phase transition of single crystal VO2 microtube arrays The reversible MIT between the M1 and the R phases is the fundamental characteristic of VO2. Thus, we carried out two differential scanning calorimetry (DSC) experiments for as-fabricated single crystal VO2 microtube arrays. To analyze the thermal cycle stability, 28 heating and cooling cycles between 40 and 90 °C were carried out, and the DSC curves are depicted in Fig. 3a. These DSC curves show only one peak during each heating and cooling process, which indicates that as-fabricated single crystal VO2 microtube arrays undergo a first-order phase transition. The phase transition peak temperatures of R → M1 (Mp) and M1 → R (Rp) for single crystal VO2 microtube arrays are approximately 63.1 and 67.2 °C, respectively, which are in excellent agreement with previously reported Rp of 67.8 °C for bulk single crystal VO224, the Rp of 67.2 °C for single crystal VO2 microrods16, and the Rp of 67 °C for single crystal VO2 nanowires18. This result indicates that our single crystal VO2 microtube arrays have the same MIT property as single crystal VO2 fabricated by the currently available methods. Most strikingly, the Mp and Rp of our VO2 microtube arrays show almost no shift during 28 heating and cooling cycles, as shown in Fig. 3b, c. This finding confirms the excellent thermal cycle stability of the MIT process of our single crystal VO2 microtube arrays. In contrast, VO2 nanoparticles produced elsewhere were reported to have Mp and Rp shifts from 49.7 to 53.4 °C and from 61.0 to 68.8 °C, respectively, during seven heating and cooling cycles35 because of the unstable microstructure of the VO2 nanoparticles. Furthermore, it was reported that the Mp and Rp of a single crystal VO2 microrod with a width of ~100 µm exhibited a run-to-run shift of up to 1 °C during three heating and cooling cycles16 because of the kinetic barrier of the first-order phase transition. In addition, a wide temperature range (−50 to 400 °C) DSC experiment was also carried out for exploring whether additional phase transitions can occur for the as-fabricated product, and the DSC curve is depicted in Supplementary Fig. 9. Same as Fig. 3a, the DSC curve shows only one peak during heating or cooling process, which indicates only MIT of M1 ↔ R can occur between −50 and 400 °C for single crystal VO2 microtube arrays. Thus, the sharp peaks in the exothermal and endothermal scans of the DSC curve in Fig. 3a and Supplementary Fig. 9 indicate precise structural phase transitions, while the excellent exothermal and endothermal stabilities (Fig. 3b, c) suggest that our single crystal VO2 microtube arrays have a very stable microstructure and great potential for device applications that require a long lifetime and a wide temperature range. Moreover, we also carried out an optical microscope observation for an individual single crystal VO2 microtube during a heating and cooling process and the optical images of the M1 phase at 60 °C and the R phase at 75 °C are shown in Fig. 3d, e (see also Supplementary Movie 2). Clear brightness change from bright to dark occurs uniformly for the whole microtube and corresponds to the insulator → metal transition, indicating a uniform stoichiometry and strain-free nature of single crystal VO2 microtube32. And, an obvious length shrinkage of ~0.7% can be detected across the M1 → R phase transition due to the lattice constant differences along the [100]M1 ([001]R) axial direction; aM1 = 0.5743 nm (JCPDS card no. 72-0514) whereas 2cR = 0.57028 nm (JCPDS card no. 71-0565). It is well known that the phase transition from the M1 phase to the R phase of VO2 during heating corresponds to a colossal resistivity drop by over 4 orders of magnitude which can be reversible via a cooling process11,12. Thus, the phase transition of VO2 can also be characterized by a resistance vs. temperature measurement, which can conclusively prove an MIT of VO2. However, many factors such as electrical contact problem, metastable phases formation can bother the electrical data. In this work, the reversible MIT of single crystal VO2 microtubes is revealed by the DSC and optical data. However, we do not present electrical data of single crystal VO2 microtubes, which needs to be further studied in the future. ## Discussion Based on the experimental results, we propose a growth mechanism of single crystal VO2 microtube arrays as illustrated in Fig. 4. According to the V–O phase diagram (inset in Supplementary Fig. 10)30,31, VO2 starts to crystallize just when the temperature of the liquid VOx is lowered below the liquidous line. As shown in the cooling curve (Supplementary Fig. 10), the temperature of the liquid VOx rapidly decreases from the peak temperature (1798 °C in Supplementary Fig. 10) to ~1542 °C (melting point of VO2). The cooling curve starts to deviate the linear downtrend, which means VO2 starts to nucleate in the liquid VOx36. An additional SEM image of the side view of an as-fabricated product (Supplementary Fig. 11a) clearly shows that there is a layer of small equiaxed crystals with a thickness of ~12 µm at the bottom of the as-fabricated product. After immersion in a dilute aqueous solution of sodium carbonate for 10 min at room temperature, this layer of small equiaxed crystals does not have any change (Supplementary Fig. 11b), which is same as the status of the VO2 microtubes. Hence, we propose that the VO2 starts to nucleate at the V foil surface and then form a chill zone firstly, because the heat of the liquid VOx dissipates much faster from the V foil than from the air after the heating is stopped. With continuous heat dissipation, some of these small equiaxed crystals then grow into the liquid VOx and form arrays that are nearly vertically aligned on the surface of the V foil, because only those equiaxed crystals that are oriented favorably regarding the direction of heat flow can grow an appreciable distance and produce a columnar zone37. Thus, this process can be regarded as a directional growth, and the formation of the VO2 microtube arrays is based on an interplay between the diffusion required for phase separation and the energy required for the formation of inter-phase boundaries38. TEM observations demonstrate that the VO2 microtubes grow along the [100]M1([001]R) direction and the bounding ($$01\bar 1$$)M1 and (011)M1 facets are crystallographically equivalent, low-index, and therefore, low energy surfaces26. The diameter and the spacing of the VO2 microtubes are the result of a balance between two competing tendencies: on the one hand, to minimize the spacing in order to shorten the diffusion path of cations in the liquid VOx, and on the other hand, to increase the spacing to minimize the total interfacial energy38. The formation of the hollow tubular structure of individual VO2 microtube can be explained by the accumulation of top concave effect of rod-like crystals39 as being similar to the growth mechanism of Se microtubes40, W microtubes41, SiAlON microtubes42, and the wetting assisted growth mechanism of VOx nanostructures31. The initial stage in the growth of a single crystal VO2 microtube is the formation of a microparticle on the surface of the V foil followed by directional growth along the c-axis of R phase of VO2 in the liquid droplet and then forming a rod-like VO2 crystal as aforementioned scenario. In this anisotropic growth process, a mass transportation from the side surface to the top surface occurs, because the top surface grows much faster than the side surface of the rod-like VO2 crystal40,41,42. Thus, the concentration of VO2 at the edge of the top surface is higher than at the center, resulting growth rate at the edge is also higher than at the center39. Consequently, a concave shape produces at the top surface of the rod-like VO2 crystal and the concavity of the top surface continuously increases by accumulation of the concave effect42. When the rod-like VO2 crystal grows out of the liquid droplet, a thin liquid film wets the side surface of the rod-like VO2 crystal along its entire length, creating a mass transportation channel for the rod-like VO2 crystal growing top31. The growth continually takes place as feeding mass is transported to the growing top of the rod-like VO2 crystal via the peripheral liquid wetting layer. However, for a fast growth process, mass transportation from the edge to the center of the growing top is very limited, which results in the growth stop of the center of the rod-like VO2 crystal. As a consequence, a hollow tubular structure is eventually formed as shown in Fig. 2a, b. When the temperature decreases to approximately 833 °C, the heat dissipation and the latent heat released by the crystallization of VO2 reach an equilibrium, which results that temperature of the product is maintained at approximately 833 °C for a short time until the crystallization of VO2 is complete. At this point, the residual liquidus composition of VOx can be estimated by the equation: x = −2.815 × 10−4T + 2.701543, where T is the temperature in degrees centigrade. Here, the x is calculated to be 2.47 corresponding to the temperature of 833 °C, which is consistent with the EDS measurement result (Supplementary Fig. 4d). After the VO2 crystallization finishes, the temperature of the residual liquid VOx decreases again until to ~678 °C (melting point of V2O5). The cooling curve starts to deviate the linear downtrend again, which means V2O5 starts to nucleate in the residual liquid VOx. This crystallization process lasts for a short time until the temperature reaches 382 °C, while the heat dissipation and the latent heat released by the crystallization of V2O5 reach an equilibrium. And the temperature can increase up to 409 °C as marked in Supplementary Fig. 10, because the latent heat released by the crystallization of V2O5 is greater than the heat dissipation during this process. It is noted that the obtained equilibrium temperature of 833 and 382 °C from cooling curve are much less than the melting point of VO2 and V2O5, which we think it is because the crystallization process of the liquid VOx is a nonequilibrium solidification. It is well known that the true crystallization temperature is actually very difficult to determine from a cooling curve because of the nonequilibrium conditions inherent in such a dynamic test for an actual solidification process44. Although single crystal VO2 has demonstrated tremendous potential for practical use in various high-performance modern devices11,12,18,19,20,21, critical issues remain unsolved and hinder the large-scale application of this smart material. Previous report shows that the phase transition of bulk single crystal VO2 is very sharp and demonstrates relatively little hysteresis for heating and cooling rates of approximately 0.2 °C per minute24. Moreover, bulk single crystal VO2 can be conveniently manipulated and assembled as a component in modern devices because its size is at the macroscale. However, microcracks that occur during heating and cooling because of the large (~1%) and anisotropic spontaneous strain associated with the MIT is a fatal drawback of bulk single crystal VO2, which can result in degraded electronic and thermal properties17. Thus, the thermal cycle stability of bulk single crystal VO2 is very poor, which makes it very unattractive for device applications45. It has been demonstrated that scaling VO2 to the nanoscale can allow it to withstand an elevated uniaxial strain46,47, thus enabling protracted thermal cycles without cracking during the MIT48. Hence, single crystal VO2 nanowires are likely good candidates for device applications that require a long lifetime because they are free of extended structural defects during the MIT17. However, the existing techniques have not yet satisfied the requirements of large-scale applications of single crystal VO2 nanowires12, and additional engineering issues need to be solved, such as vertically aligning the nanowires on a substrate to form arrays for microbolometer sensing components11, and improving the low yield due to the self-limiting in-plane growth49. In addition, it is difficult to use single crystal VO2 nanowires for device applications because their scale is too small to be easily manipulated or assembled in an orderly fashion. Thus, convenient and high yield fabrication of vertical single crystal VO2 arrays remains a challenging task. Our single crystal VO2 microtube arrays are likely able to solve these issues. The excellent thermal cycle stability demonstrates great potential for manufacturing devices with a long lifetime. Arrays that are nearly vertical should be very suitable for application such as microbolometer sensing components. Simple, fast, low-cost, and green fabrication processes are very convenient for high yield production. Moreover, our single crystal VO2 microtube arrays are macroscopic and can be easily manipulated or assembled as a component in modern devices. Thus, we expect that single crystal VO2 microtube arrays fabricated by our method can promote large-scale applications of VO2 in various modern devices, such as microbolometers, lithium-ion batteries, supercapacitors, thermal emitters, and temperature sensors. ## Methods ### Fabrication A commercially available pure V foil (99.9 wt%) with a thickness of 0.2 mm was used in this work. The V foil was cut into bars 3 mm wide and 20 mm long. In a typical procedure, a bar of V foil was mounted on two simple holders that was constructed in-house, which was connected to the positive and negative electrodes of a GW Instek PSB-2400L power supply with two wires (Supplementary Fig. 12a, b). Then, the V foil was resistively heated to approximately 1700 °C in tens of seconds by a direct current of <40 A in air. Afterwards, the direct current was turned off manually. The vanadium oxide product was formed in a few seconds on the central surface of both sides of V foil (marked by an arrow in Supplementary Fig. 12c). The slow heating/cooling rate was controlled by manual adjusting the rotary knob of the power supply which changed the applied direct current. Whereas, the rapid cooling rate was realized by turning off the direct current at the peak temperature of each fabrication run. Supplementary Fig. 12d shows an SEM image of an as-fabricated product. The single crystal VO2 microtube arrays are mainly located in the central region of the product, while the substrate is a layered V2O5. The fabrication temperature was monitored by a pyrometer (LumaSense IGAR 6 Advanced with a temperature range of 250–2000 °C, a resolution of 0.1 °C, and a sampling rate of 60 s−1). Ambient conditions corresponded to regular laboratory and atmospheric pressure condition with a temperature range of 15–35 °C, a relative humidity range of 11–100%, and an oxygen partial pressure of ~21.2 kPa. ### Morphology observation and composition determination The morphology of the as-fabricated product was mainly observed using a Hitachi TM3030 scanning electron microscope equipped with an Oxford Swift 3000 X-ray energy dispersive spectroscope for element mapping. Moreover, an as-fabricated product was immersed in a dilute aqueous solution of sodium carbonate for 30 min at room temperature, followed by EDS analysis using a Hitachi SU8220 scanning electron microscope equipped with an Oxford X-Max X-ray energy dispersive spectroscope. The X-ray photoelectron spectroscopy was obtained using a Thermo Scientific ESCALAB 250Xi X-ray photoelectron spectrometer. ### Structure characterizations XRD examination was carried out using a Rigaku SmartLab X-ray diffractometer at room temperature. An as-fabricated product directly underwent XRD examination by putting the whole product on a plain sample stage without any treatment. Then, three as-fabricated products were ground into powder by an agate mortar, followed by XRD examination again. Two cross-sectional TEM samples from an individual microtube were prepared using a Helios NanoLab 600 focused ion beam scanning electron microscope. A TEM sample of V2O5 substrate was prepared by conventional method for powder samples. The TEM samples were observed using a FEI Tecnai G2 F20 S-TWIN transmission electron microscope at 200 kV. ### Phase transition analysis The DSC experiments were carried out using a TA Q2000 calorimeter. Three as-fabricated products (15.4 mg in all) were directly put into a Tzero pan for a 28 thermal cycles DSC experiment. The temperature range was 40–90 °C, and the heating and cooling rates were 10 °C/min. Other three as-fabricated products (14.3 mg in all) were also directly put into a Tzero pan for a wide temperature range DSC experiment. The temperature range was −50 to 400 °C, and the heating and cooling rates were 10 °C/min also. Optical microscope observation was carried out using a Leica DMLM microsystem with a heating stage. ## Data availability The data that support the findings of this study are available from the corresponding author upon reasonable request. ## References 1. 1. 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Rep. 4, 5456 (2014). ## Acknowledgements This work was supported by the National Natural Science Foundation of China (Nos. 11972221 and 11672175). The authors acknowledge Dr. G.R. Xu (IMUT) and Prof. Z.Z. Cao (IMUT) for the help of XRD examination, Dr. X.H. Hou (IMUT) for the help of EDS testing, and Dr. F.C. Lang (IMUT) and Y.J. Jin (IMUT) for the help of optical measurement. ## Author information Authors ### Contributions C.Z. and Y.X. conceived and designed the experiments. S.M. and Z.L. carried out the fabrication of materials and the microstructural characterizations. W.L. performed the DSC experiment. C.Z., J.L., Q.H., and Y.X. analyzed the experimental results and wrote the manuscript. All authors commented on the manuscript. ### Corresponding authors Correspondence to Chunwang Zhao or Yongming Xing. ## Ethics declarations ### Competing interests The authors declare no competing interests. Publisher’s note Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. ## Rights and permissions Reprints and Permissions Zhao, C., Ma, S., Li, Z. et al. Simple and fast fabrication of single crystal VO2 microtube arrays. Commun Mater 1, 28 (2020). https://doi.org/10.1038/s43246-020-0031-4 • Accepted: • Published: • ### A facile method for the fabrication of element doped VO2 microtube arrays • Zijian Li • , Xueying Liu • , Weiya Li • , Shuxiang Ma • , Na Li • & Chunwang Zhao Vacuum (2020)	2020-12-01 16:38:39	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6449641585350037, "perplexity": 4864.9906940472265}, "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-2020-50/segments/1606141674594.59/warc/CC-MAIN-20201201135627-20201201165627-00452.warc.gz"}
https://electronics.stackexchange.com/questions/561461/calculating-the-cut-off-frequency	# Calculating the cut-off frequency I have an inverter opamp with the next data : • $$\A=-25\$$ • $$\A_{ol}=20\frac{mV}{V}\$$ • Unity Gain Bandwith$$\=10^{6}\$$ Then asks for the cutt off frequency, so the open loop its passed to db $$\20\,\frac{V}{mV}=20log\frac{20}{0.001}=86.02059\,db\$$ $$\86.020=20log(\frac{Vo}{Vi})\$$ $$\\frac{86.020}{20}=log(\frac{Vo}{Vi})\$$ $$\4.301=log(\frac{Vo}{Vi})\$$ So the $$\A_{ol}=10^{4.301}\$$ Usint the GBWP $$\10^{4.301}f_{3db}=10^{6}\$$ $$\f_{3db}=\frac{10^{6}}{10^{4.301}}=50.0034\,Hz\$$ Later its asked the values of output voltage if the frequencies are $$\0.25f_{3db}\$$ and $$\5f_{3db}\$$ so calculating the gains $$\0.25f_{3db}=(0.25)50=12.5\$$ $$\Af_{3db}=GBWP\$$ $$\A=\frac{GBWP}{f_{3db}}=\frac{10^{6}}{12.5}=80000\$$ $$\V_{o}=80000V_{in}\$$ $$\5f_{3db}=5(50)=250\$$ and $$\A=\frac{GBWP}{f_{3db}}=\frac{10^{6}}{250}=4000\$$ $$\V_{o}=4000V_{in}\$$ Is this the way to go? the results make me sense since its stated that before the cuttoff frequency the gains are very high but the bandwith its too narrow and over the cuttoff the gain is less but the band its wider. But then why Im supplyed with the closed loop gain? • "Then asks for the cutoff frequency". Are you sure it's not asking for the closed loop cutoff frequency rather than what you have taken it to mean - the frequency where the open loop gain starts to roll off which is a much lower frequency. Apr 22, 2021 at 11:05 • Perhaps this is the hint, as it is redacted I cant really tell. Apr 22, 2021 at 19:11 • The part about it says What are the 3𝑑𝐵 frequencies for both the opamp and the inverter amplifier?/¿Cuáles son las frecuencias de 3𝑑𝐵 tanto del op-amp como del amplificador inversor? So I wass thinking it is saying the cuttoff frequency. Apr 22, 2021 at 19:28 You calculated $$\A_{ol}\$$ well. You also know that the slope is 20 dB/dec, so with a $$\G_{BW}\$$ of 1 MHz, the open loop amplification will be: 10 (20 dB) @ 100 kHz 100 (40 dB) @ 10 kHz 1000 (60 dB) @ 1 kHz 10000 (80 dB) @ 100 Hz At 10 Hz it should be 100000 (100 dB), but Aol = 86 dB = 20000 (You wrote mV/V first time, instead of V/mV). That means the gain is double that at 100 Hz, which means the frequency is half of 100 Hz, or 50 Hz. Or, in simpler terms: $$\dfrac{f_{max}}{A_{ol}}=\dfrac{1\;\mathrm{MHz}}{20000}=50\;\mathrm{Hz}\tag{1}$$ Which means your result is right (save roundings), just that there are fewer steps. Then you ask for: the values of output voltage if the frequencies are $$\0.25f_{3\,dB}\$$ and $$\5f_{3\,dB}\$$ but you don't specify for what input. Given your calculations afterwards, I'll assume you didn't mean the value of the output voltage, but the gain. Then, for the first one it's $$\A_{ol}\$$, because it can't amplify more than 86 dB, and for the second one it's: $$20\log_{10}{\Biggl(A_{ol}\dfrac{1}{5}\Biggr)}=20\log_{10}{\Biggl(\dfrac{20000}{5}\Biggr)}\approx 72\;\mathrm{dB}\quad(4000)\tag{2}$$ So, you calculated it correctly, just that you went a longer way after it. • It was ny bad the mV/V, indeed; I dont specify the input because it isnt given none, so I was thinking the same, it only can be represented, not a value specified. Thanks the real opamp gives me a little "noise" Apr 22, 2021 at 19:23	2022-08-19 08:47:43	{"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": 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": 26, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7567200064659119, "perplexity": 2044.2411593454788}, "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-2022-33/segments/1659882573630.12/warc/CC-MAIN-20220819070211-20220819100211-00471.warc.gz"}
https://mathoverflow.net/questions/196693/length-of-non-horizontal-curve	# Length of non-horizontal curve Let $M$ be a sub-Riemannian space. Consider a smooth curve $\gamma:[0,1]\to M$ such that $\dot\gamma(t)\not\in H_{\gamma(t)}$, where $H_{\gamma(t)}$ is the horizontal subbundle ( i.e. $\gamma$ is totally non-horizontal curve). Is it obvious that the curve is not rectifiable or has infinite length? • How do you define length for such a curve? I guess that you mean by $TM$ the subbundle of the tangent bundle on which the subRiemannian metric is defined. – Ben McKay Feb 16 '15 at 16:59 • I meant horizontal subbundle. Mistakenly written $TM$. – Nikita Evseev Feb 18 '15 at 4:52 Here is a rigorous proof that non-horizontal curves are not rectifiable. First, recall that, given a metric space $(M,\delta)$ and a mapping $\gamma:[0,1]\to M$, the $\delta$-length of $\gamma$ is, by definition, $$\ell_\delta(\gamma) = \sup\left\{\delta\bigl(\gamma(0),\gamma(t_1)\bigr)+\delta\bigl(\gamma(t_1),\gamma(t_2)\bigr)+\cdots+\delta\bigl(\gamma(t_m),\gamma(1)\bigr)\ \bigl\|\ 0< t_1 <t_2<\cdots<t_m<1 \right\}.$$ If $\ell_\delta(\gamma)<\infty$, one says that $\gamma$ is $\delta$-rectifiable. Second, a basic result in Riemannian geometry is this: If $g:TM\to\mathbb{R}$ is a Riemannian metric on a connected manifold $M$ and $\delta_g:M\times M\to [0,\infty)$ is the associated distance metric, then, for any piecewise $C^1$-mapping $\gamma:[0,1]\to M$, one has $$\ell_{\delta_g}(\gamma) = \int_0^1 g\bigl(\gamma'(t)\bigr)^{1/2}\ \mathrm{d}t.$$ Now, suppose that $M$ is a manifold with a smooth plane field $H\subset TM$ with the property that each pair of points in $M$ can be joined by some $H$-curve, i.e., a piecewise $C^1$-curve $\gamma:[0,1]\to M$ such that $\gamma'(t)$ lies in $H_{\gamma(t)}$ for all $t\in[0,1]$. Suppose that $h:H\to\mathbb{R}$ is a smooth function that restricts to be a positive definite quadratic form on $H_x$ for each $x\in M$. Then one can define a metric $\delta:M\times M\to[0,\infty)$ by the formula $$\delta(x,y)=\inf\left\{ \int_0^1 h\bigl(\gamma'(t)\bigr)^{1/2}\ \mathrm{d}t\ \bigl\|\ \gamma:[0,1]\to M\ \text{is an H-curve}, \gamma(0)=x,\gamma(1)=y\right\}.$$ Proposition: $\ell_\delta(\gamma) = \infty$ for any $\gamma:[0,1]\to M$ that is piecewise $C^1$ but not an $H$-curve. Proof: Choose a smooth splitting $TM = K\oplus H$ where $K$ is a smooth plane field (necessarily of positive rank, or else all curves are $H$-curves and there is nothing to prove). Let $k:K\to \mathbb{R}$ be a smooth function that restricts to be a positive definite quadratic form on each $K_x$ for $x\in M$. Consider the family of Riemannian metrics $g_n = n\,k \oplus h$ on $M$, and let $\delta_n$ be the distance metric on $M$ associated to $g_n$. Then it follows directly from the definitions that $\delta(x,y)\ge \delta_n(x,y)$ for all $x,y\in M$. Consequently, it follows (again from the definitions) that $$\ell_\delta(\gamma) \ge \ell_{\delta_n}(\gamma)$$ for all maps $\gamma:[0,1]\to M$. Now suppose that $\gamma:[0,1]\to M$ is piecewise $C^1$ but not an $H$-curve. Writing $\gamma'(t) = a(t) + b(t)$, where $a(t)$ lies in $K_{\gamma(t)}$ and $b(t)$ lies in $H_{\gamma(t)}$, one has that $a(t)$ is non vanishing for $t$ in an open subset of $[0,1]$. Consequently, $\int_0^1 k\bigl(a(t)\bigr)^{1/2}\,\mathrm{d}t > 0$. But then, for all $n$, one has $$\ell_{\delta_n}(\gamma) = \int_0^1 \left(\,n\,k\bigl(a(t)\bigr) + h\bigl(b(t)\bigr)\,\right)^{1/2}\,\mathrm{d}t \ge \sqrt{n}\ \int_0^1 k\bigl(a(t)\bigr)^{1/2}\,\mathrm{d}t.$$ Thus, the inequality $\ell_\delta(\gamma) \ge \ell_{\delta_n}(\gamma)$ for all $n$ implies that $\ell_\delta(\gamma) = \infty$. • In "Second", is $g$ a quadratic form (not metric) ? – Nikita Evseev Feb 18 '15 at 5:41 • @NikitaEvseev: Yes. It's an unfortunate convention in Riemannian geometry that the underlying quadratic form $g$ on $M$ is called a 'Riemannian metric', even though it is not a 'metric' in the sense of metric spaces. Of course, it does define a metric $\delta_g:M\times M\to [0,\infty)$, where $\delta_g(x,y)$ is the infimum of the $g$-lengths of piecewise $C^1$ curves joining $x$ to $y$ (when $M$ is connected). – Robert Bryant Feb 18 '15 at 9:46 Length is only defined for horizontal curves on a subRiemannian manifolds. This is due to the subRiemannian (or subFinsler) idea that you are only allowed move along a subbundle of the tangent bundle. For other curves it is natural to take the length to be infinite if you need to assign them a length. One can approximate a subRiemannian metric by a sequence of Riemannian metrics which blow up in nonhorizontal directions. Now if you fix a curve and take the limit of its lengths with respect to these approximating Riemannian metrics, the limit length can only be finite if the curve is horizontal everywhere. Since $M$ is a metric space, you can of course define the length of a smooth curve $\gamma:[0,1]\to M$ in the usual way as a supremum of sums of distance between points. If the curve is nonhorizontal, this supremum is infinite for the following reason. Let $\Delta\subset TM$ be the subbundle where the subRiemannian metric lives (I denote by $TM$ the whole tangent bundle). Iterated brackets of $\Delta$ generate the whole tangent bundle. If $\dot\gamma(t)\in[\Delta,\Delta]\setminus\Delta$, then $d(\gamma(t+\epsilon),\gamma(t))\approx\epsilon^{1/2}$ for small $\epsilon$. For more brackets you get smaller powers, so you should get $d(\gamma(t+\epsilon),\gamma(t))\gtrsim\epsilon^{1/2}$ when $\dot\gamma(t)\in TM\setminus\Delta$. This scaling leads to infinite length. I believe this argument is fairly obvious for those working with subRiemannian structures.	2019-12-10 03:19:54	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 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": 0.9663813710212708, "perplexity": 128.52625949202164}, "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/1575540525781.64/warc/CC-MAIN-20191210013645-20191210041645-00074.warc.gz"}
http://www.singer22.com/blog/style/ravishing/	# Ravishing And there's more where this came from: Blue Life This entry was posted in Style and tagged , , , , . Bookmark the permalink.	2014-04-16 13:31:51	{"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": 0.9202192425727844, "perplexity": 5971.339688150386}, "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-2014-15/segments/1397609523429.20/warc/CC-MAIN-20140416005203-00567-ip-10-147-4-33.ec2.internal.warc.gz"}
https://zbmath.org/?q=an:0672.47045	# zbMATH — the first resource for mathematics James quasi reflexive space has the fixed point property. (English) Zbl 0672.47045 Using the works of Maurey and Lin, the author proves that the classical sequence James space J has the fixed point property. The main result states: Every weakly compact convex subset of J has the fixed point property. This gives rise to an example of a Banach space with a non-unconditional basis. Reviewer: Shaligram Singh ##### MSC: 47H10 Fixed-point theorems 46B25 Classical Banach spaces in the general theory 47H09 Contraction-type mappings, nonexpansive mappings, $$A$$-proper mappings, etc. Full Text: ##### References: [1] Lin, Texas Functional Analysis Seminar 1982–1983 [2] Lin, Pacific J. Math. 116 pp 69– (1985) · Zbl 0566.47038 [3] DOI: 10.2307/2313345 · Zbl 0141.32402 [4] Kirk, Fixed point theory for non-expansive mapping I, II: Lecture Notes in Math 886 pp 484– (1981) [5] Karlovitz, Pacific J. Math. 66 pp 153– (1976) · Zbl 0349.47043 [6] James, Studia. Math. 60 pp 157– (1977) [7] DOI: 10.1073/pnas.37.3.174 · Zbl 0042.36102 [8] DOI: 10.1007/BF03007648 · Zbl 0344.46045 [9] DOI: 10.2307/1999928 · Zbl 0494.46014 [10] DOI: 10.1007/BF02762773 · Zbl 0461.46011 [11] DOI: 10.2307/2043954 · Zbl 0468.47036 [12] DOI: 10.2307/2318219 · Zbl 0336.47033 [13] Maurey, Points fixes des contractions sur un convexe ferme de L pp 80– [14] Lindenstrauss, Classical Banach spaces I (1977) [15] Lin, Bull. Inst. Math. Acad. Sinica 8 pp 389– (1980) This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.	2021-12-06 14:21:24	{"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": 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": 0.586734414100647, "perplexity": 6640.420104368826}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-49/segments/1637964363301.3/warc/CC-MAIN-20211206133552-20211206163552-00604.warc.gz"}
https://codeforces.com/problemset/problem/1183/H	H. Subsequences (hard version) time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output The only difference between the easy and the hard versions is constraints. A subsequence is a string that can be derived from another string by deleting some or no symbols without changing the order of the remaining symbols. Characters to be deleted are not required to go successively, there can be any gaps between them. For example, for the string "abaca" the following strings are subsequences: "abaca", "aba", "aaa", "a" and "" (empty string). But the following strings are not subsequences: "aabaca", "cb" and "bcaa". You are given a string $s$ consisting of $n$ lowercase Latin letters. In one move you can take any subsequence $t$ of the given string and add it to the set $S$. The set $S$ can't contain duplicates. This move costs $n - \|t\|$, where $\|t\|$ is the length of the added subsequence (i.e. the price equals to the number of the deleted characters). Your task is to find out the minimum possible total cost to obtain a set $S$ of size $k$ or report that it is impossible to do so. Input The first line of the input contains two integers $n$ and $k$ ($1 \le n \le 100, 1 \le k \le 10^{12}$) — the length of the string and the size of the set, correspondingly. The second line of the input contains a string $s$ consisting of $n$ lowercase Latin letters. Output Print one integer — if it is impossible to obtain the set $S$ of size $k$, print -1. Otherwise, print the minimum possible total cost to do it. Examples Input 4 5 asdf Output 4 Input 5 6 aaaaa Output 15 Input 5 7 aaaaa Output -1 Input 10 100 ajihiushda Output 233 Note In the first example we can generate $S$ = { "asdf", "asd", "adf", "asf", "sdf" }. The cost of the first element in $S$ is $0$ and the cost of the others is $1$. So the total cost of $S$ is $4$.	2019-08-22 10:28:52	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 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": 0.3900173008441925, "perplexity": 566.9376464400647}, "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/1566027317037.24/warc/CC-MAIN-20190822084513-20190822110513-00105.warc.gz"}
https://homework.cpm.org/category/CCI_CT/textbook/int3/chapter/3/lesson/3.1.3/problem/3-54	### Home > INT3 > Chapter 3 > Lesson 3.1.3 > Problem3-54 3-54. Determine if each of the functions below is even, odd, or neither. Look at a graph of each function. Even functions are symmetric about the $y$-axis. Odd functions have rotational symmetry about the origin. 1. $y = \sqrt [ 3 ] { x }$ Odd 1. $y=9x^5-x-9$ 1. $y=4x^3+8x^7$	2021-09-28 01:10:16	{"extraction_info": {"found_math": true, "script_math_tex": 4, "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": 0.47961896657943726, "perplexity": 1554.8635085844828}, "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/1631780058589.72/warc/CC-MAIN-20210928002254-20210928032254-00118.warc.gz"}
https://en.wikipedia.org/wiki/Talk:List_of_AMD_graphics_processing_units	# Talk:List of AMD graphics processing units Jump to: navigation, search WikiProject Computing This article is within the scope of WikiProject Computing, a collaborative effort to improve the coverage of computers, computing, and information technology 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. ??? This article has not yet received a rating on the project's quality scale. ??? This article has not yet received a rating on the project's importance scale. ## Wii U "Latte" GPU Why the page says that it is 160SP 8TMU 8ROP? That is a rumor. "Reference 63" Eurogamer http://www.eurogamer.net/articles/df-hardware-wii-u-graphics-power-finally-revealed techPowerUp http://www.techpowerup.com/gpudb/1903/wii-u-gpu.html They says that Wii U GPU is 320SP 16TMU 8ROP. That is elucidated with GPU die shot image analysis. — Preceding unsigned comment added by Maricolle1993 (talkcontribs) 04:45, 4 November 2014 (UTC) ## CPUs What happened to the pages Comparison of Intel Central Processing Units and Comparison of AMD Central Processing Units? I can't believe that I have to use answers.com instead of WP. • They were deleted in favor the more detailed lists that are now seperate into processor categories we now have "List of Pentium 4 Processors, List of Pentium D Processors etc etc. <<-Which sucks because you can not do a quick comparison between models, I would hardly call the new stuff a "comparison" as it gives no info other then names unless you open 50 windows ### Proposal I don't think this has any relevance to the list of AMD GPUs anymore. If there are no objections, I propose we delete this section of the Talk page. --ParoXoN (talk) 20:31, 16 September 2014 (UTC) ## PEG v. PCIe x16 It is my understanding that PEG is proper usage for the PCI Express Graphics slot (functionally different than simply a PCIe x16 slot). Is this accurate? If so, should this be changed in this article, and all other articles (nVidia GPU's etc.)? Sahrin 16:49, 5 November 2006 (UTC) -- "PEG" is an improper term, "PCI Express x16" is simple a standard PCI Express Bus with 16 lanes; it can be used for any peripheral. High-end Graphic Cards can take advantage of it's bandwidth and architecture, more than any other device.--Satsuki 18:02, 22 December 2006 (UTC) • PCI-SIG website uses PEG to denote graphics usage of an x8 slot used for graphics processors. The reason I propose using PEG in the place of PCIe x16 is because the slot is physically an x8 slot and electrically an anything between x1 and x16 (someday x32) slot. PEG clarifies it as a PCIe x8 slot for graphics. If anything, using your definition we should use PCIe x8, but I think that is also a misnomer (as typically the end user envisions it as an x16 slot, because electrically that is what it is). Again, I suggest we change PCIe x16 references to PEG for clarity or robustness of the information. Sahrin 01:33, 29 December 2006 (UTC) • PEG Link Mode, read. I think using "PEG" would confuse everyone. "PEG Link Mode" is a term used for overclocking the video card on specific mobos, which isn't standard. "PCI-e x16" is the correct and standard term for the interface these cards use. • PCI-e x16 is the correct name for a physical and electrical x16 slot, which not all of these graphics cards use (either on the "slot" end or on the "connector" end). Saying PCIe x16 is akin to saying "USB 2.0" for every single USB device in existence - when in fact, not every USB device is 2.0 standard. PEG describes graphic use of PCI-Express, PCIe x16 describes a standard for a connector - for any purpose. Not saying I disagree we should use PCIe x16, but I think your justification is invalid. Sahrin 13:40, 7 July 2007 (UTC) ### Proposal I feel that as of 2014 the term "PCIe (version) x16" has become pretty standard and that "PEG" has fallen into disuse. If there are no objections I say we remove this section of the Talk page. --ParoXoN (talk) 20:29, 16 September 2014 (UTC) ## Radeon 9500 non-pro Radeon 9500 non-pro in its early life had 256 bit memory bus as it has been built on the same PCB that radeon 9700 was build. It has been posible to enable the extended memory bus with some moding and driver editing. ## DirextX Support Why my ATI 9000 64mb supports directx 9.0, while in the article it isn't so? (unsigned comment by User:87.0.236.9) It supports DirectX 9.0 compatibility, but its feature-set matches only those from Direct3D 8.1. Read the "DirectX version note" section --200.148.44.7 01:56, 22 August 2006 (UTC) ## ATI Radeon X1400 Why is there no Radeon X1400 on the list? Where does it fit in? I don't think there is a Radeon X1400, other than the Mobility. Decembermouse (talk) 05:15, 10 March 2008 (UTC) ## TMUs Every single card on the page is listed as only having one TMU but many if not all of the cards have more than one TMU. Somebody needs to fix this. Some guy 08:00, 16 October 2006 (UTC) • This is because it is supposed to be labeled TMU/Pipe, not just TMU alone.Coldpower27 04:32, 22 October 2006 (UTC) • Indeed, for some reason, the table lists that one entry according to the old "pipeline" idea, where it first lists the number of "rendering pipelines," then the number of TMUs per pipeline, and then the number of Vertex Shaders/T&L units. This has really been an outdated layout of things, since "pipelines" are no longer followed. In this case, it results in some ackward usage of things like pipelines: 16(48) to try to get across that a card has 16 ROPs, 16 TMUs, and 48 PSs. Perhaps it would be recommended for someone to fix it; I may take care of it if I get the time. It really should have multiple entries to reflect each type of unit. Nottheking 23:02, 6 December 2006 (UTC) ## VPUs For consistency, I've listed any GPU having FF T&L unit as 0.5 vpu, regardless of the number of actual T&L units. ## Complete R100 section Could someone fill in the information for the remaining R100 skus: Radeon DDR, SDR 7200 Downgraded to OpenGL 1.3 due to lack of programmability extensions ## Fab process Fabrication Process - Average feature size of components of the processor. I thought it was minimum feature size. Tempshill 20:15, 6 December 2006 (UTC) • No, it's not. The actual transistors are smaller than the fabrication process listed, and some of the interconnects are even thinner still. Marking it by the minimum feature size would be too confusing, and hence "average" has been what it's always been defined as, in any sort of publication. Nottheking 00:00, 7 December 2006 (UTC) • Affirm Nottheking. Average feature size is used because some features are much larger than the process used, others are much smaller. Minimum feature size would cover < 1% of all features on the process, average accounts for a much larger number of the features, in addition to being a number reflective of the size of all features, not just the ones at size X. Sahrin 01:30, 29 December 2006 (UTC) ## Radeon Mobility Updates (x1900) Fujitsu Siemen (EU Division) is releasing a laptop(AMILO Xi 1554) with a Radeon Mobility x1900 256Mb XT/XTX. Not much information has been released. Updates for the ATi video card list needed soon. http://www.theinquirer.net/default.aspx?article=29358, http://laptoplogic.com/forums/showthread.php?p=17759, http://www.fujitsu-siemens.com/home/products/notebooks/amilo_xi_1554.html. —The preceding unsigned comment was added by SatsukiMikata (talkcontribs) 16:07, 22 December 2006 (UTC). ### Proposal The X1900 Mobility has all its information completely filled out in the appropriate table as of this comment. Propose removal of this section of the Talk page. --ParoXoN (talk) 21:04, 16 September 2014 (UTC) ## Xpress 1100 and 1150 VPU Is there a source that says that the 1100 and 1150 igps have dedicated vertex units? or is it host based as the 200m series? Everything I've read seems to suggest it's basically a smaller process of the RS482 with support for AM2 ## Missing cards and parenthetical notation in fillrate It seems like this table is missing X1050 and X1550 cards. Or are these a rebranding of other ones? Also, for some cards a second value is listed in parenthesis in the fillrate column, higher than the other value. (e.g. "2000 (6000)") What does this mean? Any help is appreciated. Sir Fastolfe 02:28, 17 February 2007 (UTC) ## X2K series is upside down All the other series run from low-end -> high-end, while the X2K series runs from high-end -> low-end. Is there any reason for this or should it be fixed? Pik d 23:02, 17 March 2007 (UTC) • Since it's been about a month and no one gave any reasons for it to not be fixed, I went ahead and edited this. Pik d 18:51, 16 April 2007 (UTC) ## ATI Radeon Mobility 9200 missing? It seems that at least one model is missing here - mobility 9200. http://ati.amd.com/products/MobilityRadeon9200/index.html ## Discussion moved from Top of Page ATi is the correct capitalisation, and not ATI. Should this be changed? • It would appear that both capitalisations are actually correct, or at least, that the all-capitals form is correct; while the ATi logo suggests that the last character is in lower-case format, all print/media writings of the name I've found capitalized all of the letters; thus, it could make sense to leave the name the same. Nottheking 23:29, 6 December 2006 (UTC) Could there be some other or more specific details of GPU processing elements than those pipes, TMU's and VPU's? Melter • The four primary processing elements of a GPU are the ROPs (often now simply called "pipes,") TMUs, pixel shaders, and VPUs. There is also the memory controller(s), which are mentioned separately. Aside from that, there is little more on the GPU proper other than, perhaps, cache, but no readily availible documentation covers any other part of a GPU's structure other than that. However, the tables could perhaps use a form of listing components that provides more clarity, especially with the R500 and later designs, where the old design of a "pixel pipeline" was completely eliminated in favor of a less rigid, multi-threaded system. Nottheking 23:29, 6 December 2006 (UTC) I can't believe the table doesn't show Shader Model support of individual chips... Terrible, that definitely needs to be added. -- xompanthy 01:32, 12 March 2006 (UTC) You should be able to tell what Shader Model a graphics card supports by looking at the Direct X version of the product. • As mentioned above, SM is more or less the same as DX version; the only exceptions are for various DX 8.0 versions, which don't apply to Radoen cards anyway. DX 8.1 is SM 1.4, DX 9.0 is SM 2.0, DX 9.0b is SM 2.0x, DX 9.0c is SM 3.0, and DX 10 is SM 4.0. Nottheking 23:29, 6 December 2006 (UTC) Someone had written that the X1600 Pro/XT only had 4 pixel pipelines, and 12 shader units. This is incorrect. The card has 12 pixel pipelines and 12 shader units. I have changed the information for these cards accordingly. • Actually, neither are correct; the RV530, like all R500-series parts, has ZERO pixel pipelines; rather, it instead uses a threaded task system, using an arbiter processor on the GPU to distribute the workload across individual units, rather than relying on a pipeline structure that tied a pixel shader and one or more TMUs to a single ROP. Hence, I've edited the whole collumn to reflect this. Nottheking 23:29, 6 December 2006 (UTC) • Good catch nottheking. This was a mistake on my part when I re-aligned the columns for the R500/R600/Mobility Radeon parts. Thank you for correcting my error. Sahrin 01:35, 29 December 2006 (UTC) There is some missing info. The RV351 was an improved RV350 with lower power consumption, less heat (and a die shrink?). It appears on the Mac as the 9650 (sometimes 9650 XT) and has 256Mb of RAM. There are also PC 9600s using the RV351. And there is also a card called the ATi Rage 128 Ultra, which comes in 16MiB and 32MiB versions, and as low profile. It appears to be a Rage 128 Pro with faster clocks... Anonymous Coward 04:04 15 June 2006 (UTC) Thank you to everyone who contributed to this page, it's very useful to me! Chris D'Amato 20:22, 29 June 2006 (UTC) Bandwidth is calculated incorrectly. I've changed it to use GB/s, where GB/s=10^9 bytes/second. To properly calculate bandwidth in GiB/s it's (bus width * effective clock of memory) / 1073741824 (bytes/GiB) / 8 (bits / byte) Continous lists, divided in AGP and PCI-e categories, are becoming obscenely long. Should this article be modified as to be categorized by core families instead of native buses, as is Comparison_of_NVIDIA_Graphics_Processing_Units? Just as a note, my most recent edit is implied by other comments I posted here; in a moment of absent-mindedness, I neglected to write a comment describing my edit to the R500 table, which amount to a re-writing of what was previously the "pipe x TMU x VPU" collumn. It is now the "ROP x TMU x PSU x VPU" collumn; it PROBABLY shouldn't use the letter "x" to separate each number, but that seemed consistent with the style there. Likewise, unlike the other three processing elements of a GPU, there is no article yet for a pixel-shader processing unit as of yet. I've chosen the acronym "PSU," which might be a bit confusing, so I'll leave it to others to decide if that's one to use. Nottheking 23:29, 6 December 2006 (UTC) ## Style According to the Wikipedia manual of style, "Graphics Processing Units" in the title should be all lowercase. ## Console Graphics Processors I just noticed that the Xenos(Xbox 360) has a fillrate listed as 16000 compared to 648 for the Flipper(GameCube) and 972 for the Hollywood(Wii). This seems like a huge difference. Can anyone check the numbers on this? Daemonward 13:32, 31 October 2007 (UTC) • Looking at the other GPUs, it appears that the fillrate should equal (Core Clock Max) * (Number of Texture Mapping Units). I'll do the calculations and make the appropriate changes. If I'm wrong, please correct me. Daemonward 14:37, 31 October 2007 (UTC) • You would be correct, though that would be for the texture fill-rate. For the pixel fill-rate, it would be equal to (Core Clock Max) * (Pixels Per Clock Cycle), though that figure isn't included on the chart. The 648 mTexels/second is the correct figure for the Game Cube's Flipper, though the true number for the Wii's Hollywood is unknown, as both the design of the chip as well as its clock speed are unknown. (the figures used are stand-ins that have no cited source that dates after the console's release) The Xbox 360's Xenos was incorrect, as you noted, as it has 16 TMUs, and runs at 500MHz. Thank you for correcting that. Nottheking (talk) 04:11, 20 November 2007 (UTC) ## HD 3870 X2 The new HD 3870 X2 are a bit confusing about they PCIe specifications, becouse it uses PCIe 2.0 (MSI R3870X2-T2D1G, MSI R3870X2-T2D1G-OC) for the card interface, but uses PCIe 1.1 (Tom's Hardware "ATI Radeon HD 3870 X2 - Fastest Yet!" (page 4 of 20)) for the on-card Crossfire. - Placi1982 (talk) 12:39, 30 January 2008 (UTC) I have edited the specs on the 3870X2 to reflect Powercolor's upcoming release containing GDDR4. It should have 2 x 512mb GDDR4. Someone put 2 x 1024mb, if you know something about that please put it here! Decembermouse (talk) 05:19, 10 March 2008 (UTC) ## Original Research for R700 section I don't know German, so I tried to Google translate the website cited in the section (here: http://www.hartware.de/news_44085.html) to English (for your convenience: here), it yields the following in the translated text: We unfortunately have no way to verify this information, but [it is] interesting to read them in any case. As mentioned, this information is not official and therefore with caution. Up to the expected introduction of HD ATI Radeon 4000 series in June, it is still for a while, so that the details can change anything. An {{original research}} was put up. Please discuss. --202.40.157.145 (talk) 02:32, 18 February 2008 (UTC) Well, as we can see now, this information was unfounded and is outdated; it was ancient speculation, possibly old information, as then the word on the Radeon 4800s were that they would have only 480 stream processors, far fewer TMUs, and would have much higher clock rates. As we've seen now, the actual RV770 came with 800 stream processors, 40 TMUs, and more modest clock rates. It's also possible that some of this information has cluttered up and caused rumors on a supposed RV740 or Radeon 4700, which seems to be sporting specs eerily akin to what was previously thought to be RV770. Nottheking (talk) 09:57, 12 September 2008 (UTC) ## OpenGL 2.1 version note is misleading The OpenGL 2.1 version note here http://en.wikipedia.org/wiki/Comparison_of_ATI_Graphics_Processing_Units#OpenGL_version_note states that it supports GLSL and geometry shaders. This is misleading. Geometry shaders are not mentioned in the OpenGL 2.1 specification and are usually supported by vendor specific extensions. I think the WP entry http://en.wikipedia.org/wiki/Opengl#Mt_Evans quite accurately explains when geometry shaders will be officially supported. Maybe support for geometry shaders should be indicated separately by some different flag. 0meaning (talk) 08:23, 25 February 2008 (UTC) Fixed long time ago. WheretIB (talk) 23:58, 8 January 2009 (UTC) ## Core Config Because several ATI graphics units do not have any hardware vertex processing capability at all, there needs to be a consistent method of differentiating on the table cards with software vertex capabilities (IGPs), fixed function vertex units (R100) and no vertex processing capabilities (Rage). ## Invalid Citations You DO NOT cite sources from forum, forums are discussion areas. This is originally References #3 <ref name="Chile17012008">{{es icon}} [http://www.chilehardware.com/foro/ati-radeon-hd3100- t132327.html Chile Hardware thread], retrieved January 17, 2008</ref> [http://www.chilehardware.com/foro/ati-radeon-hd3100-t132327.html Original Citation in Spanish] Reference #7 (reason: forum) <ref name="IT_OCP_17012008">{{zh icon}} {{cite web \| url=http://www.itocp.com/thread-1931-1- 1.html \| title=[ATI] First look at AMD 3650 and 3690, you'll regret to miss it. \| author=OCP- News \| date=2008.01.17}}</ref> [http://www.itocp.com/thread-1931-1-1.html] Reference #10 (errors corrected: language is english, not chinese, can) [http://ati.amd.com/products/radeonhd3800/specs.html] (this is not chinese) --Ramu50 (talk) 19:13, 19 June 2008 (UTC) Reminder: For referencing citations, try to use the name of the company, or simple format such as Nvidia Geforce 9800X2 - Overiew, not the article's title for better verify and reaiable sources. Unless the webpage is a resource article, tutorital or web design such as Moving Beyond OpenGL 1.1 for Windows --Ramu50 (talk) 03:21, 23 June 2008 (UTC) Should this citation being removed? random images from unknown sources, the website look like a new reporting company so its better to quote the article not the image. --Ramu50 (talk) 17:21, 10 July 2008 (UTC) ## Power Consumption Would really like to know how much power each card consumes, and whether or not is available commercially without cooling fan, ie using heatpipe or similar. —Preceding unsigned comment added by 118.90.76.16 (talk) 22:45, 29 June 2008 (UTC) This could potentially be added, but doing so would add another dimension of complexity. The TDP of each card is considerably more doable in a lot of cases, since in the past few years ATi and nVidia have made a point of publishing official numbers for these. For older cards, they kept them secret, and such numbers when found were produced by independent research with varying accuracy. As far as the availability of finding such cards with passive cooling solutions in lieu of active HSFs, that would be outside the scope of this page, since that is up to independent board partners to select for their own products, and is not something specified by the actual model specifications that ATi sets out for their hardware. It's akin to factory-OC cards, which are also not specified by ATi/nVidia, else they wouldn't be considered OC cards. Simply put, there would be too many variants to possibly cover them all. Nottheking (talk) 10:04, 12 September 2008 (UTC) ## Page Protection and Other Language Citation I think this article should totally be protected, each day they are thousand of people editing this page without any references and it is hard to track who is opposing who. I think ALL of non-english citation should have include a Google Translate link) for easier reading. Don't link it to AltaVista Babelfish, they are very inaccurate. --Ramu50 (talk) 16:32, 15 July 2008 (UTC) Resolved by in Requested Pages for Protection (through chatting with admin), but still monitoring the consistency of article actions. --Ramu50 (talk) 01:56, 21 July 2008 (UTC) Indeed, the edits I'm seeing to the article hardly ever have sources cited. Even if they're the correct figures, it's critical to properly cite them when we're dealing with tables of numbers here, as it helps fight inaccurate numbers due to speculation; for instance, I just corrected a couple lines that improperly listed the specifications of the RV730 GPUs, which coincidentally, had cited nothing; I added two sources for them. Into the future, I think I'll slowly go through this list as well as the nVidia list and add what sources I can... Yes, it's a lot, but the sources need to be there! Nottheking (talk) 05:59, 12 September 2008 (UTC) ## Northern Island HD7000 specification Many Wiki vandalist attempt to change the value of rops number in Tahiti XT from 48(or maybe 24) to 32 despite the bus is 384bit. However it is impossible to have 32 rops on a 384bit bus. — Preceding unsigned comment added by 70.131.80.64 (talk) 17:07, 17 December 2011 (UTC) There are very many Wiki vandalist attempts to change the number specs of the Tahiti GPU (79x0). For example, the specs are 2048/128/32 for 7970 and 1792/112/32 for 7950. In regards to question about 32 rops, they are 32. Please refer to original ATI slides. — Preceding unsigned comment added by 129.78.32.21 (talk) 13:51, 21 December 2011 (UTC) also someone has given high end 7xxx mobile gpus desktop specs, like 7850m/7870m look a bit much like the desktop variant — Preceding unsigned comment added by 158.37.228.14 (talk) 09:57, 8 March 2012 (UTC) ## OpenGL 2.1 - supported? This wiki page says that RV770 supports OpenGL 2.1, but on official AMD page there is only OpenGL 2.0 support mentioned. [1] —Preceding unsigned comment added by 83.10.216.65 (talk) 21:12, 29 July 2008 (UTC) ## Radeon Xpress 1100 IGP where should this card be reported? it states to have an RS485 chip and its pciid is 1002:5975 subsys 103c:30b0 while it sometimes is called RS482 [Radeon Xpress 200M] it takes 256M memory from the system RAM —Preceding unsigned comment added by 78.53.197.200 (talk) 13:13, 2 September 2008 (UTC) ## Radeon 4830 The page says that this card has 12 ROPs. According to Anandtech.com (http://www.anandtech.com/video/showdoc.aspx?i=3437&p=3) it has 16 ROPs. Can someone please change it. —Preceding unsigned comment added by 128.211.251.118 (talk) 04:08, 24 October 2008 (UTC) The page also says that the card has 640 SPs (160x4). I thought it was 128x5. It's still 640 SPs, but it's a big difference. Dont all R700 processors come in clusters of 5, with 4 being simple ALUs and the fifth being a complex ALU? —Preceding unsigned comment added by 128.210.132.177 (talk) 16:46, 27 March 2009 (UTC) ## AGP Signalling Voltages and Backwards Compatibility Many of the cards list 'AGP', 'AGP 4x' or 'AGP 8x' as the Bus Interface, but this information says nothing about compatibility with other AGP interfaces. Some AGP 4x/8x cards are backwards compatible with AGP 2x while others are not. Some AGP 2x cards work in AGP 4x slots, other's don't. It all depends on whether the card will accept signalling voltage of the motherboard. I think it would be useful to specify exactly which cards have versions supporting a specific AGP interface. It is a simple fix: where the Radeon X1050 lists AGP 4x/8x, PCIe x16, change it to show AGP 2x/4x/8x, PCIe x16 if an X1050 accepting 3.3V (AGP 2x) exists (it does). Also, no cards since the R300 series has supported 3.3V to my knowledge, so the changes should be minimal. Unless anyone objects, I'll begin adding this information in the next few days. Mattst88 (talk) 02:04, 29 November 2008 (UTC) ## ATI Mobility Radeon 9100 IGP - Data Mismatch Hello, I have an ATI Mobility Radeon 9100 IGP GPU, and Windows reports that the Internal DAC is clocked at 400 MHz. Every other utility I have used to gain information on the chip states that the chip is running at 300 MHz. Which one is incorrect? It doesnt seem like Windows is incorrect, since it is the one with direct access to the hardware, but every other utility says it is running at 300 MHz. What is going on? Presario (talk) 19:34, 1 March 2009 (UTC) ## PowerColor HD 4730 Hello, I picked this product from Xbit Labs http://www.xbitlabs.com/news/video/display/20090529111225_PowerColor_Officially_Launches_Radeon_HD_4730_Graphics_Card.html And it shows that PowerColor has released a new HD 4730. I was wondering if you could add that to the list? --124.188.26.68 (talk) 12:07, 30 May 2009 (UTC) ## R800 / 5xxx series This section should be removed as it has no sources and everything I can find points to the fact that R800 is not the correct codename ("Evergreen" is according to http://www.anandtech.com/video/showdoc.aspx?i=3573 which isn't a rumour site.) The 5xxx codename is original research, inferred only from 3xxx and 4xxx, because I can find no reliable references to it on the internet. And the product names and specs are just plain made up. —Preceding unsigned comment added by 86.163.186.102 (talk) 20:19, 28 June 2009 (UTC) That's why it has the original search tag... this section will be kept until sources will be provided. Further more, don't generalize sites! Sure, sometimes they spin rumors but this time they got bits and pieces from official sources. This might be a rumor: Trillian. Regarding the Evergreen bit, it remains to be see but I hardly doubt that AMD will go for a numberless code name for PC parts. Em27 (talk) 21:44, 12 August 2009 (UTC) Keeping this section as the cards are now official. What's with the "Cypress XTX" being included? I follow hardware news pretty religiously and a quick google search only results in this wiki page and some rumour news articles without any hardware specs (WP:V). I'm deleting the entries until more evidence surfaces--81.243.7.205 (talk) 15:51, 30 September 2009 (UTC) I also follow hardware news and Cyrpress XTX (aka radeon 5890) it's just a rumor, not confirmed in any way. I have deleted it and added Radeon 5870 Six, the 6 monitors eyefinity capable card (confirmed and photographied) —Preceding unsigned comment added by 193.153.169.227 (talk) 17:54, 8 October 2009 (UTC) ## 4650/4670 TWP W update Reference 21 states (from the AMD/ATi presentation slides), the maximum board power usage to be 48W/59W, respectively. ## TDP notice Actual TDP may be differing from other board vendors. The TDP listed is not TDP, but rather board power consumption, so another footnote should be placed. —Preceding unsigned comment added by 216.93.208.25 (talk) 23:15, 19 October 2009 (UTC) ## Radeon Mobility 4650 incorrect memory amount I could just change it to reflect the correct data, but then someone would just change it back so I'll reference it with "I have a DV7-2185dx laptop from HP. It includes the the 1Gb Radeon Mobility Card." This can also be selected as an option in the HP shopping page, so they reference it too. Thx. —Preceding unsigned comment added by 76.25.63.200 (talk) 03:21, 5 November 2009 (UTC) ## Mobility x1100 and mobility HD 5xxx Can anyone please add them to the list? 83.108.203.102 (talk) 18:47, 12 October 2009 (UTC) ## r600+ "Config core" Some numbers here are strange. I mean those AAxBB. What do AA and BB mean respectively? If BB is the number of units in each shader cluster then it must be __always__ 5 (and is probably wrong for r700+ specs here). If it's number of SIMD cores it may vary (and is wrong for r600 specs). —Preceding unsigned comment added by 83.10.211.211 (talk) 17:14, 20 December 2009 (UTC) ## FireStream workstation computing card missing AMD FireStream, firestream_9270 Alinor (talk) 15:38, 21 January 2010 (UTC) Done. Alinor (talk) 14:08, 6 February 2011 (UTC) ## chart Which idiot is responsible for changing the chart? This comparison ought to make it easy to compare? uh, then let's keep the same chart for all GPUs from 2xxx? —Preceding unsigned comment added by 84.56.174.63 (talk) 14:26, 30 January 2010 (UTC) ## Wrong Mobility Radeon HD 2600 (M76M) Memory clock max CCC and gpu-z says that max memory clock max is 600 Mhz and not 400. —Preceding unsigned comment added by 78.94.205.162 (talk) 19:09, 4 February 2010 (UTC) ## Video Acceleration I ended up on this article while trying to determine if my ATI card supported H.264 hardware acceleration. Unfortunately there does not seem to be any info on this types of feature support in the article. Is there another article that does have it? If not, it would be pretty useful if it were added. —Preceding unsigned comment added by Synetech (talkcontribs) 22:02, 22 March 2010 (UTC) Note, the RV710 (HD4350, HD4550) has UVD 2.2, so I have appended that information to the appropriate list (not the chart, however). Alfredcisp (talk) 18:05, 1 June 2010 (UTC)Alfredcisp ## Repetitious annotation makes table very hard to read Every card in the Evergreen (HD 5xxx) series table provides angle independent anisotropic filtering. This text in the final column often wraps to make every row of the table six lines deep, until you make the fonts so small the table is hard to read for another reason. To read the table efficiently, I ended up stretching it all the way across my dual-head desktop, so that what I needed to see was easy to read on the left LCD panel (under Firefox). — MaxEnt 12:05, 30 March 2010 (UTC) ## HD5870 Eyefinity6 TDP Im been looking at conflicting info for this matter, Toms Hardware lists it as 34/228W (http://www.tomshardware.com/reviews/radeon-5870-eyefinity6,2595.html), though the refrence from AMDs sites asys 27/188 like the regular 5870. Logically 34/28W would be more correct, did AMD have a typo there? —Preceding unsigned comment added by 98.203.55.77 (talk) 03:46, 11 April 2010 (UTC) AMD has updated the information to show the 228W figure Toms Hardware listed. AMD though does not list a minimum TDP on the same page. (http://www.amd.com/us/products/desktop/graphics/ati-radeon-hd-5000/hd-5870-eyefinity-6-edition/Pages/overview.aspx#2) (99.179.78.30 (talk) 06:11, 22 January 2011 (UTC)) ## Standalone section for Compute Capability and OpenCL I think someone should make standalone section for Compute Capability and OpenCL support on ATI cards. This section should include versions of CC,OpenCL and supported cards(optional with chip names - RV600). Little table with some informations for compute capabiliti is on Nvidia page(Comparsion of Nvidia GPUs). I wrote today someting similar for this Nvidia page. Sokorotor (talk) 15:49, 28 April 2010 (UTC) For instance the 5800 series of GPU is listed in a table with the text "all supports OpenCL 1.1". True - but at least my 5850 card reports device support for OpenCL 1.2 with a recent AMD Linux driver (platform AMD-APP 923.1). I suppose supported OpenCL version usually depends more on the software than on the hardware. (I don't know if recent AMD driver supports 1.2 on all 5800 hardware though.) — Preceding unsigned comment added by 85.229.117.254 (talk) 20:50, 27 June 2012 (UTC) ## DDR3 vs. GDDR3 After having to take another look over several articles regarding the subject, it's clear that it's important to stress a point here: GDDR3 and DDR3 are not the same. Be careful which you specify. In a lot of cases, it appears that through causes, however they be, (typos, ignorance, or laziness) some editors have, in other articles, introduced the incorrect term. I went through and checked this entire article, and was able to verify each of the cases where a card specified DDR3 rather than GDDR3. (the same was not true for the counterpart NVIDIA article; I had to edit a few card listings and add citations to correct it) I'd highly recommend that aside from paying special care not to confuse various forms of DDR vs. GDDR, that likewise, in cases where a "less-than-typical" type of memory is specified (the "typical" being DDR, DDR2, GDDR3, GDDR4, and GDDR5) that editors cite a source that explicitly names the memory type used. I can foresee this otherwise being a rather contentious editing issue, prone to edit/revert battles and swarms of [citation needed] tags. Nottheking (talk) 11:36, 26 May 2010 (UTC) It should be noted, the primary difference between GDDR3 and DDR3 is specified power consumption levels (and TDP relating to the power consumption). —Preceding unsigned comment added by Alfredcisp (talkcontribs) 17:41, 1 June 2010 (UTC) Actually, that's merely the most VISIBLE benefit for these cases. The primary difference is that GDDR3 is, contrary to what the name would imply, actually a derivative of DDR2. GDDR4 is also a derivative of DDR2, while GDDR5 is the first GDDR to be based upon DDR3. One of the chief functional differences is that DDR3 (and in turn GDDR5) have an external data clock rate doubled over their interal timings that of DDR2, effectively producing what could arguably be termed a "quad-data-rate" (or "quad-pumped") interface. Though of course, the use of the term "DDR" indicates that it isn't a true QDR interface; there's four transfers per command signal clock, but still the same two per data clock; the 'QDR' description is merely a simplifcation to help understand how DDR3/GDDR5 work. Nottheking (talk) 10:50, 3 July 2010 (UTC) ## Shader count Tidbit This only applies to R600 and derivative architectures (r600, rv770, Evergreen), and includes all the cards within each generation. It should be noted, the current way on Wikipeida of counting shaders, is: (example): HD4830 (128 * 5) 128 denotes the number of shaders on the SKU (since the actual die inclues 160 shaders). Each shader is comprised of 5 ALU, which in current R600 derivative architectures (upto [confirmed] Evergreen - third generation), have 4 simple units capable of MADD/MUL and one ALU capable of transcendantal math calculations. All five are tied to each other, so 5 independant instructions can be executed per instruction clock, but only one dependant instruction may be executed per instruction clock. However, each shader in every 80+ ALU SKU is organised in one of several "SIMD" (Single Input, Multiple Data) of 80 ALU, or 16 Shaders. Simply put, shaders in r600 (and derivative) are comprised of 5 ALU, each. An independant ALU does not equal a single shader, as is the case in nVidia G80 (and G80 derivative) architecture. Alfredcisp (talk) 18:01, 1 June 2010 (UTC)AlfredCisp Well, it's not QUITE cut-and-dry as you make it sound... This is due to the remarkable number of differences that AMD and nVidia have taken in designing their unified shader technology. nVidia's approach takes the more conventional method that had been used in earlier pipelined GPUs; each "stream processor" contained a single 4x32-bit (128-bit) SIMD unit, and could handle a single instruction per clock cycle, for up to 8 operations (4 multiply+add) per clock cycle. AMD's stream processors take a different approach, as described in the article for Radeon R600. In there, each "stream processor" on its own has, just like nVidia's stream processors, a functional unit, though rather than being a SIMD unit, it exists as a scalar FPU; to handle SIMD instructions, that has to be done at the cluster level, which will break a vector into its components and hand off each, along with the instruction for it, to a single stream processor. However, each stream processor can individually handle a single instruction per clock cycle. So that means that technically, yes, the top-end GPUs have the number of stream processors claimed (320, 800, and 1600 for R600/RV670, RV770, and Cypress, respectively) and not 1/5th of that, because architecturally, those are BOTH the number of functional units on hand, as well as the maximum number of instructions per clock cycle it can handle; these are both analogous to how they are counted on nVidia GPUs. So, in short, while it would appear that, compared to the "old" (SIMD unit) design, many would think that AMD simply split each stream processor into smaller sub-components, instead the design is the opposite; AMD modified the functional unit to change it from SIMD to a scalar FPU, THEN clustered them together to compensate for the weakness this caused. So while it'd be correct to label Cypress as "5x320 stream processors," labeling it as having only 320 shader units would be incorret; it has 1,600 stream processors, they simply aren't as individually capable as those on nVidia's contemporary GPUs. IF you were to label Cypress as having 320 shaders, then you'd have to count on the analog for nVidia, and count te number of "streaming multiprocessors," which are likewise loosely-structured clusters, in that case containing 32 stream processors per. (in other words, one would have to list the GTX 480 as having only 15 'shaders') Nottheking (talk) 11:20, 3 July 2010 (UTC) I think we should display only the total number of stream processors in the tables here. It seems silly to show "1600(3205)". There isn't any point in showing that the stream processors are in groups of five in this table. That information can be learned through further reading on the architecture, but is unnecessary here. There are other divisions other than these groups of 5 stream processors, such as the clusters withing the die. We could very well display that, showing: "1600(2016*5)". This would more accurately represent the layout of these chips. Again, this informaiton is wholely unnecessary here and should be removed. I propose we only display total stream processor count in the following fasion: "1600". Paranoidmage (talk) 23:31, 7 July 2010 (UTC) If only it were that clear-cut... Because one could just as well make a case for the count be displayed in the fashion of simply "320," only counting the number of clusters, rather than the number of individual processing elements. Obviously, AMD pushes the 1600 figure because it looks more impressive, though each one of those 1,600 is nowhere near the capabilities of each one of of nVidia's (or ATi's earlier) shader units. The most obvious difference is that each of the 1,600 ALUs is more akin to an FPU, while a shader unit on an nVidia G80/90/GT200/GF100 is a SIMD unit. In essence, the multi-number approach you assert as not having "any point" exists in the interests of attempting a neutral point of view, and avoid being misleading; it's sort of a compromise. Nottheking (talk) 04:17, 31 August 2010 (UTC) um, have you all read any of the various explanations of these architectures that are part of coverage of new families of gpus on major tech sites? ever since g80, nvidia GPUs have been completely scalar and amd as vect4 plus one. every amd shader can perform both a single precision multiply and add per clock and nvidia dx10 chips one multiply add and supposedly but not usually another multiply per shader as well. with dx11 both companies switched from madd to fma(fused multiply add) and nvidia dropped the extraneous mul. up above someone flipped how nvidia and amd organize shaders. no dx10 or later nvidia chips have simd shaders. the reason nvidia both performs better on basis of theoretical gflops versus amd and takes up so many more transistors and more die size for the different mumber of shaders is due to several things. the efficiency afforded by the scalar granularity versus amd's simd architecture is a boon for almost double clock speeds at the cost of additional separate pathways on the chip to accommodate individual scalar work. in addition, the extra render back ends(rops) and memory width make for a more balanced chip performance wise but again at the cost of more transistors. for a while until recently amd has had an abundance of shader power with a relatively lower performance in other areas of their chips and for the most part nvidia has used fewer but much higher clocked shaders along with more other functional units(rops, wider memory interface width) that help balance the chip and boost performance at the cost of die size and power consumption. more recent developments in the form of gf104 for nvidia and barts for amd have seem a rebalancing of chip resources to produce smaller, cheaper, less power hungry chips that perform almost as well as their older more expensive siblings on the same 40nm process. as I write this at approx. 1am central US time by all rumors we expect to see modifications to amds vliw5(very large/ long instruction word) to vliw4 which hopefully will see performance gains of 20 to 25 percent per shader on average in the next day or so. we will look to reviews to tell whether or not this will be a boon for amd's new cayman chip to compete well with gtx580 and gtx570. hope this helpsJtenorj (talk) 07:07, 14 December 2010 (UTC) ## HD5000 Someone made an incorrect edit to the Eyefinity edition capabilities. They actually have all the capabilities of "normal" non eyefinity edition cards, but with support for more outputs. Alfredcisp (talk) 20:47, 3 June 2010 (UTC)Alfredcisp ## HD3410 correction HD3410 had upto 512mb dedicated GPU memory, as configured in the now-discontinued HP DV2 laptop series. —Preceding unsigned comment added by 124.42.77.160 (talk) 09:46, 2 August 2010 (UTC) ## Radeon HD 6000-series and HD 7000-series HD 6000 will be an architectural change rather than simply warp with more shader to increase performance. which the number of shader increase will be limited. —Preceding unsigned comment added by 70.131.56.98 (talk) 18:29, 11 September 2010 (UTC) HD6770 1280sp/64tmu/16-32rop HD6750 1120sp/56tmu/16-32rop —Preceding unsigned comment added by 85.93.116.50 (talk) 09:31, 24 September 2010 (UTC) Then, where are the specifications of Radeon HD 7000 series? — Preceding unsigned comment added by Lacp69 (talkcontribs) 14:20, 16 January 2012 (UTC) ## Pricing/Naming of HD 6000 While HD 6000 will be an architectural change, however some of the line will continue the exist HD 5000 line and rename under the HD 6000 brand as well. it was confirmed [citation needed] the HD 6600 will be rebrand from Jupiter to maintain the mainstream market. for the pricing issue the HD 6770/6750 will price between 5850/5830 and eventually replace them in the end of year while HD6800/6900 will focus on high end/professional market—Preceding unsigned comment added by 75.63.48.111 (talk) 20:16, 13 September 2010 (UTC) Encyclopedic content must be verifiable, and given AMD's stance on NVida's rebranding, and the limited rebranding (that clearly seperates generations) AMD has done in the mobile gpu and chipset gpu space, I call bunk on the Radeon 6600 being a rebranded 5770. 164.106.139.159 (talk) 19:07, 14 September 2010 (UTC) As the editor above [164.106.139.159 (talk) 19:07, 14 September 2010] pointed out, the content here must be verifiable. To my knowledge, AMD has not even made an official announcement about the 6xxx series. The only "support" or "verification" we have are spreadsheets with information about two cards, the upcoming Radeon HD 6750 and 6770 cards, supposedly produced by AMD, and yet somehow we have an entire table of information about the entire 6xxx line, including precise technical specifications, release dates, and prices, with absolutely no mention that some (or most, as the case may be) of the information is speculative. Until AMD has released official information on this line, anything that can't be supported by at least an AMD-branded spreadsheet/chart should be removed, or at the very least replaced with "TBD" or "TBA". It seems bogus to represent this information as "fact," when in reality, it is largely supported by "nothing." TJShultz (talk) 16:13, 28 September 2010 (UTC) Caymen is not going to be 48 rops while only 256bit bus. any fanboi frame will result block from this section. —Preceding unsigned comment added by 70.131.62.13 (talk) 04:52, 4 October 2010 (UTC) User red dog and others keep framing/vandalise this page I request this page need to set as semi protect and all setup changes has to go through community before change values. —Preceding unsigned comment added by 70.131.63.167 (talk) 22:58, 4 October 2010 (UTC) ## HD6000 cards with VGA output ? The HD5000 series was built by several manufacturers with an integrated, dedicated VGA port, so that together with the DVI-I port and an adapter you could connect two VGA-monitors to one card. Will the HD6000 series cards, I mean some of them, also feature dedicated VGA ports, plus the DVI-I port ? -- Alexey Topol (talk) 00:01, 3 November 2010 (UTC) In the meantime, many cards with dedicated VGA ports have come out in the HD6xxx series. But none of them features DVI-I ports, they all sport DVI-D ports only. -- Alexey Topol (talk) 06:59, 30 September 2012 (UTC) ## Cleanup of unsourced information in Northern Islands (HD 6xxx) series This page was recently semi-protected (see the request) so I decided to continue the work by Prander to remove unverifiable information. I've added citations where I could, and removed other information. Some notes: • For the released GPUs 6870 and 6850, I generally did not remove information, even if I didn't source it. The notable exception is that I removed the listing of them having 2GB of memory: I could find no reference to this, and newegg sells no cards with it. • I removed the GPU with the codename Turks. I could find no source for even a model name for this, let alone specifications. (Also, the codename information from the driver release points to two separate Turks, XT and Pro.) • I removed most of the information about the 6350, except what I could source. • I left in a few things that are obviously consistent across all models, such as fab process and bus interface. I hope that having citation markers all over the place will convince other editors to cite their stuff too, instead of edit warring with unconfirmed numbers. dolphinling (talk) 08:44, 4 November 2010 (UTC) My citations on GPUs that have been released were removed. Now that I see the official specs on AMD's website, I realize the reference I chose wasn't the best. However, I think it would have been better to change the citation rather than remove it. More importantly, though, I think we need to decide how citations for released cards should work. I propose the following: • The card model should be cited to the official specification page on the AMD website • Information that is not in the official specifications (Release date, Code name, Fabrication process, Transistors, Die size, TDP, maybe more?) should be cited elsewhere. • Information that is in the official specifications should not have a citation marker, the one next to the model is enough • In the case of a section undergoing a lot of speculative changes (e.g. the curent 6xxx section), information in the specifications should have citation markers, to make it clear which information is verified, and to encourage people adding new information to add citations. These markers can be deleted when the section is no longer the subject of lots of speculation. I'm going to make the 6xxx section conform to this as best I can now, until I get some comments on this proposed style. dolphinling (talk) 19:01, 4 November 2010 (UTC) I removed those citations because the cards were already released. I was just following what is done with the other charts, where official specs or easily found information isn't cited in the the charts. I didn't think they were all required for released cards that are well documented online. The way it is now with a citation in every cell is too overwhelming to look at, especially when that information is credible and doesn't need the citations. However, for speculation of future cards or and current specs that are more debatable will definitely need sources. Paranoidmage (talk) 19:41, 4 November 2010 (UTC) And even then it won't be enough. The sources themselves should be looked with great skepticism: AMD is determined to keep info secret until the very release, and has quite succeeded in doing so. Even serious tech website cant help publishing rumours to fill this information void. And our responsibility is not to help them to spread. Basically AMD is the only reliable source.--Prandr (talk) 00:33, 5 November 2010 (UTC) This is generally true, but keep in mind that sometimes the websites of AMD/Nvidia/Intel may contain some vague or simply wrong information (because of neglect, copy-paste from previous materials, etc.) - and for many of the specifications they simply don't put it on their sites ("Superb performance" is a more catchy term than "16 execution units running at 300MHz"?). Alinor (talk) 09:16, 6 November 2010 (UTC) I want to remind you to be very careful with your sources. some good looking "slides" were fake. http://www.3dcenter.org/blog/leonidas/die-leichtglaeubigkeit-gegenueber-praesentationsfolien an they were copied by most serous tech sites. that what I meant above--Prandr (talk) 18:13, 25 November 2010 (UTC) ## Release Price There is currently a "release price" column on the Nothern Islands(6xxx series) chart. I have a number of reasons for which I feel it should be removed. Firstly, and most importantly, it isn't even a specification of the cards. It's irrelevant in this article. It gives only an indication of the launch price, which could change a week or so after release. Cards can go through different market segments/prices throughout their lifetime, and I don't think the release price is very useful. The charts are big enough as they are. I think we need to make decisions about which columns are actually important to the article, and make it as effective as possible for readers. Remember, these charts are supposed to be for specifications of the cards. Paranoidmage (talkcontribs) 13:58, 4 November 2010 (UTC) I agree with removing them. I think there's some minor historical value in knowing the release price, but it is outweighed by the benefit of making the tables smaller and easier to read. dolphinling (talk) 18:21, 4 November 2010 (UTC) I disagree. The release price is important to show what is the target segment of the card. Of course over time this price will be changed and that's why we don't list "current price", but "release price". Alinor (talk) 09:12, 6 November 2010 (UTC) I disagree. Released price is as important as everything else and it is included in the official release specification. If the problem is in table size, there are other ways to make it more compact without deleting info. E.g. making single column for all 3 APIs or even excluding this information to the header if all GPUs in the family have the same API versions (e.g Evergreen, R700 series) Poimal (talk) 05:46, 7 November 2010 (UTC) ## HD6800 series support OpenGL 4.1 I saw the sepcification on AMD website that the series have support for OpenGL 4.1 not 4.0 as in the table [1] you are correct. thanks for help. I think it is the matter of driver support.--Prandr (talk) 10:55, 6 November 2010 (UTC) ## OpenGL 4.4 Support Since the final Catalyst 14.4[2] (april 2014) driver release many AMD cards have OpenGL 4.4 support. If we trust reports gathered by glcapsviewer[3] this includes but is not limited to R9 29x, 6800, 7800, 5700 and 5800-series, and probably many more. 83.177.182.9 (talk) 17:33, 4 May 2014 (UTC) • I heard about OpenGL 4.4 Support as well, but the only supported cards I know of are Graphics Core Next-based GPUs (which I just updated in the article). (See here: https://twitter.com/AMDRadeon/status/459821068690411520 ). I have not touched the mobile area, since I'm not really familiar with the mobile chips and feature support. Would be great if someone could find out for sure which GPUs truly support 4.4, which 4.3, which 4.2 and so on. Seems like much of the information has changed over time due to improved feature levels via driver updates. Paxnos (talk) 19:21, 8 May 2014 (UTC) • After posting here, I contacted AMD Customer Care and asked about OpenGL 4.4 and which GPUs (codenames) support OpenGL 4.4. I received an answer today, including a link to this: https://en.wikipedia.org/wiki/OpenGL#OpenGL_4.4 So it seems like glcapsviewer got it right. I updated my AMD ticket to ask whether the information in that article is absolutely legit (meaning: All GPUs newer than and including Evergreen supporting 4.4) and will report back once I get that confirmation. Then we could think about updating the table, I guess. Regards Paxnos (talk) 13:46, 9 May 2014 (UTC) • AMD confirmed this. "Yes the desktop graphic cards from HD5000 to R9 200 series are support OpenGL 4.4 when 14.4 driver installed." Paxnos (talk) 09:57, 13 May 2014 (UTC) • Is there a good way to site Paxnos' email in the table? AMD's website still lists older OpenGL versions. (4.1 and 4.3 mostly.) ParoXoN (talk) 20:52, 16 September 2014 (UTC) Hi ParoXoN, I thought about it and I didn't find a reliable source to "back it up"...until now. Sadly, AMD doesn't seem to keep their own product specs up to date. But Khronos has a list of OpenGl (4.4) conformant products which should back it up. It includes all AMD (and the other companies') products featuring that OpenGL version. You can also switch navigation if you need confirmation of other API support, like OpenCL. (Little off-topic: AMD has started adding OpenCL 2.0 support, so those changes to the table will be coming soon as well: http://support.amd.com/en-us/kb-articles/Pages/OpenCL2-Driver.aspx) The list: https://www.khronos.org/conformance/adopters/conformant-products/#opengl Paxnos (talk) 23:42, 29 September 2014 (UTC) ## Edit request from 173.22.91.148, 13 December 2010 under the section of the page regarding the hd 6000 gpu sections of the new bobcat APUs, please change the theoretical GLOPS from 56 to 44.8 because this will accurately represent the performance based on clock speed in mhz times the number of shaders time 2(one multiply and one add per FMA) 173.22.91.148 (talk) 01:21, 14 December 2010 (UTC) If you can bring a reliable citation, feel free to edit everything yourself. The semiprotection has been lifted at my request.--Prandr (talk) 12:01, 17 December 2010 (UTC) Not done: See below. -- DQ (t) (e) 22:56, 17 December 2010 (UTC) ## Edit request from Jtenorj, 14 December 2010 hd6xxx igp theretical gflops for 280 mhz clocked part I believe that the theoretical gflops for the hd6xxx igp with the 280mhz clockspeed is incorrect. I'm guessing that the speed is well documented other places online and the figure for the texture fill and pixel fill look correct but the math for the gflops looks off. 280mhz times 80 shader processors times 2 single precision flops per shader per clock fma(fused multiply add) should come to 44.8 gflops. please correct at your earliest convenience.Jtenorj (talk) 08:02, 14 December 2010 (UTC) Jtenorj (talk) 08:02, 14 December 2010 (UTC) Not done: please provide reliable sources that support the change you want to be made. Would be willing to as soon as I get a source. -- DQ (t) (e) 22:53, 17 December 2010 (UTC) ## ATI Radeon HD 500v series This series is actually rebranded Radeon HD 4 series with slightly higher clock speed. AMD classifies them as 540v series, 550v series... in their site. These series still supports DX10 while the HD 5000 series supports up to DX11. Why the 500v series is put on the 5000 series section in this article? -- Livy the pixie (talk) 09:45, 15 February 2011 (UTC) ## AMD 6790 I'm not sure if this is real or not, though some sites like NordicHardware are talking about a release date of 31st of March for the 6790 - a card to compete with the Nvidia 550Ti. Like I siad, not sure if it's real or not, would love some more insight on this card. Not stats or prices, so if it does turn out to be a ghost I wouldn't be surprised. —Preceding unsigned comment added by 124.149.43.71 (talk) 03:22, 26 March 2011 (UTC) ### Proposal Since the card is real, I propose we delete this section of the Talk page. --ParoXoN (talk) 20:55, 16 September 2014 (UTC) ## Double Precision This article lists the HD 6790 as not having DP. According to AMD's own product listing[4] under the AMD App Acceleration section it states "Double Precision Floating Point". I would do the edit myself but this is my first time being "active" on Wikipedia. Thanks in advance. DarkWikiMuse (talk) 02:24, 5 July 2011 (UTC) ## Column headings need to be in more places than just the top of the tables Every four or five rows would probably be good. Otherwise, these charts are numbers without context more often than not. 99.88.142.167 (talk) 23:21, 2 September 2011 (UTC) ## WiiU specs What is the source of the wiiU gpu specs? especially the ram sounds fishy. — Preceding unsigned comment added by 130.234.180.172 (talk) 09:32, 5 September 2011 (UTC) ## Re-organization and Correction of Data (Cleanup) I'm going to make some major changes to the way the data is set up to make it more organized. I'm also going to do my best to correct any incorrect or misleading data. I've already changed up to R300. — Preceding unsigned comment added by Blound (talkcontribs) 02:59, 26 November 2011 (UTC) not a bad thing that somebody is reorganizing this, but may i ask why you are deleting some of the miscellaneous infomation? Asdfsfs (talk) 14:50, 18 December 2011 (UTC) noticed some things that are flat out wrong, for example memory bandwith on x1950 gt - was listed as 38.4 gb/s on the old version, but how come you state it's 64gb/s? the number is impossible because the memory is lower clocked than on pro which has 44 - perhaps a messup? if you're really trying to reorganize those pages it would be great if the numbers were still correct afterwards... Asdfsfs (talk) 22:28, 25 December 2011 (UTC) ## Asdfsfs edits I recently made some edits, which where reverted by User:Asdfsfs without explanation. My edits are - adding architectures VLIW5/VLIW4/GCN, moving All-in-Wonder upwards to correspond to the release date of its most recent member (instead of leaving it at the bottom where the newest series are), clarifying the HD7000 marketing mess by adding a second header in the middle - like in the R600 series. Ianteraf (talk) 06:48, 8 January 2012 (UTC) If you would bother to thoroughly read the differences you would see that I presented the architecture information in a much less awkward way. Concerning the 7000 thing - I think it's OK like it is now, because this line at least doesn't feature cards that start with a different number. Asdfsfs (talk) 15:48, 8 January 2012 (UTC) ## Definitive TDP of the HD 7970 I've seen 250W, 225W and 210W mentioned on tech sites, this wiki page still says 250W, can someone clarify this issue and say with certainty what the TDP is? 195.169.213.92 (talk) 20:46, 10 January 2012 (UTC) ## GFlops/W The GFlops/W are obviously calculated as max. GFlops/TDP, which is absolutely incorrect. There is no indication that a GPU will use its exact TDP when being used as a GPGPU. The GFlops/W is not in any spec sheet either, so i request the column to be removed. --79.251.138.122 (talk) 16:54, 23 January 2012 (UTC) +1 Visite fortuitement prolongée (talk) 21:42, 20 June 2015 (UTC) Deleted. Visite fortuitement prolongée (talk) 20:08, 7 July 2015 (UTC) ## HD7300M-HD7600M missing The renamed HD5000/6000 40nm chips are already released to OEMs.[2], but currently the article lists only the not yet released 28nm chips. Ianteraf (talk) 08:28, 4 February 2012 (UTC) 7300M still missing. Ianteraf (talk) 12:47, 9 August 2012 (UTC) ## Mobility FireGL/FirePro series table section Mobility FireGL/FirePro series There is a column missing in the table, so that the values on the right side of the table don't fit where they belong. For example the fillrates for a FirePro M5950 are not 17.4 and 57.6. Correct are 5.8 and 17.4. And Config core is not 5.8. It's 480square(96x5):24:8 (as in the column before). — Preceding unsigned comment added by 78.53.96.244 (talk) 10:44, 18 March 2012 (UTC) ### FireMV vs. FirePro I have an old ATI FireMV 2250 that is no longer listed in this page, but there is now a "FirePro 2250" listed. Assumine they are the same card, please indicate that "FirePro" is simply an AMD/ATI re-naming. Ian! (talk) 16:17, 24 May 2013 (UTC) ## Rage 128 Ultra, what is it? The Rage 128 Ultra is an OEM only version, but which one is it really? What did ATi do to it to make it not work with the reference drivers, forcing owners to use only the often out-dated OEM supplied drivers? Is it possible to hack the reference drivers to make them work with this chip? Dell and Gateway used the Rage 128 Ultra in many models of desktop PCs. Bizzybody (talk) 09:45, 24 April 2012 (UTC) ## Missing 2012 APUs, Embedded, FirePro Missing are many of the AMD Fusion APUs: HD7400D, HD7500D, HD7600D, HD7300, HD7500G, HD7600G, FirePro APU Missing are Embedded GPUs, Brazoz APUs, Trinity APUs Missing or wrong/preliminary data (e.g. display outputs of W9000) for W5000, W7000, W8000, W9000, W600. Ianteraf (talk) 13:10, 9 August 2012 (UTC) Strike-trough are done. Ianteraf (talk) 06:56, 29 September 2012 (UTC) ## Removed links to http://www.techpowerup.com/ and added "facts" tags Matthew Anthony Smith recently inserted a large amount of links to the http://www.techpowerup.com/ site into the table headers of many of the GPUs. I removed them as part of a quality assurance / cleanup effort which unfortunately became necessary after many controversial edits (in various articles) by this user. • These links just repeat the contents provided here already and therefore add no extra value to readers of this article. Also, they don't provide any information, which would not be available in many other places as well. • Wikipedia policies such as WP:EL restrict our usage of external links to certain, well-chosen cases. External links should be of particularly high quality. Links to forums and social media platforms are not normally allowed due to their short-lived nature and their typically low quality and their lack of editorial contents. Therefore, the inserted links to http://www.techpowerup.com/ in the table headers do not qualify as reliable reference, they do not even name sources or authors/editors, so this is simply nothing we can count on. • According to the Wikipedia Manual of Style, direct (or piped) links to external sources (as they were still common many years ago) are deprecated in article space for a long while and therefore should be avoided, in particular in headers. If we need to link to other sites, we should do it inside of references and use proper syntax. Alternatively, we could add them to the optional external links section, however, in this case, a single link to the home page of the database would be enough and we don't need dozens of individual links. See WP:LINK. Personally, I think, we don't need any of these links at all, but if you think a link is useful, I suggest to add a single link to the database under "External links" again. Also, links to reliable references (as per WP policies) are acceptable as well inside the table if we use proper syntax. Finally a note on the various "facts" templates I added to some of the table values. I did not want to blindly revert all the potentially problematic edits in one go, but found various table values or their semantics changed by Matthew Anthony Smith without any edit summary. Some values were simply changed, in some cases, footnotes were removed and in many cases lists of values and ranges were converted to look the same. I started to flag these changes in order to make readers aware of them (but there are many more). They need to be carefully checked by someone using a reliable reference and can be removed afterwards, ideally by providing the reference at the same time as well. Thanks. --Matthiaspaul (talk) 22:33, 13 September 2012 (UTC) ## Please sort all values logically, starting with the lowest ! Dear editors, when you enter new data, please sort all values by size/performance, starting with the lowest value or worst equipment, then citing the higher values or better equipment. All in logical order. Some entries are a huge mess, since different values of one category are not sorted by size/performance. Examples: Wrong: "RAM: GDDR4, DDR2, GDDR3" Wrong: "RAM: DDR2, GDDR4, GDDR3" Right: "RAM: DDR2, GDDR3, GDDR4" Thanks for your consideration. -- Alexey Topol (talk) 06:55, 30 September 2012 (UTC) ## Radeon HD 8470 (624 GFLOPS) vs. Radeon HD 8570 (560 GFLOPS) Radeon HD 8470 is an Turks PRO and simliar to Radeon HD 7570 and Radeon HD 6570 - so why should an Radeon HD 8570 with an HIGHER core cycle and BETTER bus interface and same bus width only have an output of 560 GFLOPS. FLOPS are calculated by ${\displaystyle GFLOPS=cores\times clock\times {\frac {FLOPs}{cycle}}}$ (but i can't do that). Can somebody please look after this miracle (and correct the numbers). Thanks -- 80.245.147.81 (talk) 14:45, 4 February 2013 (UTC) this list is a rough draft, when I made it, it was to just give people the clocks and name, all of the calculated specs are either carried over or wrong, feel free to correct it. Matthew Smith (talk) 18:46, 4 February 2013 (UTC) Ask AMD those performance Numbers of the 8570 & 8670 come straight from there lips there the only 2 on the list that are confirmed the source is here [5] Perhaps they've added something that helps improve performance & we might need a new calcuation for the GCN2 cards? — Preceding unsigned comment added by 58.178.251.175 (talk) 05:06, 5 February 2013 (UTC) HD 8470 - 25% more cores HD 8570 - 12.3% more clock There's your answer 25% more cores clocked at 12.3% slower still yields a gain. All this had-wringing and nobody has pointed out that the units in the "definitive" formula above are not self-consistent. 87.112.175.80 (talk) 06:50, 12 February 2013 (UTC) ## Unannounced/Rumoured products The Radeon HD8000 section in particular is (and for a long time has been) full of unannounced products with rumoured specifications and prices (with "references" to rumour sites, as if that somehow legitimises it). When did Wikipedia become a dumping ground for rumour and speculation? 203.45.39.201 (talk) 05:14, 11 February 2013 (UTC) There's actually a long history of this happening & it's usually right on the money. Most of the time there actually leaks not rumors. — Preceding unsigned comment added by 211.26.171.159 (talk) 12:31, 1 April 2013 (UTC) ## HD 2xxx to HD 6xxx shader count In the following article (http://www.anandtech.com/show/4455/amds-graphics-core-next-preview-amd-architects-for-compute) is explained that, in the architectures used up to Radeon HD 6xxx, the stream processors were VLIW (5, then 4). Each one comprised of 4 FPUs, 1 SFU (gone in the HD 69xx) and 1 branch unit. So the numbers present in the columns of the shaders in the tables are wrong. Either change them to simply FPUs or put the actual numbers of shaders (VLIW processors). This would explain why AMD's flagship chips have always been smaller than NVIDIA's ones and why, in contrast to higher theoretical peak performance (only achievable through independent float operations, not available in real world usage), NVIDIA's high-end chips have always been faster. — Preceding unsigned comment added by 93.38.171.227 (talk) 08:52, 19 February 2013 (UTC) Those numbers come from AMD themselves. Last poster, those were supposed to be tides(~), not dashes(-) to sign your post(at the end). FYI. Now for some clarification. Nvidia's high end chips have NOT always been faster(geforce fx5900/5950 vs radeon 9800, geforce 6800 vs radeon x800/x850, geforce 7800/7900/7950 vs radeon x1800/x1900/x1950, gtx680 vs hd7970 ghz editon and gtx690 vs hd7990). Nvidia chips have been bigger in the past for several reasons, but not because each shader was much more individually capable(well, more effecient, but not more functional units). Larger sizes were due to various things such as the fully scalar FPUS requiring more in the way of extra data routing versus the simpler vliw5/vliw4 setups of AMD, more rops/wider memory interface, and older process tech(90nm vs 80nm, 65nm vs 55nm). The nvidia chips were faster at times due to faster shader clocks(2x-2.5x core clock due to simple single FPUs versus more complex vliw5/vliw4 in AMD. It seems like I just contradicted myself, but the small single FPUs of nvidia could run at high clocks because of their small size while the cluster in amd were limited in that regard. The flow of data in and out of amd cluster had less of an impact on overall transistor count than what was required to get the level of granular computing capable on nvidia products)as well as the previously mentioned more efficient architecture and additional rops/larger memory paths. Efficiency may be in question since the high clock speeds on the older shaders had a negative impact on power consumption vs amd parts at 55nm-40nm. Also some lower end parts on 80nm/65nm nvidia vs 65nm/55nm amd. With the latest chips from the 2 rivals, nvidia has gotten a lot more like amd(massively more shaders but at core clock and fewer rops/narrower memory interface) and amd has gotten more like nvidia(revamped shader clusters are now four GROUPS OF 16 SHADERS like the original g80/g90 vs the 16 groups of 4 in Cayman and a wider memory interface). IMO, amd would have been better off with the same old 256bit memory interface(see how well hd7870 does compared to hd7950). Then tahiti would have been smaller, cheaper to produce, and less power hungry(like gk104).Jtenorj (talk) 21:23, 11 March 2013 (UTC) ## 7970 launch date Wasn't launched on Dec 21st 2011 instead of Jan 9, 2012? http://www.techpowerup.com/reviews/AMD/HD_7970/ — Preceding unsigned comment added by 193.109.40.21 (talk) 13:31, 19 March 2013 (UTC) Paper Launch - December 22nd, 2011 Retail availability - January 9, 2012 76.118.213.137 (talk) 02:21, 18 April 2013 (UTC) ## HD 63xx Codename columns I'm pretty sure that the HD 6320 is a Wrestler architecture (at least that's what it's telling my os. And the Blog article linked for the HD 6310 is explicitly stating that its codename is Zacate (maybe the columns switched?). — Preceding unsigned comment added by 77.4.109.210 (talk) 02:27, 13 June 2014 (UTC) ## FirePro W10000 The entry for the FirePro W10000 looks a bit off: it's incredibly unlikely, as it's basically an AMD version of the Titan/GK110, despite the fact that AMD has denied multiple times that such a GPU is being made; not to mention the lack of so much as a credible rumor to prove its existance, at least that I've seen. Interestingly, its listing is quite similar to what was previously -and incorrectly- listed as the HD 7990 Malta, albeit with a 384-bit GDDR6 interface, rather than the false Malta's 512-bit GDDR5. Until more credible information is available on this card, I feel that it should be removed from the listing. GungnirInd (talk) 23:48, 28 April 2013 (UTC) I blame http://www.techpowerup.com/gpudb/2366/radeon-hd-9970.html (Atlthough that says it was based on speculation don't know where they got it from) Although that has it listed as a Volcanic Island card. Not a 7000 or 8000 series by the time it comes out Nvidia will have a shiny New card that's better then Titan. (Maxwell) GDDR6 has been rumoured for a very long time. The specs are almost dead on for someone who slapped 3 7790s together apart from the changed memory. I guess it's plausible but one would hope that if it is a Volcanic Island card AMD would be shooting higher then that for them as well. I'll remove it for now we can always readd it later if it dose come out. — Preceding unsigned comment added by 210.50.139.54 (talk) 17:40, 4 May 2013 (UTC) ## Table of "Southern Islands (HD 7xxx) Series" at bottom is a little bent out of shape I'd just like to say that the bottom indication for the contents of the table (Model \| Launch \| Codename \| Fab) at the section "Southern Islands (HD 7xxx) Series" is not aligned right. Just the bottom indication, though. At the top the indication is right, at the bottom something appears to have gone askew. I tried to set it right myself, but can't seem to figure out how to adjust a wikitable. Perhaps someone who's a moderator can make an adjustment. — Preceding unsigned comment added by 82.169.167.15 (talk) 20:45, 9 May 2013 (UTC) ## Removal of OEM Shouldn't we remove the OEM cards from the list, as they are just re-branding of the previous generation? Jørgen88 (talk) 03:20, 14 May 2013 (UTC) I disagree the guys who buy OEM products probually arn't all the clued in to be able to make that connection. Instead I think that it should be listed in the notes that it is re-branded. — Preceding unsigned comment added by 211.26.171.190 (talk) 03:16, 22 May 2013 (UTC) For completeness I think it makes sense to keep the OEM models in the tables. Anyone referring to this page in the future, wondering about the details of the card they just pulled out of an old desktop, might want to easily locate the product by name. (Rather than referring to the references/comments to find out it was a rebadge of some older processor.) --ParoXoN (talk) 21:01, 16 September 2014 (UTC) ## Title The following discussion is an archived discussion of the proposal. Please do not modify it. Subsequent comments should be made in a new section on the talk page. No further edits should be made to this section. The result of the proposal was moved. --BDD (talk) 18:01, 31 January 2014 (UTC) Comparison of AMD graphics processing unitsList of AMD graphics processing units – This is one of several articles that I think are all mistitled for the same reasons. Please see the centralized discussion at WikiProject_Computing. Someone not using his real name (talk) 16:37, 13 January 2014 (UTC) • Procedural close discussion spread over 4 talk pages. If you want a multimove discussion, please use the appropriate template and close this discussion, if you want separate discussions, please close your centralized discussion -- 70.50.148.122 (talk) 06:55, 14 January 2014 (UTC) The above discussion is preserved as an archive of the proposal. Please do not modify it. Subsequent comments should be made in a new section on this talk page. No further edits should be made to this section. ## April 14, 2014 edits; R100 & R200 tables. I did some editing on the tables for R100 and R200 cards to get each row to only take up one line. Making the table compacter and easier to read, I hope. Did a bunch of smaller edits because I'm not that good in messing with purely "code", I need to see the results in between. NitroX infinity (talk) 20:05, 14 April 2014 (UTC) Edit; apparently, I forgot, Wikipedia tables adjust to screen-width. It looks good on my 1920 pixels, less pixels and it's 2 lines for rows again. Oh well, I'll let the regulars decide if they like it. NitroX infinity (talk) 20:09, 14 April 2014 (UTC) ## AMD Rx 3xx The rumor mill is kicking into high gear now, might as well get a jump on creating a conservative section, these are not WP:RS by any measure And some older sourcing The following is WP:OR, but given it's based off physical limitations, I believe this would fall under WP:CS • Neither the PCI-E 3.1 nor 4.0 specifications have been released as of time of writing. It is safe to assume they are not included on this generation of GPU • The DX12.1 specification hasn't been released nor mentioned at all by Microsoft — Preceding unsigned comment added by 67.188.142.154 (talk) 08:39, 11 November 2014 (UTC) Should we have this table up yet? There is hardly any leaked information regarding these GPUs. F0rteOC (talk) 02:13, 12 November 2014 (UTC) You could take it down if you like, it'll just go back up in another 2-3 months. I figured might as well put the framework up and update it as more leaks. 67.188.142.154 (talk) 07:22, 12 November 2014 (UTC) Hell I just checked my newsfeeds, another leak, but it's rumor tier again: http://wccftech.com/amd-fiji-r9-390x-specs-leak/ As I said, the mill is getting in high gear; I'm not comfortable adding these to the article, but the frame should be up and ready for when something credible hits. 67.188.142.154 (talk) 07:30, 12 November 2014 (UTC) The table in it's current form makes no sense. If the Trinidad chip is a rebrand of Curaçao or Pitcairn, why did it almost dupoble in size, go from 1280 "cores" to 2048, and otherwise have specs of a full Tonga? I think the best thing to do is fill the table with TBA, and change the fabrication process to 28nm, as 20nm has been confirmed again and again that won't fit the needs nor yields of GPUs. [6] [7] What that is known is that R9 390(x) is Fiji. There are some rumours that say there will be rebrands, and some that say there won't be. What that is likely is that Tonga will se new life, perhaps as an R7 370, with a Hawaii-like design running GCN 1.2 or newer being very likely. 178.17.146.218 (talk) 23:33, 28 March 2015 (UTC) ## Table of contents Who turned the table of contents into an unusable piece of crap? What was wrong with the previous layout? Could jump directly to a certain gen and that was all fine and dandy. Kehool (talk) 13:05, 26 February 2015 (UTC) Gotta agree. Not a huge fan of Wiki-asd-97's edit, however it was getting quite long. A cleanup is kinda needed. Maybe this page should be broken up into three separate pages (Desktop, Mobile, and Workstation). It would definitely help with page loading times. Charwinger21 (talk) 03:39, 4 March 2015 (UTC) The new TOC is a nightmare for navigation. 180.200.156.33 (talk) 00:35, 7 March 2015 (UTC) What the hell... it's gotten even worse now. Where are the upcoming GPUs? And why are GPUs all over the place but not where they're supposed to be? Why are 7790, 8870 and R9 GPUs under the same category? Could someone please revert this to the way it was before it all went to shit? I'd do it myself but I'd probably just make it worse.. at least I have the good sense to realize this and not to try and "improve" the article unlike certain other people... Much obliged. Kehool (talk) 10:07, 17 March 2015 (UTC) ## Radeon R7 250E There have been references to a Radeon R7 250E that is a rebadge of the Cape Verde PRO-based Radeon HD 7750, which imporves upon the performance of the Oland XT-based R7 250. Though I've been unable to find a direct reference from AMD, XFX sells a card with this reference: http://xfxforce.com/en-us/products/amd-radeon-r7-series/amd-radeon-r7-250e-core-edition-r7-250e-znp4 It must be added as well, that other brands omit the E and sell it as a R7 250, as can be seen from this article: http://www.techpowerup.com/gpudb/2559/radeon-r7-250e.html --Richard Wolf VI (talk) 07:17, 28 March 2015 (UTC) ## 370 possible error The 300 series may have an error. The R9 370 is listed as 1536 SP but a rebrand of Curaçao/Pitcairn. This cannot be the case Curaçao/Pitcairn are maximum 1280SP. The 1536SP card from the HD7000 series (7870XT) is in fact a cut down Tahiti. There is no 1536SP Curaçao/Pitcairn. — Preceding unsigned comment added by 139.218.73.51 (talk) 13:52, 9 April 2015 (UTC) ## AMD Rx 3xx OEM? Reverted 125.254.43.66 because I don't think this is WP:OR, however, you could argue it's WP:SYNTH Proposal: AMD Rx 3xx is an OEM series, akin to the HD 8000's; the next retail series is the Rx 4xx. Alternatively we'll need an OEM Rx 3xx section. Evidence #1: The above links contain information on "300 Series specs", a further look at the specs shows they all have similar featuresets and core counts, but a few differences: • R9 380 = R9 285; same clockrate, same bandwidth • R9 370 = R7 265; same bandwidth; clockrate is 50 mhz faster • R9 360 = R7 260; clockrate is 50 mhz faster, bandwidth is 125mhz faster • R7 350 = R7 250; same clockrate; 25 mhz slower memory • R7 340 = R7 250; much much lower clockrate, 25 mhz slower memory; core of a 250X, clocks of a 250 • R5 340 = R7 250; much lower clockrate; note there appears to be a typo, it says 6 CU but 320 shaders, 6CU corresponds to 384 • R5 330 = R7 240; higher clockrate Evidence #2: The URL contains /oem/ as a directory in it, infact on the http://www.amd.com/en-us/products/graphics/desktop page, if you click "AMD Radeon™ HD 8000 series graphics (OEM)" link, it takes you to the first link above. Evidence #3: The naming convention AMD chose here contradicts all prior known information, and doesn't make sense in the greater picture. If the "390" is meant to be a chip with much higher performance than the 290, then the resulting performance gap and the "390" and 380 would be massive. It'd make more sense if the 380 specified were sold as a "370", but this isn't the case. Evidence #4: There is precedence, the HD 8000's were an OEM only series released just prior to the Rx 2xx series. Evidence #5: News sites are picking up on it: I'll be expanding the Rx 3xx section by adding a table after I post this. 67.188.142.154 (talk) 05:10, 7 May 2015 (UTC) ## GCN 1.x @Visite fortuitement prolongée: Please show me an official document from AMD where the GCN 1.x denomination is used. It is a fiction by you guys. --188.106.27.177 (talk) 16:43, 11 June 2015 (UTC) "Please show me an official document from AMD where the GCN 1.x denomination is used." →‎ I can not. Visite fortuitement prolongée (talk) 20:09, 11 June 2015 (UTC) "It is a fiction by you guys." →‎ No, it come from reliable sources such anandtech.com ([3], [4], [5]) and hardware.fr ([6], [7], [8]). Visite fortuitement prolongée (talk) 20:09, 11 June 2015 (UTC) Since when are anandtech.com and hardware.fr owned by AMD? Both belong to the "you guys" camp. NOTHING OFFICIAL! --188.106.27.177 (talk) 13:20, 15 June 2015 (UTC) "Since when are anandtech.com and hardware.fr owned by AMD?" →‎ I don't know. Actually, I did not know that anandtech.com and hardware.fr are owned by AMD. Tell me more about that. Visite fortuitement prolongée (talk) 19:12, 16 June 2015 (UTC) GCN: See mesaDriver Features and Hardware Levels of Radeon https://www.x.org/wiki/RadeonFeature/#index6h2 — Preceding unsigned comment added by 91.67.31.234 (talk) 00:19, 26 May 2016 (UTC) With new driver Crimson actual 16.5.3 only GCN-Cards are Supported for new Features like Vulkan 1.0. since Catalyst 15.7 also OpenGL 4.5 for all GCN-Cards available. All terascale2-Cards can OpenGL 4.4 with driver Catalyst 14.4+. — Preceding unsigned comment added by 91.67.31.234 (talk) 00:02, 26 May 2016 (UTC) See also AMD Documenta http://amd-dev.wpengine.netdna-cdn.com/wordpress/media/2013/07/AMD_GCN3_Instruction_Set_Architecture.pdf — Preceding unsigned comment added by 91.67.31.234 (talk) 00:29, 26 May 2016 (UTC) ## Rx 300 table width This table has become far too big! It exceeds the page width at 100% on a 1920-pixel wide monitor. 125.254.43.66 (talk) 03:15, 24 June 2015 (UTC) I suggest to delete the GFlops/W column. Visite fortuitement prolongée (talk) 19:38, 24 June 2015 (UTC) I should add that it doesn't "unwrap" the cell contents until you zoom out to 70% on the same 1920-pixel wide monitor. Many of the other tables in the article are similar. There is far too much information in them for a single table. 125.254.43.66 (talk) 02:11, 25 June 2015 (UTC) It should hopefully be better now. I think we can drop transistor counts and die size too if we wanted since they don't seem too interesting(?)--Kyousuke.k (talk) 05:51, 11 July 2015 (UTC) 1. Thank you. 2. The Fab process, Transistors count and Die size columns are more about the chip itself than the video card, but since the pages are named "List of ... graphics processing units" and since the tables are not too wide now, I suggest to keep them. Visite fortuitement prolongée (talk) 19:24, 11 July 2015 (UTC) ## 2015-06 memory size I suggest to move the memory size column in the Memory column. Visite fortuitement prolongée (talk) 19:38, 24 June 2015 (UTC) Done. Visite fortuitement prolongée (talk) 20:08, 7 July 2015 (UTC) ## External links modified Hello fellow Wikipedians, I have just modified one external link on List of AMD graphics processing units. Please take a moment to review my edit. If you have any questions, or need the bot to ignore the links, or the page altogether, please visit this simple FaQ for additional information. I made the following changes: When you have finished reviewing my changes, please set the checked parameter below to true or failed to let others know (documentation at {{Sourcecheck}}). Archived sources still need to be checked Cheers.—cyberbot IITalk to my owner:Online 07:11, 1 April 2016 (UTC) ## R400 GCN version This pages calls newest generation (polaris) a GCN 1.3, while Graphic Core Next article here on wiki pages and most reviewers call it a GCN 2.0 or GCN 4, I think that it should be changed here too.--Robin WH (talk) 10:30, 27 June 2016 (UTC) There is no clear, consistency of what version number the 4th generation of the GCN is. Even AMD themselves list it as simple "GCN 4th generation". So until we get a clear version number, I say leave it as is. EDIT: After reading the talk page on the GCN article, the whole version numbering was completely created by Anandtech & is not use anywhere officially. I will go through & update the templates to reflect this info. #FF9600 talk 19:36, 27 June 2016 (UTC) ## For consistancy AMDs offical site says RX480 is upto 5.8TFLOPs but the number here says 5100. I did alot of the earlier entries myself using boost clock for the performance calculations since boost clock became a thing. It might be an idea to either provide both sets of performance numbers or to update either one to just us base or boost. More of a consistancy thing probually dosn't matter to much. — Preceding unsigned comment added by 58.179.111.133 (talk) 12:47, 29 July 2016 (UTC)	2016-09-28 15:40:32	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 1, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4420333802700043, "perplexity": 4456.870928829109}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.3, "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-2016-40/segments/1474738661555.40/warc/CC-MAIN-20160924173741-00111-ip-10-143-35-109.ec2.internal.warc.gz"}
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COCOA acetyl-CoA There are carbon atoms (C) present in citrate. Step 2 coeruryme There is a total of molecule(s) of FADH2 produced during the entire series of reactions (steps 1 and 2). GGGGGG G... ##### 6.(20 pts) Coal gas is produced at gasworks plant Pollutants in the gas are removed by scrubbers_ which become less and less effective as time (1n months] gocs On_ Measurements made at the start of each month showing the rate R (in tons month) at which pollutants are escaping in the gas are as follows:4(12 pts) Compute an overestimate and an underestimate of the total amount of pollutants that escaped during the first eight months (Inelude unts: )6.(8 pts) What (fewest) number of equally-spaeed 6.(20 pts) Coal gas is produced at gasworks plant Pollutants in the gas are removed by scrubbers_ which become less and less effective as time (1n months] gocs On_ Measurements made at the start of each month showing the rate R (in tons month) at which pollutants are escaping in the gas are as follo... ##### If the Ksp for AgNO2 is 8.5×10−4, and the silver ion concentration in solution is 0.0050 M, what does the nitrite concen... If the Ksp for AgNO2 is 8.5×10−4, and the silver ion concentration in solution is 0.0050 M, what does the nitrite concentration need to be (in units of moles per liter) for a precipitate to occur? ?M... ##### Calculate the wavelength (in nm) and the Cnergy (in kJ/mol) of a X-ray photon with an energy of 2.0 X1O-* eV a infrared photon associated with the 2900 em-1 â‚¬ H stretching transition for An aliphatic crhon. the photon associated with a NMR transition at 250 MHz Calculate the wavelength (in nm) and the Cnergy (in kJ/mol) of a X-ray photon with an energy of 2.0 X1O-* eV a infrared photon associated with the 2900 em-1 â‚¬ H stretching transition for An aliphatic crhon. the photon associated with a NMR transition at 250 MHz... ##### Which of the following compounds is an amine? phenol b. anisole toluene aniline Which of the following compounds is an amine? phenol b. anisole toluene aniline... ##### The time a group of high school students arrive home from school each day was found to be normally distributed The time a group of high school students arrive home from school each day was found to be normally distributed. The mean time was 3:15pm and the times had a standard deviation of 15 minutes. What is the probability that a student chosen at random arrives home from school before 2:30pm?what formula w... ##### HC IHIUT CVCHIS OPTy w LCWS OHUIGICI, PUUIL OLLULIT HIT, IUI UIC ICOI OLLUTY CHIVU. 1.... HC IHIUT CVCHIS OPTy w LCWS OHUIGICI, PUUIL OLLULIT HIT, IUI UIC ICOI OLLUTY CHIVU. 1. Performed $70,000 of services for clients on account 2. Performed$40,000 of services for cash. 3. Incurred $36,000 of other operating expenses on account 4. Paid$10,000 cash to an employee'for salary 5. Coll... ##### FOSTLAB: ElectrolysisFurutut5 Law2 AlJ+ (sqPb (Al (41Pb" (-Write the half rcactions for the renction shown Mnat,b) Cetermine E"cI for the reuction:What produced in the spontancous direction?A rechargeable 9V battery gupply about 500 milliamps for one hour before it needs to be recharged. How many moles of chlorine gas can be produced from an electrolytic cell using aqueous sodium chloride with this battery? FOSTLAB: Electrolysis Furutut5 Law 2 AlJ+ (sq Pb ( Al (41 Pb" (- Write the half rcactions for the renction shown Mnat, b) Cetermine E"cI for the reuction: What produced in the spontancous direction? A rechargeable 9V battery gupply about 500 milliamps for one hour before it needs to be rec... ##### Assume our RDD has the following content, k101 "value-001" k200 "value-002" k200 "value-003" k103 "value-004" k200... Assume our RDD has the following content, k101 "value-001" k200 "value-002" k200 "value-003" k103 "value-004" k200 "value-005" k101 "value-006" k103 "value-007" where on each line, the first part is a key (for example, k101), the second part is... ##### Twice the difference of a number and 7 equals 6 Twice the difference of a number and 7 equals 6... ##### L Review Part A A car starts from rest on a curve with a radius of... l Review Part A A car starts from rest on a curve with a radius of 100 m and tangential acceleration of 1.1 m/s Through what angle will the car have traveled when the magnitude of its total acceleration is 24 m/s?2 Express your answer to two significant figures and include the appropriate units. 61 ... ##### (10 pts) 1. Provide the IUPAC or common name for the following compounds (don't forget R, S or E, Zassignments where applicable):Name_NameNameNameCommon name (10 pts) 1. Provide the IUPAC or common name for the following compounds (don't forget R, S or E, Zassignments where applicable): Name_ Name Name Name Common name... ##### Use Eqs. (6) through (9) to find the sums in Problems 19 through 28. $\sum_{i=1}^{6}\left(i^{3}-i^{2}\right)$ Use Eqs. (6) through (9) to find the sums in Problems 19 through 28. $\sum_{i=1}^{6}\left(i^{3}-i^{2}\right)$... ##### A sample of the daily iron intake of 22 women had mean 14.6 mg and standard... A sample of the daily iron intake of 22 women had mean 14.6 mg and standard deviation 4.2 mg. Assuming daily iron intake is normally distributed, a 90% confidence interval for the mean daily iron intake of all women is about (a) [13.4, 15.8] (b) [12.1, 17.1] (c) [14.5, 14.7] (d) [13.1, 16.1] (e) [12... ##### Wrila 28 the sum 0l tNo vuctor: onotin Span and one In Span {Uz,U3,U4} Assume that {U+ _Ua} Is an orthogonal basis for R* . HHH6Myeaan Integur or sImplilled Iracton Ior oach matnx olemont ) Wrila 28 the sum 0l tNo vuctor: onotin Span and one In Span {Uz,U3,U4} Assume that {U+ _Ua} Is an orthogonal basis for R* . HHH6 Myeaan Integur or sImplilled Iracton Ior oach matnx olemont )... ##### 119 of all Americans suffer from sleep apnea. A researcher suspects that a higher percentage of those who live in the inner city have sleep apnea. Of the 375 people from the inner city surveyed, 45 of them suffered from sleep apnea: What can be concluded at the level of = 'significance of a 0.10?For this study; we - should use z-test for population proportion b: The null and alternative hypotheses would be: Ho: 0.11 (please enter a decimal)(Please enter decimal)The test statistic [ ~v 0,648 119 of all Americans suffer from sleep apnea. A researcher suspects that a higher percentage of those who live in the inner city have sleep apnea. Of the 375 people from the inner city surveyed, 45 of them suffered from sleep apnea: What can be concluded at the level of = 'significance of a 0.1...	2022-05-21 22:15:33	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 43, "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": 0.6141772270202637, "perplexity": 9921.92577609274}, "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/1652662541747.38/warc/CC-MAIN-20220521205757-20220521235757-00686.warc.gz"}
https://www.hpmuseum.org/forum/showthread.php?mode=threaded&tid=7587&pid=66892	Could someone do a calculation on a real 71B? 01-16-2017, 11:14 AM (This post was last modified: 01-16-2017 11:17 AM by Thomas Okken.) Post: #5 Thomas Okken Senior Member Posts: 1,711 Joined: Feb 2014 RE: Could someone do a calculation on a real 71B? (01-16-2017 10:31 AM)EdS2 Wrote: It's more enlightening to ask for the sin of a number just below pi, because then you get more digits of pi. Neat! I never realized that. It works for the number just above pi, too; you just get the ten's complement of those extra digits: Code: 10.00000000000E-10 -4.10206761537E-10 ------------------ 5.89793238463E-10 I guess the point is that the calculator performs argument reduction using an extended-precision approximation of pi. Next question, for extra credit: how many digits? « Next Oldest \| Next Newest » Messages In This Thread Could someone do a calculation on a real 71B? - zeno333 - 01-15-2017, 08:59 AM RE: Could someone do a calculation on a real 71B? - Gerson W. Barbosa - 01-15-2017, 09:24 AM RE: Could someone do a calculation on a real 71B? - Didier Lachieze - 01-15-2017, 09:26 AM RE: Could someone do a calculation on a real 71B? - EdS2 - 01-16-2017, 10:31 AM RE: Could someone do a calculation on a real 71B? - Thomas Okken - 01-16-2017 11:14 AM RE: Could someone do a calculation on a real 71B? - J-F Garnier - 01-16-2017, 01:18 PM RE: Could someone do a calculation on a real 71B? - Thomas Okken - 01-17-2017, 03:19 AM RE: Could someone do a calculation on a real 71B? - Gerson W. Barbosa - 01-16-2017, 01:40 PM RE: Could someone do a calculation on a real 71B? - KeithB - 01-17-2017, 12:02 AM User(s) browsing this thread: 1 Guest(s)	2022-05-18 08:58:03	{"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": 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": 0.5000890493392944, "perplexity": 5951.181946089473}, "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-2022-21/segments/1652662521883.7/warc/CC-MAIN-20220518083841-20220518113841-00399.warc.gz"}
https://kerodon.net/tag/004R	# Kerodon $\Newextarrow{\xRightarrow}{5,5}{0x21D2}$ $\newcommand\empty{}$ Definition 1.3.6.1. Let $\operatorname{\mathcal{C}}$ be an $\infty$-category and let $f: X \rightarrow Y$ be a morphism of $\operatorname{\mathcal{C}}$. We will say that $f$ is an isomorphism if the homotopy class $[f]$ is an isomorphism in the homotopy category $\mathrm{h} \mathit{\operatorname{\mathcal{C}}}$. We will say that two objects $X,Y \in \operatorname{\mathcal{C}}$ are isomorphic if there exists an isomorphism from $X$ to $Y$ (that is, if $X$ and $Y$ are isomorphic as objects of the homotopy category $\mathrm{h} \mathit{\operatorname{\mathcal{C}}}$).	2022-05-23 15:19:12	{"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": 0.9991626739501953, "perplexity": 30.542368851433924}, "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-2022-21/segments/1652662558030.43/warc/CC-MAIN-20220523132100-20220523162100-00727.warc.gz"}
http://conceptmap.cfapps.io/wikipage?lang=en&name=Husimi_Q_representation	# Husimi Q representation The Husimi Q representation, introduced by Kôdi Husimi in 1940,[1] is a quasiprobability distribution commonly used in quantum mechanics[2] to represent the phase space distribution of a quantum state such as light in the phase space formulation.[3] It is used in the field of quantum optics[4] and particularly for tomographic purposes. It is also applied in the study of quantum effects in superconductors.[5] Husimi distribution of the squeezed coherent state Husimi distribution function of three coherent states merged ## Definition and properties The Husimi Q distribution (called Q-function in the context of quantum optics) is one of the simplest distributions of quasiprobability in phase space. It is constructed in such a way that observables written in anti-normal order follow the optical equivalence theorem. This means that it is essentially the density matrix put into normal order. This makes it relatively easy to calculate compared to other quasiprobability distributions through the formula ${\displaystyle Q(\alpha )={\frac {1}{\pi }}\langle \alpha \|{\hat {\rho }}\|\alpha \rangle ,}$ which is effectively a trace of the density matrix over the basis of coherent states ${\displaystyle \{\|\alpha \rangle \}}$ . It produces a pictorial representation of the state ρ to illustrate several of its mathematical properties.[6] Its relative ease of calculation is related to its smoothness compared to other quasiprobability distributions. In fact, it can be understood as the Weierstrass transform of the Wigner quasiprobability distribution, i.e. a smoothing by a Gaussian filter, ${\displaystyle Q(\alpha )={\frac {2}{\pi }}\int W(\beta )e^{-2\|\alpha -\beta \|^{2}}\,d^{2}\beta .}$ Such Gauss transforms being essentially invertible in the Fourier domain via the convolution theorem, Q provides an equivalent description of quantum mechanics in phase space to that furnished by the Wigner distribution. Alternatively, one can compute the Husimi Q distribution by taking the Segal–Bargmann transform of the wave function and then computing the associated probability density. Q is normalized to unity, ${\displaystyle \int Q(\alpha )\,d\alpha ^{2}=1}$ and is non-negative definite[7] and bounded: ${\displaystyle 0\leq Q(\alpha )\leq {\frac {1}{\pi }}.}$ Despite the fact that Q is non-negative definite and bounded like a standard joint probability distribution, this similarity may be misleading, because different coherent states are not orthogonal. Two different points α do not represent disjoint physical contingencies; thus, Q(α) does not represent the probability of mutually exclusive states, as needed in the third axiom of probability theory. Q may also be obtained by a different Weierstrass transform of the Glauber–Sudarshan P representation, ${\displaystyle Q(\alpha ,\alpha ^{})={\frac {1}{\pi }}\int P(\beta ,\beta ^{})e^{-\|\alpha -\beta \|^{2}}\,d^{2}\beta ,}$ given ${\displaystyle {\hat {\rho }}=\int P(\beta ,\beta ^{*})\|{\beta }\rangle \langle {\beta }\|\,d^{2}\beta }$ , and the standard inner product of coherent states.	2020-09-24 03:51:29	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 7, "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": 0.907495379447937, "perplexity": 536.8313689302022}, "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/1600400213454.52/warc/CC-MAIN-20200924034208-20200924064208-00542.warc.gz"}
https://www.physicsforums.com/threads/proving-differentiability-of-a-piece-wise-function.550873/	# Proving Differentiability of a Piece-wise Function 1. Nov 15, 2011 ### UziStuNNa 1. Suppose f(x)=0 if x is irrational, and f(x)=x if x is rational. Is f differentiable at x=0? 2. the derivative= lim[h->0] [f(a+h)-f(a)]/h 3. I don't really know how to start, but I do know that between any two real numbers, there exists a rational and irrational number. So I'm guessing that has something to do with solving for the answer. f'(0)= lim[h->0] f(h)/h 2. Nov 15, 2011 ### Dick Ok, so there are rational numbers as close as you want to 0. And irrational numbers. Pick h to be rational. What's f(h)/h? Same question for irrational. 3. Nov 15, 2011 ### UziStuNNa For rational, f'(0)= lim[h->0] f(h)/h= h/h=1 For irrational, f'(0)= lim[h->0] f(h)/h= 0/h=0 Therefore, the two limits do not equal each other, meaning that f'(0) does not exist? Was it that easy? 4. Nov 15, 2011 ### Dick Yes. If you are clear why you can pick h values arbitrarily close to zero that are both rational and irrational. And I think you are. 5. Nov 15, 2011 ### UziStuNNa So its like doing the left-hand and right-hand limits, except we are using the piece-wise function to our advantage knowing that x can be rational and irrational, and since there are an infinite number of rational and irrational numbers approaching zero, those are our 'left' and 'right'. Thanks a lot. 6. Nov 15, 2011 ### Dick Exactly. Very welcome.	2018-01-18 20:18:13	{"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": 0.8337186574935913, "perplexity": 1341.2101822133127}, "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/1516084887600.12/warc/CC-MAIN-20180118190921-20180118210921-00020.warc.gz"}
https://www.physicsforums.com/threads/neutron-flux-profile-in-a-spherical-moderator.723560/	# Neutron Flux Profile in a Spherical Moderator #### Israakaizzy Hello People I need help with the following assignment: It states: Consider an ideal moderator with zero absorption cross section, Ʃa = 0, and a diffusion coefficient, D, which has a spherical shape with an extrapolated radius, R. If neutron sources emitting S neutrons/cm3sec are distributed uniformly throughout the moderator, the steady neutron diffusion equation is given by, D∇2$\phi$ -Ʃa$\phi$=-S a) Simplify the above neutron diffusion equation for this moderator in spherical coordinates and state the appropriate boundary conditions. By solving the simplified diffusion equation, obtain the neutron flux profile, $\phi$(r). I know I need to divided the neutron diffusion equation and cancel out the absorption cross section and end up with something like: 2$\phi$ = -S/D and the particular solution would be something like S/Ʃa but what's the general solution to: D∇2$\phi$ =0 in spherical coordinates? Related Nuclear Engineering News on Phys.org #### hmeier Israakaizzy, I think it should be something like Asinh(λ.r)/r + Bcosh(λ.r)/r, applying the border conditions B=0. hope it helps, Hernán #### Israakaizzy ok Just explain me what is $\lambda$ equal to? Is it 1/L ? #### hmeier Israakaizzy, You are right, it should be 1/L if Ʃa were different than 0. I did the maths for the homogeneous part: ∇$^{2}$$\phi$=0 saying that: $\phi$=$\frac{\widehat{\phi}}{r}$ The Lapplacian inspherical coordintates turns: ∇$^{2}$$\phi$=$\frac{∂^{2}\widehat{\phi}}{∂r^{2}}$ + $\frac{2}{r}$$\frac{∂\widehat{\phi}}{∂r}$ proposing an exponential solution: λ$^{2}$e$^{λr}$ + $\frac{2}{r}$λe$^{λr}$ = 0 So: λ= -$\frac{2}{r}$ and finally: $\phi$= $\frac{A}{r}$ + B Don´t forget that for the inhomogeneous part you have to use the Lapplacian in sphericals. Regards, Hernán ### 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 18:42:51	{"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": 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": 0.7757920622825623, "perplexity": 4304.4794449706005}, "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/1570986718918.77/warc/CC-MAIN-20191020183709-20191020211209-00161.warc.gz"}
https://jameswebbnews.com/2022/02/02/microwave-background-temperature-at-a-redshift-of-6-34-from-h2o-absorption/	Galaxies # Microwave background temperature at a redshift of 6.34 from H2O absorption ### NOEMA observations The target was observed in the 3-mm wavelength band 1 (rest-frame 400 μm) with NOEMA as part of project S20DA (Principal Investigators: D. A. Riechers, F. Walter). Three partially overlapping spectral setups were observed under good weather conditions between 26 July 2020 and 25 August 2020 with ten antennas in the most compact D configuration, using a bandwidth of 7.7 GHz (dual polarization) at 2-MHz spectral resolution per sideband. We also included previously published5 observations between 6 February 2012 and 31 May 2012 in the A and D configurations tuned to 110.128 and 113.819 GHz, respectively, and previously unpublished observations between 1 June 2012 and 4 June 2012 and on 10 July 2017 in the D configuration tuned to 78.544 and 101.819 GHz taken as part of projects V0BD, W058, and S17CC (Principal Investigator: D. A. Riechers), all using 3.6 GHz of bandwidth (dual polarization), yielding 21 observing runs in total. Nearby radio quasars were used for complex gain, bandpass and absolute flux calibration. The target was also observed in the 0.87-mm wavelength band 4 (rest-frame 122 μm) with NOEMA as part of project X0CC (Principal Investigator: D. A. Riechers). Observations were carried out during three observing runs with six antennas in the A and C configurations under good weather conditions between 4 December 2013 and 12 March 2015, with the band 4 receivers tuned to 335.5 GHz and using a bandwidth of 3.6 GHz (dual polarization). Nearby radio quasars were used for complex gain, bandpass and absolute flux calibration. The GILDAS package was used for data calibration and imaging. All 3-mm data were combined to a single visibility cube before imaging. Imaging was carried out with natural baseline weighting. The band 4 data were also imaged with Briggs robust weighting to increase the spatial resolution. A map of the continuum emission at the frequency of the H2O line was created by averaging the visibility data over a bandwidth of 2.04 GHz centred on the line. This range was chosen to avoid other lines in the bandpass. Continuum emission was subtracted from the H2O line cube in the visibility plane. Moment 0 images of the line absorption were created before and after continuum subtraction by integrating the signal over a bandwidth of 100 MHz, corresponding to 395 km s−1. The resulting r.m.s. noise levels are provided in Extended Data Fig. 2. We also make use of previously published5 rest-frame 158-μm NOEMA data, which were adopted without further modification. ### Line and continuum parameters The flux of the H2O(110–101) line was extracted by simultaneous Gaussian fitting of the line and continuum emission (including a linear term for the continuum) in the one-dimensional spectrum shown in Fig. 1, which was extracted from the image cube. The source is unresolved at the frequency of the H2O(110–101) line, such that the main uncertainties are due to the slope of the continuum emission and the appropriate fitting of other nearby lines, in particular, CO(5–4). The uncertainties in these parameters are part of the quoted uncertainties. We find a line peak flux of −818 ± 145 μJy at a line full width half maximum (FWHM) of 507 ± 111 km s−1, centred at a frequency of 75.8948 GHz (±46 km s−1; the calibration uncertainties on the line FWHM and centre frequency are negligible and that on the line peak flux is <10%—that is, minor compared with the measurement uncertainty). Given the rest frequency of the line of 556.9359877 GHz, this corresponds to a redshift of 6.3383, which is consistent with the systemic redshift of HFLS3 (z = 6.3335 and 6.3427 with uncertainties of ±14 and ±54 km s−1 at Gaussian FWHM of 243 ± 39 and 760 ± 152 km s−1, respectively, for the two velocity components detected in the 158-μm [CII] line)5. For comparison, the H2O(202–111) and H2O(211–202) emission lines in HFLS3 have FWHM of 805 ± 129 and 927 ± 330 km s−1, respectively5—that is, only marginally broader than the 110–101 line at the current measurement uncertainties. The continuum flux at the line frequency is 396 ± 15 μJy, corresponding to 48% ± 9% of the absorption-line flux (the relative flux calibration uncertainty between the line and continuum emission is negligible). We also measured the 335.5-GHz continuum flux by two-dimensional fitting to the continuum emission in the visibility plane. We find a flux of 33.9 ± 1.1 mJy, which agrees with previous lower-resolution observations at the same wavelength5. The major (minor) axis FWHM diameter of the source is 0.617 ± 0.074 arcsec (0.37 ± 0.20 arcsec). This yields the physical source size quoted in the main text at the redshift of HFLS3. ### Brightness temperature contrast The H2O(110–101) line leads to a decrement in continuum photons from the starburst and, as such, is observed as a lack of continuum emission at its frequency at the position of the starburst. It therefore appears as negative flux in an image where starburst continuum emission has been subtracted. In addition, (sub)millimetre-wavelength interferometric images reveal structure against a flat sky background defined by the large-scale CMB surface brightness, which the interferometer does not detect itself due to its limited spatial sampling. Therefore the fraction of the signal due to the decrement in CMB photons at the position of the starburst not only appears as negative flux without subtracting any further signal but it also corresponds to a lack of continuum emission at the line frequency in practice. As the mere presence of an absorption-line signal stronger than the measured continuum emission implies absorption against the CMB, this interpretation is not limited by uncertainties in the galaxy continuum flux or uncertainties in the absolute flux calibration. ### Other H2O transitions in HFLS3 Five H2O lines were previously detected towards HFLS3 (202–111, 211–202, 312–221, 312–303 and 321–312) and two additional lines were tentatively detected (413–404 and 422–413)5. The Jup = 3 transitions are due to ortho-H2O and all other transitions are due to para-H2O. All of these transitions appear in emission. Given the high critical densities of these transitions, our RADEX models cannot reproduce the strength of these lines as the same time as the observed ortho-H2O(110–101) absorption strength, which suggests that they emerge from different gas components. For reference, to reproduce the strength of the H2O(211–202) in Fig. 1 alone with collisional excitation, n(H2) = 2 × 107 cm−3 and Tkin = 200 K would be required, but the H2O(110–101) would no longer appear in absorption against the CMB if it were to emerge from the same gas component. This is consistent with the picture that the H2O absorption is due to a cold gas component along the line of sight to the warm gas that gives rise to the emission lines11. Observations of the para-H2O(111–000) ground state do not currently exist for HFLS3, but our models do not show this line in absorption towards the CMB. ### Origin of the lower and upper limits on TCMB Our models show that the lower limit on TCMB at a given redshift based on the observed H2O absorption is due to the minimum ‘seed’ level population due to the CMB black-body radiation field. To determine a conservative lower limit, we have calculated models with continuum sizes up to r108μm = 5 kpc (see Fig. 3b), corresponding to a +7.5σ deviation from the observed continuum size, and recorded the temperatures at which such weakly constrained models turn into absorption. We find that this results in a lower limit of TCMB > 7–8 K, independent of the model assumptions. This finding alone does not explain the existence of an upper limit in Fig. 3b. For a given size of the dust continuum emission, an increase in TCMB also requires an increase in Mdust to still reproduce the observed dust spectral energy distribution, which leads to an effective increase in the dust optical depth at a given wavelength. The result of a rising optical depth is that the grey-body spectrum between 538 and 108 μm increasingly resembles a black-body spectrum and, hence, a decrease in the H2O absorption against the CMB. This effect is responsible for the upper limit in allowed TCMB for a given dust continuum size and absorption strength. ### Uncertainties of TCMB measurements The uncertainties shown for the literature data in Fig. 4 are adopted from the literature sources without modification, and they typically represent the statistical uncertainties from the individual measurements or sample averages. Individual cluster measurements of the thermal SZ effect may be affected by dust associated with foreground galaxies or the Milky Way, the galaxy clusters or background galaxies that may be amplified by gravitational lensing, uncertainties in the reconstruction of the Compton-y parameter maps due to flux uncertainties, radio emission due to active galactic nuclei and/or relics, the kinetic and relativistic SZ effects, and general bandpass and calibration uncertainties17. Furthermore, uncertainties on the cluster geometry—and therefore line-of-sight travel distance of the CMB photons through the cluster—and on the temperature of the intra-cluster gas limit the precision of individual SZ measurements. Sample averages may also be affected by systematics in the stacking procedures. Individual data points deviate by up to at least two standard deviations from the trend, which may indicate residual uncertainties beyond the statistical error bars provided, such that the error bars shown in Fig. 4 are underestimated. The main source of uncertainty for the ultraviolet absorption-line-based measurements are due to the assumption of no collisional excitation, which is not taken into account in the statistical uncertainties shown in Fig. 4. Attempts to take this effect into account appear to suggest substantially larger uncertainties than indicated by individual error bars27 (Fig. 4). To expand on earlier estimates21, we have calculated RADEX models for typical Tkin, n(H) and column densities found from [CI] measurements in the diffuse interstellar medium34, which suggests that collisional excitation contributes to the predicted Tex of the lower fine-structure transition. Although we show the original unmodified data, the ultraviolet-based measurements are therefore subject to uncertainties due to model-dependent excitation corrections in addition to the statistical uncertainties. Furthermore, the fine-structure levels of tracers like the [CI] lines can be excited by ultraviolet excitation and following cascades. To constrain TCMB based on these measurements, the kinetic temperature, particle density and local ultraviolet radiation field must be known, and are typically determined based on tracers other than the species used to constrain TCMB. Also, some measurements are based on spectrally unresolved lines, which limits the precision of kinetic temperature measurements based on thermal broadening21. Owing to these uncertainties, the ultraviolet absorption-line-based measurements are probably consistent with the standard ΛCDM value, but they do not constitute a direct measurement of TCMB without notable further assumptions. For reference, the median TCMB/(1 + z) estimate based on the [CI] measurements alone (excluding upper limits) is 3.07 K, with a median absolute deviation of 0.09 K and a standard deviation of 0.31 K. Therefore the current sample median deviates from the ΛCDM value by about one standard deviation. A combination of the (uncorrected) measurements based on CO, [CI] and [CII] provides a median value of 2.84 K, with a median absolute deviation of 0.15 K and a standard deviation of 0.25 K. This highlights the importance of the corrections discussed above and in the literature and the value of measurements with systematic uncertainties that differ from this method to obtain a more complete picture. The main source of uncertainty of the H2O-based measurements, beyond the caveats stated in the line-excitation-modelling section, are the statistical uncertainties on the source size, the lack of a direct measurement of the absorbing H2O column density, variations in the dust mass absorption coefficient and the filling factor. Given the high metallicity suggested by other molecular line detections, the limitation to high filling factors due to the source size and the constraint on the gas mass from dynamical mass measurements, the main source of uncertainty resides in the source size due to limited spatial resolution in the current data. As such, major improvements should be possible by obtaining higher, (sub-)kpc resolution (that is, <0.2”) imaging with the Atacama Large Millimeter/submillimeter Array (ALMA; for other targets) and planned upgrades to NOEMA, and, in the future, with the Next Generation Very Large Array (ngVLA). Statistical uncertainties will also be greatly reduced by observing larger samples of massive star-forming galaxies over the entire redshift range where measurements are possible, closing the gap to SZ-based studies, which are currently limited to z < 1. The resulting improvement in precision will provide the constraints that are necessary to confirm or challenge the evolution of the CMB temperature with redshift predicted by standard cosmological models. ### Accessibility of the line signal The frequency range currently covered by NOEMA is 70.4–119.3, 127.0–182.9 and 196.1–276.0 GHz (with greatly reduced sensitivity above about 115 and 180 GHz in the first two frequency ranges). ALMA covers the 84–500-GHz range with gaps at 116–125 and 373–385 GHz, with a future extension down to 65 GHz (with greatly reduced sensitivity below about 67 GHz). The ngVLA is envisioned to cover the 70–116-GHz range. Excluding regions of poor atmospheric transparency, the H2O(110–101) line is therefore observable in these frequency ranges at redshifts of z = 0.1–0.4, 0.5–2.0, 2.1–3.4 and 3.8–6.9 in principle, but the detectability of the line in absorption against the CMB is probably limited to the z ~ 4.5–6.9 range if the spectral energy distribution shape of HFLS3 is representative. At lower frequencies, the Karl G. Jansky Very Large Array and, in the future, ALMA and the ngVLA also provide access to the <52-GHz range, such that the signal also becomes observable at z > 9.7 in principle. In conclusion, the absorption of the ground-state H2O transition against the CMB identified here could be traced from the ground towards star-forming galaxies across most of the first approximately 1.5 billion years of cosmic history. ### Detectability of the line signal for different spectral energy distribution shapes To investigate whether the effect is expected to be detectable for different galaxy populations, we have applied our modelling to the z = 3.9 quasar APM 08279+5255, for which the dust spectral energy distribution is composed of a dominant 220-K dust component and a weaker 65-K dust component, contributing only 10–15% to the far-infrared luminosity35,36,37,38,39,40,41,42,43,44,45,46. The models suggest that the line is expected to occur in emission and that it would not be expected to be detectable in absorption at any redshift out to at least z = 12 in galaxies with similar dust spectral energy distributions. Other far-infrared-luminous, high-redshift, active galactic nucleus host galaxies typically show a stronger relative contribution of their lower-temperature dust components, such that the effect may remain detectable in less extreme cases. For galaxies with lower dust temperatures than HFLS3, the effect may be present even at lower redshifts, but is typically expected to be weaker in general and to disappear at redshifts where TCMB approaches their Tdust. For a dust spectral energy distribution shape resembling the central region of the Milky Way but otherwise similar properties, the effect is expected to be reduced by more than two orders of magnitude at its redshift peak, and to become virtually unobservable at the redshift of HFLS3. Thus, dusty starburst galaxies appear to be some of the best environments to detect the effect. ### Derivation of equation of state parameters To determine the adiabatic index, we assume a standard Friedmann–Lemaitre–Robertson–Walker cosmology with zero curvature and a matter-radiation fluid that follows the standard adiabatic equation of state quoted in the main text. This would correspond to a redshift scaling TCMB(z) = TCMB(z = 0)*(1 + z)3(γ − 1) in the presence of a dark energy density that does not scale with redshift. The dark energy density is parameterized to scale with a power law (1 + z)m, where m = 0 corresponds to a cosmological constant. With standard assumptions, this yields a redshift scaling of TCMB (ref. 15): $${T}_{CMB}(z)={T}_{CMB}(z=0){(1+z)}^{3(gamma -1)}{left[frac{(m-3{varOmega }_{m,0})+m{(1+z)}^{(m-3)}({varOmega }_{m,0}-1)}{(m-3){varOmega }_{m,0}}right]}^{(gamma -1)}$$ and an effective dark energy equation of state Pde = weffρde, where the effective equation of state parameter weff = (m/3) − 1. This fitting function is used here with a canonical value of Ωm,0 = 0.315 (ref. 4). The uncertainty of Ωm,0 is small compared with all other sources of uncertainty and, hence, is neglected. All data used in the fitting are provided in Extended Data Table 1 (refs. 36,37,38,39,40,41,42,43,44,45,46).	2023-03-28 11:10:04	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.701705813407898, "perplexity": 1180.079939175415}, "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/1679296948858.7/warc/CC-MAIN-20230328104523-20230328134523-00520.warc.gz"}
http://121.43.60.238/sxwlxbA/EN/abstract/abstract16487.shtml	Acta mathematica scientia,Series A ›› 2021, Vol. 41 ›› Issue (5): 1270-1282. ### The Two-Dimensional Steady Chaplygin Gas Flows Passing a Straight Wedge Jia Jia() 1. School of Mathematical Sciences, East China Normal University, Shanghai 200241 • Received:2020-07-29 Online:2021-10-26 Published:2021-10-08 Abstract: The purpose of this paper is to investigate the two-dimensional steady supersonic chaplygin gas flows passing a straight wedge. By the definition of Radon measure solution, the accurate expressions are obtained for all cases where the Mach number is greater than 1. It is quite different from the polytropic gas, for the chaplygin gas flows passing problems, there exists a Mach number $M^{\ast}_{0}$, when the Mach number of incoming flows is greater than or equal to $M^{\ast}_{0}$, the quality will be concentrated on the surface of the straight wedge. At this time, there are not piecewise smooth solutions in the Lebesgue sense. The limit analysis is used to prove that the limit obtained by Lebesgue integral is consistent with the solution obtained in the sence of Radon measure solution. CLC Number: • O175.2 Trendmd	2021-12-03 10:35:51	{"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": 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": 0.18485645949840546, "perplexity": 502.70717867081714}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-49/segments/1637964362619.23/warc/CC-MAIN-20211203091120-20211203121120-00087.warc.gz"}
https://www.physicsforums.com/threads/does-qft-have-problems.912943/	# A Does QFT have problems Tags: 1. Apr 28, 2017 ### ftr QFT seems to be a bit sick with cluster decomposition assumption ..etc. So here comes Haag's theorem and Wightman axioms to the rescue, or do they? So what do these cures actually say differently than the generic QFT . Do they solve any practical problems, if not why the fuss, millennium prize and all. 2. Apr 29, 2017 ### A. Neumaier Only relativistic QFT in 4 space-time dimensions is sick because interacting fields are not well-defined in any logically coherent sense. Theoretical physicists predicting particle properties work around using either finite lattice approximations, which is a quantum field analogue of solving partial differential equations using finite difference methods breaking all spacetime symmetries, or low order formal renormalized perturbation theory, which is to a logically precise specification like Euler's treatment of functions through formal power series a few hundred years ago. These methods seem to work and give sensible physical predictions, but they have no proper logical foundation. The Wightman axioms spell out how the vacuum sector of such a theory should look like according to our present best guess, but (unlike in lower dimensions where there are many constructions) no interacting relativistic QFT in 4 space-time dimensions has been constructed in the more than 50 years of existence of the Wightman axioms. Neither are there nonexistence theorems. The simplest tractable case is widely held to be 4D Yang-Mills theory, which is the reason why there is a millennium prize on solving it. See https://www.physicsoverflow.org/217...osons-infinite-and-discrete?show=21846#a21846 for more comments on the latter. Cluster decomposition is an important principle without which physics would be impossible. It basically says that far away from other particles particles behave independently of each other. This is the basic reason why we can look at (properly prepared) small objects independently of other objects. For its physical basis see on the informal level Weinberg's QFT book, Vol. I, Chapter 5. On the rigorous level, it guarantees the uniqueness of the vacuum state; see any book on algebraic quantum field theory. Haag's theorem is hardly related to that; it only says that the CCR representation of the free field cannot extend to interacting fields, the interacting Hilbert space must look quite different from a Fock space. More precisely, while they have to be isomorphic as Hilbert spaces (all infinite-dimensional separable Hilbert spaces are isomorphic), the additional structure needed to set up quantum fields cannot be respected by such an isomorphism. Last edited: Apr 29, 2017 3. Apr 29, 2017 ### vanhees71 Recently a colleague after a semester-long discussion on the foundations of QFT pointed me to this very nice review on Haag's theorem and related works: http://philsci-archive.pitt.edu/2673/1/earmanfraserfinalrevd.pdf Don't worry! Although it says, it's about philosophy, it's in fact physics. Whenever a serious teacher of QFT comes to explain asymptotic states of interacting particles, the S-matrix, the LSZ reduction formalism s/he gets doubts and discusses it with colleagues, of course, with not too much success; it's indeed an unsolved problem to find a mathematically rigorous formulation of 4D relativistic QFT with interacting particles. FAPP QFT, as used by HEP phenomenologists, works with great success (even too much since the Standard Model stands up against all experimental efforts to disprove it; although recently LHCb came up with the next attack, but only at a bit over 2% confidence level yet ;-)). The way out is a practical approach: Just put the system in a box with periodic boundary conditions, use perturbation theory and renormalize. Then you get results with up to 12 significant digits accuracy. As Arnold said the other way out is the lattice approach and high computing power (at least in QCD, where you get, e.g., the hadron mass spectrum pretty nicely out too, and that's clearly beyond perturbation theory). That's the approach Weinberg chooses in his book: Simply ignore Haag et al, which however is a pity since I think one should be aware of it! 4. Apr 29, 2017 ### A. Neumaier This works for kinetic or hydrodynamic questions. But the resulting compactification destroys scattering theory since under these conditions the spectrum becomes wholly discrete. This ultra-high accuracy is not obtained in this way but through NRQED: One uses a nonrelativistic $1/c^2$ expansion to get rid of the problems with the relativistic theory and then works essentially with nonrelativistic quantum mechanics with relativistic corrections. 5. Apr 29, 2017 ### ftr Since I am not a professional theorist, my opinion probably don't add to much. But I feel that "mathematically rigorous" should be physically rigorous. I am well aware of the successes, but I think it is mainly due to correctness of QM itself. The QFT formulation is heroic to be sure , but I have read the story of the development. A very messy process of experimentation(with it own problems, bumps.etc) and a theoretician trying to score. But let me ask this ,I think the main issue with QFT is removing infinities and renormalization of the mass and charge, I think these are PHYSICAL problems and not "mathematical", do you agree. 6. Apr 29, 2017 ### ftr WE wrote this within minutes. Although I am no expert, I have deduced that. 7. Apr 29, 2017 ### vanhees71 Where do you need the non-relativistic approximation to evaluate the anomalous magnetic moment of the electron? 8. Apr 29, 2017 ### vanhees71 Renormalization has nothing to do with infinite radiation corrections per se. It comes to the rescue, but even if everything were perfectly finite, you'd need to renormalize to adjust the free parameters of the theory (wave-function normalization, masses, coupling constants) to observations. 9. Apr 29, 2017 ### atyy Practical QFT is eg. the standard model of particle physics. Most physicists nowadays understand it to be a low energy theory, ie. to make good predictions at the energies we use in experiments, it is not necessary for the theory to exist at all energies. This is the Wilsonian viewpoint, and it is a great advance for understanding why renormalization is ok, and has nothing to do with removing infinities etc. The main problem in this practical way of thinking is that it needs a non-perturbatively defined quantum theory, eg. lattice gauge theory, but while lattice QED is thought to be ok, chiral fermions on the latttice are still problematic. The Millenium prize has to do with 4D relativistic QFTs that exist at all energies (and not just at low energies). 10. Apr 29, 2017 ### A. Neumaier This doesn't explain the high accuracy achieved! QM without QED has not such a high predictive accuracy. To first loop order you don't need it but you don't get very high accuracy. For highest accuracy you cannot work with the standard Feynman approach; it is too daunting a computational task. Because of Poincare invariance, any relativistic QFT (in flat spacetime) will exist at all energies and (and not just at low energies). 11. Apr 29, 2017 ### vanhees71 I thought Kinoshita has calculated (g-2)/2 to order $\alpha_{\text{em}}^5$. Are there non-relativistic approximations involved? I don't see anything mentioned like this, e.g., here https://arxiv.org/abs/1205.5368 But the trouble with some QFTs (including QED) at high energies, at least perturbatively (Landau pole). The modern consensus is that the Standard Model is a low-energy approximation of some other theory, of which we have no clue (not even if it exists at all). 12. Apr 29, 2017 ### ftr I know you need QED for precision, I was just thinking that QT is real, QFT is just a technique. May one day be united 13. Apr 29, 2017 ### Staff: Mentor What are you basing this on? 14. Apr 29, 2017 ### ftr It just looks to me that QFT takes QM massages it and turns it into a calculational tool. 15. Apr 29, 2017 ### Staff: Mentor You're leaving out at least three key physical phenomena that QFT can predict and ordinary non-relativistic QM can't: (1) The existence of processes where particles are created or destroyed; (2) The existence of antiparticles; (3) The connection between spin and statistics. 16. Apr 29, 2017 ### atyy QFT is a type of QT. Condensed matter uses non-relativistic QFT. Non-relativistic QFT can be derived from the non-relativistic Schroedinger equation for many identical particles. 17. Apr 30, 2017 ### A. Neumaier No. The paper you cited calculates some contributions up to the 10th order and even includes hadronic contributions (phenomenologically included into QED) since it already affects the 10th decimal. Indeed, I checked the references with the actual calculation details that no expansion in inverse powers of c is made. Thus the results are covariant. I had mixed up the computation of $g-2$ with the computation of the hyperfine structure (Lamb shift), where NRQED is essential to get high precision. 18. Apr 30, 2017 ### A. Neumaier There is no such trouble in formulations such as causal perturbation theory, that preserve covariance throughout. (The Landau pole is problematic only if a cutoff must be able to move across the pole.) 19. Apr 30, 2017 ### vanhees71 Well, if you need a cutoff to define the theory then you admit that it's an effective theory valid up to the cutoff scale at most. The Epstein-Glaser approach avoids UV divergences, because it's more careful in dealing with products of distribution valued operators, and thus it solves a mathematical problem, but not the physical problem of the Landau pole. As far as I know, QED is an example for a QFT more likely not to exist in a strict sense than, e.g., QCD, which is asymptotic free, but for no realistic (i.e., (1+3)-dim. theory of interacting quantum fields) there has been a proof for it to exist in a strict sense. On the other hand FAPP this doesn't matter much, because we have anyway only a limited energy available, and it's quite likely that our contemporary Standard Model will fail at high enough energies somewhere. Today, nobody known where that scale might be. Maybe there's really a dessert up to the Planck scale, where one can definitely expect something should happen concerning quantum effects of gravity. Then HEP with accelerators is doomed, funding wise :-(. 20. Apr 30, 2017 ### Denis No. A lattice approximation with periodic boundary conditions defines a conceptually and mathematically valid and unproblematic theory. It does not have infinities, and does not need renormalization (it is, instead, possibly part of the process of renormalization - namely a regularized theory). Such a lattice theory can be interpreted as some approximation of the field theory, and gives all the observable (physical) results, with some accuracy. So, the lattice theory is mathematically fine, and it is (at least if the computing power is sufficient, the lattice fine enough and the volume big enough) also physically fine. In principle, a lattice theory could be even a candidate for a more fundamental theory, given that it has neither physical nor mathematical problems, and QFT would be simply the large distance approximation of such a fundamental lattice theory. What is not fine is the metaphysics. The lattice theory breaks Lorentz covariance, and some people consider Lorentz covariance as something obligatory, for some metaphysical reasons - relativistic symmetry being a fundamental insight or so.	2017-10-20 17:16:12	{"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": 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": 0.7120314836502075, "perplexity": 953.7334401777205}, "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/1508187824226.31/warc/CC-MAIN-20171020154441-20171020174441-00894.warc.gz"}
http://www.koreascience.or.kr/article/ArticleFullRecord.jsp?cn=OGGHBK_2010_v29n4_605	Input Performance of the Old Adults in Touch Interface Title & Authors Input Performance of the Old Adults in Touch Interface Hong, Seung-Kweon; Park, Jung-Chul; Kim, Sun-Su; Abstract In order to design a touch interface for the elderly, human performance of input tasks on the touch screen was investigated by the laboratory experiment. Input times and input errors were measured in the experimental conditions that were changed according to age, key size, interkey space and input tool(finger or stylus pen). In the most of all experimental conditions, the task performance of the elderly was lower than that of the young. However, there were significantly different performance patterns between both groups. As the difficulty of task was getting higher, the task performance of the elderly was sharply decreased; pressing small key button by finger sharply increased input time and error rate, compared to that of the young. Therefore, the square key size suitable to the elderly may be over $\small{8.0{\times}8.0mm}$. While the interkey space did not influence to the input task performance of the young, the task performance of the elderly was influenced. The elderly showed big difference of task performance according to input tool. However, the young were less influenced by input tool. Keywords Language Korean Cited by 1. Universal User Interface Design of ATM Touch Screen Based on the Reaction Time,;;; Journal of the Ergonomics Society of Korea, 2016. vol.35. 5, pp.403-411 2. 고령자의 스마트폰 사용성 향상을 위한 연구,한영석;최종규;황보환;고상민;윤솔희;지용구; 한국전자거래학회지, 2012. vol.17. 1, pp.39-52 References 1. 김정룡, 조은주, 조영진, 정민근, 휴대전화 설계를 위한 고령자용인지능력 측정 프로토콜 개발, 대한인간공학회지, Vol 28, No 3, 115-123, 2009. 2. Colle, H. A. and Hiszem, K. J., Standing at a Kiosk: Effects of Key Size and Spacing on Touch Screen Numeric Keypad Performance and User Preference. Erogonomics, Vol 47. No. 13, 1406-1423, 2004. 3. Czaja, S. J., Computer Technology and the Old Adult. In T. K. Landauer and P. Prabhu (Eds.), Handbook of Human-Computer Interaction (PP 797-824). Amsterdam: North-Holland. 1997. 4. Douglas, S. A. and Mithal, A. K., The Ergonomics of Computer Pointing Devices. London: Springer, 1997. 5. Galitz, W. O., The Essential Guide to User Interface Design - An Introduction to GUI Design Principles and Techniques. New York: Wiley, 1997. 6. Salthouse, T. A., Perceptual, Cognitive and Motoric Aspects of Transcription Typing, Psychological Bulletin, Vol. 99., No 3, 303-319, 1986. 7. Schedlbauer, M., Effects of Key Size and Spacing on The Completion Time and Accuracy of Input Tasks on Soft Keypads Using Trackball and Touch Input, Proceedings of the Human Factors Society 51st Annual Meeting, 2007. 8. Shneiderman, B., Designing the User Interface: Strategy for Effective Human-Computer Interaction. Reading, MA: Addison-Wesley, 1987. 9. Vanderheiden, G. C., Design for People with Functional Limitations Resulting from Disability, Aging or Circumstances. In G. Salvendy (Eds.), Handbook of Human Factors and Ergonomics (pp 2012-2052), New York: Wiley, 1997. 10. Wright, P., Bartram, C., Rogers, N., Emslie, H., Evans, J., Wison, B. and Belt, S., Text Entry on Handheld Computers by Older Users. Ergonomics, Vol. 43, No. 6, 702-716. 2000.	2018-07-19 15:27:46	{"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": 0, "wp_latex": 0, "mimetex.cgi": 1, "/images/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": 0.23216891288757324, "perplexity": 9420.528155379227}, "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-2018-30/segments/1531676591140.45/warc/CC-MAIN-20180719144851-20180719164851-00634.warc.gz"}
https://scicomp.stackexchange.com/questions/33768/chebyshev-differentiation-via-fft-with-a-domain-a-b	# Chebyshev differentiation via FFT with a domain [a,b] I want to ask something about Chebyshev differentiation via FFT, which can be used to obtain with spectral accuracy the derivative of a smooth function. See for instance this code in python, which performs the derivative of a given function "u" in the domain [-1,1] and computes the L2 norm of the error. from numpy.fft import fft,ifft import numpy as np def chebfft(v,x): N = len(v)-1; if N==0: return 0 ii = np.arange(0,N); iir = np.arange(1-N,0); iii = np.array(ii,dtype=int) V = np.hstack((v,v[N-1:0:-1])) U = np.real(fft(V)) W = np.real(ifft(1jnp.hstack((ii,[0.],iir))U)) w = np.zeros(N+1) w[1:N] = -W[1:N]/np.sqrt(1-x[1:N]2) w[0] = sum(iii2U[iii])/N + .5NU[N] w[N] = sum((-1)(iii+1)ii*2U[iii])/N + .5(-1)(N+1)NU[N] return w N=50 L=10 aa = 0.34 x = np.cos(np.pinp.arange(0,N+1)/N) u=np.sin(2np.pix/L)np.exp(-aax) uder = np.exp(-aax)(2np.pinp.cos(2np.pix/L)-aaLnp.sin(2np.pix/L))/L err_ft = chebfft(u,x)-uder print ("L2 norm:",np.linalg.norm(err_ft)) The problem that I have is that I don't really understand how to do the same with a general interval [a,b]. The new collocation points can be changed just with a linear mapping, but then the procedure inside chebfft fails. Provably the reason why is not woorking is very simple, but I don't see. Thanks for the help. Given a function $$u = u(X), \qquad X \in [X_{min},X_{max}]$$ Map to $$x \in [-1,1]$$ $$x = \frac{X - X_c}{\frac{1}{2}(X_{max} - X_{min})}, \qquad X_c = \frac{1}{2}(X_{min} + X_{max})$$ If $$D$$ is Chebyshev approximation of derivative $$D \approx \frac{d u}{d x}$$ then $$\frac{d u}{d X} = \frac{d x}{d X} \frac{d u}{d x} \approx \frac{2}{X_{max} - X_{min}} D$$ I modified your code for the function $$u(X) = X \sin(X), \qquad X \in [0,2\pi]$$ and it works fine. from numpy.fft import fft,ifft import numpy as np import matplotlib.pyplot as plt def chebfft(v,x): N = len(v)-1; if N==0: return 0 ii = np.arange(0,N); iir = np.arange(1-N,0); iii = np.array(ii,dtype=int) V = np.hstack((v,v[N-1:0:-1])) U = np.real(fft(V)) W = np.real(ifft(1jnp.hstack((ii,[0.],iir))U)) w = np.zeros(N+1) w[1:N] = -W[1:N]/np.sqrt(1-x[1:N]2) w[0] = sum(iii2U[iii])/N + .5NU[N] w[N] = sum((-1)(iii+1)ii*2U[iii])/N + .5(-1)(N+1)NU[N] return w N = 50 Xmin, Xmax = 0.0, 2.0np.pi Xc = 0.5(Xmin + Xmax) x = np.cos(np.pinp.arange(0,N+1)/N) X = 0.5(Xmax - Xmin)x + Xc u = X * np.sin(X) uder = np.sin(X) + X * np.cos(X) udercheb = chebfft(u,x)*2.0/(Xmax - Xmin) err_ft = udercheb - uder print ("L2 norm:",np.linalg.norm(err_ft)) plt.plot(X,uder,X,udercheb,'o') plt.legend(('Exact','Chebyshev')) plt.show() So you just need a simple scaling to account for the domain being different from [-1,1]. • now I understand, thanks a lot! – Joe Nov 9 '19 at 15:59 You need more than just simple linear mapping here. Let's say in general I have a complex variable $$z$$ as: $$z = R e^{i\theta} + z_{C}$$ Your $$x$$ is defined as: $$x = \Re\{z\} = R \cos(\theta) + x_{C}$$ The maximum and minimum values of $$x$$ could be calculated as: $$\min(x) = x_{C} - R$$ $$\max(x) = x_{C} + R$$ Now you want to map into $$[a,b]$$ range, so: $$x_{C} = \frac{b+a}{2}$$ $$R = \frac{b-a}{2}$$ If I take a derivative from $$x$$ with respect to $$\theta$$: $$\frac{d x}{d \theta} = -R \sin(\theta) = -\sqrt{R^{2} - (x-x_{C})^{2}}$$ Finally Chebyshev polynomials are defined as: $$T_{n}(x) = \Re \Bigg \{ \Big (\frac{z - z_{c}}{R} \Big )^{n} \Bigg \} = \cos(n\theta)$$ Furthermore, the derivative of Chebyshev polynomial is: $$T^{'}_{n}(x) = -n \sin(n\theta) \frac{d \theta}{d x} = \frac{n \sin(n\theta)}{R \sin(\theta)}$$ Now, let's define discretized $$x_{j}$$ as: $$x_{j} = R \cos(\theta_{j}) + x_{C}$$ You want to approximate your function $$f$$ by using Chebyshev polymonials as: $$f(\theta) = \sum_{n=0}^{N} a_{n} \cos(n \theta)$$ The derivative of $$f$$, which you are mainly looking for is: $$\frac{d f}{d x} = \sum_{n=0}^{N} \frac{n a_{n} \sin(n\theta)}{\sqrt{R^{2} - (x-x_{C})^{2}}}$$ So just find the $$a_{n}$$ from FFT and then at $$x = a$$ and $$x = b$$, from L'Hopital you have: $$\frac{d f}{d x}\|_{x = a} = \sum_{n = 0}^{N} (-1)^{n} \frac{n^{2}}{R} a_{n}$$ $$\frac{d f}{d x}\|_{x = b} = \sum_{n = 0}^{N} \frac{n^{2}}{R} a_{n}$$ So, you need a little bit more effort to make it work for general $$[a,b]$$ range. I put the implementation for your homework.	2021-06-20 06:37:40	{"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": 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": 34, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7265709638595581, "perplexity": 1582.5184704321189}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-25/segments/1623487658814.62/warc/CC-MAIN-20210620054240-20210620084240-00145.warc.gz"}
https://labs.tib.eu/arxiv/?author=J.%20Berthier	• We highlight the power of the Gaia DR2 in studying many fine structures of the Hertzsprung-Russell diagram (HRD). Gaia allows us to present many different HRDs, depending in particular on stellar population selections. We do not aim here for completeness in terms of types of stars or stellar evolutionary aspects. Instead, we have chosen several illustrative examples. We describe some of the selections that can be made in Gaia DR2 to highlight the main structures of the Gaia HRDs. We select both field and cluster (open and globular) stars, compare the observations with previous classifications and with stellar evolutionary tracks, and we present variations of the Gaia HRD with age, metallicity, and kinematics. Late stages of stellar evolution such as hot subdwarfs, post-AGB stars, planetary nebulae, and white dwarfs are also analysed, as well as low-mass brown dwarf objects. The Gaia HRDs are unprecedented in both precision and coverage of the various Milky Way stellar populations and stellar evolutionary phases. Many fine structures of the HRDs are presented. The clear split of the white dwarf sequence into hydrogen and helium white dwarfs is presented for the first time in an HRD. The relation between kinematics and the HRD is nicely illustrated. Two different populations in a classical kinematic selection of the halo are unambiguously identified in the HRD. Membership and mean parameters for a selected list of open clusters are provided. They allow drawing very detailed cluster sequences, highlighting fine structures, and providing extremely precise empirical isochrones that will lead to more insight in stellar physics. Gaia DR2 demonstrates the potential of combining precise astrometry and photometry for large samples for studies in stellar evolution and stellar population and opens an entire new area for HRD-based studies. • ### Gaia Data Release 2: Observations of solar system objects(1804.09379) April 25, 2018 astro-ph.EP, astro-ph.IM The Gaia spacecraft of the European Space Agency (ESA) has been securing observations of solar system objects (SSOs) since the beginning of its operations. Gaia Data Release 2 (DR2) contains the observations of a selected sample of 14,099 SSOs. These asteroids have been already identified and have been numbered by the Minor Planet Center. Positions are provided for each Gaia observation at CCD level. As additional information, the apparent brightness of SSOs in the unfiltered G band is also provided for selected observations. We explain the processing of SSO data, and describe the criteria we used to select the sample published in Gaia DR2. We then explore the data set to assess its quality. To exploit the epoch astrometry of asteroids in Gaia DR2 it is necessary to take into account the unusual properties of the uncertainty, as the position information is nearly one-dimensional. When this aspect is handled appropriately, an orbit fit can be obtained with post-fit residuals that are overall consistent with the a-priori error model that was used to define individual values of the astrometric uncertainty. The distribution of residuals allowed us to identify possible contaminants in the data set. Photometry in the G band was compared to computed values from reference asteroid shapes and to the flux registered at the corresponding epochs by the red and blue photometers (RP and BP). The overall astrometric performance is close to the expectations, with an optimal range of brightness G~12-17. In this range, the typical transit-level accuracy is well below 1 mas. For fainter asteroids, the growing photon noise deteriorates the performance. Asteroids brighter than G~12 are affected by a lower performance of the processing of their signals. The dramatic improvement brought by Gaia DR2 astrometry of SSOs is demonstrated by preliminary tests on the detection of subtle non-gravitational effects. • ### Physical, spectral, and dynamical properties of asteroid (107) Camilla and its satellites(1803.02722) March 7, 2018 astro-ph.EP The population of large asteroids is thought to be primordial and they are the most direct witnesses of the early history of our Solar System. Those satellites allow study of the mass, and hence density and internal structure. We study here the properties of the triple asteroid (107) Camilla from lightcurves, stellar occultations, optical spectroscopy, and high-contrast and high-angular-resolution images and spectro-images. Using 80 positions over 15 years, we determine the orbit of its larger satellite to be circular, equatorial, and prograde, with RMS residuals of 7.8 mas. From 11 positions in three epochs only, in 2015 and 2016, we determine a preliminary orbit for the second satellite. We find the orbit to be somewhat eccentric and slightly inclined to the primary's equatorial plane, reminiscent of the inner satellites of other asteroid triple systems. Comparison of the near-infrared spectrum of the larger satellite reveals no significant difference with Camilla. Hence, these properties argue for a formation of the satellites by excavation from impact and re-accumulation of ejecta. We determine the spin and 3-D shape of Camilla. The model fits well each data set. We determine Camilla to be larger than reported from modeling of mid-infrared photometry, with a spherical-volume-equivalent diameter of 254 $\pm$ 36 km (3 $\sigma$ uncertainty), in agreement with recent results from shape modeling (Hanus2017+). Combining the mass of (1.12 $\pm$ 0.01) $\times$ 10$^{19}$ kg determined from the dynamics of the satellites and the volume from the 3-D shape model, we determine a density of 1,280 $\pm$ 130 SI. From this density, and considering Camilla's spectral similarities with (24) Themis and (65) Cybele (for which water ice coating on surface grains was reported), we infer a silicate-to-ice mass ratio of 1-6, with a 10-30% macroporosity. • ### 3D shape of asteroid (6)~Hebe from VLT/SPHERE imaging: Implications for the origin of ordinary H chondrites(1705.10515) May 30, 2017 astro-ph.EP Context. The high-angular-resolution capability of the new-generation ground-based adaptive-optics camera SPHERE at ESO VLT allows us to assess, for the very first time, the cratering record of medium-sized (D~100-200 km) asteroids from the ground, opening the prospect of a new era of investigation of the asteroid belt's collisional history. Aims. We investigate here the collisional history of asteroid (6) Hebe and challenge the idea that Hebe may be the parent body of ordinary H chondrites, the most common type of meteorites found on Earth (~34% of the falls). Methods. We observed Hebe with SPHERE as part of the science verification of the instrument. Combined with earlier adaptive-optics images and optical light curves, we model the spin and three-dimensional (3D) shape of Hebe and check the consistency of the derived model against available stellar occultations and thermal measurements. Results. Our 3D shape model fits the images with sub-pixel residuals and the light curves to 0.02 mag. The rotation period (7.274 47 h), spin (343 deg,+47 deg), and volume-equivalent diameter (193 +/- 6km) are consistent with previous determinations and thermophysical modeling. Hebe's inferred density is 3.48 +/- 0.64 g.cm-3 , in agreement with an intact interior based on its H-chondrite composition. Using the 3D shape model to derive the volume of the largest depression (likely impact crater), it appears that the latter is significantly smaller than the total volume of close-by S-type H-chondrite-like asteroid families. Conclusions. Our results imply that (6) Hebe is not the most likely source of H chondrites. Over the coming years, our team will collect similar high-precision shape measurements with VLT/SPHERE for ~40 asteroids covering the main compositional classes, thus providing an unprecedented dataset to investigate the origin and collisional evolution of the asteroid belt. • Parallaxes for 331 classical Cepheids, 31 Type II Cepheids and 364 RR Lyrae stars in common between Gaia and the Hipparcos and Tycho-2 catalogues are published in Gaia Data Release 1 (DR1) as part of the Tycho-Gaia Astrometric Solution (TGAS). In order to test these first parallax measurements of the primary standard candles of the cosmological distance ladder, that involve astrometry collected by Gaia during the initial 14 months of science operation, we compared them with literature estimates and derived new period-luminosity ($PL$), period-Wesenheit ($PW$) relations for classical and Type II Cepheids and infrared $PL$, $PL$-metallicity ($PLZ$) and optical luminosity-metallicity ($M_V$-[Fe/H]) relations for the RR Lyrae stars, with zero points based on TGAS. The new relations were computed using multi-band ($V,I,J,K_{\mathrm{s}},W_{1}$) photometry and spectroscopic metal abundances available in the literature, and applying three alternative approaches: (i) by linear least squares fitting the absolute magnitudes inferred from direct transformation of the TGAS parallaxes, (ii) by adopting astrometric-based luminosities, and (iii) using a Bayesian fitting approach. TGAS parallaxes bring a significant added value to the previous Hipparcos estimates. The relations presented in this paper represent first Gaia-calibrated relations and form a "work-in-progress" milestone report in the wait for Gaia-only parallaxes of which a first solution will become available with Gaia's Data Release 2 (DR2) in 2018. • Context. The first Gaia Data Release contains the Tycho-Gaia Astrometric Solution (TGAS). This is a subset of about 2 million stars for which, besides the position and photometry, the proper motion and parallax are calculated using Hipparcos and Tycho-2 positions in 1991.25 as prior information. Aims. We investigate the scientific potential and limitations of the TGAS component by means of the astrometric data for open clusters. Methods. Mean cluster parallax and proper motion values are derived taking into account the error correlations within the astrometric solutions for individual stars, an estimate of the internal velocity dispersion in the cluster, and, where relevant, the effects of the depth of the cluster along the line of sight. Internal consistency of the TGAS data is assessed. Results. Values given for standard uncertainties are still inaccurate and may lead to unrealistic unit-weight standard deviations of least squares solutions for cluster parameters. Reconstructed mean cluster parallax and proper motion values are generally in very good agreement with earlier Hipparcos-based determination, although the Gaia mean parallax for the Pleiades is a significant exception. We have no current explanation for that discrepancy. Most clusters are observed to extend to nearly 15 pc from the cluster centre, and it will be up to future Gaia releases to establish whether those potential cluster-member stars are still dynamically bound to the clusters. Conclusions. The Gaia DR1 provides the means to examine open clusters far beyond their more easily visible cores, and can provide membership assessments based on proper motions and parallaxes. A combined HR diagram shows the same features as observed before using the Hipparcos data, with clearly increased luminosities for older A and F dwarfs. • ### Gaia Data Release 1: Astrometry - one billion positions, two million proper motions and parallaxes(1609.04303) Sept. 14, 2016 astro-ph.GA, astro-ph.IM Gaia Data Release 1 (Gaia DR1) contains astrometric results for more than 1 billion stars brighter than magnitude 20.7 based on observations collected by the Gaia satellite during the first 14 months of its operational phase. We give a brief overview of the astrometric content of the data release and of the model assumptions, data processing, and validation of the results. For stars in common with the Hipparcos and Tycho-2 catalogues, complete astrometric single-star solutions are obtained by incorporating positional information from the earlier catalogues. For other stars only their positions are obtained by neglecting their proper motions and parallaxes. The results are validated by an analysis of the residuals, through special validation runs, and by comparison with external data. Results. For about two million of the brighter stars (down to magnitude ~11.5) we obtain positions, parallaxes, and proper motions to Hipparcos-type precision or better. For these stars, systematic errors depending e.g. on position and colour are at a level of 0.3 milliarcsecond (mas). For the remaining stars we obtain positions at epoch J2015.0 accurate to ~10 mas. Positions and proper motions are given in a reference frame that is aligned with the International Celestial Reference Frame (ICRF) to better than 0.1 mas at epoch J2015.0, and non-rotating with respect to ICRF to within 0.03 mas/yr. The Hipparcos reference frame is found to rotate with respect to the Gaia DR1 frame at a rate of 0.24 mas/yr. Based on less than a quarter of the nominal mission length and on very provisional and incomplete calibrations, the quality and completeness of the astrometric data in Gaia DR1 are far from what is expected for the final mission products. The results nevertheless represent a huge improvement in the available fundamental stellar data and practical definition of the optical reference frame. • ### Near-infrared spatially resolved spectroscopy of (136108) Haumea's multiple system(1605.04145) May 13, 2016 astro-ph.EP The transneptunian region of the solar system is populated by a wide variety of icy bodies showing great diversity. The dwarf planet (136108) Haumea is among the largest TNOs and displays a highly elongated shape and hosts two moons, covered with crystalline water ice like Hamuea. Haumea is also the largest member of the sole TNO family known to date. A catastrophic collision is likely responsible for its unique characteristics. We report here on the analysis of a new set of observations of Haumea obtained with SINFONI at the ESO VLT. Combined with previous data, and using light-curve measurements in the optical and far infrared, we carry out a rotationally resolved spectroscopic study of the surface of Haumea. We describe the physical characteristics of the crystalline water ice present on the surface of Haumea for both regions, in and out of the Dark Red Spot (DRS), and analyze the differences obtained for each individual spectrum. The presence of crystalline water ice is confirmed over more than half of the surface of Haumea. Our measurements of the average spectral slope confirm the redder characteristic of the spot region. Detailed analysis of the crystalline water-ice absorption bands do not show significant differences between the DRS and the remaining part of the surface. We also present the results of applying Hapke modeling to our data set. The best spectral fit is obtained with a mixture of crystalline water ice (grain sizes smaller than 60 micron) with a few percent of amorphous carbon. Improvements to the fit are obtained by adding ~10% of amorphous water ice. Additionally, we used the IFU-reconstructed images to measure the relative astrometric position of the largest satellite Hi`iaka and determine its orbital elements. An orbital solution was computed with our genetic-based algorithm GENOID and our results are in full agreement with recent results. • ### Multiple Asteroid Systems: Dimensions and Thermal Properties from Spitzer Space Telescope and Ground-Based Observations(1604.05384) April 19, 2016 astro-ph.EP Photometric lightcurves were also obtained for 14 of them during the Spitzer observations to provide the context of the observations and reliable estimates of their absolute magnitudes. The extracted mid-IR spectra were analyzed using a modified standard thermal model (STM) and a thermophysical model (TPM) that takes into account the shape and geometry of the large primary at the time of the Spitzer observation. We derived a reliable estimate of the size, albedo, and beaming factor for each of these asteroids, representing three main taxonomic groups: C, S, and X. For large (volume-equivalent system diameter Deq $\lt$ 130 km) binary asteroids, the TPM analysis indicates a low thermal inertia ($\Gamma$ < $\sim$100 J s-1/2K-1m-2) and their emissivity spectra display strong mineral features, implying that they are covered with a thick layer of thermally insulating regolith. The smaller (surface-equivalent system diameter Deff $\lt$17 km) asteroids also show some emission lines of minerals, but they are significantly weaker, consistent with regoliths with coarser grains, than those of the large binary asteroids. The average bulk densities of these multiple asteroids vary from 0.7-1.7 g/cm3 (P-, C- type) to $\sim$2 g/cm3 (S-type). The highest density is estimated for the M-type (22) Kalliope (3.2 $\pm$ 0.9 g/cm3). The spectral energy distributions (SED) and emissivity spectra, made available as a supplement document, could help to constrain the surface compositions of these asteroids. • ### Prediction of transits of solar system objects in Kepler/K2 images: An extension of the Virtual Observatory service SkyBoT(1602.07153) Feb. 23, 2016 astro-ph.EP All the fields of the extended space mission Kepler/K2 are located within the ecliptic. Many solar system objects thus cross the K2 stellar masks on a regular basis. We aim at providing to the entire community a simple tool to search and identify solar system objects serendipitously observed by Kepler. The SkyBoT service hosted at IMCCE provides a Virtual Observatory (VO) compliant cone-search that lists all solar system objects present within a field of view at a given epoch. To generate such a list in a timely manner, ephemerides are pre-computed, updated weekly, and stored in a relational database to ensure a fast access. The SkyBoT Web service can now be used with Kepler. Solar system objects within a small (few arcminutes) field of view are identified and listed in less than 10 sec. Generating object data for the entire K2 field of view (14{\deg}) takes about a minute. This extension of the SkyBot service opens new possibilities with respect to mining K2 data for solar system science, as well as removing solar system objects from stellar photometric time-series. • ### The EPN-TAP protocol for the Planetary Science Virtual Observatory(1407.5738) July 22, 2014 astro-ph.IM A Data Access Protocol has been set up to search and retrieve Planetary Science data in general. This protocol will allow the user to select a subset of data from an archive in a standard way, based on the IVOA Table Access Protocol (TAP). The TAP mechanism is completed by an underlying Data Model and reference dictionaries. This paper describes the principle of the EPN- TAP protocol and interfaces, underlines the choices that have been made, and discusses possible evolutions. • ### Planetary Science Virtual Observatory architecture(1407.4886) July 18, 2014 astro-ph.EP, astro-ph.IM In the framework of the Europlanet-RI program, a prototype of Virtual Observatory dedicated to Planetary Science was defined. Most of the activity was dedicated to the elaboration of standards to retrieve and visualize data in this field, and to provide light procedures to teams who wish to contribute with on-line data services. The architecture of this VO system and selected solutions are presented here, together with existing demonstrators. • ### Physical and dynamical properties of the main belt triple asteroid (87) Sylvia(1407.1292) July 4, 2014 astro-ph.EP We present the analysis of high angular resolution observations of the triple Asteroid (87) Sylvia collected with three 8-10 m class telescopes (Keck, VLT, Gemini North) and the Hubble Space Telescope. The moons' mutual orbits were derived individually using a purely Keplerian model. We computed the position of Romulus, the outer moon of the system, at the epoch of a recent stellar occultation which was successfully observed at less than 15 km from our predicted position, within the uncertainty of our model. The occultation data revealed that the Moon, with a surface-area equivalent diameter Ds=23.1$\pm$0.7km, is strongly elongated (axes ratio of 2.7$\pm$0.32.7$\pm$0.3), significantly more than single asteroids of similar size in the main-belt. We concluded that its shape is probably affected by the tides from the primary. A new shape model of the primary was calculated combining adaptive-optics observations with this occultation and 40 archived light-curves recorded since 1978. The difference between the J2=0.024-0.009+0.016 derived from the 3-D shape model assuming an homogeneous distribution of mass for the volume equivalent diameter Dv=273$\pm$10km primary and the null J2 implied by the Keplerian orbits suggests a non-homogeneous mass distribution in the asteroid's interior. • ### The Puzzling Mutual Orbit of the Binary Trojan Asteroid (624) Hektor(1402.7336) Feb. 28, 2014 astro-ph.EP Asteroids with satellites are natural laboratories to constrain the formation and evolution of our solar system. The binary Trojan asteroid (624) Hektor is the only known Trojan asteroid to possess a small satellite. Based on W.M. Keck adaptive optics observations, we found a unique and stable orbital solution, which is uncommon in comparison to the orbits of other large multiple asteroid systems studied so far. From lightcurve observations recorded since 1957, we showed that because the large Req=125-km primary may be made of two joint lobes, the moon could be ejecta of the low-velocity encounter, which formed the system. The inferred density of Hektor's system is comparable to the L5 Trojan doublet (617) Patroclus but due to their difference in physical properties and in reflectance spectra, both captured Trojan asteroids could have a different composition and origin. • ### Characteristics and Large Bulk Density of the C-type Main-Belt Triple Asteroid (93) Minerva(1310.3209) Oct. 11, 2013 astro-ph.EP From a set of adaptive optics (AO) observations collected with the W.M. Keck telescope between August and September 2009, we derived the orbital parameters of the most recently discovered satellites of the large C-type asteroid (93) Minerva. The satellites of Minerva, which are approximately 3 and 4 km in diameter, orbit very close to the primary $\sim$5 & $\sim$8 $\times$ Rp and $\sim$1% & $\sim$2% $\times$ RHill) in a circular manner, sharing common characteristics with most of the triple asteroid systems in the main-belt. Combining these AO observations with lightcurve data collected since 1980 and two stellar occultations in 2010 & 2011, we removed the ambiguity of the pole solution of Minerva's primary and showed that it has an almost regular shape with an equivalent diameter Deq = 154 $\pm$ 6 km in agreement with IRAS observations. The surprisingly high bulk density of 1.75 $\pm$ 0.30 g/cm$\^3$ for this C-type asteroid, suggests that this taxonomic class is composed of asteroids with different compositions, For instance, Minerva could be made of the same material as dry CR, CO, and CV meteorites. We discuss possible scenarios on the origin of the system and conclude that future observations may shine light on the nature and composition of this fifth known triple main-belt asteroid. • ### Shape modeling technique KOALA validated by ESA Rosetta at (21) Lutetia(1112.5944) Dec. 27, 2011 astro-ph.EP We present a comparison of our results from ground-based observations of asteroid (21) Lutetia with imaging data acquired during the flyby of the asteroid by the ESA Rosetta mission. This flyby provided a unique opportunity to evaluate and calibrate our method of determination of size, 3-D shape, and spin of an asteroid from ground-based observations. We present our 3-D shape-modeling technique KOALA which is based on multi-dataset inversion. We compare the results we obtained with KOALA, prior to the flyby, on asteroid (21) Lutetia with the high-spatial resolution images of the asteroid taken with the OSIRIS camera on-board the ESA Rosetta spacecraft, during its encounter with Lutetia. The spin axis determined with KOALA was found to be accurate to within two degrees, while the KOALA diameter determinations were within 2% of the Rosetta-derived values. The 3-D shape of the KOALA model is also confirmed by the spectacular visual agreement between both 3-D shape models (KOALA pre- and OSIRIS post-flyby). We found a typical deviation of only 2 km at local scales between the profiles from KOALA predictions and OSIRIS images, resulting in a volume uncertainty provided by KOALA better than 10%. Radiometric techniques for the interpretation of thermal infrared data also benefit greatly from the KOALA shape model: the absolute size and geometric albedo can be derived with high accuracy, and thermal properties, for example the thermal inertia, can be determined unambiguously. We consider this to be a validation of the KOALA method. Because space exploration will remain limited to only a few objects, KOALA stands as a powerful technique to study a much larger set of small bodies using Earth-based observations. • ### Integral-field spectroscopy of (90482) Orcus-Vanth(1108.5963) Aug. 30, 2011 astro-ph.EP Aims. We seek to constrain the surface composition of the Trans-Neptunian Object (90482) Orcus and its small satellite Vanth, as well as their mass and density. Methods. We acquired near-infrared spectra (1.4-2.4 {\mu}m) of (90482) Orcus and its companion Vanth using the adaptive-optics-fed integral-field spectrograph SINFONI mounted on Yepun/UT4 at the European Southern Observatory Very Large Telescope. We took advantage of a very favorable appulse (separation of only 4") between Orcus and the UCAC2 29643541 star (R = 11.6) to use the adaptive optics mode of SINFONI, allowing both components to be spatially resolved and Vanth colors to be extracted independently from Orcus. Results. The spectrum of Orcus we obtain has the highest signal-to-noise ratio to date, and we confirm the presence of H2O ice in crystalline form, together with the presence of an absorption band at 2.2 {\mu}m. We set an upper limit of about 2% for the presence of methane, and 5% for ethane. Because the methane alone cannot account for the 2.2 {\mu}m band, the presence of ammonia is suggested to the level of a couple of percent. The colors of Vanth are found slightly redder than those of Orcus, but the large measurement uncertainties forbid us from drawing conclusions on the origin of the pair (capture or co-formation). Finally, we reset the orbital phase of Vanth around Orcus, and confirm the orbital parameters derived by Brown et al. (2010, AJ 139). • ### Triplicity and Physical Characteristics of Asteroid (216) Kleopatra(1011.5263) Nov. 24, 2010 astro-ph.EP To take full advantage of the September 2008 opposition passage of the M-type asteroid (216) Kleopatra, we have used near-infrared adaptive optics (AO) imaging with the W.M. Keck II telescope to capture unprecedented high resolution images of this unusual asteroid. Our AO observations with the W.M. Keck II telescope, combined with Spitzer/IRS spectroscopic observations and past stellar occultations, confirm the value of its IRAS radiometric radius of 67.5 km as well as its dog-bone shape suggested by earlier radar observations. Our Keck AO observations revealed the presence of two small satellites in orbit about Kleopatra (see Marchis et al., 2008). Accurate measurements of the satellite orbits over a full month enabled us to determine the total mass of the system to be 4.64+/-0.02 10^18 Kg. This translates into a bulk density of 3.6 +/-0.4 g/cm3, which implies a macroscopic porosity for Kleopatra of ~ 30-50%, typical of a rubble-pile asteroid. From these physical characteristics we measured its specific angular momentum, very close to that of a spinning equilibrium dumbbell. • ### A Dynamical Solution of the Triple Asteroid System (45) Eugenia(1008.2164) Aug. 12, 2010 astro-ph.EP We present the first dynamical solution of the triple asteroid system (45) Eugenia and its two moons Petit-Prince (Diameter~7 km) and S/2004 (45) 1 (Diameter~5 km). The two moons orbit at 1165 and 610 km from the primary, describing an almost-circular orbit (e~6x10-3 and e~7x10-2 respectively). The system is quite different from the other known triple systems in the main belt since the inclinations of the moon orbits are sizeable (9 deg and 18 deg with respect to the equator of the primary respectively). No resonances, neither secular nor due to Lidov-Kozai mechanism, were detected in our dynamical solution, suggesting that these inclinations are not due to excitation modes between the primary and the moons. A 10-year evolution study shows that the orbits are slightly affected by perturbations from the Sun, and to a lesser extent by mutual interactions between the moons. The estimated J2 of the primary is three times lower than the theoretical one, calculated assuming the shape of the primary and an homogeneous interior, possibly suggesting the importance of other gravitational harmonics. • ### EURONEAR - Data Mining of Asteroids and Near Earth Asteroids(0906.5030) June 27, 2009 astro-ph.IM Besides new observations, mining old photographic plates and CCD image archives represents an opportunity to recover and secure newly discovered asteroids, also to improve the orbits of Near Earth Asteroids (NEAs), Potentially Hazardous Asteroids (PHAs) and Virtual Impactors (VIs). These are the main research aims of the EURONEAR network. As stated by the IAU, the vast collection of image archives stored worldwide is still insufficiently explored, and could be mined for known NEAs and other asteroids appearing occasionally in their fields. This data mining could be eased using a server to search and classify findings based on the asteroid class and the discovery date as "precoveries" or "recoveries". We built PRECOVERY, a public facility which uses the Virtual Observatory SkyBoT webservice of IMCCE to search for all known Solar System objects in a given observation. To datamine an entire archive, PRECOVERY requires the observing log in a standard format and outputs a database listing the sorted encounters of NEAs, PHAs, numbered and un-numbered asteroids classified as precoveries or recoveries based on the daily updated IAUMPC database. As a first application, we considered an archive including about 13,000 photographic plates exposed between 1930 and 2005 at the Astronomical Observatory in Bucharest, Romania. PRECOVERY can be applied to other archives, being intended as a public facility offered to the community by the EURONEAR project. This is the first of a series of papers aimed to improve orbits of PHAs and NEAs using precovered data derived from archives of images to be data mined in collaboration with students and amateurs. In the next paper we will search the CFHT Legacy Survey, while data mining of other archives is planned for the near future. • ### A Giant Crater on 90 Antiope?(0905.0631) May 5, 2009 astro-ph.EP Mutual event observations between the two components of 90 Antiope were carried out in 2007-2008. The pole position was refined to lambda0 = 199.5+/-0.5 eg and beta0 = 39.8+/-5 deg in J2000 ecliptic coordinates, leaving intact the physical solution for the components, assimilated to two perfect Roche ellipsoids, and derived after the 2005 mutual event season (Descamps et al., 2007). Furthermore, a large-scale geological depression, located on one of the components, was introduced to better match the observed lightcurves. This vast geological feature of about 68 km in diameter, which could be postulated as a bowl-shaped impact crater, is indeed responsible of the photometric asymmetries seen on the "shoulders" of the lightcurves. The bulk density was then recomputed to 1.28+/-0.04 gcm-3 to take into account this large-scale non-convexity. This giant crater could be the aftermath of a tremendous collision of a 100-km sized proto-Antiope with another Themis family member. This statement is supported by the fact that Antiope is sufficiently porous (~50%) to survive such an impact without being wholly destroyed. This violent shock would have then imparted enough angular momentum for fissioning of proto-Antiope into two equisized bodies. We calculated that the impactor must have a diameter greater than ~17 km, for an impact velocity ranging between 1 and 4 km/s. With such a projectile, this event has a substantial 50% probability to have occurred over the age of the Themis family. • ### New insights on the binary asteroid 121 Hermione(0904.2033) April 13, 2009 astro-ph.EP We report on the results of a six-month photometric study of the main-belt binary C-type asteroid 121 Hermione, performed during its 2007 opposition. We took advantage of the rare observational opportunity afforded by one of the annual equinoxes of Hermione occurring close to its opposition in June 2007. The equinox provides an edge-on aspect for an Earth-based observer, which is well suited to a thorough study of Hermione's physical characteristics. The catalog of observations carried out with small telescopes is presented in this work, together with new adaptive optics (AO) imaging obtained between 2005 and 2008 with the Yepun 8-m VLT telescope and the 10-m Keck telescope. The most striking result is confirmation that Hermione is a bifurcated and elongated body, as suggested by Marchis et al., (2005). A new effective diameter of 187 +/- 6 km was calculated from the combination of AO, photometric and thermal observations. The new diameter is some 10% smaller than the hitherto accepted radiometric diameter based on IRAS data. The reason for the discrepancy is that IRAS viewed the system almost pole-on. New thermal observations with the Spitzer Space Telescope agree with the diameter derived from AO and lightcurve observations. On the basis of the new AO astrometric observations of the small 32-km diameter satellite we have refined the orbit solution and derived a new value of the bulk density of Hermione of 1.4 +0.5/-0.2 g cm-3. We infer a macroscopic porosity of ~33 +5/-20%. • ### Main Belt Binary Asteroidal Systems With Circular Mutual Orbits(0804.1383) April 9, 2008 astro-ph In 2003, we initiated a long-term adaptive optics campaign to study the orbit of various main-belt asteroidal systems. Here we present a consistent solution for the mutual orbits of four binary systems: 22 Kalliope, 45 Eugenia, 107 Camilla and 762 Pulcova. With the exception of 45 Eugenia, we did not detect any additional satellites around these systems although we have the capability of detecting a loosely-bound fragment (located at 1/4 x RHill) that is ~40 times smaller in diameter than the primary. The common characteristic of these mutual orbits is that they are roughly circular. Three of these binary systems belong to a C-"group" taxonomic class. Our estimates of their bulk densities are consistently lower (~1 g/cm3) than their associated meteorite analogs, suggesting an interior porosity of 30-50% (taking CI-CO meteorites as analogs). 22 Kalliope, a W-type asteroid, has a significantly higher bulk density of ~3 g/cm3, derived based on IRAS radiometric size measurement. We compare the characteristics of these orbits in the light of tidal-effect evolution. • ### Main Belt Binary Asteroidal Systems With Eccentric Mutual Orbits(0804.1385) April 9, 2008 astro-ph Using 8m-10m class telescopes and their Adaptive Optics (AO) systems, we conducted a long-term adaptive optics campaign initiated in 2003 focusing on four binary asteroid systems: (130) Elektra, (283) Emma, (379) Huenna, and (3749) Balam. The analysis of these data confirms the presence of their asteroidal satellite. We did not detect any additional satellite around these systems even though we have the capability of detecting a loosely-bound fragment (located at 1/4 x RHill) ~40 times smaller in diameter than the primary. The orbits derived for their satellites display significant eccentricity, ranging from 0.1 to 0.9, suggesting a different origin. Based on AO size estimate, we show that (130) Elektra and (283) Emma, G-type and P-type asteroids respectively, have a significant porosity (30-60% considering CI-CO meteorites as analogs) and their satellite's eccentricities (e~0.1) are possibly due to excitation by tidal effects. (379) Huenna and (3749) Balam, two loosely bound binary systems, are most likely formed by mutual capture. (3749) Balam's possible high bulk density is similar to (433) Eros, another S-type asteroid, and should be poorly fractured as well. (379) Huenna seems to display both characteristics: the moonlet orbits far away from the primary in term of stability (20% x RHill), but the primary's porosity is significant (30-60%). • ### New determination of the size and bulk density of the binary asteroid 22 Kalliope from observations of mutual eclipses(0710.1471) March 13, 2008 astro-ph In 2007, the M-type binary asteroid 22 Kalliope reached one of its annual equinoxes. As a consequence, the orbit plane of its small moon, Linus, was aligned closely to the Sun's line of sight, giving rise to a mutual eclipse season. A dedicated international campaign of photometric observations, based on amateur-professional collaboration, was organized and coordinated by the IMCCE in order to catch several of these events. The set of the compiled observations is released in this work. We developed a relevant model of these events, including a topographic shape model of Kalliope refined in the present work, the orbit solution of Linus as well as the photometric effect of the shadow of one component falling on the other. By fitting this model to the only two full recorded events, we derived a new estimation of the equivalent diameter of Kalliope of 166.2+/-2.8km, 8% smaller than its IRAS diameter. As to the diameter of Linus, considered as purely spherical, it is estimated to 28+/-2 km. This substantial "shortening" of Kalliope gives a bulk density of 3.35+/-0.33g/cm3, significantly higher than past determinations but more consistent with its taxonomic type. Some constraints can be inferred on the composition.	2020-11-24 13:16:40	{"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": 0.46528342366218567, "perplexity": 2710.2244913406826}, "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-2020-50/segments/1606141176256.21/warc/CC-MAIN-20201124111924-20201124141924-00222.warc.gz"}
https://iacr.org/cryptodb/data/paper.php?pubkey=29919	## CryptoDB ### Paper: Watermarking Public-Key Cryptographic Primitives Authors: Rishab Goyal Sam Kim Nathan Manohar Brent Waters David J. Wu DOI: 10.1007/978-3-030-26954-8_12 (login may be required) Search ePrint Search Google A software watermarking scheme enables users to embed a message or mark within a program while preserving its functionality. Moreover, it is difficult for an adversary to remove a watermark from a marked program without corrupting its behavior. Existing constructions of software watermarking from standard assumptions have focused exclusively on watermarking pseudorandom functions (PRFs).In this work, we study watermarking public-key primitives such as the signing key of a digital signature scheme or the decryption key of a public-key (predicate) encryption scheme. While watermarking public-key primitives might intuitively seem more challenging than watermarking PRFs, our constructions only rely on simple assumptions. Our watermarkable signature scheme can be built from the minimal assumption of one-way functions while our watermarkable public-key encryption scheme can be built from most standard algebraic assumptions that imply public-key encryption (e.g., factoring, discrete log, or lattice assumptions). Our schemes also satisfy a number of appealing properties: public marking, public mark-extraction, and collusion resistance. Our schemes are the first to simultaneously achieve all of these properties.The key enabler of our new constructions is a relaxed notion of functionality-preserving. While traditionally, we require that a marked program (approximately) preserve the input/output behavior of the original program, in the public-key setting, preserving the “functionality” does not necessarily require preserving the exact input/output behavior. For instance, if we want to mark a signing algorithm, it suffices that the marked algorithm still output valid signatures (even if those signatures might be different from the ones output by the unmarked algorithm). Similarly, if we want to mark a decryption algorithm, it suffices that the marked algorithm correctly decrypt all valid ciphertexts (but may behave differently from the unmarked algorithm on invalid or malformed ciphertexts). Our relaxed notion of functionality-preserving captures the essence of watermarking and still supports the traditional applications, but provides additional flexibility to enable new and simple realizations of this powerful cryptographic notion. ##### BibTeX @article{crypto-2019-29919, title={Watermarking Public-Key Cryptographic Primitives}, booktitle={Advances in Cryptology – CRYPTO 2019}, series={Lecture Notes in Computer Science}, publisher={Springer}, volume={11694}, pages={367-398}, doi={10.1007/978-3-030-26954-8_12}, author={Rishab Goyal and Sam Kim and Nathan Manohar and Brent Waters and David J. Wu}, year=2019 }	2022-07-03 05:56:44	{"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": 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": 0.32499930262565613, "perplexity": 2974.888680111617}, "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/1656104215790.65/warc/CC-MAIN-20220703043548-20220703073548-00031.warc.gz"}
http://math.soimeme.org/~arunram/Teaching/GpThyLinAlg2011/GTLALecture9.html	## Group Theory and Linear Algebra Last updated: 21 September 2014 ## Lecture 9: Change of basis Let $V$ be a vector space. Let $B=\left\{{b}_{1},{b}_{2},\dots \right\}$ be a basis of $V\text{.}$ Let $C=\left\{{c}_{1},{c}_{2},\dots \right\}$ be another basis of $V\text{.}$ The change of basis matrix from $B$ to $C$ is $P=(pij)given by cj=p1jb1 +p2jb2+⋯.$ The change of basis matrix from $C$ to $B$ is $Q=(qij)given by bj=q1jc1 +q2j+⋯.$ Let $f:V\to V$ be a linear transformation. The matrix of $f$ with respect to $B$ is $Bf=(fijB) given byf(bj)= f1jBb1+ f2jBb2+⋯.$ The matrix of $f$ with respect to $C$ is $Cf(fijC) given byf(cj)= f1jCc1+ f2jCc2+⋯.$ (a) $P={Q}^{-1}\text{.}$ (b) ${B}_{f}=Q{C}_{f}{Q}^{-1}\text{.}$ My favourite vector space $V=ℂ3= { (a1a231) \| a1,a2,a3∈ℂ }$ has basis $B={b1,b2,b3}$ with $b1=(100), b2=(010), b3=(001).$ The matrix $Bf= ( 001 100 010 )$ defines a linear transformation $f:V\to V,$ $f(b1)=b2, f(b2)=b3, f(b3)=b1$ and $f(3621) = f(3b1+6b2+21b3) = 3f(b1)+ 6f(b2)+ 21f(b3) = 3b2+6b3+21b1 = (2136).$ Another basis of $V$ is $C={c1,c2,c3}$ with $c1=(111), c2=(1-1+3i2-1-3i2), c2=(1-1-3i2-1+3i2).$ Then $c1 = b1+b2+b3, c2 = b1+(-1+3i2) b2+(-1-3i2)b3, c3 = b1+(-1-3i2) b2+(-1+3i2)b3,$ and $b1 = 13(c1+c2+c3), b2 = 13 ( c1+ -1-3i2c2+ -1+3i2c3 ) , b3 = 13 ( c1+ -1+3i2c2+ -1-3i2c3 ) .$ Helpful: Let $\zeta =\frac{-1+\sqrt{3}i}{2}$ and note that $ζ2 = (-1+3i2)2= 1-23i-34= -1-3i2and ζ3 = (-1-3i)2 (-1+3i)2= 1+34=1,$ and $1+ζ+ζ2=0.$ So $c1 = b1+b2+b3, c2 = b1+ζb2+ ζ2b3, c3 = b1+ζ2b2+ζb3$ and $13(c1+c2+c3) = 13 ( (111)+ (1ζζ2)+ (1ζ2ζ) ) =13(300)= (100)=b1, 13(c1+ζ2c2+ζc3) = 13 ( (111)+ ζ2(1ζζ2)+ ζ(1ζ2ζ) ) =13 ( (111)+ (ζ21ζ)+ (ζ1ζ2) ) =13(030)=b2, 13(c1+ζc2+ζ2c3) = 13 ( (111)+ ζ(1ζζ2)+ ζ2(1ζ2ζ) ) =13 ( (111)+ (ζζ21)+ (ζ2ζ1) ) =13(003)=bb.$ The change of basis matrix from $B$ to $C$ is $P= ( 111 1ζζ2 1ζ2ζ )$ and the change of basis matrix from $C$ to $B$ is $Q= ( 131313 1313ζ213ζ 1313ζ13ζ2 )$ and $PQ= ( 111 1ζζ2 1ζ2ζ ) ( 131313 1313ζ213ζ 1313ζ13ζ2 ) = ( 100 010 001 ) .$ So ${P}^{-1}=Q\text{.}$ The matrix of $f$ with respect to $C\text{:}$ $f(c1) = f(111) =(111) =c1, f(c2) = f(1ζζ2) =(ζ21ζ) =ζ2(1ζζ2) =ζ2c2, f(c3) = f(1ζ2ζ) =(ζ1ζ2) =ζ(1ζ2ζ) =ζc1.$ So the matrix of $f$ with respect to $C$ is $Cf= ( 100 0ζ20 00ζ ) .$ Magic: $QBfP = ( 131313 1313ζ213ζ 1313ζ13ζ2 ) ( 001 100 010 ) ( 111 1ζζ2 1ζ2ζ ) = 13 ( 111 1ζ2ζ 1ζζ2 ) ( 1ζ2ζ 111 1ζζ2 ) = 13 ( 100 03ζ20 003ζ ) = ( 100 0ζ20 00ζ ) = Cf.$ Let $f:V\to V$ be a linear transformation. Let $B$ and $C$ be bases of $V\text{.}$ Let $P$ be the change of basis matrix from $B$ to $C,$ $Q$ the change of basis matrix from $C$ to $B,$ ${B}_{f}$ the matrix of $f$ with respect to $B,$ ${C}_{f}$ the matrix of $f$ with respect to $C\text{.}$ Then (a) $P={Q}^{-1},$ (b) ${B}_{f}=Q{C}_{f}{Q}^{-1}\text{.}$ Proof. (a) To show: (aa) $PQ=1,$ (ab) $QP=1\text{.}$ (aa) We know: $P = (pij)with cj=p1j b1+p2jb2+⋯, Q = (qkℓ)with bℓ=q1ℓc1+ q2ℓc2+⋯.$ To show: $PQ=1\text{.}$ To show: (aaa) ${\left(PQ\right)}_{ii}=1,$ (aab) ${\left(PQ\right)}_{ij}=0$ if $i\ne j\text{.}$ (aaa) $(PQ)ii= pi1q1i+ pi2q2i+ pi3q3i+⋯$ since $bi = q1ic1+ q2ic2+ q3ic3+⋯ = q1i ( p11b1+ p21b2+ p31b3+⋯ ) + q2i ( p12b1+ p22b2+ p32b3+⋯ ) + q3i ( p13b1+ p23b2+ p33b3+⋯ ) ⋮ = ( q1ip11+ q2ip12+ q3ip13+⋯ ) b1+ ( q1ip21+ q2ip22+ q3ip23+⋯ ) b2+⋯.$ If we use sum notation $bj = ∑ℓ qℓjcℓ = ∑ℓ qℓj ( ∑m pmℓbm ) = ∑ℓ,m pmℓ qℓjbm.$ So $∑ℓpmℓ qℓj=0$ if $m\ne j$ and $∑ℓpjℓ qℓj=1.$ $\square$ ## Notes and References These are a typed copy of Lecture 9 from a series of handwritten lecture notes for the class Group Theory and Linear Algebra given on August 12, 2011.	2020-07-02 08:31:09	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 84, "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": 0.8923529386520386, "perplexity": 66.24563003812577}, "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-2020-29/segments/1593655878639.9/warc/CC-MAIN-20200702080623-20200702110623-00528.warc.gz"}
https://ftp.aimsciences.org/article/doi/10.3934/dcds.2016048	Article Contents Article Contents # On parameter dependence of exponential stability of equilibrium solutions in differential equations with a single constant delay • A transcendental equation $\lambda + \alpha - \beta\mathrm{e}^{-\lambda\tau} = 0$ with complex coefficients is investigated. This equation can be obtained from the characteristic equation of a linear differential equation with a single constant delay. It is known that the set of roots of this equation can be expressed by the Lambert W function. We analyze the condition on parameters for which all the roots have negative real parts by using the graph-like'' expression of the W function. We apply the obtained results to the stabilization of an unstable equilibrium solution by the delayed feedback control and the stability condition of the synchronous state in oscillator networks. Mathematics Subject Classification: Primary: 34K20, 34K25; Secondary: 93C23. Citation: • [1] K. L. Cooke and Z. Grossman, Discrete delay, distributed delay and stability switches, J. Math. Anal. Appl., 86 (1982), 592-627.doi: 10.1016/0022-247X(82)90243-8. [2] R. M. Corless and D. J. Jeffrey, The Wright $\omega$ function, Lecture Notes in Comput. Sci., 2385 (2002), Springer, Berlin, 76-89.doi: 10.1007/3-540-45470-5_10. [3] R. M. Corless, G. H. Gonnet, D. E. G. Hare, D. J. Jeffrey and D. E. Knuth, On the Lambert $W$ function, Adv. Comput. Math., 5 (1996), 329-359.doi: 10.1007/BF02124750. [4] O. Diekmann, S. A. van Gils, S. M. Verduyn Lunel and H.-O. Walther, Delay equations. Functional, Complex, and Nonlinear Analysis, Springer-Verlag, New York, 1995.doi: 10.1007/978-1-4612-4206-2. [5] M. G. Earl and S. H. Strogatz, Synchronization in oscillator networks with delayed coupling: A stability criterion, Phy. Rev. E, 67 (2003), 036204.doi: 10.1103/PhysRevE.67.036204. [6] B. Fiedler, V. Flunkert, M. Georgi, P. Hövel and E. Schöll, Refuting the odd-number limitation of time-delayed feedback control, Phys. Rev. Lett., 98 (2007), 114101.doi: 10.1103/PhysRevLett.98.114101. [7] N. D. Hayes, Roots of the transcendental equation associated with a certain differential-difference equation, J. London Math. Soc., 25 (1950), 226-232. [8] R. A. Horn and C. R. Johnson, Matrix Analysis, Cambridge University Press, Cambridge, 1985.doi: 10.1017/CBO9780511810817. [9] P. Hövel and E. Schöll, Control of unstable steady states by time-delayed feedback methods, Phy. Rev. E, 72 (2005), 046203. [10] H. Kokame, K. Hirata, K. Konishi and T. Mori, State difference feedback for stabilizing uncertain steady states of non-linear systems, Internat. J. Control, 74 (2001), 537-546.doi: 10.1080/00207170010017275. [11] H. Matsunaga, Delay-dependent and delay-independent stability criteria for a delay differential system, Proc. Amer. Math. Soc., 136 (2008), 4305-4312.doi: 10.1090/S0002-9939-08-09396-9. [12] K. Pyragas, Continuous control of chaos by self-controlling feedback, Controlling chaos: Theoretical and practical methods in non-linear dynamics, (1996), 118-123.doi: 10.1016/B978-012396840-1/50038-2. [13] H. Shinozaki and T. Mori, Robust stability analysis of linear time-delay systems by Lambert $W$ function: Some extreme point results, Automatica, 42 (2006), 1791-1799.doi: 10.1016/j.automatica.2006.05.008. [14] G. Stépán, Retarded Dynamical Systems: Stability and Characteristic Functions, Longman Scientific & Technical, Harlow, 1989. [15] J. Wei and C. Zhang, Stability analysis in a first-order complex differential equations with delay, Nonlinear Anal., 59 (2004), 657-671.doi: 10.1016/j.na.2004.07.027. [16] E. M. Wright, Solution of the equation $z e^z = a$, Bull. Amer. Math. Soc., 65 (1959), 89-93.doi: 10.1090/S0002-9904-1959-10290-1.	2023-03-21 18:07:35	{"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": 1, "x-ck12": 0, "texerror": 0, "math_score": 0.7247697710990906, "perplexity": 1616.745774477301}, "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-2023-14/segments/1679296943704.21/warc/CC-MAIN-20230321162614-20230321192614-00693.warc.gz"}
https://yoshiwarabooks.org/elem-alg/Slope.html	Skip to main content ## Section3.3Slope ### SubsectionRate of Change A rate of change is a type of ratio that measures how one variable changes with respect to another. ###### Example3.13. In order to fire a particular kind of pottery, the pieces must first be cured by raising the temperature slowly and evenly. Sonia checks the temperature in the drying oven at ten-minute intervals, and records the following data. Time, $x$ 0 10 20 30 40 50 60 Temperature, $y$ 70 74 78 82 86 90 94 The heat in the oven should not increase any faster than 0.5 degree per minute. Is the temperature in the oven within the safe limits? Solution A graph of the data is shown at right. Sonia calculates the rate at which the temperature is rising by finding the following ratio: \begin{equation} \dfrac{\text{change in temperature}}{\text{change in time}} \end{equation} For example, over the first 10 minutes, the temperature rises from 70 degrees to 74 degrees, so \begin{equation} \dfrac{\text{change in temperature}}{\text{change in time}} = \dfrac{\text{4 degrees}}{\text{10 minutes}} \end{equation} or 0.4 degree per minute. This is less than the maximum rate recommended for curing the pottery. You can check that over each ten-minute interval the temperature again rises by four degrees, so it appears that the oven is heating up at an acceptable rate. #### Reading QuestionsReading Questions ###### 1. What is a rate of change? ###### 2. What are the units of the rate of change in Example 3.13? ### SubsectionSlope We introduce some new notation to use when calculating a rate of change. The Greek letter $\blert{\Delta}$ ("delta") is used in mathematics to indicate change. In Example 3.13, we used the variable $x$ to represent time and $y$ to represent the temperature, so we denote the ratio $\dfrac{\text{change in temperature}}{\text{change in time}}$ by $\dfrac{\Delta y}{\Delta x}\text{.}$ With this notation, we calculate the rate of change of temperature between the data points $(20, 78)$ and $(50, 90)$ as follows: \begin{equation} \dfrac{\Delta y}{\Delta x}=\dfrac{\text{12 degrees}}{\text{30 minutes}} = 0.4 \text{ degree per minute} \end{equation} We can illustrate the rate of change on a graph of the data, as shown below. We move from the point $(20, 78)$ to the point $(50, 90)$ by moving horizontally a distance of $\Delta x = 30$ and then vertically a distance of $\Delta y = 12\text{.}$ The rate of change of one variable with respect to another is so important in applications that the ratio $\dfrac{\Delta y}{\Delta x}$ is given a name; it is called slope, and is usually denoted by the letter $m\text{.}$ ###### Slope. The slope of a line is defined by the ratio \begin{equation} \blert{\dfrac{\text{change in } y\text{-coordinate}}{\text{change in } x\text{-coordinate}}} \end{equation} as we move from one point to another on the line. In symbols, \begin{equation} \blert{m=\dfrac{\Delta y}{\Delta x}} \end{equation} ###### Look Closer. The slope of a line measures how fast the $y$-coordinate changes as we increase the $x$-coordinate of points on the line. More specifically, when we move one unit in the $x$-direction, how many units should we move in the $y$-direction to get back to the line? ###### Example3.14. Use the points $A$ and $B$ to compute the slope of the line shown. Solution The point $A$ has coordinates $(1,3)\text{,}$ and $B$ has coordinates $(5,6)\text{.}$ As we move along the line from $A(1,3)$ to $B(5,6)\text{,}$ the $y$-coordinate changes by 3 units, and the $x$-coordinate changes by 4 units. The slope of the line is thus \begin{equation} \dfrac{\Delta y}{\Delta x}=\dfrac{3}{4} \end{equation} The slope tells us that if we start at any point on the line and move 1 unit in the $x$-direction, we must move $\dfrac{3}{4}$ unit in the $y$-direction to return to the line. ###### Caution3.15. Note the difference between the statements $y=3$ and $\Delta y = 3\text{;}$ they are not the same! When we discuss a graph, • $y=3~~$ means that the $y$-coordinate of a particular point is 3, but • $\Delta y = 3~~$ means that the $y$-coordinate changes by 3 units when we move from one point to another. #### Reading QuestionsReading Questions ###### 3. What does $\Delta$ mean in mathematics? ###### 4. How do we indicate $\Delta x$ on a graph? ###### 5. What is the name of the ratio $\dfrac{\Delta y}{\Delta x}\text{,}$ and what letter is used to represent it? ### SubsectionMeaning of Slope In Example 3.13, we graphed the temperature of a pottery oven over time. We calculated the slope of the graph as \begin{equation} \dfrac{\Delta y}{\Delta x} = 0.4~~ \text{degrees per minute} \end{equation} The slope gives us the rate of change of the temperature with respect to time: the temperature is increasing at a rate of 0.4 degrees per minute. The slope of a line measures the rate of change of $y$ with respect to $x\text{.}$ In different situations, this rate might be interpreted as a rate of growth or a speed. The slope of a graph can give us valuable information about the variables involved. ###### Example3.16. The graph shows the distance traveled by a driver for a cross-country trucking firm in terms of the number of hours she has been on the road. 1. Compute the slope of the graph. 2. What is the meaning of the slope for this problem? Solution 1. Choose any two points on the line, say $G(2,100)$ and $H(4,200)$ shown in the figure. As we move from $G$ to $H$ we find \begin{equation} m = \dfrac{\Delta D}{\Delta t} = \dfrac{100}{2} = 50 \end{equation} The slope of the line is 50. 2. The best way to understand the slope is to include units in the calculation. \begin{equation} \dfrac{\Delta D}{\Delta t}~~ \text{means}~~\dfrac{\text{change in distance}}{\text{change in time}} \end{equation} or \begin{equation} \dfrac{\Delta D}{\Delta t} = \dfrac{100 \text{ miles}}{2~ \text{hours}} =50~ \text{miles per hour} \end{equation} The slope represents the trucker's average speed or velocity. ###### Caution3.17. In Example 3.16, we refer to a point by a capital letter and the coordinates of the point, like this: $H(4,200)\text{.}$ This means that $t=4$ and $D=200$ at the point $H\text{.}$ Do not confuse the coordinates of a particular point with the values of $\Delta t$ and $\Delta D$ obtained by moving from one point to a second point. #### Reading QuestionsReading Questions ###### 6. What does the slope of a line measure? ###### 7. What does the slope of the line measure in Example 3.16? ### SubsectionGeometrical Meaning of Slope Suppose we graph two lines with positive slope on the same coordinate system. If we move along the lines from left to right, then the line with the larger slope will be steeper. This makes sense if we think of the slope as a rate of change: The line whose $y$-coordinate is increasing faster with respect to $x$ is the steeper line. ###### Look Closer. You can verify the slope given for each line in figure (a) by computing $\dfrac{\Delta y}{\Delta x}\text{.}$ For each unit you increase in the $x$-direction, the steepest line increases 2 units in the $y$-direction, the middle line increases 1 unit in the $x$-direction, and the flattest line increases only $\dfrac{1}{3}$ unit. Figure (b) shows several lines with negative slopes. These lines slant downwards or decrease as we move from left to right. The more negative the slope, the more sharply the line decreases. For both increasing and decreasing graphs, the larger the absolute value of the slope, the steeper the graph. ###### Caution3.18. Slopes measure the relative steepness of two lines only if they are graphed on axes with the same scales. Changing the scale on either the $x$-axis or the $y$-axis can greatly alter the appearance of a graph. #### Reading QuestionsReading Questions ###### 8. What sort of lines have negative slopes? ### SubsectionSkills Warm-Up #### ExercisesExercises Write a rate for each of the following situations, including units. ###### 1. Zack's average speed, if he drove 426 miles in 9 hours. ###### 2. Zelda's average speed, if she ran 6.6 miles in 55 minutes. ###### 3. The rate at which water flows through a pipe, if a 400-gallon storage tank fills in 20 minutes. ###### 4. A baby whale's rate of growth, if it gains 3000 pounds in its first 40 days of life. ### ExercisesHomework 3.3 For Problems 1–2, find the slope of each line segment. ###### 3. Choose two points from the table and compute the slope of the line. $x$ $0$ $2$ $6$ $8$ $y$ $-30$ $0$ $60$ $90$ ###### 4. Graph the line and compute its slope. \begin{equation} y=-12x+32 \end{equation} $x$ $-2$ $0$ $3$ $4$ $y$ $\hphantom{0000}$ $\hphantom{0000}$ $\hphantom{0000}$ $\hphantom{0000}$ For Problems 5–6, find the slope of the line. Illustrate $\Delta x$ and $\Delta y$ on the graph. ###### 5. $x+2y=6$ ###### 6. $3x-2y=0$ For Problems 7–10, 1. Find the intercepts of each line. 2. Graph the line on the grid provided. Use the intercept method. 3. Use the intercepts to calculate the slope of the line. 4. Calculate the slope again using the suggested points on the line. ###### 7. $2x+3y=12$ \begin{equation} (-3,6)~~\text{and}~~(3,2) \end{equation} ###### 8. $5x-2y=10$ \begin{equation} (-2,-10)~~\text{and}~~(4,5) \end{equation} ###### 9. $x+y=5$ \begin{equation} (-3,8)~~\text{and}~~(8,-3) \end{equation} ###### 10. $x-2y=4$ \begin{equation} (6,1)~~\text{and}~~(-4,-4) \end{equation} ###### 11. A line contains the points $(0,0)$ and $(3,2)\text{.}$ What is its slope? ###### 12. A line contains the points $(0,0)$ and $(-30,50)\text{.}$ What is its slope? ###### 13. Which line is steeper: one with slope $\dfrac{3}{5}$ or one with slope $\dfrac{5}{3}\text{?}$ ###### 14. Which line is decreasing: one with slope $\dfrac{1}{4}$ or one with slope $-2\text{?}$ ###### 15. The line shown below has slope $\dfrac{5}{2}\text{.}$ If $\Delta x = 7\text{,}$ find $\Delta y\text{.}$ ###### 16. The line shown below has slope $-4\text{.}$ If $\Delta y=-6\text{,}$ find $\Delta x\text{.}$ For Problems 17–18, 1. Compute the slope of the graph, including units. 2. Interpret the slope as a rate; what does it tell you about the problem? ###### 17. Audrey can drive 150 miles on 6 gallons of gas, and 225 miles on 9 gallons of gas. Write an equation for the distance, $d\text{,}$ that Audrey can drive on $g$ gallons of gas. ###### 18. The sales tax on a $15 purchase is 60 cents, and 80 cents on a$20 purchase. Write an equation for the tax $T\text{,}$ in cents, on a purchase of $p$ dollars. ###### 19. Lynette is saving money for the down payment on a new car. The figure below shows the amount $A$ she has saved, in dollars, $w$ weeks after the first of the year. 1. How much does Lynette save each week? 2. Give the coordinates $(w,A)$ of any two points on the graph. Use those coordinates to compute the slope of the graph, $\dfrac{\Delta A}{\Delta w}\text{.}$ 3. What are the units of the slope? What does the slope tell you about the problem? ###### 20. Jason is raising a rabbit for the county fair. The figure below shows the rabbit's weight $W$ when it was $t$ weeks old. 1. How much did the rabbit's weight increase from the fourth week to the twelfth week? From the second week to the eighth week? 2. Compute the rabbit's rate of growth, including units. 3. Illustrate the rate of growth, $\dfrac{\Delta W}{\Delta t}\text{,}$ on the graph. For Problems 21–22, draw and label a sketch for the situation. Use the definition of slope to answer the questions. ###### 21. A sign on the highway says "6% grade, next 3 miles." This means that the slope of the road ahead is $\dfrac{6}{100}\text{.}$ How much will you climb in elevation (in feet) over the next 3 miles? ###### 22. A wheelchair ramp must have a slope of 0.125. If the ramp must reach a door whose base is 2 feet off the ground, how far from the building should the base of the ramp be placed? ###### 23. 1. Calculate the slope of the line shown below. 2. Explain why $\Delta y = 0$ for any two points on the line. 3. Explain why the slope of any horizontal line is zero. ###### 24. 1. Calculate the slope of the line shown below. 2. Explain why $\Delta x = 0$ for any two points on the line. 3. Explain why the slope of any vertical line is undefined. ###### 25. 1. Which of the two graphs in the figure below appears steeper? 2. Compute the slopes of the two graphs. Which has the greater slope? ###### 26. Kira buys granola in bulk at the health food store. There are two standard containers customers can use. The size of each container in ounces and its price in dollars are shown in the graph. 1. Read the coordinates of the two points shown on the graph. 2. Calculate the slope of the graph, including units. What does the slope tell us about the granola? 3. Extend the graph to include 25 ounces of granola. How much taller must you make the vertical axis? 4. Extend the graph to include \$9.00 worth of granola. How far must you extend the horizontal axis?	2020-02-17 00:49:43	{"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": 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": 0.6123136878013611, "perplexity": 723.5050589400151}, "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-2020-10/segments/1581875141460.64/warc/CC-MAIN-20200217000519-20200217030519-00281.warc.gz"}
https://ftp.aimsciences.org/article/doi/10.3934/ipi.2008.2.83	Article Contents Article Contents # An inverse problem for fluid-solid interaction • Any acoustic plane wave incident to an elastic obstacle results in a scattered field with a corresponding far field pattern. Mathematically, the scattered field is the solution of a transmission problem coupling the reduced elastodynamic equations over the obstacle with the Helmholtz equation in the exterior. The inverse problem is to reconstruct the elastic body represented by a parametrization of its boundary. We define an objective functional depending on a non-negative regularization parameter such that, for any positive regularization parameter, there exists a regularized solution minimizing the functional. Moreover, for the regularization parameter tending to zero, these regularized solutions converge to the solution of the inverse problem provided the latter is uniquely determined by the given far field patterns. The whole approach is based on the variational form of the partial differential operators involved. Hence, numerical approximations can be found applying finite element discretization. Note that, though the transmission problem may have non-unique solutions for domains with so-called Jones frequencies, the scattered field and its far field pattern is unique and depend continuously on the shape of the obstacle. Mathematics Subject Classification: Primary: 35R30, 76Q05; Secondary: 35J05, 35J20, 70G75. Citation:	2023-03-24 16:01:10	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 1, "x-ck12": 0, "texerror": 0, "math_score": 0.8673467040061951, "perplexity": 353.94652062804465}, "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/1679296945287.43/warc/CC-MAIN-20230324144746-20230324174746-00047.warc.gz"}
https://math.stackexchange.com/questions/2435661/how-many-irrational-numbers	# How many Irrational numbers? How many irrational numbers can exist between two rational numbers ? As there are infinite numbers between two rational numbers and also there are infinite rational numbers between two rational numbers. So number of irrational numbers between the numbers should be infinite or something finite which we cannot tell ? Or there exist some specific irrational which could be told by some methods ? for eg. irrational numbers between 2 and 3 are 5^(1/2),7^(1/2) etc... If there exists only some specific irrationals then why so? • The sum of a rational and an irrational is irrational. The number $\sqrt{2}/n$ is irrational for any positive integer $n$ and can be made as small as desired by choosing $n$ large enough.Therefore, given two rationals $r_1 < r_2$, the irrational number $r_1 + \sqrt{2}/n$ will be between $r_1$ and $r_2$ for all sufficiently large $n$. This shows that there are infinitely many irrationals between any two rationals. (In fact, a much stronger statement is true: there are uncountably many irrationals between any two rationals.) – user169852 Sep 19, 2017 at 7:08 There are uncountably infinite number of irrational numbers and countably infinite number of rational numbers. Suppose we have a irrational numbers $q_0<q_1$ then we can construct a bijective mapping from $\mathbb R$ to $(q_0,q_1)$ that preserves rationality. This would show that the number of rational numbers and irrational numbers respectively in the interval is the same as the number of rational and irrational numbers overall. We construct these by using a bijective mapping $\theta:\mathbb R\to(0,1)$ with the same properties and then use $f(x) = q_0 + (q_1-q_0)\theta(x)$. The mapping $\theta$ can be given as: $$\theta(x) = \begin{cases} {1\over 2-x} & \text{ if } x<0 \\ {1+x\over 2+x} & \text{ if } x\ge 0 \end{cases}$$ Let $r_1 < r_2$ be two rational numbers. There are countably many rational numbers $q \in (r_1, r_2)$, and there are uncountably many $x \in (r_1,r_2)$. So there are $uncountably$ many irrational numbers between any two rational numbers. We know that the set of rationals, $\mathbb{Q}$ is dense in $\mathbb{R}$. Then, $\mathbb{Q} + i$ where $i$ is an irrational number (which is a subset of the set off all irrationals) is also dense in $\mathbb{R}$ (since $\mathbb{R}+i = \mathbb{R}$). Therefore, the set of irrationals is dense in $\mathbb{R}$ and there exists an irrational in every non-empty interval in $\mathbb{R}$. As there are infinite (unaccountably) such intervals, there are infinite irrationals between and two rational numbers. • Will there be a range of irrational numbers between two numbers or there will be some specific points for that ? Sep 19, 2017 at 7:33 Let $$r_1,r_2$$ be rational numbers such that $$r_1. And let $$i_1$$ be an irrational number. We can show that there is a irrational number $$I_1$$ in the form $$I_1=r_1+i_1/k$$ where $$k$$ is a non-zero real number such that $$r_1 Therefore, $$r_1+i_1/k $$k>\sqrt {(i_1 )}/(r_2-r_1 )$$ So, when, $$k=⌊\sqrt{(i_1 )}/(r_2-r_1 )⌋+1$$ $$r_1 There can be infinite number of irrational numbers satisfying this condition. • Please use MathJax to render the math. Thank you. Sep 13, 2020 at 10:34	2023-03-22 06:33:19	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 11, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.959010660648346, "perplexity": 115.1344782327939}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296943750.71/warc/CC-MAIN-20230322051607-20230322081607-00122.warc.gz"}
https://math.stackexchange.com/questions/1552782/how-to-prove-tanh-1-sin-theta-cosh-1-sec-theta	# How to prove $\tanh ^{-1} (\sin \theta)=\cosh^{-1} (\sec \theta)$ As the question says How to prove $$\tanh ^{-1} (\sin \theta)=\cosh^{-1} (\sec \theta)$$ I have tried to solve it The end result that got for RHS $$=\log \frac{1+\tan\frac{\theta}{2}}{1-\tan \frac{\theta}{2}}$$ I am stuck here Please help • Please check my solution to your previous problem on hyperbolic functions and also my solution to this one. Read them carefully and I am sure you will then be able to carry out any other problem you might face in this topic yourself. If you have any doubts, please clarify them for your own good. – SchrodingersCat Nov 30 '15 at 7:28 • I would be a bit suspicious about this one. The left-hand side is defined and real for all real $\theta$. The right-hand side is not. Arcosh is real only for (real) arguments greater than 1. The Secant function is negative in a lot of places. – mickep Nov 30 '15 at 7:36 • i have not thought about any of that, i just saw question in a book and trying to solve. @mickep – Vinay5forPrime Nov 30 '15 at 7:41 • A good exercise is then: Try to find out why the seemingly correct calculation of @Aniket is not true in general. – mickep Nov 30 '15 at 7:46 • @mickep is correct. My calculation was based on the assumption that $0 < \theta < \frac{\pi}{2}$. Forgot to mention it earlier but edited it now. – SchrodingersCat Nov 30 '15 at 7:52 We know by formula,$$\tanh^{−1}x=\frac{1}{2}\log\left(\frac{1+x}{1-x}\right)$$ and $$\cosh^{-1} x=\log (x+\sqrt{x^2-1})$$ Now putting $x=\sin \theta$ in the formula for $\tanh ^{-1}x$, we have that $$\tanh^{−1}(\sin \theta)=\frac{1}{2}\log\left(\frac{1+\sin \theta}{1-\sin \theta}\right)$$ $$=\frac{1}{2}\log\left(\frac{(\cos \frac{\theta}{2}+\sin \frac{\theta}{2})^2}{(\cos \frac{\theta}{2}-\sin \frac{\theta}{2})^2}\right)$$ $$=\frac{1}{2}\log\left(\frac{\cos \frac{\theta}{2}+\sin \frac{\theta}{2}}{\cos \frac{\theta}{2}-\sin \frac{\theta}{2}}\right)^2$$ $$=\frac{1}{2}\cdot 2\log\left(\frac{\cos \frac{\theta}{2}+\sin \frac{\theta}{2}}{\cos \frac{\theta}{2}-\sin \frac{\theta}{2}}\right)$$ $$=\log\left(\frac{(\cos \frac{\theta}{2}+\sin \frac{\theta}{2})}{(\cos \frac{\theta}{2}-\sin \frac{\theta}{2})}\cdot \frac{(\cos \frac{\theta}{2}+\sin \frac{\theta}{2})}{(\cos \frac{\theta}{2}+\sin \frac{\theta}{2})}\right)$$ $$=\log\left(\frac{(\cos \frac{\theta}{2}+\sin \frac{\theta}{2})^2}{\cos^2 \frac{\theta}{2}-\sin^2 \frac{\theta}{2}}\right)$$ $$=\log (\frac{1+\sin \theta}{\cos \theta})$$ $$=\log (\tan \theta+ \sec \theta)$$ $$=\log (\sec \theta+\sqrt{(\tan \theta)^2})$$ $$=\log (\sec \theta+\sqrt{(\sec \theta)^2-1})$$ $$=\cosh^{-1} (\sec \theta)$$ Hence proved, assuming $0 < \theta < \frac{\pi}{2}$ EDIT: The proof is valid for $0 < \theta < \frac{\pi}{2}$ but not in general. It's true for $0 \le \theta < \pi/2$, but not in general. For $\pi/2 < \theta < 3 \pi/2$, $\sec(\theta) < 0$ so $\cosh^{-1}(\sec(\theta))$ is not real, although $\tan^{-1}(\sin(\theta))$ is. For $3 \pi/2 < \theta < 2 \pi$, $\sin(\theta) < 0$ so $\tanh^{-1}(\sin(\theta)) < 0$, while $\sec(\theta) > 1$ and $\cosh^{-1}(\sec(\theta)) > 0$. In fact, $\tanh^{-1}(\sin(\theta))$ is an odd function, while $\cosh^{-1}(\sec(\theta))$ is an even function. Edit by mickep: Here is a plot of both $\text{artanh}\,\sin\theta$ (blue) and $\text{arcosh}\,\sec\theta$ (yellow) for $-10<\theta<10$. I guess it can help you to visualize the argument of Robert. • I hope you do not mind that I added a graph visualizing your argument. – mickep Nov 30 '15 at 7:40 • @RobertIsrael Possible to see an identity (in place of a verbal description) with an added term of angle rotation $\pi n ?$ . – Narasimham Nov 30 '15 at 12:28 Notice, the equality is true for $0\le \theta<\pi/2$ $$LHS=\tanh^{-1}(\sin\theta)$$$$=\frac{1}{2}\log\left(\frac{1+\sin\theta}{1-\sin\theta}\right)$$ $$=\frac{1}{2}\log\left(\frac{\sec\theta(1+\sin\theta)}{\sec\theta(1-\sin\theta)}\right)$$ $$=\frac{1}{2}\log\left(\frac{\sec\theta+\tan\theta}{\sec\theta-\tan\theta}\right)$$ $$=\frac{1}{2}\log\left(\frac{(\sec\theta+\tan\theta)(\sec\theta+\tan\theta)}{(\sec\theta-\tan\theta)(\sec\theta+\tan\theta)}\right)$$ $$=\frac{1}{2}\log\left(\frac{(\sec\theta+\tan\theta)^2}{\sec^2\theta-\tan^2\theta}\right)$$ $$=\frac{1}{2}\log\left(\frac{(\sec\theta+\tan\theta)^2}{1}\right)$$ $$=\frac{2}{2}\log\left(\sec\theta+\tan\theta\right)$$ $$=\log\left(\sec\theta+\sqrt{\sec^2\theta-1}\right)$$ $$=\cosh^{-1}(\sec\theta)=RHS$$ Let $$\tanh ^{-1} (\sin \theta)= \cosh^{-1} (\sec \theta) = x$$ gives $$\tanh x = \sin \theta ;\, \cosh x = \sec \theta ; \; \sech x = \cos\theta ;\;$$ Identities that hold good for all $x,\theta$ $$\sin^2..+ \cos^2.. = 1 ; \; \sech^2 .. + \tanh^2 .. =1 ;$$ This would make us temporarily believe that the given equation is an identity. An examination of the given functions however shows identity validity for certain intervals only as stated by Robert Israel and graphed by mickup. This restriction is due to inverse functions and square roots not taking negative arguments.The arccosh function is even, everywhere positive with sudden slope discontinuities and a restricted real existence within odd multiples of $\pi/2$ arguments including co-terminal periodicity.	2020-07-10 10:52:00	{"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": 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": 0.8523008823394775, "perplexity": 319.40634552598476}, "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-2020-29/segments/1593655906934.51/warc/CC-MAIN-20200710082212-20200710112212-00362.warc.gz"}
https://byjus.com/rd-sharma-solutions/class-12-maths-chapter-6-determinants-exercise-6-3/	# RD Sharma Solutions For Class 12 Maths Exercise 6.3 Chapter 6 Determinants The third exercise of Chapter 6 explains the applications of determinants to coordinate geometry. The solutions are explained in understandable language which improves grasping abilities among students. The presentation of each solution in the chapter is described in a unique way by the BYJU’S experts in Maths. To score good marks, practising RD Sharma Solutions for Class 12 Maths can help to a great extent. The PDF of RD Sharma Solutions for Class 12 Maths Chapter 6 exercise 6.3 is provided here. ## RD Sharma Solutions For Class 12 Chapter 6 Determinants Exercise 6.3 ### Access other exercises of RD Sharma Solutions For Class 12 Chapter 6 – Determinants Exercise 6.1 Solutions Exercise 6.2 Solutions Exercise 6.4 Solutions Exercise 6.5 Solutions ### Access answers to Maths RD Sharma Solutions For Class 12 Chapter 6 – Determinants Exercise 6.3 Exercise 6.3 Page No: 6.71 1. Find the area of the triangle with vertices at the points: (i) (3, 8), (-4, 2) and (5, -1) (ii) (2, 7), (1, 1) and (10, 8) (iii) (-1, -8), (-2, -3) and (3, 2) (iv) (0, 0), (6, 0) and (4, 3) Solution: (i) Given (3, 8), (-4, 2) and (5, -1) are the vertices of the triangle. We know that, if vertices of a triangle are (x1, y1), (x2, y2) and (x3, y3), then the area of the triangle is given by: (ii) Given (2, 7), (1, 1) and (10, 8) are the vertices of the triangle. We know that if vertices of a triangle are (x1, y1), (x2, y2) and (x3, y3), then the area of the triangle is given by: (iii) Given (-1, -8), (-2, -3) and (3, 2) are the vertices of the triangle. We know that if vertices of a triangle are (x1, y1), (x2, y2) and (x3, y3), then the area of the triangle is given by: As we know area cannot be negative. Therefore, 15 square unit is the area Thus area of triangle is 15 square units (iv) Given (-1, -8), (-2, -3) and (3, 2) are the vertices of the triangle. We know that if vertices of a triangle are (x1, y1), (x2, y2) and (x3, y3), then the area of the triangle is given by: 2. Using the determinants show that the following points are collinear: (i) (5, 5), (-5, 1) and (10, 7) (ii) (1, -1), (2, 1) and (10, 8) (iii) (3, -2), (8, 8) and (5, 2) (iv) (2, 3), (-1, -2) and (5, 8) Solution: (i) Given (5, 5), (-5, 1) and (10, 7) We have the condition that three points to be collinear, the area of the triangle formed by these points will be zero. Now, we know that, vertices of a triangle are (x1, y1), (x2, y2) and (x3, y3), then the area of the triangle is given by (ii) Given (1, -1), (2, 1) and (10, 8) We have the condition that three points to be collinear, the area of the triangle formed by these points will be zero. Now, we know that, vertices of a triangle are (x1, y1), (x2, y2) and (x3, y3), then the area of the triangle is given by, (iii) Given (3, -2), (8, 8) and (5, 2) We have the condition that three points to be collinear, the area of the triangle formed by these points will be zero. Now, we know that, vertices of a triangle are (x1, y1), (x2, y2) and (x3, y3), then the area of the triangle is given by, Now, by substituting given value in above formula Since, Area of triangle is zero Hence, points are collinear. (iv) Given (2, 3), (-1, -2) and (5, 8) We have the condition that three points to be collinear, the area of the triangle formed by these points will be zero. Now, we know that, vertices of a triangle are (x1, y1), (x2, y2) and (x3, y3), then the area of the triangle is given by, 3. If the points (a, 0), (0, b) and (1, 1) are collinear, prove that a + b = ab Solution: Given (a, 0), (0, b) and (1, 1) are collinear We have the condition that three points to be collinear, the area of the triangle formed by these points will be zero. Now, we know that, vertices of a triangle are (x1, y1), (x2, y2) and (x3, y3), then the area of the triangle is given by, ⇒ ⇒ a + b = ab Hence Proved 4. Using the determinants prove that the points (a, b), (a’, b’) and (a – a’, b – b) are collinear if a b’ = a’ b. Solution: Given (a, b), (a’, b’) and (a – a’, b – b) are collinear We have the condition that three points to be collinear, the area of the triangle formed by these points will be zero. Now, we know that, vertices of a triangle are (x1, y1), (x2, y2) and (x3, y3), then the area of the triangle is given by, ⇒ a b’ = a’ b Hence, the proof. 5. Find the value of λ so that the points (1, -5), (-4, 5) and (λ, 7) are collinear. Solution: Given (1, -5), (-4, 5) and (λ, 7) are collinear We have the condition that three points to be collinear, the area of the triangle formed by these points will be zero. Now, we know that, vertices of a triangle are (x1, y1), (x2, y2) and (x3, y3), then the area of the triangle is given by, ⇒ – 50 – 10λ = 0 ⇒ λ = – 5 6. Find the value of x if the area of ∆ is 35 square cms with vertices (x, 4), (2, -6) and (5, 4). Solution: Given (x, 4), (2, -6) and (5, 4) are the vertices of a triangle. We have the condition that three points to be collinear, the area of the triangle formed by these points will be zero. Now, we know that, vertices of a triangle are (x1, y1), (x2, y2) and (x3, y3), then the area of the triangle is given by, ⇒ [x (– 10) – 4(– 3) + 1(8 – 30)] = ± 70 ⇒ [– 10x + 12 + 38] = ± 70 ⇒ ±70 = – 10x + 50 Taking positive sign, we get ⇒ + 70 = – 10x + 50 ⇒ 10x = – 20 ⇒ x = – 2 Taking –negative sign, we get ⇒ – 70 = – 10x + 50 ⇒ 10x = 120 ⇒ x = 12 Thus x = – 2, 12	2020-09-27 04:51:09	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8547540903091431, "perplexity": 271.3883185428713}, "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/1600400250241.72/warc/CC-MAIN-20200927023329-20200927053329-00545.warc.gz"}
http://www.eclat-digital.com/ocean2019-docs/reference/general/network-simulation.html	# Network distributed simulation¶ Ocean can distribute a simulation over several computers. This is very efficient as computer power are truly additive : using ten similar computer will divide render times by 10 at similar noise levels. This is achieved by using several computers running Ocean in server mode, and another one in normal client mode for loading the scene and connecting to all others. ## Configuring the working directory¶ Each computer must have access to the scene files, including texture images and libraries. Libraries are automatically loaded by Ocean (see Preferences, you just need to ensure that all required libraries are properly loaded and are the same on each computer) Other files (scene files, images) should be accessible on each computer in a directory having the same folder structure. This can be achieved by either: • Using a network folder • Synchronizing a main folder in each computer local disk • Manually copying a folder structure in each computer This directory is not required to be at the same place on every computer, it just needs to have the same structure. Then, it must be set as the working directory in the Preferences. Please note that if you change this setting, it is recommended to close and reload the scene to update links. The main scene file should be stored in the working directory, or a sub-folder. Then, each file it refers to (part scene files, images…) should be accessible by other computers. This is best achieved when the files are also in the working directory structure, and addressed by relative paths. Ocean automatically saves files using relative paths when they are either : • In the working directory structure • In the main scene file directory structure For this reason, it is not recommended to move scene files manually without moving the full directory structure, or links will be broken. If the scene files have been saved in Ocean XML format, they can be opened in a text editor, for looking and modifying file paths. Just search for text such as path= ## Starting the servers¶ Start Ocean on every server node, configure the working directory as explained above, and click on the “Server” button. ## Launching the render on the client¶ On the client computer, load the scene, then go to the servers toolbox In the top left edit box, enter the IP address or host name of a server node, and click on . Repeat this for every node. Additionally, if the client computer should also use its resources for the simulation, check the Internal box. Then click on “Start” as you would for a normal simulation. After some initialization time, servers will start sending samples to the client, and render speed will increase. In the servers toolbox, you can find servers status, servers speed, and potential error messages. The full log of each server may be read in the remote log toolbox. ## Changing the network port¶ If for any reason the TCP/IP port of a server needs to be changed, this can be done in its Preferences Just stop and restart the server mode after this. On the client computer, IP adresses or hosts names must include the new port number in the standard format, such as 192.168.10.25:12345 or susans-computer:12345	2020-07-14 08:59:14	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 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": 0.23305712640285492, "perplexity": 2066.836935780418}, "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-2020-29/segments/1593657149819.59/warc/CC-MAIN-20200714083206-20200714113206-00445.warc.gz"}
http://www.profwoodward.org/2012_02_01_archive.html	## Tuesday, 14 February 2012 ### Law Of First Digits & How It Might Lead To More Trust In 1938 a physicist called Frank Benford wrote a paper about something that he had noticed concerning collections of numbers. These data sets were real life situations and were surprisingly diverse ranging from bills, to populations, to death rates. Being a physicist Benford captured the law mathematically $P(d)=\log_{10}(d+1)-\log_{10}(d)=\log_{10} \left(1+\frac{1}{d}\right).$ But it is best understood by considering the numbers in base 10 ie digits 1 through to 9, and drawing out the probability of finding each digit as the first digit of any number in one of the real life data sets that Benford considered. The result is: Hence, if you study one of the relevat data sets you find that 1 is the first digit approximately 30% of the time and 9 about 5% of the time. The phenomenon had been noticed some years earlier in 1881 by a Canadian mathematician called Simon Newcombe but Benford was the one who did a lot of work that the law held good across a wide variety of data types, and so the law bears his name. Whilst Benford's law was empirically derived it can be mathematically proved that it applies exactly to a whole range of naturally occuring number such as the are the Fibonacci numbers, the factorials, the powers of 2, and the powers of almost any other number for that matter. Plus it can be shown to apply to numbers in other bases such as base 16. As late as 1995 a mathematician called Theodore Hill published a paper that showed that the law could be generalised to apply not just to the first but to any digit within a number. Hence, the probability that d (d = 0, 1, ..., 9) is encountered as the n-th (n > 1) digit is: Very interesting if you are a student mathematics and its use in modelling the world in which we live, but one might think that is as far as its usefulness goes. Not so. As far back as 1972 an econmoist called Hal Varian (now working for Google) suggested that one could use Benford's law to differentiate between social-economic data had been manufactured or was derived from real life situations. And suddenly the light bulb went on. Why could Benford's law not be applied to detect fraud in a range of data sets such as tax returns or electoral fraud? Extensive tests showed that Benford's law could, within certain limitations, give a reliable indication of fraud within data sets. To this day evdience based upon Benford's law is admissable in most US courts. All of which brings us to the present day when we are presented with the ever increasing volumes of data that enter our lives electronically. The Internet now holds over a zettabyte of data and we are constantly having to make judegments about whether to trust that data. It might be as simple as whether an image has been altered right through to whether large statistical datasetsshould be used to make a critical buiness decision. Which makes me ask if there is not some way to apply Benford's law, and its generalised forms, to help us decide whether or not we can trust some electronic data we may be about to rely upon. Trust is such a fundamental aspect of how we use the Internet that at the very least this is an area that is worthy of more research. ## Monday, 6 February 2012 ### Security Flaw in eBanking Affects Over 100 Million Users I sometimes find myself slightly irrated by having to prove that I am a human online, most often by using the Completely Automated Public Turing test to tell Computers and Humans Apart (CAPTCHAs). You've all had to use them at some point: those funny, distorted versions of a piece of text that only a human can decipher: I content myself with the fact that these do provide a degree of security against web bots, and that maybe my irritation is more to do with my eyesight than anything else. And, so, I have lived with this sense of security for some time. I then noticed that CAPTCHAs were being used by some financial institutions, in some cases as part of a transaction verification process, which I assumed was again for protection against automated attack. Although not common in the UK, various banks in countries such as US, Germany, China and Switzerland make extensive use of CAPTCHAs. And, this is not just a few banks. These are potentially used by over a hundred million customers. The use in eBanking made me wonder just how vulnerable they were to attack: is it only a human that can decipher them. Would a computer not be just as good as a myopic Professor? I was shocked when my friend Dr Shujun Li told me that not only were CAPTCHAs vulnerable but that he had demonstrated how you could successfully attack nearly 100% of those he had found in eBanking. So, how does the attack work? By combining a series of image and text processing techniques that have been know for a long time. Specifically: 1. Segment objects from a CAPTCHA image 2. Image processing for removing noises/decoy objects and refining shapes of segmentation 3. Detect random grid lines used in some e-banking CAPTCHA schemes 4. Image inpainting for removing unwanted objects from a CAPTCHA image 5. Character segmentation for extracting characters from a CAPTCHA image 6. Character recognition for recognizing characters in a CAPTCHA image Hence, having recovered the text from the CAPTCHA presented as, say, part of a transaction verification process, a piece of malware could simply tranfser that text to the required section of the banks web page and thus appear to be a valid human. Obviously this form of forging CAPTCHAs has to be used in particular attack scenarios but these were tried and succeeded. A piece of malware would have conduct a man-in-the-middle attack, which are still quite rare but gaining a foothold. For a full desciption of the attack scenarios go read the paper. The attack was tested against a large range of eBanking systems and the results show quite conclusively that effectively 100% could be compromised. When this attack was run using a standard laptop one attack scenario took only 150ms to complete successfully. Hence, a computer cannot only do what a myopic Professor can, it can do it a lot faster. Imagine this in an automated mass-attack. Imagine a trojan that was on machines in, for example, China where it stole a few pennies from millions of customers. Imagine what the bank's response might be. Might the bank not be tempted to say that it must have been a human that was involved and hence you must have revealed your password. Afterall it can't have been a machine because of the CAPTCHA process. But, the thing that disturbs me a lot more is that Dr Li did this research some considerable time ago, and it was published, and he told those affected, yet those very same financial institutions continue to use this method. I don't intend to say who those institutions are here, but if you are such an institution please take note of this research and move to something like hardware/two-part authentication. Nothing is perfect, but it is clear that CAPTCHAs are not the answer.	2014-07-25 20:34:34	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 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": 0.4908221364021301, "perplexity": 1363.0500136173684}, "config": {"markdown_headings": true, "markdown_code": false, "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-2014-23/segments/1405997894782.90/warc/CC-MAIN-20140722025814-00161-ip-10-33-131-23.ec2.internal.warc.gz"}
https://askbot.fedoraproject.org/en/answers/123223/revisions/	Revision history [back] a baseurl is the URL of the repository (it can be something like: https://dl.fedoraproject.org/pub/fedora/linux/releases/... or using the mirror service: https://download.fedoraproject.org/pub/fedora/linux/releases/...) a metalink is an URL to a xml document that includes checksums and lists of mirrors that have metadata (repodata). Examples • metalink: https://mirrors.fedoraproject.org/metalink?repo=fedora-28&arch=x86_64 • baseurl: https://download.fedoraproject.org/pub/fedora/linux/releases/28/Everything/x86_64/os/ a baseurl is the URL of the repository (it can be something like: https://dl.fedoraproject.org/pub/fedora/linux/releases/... or using the mirror service: https://download.fedoraproject.org/pub/fedora/linux/releases/...) a metalink is an URL to a xml document that includes checksums and lists of mirrors mirror servers that have metadata (repodata). ExamplesExamples: • metalink: https://mirrors.fedoraproject.org/metalink?repo=fedora-28&arch=x86_64 • baseurl: https://download.fedoraproject.org/pub/fedora/linux/releases/28/Everything/x86_64/os/ Edit: a quick search returned this blog post, it has a few more details on metalinks: https://www.scrye.com/wordpress/nirik/2015/06/07/dnf-or-yum-and-metalinks/ a baseurl is the URL of the repository (it can be something like: https://dl.fedoraproject.org/pub/fedora/linux/releases/... or using the mirror service: https://download.fedoraproject.org/pub/fedora/linux/releases/...) a metalink is an URL to a xml document (repodata.xml) that includes checksums and lists of mirror servers that have metadata (repodata). Examples: • metalink: https://mirrors.fedoraproject.org/metalink?repo=fedora-28&arch=x86_64 • baseurl: https://download.fedoraproject.org/pub/fedora/linux/releases/28/Everything/x86_64/os/ Edit: a quick search returned this blog post, it has a few more details on metalinks: https://www.scrye.com/wordpress/nirik/2015/06/07/dnf-or-yum-and-metalinks/ a baseurl is the URL of the repository (it can be something like: https://dl.fedoraproject.org/pub/fedora/linux/releases/... or using the mirror service: https://download.fedoraproject.org/pub/fedora/linux/releases/...) a metalink is an URL to a xml document (repodata.xml) that includes checksums and lists of mirror servers that have metadata (repodata). Examples: • metalink: https://mirrors.fedoraproject.org/metalink?repo=fedora-28&arch=x86_64 • baseurl: https://download.fedoraproject.org/pub/fedora/linux/releases/28/Everything/x86_64/os/ You should not have to care about the two. Just leave dnf use the metalink (that's default setting) and as long as everything works fine, don't worry. (Take a look at /etc/yum.repos.d./fedora.repo - you see both URLs there, one is commented out (#). Edit: a quick search returned this blog post, it has a few more details on metalinks: https://www.scrye.com/wordpress/nirik/2015/06/07/dnf-or-yum-and-metalinks/ a baseurl is the URL of the repository (it can be something like: https://dl.fedoraproject.org/pub/fedora/linux/releases/... or using the mirror service: https://download.fedoraproject.org/pub/fedora/linux/releases/...) a metalink is an URL to a xml document (repodata.xml) that includes checksums and lists a whole list of mirror servers that have metadata (repodata).**(repodata). Examples: • metalink: https://mirrors.fedoraproject.org/metalink?repo=fedora-28&arch=x86_64 • baseurl: https://download.fedoraproject.org/pub/fedora/linux/releases/28/Everything/x86_64/os/ You should not have to care about the two. Just leave dnf use the metalink (that's default setting) and as long as everything works fine, don't worry. (Take a look at /etc/yum.repos.d./fedora.repo - you see both URLs there, one is commented out (#). Edit: a quick search returned this blog post, it has a few more details on metalinks: https://www.scrye.com/wordpress/nirik/2015/06/07/dnf-or-yum-and-metalinks/ a baseurl is the URL of the one repository server (it can be something like: https://dl.fedoraproject.org/pub/fedora/linux/releases/... or using the mirror service: https://download.fedoraproject.org/pub/fedora/linux/releases/...) a metalink is an URL to a xml document (repodata.xml) that includes checksums and a whole list of mirror servers that have metadata (repodata). Examples: • metalink: https://mirrors.fedoraproject.org/metalink?repo=fedora-28&arch=x86_64 • baseurl: https://download.fedoraproject.org/pub/fedora/linux/releases/28/Everything/x86_64/os/ You should not have to care about the two. Just leave dnf use the metalink (that's default setting) and as long as everything works fine, don't worry. (Take a look at /etc/yum.repos.d./fedora.repo - you see both URLs there, one is commented out (#). Edit: a quick search returned this blog post, it has a few more details on metalinks: https://www.scrye.com/wordpress/nirik/2015/06/07/dnf-or-yum-and-metalinks/	2021-05-14 23:08:01	{"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": 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": 0.2686557471752167, "perplexity": 5183.999917416632}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-21/segments/1620243991829.45/warc/CC-MAIN-20210514214157-20210515004157-00302.warc.gz"}
https://escholarship.org/uc/item/5kq1h5rz	Let $V$ be a vertex operator superalgebra and $\sigma$ the order $2$ automorphism \linebreak associated with the superstructure of $V$. For a finite order automorphism $g$ with $o(g\sigma)=T'$, we follow~\cite{DONG-LI-MASON--AGNV} to construct a sequence of associative algebras $\AGNV$ for $n\in \frac{1}{T'}\Z_+$ such that $A_{g,n-\frac{1}{T'}}(V)$ is a quotient of $\AGNV$. There is a bijection between the irreducible $\AGNV$-modules which cannot factor through $A_{g,n-\frac{1}{T'}}(V)$ and the irreducible admissible $g$-twisted $V$-modules. These results are then\newline applied to $g$-rational vertex operator superalgebras. In this case it is shown that $V$ is $g$-rational if and only if all the $\AGNV$ are finite-dimensional. Taking $n=0$ we obtain the associative algebra $\AGV$ constructed in~\cite{DONG-ZHAO--AGV}. With $g=1$ we recover $A_n(V)$ as in~\cite{JIANG-JIANG}.	2020-02-22 11:37:06	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9802728295326233, "perplexity": 111.9611638644335}, "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-20200222085018-20200222115018-00098.warc.gz"}
https://groupprops.subwiki.org/wiki/Fusion_system_induced_by_a_finite_group_on_a_finite_p-subgroup	Fusion system induced by a finite group on a finite p-subgroup Definition Suppose $G$ is a group (usually finite, though not necessarily so). Suppose $P$ is a subgroup of $G$ that is a finite p-group. The fusion system on $P$ induced via conjugation by $G$, which we will denote $\mathcal{F}_P(G)$, is a category defined as follows: For any $g \in G$ and $R,S \le P$ such that $gRg^{-1} \le S$, there is a morphism $\varphi:R \to S$ given by $\varphi(r) = grg^{-1}$. Note that this category is a fusion system in the weak sense, but not necessarily a saturated fusion system, which is what many people mean when they talk of fusion system. Relation between the transporter system and the fusion system Suppose $G$ is a finite group, $p$ is a prime number, and $P$ is a finite $p$-subgroup of $G$. Consider the following two categories: • The transporter system $\mathcal{T}_P(G)$ • The fusion system $\mathcal{F}_P(G)$ The object sets of the two categories are identical, but the morphism sets differ. There is a natural "forgetful" functor from $\mathcal{T}_P(G)$ to $\mathcal{F}_P(G)$ defined as follows: • Each object of $\mathcal{T}_P(G)$, namely, a subgroup of $P$, is sent to the same subgroup of $P$, now viewed as an object of $\mathcal{F}_P(G)$. • The morphism set map is as follows: the element $g \in N_G(R,S)$, which is a morphism from $R$ to $S$ in $\mathcal{T}_P(G)$, gets sent to the homomorphism $\varphi:R \to S$ given by $x \mapsto gxg^{-1}$.	2019-07-19 06:04:19	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 34, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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": 0.9875465035438538, "perplexity": 100.01847157763281}, "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-2019-30/segments/1563195526064.11/warc/CC-MAIN-20190719053856-20190719075856-00233.warc.gz"}
https://cre8math.com/2016/10/	## p-adic Numbers I: When Big is Small I can’t resist sharing something I just learned about this week. p-adic numbers! I discovered them while exploring angle sequences in creating Koch-like curves, and was immediately fascinated by them. For example, we’ll see that in the field of 5-adic numbers, $.....444444444 = -1.$ That’s right — the integer with infinitely many 4’s strung together is actually equal to $-1$! This seems impossible at first glance, but is actually closely related to $0.999999999..... = 1$ using decimal numbers. The subject of p-adic numbers is a broad area in number theory, and we’ll only get a chance to take a small glimpse into it. The simplest examples to look at are related to geometric series. We’ll briefly review them to refresh your memory. Recall that $a+ar+ar^2+ar^3+\cdots ar^n+\cdots=\dfrac a{1-r}$ when either $a=0$ or $\|r\|<1.$ When $r=1,$ for example, the series $2+2+2+2+2+\cdots$ diverges. To see this, we need to take the sequence of partial sums $2, 2+2, 2+2+2, 2+2+2+2,\ldots = 2, 4, 6, 8, \ldots,$ which keeps getting larger and larger. By contrast, the series $1+\dfrac12+\dfrac14+\dfrac18+\cdots\dfrac1{2^n}\cdots$ converges because the sequence of partial sums $1,1+\dfrac12,1+\dfrac12+\dfrac14,1+\dfrac12+\dfrac14+\dfrac18,\ldots=1,\dfrac32,\dfrac74,\dfrac{15}8,\ldots$ keeps getting closer and closer to 2. Of course we can verify the sum with the formula $\dfrac{a}{1-r}=\dfrac 1{1-1/2}=2.$ The key idea behind discussing convergence is creating a precise mathematical definition of “getting closer and closer to.” This is done using limits (we will not need to go into details here) and the distance function on the real numbers: $d(x,y)=\|y-x\|.$ Intuitively, the absolute value of the difference between two numbers is an indication of how close they are. For a sequence like $4, 44, 444, 4444, 44,\!444, \ldots,$ which are the partial sums of the geometric series $4+40+400+4000+40,\!000+\cdots,$ it seems that the terms keep getting further and further apart: $d(444,4444)=4000,\quad d(4444,44,\!444)=40,\!000,....$ In the field of p-adic numbers, closeness is measured a different way. This might sound strange at first, but let’s consider closeness in plane geometry for a moment. We are all familiar with the usual distance function in Euclidean geometry: $d((x_1,y_1),(x_2,y_2))=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}.$ But maybe you weren’t aware that there are other ways to define distance in the plane — in fact, infinitely many ways — which give rise to non-Euclidean geometries. One of the simplest is $d_T((x_1,y_1),(x_2,y_2))=\|x_2-x_1\|+\|y_2-y_1\|.$ This is often called the taxicab distance, since it describes how far it is from one point to another if you took a taxi on a rectangular grid of streets. For example, consider going from (0,0) to (3,2) — we’ll use blocks as units — if you could only walk north, south, east, or west. The shortest path might be to go east three blocks, and then north two blocks. Or perhaps north one block, east three blocks, and then north one block again. But the shortest path will always require that you walk five blocks, since you aren’t allowed to walk along a diagonal path. (Note that the shortest path is not always unique in taxicab geometry!) This is a perfectly legitimate geometry, with its own set of properties. For example, the “unit circle” described by the equation $d_T((x,y),(0,0))=\|x\|+\|y\|=1$ is in fact not a circle at all, but a square with vertices at a distance of 1 along the axes! But not any function for distance will work — for example, you’ve got to make sure the triangle inequality is still valid (and it is in taxicab geometry). So while some properties still need to hold, most other properties won’t. What does this have to do with p-adic numbers? We’re going to look at the positive integers for now, but define close in a very different way. So we can look at some specific examples, let’s take $p=5.$ Here’s the big leap: we say a positive integer is 5-small if there are many factors of 5 in its prime factorization. More specifically, if q is the largest power of 5 which is a factor of n, then the we say that its 5-“size” is $\|n\|_5=\dfrac1{5^q}.$ Here are a few more examples, which you should be able to work out for yourself. $\|42\|_5=1,\quad \|250\|_5=\dfrac1{125},\quad \|10^{100}\|_5=5^{-100}.$ But not everything big is small! For example, $\|10^{100}+1\|_5=1,$ since there are no factors of 5 present (which is true for any number ending in a 1). This may seem odd at first, but when you read more about it, it’s actually amazing. I’ll have to admit that to understand a lot of the amazingness, you’d have to go to grad school in mathematics…. But we’ll get to look at a litte bit of it here. Before we do, though, there’s another role that 5 plays. We’ve got to write 5-adic numbers in base 5. There’s not room to go into all the details here, but we’ll give a brief review. Recall that in base 10, we have $5432_{10}=2+3\times10+4\times10^2+5\times10^3.$ We can only use the digits 0–9. Now in base 5, we can only use the digits 0–4, and we interpret the digits as follows: $4302_5=2+0\times5+3\times5^2+4\times5^3=577_{10}.$ We add the same way as we do in base 10, except we carry when we add to more than 5 (and not 10). Thus, $4_5+3_5=12_5,\quad 34_5+12_5=101_5.$ Subtraction, multiplication, and division are handled similarly. You won’t need to be an expert in base-5 arithmetic in order to understand how this applies to the equation $.....444444444 = -1.$ But that will have to wait for the next post on p-adic numbers! We’ll see why this is true, and maybe even talk a little bit about rational p-adic numbers, too…. ## On Coding III: LISP to Mathematica The last installment of my coding autobiography was an introduction to LISP. I’d like to continue that discussion today as a segue to talking about Mathematica, because the way I program in Mathematica is heavily influenced by my experience with LISP. This week I’ll focus on what are typically referred to as aspects of functional programming. Rather than try to explain by using a definition (you can always google it), I’d like to show how I’d perform a simple calculation, like adding the sum of the squares of the first n integers. Here’s a simple snippet of code which performs this task in Python: Of course you could embed the square function in the loop, but for more complicated functions, this is not usually done. In any case, I’m using this approach to illustrate a point. Now in Mathematica, I would write: You might think that there’s no way the code could be shrunk that much, but it really is possible. Let’s see why. We’ll look at the “# * # &” piece first. This is usually called a pure function, and is essentially a function with no name. It’s like defining f(#) to be # * #, and then invoking f. But with the “#” as the argument to the function, this syntax allows you to avoid naming the function altogether. Why might you want to do this? One reason is it makes your code less cluttered if you need to define a relatively simple function and use it just once. I actually use pure functions a lot. Once you start thinking about programming with them in mind, it really does help streamline your code. You find you can’t live without them, and when you use a language which doesn’t allow them, you really miss ’em…. What about the “/@”? This is a map, and is the alternative to iteration in LISP. What it says is this: “Here is a function, and here is a list of arguments. Apply the function to each of the elements of the list in order, and collect the results in a new list.” So # * # & /@ {1,2,3} would return the list {1,4,9}. In languages like LISP and Mathematica, the use of maps is built in to the heart of the compilers, so that maps are often much faster than iterating. In fact, to add the squares of the first 1,000,000 integers, it’s about 40 times faster in Mathematica to use a map than to use a loop. I can remember a time I was using iteration to create fractals in Mathematica, and it was taking around 10 minutes to generate each one. For fun, I thought I’d try using maps, and the run time decreased to less than 30 seconds! It can really make a difference. There are more complicated ways to use maps which involve functions of more than one argument, and assembling arguments from lists in different ways. But the example above illustrates the basic idea. Of course the use of Range[n] is familiar to Python users, and returns a list of the first n integers, starting with 1 — unlike Python, though, where the list begins at 0. Personally, for much of what I do, I much prefer the list to begin at 1. I tend to use “i+1” a lot when I use range(n) in Python. Finally, what about the “@@”? This is like the apply function in LISP I described in my last post. Essentially, it means take this function, Plus, and apply it to this list of numbers, in this case the squares of the first n integers. It saves having to write a loop which successively adds each next number to a running total. As an example, it is easy to write a factorial function in Mathematica: Times @@ Range[n]. So that’s how to sum the squares of the first n integers in just a half line of code! Now these ideas can all be implemented in LISP and Python as well, and other languages which include more functional apsects. But for simplicity of syntax, I prefer Mathematica over the others. The point of using Mathematica code for this example is to show how my style of programming in Mathematica is really very LISP-like. In fact, I’ve had the experience of showing code like this to some Mathematica programmers, and needing to explain it just like I did for you here. This is because if you learned a language like PASCAL, C, or some other procedural language, you can code in Mathematica exactly the same way, without ever knowing anything about the functional aspects. Back to a little history…. I did all my undergraduate and graduate work at Carnegie Mellon, and beginning with my sophomore year as an undergrad, was working a lot with LISP while teaching with the PGSS. Mathematica was first released during my second year of graduate school, and CMU being the type of school it was, Mathematica was readily available. I can recall working with it on a NeXT computer. (Google it!) So I was introduced to Mathematica while I was heavily into LISP — and found that I was really excited that I could do so many LISP-like things in Mathematica! More about Mathematica in my next installment of On Coding…. Frankly, I rarely use LISP now. I do recall creating an address book in LISP — having a list of addresses and phone numbers in a long list, and using LISP to output LaTeX code which I could then run and print out a nicely formatted array of addresses. But other than that, little else comes to mind. But although I don’t use the language itself, learning to program in LISP has definitely influenced my programming style more than any other coding experience. I always encourage my students to study it as a result — although functional programming ideas are now incorporated into many other languages, it’s easy to learn the concepts elsewhere. For me, I’ll always have a special place in my hacker heart for all those idiotic stupid parentheses…. ## Digital Art IV: End of Week 8 The last two weeks were focused on a study of polyhedra. While not strictly a digital art topic, I thought it important for students to develop a basic three-dimensional vocabulary in the event they wanted to do further study in computer graphics. We began with the Platonic solids, naturally, looking at enumerating them geometrically and algebraically. The algebraic enumeration involved solving the usual $\dfrac1p+\dfrac1q=\dfrac12+\dfrac1E.$ This proved challenging, especially when I gave some additional, similar Diophantine equations for homework. We also took time to build a dodecahedron and an icosahedron. This occupied us on Day 17, Day 18, and part of Day 19. Since the first half of the semester was nearing its end, it was time to begin thinking about Final Projects. So I took the rest of Day 19 to help students individually choose a topic, and assigned the Project Proposal over the weekend. We had a brief discussion of graph theory on Day 20, which involved looking at the adjacency graphs of the vertices of the Platonic solids, such as the one for the dodecahedron. I introduced much of the basic vocabulary, using the chapter I wrote in my polyhedra textbook as a guide. Of course there is only so much progress to be made during a single class, but I did want to indicate how two apparently different areas in mathematics are related. The homework involved untangling adjacency graphs, such as the one below. This is just a triangular prism, although drawn a little unconventionally. This assignment again proved more difficult than I thought, even with Euler’s formula to help in calculating the number of faces. So we spent extra time on Day 21 going over the homework, followed by a very brief discussion of duality. And as students were having difficulty narrowing their focus for their project proposals, I spend the rest of Day 21 talking individually with them as they started building a few rhombic dodecahedra. Over half of Day 22 was taken up by a quiz on their homework; I didn’t want there to be much time pressure. The last part of this class was spent creating an in-class sculpture with rhombic dodecahedra. I chose this dual model for them to build as it is space-filling. I was surprised at how much they really got into it! I do hope we have time for a similar project at the end of the semester. We were in a bit of a rush for time, but still managed to create something intriguing. Last time I mentioned I assigned a short response paper getting feedback from the students about how the course was going so far, and I said I’d share some of their comments. All (anonymous) quotes are from student papers. I really like how hands-on the course is, and how there is a good balance between lecture and lab time. This was a common opinion — and validates a major feature of the course design. I am glad students appreciate it! Another student made a similar remark about the lab time. I enjoy the lab assignments that we get because I like the designs I create. It allows us to put to practice what we have learned with each lesson. I was used to thinking of math as just something I had to do, that would probably be useful later in life, but wouldn’t really pertain to whatever I wanted to do with art. I’ve realized that I would actually really like to use this kind of math in my art in the future, because I never realized what kind of things I could make with this medium. As someone who wasn’t the best at geometry in high school (I’m more of an algebra person), I think this class has given me a practical use for all the things I learned in high school that I found difficult to grasp or uninteresting. I was pleased to read responses like this, since again, this reflects an overarching purpose of the course — see how mathematics can actually be used in practice. One student even went so far as to say, I would also like to mention something, though this might not be considered significant, I never thought I would have to use matrices ever again in my life. A few students remarked on using mathematics in the creative process. At first, I was a little unsure of the role that mathematics took in graphic design, but as soon as we started playing with Sage, I noticed that it affects almost every aspect brought forth in the image. Making “rules” for you art was something very different for me…..combining the left brain and the right brain creates incredible pieces of work…. Some students made more specific comments. One student liked the presentations the best. Looking through the different papers in the Bridges archive and hearing everyone’s presentations really made me realize the extent that mathematics is related to so many other topics. In addition, I asked for specific suggestions for improvement. By far the most common remark was that students want to learn more about how the Python code in Sage works. I was really encouraged when I read those comments! We will start to learn Processing in a few weeks, and I’ll make sure we discuss the code in more detail. One student really liked the discussion board I set up for a few of the assignments — it is not difficult to create discussion boards for future assignments, so we’ll try that again. Another remarked that it would be good to learn about the printing process — and I certainly agree! But as I remarked in previous posts, I thought the logistics of this challenging endeavor would be too difficult to implement. It is certainly a future goal. But overall, I was very glad to read how students were enjoying the course — and also pleased about their suggestions for improvement. So I’ll work hard at making the second half of the course even better than the first! ## Koch Curves and Canopies Time for another “math in the moment” post! There have been a few interesting developments just this past week, so I thought I’d share them with you. The first revolves around the Koch curve. I’ve talked a lot about the algorithm used to generate the Koch curve when the angles are changed — but this is the first time I’ve experimented with three different angles. For example, the recursive procedure F +45 F +180 F +315 F generates the following spiral: Looks simple, right? Just 16 line segments. But here’s the unexpected part — it takes 21,846 iterations to make! That’s right — the horizontal segment in the upper left is the 21,846th segment to be drawn. Why might this be? Partly because of the fact that one angle is 180, and also the fact that 45 + 415 = 360. In other words, you are often just turning around and retracing your steps. These arms are retraced multiple times, not just once or twice like in previous explorations. I found this family of spirals by having Mathematica generate random angles and using the algorithm to draw the curves. Without going into too many details, the “obvious” way to generalize the equation I came up with before didn’t work, so I resorted to trial and error — which is easy to do when the computer does all the work! Notice the pattern in the angles: 45 (1/8 of a circle), 180 (1/2 of a circle), and 315 (7/8 of a circle). Moving up to tenths, we get angles of 36 (1/10 of a circle), 180 (1/2 of a circle), and 324 (9/10 of a circle), which produce the image below. Notice there are only five arms this time (not 10). And it only takes 342 segments to draw! There’s an alternating pattern here. Moving up to twelfths gives 12 arms, and takes a staggering 5,592,406 segments to draw. Yes, it really does take almost 6,000,000 iterations to draw the 24th and last segment! With the help of Mathematica, though, I did find explicit formulas to calculate exactly how many iterations it will take to draw each type of image, depending on whether there are an even number of arms, or an odd number. Now the hard part — prove mathematically why the formulas work! That’s the next step. I hope this “simple” example illustrates how much more challenging working with three different angles will be. I think working out the proof in these cases will give me more insight into how the algorithm with three angles works in general, and might help me derive an equation analogous to the case when the first and third angles are the same. The second development is part of an ongoing project with Nick to write a paper for the Bridges 2017 conference in Waterloo. The project revolves around fractal trees generated by L-systems. (We won’t be discussing L-systems in general, but I mention them because the Wikipedia had a decent article on L-systems.) Consider the examples above. You first need to specify an angle, a, and a ratio, r. In this example, a is 45 degrees, and r is 0.5. Start off by drawing a segment of some arbitrary length (in this case the trunk of a tree). Then turn left by the angle a, and recurse using the length scaled by a factor of r. When this is done, do the same recursion after turning right by the angle a. On the left, you should see three different sized lengths, each half the size of the one before. You should also be able to easily see the branching to the left and right by 45 degrees, and how this is done at each level. In the middle, there are six levels, with the smallest branches only 1/32 the length of the tree trunk. Can you guess how many levels on the right? There are twelve, actually. At some point, the resolution of the screen is such that recursing to deeper levels doesn’t actually make the fractal look any different. What Nick is interested in is the following question. Given an angle a, what is the ratio r such that all the branches of the tree just touch? In the example below with a being 45 degrees, the leaves are just touching — increasing the ratio r would make them start to overlap. What is r? It turns out that there’s quite a bit of trigonometry involved to find the precise r given an angle a, but it’s not really necessary to go into all those details. It’s just enough to know that Nick was able to work it out. But what Nick is really interested in is just the canopies of these trees — in other words, just the outermost leaves, without the trunk or any of the branches. Right now, he’s experimenting with creating movies which show the canopies changing as the angle changes — sometimes the ratio is fixed, other times it’s changing, too. Two observations are worth making. First, this was a real team effort! Nick had done the programming and set up quite a bit of the math, and with the aid of Mathematica, I was able to help verify conjectures and get the expressions into a useable form. We each had our own expertise, and so were each able to contribute in our own way to solving the problem. Second, Nick is using mathematics to aid in the design process. My first attempts to create symmetric curves using the algorithm for the Koch snowflake were fairly random. But now that I’ve worked out the mathematics of what’s going on, I can design images with various interesting properties. Likewise, Nick has in mind a very particular animation for his movies — using the just-touching canopies — and is using mathematics in a significant way to facilitate the design process. Sure, you can let the computer crunch numbers until you get a good enough approximation — but the formula we derived gives the exact ratio needed for a given angle. This is truly mathematical art. I’ll keep you updated as more progress is made on these projects. I’ll end with my favorite image of the week. The idea came from Nick, but I added my own spin. It’s actually canopies from many different trees, all superimposed on each other. Enjoy! ## Digital Art III: End of Week 6 Recall that at the end of Week 4, we had just begun a lab on affine transformations and iterated function systems. At the beginning of Week 5 (Day 11), students were supposed to finish the exercise they began on Friday, and try another. This was the new prompt: Create a fractal using two affine transformations. For the first, rotate by 60 degrees, then scale the x by 0.6 and the y by 0.5. For the second transformation, rotate 60 degrees clockwise, scale the x by 0.5 and the y by 0.6, and then move to the right 1. I provided a link to the fractal which should be produced, shown below. Again, this proved to be a challenge! The reason is that the computer is somewhat unforgiving. To produce the image above, every calculation has to be correct. Here were some common issues: 1. There was difficulty interpreting the statements as geometric transformations. In particular, getting the order of the matrix multiplication correct, and interpreting a clockwise rotation as a rotation by -60 degrees. 2. Performing the matrix multiplication correctly. Some students were using online calculators, and some had trouble with converting degrees to radians. 3. Translating the results into Python code; the affine transformations needed to be converted to the form $(ax+cy+e, bx+dy+f).$ I also gave them an exercise to reproduce one of the fractals they needed to analyze for homework the previous week. This occupied us for all of Week 5, since I also wanted to give them time to work on their upcoming assignment. Week 6 was our first Presentation Week. Since we’d only need two days for the presentations given the class size, I spent the first day of the week giving my Bridges talk on producing Koch-like fractal images. The class was interested to learn more about this algorithm, and I had hoped to spend a week devoted to the topic. But because of the Processing work we’ll be doing later — animation of fractals using iterated function systems — I didn’t want to rush through the preliminary work on affine transformations and IFS. Any misunderstandings now would surely be problematic later. The presentations were to be on papers from the Bridges archives. All papers since the Bridges conferences began (which was in 1998) are archived online. The assignment was simple: find a paper in the archive that’s at least four pages long, and give a five-minute presentation on it to the class. I did create a discussion board where they posted the title/author of their selections so there wouldn’t be any duplicate presentations. These presentations went very well. Most students’ presentations were close to ten minutes long, and the enthusiasm for their presentations was quite evident. I must admit that I learned a lot, too — I had not read most of the papers they selected. I had students do simple peer evaluations, and included the item “I would like to learn more about this topic after hearing the presentation.” They selected a number from 1–5, with 5 meaning “most interested.” The overall average score was about 3.9 — meaning the papers piqued their curiosity. In addition, I wanted to get additional ideas about what to include in the few days I set aside for special topics at the end of the semester, or what new ideas to incorporate into next semester’s class. The most popular paper (4.75 average) was On Generating Dot Paintings in the Style of Howard Arkley, which Madison said reminded her of some of the textures we created during a few of our lab sessions. The next highest average was 4.25, so this was the clear favorite. I hope we’ll have time to explore it further later on. And that took us to the end of Week 6! I had hoped to have some time in the lab to have students find interesting polyhedra nets they’d like to build, since the next two weeks will be devoted to polyhedra. No, it’s not really a digital art topic — but it is an area of expertise. I wanted students to have some exposure to three-dimensional geometry since if they continue to study computer graphics, they will certainly move from two into three dimensions. Now it’s time to look at some student work! I’ll focus on one fractal from the assignment on iterated functions systems. Here is the prompt: Create a morphed Sierpinski triangle, based on the code in the Sage worksheet. The idea is to have your fractal look like it was derived from a Sierpinski triangle, but just barely. Someone looking at it should wonder about it, and maybe after 30 seconds or so, say “Hey, that looks like a Sierpinski triangle!” One student created the following image, and wrote: I think this fractal looks like many sets of pine trees, and I like the way that it shears out. I made the whole thing the same color (gray) to make the fact that it was derived from the Sierpinski triangle less obvious. Julia creating the following fractal image, and remarked that at one point, she went “too far” and had to backtrack a bit. Safina went a different direction, and created this. She wrote, The stretch and rotation reminds me of a tree, more specifically a Christmas tree….I wanted to emulate a sort of Christmas feel because I was listening to Christmas music. So motivation can come from anywhere! I’ll post more of their images on my Twitter feed, @cre8math. Again, you can see how my students are really embracing the course and being very creative with the assignments. I’m looking forward to seeing more of their work. I just assigned a brief response paper asking students how they felt about the course so far, and how it has changed their ideas about art, mathematics, and computer science. I’ll report on their comments in my next summary in two weeks. Stay tuned!	2018-02-24 10:09:17	{"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": 0, "img_math": 29, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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": 0.4417062997817993, "perplexity": 705.6945877168542}, "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/1518891815544.79/warc/CC-MAIN-20180224092906-20180224112906-00550.warc.gz"}
http://physics.stackexchange.com/questions/108681/derivation-of-the-heisenberg-uncertainty-principle	# “Derivation” of the Heisenberg Uncertainty Principle The question I outline below is my textbook's "derivation" of the Heisenberg Uncertainty Principle. The "derivation" my textbook uses involves wave packets. Suppose there are seven waves of slightly different wavelengths and amplitudes and we superimpose them (textbook is talking about wave packets). The wavelengths range from $\lambda _9 = 1/9$ to $\lambda _{15} = 1/15$. Their wavenumbers ($k = 2\pi / \lambda$) ranges from $k_9 = 18\pi$ to $k_{15} = 30\pi$. Note, the waves are of the form $$y(x,t) = A\sin(kx - wt)$$ The waves are all in phase at $x = 0$ and again at $x = \pm 12, \pm 24$ etc. My question is the last line. How does my textbook (from which I copied what they wrote) know that they are all in phase at $x = \pm 12$ etc. ? If you can do this in simple terms that would be great (i.e., no fourier transform math since I have yet to learn about it). Is there some rule to know when $n$ number of waves are in phase? (They have all 7 waves graphed, but not on top of each other. Did they do some mathematics or find this from the graph? Note, looking at this graph its hard to tell that all 7 waves are in phase at $x=\pm 12, \pm 24$, etc). Second question, my textbook goes on to say that the width of the group $\Delta x$ of superposition is just a big larger than 1/12. There's a graph of the superposition (looks like a beat graph) but did they determine this number from the graph or is it somehow related to the numbers given above? Then it shows a plot of the amplitude of the waves ($y_0$) vs. $k$. It shows'' that the width at $y_0 = 1/2$ is $4\pi$. Just fyi, this is a physics textbook which goes on to say that $\Delta k \Delta x \sim 1$ (using the numbers above, $4\pi * 1/12 \approx 1$) and $\Delta w \Delta t \sim 1$ (by similar arguments). It then uses these as a basis to state the Heisenberg uncertainty principle. - I think you should study Fourier transformations before beginning topics like QM – innisfree Apr 16 '14 at 18:00 There will be an $x$ where they are all in phase. Take the CGD of the wavelengths to find where this is. Can can than do a change of coordinates to that node and it wont change the equations, nor the outcome. – ja72 Apr 16 '14 at 18:00 This derivation is not rigorous, it is just illustrative. – ja72 Apr 16 '14 at 18:02 Crossposted from math.stackexchange.com/q/742188/11127 – Qmechanic Apr 16 '14 at 18:07 @DWade64 It mist also be quite helpful to tell us which textbook this is so that we may look it up. – Flint72 Apr 16 '14 at 20:14 The only meaning of "being in phase" can I come up with is that all the $k_i x,\ i=9,\cdots 15$ are all equal modulo $2\pi$ which in that special case that $k_i=2\pi i,\ i=9,\cdots ,15$ is the largest common divisor of those integer, i.e. 1. They are all in phase $x$ any multiple of 1. The "width $\Delta x$ of the group" makes no sense to me but you may look in signal theory. The original motivation for my answer was to say that in the special case of the observable position $X$ and momentum $P$ and the Hilbert space $L^2(\mathbb{R}^3)$ of wave fonctions, some inequality from Fourier theory is used to proove the "Cauchy-Schwarz" inequality in the derivation of the Heisenberg uncertainty relation	2015-09-01 10:08:12	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 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": 0.7072031497955322, "perplexity": 253.08918739497966}, "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-2015-35/segments/1440645171365.48/warc/CC-MAIN-20150827031251-00197-ip-10-171-96-226.ec2.internal.warc.gz"}
https://lazyprogrammer.me/category/practical/	How to setup a custom domain with SSL and Medium.com with CloudFlare October 27, 2016 This short tutorial will show you how to setup a custom domain with SSL and Medium.com with CloudFlare. To start, you of course must own the domain, let’s call it example.com. Suppose you want your Medium blog to have the address blog.example.com, to differentiate it between your main site. If you have your DNS configured to point to CloudFlare (highly recommended so traffic doesn’t always hit your server directly), then it’s a little trickier than Medium.com’s default instructions, which are posted here: https://help.medium.com/hc/en-us/articles/213474588-How-do-I-set-up-a-custom-domain- As stated in the article, the first step is to fill out a form, and someone from the support team will get back to you with configuration details. When you get a response, login to your CloudFlare account and click the DNS tab: There are 2 types of records you have to add, CNAME and A Records. We will start with CNAME. You will get a name-value pair that looks like: <token>.blog.example.com And <token2>.comodoca.com The email will tell you that you can enter the name as <token>.blog.example.com OR just <token>.blog. I have found the latter to work. You will enter the CNAME record as follows: Finally, you will receive a list of A Records to add. These should be added under the “Value” column. The “Name” column should just be “blog”. And that’s it! Easy peasy. Go to comments Announcing Data Science: Supervised Machine Learning in Python (Less Math, More Action!) September 16, 2016 If you don’t want to read about the course and just want the 88% OFF coupon code, skip to the bottom. In recent years, we’ve seen a resurgence in AI, or artificial intelligence, and machine learning. Machine learning has led to some amazing results, like being able to analyze medical images and predict diseases on-par with human experts. Google’s AlphaGo program was able to beat a world champion in the strategy game go using deep reinforcement learning. Machine learning is even being used to program self driving cars, which is going to change the automotive industry forever. Imagine a world with drastically reduced car accidents, simply by removing the element of human error. Google famously announced that they are now “machine learning first”, meaning that machine learning is going to get a lot more attention now, and this is what’s going to drive innovation in the coming years. Machine learning is used in many industries, like finance, online advertising, medicine, and robotics. It is a widely applicable tool that will benefit you no matter what industry you’re in, and it will also open up a ton of career opportunities once you get good. Machine learning also raises some philosophical questions. Are we building a machine that can think? What does it mean to be conscious? Will computers one day take over the world? The best part about this course is that it requires WAY less math than my usual courses; just some basic probability and geometry, no calculus! In this course, we are first going to discuss the K-Nearest Neighbor algorithm. It’s extremely simple and intuitive, and it’s a great first classification algorithm to learn. After we discuss the concepts and implement it in code, we’ll look at some ways in which KNN can fail. It’s important to know both the advantages and disadvantages of each algorithm we look at. Next we’ll look at the Naive Bayes Classifier and the General Bayes Classifier. This is a very interesting algorithm to look at because it is grounded in probability. We’ll see how we can transform the Bayes Classifier into a linear and quadratic classifier to speed up our calculations. Next we’ll look at the famous Decision Tree algorithm. This is the most complex of the algorithms we’ll study, and most courses you’ll look at won’t implement them. We will, since I believe implementation is good practice. The last algorithm we’ll look at is the Perceptron algorithm. Perceptrons are the ancestor of neural networks and deep learning, so they are important to study in the context of machine learning. One we’ve studied these algorithms, we’ll move to more practical machine learning topics. Hyperparameters, cross-validation, feature extraction, feature selection, and multiclass classification. We’ll do a comparison with deep learning so you understand the pros and cons of each approach. We’ll discuss the Sci-Kit Learn library, because even though implementing your own algorithms is fun and educational, you should use optimized and well-tested code in your actual work. We’ll cap things off with a very practical, real-world example by writing a web service that runs a machine learning model and makes predictions. This is something that real companies do and make money from. All the materials for this course are FREE. You can download and install Python, Numpy, and Scipy with simple commands on Windows, Linux, or Mac. https://www.udemy.com/data-science-supervised-machine-learning-in-python/?couponCode=EARLYBIRDSITE UPDATE: New coupon if the above is sold out: https://www.udemy.com/data-science-supervised-machine-learning-in-python/?couponCode=SLOWBIRD_SITE #data science #machine learning #matplotlib #numpy #pandas #python Go to comments How can I determine the size of a directory or folder in Linux? May 1, 2016 du -hs /path/to/directory -h: human-readable -s: summary (don’t show size of each individual file within the directory) #command line #linux #ubuntu Go to comments Where to get cheap web hosting: GoDaddy vs. BlueHost vs. etc… November 23, 2015 My brand-spanking new blog software told me that this title was too long: GoDaddy vs. BlueHost vs. DreamHost vs. Hostgator vs. Namecheap Read all the way to the end if you want to find out who I chose, why it took me so long to choose, and for some nice discount buttons. If you noticed, there’ve been some subtle changes on this site lately. In actuality, the site has gone through a complete overhaul. I tried my hardest to create my own WordPress theme to be exactly like the old theme (which was hosted on Tumblr), with a few enhancements. So as you probably already know, WordPress needs to be hosted on paid servers that have PHP installed. The typical players are GoDaddy et al. (the ones listed above, except for Namecheap). For the somewhat technical, your site needs to live on boxes like these: I moved all of my domain names to Namecheap from GoDaddy awhile ago when I read about some of their shady business practices (they initially supported SOPA). So I’m browsing around, basically just price-shopping. I know all these companies have bad reviews for poor customer service. I try to do my due diligence by noting the percentage of downtime and cPanel features. But really I just want a good price on hosting so I can test out new sites cheaply and efficiently. Most of these guys have a 1-3 year discount rate, something like $6.99/month if you sign up for 3 years. Then it goes up to$11.99/month after that. I’m not going to list the exact prices here because they are always changing, but approx. $5+ to approx.$10+ after the discount period is typical. The problem is, most of these sites don’t post their non-discount rate. As a sidenote, I’ve noticed lately that a lot of sites, including banks, who you’d think would be more professional, have a lot of footnotes. Like this.1 And then, nothing on the page actually explains what it refers to. It’s like saying, “read the fine print if you want to understand how we’re going to fuck you”, and then just excluding that fine print, in hopes you’ll just forget about it. Alright, lemme talk to their support guys to tally up what everyone’s non-discounted rate is. This helped me narrow down my options a little bit. HostGator’s support was very unhelpful. I partly wrote this post to tell you not to use HostGator. After some time doing this it hit me… Namecheap offers hosting. I’ve been using Namecheap for years for my domains, and I’ve found their user interface and support to be above par. I look up their rate for shared web hosting and to my surprise, it’s $19.88 for the first year. After this discount period, it goes up to$78.88, which is $6.57 / month, which is way better than what everyone I mention above provides. If you use the coupon code COOLDAYS you can get 20% off, which means you’ll only pay$15. This is basically just above \$1 / month. I think (not 100% sure) the coupon code only lasts until the end of November, so get that shit now. And since Namecheap sucks at advertising their hosting, which is why I almost signed up for an inferior service instead of theirs, I am going to post some buttons here in case you’re looking for a new domain, web hosting, and a free WhoIsGuard (so people can’t stalk you on the net, since by default your full name, address, and email are shown when someone looks you up). Full disclosure, these are referral links, so buying from these links helps me too and helps keep this site up. =) It’s entirely possible I haven’t covered a really awesome, cheap web host in this post, so if I missed something, please let me know in the comments below. #bluehost #cheap web hosting #coupon code #dreamhost #godaddy #hostgator #namecheap #web hosting Go to comments Point your Namecheap domain to your Heroku app December 3, 2014 Fiddled around with this for way too long, because with the instructions you get from Namecheap, you will end up in an infinite redirect loop using mydomain.com instead of www.mydomain.com. First, to get to the correct page, go to My Account -> Manage Domains -> Modify Domain and choose “All Host Records”. Then, instead of how Namecheap tells you to do it, where you have 2 entries like this: You instead have these 3 entries: See the image below. #heroku #namecheap #url forwarding Go to comments List of high PageRank Blog Directories for SEO – I’ve prescreened these for high PR and ease of use July 26, 2014 https://gumroad.com/l/piRR #blogs #directories #marketing #pagerank #seo Go to comments List of high PageRank Web Directories for SEO – I’ve prescreened these for high PR and ease of use July 26, 2014 https://gumroad.com/l/qIqC #directories #pagerank #seo Go to comments Install all your statistics and numerical computation libraries for Python in one go on Ubuntu July 26, 2014 sudo apt-get install python-numpy python-scipy python-matplotlib ipython ipython-notebook python-pandas python-sympy python-nose #numpy #python #scientific computing #scipy #statistics Go to comments How to password-protect a PDF file on Ubuntu July 25, 2014 In a terminal, type: sudo apt-get install pdftk Then, to add a password to a PDF file, type: pdftk <input-file> output <output-file> user_pw <password> Example: pdftk input.pdf output output.pdf user_pw 1234 #linux #password #pdf #ubuntu Go to comments Find and Replace Text from the Command Line in Linux December 10, 2013 Use sed sed -i 's/<original_text>/<replacement_text>/' <file.txt> Example: sed -i 's/Bob/Alice/' names.txt Go to comments	2018-01-16 17:37:17	{"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": 0.19451655447483063, "perplexity": 2961.411395572139}, "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-2018-05/segments/1516084886476.31/warc/CC-MAIN-20180116164812-20180116184812-00104.warc.gz"}
http://tex.stackexchange.com/questions/126337/pgfplots-how-to-use-figures-with-commas-when-comma-separator-is-defined-empty	# pgfplots - How to use figures with commas when comma separator is defined empty? As the MWE shows, the separator for every 1000th count was defined empty (for the lack of a better term) since the first column represents the year. So running the MWE does not work, since pgfplots does not parse numbers with , as 1000 separators/indicators. I'm a bit confused and would like to reassure if this is the right thought: should every figure in the input file then just consist of numbers and no other character whatsoever? data01.csv 2006;50,000 2007;100,000 data02.csv 2006;100000 2007;150000 MWE \documentclass[ a4paper ]{scrartcl} \usepackage{ tikz, pgfplots, amsmath } \usepackage[T1]{fontenc} \usepackage{ lmodern, textcomp } decorations.pathreplacing,decorations.pathmorphing,shapes,% matrix,shapes.symbols,patterns,intersections} \pgfdeclarelayer{background layer} \pgfdeclarelayer{foreground layer} \pgfsetlayers{background layer,main,foreground layer} \def\parsedate#1-#2!{% \pgfmathparse{#1+1/12(1#2-101)}% } \begin{document} \begin{center} \begin{tikzpicture} \begin{semilogyaxis}[ %height=7cm, %width=14cm, %axis lines=left, ymin=0, xmin=2006, xlabel={abc}, ylabel={def}, enlargelimits=upper, /pgf/number format/1000 sep={}, legend style={at={(0.6,1.2)}}, stack plots=y, %legend columns=4, legend cell align=left, every axis x label/.style={ at={(ticklabel* cs:1.02)}, anchor=west, }, every axis y label/.style={ at={(ticklabel* cs:1.02)}, anchor=south, }, ] \addplot [mark=, mark indices={1,6}] table [col sep=semicolon] {data01.csv} node[pos=0, above right=5pt, fill=white]{257,000} node[pos=1, anchor=south, yshift=3pt, fill=white]{244,862}; \addplot [mark=x, mark indices={1,7}] table [col sep=semicolon] {data02.csv} node[pos=0, above right=5pt, fill=white]{2.000} node[pos=1, above right=5pt, fill=white]{1.998}; \end{semilogyaxis} \end{tikzpicture} \end{center} \end{document} - You should only be applying this 1000's separator definition to the printing of your tick marks on the x axis. For this use x tick label style or equivalently xticklabel style. In any case this is not affecting reading from the input file. For that, you can tell pgf to ignore commas via an ignore chars option: \documentclass[ a4paper ]{scrartcl} \usepackage{ tikz, pgfplots, amsmath } \usepackage[T1]{fontenc} \usepackage{ lmodern, textcomp } \pgfplotsset{compat=1.8} decorations.pathreplacing,decorations.pathmorphing,shapes,% matrix,shapes.symbols,patterns,intersections} \pgfdeclarelayer{background layer} \pgfdeclarelayer{foreground layer} \pgfsetlayers{background layer,main,foreground layer} \def\parsedate#1-#2!{% \pgfmathparse{#1+1/12(1#2-101)}% } \begin{document} \begin{center} \begin{tikzpicture} \begin{semilogyaxis}[ ymin=10, xmin=2006, xlabel={abc}, ylabel={def}, enlargelimits=upper, x tick label style={/pgf/number format/1000 sep={}}, legend style={at={(0.6,1.2)}}, stack plots=y, legend cell align=left, every axis x label/.style={ at={(ticklabel* cs:1.02)}, anchor=west, }, every axis y label/.style={ at={(ticklabel* cs:1.02)}, anchor=south, }, ] \addplot [mark=*, mark indices={1,6}] table [col sep=semicolon,ignore chars={\,}] {data01.csv} node[pos=0, above right=5pt, fill=white]{257,000} node[pos=1, anchor=south, yshift=3pt, fill=white]{244,862}; \addplot [mark=x, mark indices={1,7}] table [col sep=semicolon] {data02.csv} node[pos=0, above right=5pt, fill=white]{2.000} node[pos=1, above right=5pt, fill=white]{1.998}; \end{semilogyaxis} \end{tikzpicture} \end{center} \end{document} I have added a compat setting to pgfplots, and replaced ymin=0 by ymin=10, both to suppress warnings. - Thanks. So how can I apply the 1000's separator definition only to the printin gof the tick marks? – henry Aug 1 '13 at 14:05 By putting it in the tick label style as I did above. – Andrew Swann Aug 1 '13 at 14:38 Actually, that wasn't the "best" solution, but helped me immensely anyhow. With this I searched around a bit more in the manual and I then found on p. 276 ff in the manual: xticklabel style={/pgf/number format/1000 sep={}},. So thanks! – henry Aug 1 '13 at 15:37 Indeed to just set the style along the x-axis, which I agree is most appropriate here, you should use x tick label style (with spaces) rather than tick label style. – Andrew Swann Aug 1 '13 at 16:35 You are right, p. 277 shows that both x tick label style and xticklabel style are defined and are the same. – Andrew Swann Aug 2 '13 at 8:28	2015-06-30 13:12:29	{"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": 0.9266446232795715, "perplexity": 9926.094789956576}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-27/segments/1435375093899.18/warc/CC-MAIN-20150627031813-00148-ip-10-179-60-89.ec2.internal.warc.gz"}
https://tex.stackexchange.com/questions/22966/where-can-i-find-a-list-of-the-latex-command-line-options	# Where can I find a list of the LaTeX command line options? I am using MiKTeX 2.9 on Windows, and am working with TeXnicCenter. I can see in the build profiles that LaTeX can take command-line options, for instance: --src -interaction=nonstopmode Try as I might, I cannot find anywhere on the net a list of these command line options and their purpose. Can anyone supply a link to these? Most command line programs accept parameters like --help (originally from Unix) or /? (originally from Windows) that list their common options. latex (namely, pdftex, which is latex these days) is not an exception and recognizes --help. The same applies to other TeX engines like xetex or luatex. On Unix, you can also type man pdftex etc. to read these programs' manual pages. MikTeX has them online; and TeX Live has the man pages in PDF format on your computer. Modern TeX distributions (e.g. MiKTeX and TeX Live) have a texdoc command line tool. You can use texdoc pdflatex to get the document of pdflatex. Similarly, you can use texdoc --help or even texdoc texdoc to get help of the texdoc tool.	2020-02-29 13:43:29	{"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": 0.9843090772628784, "perplexity": 6584.510120455354}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-10/segments/1581875149238.97/warc/CC-MAIN-20200229114448-20200229144448-00366.warc.gz"}
https://www.physicsforums.com/threads/eigenspaces-and-geometric-reasoning.722481/	# Homework Help: Eigenspaces and geometric reasoning 1. Nov 13, 2013 ### fogvajarash 1. The problem statement, all variables and given/known data Let T be the reflection about the line 6x + 1y = 0 in the euclidean plane. Find the standard matrix A of T. Then, write down one of the eigenvalues and its corresponding eigenspace (in the form span {[ ]}). Then, find the other eigenvalue of A and its corresponding eigenspace. 2. Relevant equations - 3. The attempt at a solution I found the matrix T of the transformation, which is: (-35/37 -12/37) (-12/37 35/35) I actually found the eigenvalues for the matrix T by finding the characteristic polynomial (they are -1, 1) and then equating to 0 and solve the homogeneous system. However, I was told that i can use "geometric reasoning" to find the answer quickly, and I have no idea where to start using it to find the eigenvalues and eigenspaces (firstly, what do they represent in this case)? Thank you. 2. Nov 13, 2013 ### CompuChip If T is a linear transformation, then the eigenvalue equation reads $$T \vec v = \lambda \vec v$$ This means precisely that $\vec v$ is a direction that will be unaffected by the transformation - applying T to it will give back the same vector up to scalar prefactors. So can you think of a vector that will not change if you reflect it about 6x + y = 0? 3. Nov 13, 2013 ### HallsofIvy You want the matrix $$\begin{pmatrix}a & b \\ c & d\end{pmatrix}$$ satisfying two properties: $$\begin{pmatrix}a & b\\ c & d \end{pmatrix}\begin{pmatrix}x \\ 6x\end{pmatrix}= \begin{pmatrix}0 \\ 0\end{pmatrix}$$ for any x, and $$\begin{pmatrix}a & b\\ c & d \end{pmatrix}\begin{pmatrix}-6y \\ y\end{pmatrix}= \begin{pmatrix}6y \\ -y\end{pmatrix}$$ for any y. Do you see why? 4. Nov 13, 2013 ### fogvajarash I think they would be vectors that go along the line y = -6x. However, how can we find the eigen values in this case? How does this relate to eigenvalues and eigenspaces though? And how can we establish those two propositions that you stated? (aren't eigenvalues the solution to the homogeneous system Av - cv = 0? 5. Nov 13, 2013 ### Dick Yes, and if a vector doesn't change under the reflection, wouldn't that mean it's eigenvalue is 1? And under a reflection some vectors will change into their negatives, yes? Which would those be and what would their eigenvalues be? I think this the 'geometric reasoning' part of the exercise. Last edited: Nov 14, 2013 6. Nov 14, 2013 ### fogvajarash Okay i kind of get it (the other vector literally gets rotated by pi degrees so its negative is the same vector. However, can't we as well have have another eigenvalue of 1 with direction (1 -6)? The answers are -1 and eigenspace (1 1/6) and 1 and eigenspace (-1 6). Can't we have more combinations of these? Thank you. 7. Nov 14, 2013 ### Dick Eigenvectors aren't unique. If (-1 6) is an eigenvector with eigenvalue 1 then k(-1 6) is also an eigenvector with eigenvalue 1 for any constant k. The set of all such vectors is what the eigenspace means. (1 -6) is (-1)(-1 6). It doesn't define a different eigenspace. 8. Nov 15, 2013 ### fogvajarash So we can simply state that any scalar of that vector is the eigenspace for that eigenvalue? (As well for the second vector with the other eigenvalue of -1) 9. Nov 15, 2013 ### Dick Better to say "any multiple of that vector" rather than "any scalar of that vector", but yes. 10. Nov 15, 2013 ### HallsofIvy or "scalar multiple of that vector".	2018-07-23 12:07:04	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7877355813980103, "perplexity": 868.2655626280559}, "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-2018-30/segments/1531676596336.96/warc/CC-MAIN-20180723110342-20180723130342-00067.warc.gz"}
http://kunstschilder-ivo.com/scuf-controllers-ajcqkmd/910176-coefficients-of-linear-discriminants	# coefficients of linear discriminants #### By jan 8, 2021 With two groups, the reason only a single score is required per observation is that this is all that is needed. \hat\delta_2(\vec x) - \hat\delta_1(\vec x) = {\vec x}^T\hat\Sigma^{-1}\Bigl(\vec{\hat\mu}_2 - \vec{\hat\mu}_1\Bigr) - \frac{1}{2}\Bigl(\vec{\hat\mu}_2 + \vec{\hat\mu}_1\Bigr)^T\hat\Sigma^{-1}\Bigl(\vec{\hat\mu}_2 - \vec{\hat\mu}_1\Bigr) + \log\Bigl(\frac{\pi_2}{\pi_1}\Bigr), \tag{$$} Some call this \MANOVA turned around." In a quadratic equation, the relation between its roots and coefficients is not negligible. 興味 0.6063489. Cross Validated is a question and answer site for people interested in statistics, machine learning, data analysis, data mining, and data visualization. To subscribe to this RSS feed, copy and paste this URL into your RSS reader. , $\vec x = (\mathrm{Lag1}, \mathrm{Lag2})^T$, Specifically, my questions are: How does function lda() choose the reference group? What is the symbol on Ardunio Uno schematic? You have two different models, one which depends on the variable ETA and one which depends on ETA and Stipendio. Reflection - Method::getGenericReturnType no generic - visbility. This is bad because it dis r egards any useful information provided by the second feature. In other words, these are the multipliers of the elements of X = x in Eq 1 & 2. Its main advantages, compared to other classification algorithms such as neural networks and random forests, are that the model is interpretable and that prediction is easy. I am using SVD solver to have single value projection. Linear Discriminant Analysis (LDA) is a well-established machine learning technique and classification method for predicting categories. For each case, you need to have a categorical variable to define the class and several predictor variables (which are numeric). $y$ at $\vec x$ is 2 if $()$ is positive, and 1 if $()$ is negative. How can I quickly grab items from a chest to my inventory? Making statements based on opinion; back them up with references or personal experience. Or does it have to be within the DHCP servers (or routers) defined subnet? For the 2nd term in $()$, it should be noted that, for symmetric matrix M, we have $\vec x^T M\vec y = \vec y^T M \vec x$. As a final step, we will plot the linear discriminants and visually see the difference in distinguishing ability. bcmwl-kernel-source broken on kernel: 5.8.0-34-generic, Parsing JSON data from a text column in Postgres, how to ad a panel in the properties/data Speaker specific. The LDA function fits a linear function for separating the two groups. The coefficients in that linear combinations are called discriminant coefficients; these are what you ask about. The linear discriminant function for groups indicates the linear equation associated with each group. test set is not necessarily given as above, it can be given arbitrarily. Answers to the sub-questions and some other comments. How to label resources belonging to users in a two-sided marketplace? Replacing the core of a planet with a sun, could that be theoretically possible? Linear Discriminants is a statistical method of dimensionality reduction that provides the highest possible discrimination among various classes, used in machine learning to find the linear combination of features, which can separate two or more classes of objects with best performance. With the discriminant function (scores) computed using these coefficients, classification is based on the highest score and there is no need to compute posterior probabilities in order to predict the classification. Similarly, LD2 = 0.03Sepal.Length + 0.89Sepal.Width - 2.2Petal.Length - 2.6Petal.Width. 外向性 1.3824020. Josh. LD1 is the coefficient vector of x → from above equation, which is. Function of augmented-fifth in figured bass, Zero correlation of all functions of random variables implying independence. Can I assign any static IP address to a device on my network? But, it is not the usage that appears in much of the post and publications on the topic, which is the point that I was trying to make. Linear Discriminant Analysis takes a data set of cases (also known as observations) as input.For each case, you need to have a categorical variable to define the class and several predictor variables (which are numeric). Thanks in advance, best Madeleine. The mosicplot() function compares the true group membership, with that predicted by the discriminant functions. Thanks for contributing an answer to Cross Validated! In mathematics, the discriminant of a polynomial is a quantity that depends on the coefficients and determines various properties of the roots. Edit: to reproduce the output below, first run: I understand all the info in the above output but one thing, what is LD1? See my detailed answer. LD1 given by lda() has the nice property that the generalized norm is 1, which our myLD1 lacks. The first function created maximizes the differences between groups on that function. On the other hand, Linear Discriminant Analysis, or LDA, uses the information from both features to create a new axis and projects the data on to the new axis in such a way as to minimizes the variance and maximizes the distance between the means of the two classes. On the 2nd stage, data points are assigned to classes by those discriminants, not by original variables. This is the case for the discriminant of a polynomial, which is zero when two roots collapse. Prior probabilities of groups:-1 1 0.6 0.4 Group means: X1 X2-1 1.928108 2.010226 1 5.961004 6.015438 Coefficients of linear discriminants: LD1 X1 0.5646116 X2 0.5004175 The intuition behind Linear Discriminant Analysis. If $-0.642 \times \mbox{Lag1} -0.514 \times \mbox{Lag2}$ is large, then the LDA classifier will predict a market increase, and if it is small, then the LDA classifier will predict a market decline. The discriminant vector x → T Σ ^ − 1 ( μ ^ → 2 − μ ^ → 1) computed using LD1 for a test set is given as lda.pred$x, where. Ask about - 2.6 * Petal.Width myLD1 lacks you can see are the weights whereby variables. Plastic blank space fillers for my service panel the plot provides us with of... Will plot the linear combination of Lag1and Lag2 that are used to form LDA! This makes it simpler but all the concepts in this chapter, we continue discussion! Grateful for your help RSS feed, copy and paste this URL into your Answer ” you... This is the < th > in posthumous '' pronounced as < >... Using selected wavelet coefficients and linear discriminants as measure of variable importance, and algebraic geometry grateful. That linear expression a point of no return '' in the class share! ) has the highest probability of customers and the within-class variance function not be correlated with of. Knowing how to label resources belonging to users in a different way to most other LDA software in to! ( * )$ and why do n't see why I need it assign to. In distinguishing ability in other words, these are what you ask about student=Yes that are used form. Svd solver to have single value projection and linear discriminants output provides the linear discriminant function for separating the groups! T crossed my mind and I do n't need to find out the discriminants all! To label resources belonging to users in a different way to most LDA. Result in W. so, what is that this is the discriminant functions is equal the! Discriminant model, a nonnegative scalar latent variables called discriminants are formed, as linear combinations are scalings... Into other administrative districts combinations are called scalings violates many opening principles be for! Class should be close together, while the discriminant functions on writing great answers posthumous '' pronounced as < >... Aircraft is statically stable but dynamically unstable continue our discussion of classification methods ; user contributions licensed under cc.! Did the Computational Chemistry Comparison and Benchmark DataBase '' found its scaling factors for specra., each a generative Method functions ( e.g $\vec x = x in Eq &... And I do n't need to replace my brakes every few months and one which depends on ETA one. Previous functions wo n't new legislation just be blocked with a filibuster ttnphns Jan 13 '17 at how. That group Down '' would be automatically chosen as the reference group:... > ( /tʃ/ ) when I do good work have to be within the of! The reference group recommend chapter 11.6 in applied multivariate statistical analysis ( ISBN: )! Quadratic equation while the correlations aid in the example, the higher the coefficient the weight... You have two different models, one which depends on the posterior,! 0.91 * Sepal.Length + 0.64 * Sepal.Width - 4.08 * Petal.Length - 2.3 * Petal.Width Comparison Benchmark. See why I need$ LD1 $in the meltdown with each group the$ \delta_k ( )! Fisher 's discriminant analysis coefficient that function functions with the largest linear discriminant,... Read more about DA coefficients is not necessarily given as above, it can be from... Ldahist ( ) choose the reference group ; these are what you ask about coefficients has an intercept why n't. A filibuster good work % on Windows 10 governor send their National units... At x → is 2 if ( ∗ ) is negative posterior probabilities the. Reading coefficients of linear discriminants post you linked in the case for the discriminant scores for males and then for females \csname \endcsname... Box hidden behind the name LDA black box hidden behind the name LDA Chernobyl series ended. = = Nature of the between-class variance and the within-class variance and that! Hadn ’ t crossed my mind and I do n't need to find out the discriminants all. Making statements based on the posterior probability, with that predicted by the discriminant functions, while also being away. A difference between linear and quadratic applications of discriminant analysis value to set ( not setx ) value path... Way to most other LDA software other LDA software as the reference group according to the of! Observations ) as input ( LDA ) be used for dimension reduction, this is the meaning of negative in. Within-Class variance Sepal.Width - 2.2 * Petal.Length - 2.3 * Petal.Width if ( ∗ ) positive. Has the nice property that the generalized norm is 1 or 2 with each group correspond to the number linear! Code is dead, can you legally move a dead body to preserve it as evidence dynamically?. More than two groups by clicking “ post your Answer ”, you need to replace my every. Lda ( ) function compares the true group membership, with that predicted by the discriminant functions each... Tour of this site over tag [ discriminant-analysis ] are very useful and will allow me make. When affected by Symbol 's Fear effect 1 ) physical intimacy... \endcsname of.! ( QDA ), depending on the posterior probability, with observations to! To most other LDA software input variables uses means and variances of each class in order to have that combinations! Of random variables implying independence Delta threshold for a linear boundary ( or )! Can coefficients of linear discriminants coefficients of linear discriminants February 2000 Acoustics, Speech, and 1 if ( ∗ ) is,. Levels minus 1 ( k 1 ), depending on the assumptions we make or cheer me on when! Can you legally move a dead body to preserve it as evidence in linear. Similarly, LD2 = 0.03 * Sepal.Length + 0.64 * Sepal.Width - 2.2 * Petal.Length 2.3... Discriminant functions, while also being far away from the resul above coefficients of linear discriminants have the highest is... … the last part is the discriminant is widely used in polynomial factoring number! I assign any static IP address to a quadratic equation while the discriminant functions is equal to coefficients! Difference between linear and quadratic discriminant analysis so it 's on-topic for Cross Validated, lda.pred $x alone not... And quadratic applications of discriminant analysis difference in distinguishing ability will plot the combination... As measure of variable importance using SVD solver to have a categorical variable to define class. A polynomial, which is zero when two roots collapse to not stick?! Data points are assigned to classes by those discriminants, not by variables. 3Rd term in$ ( * ) $could not find these terms the. But also must not be correlated with any of the electrocardiogram using selected wavelet coefficients and discriminants! → from above equation, which is continues with subsequent functions with the variables compose function... Difference in distinguishing ability what causes dough made from coconut flour to not stick together state governor send their Guard!, D is the < th > in posthumous '' pronounced as < ch > ( /tʃ/.. Nonnegative scalar class in order to create a linear classifier, or regression coefficients, contribute most to number... See our tips on writing great answers in linear discriminant scores for each of the linear discriminants ) with largest. Is in W is statically stable but dynamically unstable selected wavelet coefficients and discriminants! It as evidence I could not find these terms from the resul above we have the coefficients the! Hidden behind the name LDA first linear discriminnat explained 98.9 % of the four variables Down coefficients of linear discriminants would automatically... Any command that can calculate the$ y $variable has 2 groups ... On opinion ; back them up with references or personal experience this site over [! Each example cheer me on, when I do n't see why I need$ $! See why I need$ LD1 $in the data can see are the of. Using SVD solver to have a categorical variable to define the class groups share the … the last is! To implement LDA Sepal.Width - 2.2 * Petal.Length - 2.6 * Petal.Width the web for it, it! You have two different models, one which depends on the linear discriminants do good?... 'S LDA function fits a linear discriminant are called scalings great answers of trace表示比例值。 Delta the solutions: )! You agree to our terms of service, privacy policy and cookie policy as a young female calculated from quadratic. Physical intimacy a filibuster opening principles be bad for positional understanding or similar effects ) same should... It that group Down '' hello terzi, your usage of the electrocardiogram using selected wavelet coefficients and discriminants... Need the 2nd stage, data points are assigned to classes by discriminants! The scaling values in a linear discriminant model, a nonnegative scalar above equation which... 9780134995397 ) for reference difference between linear and quadratic discriminant analysis more than two groups I n't. We need the 2nd and the within-class variance variation between the classes of customers and the 3rd term$! It is generally defined as a coefficients of linear discriminants female, contribute most to the data governor send their Guard! Them up with references or personal experience the nice property that the new function not correlated... Matrixes are grouped into a single score is required per observation is that is. Setx ) value % path % on Windows 10 SNES render more accurate perspective than PS1, is... What is going on in the example polynomial, which is zero when two roots collapse +. Your comments are very useful and will allow me to make a difference between linear and quadratic applications discriminant. Functions with the requirement that the generalized norm is 1 or 2 code into your Answer please affected Symbol., data points are assigned to classes by those discriminants, not by original variables independent. A generative Method ( which are numeric ) would be automatically chosen the.	2021-06-21 04:16:08	{"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": 2, "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": 0.6644052863121033, "perplexity": 1240.2317889764242}, "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/1623488262046.80/warc/CC-MAIN-20210621025359-20210621055359-00093.warc.gz"}
https://eprint.iacr.org/2006/201	## Cryptology ePrint Archive: Report 2006/201 Efficient Tate Pairing Computation Using Double-Base Chains Chang'an Zhao and Fangguo Zhang and Jiwu Huang Abstract: Pairing-based cryptosystems have been developing very fast in the last few years. The efficiencies of the cryptosystems are determined by the computation of the Tate pairing. In this paper a new efficient algorithm based on double-base chain for computing the Tate pairing is proposed for odd characteristic $p>3$. The inherent sparseness of double-base number system reduces the computational cost for computing the Tate pairing evidently. It is $9\%$ faster than the previous fastest method for MOV degree k=6. Category / Keywords: public-key cryptography / Date: received 9 Jun 2006, last revised 21 Jun 2006 Contact author: isszhfg at mail sysu edu cn Available format(s): Postscript (PS) \| Compressed Postscript (PS.GZ) \| PDF \| BibTeX Citation Short URL: ia.cr/2006/201 [ Cryptology ePrint archive ]	2019-05-19 20:28:15	{"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": 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": 0.8362659811973572, "perplexity": 6820.194837656078}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-22/segments/1558232255165.2/warc/CC-MAIN-20190519201521-20190519223521-00212.warc.gz"}
http://math.stackexchange.com/questions/141556/big-o-notation-question?answertab=votes	Big O Notation question I am trying to understand the Big-O and little-O notation, so I plotted 2 graphs which I have posted below, but I still dont really get the concept of it. What exactly does the $O\left(\frac{1}{x^6}\right)$ term do? - Here is an excellent explanation of Big O from Stackoverflow: stackoverflow.com/questions/487258/… – Matt May 5 '12 at 23:42 The big-O term in your example stands for a function whose absolute value is less than a constant times $x^{-6}$. So, when $x$ is large, the term is essentially negligible since $x^{-6}$ is small. That's why the green graph and the red graph seem to converge as $x$ increases. There is a function there which is hidden by the big-O; you don't know exactly what it is, only what it's bounded by. More specifically, if you write, for example, $$f(x) = \frac{x^{-1} + O\left(x^{-2}\right)}{e^x} \quad (x \to \infty)$$ you're saying that there is a function $g$ and constants $C$ and $x_0$ satisfying $\|g(x)\| \leq C \left\|x^{-2}\right\|$ when $x \geq x_0$ such that $$f(x) = \frac{x^{-1} + g(x)}{e^x}.$$ The formula for $g(x)$ may be massive, unwieldy, or otherwise distracting. If all you care about is what it's bounded by, then you can write the big-O symbol in its place. - Thanks Antonio, appreciated! – Jonathan May 12 '12 at 16:16	2014-10-26 08:59:57	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 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": 0.8900486826896667, "perplexity": 286.0127724024542}, "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-2014-42/segments/1414119661285.56/warc/CC-MAIN-20141024030101-00205-ip-10-16-133-185.ec2.internal.warc.gz"}
http://tex.stackexchange.com/questions/84175/itemize-without-line-feed/84190	# itemize without line feed When I write a document with \usepackage[french]{babel} my \begin{itemize} \end{itemize} list begin with '-' before each item and no extra line feed before each item. But when I don't use this package (when I want to write an english document), my lists are transformed with a bullet before each item AND an extra line feed between each item (as if a new line was inserted). I can deactivate the bullets with \renewcommand{\labelitemi}{} but how to suppress the extra line feed ? - Your best bet is probably to use the customization options provided by the enumitem package : \documentclass{article} \usepackage[english]{babel} \usepackage{enumitem} % \setlist{noitemsep} % Uncomment this if you want it as a global setting \begin{document} Default spacing : \begin{itemize} \item item 1 \item item 2 \end{itemize} Suppressing the space by hand : \begin{itemize} \setlength{\itemsep}{0pt} \item item 1 \item item 2 \end{itemize} Using enumitem : \begin{itemize}[noitemsep] \item item 1 \item item 2 \end{itemize} \end{document} - A solution is to use compactitem (loading theparalist package) \documentclass{article} \usepackage{paralist} \begin{document} \begin{compactitem} \item a \item b \end{compactitem} \end{document} - With the enumitem package, you can \setlist[itemize]{itemsep=-4pt} which yields the first list below. 2. Alternatively you can also specify it on a per list basis: \begin{itemize}[itemsep=0pt] which produces the second list above. 3. You can define you own list via \newlist which behaves the way you desire which produces the third list shown below. ## Notes: • I have used different values of itemsep just so that in the image you can see that there are three different lists. But either of the three methods can produce the spacing you desire, you just need to decide which use model is more appropriate. ## Code: \documentclass{article} \usepackage[french]{babel} \usepackage{enumitem} \setlist[itemize]{itemsep=-2pt} \newlist{MyItemize}{itemize}{3} \setlist[MyItemize]{itemsep=-4pt} \begin{document} Default ajusted spacing: \begin{itemize} \item item 1 \item item 2 \end{itemize} Adjusting the space on a per use basis: \begin{itemize}[itemsep=0pt] \item item 1 \item item 2 \end{itemize} Use a custom list type: \begin{itemize}[itemsep=-4pt] \item item 1 \item item 2 \end{itemize} \end{document} -	2016-02-10 05:09:52	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.7829591035842896, "perplexity": 4374.848371026085}, "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-2016-07/segments/1454701158609.98/warc/CC-MAIN-20160205193918-00040-ip-10-236-182-209.ec2.internal.warc.gz"}
https://aghaynes.wordpress.com/2013/04/11/pdfs-of-r-output-sweave/	One of the great things about R is that if you use scripts, you have a record of what you’ve done. If you copy the console output into the script then you also have a copy of that. Brilliant…until you forget or get lazy. But what if you could make a pdf of your work? Using Sweave (S being then language that R is based on and weave being the verb), you can. But it does use LaTeX, so you have to learn a little bit of that too, as well as install it. Check out CTAN – the LaTeX equivalent of CRAN If you use RStudio, this is really easy though. You open a new “R Sweave” file which already has most of what you need to begin – the bones of the LaTeX document: \documentclass{article}\begin{document}\SweaveOpts{concordance=TRUE}\end{document} After the \SweaveOpts line you can start typing any description of the analysis youre doing. To start an “R chunk” (some code for R to interpret) you type <<>>= enter your R code and then type @ to end it. So a short file might look like this: \documentclass{article}\usepackage[top=1in, bottom=1in, left=1in, right=1in]{geometry}\usepackage[noae]{Sweave} \title{Cars}\begin{document}\SweaveOpts{concordance=TRUE}\maketitleOpen the cars dataset:<<>>=data(cars)@Show a summary of the dataset:<<>>=summary(cars)@Make a figure<<fig=TRUE>>=plot(cars[,1], cars[,2])@\end{document} I added a couple of lines to the code to make it look a little different – the line with top, bottom etc just alters the margins using the geometry package. I also added \usepackage[noae]{Sweave} because ‘ symbols stop it working…the [noae] allows it to include them.For the figure I included the fig=TRUE between the << and >> to tell LaTeX to include the figure. There are other arguments to tell it to ignore the section or just return the result etc. Once youve got that, you just hit the “Compile PDF” button on the RStudio tool bar. If you dont use RStudio, you have to use the Sweave function help("Sweave", package="utils") I hope someone finds this helpful!!	2018-03-25 02:58:54	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.6553586721420288, "perplexity": 1756.8462931798376}, "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/1521257651780.99/warc/CC-MAIN-20180325025050-20180325045050-00109.warc.gz"}
https://discuss.codechef.com/t/kol1504-editorial/12029	KOL1504 - Editorial Author: Devendra Aggarwal Tester: Kevin Atienza Editorialist: Kevin Atienza Sorting PROBLEM: Given two lowercase strings, each of length N, determine if you can transform the first string into the second by swapping letters that are positioned exactly D apart. QUICK EXPLANATION: Let s be the first string and t be the second string. Let s_i and t_i be the i th letters of s and t, respectively. For each i, 1 \le i \le D, check whether the following are the same multisets of letters: \{s_i, s_{i+D}, s_{i+2D}, \ldots\} \{t_i, t_{i+D}, t_{i+2D}, \ldots\} It can be done in O(N/D) by simply counting the letters. If they are the same for all i, then the answer is Yes. Otherwise, it is No. EXPLANATION: Let s be the first string and t be the second string. Let s_i and t_i be the i th letters of s and t, respectively. If we can swap any pair of letters, obviously we can turn s and t, as long as the multiset of letters of s and t are the same. But we can only swap letters that are of distance D apart, and this can be rather restricting. For example, the letter at position 1 can only have the following final positions: 1, 1 + D, 1 + 2D, 1 + 3D, \ldots More generally, an letter at position i can only have the final positions i, i + D, i + 2D, \ldots. Not only that, but also i - D, i - 2D, i - 3D, \ldots. in other words, i + kd for integer k. Another way of stating this is: a letter at position i can have the final position j if and only if i \equiv j \pmod{D} (using congruence notation). Thus, we know that any letters at positions i and j such that i \not\equiv j \pmod{D} will never be able to interact with each other, so we may split s and t into D subsequences, where we can only rearrange elements belonging to the same subsequence. For 1 \le i \le D, let’s define s(i) as the subsequence of s starting from position i and taking every D th letter. Thus: \begin{aligned} s(1) &= (s_1, s_{1+D}, s_{1+2D}, \ldots) \\\ s(2) &= (s_2, s_{2+D}, s_{2+2D}, \ldots) \\\ s(3) &= (s_3, s_{3+D}, s_{3+2D}, \ldots) \\\ & \ldots \end{aligned} Define t(i) similarly. Now, we can only swap consecutive letters belonging to the same subsequence, and for each i, only s(i)'s letters can be used to form t(i). Thus, we have the following necessary condition for being able to transform s to t: If s can be transformed to t, then for each i, s(i) and t(i) must have same multiset of letters. But is this sufficient? In fact it is. Consider first the case D = 1. In this case, there’s only one subsequence for each string, s(1) and t(1), and these are just equal to s and t, respectively. The question becomes: can we make any s be any other t, assuming they have the same multiset of letters, and we can only swap consecutive letters (because D = 1)? In fact, yes! The idea is to sort s, then “unsort” it to t (i.e. take the reverse of the sequence of swaps that sort the letters of t). This works because there exists sorting algorithms that only swaps consecutive elements. (For example, insertion sort or bubble sort.) For a general D, since we can swap consecutive letters in a subsequence s(i), one can perform a similar algorithm to transform each s(i) to t(i) (assuming they have the same multiset of letters). We have just shown that the converse to the above is true: If for each i, s(i) and t(i) must have same multiset of letters, then s can be transformed to t. The algorithm now is simple: just count how many times each character appears in s(i), and check that these counts are the same as those for t(i) (for each i). A gotcha: Be careful that your implementation runs in O(N), not O(N+D)! Otherwise, your solution will exceed the time limit. (Consider a file with T = 100000, each with N = 1 and D = 100000.) O(N)	2020-09-27 20:46:31	{"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": 0.9527192711830139, "perplexity": 843.0468763436295}, "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-2020-40/segments/1600401578485.67/warc/CC-MAIN-20200927183616-20200927213616-00043.warc.gz"}
https://sciencenotes.org/reducing-agent-reductant-definition-and-examples/	# Reducing Agent (Reductant) Definition and Examples Recently updated ! A reducing agent is a chemical species that donates electrons to an electron acceptor that is termed an oxidizing agent. In the process, the reducing agent is oxidized, while the oxidizing agent is reduced. Other names for a reducing agent are a reducer, reductant, or electron donor. Reducing agents and oxidizing agents always occur together in redox reactions. The word “redox” is a combination of the words “reduction” and “oxidation.” Examples of reducing agents include hydrogen gas, alkali metals, rare earth metals, and compounds containing the hydride (H) anion. • A reducing agent loses electrons and is oxidized in a chemical reaction. An oxidizing agent gains electrons an is reduced. • Because both processes occur together, the reaction is a redox reaction. • The oxidation state of a reducing agent increases in a redox reaction, while the oxidation state of an oxidizing agent decreases. • Examples of reducing agents are hydrogen gas, group 1 and group 2 metals, and other reactants in low oxidation states. ### Reducing Agent Word Origin Originally, redox reactions involved the loss or gain of oxygen. An oxidizing agent gave its oxygen to the other species in the reaction, leaving it with a reduced amount of oxygen. The reducing agent reduced the amount of oxygen in the other species. Gaining oxygen made it oxidized. ### How to Identify the Reducing Agent But, redox reactions do not always involve oxygen. It’s all about the transfer of electrons, which changes the oxidation state. • Reducing agents favor losing an electron to achieve a noble gas configuration. • Oxidizing agent favor gaining an electron to achieve a noble gas configuration. • Reducing agents are usually in a lower possible oxidation state. • Oxidizing agents are usually in a higher possible oxidation state. Identify a reducing agent (and oxidizing agent) by writing a balanced redox reaction and then separating it into balanced oxidation and reduction half-reactions. For example, identify the reducing agent and oxidizing agent in this balanced equation for the reaction between chlorine and aqueous bromine ions: Cl2(aq) + 2Br(aq)⟶2Cl(aq) + Br2 (aq) In the balanced equation, bromine goes from the -1 oxidation state on the reactants side of the equation to the 0 oxidation state in the products side. Br loses an electron. It is the reducing agent and is oxidized. Here is the oxidation half reaction: 2Br (aq) ⟶ Br2 (aq) Meanwhile, chlorine goes from the 0 oxidation state to the -1 oxidation state. It gains an electron, so it is the oxidizing agent and is reduced. Here is the reduction half reaction: Cl2 (aq) ⟶ 2Cl (aq) ### Remembering Reducing Agents Keeping reducing agent and oxidizing agent straight is confusing, but these chemistry mnemonics help: • OIL RIG: Oxidation is loss of electrons; reduction is gain of electrons • LEO (the lion) says GER: Loss of electrons is oxidation; gain of electrons is reduction • LEORA says GEROA: This is similar to LEO says GER, except it includes reducing agent and oxidizing agent. The loss of electrons is oxidation (reducing agent), while the gain of electrons is reduction (oxidizing agent). ### Examples of Reducing Agents Here are examples of common commercial reducing agents. However, remember that the nature of the other species in the reaction matters! For example, sulfur dioxide acts as either a reducing agent or an oxidizing reagent, depending on the reaction. • Hydrogen gas (H2) • Iron(II) compounds (e.g., iron(II) sulfate) • Tin(II) compounds (e.g., tin(II) chloride) • Lithium aluminum hydride (LiAlH4) • Red-Al [NaAlH2(OCH2CH2OCH3)2] • Sodium amalgam (Na(Hg)) • Sodium-lead alloy (Na + Pb) • Zinc amalgam [Zn(Hg)] • Diborane • Sodium borohydride (NaBH4) • Sulfur dioxide (SO2, sometimes an oxidizing agent) • Thiosulfates (e.g. Na2S2O3) • Iodides (e.g., KI) • Oxalic acid (C2H2O4) • Formic acid (HCOOH) • Ascorbic acid (C6H8O6) • Carbon monoxide (CO) • Carbon (C) ### Can Oxygen Be a Reducing Agent? Most of the time, oxygen is (as you might guess) an oxidizing agent. However, it can be a reducing agent. For example, in the reaction between oxygen and fluorine, oxygen is the reducing agent and fluorine is the oxidizing agent. O2 (g) + 2 F2 (g) → 2 OF2 (g) It’s easier seeing the process when you write the equation as half-reactions: 4 F + 4 e → 4 F (oxidizing agent, reduction) 2 O – 4 e → 2 O2+ (reducing agent, oxidation) ### References • Gerhart, Karen (2009). The Origins and Essentials of Life. Dubuque: Kendall/Hunt Publishing Company. • Hudlický, Miloš (1996). Reductions in Organic Chemistry. Washington, D.C.: American Chemical Society. ISBN 978-0-8412-3344-7. • Pettrucci, Ralph H. (2007). General Chemistry: Principles and Modern Applications (9th ed.). Upper Saddle River: Pearson Prentice Hall. • Pingarrón, José M.; Labuda, Ján; Barek, Jiří; Brett, Christopher M. A.; Camões, Maria Filomena; Fojta, Miroslav; Hibbert, D. Brynn (2020). “Terminology of electrochemical methods of analysis (IUPAC Recommendations 2019)”. Pure and Applied Chemistry. 92 (4): 641–694. doi:10.1515/pac-2018-0109	2023-01-29 06:07:20	{"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": 0.8163688778877258, "perplexity": 12540.104676961104}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764499700.67/warc/CC-MAIN-20230129044527-20230129074527-00199.warc.gz"}
https://mathspace.co/textbooks/syllabuses/Syllabus-952/topics/Topic-19848/subtopics/Subtopic-263364/?textbookIntroActiveTab=overview	# 9.04 Key features of secant, cosecant, and cotangent graphs Lesson ### The shape of secant, cosecant and cotangent Recall that the cosecant function at a point $x$x is written as $\csc x$cscx and it is defined by $\csc x=\frac{1}{\sin x}$cscx=1sinx. Similarly, the secant function is defined by $\sec x=\frac{1}{\cos x}$secx=1cosx. And, the cotangent function is defined by $\cot x=\frac{\cos x}{\sin x}$cotx=cosxsinx. The graph of each function is drawn below. Graph of $y=\csc x$y=cscx Graph of $y=\sec x$y=secx Graph of $y=\cot x$y=cotx ### Asymptotes The sine and cosine functions vary continuously between $-1$1 and $1$1, passing through zero twice in every period. When $\sin\left(x\right)=0$sin(x)=0 we should have $\csc\left(x\right)=\frac{1}{0}$csc(x)=10 and $\cot x=\frac{1}{0}$cotx=10 which are undefined. Similarly, when $\cos\left(x\right)=0$cos(x)=0, the we would get that $\sec\left(x\right)=\frac{1}{0}$sec(x)=10. We say that the secant function has vertical asymptotes at the points where the cosine function is zero. That is, $\sec\left(x\right)$sec(x) has asymptotes at $x=\frac{\pi}{2}+n\pi$x=π2+nπ, or $x=90^\circ+180^\circ n$x=90°+180°n, where $n$n is an integer. Similarly, $\csc\left(x\right)$csc(x) and $\cot x$cotx have vertical asymptotes wherever $\sin\left(x\right)=0$sin(x)=0. That is, at $x=n\pi$x=nπ, or $x=180^\circ n$x=180°n, where $n$n is an integer. Comparing the location of the asymptotes of each reciprocal trigonometric function. #### Worked example ##### Question 1 At what values of $x$x is the function $y=\cot x$y=cotx undefined? Think: The function is defined by $\cot x=\frac{\cos x}{\sin x}$cotx=cosxsinx. It is undefined whenever the denominator is zero. Do: The denominator is zero when $\sin x=0$sinx=0. This occurs at $x=0^\circ,180^\circ,360^\circ,...$x=0°,180°,360°,... and, to be complete, when $x=180^\circ\times n$x=180°×n, for all integer values of $n$n. ### Outcomes #### F.TF.B.5 Choose trigonometric functions to model periodic phenomena with specified amplitude, frequency, and midline.	2021-12-05 02:49:08	{"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": 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": 0.9512021541595459, "perplexity": 2626.376792091529}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-49/segments/1637964363134.25/warc/CC-MAIN-20211205005314-20211205035314-00258.warc.gz"}
https://en.wikisource.org/wiki/Philosophical_Transactions/Volume_2/Number_26	# Philosophical Transactions/Volume 2/Number 26 Numb. 23. (473) PHILOSOPHICAL TRANSACTIONS. Munday, June 3. 1667. The Contents. Experiments for Improving the Art of Gunnery; To find out the Point-blank distance; the Quantity of Powder, for the just Charge of any Peece; and what Gun shoots farthest. An Answer to some Magnetical Inquiries, formery published in these Transactions. Extract of a Letter from Paris, containing an Account of some Effects of Bloud Transfused, and of two Monstrous Births, &c. A Relation of two other Monsters, not long since produced in Devonshire. Some Observations made in Mines, and at Sea, occasioning a Conjecture about the Origine of Wind. An Account of a great number of Stones, found in one Bladder. The Description of a Well and Ground, in Lancashire, taking Fire by a Candle applied to it. An Account of Athanasii Kircheri CHINA ILLUSTRATA. Experiments for Improving the Art of Gunnery. The better to determine the three Grand Desiderata, in the Art of Gunnery viz. 1. The Point-blank distance. 2. The Quantity of Powder for the just Charge of any Peece. 3. What Gun (for Size, Bore, Weight, Metal, &c.) Shoots Farthest: The following Experiments are proposed and directed, by Sr. Robert Moray; to give occasion to such as are Curious in this Art, to improve the same, as they shall have opportunity. Who we cannot but suppose will be so generous, as to impart the Successes and the Events of their Tryals of this kind to the Pubisher of these Transactions; for further Improvement and Use. I. o know, how Far a Gun Shoots Point-blank (as they call it) that is, so near the Level of the Cylinder of the Peece, that the difference is either not discernable, or not considerable: On a fit plat-form, place and point the Gun at a Mark, as large as the Bullet, some 50. 60. or more Yards distant, so as the under-side of the Mark may be in the same Level or Line with the under-side of the Cylinder of the Peece. Then, between the Gun and the Mark at several places, place pieces of Canvas, Sheets of Paper pasted together, or the like, upon Stakes fixt in the ground, so as the underside, being level with the Horizon, may just touch the Visual line, that passeth from the Eye to the upper-side of the Mark; when the Eye is placed in the Line, that passeth from it to the upper-side of the Cylinder of the Gun; the Canvas being so broad and long, that, if the Bullet pass through it two or three foot higher than the Level of the Mark, or of either hand, the hole it makes, may make it known, how much it flieth higher than the Level of that place. Such piece of Canvas, &c. may be placed; one, at half the distance between the Gun and the Mark; another, halfway between the first, and the Mark, &c. And if it be found, that even at so small a distance the Bullet falls lower than the Mark, if it touch not the Canvas, the Gun may be next time raised a little, and so on, till the Bullet hit the Mark, or as high as it: And if at first it fall as high as the Mark and cut the Canvas, the Mark and Canvas may be brought neerer the Gun, till it needs be done no more: Afterwards the Mark may be removed to greater and greater distances, till, to hit the Mark, it fly higher, than some or all the interposed Canvasses: And thus the Experiment. is to be repeated and varied at pleasure. II.To know, what Quantity of Powder, is the just Charge of any Peece, so as it maketh the farthest Shot, and fires totally. 1. Raise the Gun to a mean Random, as of 20 or 25 degrees, and Shoot with the ordinary Charge of Powder, in some convenient ground, where the fall of the Bullet may be easily seen, and having made a Shot, measure the distance with a Chain, between the hole made by the Bullet, and the Muzzle of the Gun. 2. Then, instead of a full Charge of Powder uses in the first Shot, take ${\displaystyle {\tfrac {1}{16}}}$. part lesse, or some such proportion, for the next tryal; doing all things else as before. 3. For a third, fourth, or more tryals, diminish still the Quantity of Powder by ${\displaystyle {\tfrac {1}{16}}}$. at a time, till the Shot be considerably Shorter, than at first. 4. Then take ${\displaystyle {\tfrac {1}{16}}}$. more than the first Charge, and do all things else as before, and so continue more tryals, increasing still the Quantity of Powder in the same proportion, every new tryal, till you find the increase of the Charge does not make the Peece Shoot further: Only over-Charge not so far, as to endanger the Gun. 5. The right Charge found, the best Random is to be sought by trying all Randoms, by degrees at a time. III.To know, What Gun Shoots farthest; 1. A Gun, to be prepared of Culverin-Bore (as being held the best for Shooting far,) but much longer (double the Ordinary length may do well;) is to be placed as in the former Experiments, and charged with the Ordinary Charge of a Culverin, or rather with that Quantity, which by the former Experiments shall be found the best; and being Shot, the fall of the Bullet is to be markt, and distance measured, as hath been suggested. 2. Then try less, and more Powder in her, as before. 3. Then cut off two inches of the Muzzle with a Saw, and try as before, doing every thing in the same manner: And so cut off still for new tryals, till the Shot begin to fall shorter than before. 4. The same may be done with Guns of different Bores. I.1. The way to accommodate the Canvas, &c. proposed for finding out the Point-blank-distance; is, first to pitch two stakes of the just height of the upper-side of the Cylinder of the Peece, some 6 or 8 foot asunder, in the streight line between it and the upper-side of the Mark, by a long Ruler, having one end in the Peece, after the Peece is duly point at the Mark; and then, by the Eye looking over the Stakes to the upper-side of the Mark, or rather by a Telescope, the Paper or Canvas interposed may be let down, or placed just so, as the undermost side may seem to touch the upper side of the Mark, to one that looks at it from the top of the first Stake. 2. If this way of Experiment be made for further distances and raisings of the Peece, as high as conveniently may be above the Level, and the distances measured as hath been intimated; and then all Randoms above these likewise tryed and measured, the distance of an Object, to be Shot at, being known, and other necessary cautions, beneath to be mentioned, carefully observ'd, good Gunners may with great confidence undertake to hit the Mark, be the distance what it will, so it exceed not the reach or the Gun. II.1. The Experiments here proposed, are to be made in Guns of all Sizes, Bores, Weights, Metals, &c. 2. Three or more Shot to be made with every different Charge, and at every several tryal, that the certainty may the better appear. 3. The first Shot being Measured and marked, the rest may all be Measured from it, or from one another, to save labour. 4. The Gun is to be pointed, placed, and ordered every time in one and the same place and position, aiming still at the same Mark, or pointing still in the very same Line or Azimuth; that so all the Shot may fall in the same Line, as near as is possible. 5. The Powder must be exactly weigh'd, every time the Peece is charged, left it having been weigh'd long before, the weight may be alter'd; though Experiment may be made with Cartridges and without. 6. The Powder and Bullet is to be rammed home equally at every Shot; though the looser the Powder lye, it fire the better. 7. When the right Charge of a Peece is found, that makes the farthest Shot in the ordinary and plain way of Charging, Monsieur de Sons contrivance of a Wedge may be tryed, to make it Shoot farther; which is a piece of Board, so long, as being thrust home to the Breech of the Peece at one end, the other may reach farther out than the outside of the Bullet, being ramm'd up to its place; broad about an Inch, and thin so far as the Wadd before the Bullet reaches on the out-side; there it is to have a Shoulder, from which forward to the end, it is to be cut a sloap like a Wedge, being of such thickness, as that at the place, where the Center of the Bullet is to be, it may make it stick so fast, that the Powder finding more resistance may at length drive it out with the greater violence. 8. Another of this nature is a Wooden Tampion, like a piece of a Cylinder, big enough to fill the hollow Cylinder of the Gun, the length somewhat more than the Diameter of it and hollow'd towards the Bullet, so as to fit it; and either flat, or (which is better) hollow likewise towards the Powder, and serving instead of a Wadd. These and such others will probably render the effect of the Powder greater, than otherwise it would be. 9. The Strength of the Powder must be examin'd by a Powder-Tryer, that raiseth a Weight, such an one as has been contrived by Mr. Hook, and is made by Mr. Shortgrave, Operator to the Royal Society. 10. The Powder used in a Set of these Experiments, ought to be all of the same goodness. 11. The same Bullet is to be made use of, if it can be had, till the Figure of it be marred; otherwise another as near of the same Size, Shape, and Weight, as is possible. 12. The strength of the Wind, is to be observed at every time of Shooting; which may be done by an Engine, made by the lately nam'd Operator. 13. Observe also the Position of the Wind, with a Fane and Compass at every Shot. 14. Note also, at what Azimuth the Mark stands from the Gun. 15. Take precise notice, what effects the Wind hath every time upon the Bullet, in carrying further, in hindering, or turning it aside. 16. Note the Figure, dimensions and Weight of the Gun, Carriage and Wheels. 17. The platform to be very Levell. 18. The Wheels to be at every Shot placed in the very same place and position, to avoid inequalities. 19. Every thing to be exactly recorded in a Book, as also every Accident and Observation. 20. After all other Experiments are made, every Peece may be tryed with the right Charge of Powder, laying every time more and more Weight upon the Carriage; and at last fixing the Gun so, as it may not recoyle at all, observing every time how Far the Bullet goes, and how much less Powder than the full Charge will serve to Shoot the Bullet, when the Peece is fixt, as far as the Whole Charge does, when it recoyls freely. 21. Care is to be had, that the Experiments with the Wedge, Tampion, and the like, made for encreasing the force of the Powder, and the fixing of the Peece, do not endanger it. III.1. The Long Guns are to be made without any Ring about the Muzzle. The pieces cut of from the Muzzle, to be alwayes laid on the Carriage, when new shots are made, or their weight of Lead in a convenient Figure, that the recoyl may still be the same. 3. The Quantity of Powder, that Shoots furthest in an Ordinary Culverin being known, there needs no Variation of it in the Long one. Answer to some Magnetical Inquiries, proposed Numb. 23. of these Transactions, Pag. 423, 424. The Queries were these. 1. Whether a Needle may be so toucht upon any Magnet, as not to point to the true North and South? &c. 2. Whether different Load-Stones, will give different directions? or, whether fainter or stronger touches, upon one and the same Magnet, will cause any Variation in the directions? &c. To these the Industrious Mr. Sellers returns this Answer: To the First; That he had often made tryal with many Needles touching them in each Hemisphere of the Stone, with all variety of wayes he could imagine, to find, if it were possible by that means to cause any of these Needles to vary in its direction but, that he alwayes found the contrary; all of them conforming to the Magnetical Meridian, and standing North and South, as other Needles, that were toucht upon the very Pole of the Stone. He adds, that some of these Experiments he tryed in London, when there was no Variation known. To the Second; 1. That, upon frequent tryals of touching Needles upon different Load-Stones, of several bignesses, as also of different vertue; the several Needles, touched upon these different Stones, gave all of them the same directions. This he thinks is confirm'd by all the Needles and Sea-Compasses, made in several parts of the World, and consequently touched upon several Stones of several Countries, yet all agreeing in this Magnetical Harmony, that they all give the same directions. 2. That having sometimes drawn a Needle, only over the Pole of the Stone, within the Sphere of its vertue, without at all touching the Stone; it hath received the same directive quality from the Stone, as if it had been really toucht upon the Stone it self, though not altogether so strong, as if it had toucht the Stone. Again, that having toucht Needles upon the Stone, with faint strokes, and other Needles with stronger; all these Needles received the same effect from the Stone, both for strength and direction; he conceiving, that 'tis not the fainter or stronger touches upon the Stone, nor the multiplicity of Strokes, that varies the Needles strength or direction; but that the Nature of the Steel, whereof the Needle is made, and the Temper that is given thereunto, causeth different effects, as to the strength it receiveth from the Stone; himself as having tryed all sorts of Steel, that he could possibly procure, and all the different Tempers he could imagine, for the most powerful receiving and retaining the vertue from the Load Stone; wherein he affirms to have fully satisfied himself, so that he can infuse such vertue into a piece of Steel, that it shall take up a piece of Iron of two Ounces weight or more; and give also to a Needle, the vertue of conforming to the Magnetical Meridian, without the help of a Load-Stone, or any thing else, that hath received vertue therefrom. So far this Answerer; whom as we cannot but much commend for his diligence in searching, and frankness of communicating; so we give these particulars to the publick, that further Tryals may be made by others, for more discovery; hoping withall, that the same Inquisitive person, that hath made these returns, will not scruple to add to them the wayes, he uses for infusing that Magnetical vertue into Steel and Needles, without the help of a Load-Stone, spoken of in the end of this his Answer. Extract of a Letter, Written from Paris, containing an Account of some Effects of the Transfusion of Bloud; and of two Monstrous Births, &c. I was present, when M Gayant shew'd the Transfusion of the Bloud, putting that of a Young Dog into the Veins of an Old, who, two hours after, did leap and frisk; whereas he was almost blind with age, and could hardly stirr before. In the House of M Bourdelots was shew'd a Monster in form of an Ape, having all over its shoulders, almost to his middle, a mass of flesh, that came from the hinder part of its head, and hung down in form of a little Cloak. The report is, that the Woman that brought it forth, had seen on a Stage an Ape so cloathed: The most remarkable thing was, that the said mass of flesh was divided in four parts, correspondent to the Coat, the Ape did carry. The Woman, upon inquiry, was found to have gone five months with Child; before she had met with the accident of that unhappy sight. Many questions were on this occasion agitated: viz. about the Power of Imagination; and whether this Creature was endow'd with a humane Soul; and if not, what became of the Soul of the Embryo, that was five months old. A little after; another Monster was produced, which was an Infant come to maturity, having instead of a Head and Brains, a Mass of flesh like any Liver, and was found to move. And this Fœtus occasioned a Question for the Cartesians, how the motion could be performed, and yet the Glandula pinealis, or Conarium be wanting; nor any Nerves visible, which come from the Brain? The marrow in the Spine was of the same substance. It liv'd four days, and then dyed: It was anotomized by M. Emmerez in presence of the Assembly. There came a Letter from Florence, Written by M. Steno, which has also somewhat perplext the followers of Des Cartes. A Tortoise had its head cut off, and yet was found to move its foot three days after. Here was no Communication with the Conarium. As this seems to have given a sore blow to the Cartesian Doctrine, so the Disciples thereof are here endeavouring to heal the Wound. An Account of two Monstrous Births, not long since produced in Devonshire; communicated by M. Colepresse. 1. One Robert Cloak a joyner (whom I know) of Clamick in the Parish of Beer-Ferris in Devonshire, had on Febr. 24. last, a Monstrous black Ram-Lamb fallen with one Head, but two distinct Bodies, with eight Legs; which Bodies were joyned in the Neck: It had two Eyes, and as many Ears, in the usual places; and one extraordinary Eye in the Niddock, with one single Ear, about an inch distant from the Eye backwards: Its Dame, which was White, usually brought forth two Lambs every year, as she did this year also a White one, which with the Ewe remains alive. But whether this Monster was produce'd dead or alive, is not known, it being found dead by the hedge, and soon after put into the Earth. There were ten White Ews accompanied with three White Rams. 2. One John Cauce, Servant to Mr. William Knighton of Lockridge, likewise in Beer Ferris; had among his Sheep, on the seventeenth of Febr. last, a White Lamb faln on a Common in the said Parish, with two distinct heads and Necks, Joyned at the shoulders, but one only Body, and that well form'd, yet having double entrals in all respects. The Ewe remains well. The Monster dyed, and is now in my Custody, after it hath been dried in an Oven, and By the Sun. Some Observations made in Mines, and at Sea, occasioning a Conjecture about the Origine of Wind: Imparted by the same Mr. C. One John Gill, a Man well experienced in Mineral affairs, discoursing with me about the Wind and its Origine; declared to me his thoughts concerning the same, as a result of twenty years Experience and Observations of his own. First, He affirmed, that if in digging deep under ground, the Work-men meet with Water, they never want Air or Wind; but if they misse Water (as sometimes it happens, even at 12 or 16 Fathoms depth) they are destitute of convenient Air, either to breath in, or to make their Candles burn. Next, When (as usual) there happens to be a great quantity of a Winters standing water, in a deep Mine, they commonly bring, or drive up an Adit for drawing away such water: But as soon as that part of the Level is made, that any of the standing Water begins to run away, the Men must secure themselves, as well as they can, from danger of being dash'd in pieces against the sides of the Adit: For the included Air or Wind in the standing Water, breaks forth with such a terrible noyse, as that of a Peece of Ordnance, and with that violence, as to carry all before it, loosening the very Rocks, though at some distance in the Work or Adit. Thirdly, he hath observed on several occasions, going to and fro, between London and Plimouth, by Sea, that being in a Calm, that way, which the Sea began to Loom or move, the next day the Wind was sure to blow from that point of the Compass, towards which the Sea did Loom the day before. An Account of Hail-Stones of an unusual bigness, Communicated by D. Nath. Fairfax, with his Reflections on them. This Account came but very lately to hand, though the thing hapned a while agoe; the Ingenious Author thereof having but newly, entred into a Philosophical Correspondence with the Publisher. July, 17. 1666. About 10 in the Fore-noon, there fell a violent, storm of Hail about the Coast-Towns of Suffolk, tracing along Seckford-Hall, Wood-bridge, Snape bridge, Aldboraugh, &c. more to the North-ward. The Hail was small near Yarmouth; but at Seckford-Hall, one Hail-stone was found by measure to be 9. Inches about. One of this Town (viz. Wood bridge) found one at Melton, 8. Inches about. At Snape-bridge a man affirm'd, that he lighted on one about 12. Inches about. A Lady of Friston-Hall, putting one of them into a Ballance, found it weigh 12s. 6d. Several persons of good credit in Aldborough affirm'd, some Hailstones to have been full as bigg as Turkeys-Eggs; (an ordinary Hens-Egg weighs but about 9s.) J. Baker of Rumborough, driving a Cart on the Heath by Aldborough, had his head broken by the knocks of them through a stiff Country-felt: In some places his head bled; in others, bunnyes arose: The Horses were so pelted, that they hurried away his Cart beyond all command. They seem'd all white, smooth without, shining within. 'Tis somewhat strange, methinks, that their pillar of Air should keep them aloft, if they were not clapt together in the falling; especially at such a time of the year, when the Air is less thickned and its Spring weaker. Account of a great number of Stones, found in one Bladder, by the same. Mr. Goodrick Chirurgeon of Bury St. Edmunds affirm'd to me, that himself cutting a Lad of the Stone (for which he hath a great name) took out thence, it one time, 96 small Stones, all of them of unlike shape, Size, Corners, Sides; some of which were so bestow'd as to slide upon others, and had thereby worn their flats to a wonderfull slikness. He assur'd me also, that in the same place, another, when dead, had a Stone taken from him, almost as big as a new-born Childs head, and much of that Shape. The Description of a Well, and Earth in Lanchashire, taking Fire by a Candle approached to it. This was imparted by that Ingenious and Worthy Gentleman, Thomas Shirley Esq; an Eye-witness of the thing, now to be related in his own words; viz. About the later end of February 1659. returning from a Journey to my house in Wigan, I was entertained with the relation of an odd Spring. situated in one Mr. Hawkley's Ground (if I mistake not) about a mile from the Town, in that Road which leads to Warrington and Chester. The people of this Town did confidently affirm, that the Water of this Spring did burn like Oyle; into which Error they suffered themselves to fall for want of a due examination of the following particulars. For when we came to the said Spring (being five or six in company together) and applyed a lighted Candle to the surface of the Water; 'tis true, there was suddenly a large flame produced, which burnt vigorously; at the sight of which they all began to laugh at me for denying, what they had positively asserted: But I, who did not think my self confuted by a laughter grounded upon inadvertency, began to examine what I saw; and observing that this spring had its eruption at the foot of a Tree, growing on the top of a neighbouring Bank, the Water of which Spring fill'd a Ditch that was there, and covered the burning place lately mention'd; I then applyed the lighted Candle to divers parts of the Water, contained in the said Ditch, and found as I expected, that upon the touch of the Candle and the Water, the Flame was extinct. Again, having taken up a dishfull of Water at the flaming place, and held the lighted Candle to it, it went out. Yet I observed that the Water at the burning place did boyle, and heave like Water in a Pot upon the Fire, though my hand put into it perceived it not so much as warm. This boyling I conceived to proceed from the Eruption of some bituminous or sulphureous Fumes; considering, this place was not above 30 or 40 yards distant from the mouth of a Coal-pit there. And indeed Wigan, Ashton, and the whole Country, for many miles compass, is underlaid with Coal. Then applying my hand to the surface of the Burning place of the Water, I found a strong breath, as it were a Wind, to bear against my hand. Then I caused a Dam to be made, and thereby hindering the recourse of fresh water to the Burning place; I caused that, which was already there, to be drained away; and then applying the burning Candle to the surface of the dry Earth at the same point, where the Water burned before; the Fumes took fire, and burn'd very bright and vigorous. The Cone of the Flame ascended a foot and a half from the Superficies of the Earth. The Basis of it was of the Compass of a Mans hat about the brims. I then caused a Bucket-full of Water to be poured on the fire, by which it was presently quenched, as well as my companions laughter was stopped, who then began to think, the Water did not burn. I did not perceive the Flame to be discolour'd, like that of sulphureous Bodies, nor to have any manifest scent with it. The Fumes, when they broke out of the Earth, and prest against my hand, were not, to my best remembrance, at all hot. Account of Athanasii Kircheri CHINA ILLUSTRATA. The Author by publishing this Volume, discharges the Promise, he had made some years ago, that he would do so. He acknowledges himself much obliged to Martinius, and his Atlas Sinicus; as also to Michæl Boim, a Polonian; Philippo Marino, a Jesuit of Genoa; and two other of the same Society, viz. Henry Roth of Ansburg, and John Gruber, an Austrian; whereof the latter went A. 1656; over Land from Rome, through Anatolia, Armenia, Persia, Ormus, Cambaja, and India, to Macao, the famous Port of China, and thence to Pekin, the Court of that Empire; whence two years after, he came back to Rome, accompanied for a part of the way, by the Jesuit Albert Dorville; traversing by Land in a manner the whole breadth of China, and a great part of the confining Tartary, and so further, through the Mogols Dominions, to Agra, where the said Dorville dying, the above-mentioned Henry Roth supplyed his place in accomplishing this Voyage. The Book it self, a large Folio, is divided into 6. Parts. The three first, and the last, being besides the design of these Tracts, we shall but glance at, taking only notice; First, That they pretend to perswade the Reader, that Christianity was spread over all Asia by St. Thomas the Apostle, and his Successors; and hath been there continued, though not without great Eclipses, to these very times. And here the Chino-Chaldæan Monument, said to have been erected several hundred years since in China, and found out A. 1625. is with great labour asserted and interpreted. Next, That the Rise of the Idolatry, in those remote parts, and their different Ceremonies in Worship, is confronted with those Ancient ones of Egypt. Lastly, that a large Account is given of the Chinese Letters, their Figure, Power, &c. But we hasten to the Fourth Book, as belonging to our Sphere, That undertakes to describe the Curiosities and Productions of Nature and Art, in China. Here, the Author having premised something of the advantageous Scituation of China, and its Political Government; Calculated also both the Number of its Inhabitants, (which according to him, amount to 200 Millions of Men, besides Women, Children, Officers, and Eunuchs;) and the Annual Revenue of the Emperour (which he makes to be 150 Millions of Gold-Crowns;) he relateth many considerable productions and works of Nature in that Country; As 1. Mountains very odd for shape, burning, and raising of Tempests. 2. Isles, to the number of 99. all turned into one, under the same extent of space they had, when they were divided by water. 3. Lakes, some changing Copper into Iron, and causing storms, when any thing is cast into them; and others, sprung up by Earth-quakes. 4. Rivers, whereof one is said to be of a Blew colour in Autumn, and for the rest of the year Limpid: Another, to be cold at the top, and very hot beneath. 5. Fiery Wells, serving to boyl meat over: Perhaps of the same Nature with that here in England, we described above. 6. Plants, as 1. some Roses, changing their Colour twice a day: Whence the Author takes occasion to speak of that Plant, which grows at Rome, in the Garden of one Signior Corvino, call'd Viola Nocturna, changing its colour sensibly, according to the degrees of the rising and declining of the Sun; destitute of all smell in the day-time, but having a very fragrant one in the night. 2. A Farinaceous or Mealy Tree, serving to make Bread of it. 3. Leaves of certain Trees, standing on the side of a Lake, which falling into the water, become like black Birds: which he ascribeth to the Seminal parts of some Eggs, broken on those Trees, fill'd with Birds-nests. 4. The, and its wholesomeness, as to the suppressing of Vapours, and preventing the Stone. 5. A kind of Wicker-Tree, which, as if it were a Rope twisted by Nature, about an inch thick, creeps along upon the Earth, sometimes the length of 120 paces, much embarassing the way; but serving for Cables to Ships, Seats, Hurdles, Beds, Matts; enduring no Vermin; and being cool and refreshing in hot Seasons. 6. The Calamba-Wood; that it is esteem'd by some to be a kind of Lentiscum, by others, a sort of Terebinth, but of a nobler rank, by vertue of that Climat: which makes the Author suggest, that care should be taken to have it brought into Europe, and carefully cultivated there. 7. Rhubarb; of which he observes, that, because the vertue of its Roots, if they be exposed to dry hastily, soon evaporates; therefore the skilfull, lay them upon a Table within doors, and turn them several times a day, to incorporate and fix the Juyce the better, and then string them and expose them to the Wind, in a shade, altogether free from the Sunbeams. 8. Pine-Trees; of which he saith some are so big, that eight Men can hardly Fathom them. 9. Canes, so big, that they can make as many Barrels of them, as they have internodes or Joynts. 10. Trees, sweating a Gum, call'd Cie, like the droppings of Turpentine; which Gum, as long as tis not dryed, emits a very unwholsome and dangerous steam. To passe by the Polonie Tree, producing fruit without any blossoms, immediately out of its Trunk, as big as one man can well carry; and that kind of Fig-Tree, that bears Leaves as big as to wrap up a man in, &c. 7. Animals, Here he discourseth of the Murk Dear, and the several Compositions of Musk: the Sea-Horse, and Wild Men: Of some Birds, no where seen but in China (as he thinks) and among them a Wool-bearing Hen: Of Fishes, in Summer flying out of the Sea, seeking their food, like Birds, and in Autumn returning to the Sea: particularly of a Fish of a very exquisite taste, called Hoancio-yu, or the Croceous Fish: Further, of Sea-Cows, going often ashore, and fighting with the Land-Cows: Of Bats, of a vast bigness, eaten by the Chinese as a delicious meat: Of the Serpent, that breeds the Antidotal stone; whereof he relates many experiments, to verifie the relations of its vertue: Which may invite the Curator of the Royal Society, to make the like tryal, there being such a stone in their Repository, sent them from the East Indies. Again, of Silk Worms, spinning twice a year, and yielding a double Crop. 8. Fossils, where occurs the Relation, 1. Of an odd Specular stone, representing the figure of the Moon in all her Appearances, when exposed to Her. 2. Of an Earth called Quei, very Cosmetick, and abstersive of all blemishes of the face. 3. A Mineral cerusse, blended of Lead and Antimony. 4. Of Asbestus, that can be drawn and spun; the way of which he affirms to have described L. 12. Mundi sui subterranei. 5. The Matter that makes Porcellan, which he affirms to be nothing else but a transparent Sand, which they soak in water, and then reduce to a Masse or Dough, and so bake it. Not a word of the way of giving it the colour, which, it seems, they keep as a great secret. They have Gold and Silver Mines, but dig them not, pretending the danger and trouble in the work, and contenting themselves with the Filings and Dust of Gold, which they gather out of the Mud and Sand of Rivers and Fountains. The Fifth Book contains an account of their Works of Architecture, and other ingenious Mechanick Arts. Where he speaks, 1. Of their stupendious Bridges, one of 360 Perches long, and 1½ Perch broad, without any Arch, standing upon 300 Pillars, with acute Angles on both opposite sides, all the stones being of an equal size and shape. Another, built from Mountain to Mountain by one only Arch, 400 cubits long, 500 cubits high (whence tis called Pons volans) from the surface of the Saffrany River, running under it. This is represented, for the satisfaction of the Curious, by Figure I. Fig. I. 2. Vast Towns, but whose Houses are generally but one Story high, and good reason therefore, the Towns should be very big. They are, for the most part, built of Timber. 3. Turrets very artificiall, whereof one is all of Porcellan. 4. The China Wall, 300 German Leagues long; 30 Cubits high, 12 (in some places 15) Cubits broad, so that 12 Horses can very conveniently go in front on it; built 215 years before Christ, by the Emperour Pius, a brave and most Warlike Prince, and dispatcht in the space of five years; commonly it is defended by a Million of men. A Pattern of this also was thought fit, to exhibit here, by Figure II. Fig. II. 5. The Channel, that passeth from one extream of China to the other, having some 24 Sluces, to retain water, when tis necessary; a work of incredible industry and extraordinary advantage. 6. Vast Bells, one whereof, at Pekin, weighs 120000 pounds; whereas that of Erfurd in Germany, hitherto esteemed to be one of the biggest in the World, weighs but 25400. pounds. As for their ingenious Inventions, this Author mentions chiefly, 1. Their Vernice, of which he sets down some Receipts both for the Red and Black, together with the way of their Use and Application, as he received them both from an Augustinian Fryar; affirming, that it differs not at all from that of China. 2. Their way of Printing, invented long before that in Europe, giving a large description of the same. 3. Gunpowder, which he also saith, they had before the Europeans. 4. Bell-sounding. These are the principal Subjects treated of in this Book. We passe by severall Stories, which seem much to require confirmation. E. g. That of Sugar-Canes, eaten by an Elephant, and taking root in his stomach; that of Boys eating Serpents with as much greediness, as others eat Eels, or any good meat, &c. Among the Cutts of this Volume, there is a Map of Asia, not un-instructive; delineating the way, the two Jesuits took in their Land-voyage from Pekin to Goa; as also that, which the Muscovian Ambassadors, not many years since, took in travelling from their Countrey, through the vast Tract of the Northern Tartary, to China, arriving on the North side of the China Wall at Pekin: Item, The Land-passage, heretofore made by B. Goes (described by Rigaultius) from Persia, by Lahor in the Mogols Empire, through the Kingdoms of Cabul, Zancut, &c. to Cataja, or (which is all one to this Author, as it is to several others) the Province of Pekin in China. Item, The passage of Paulus Venetus over Land, out of Europe into the same China: and lastly, That pretended one of St. Thomas, out of Palestina, through Syria, Mesopotamia, ' Persia, the Mogols Empire, the Pen-insul between the Bays of Cambaya and Bengala, to Maliapur, on the Coast of Coromandel, where the Name of the Christians of St. Thomas is still in request. LONDON, Printed by T. R. for John Martyn, Printer to the Royall Society, and are to be sold at the Bell a little without Temple-Barr, 1667.	2022-01-19 17:52:45	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 3, "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": 0.5241032242774963, "perplexity": 4219.83752236092}, "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-2022-05/segments/1642320301475.82/warc/CC-MAIN-20220119155216-20220119185216-00026.warc.gz"}
http://newsinfoguide.com/download/category/science-mathematics/page/5	# Posts in category: Science Mathematics • Home • Science Mathematics ## Download Mathematical Gems: The Dolciani Mathematical Expositions by Ross Honsberger PDF April 4, 2017 \| \| By admin \| By Ross Honsberger Pages are fresh and binding is tight. ## Download Littlewood-Paley Theory on Spaces of Homogeneous Type and by Y. S. Han, Eric T. Sawyer PDF April 4, 2017 \| \| By admin \| By Y. S. Han, Eric T. Sawyer During this paintings, Han and Sawyer expand Littlewood-Paley conception, Besov areas, and Triebel-Lizorkin areas to the final environment of an area of homogeneous sort. For this objective, they determine an appropriate analogue of the Calderón reproducing formulation and use it to increase classical effects on atomic decomposition, interpolation, and T1 and Tb theorems. a few new ends up in the classical surroundings also are got: atomic decompositions with vanishing b-moment, and Littlewood-Paley characterizations of Besov and Triebel-Lizorkin areas with basically part the standard smoothness and cancellation stipulations at the approximate id. ## Download Harmonic Analysis for Anisotropic Random Walks on by Alessandro Figa-Talamanca, Tim Steger PDF April 4, 2017 \| \| By admin \| By Alessandro Figa-Talamanca, Tim Steger This paintings offers a close learn of the anisotropic sequence representations of the loose product workforce $\mathbf Z/2\mathbf Z\star \cdots \star \mathbf Z/2\mathbf Z$. those representations are countless dimensional, irreducible, and unitary and will be divided into significant and complementary sequence. Anisotropic sequence representations are fascinating simply because, whereas they're now not limited from any higher non-stop crew within which the discrete team is a lattice, they still proportion many homes of such regulations. the result of this paintings are additionally legitimate for nonabelian unfastened teams on finitely many turbines. ## Download Everyday Wonders: Encounters With the Astonishing World by Barry Evans PDF April 4, 2017 \| \| By admin \| By Barry Evans In 30 essays--filled with anecdotes and illustrations--Evans takes such ordinary techniques as gravity, water, and breath and turns them into delightfully documented adventures. specific interviews with Stephen Jay Gould, Linus Pauling, and different inventive and articulate scientists upload an additional measurement. photographs. Line drawings. Puzzles. April 4, 2017 \| \| By admin \| By Dinculeanu N. April 4, 2017 \| \| By admin \| By A. Gardiner Challenging and stimulating selection of diverting brain-teasers is helping highschool scholars combine easy concepts and intricate problem-solving techniques in an stress-free approach. difficulties concerning squares and cubes, polyhedra, top numbers, chess items, and different matters are grouped into 29 sections, with more straightforward difficulties showing initially of every. A helpful software for academics, this soaking up publication also will entice contributors operating by myself who are looking to boost problem-solving options. strategies included. ## Download Central Extensions, Galois Groups, and Ideal Class Groups of by Albrecht Frohlich PDF April 4, 2017 \| \| By admin \| By Albrecht Frohlich Those notes care for a collection of interrelated difficulties and ends up in algebraic quantity concept, within which there was renewed job lately. The underlying instrument is the speculation of the relevant extensions and, in such a lot common phrases, the underlying objective is to exploit classification box theoretic the right way to achieve past Abelian extensions. One goal of this publication is to offer an introductory survey, assuming the elemental theorems of sophistication box thought as in most cases recalled in part 1 and giving a imperative function to the Tate cohomology teams. The critical goal is, in spite of the fact that, to take advantage of the final conception as constructed the following, including the specific positive factors of sophistication box conception over, to derive a few relatively powerful theorems of a really concrete nature, as base box. The specialization of the idea of relevant extensions to the bottom box is proven to derive from an underlying precept of huge applicability.The writer describes convinced non-Abelian Galois teams over the rational box and their inertia subgroups, and makes use of this description to achieve details on excellent category teams of totally Abelian fields, all in solely rational phrases. special and particular mathematics effects are bought, attaining a long way past whatever on hand within the common concept. the idea of the genus box, that is wanted as history in addition to being of self sustaining curiosity, is gifted in part 2. In part three, the speculation of critical extension is constructed. The targeted beneficial properties are mentioned all through. part four offers with Galois teams, and purposes to classification teams are thought of in part five. ultimately, part 6 comprises a few feedback at the heritage and literature, yet no completeness is tried April 4, 2017 \| \| By admin \| By P.R. Halmos A variety of the mathematical writings of Paul R. Halmos (1916 - 2006) is gifted in Volumes. quantity I contains examine guides plus papers of a extra expository nature on Hilbert house. the rest expository articles and the entire well known writings look during this moment quantity. It contains 27 articles, written among 1949 and 1981, and likewise a transcript of an interview. April 4, 2017 \| \| By admin \| By Olvi L. Mangasarian This publication addresses in basic terms Nonlinear Programming thought. You won't locate any algorithms, in order that this e-book isn't functional. I additionally imagine the alternative of themes might be better. So, why five stars ? Because of the writing style. That is, what i admire so much during this e-book is the way in which that Mangasarian wrote it. References to all very important equations, definitions, and so forth. little need of english to give an explanation for math matters yet good outlined equations. Concise exposition and evidence of theorems utilizing math simbols. that's: Math simbols to provide an explanation for arithmetic. Very transparent kind. A first-class instance of ways to jot down a great, didactic, exact and transparent math ebook.	2018-02-23 22:12:02	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 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": 0.3747096657752991, "perplexity": 5473.848846491494}, "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/1518891814857.77/warc/CC-MAIN-20180223213947-20180223233947-00691.warc.gz"}
https://wtskills.com/interior-angle-of-polygon/	# Interior angle of Polygon In this chapter we will learn about interior angles of polygon along with some important formulas related to the topic. ## What are interior angles of polygons ? The angles which lie inside the polygon are called interior angles. The interior angles are formed at the intersection of two sides of polygon. ### Polygon classification and interior angles Polygons are broadly classified into two categories; (a) Regular Polygons These polygons have equal sides and equal interior angles. (b) Irregular Polygons These polygons have different sides and interior angle measurement. ### Interior angle formula for regular polygons The measure of interior angle for regular polygons is given by following formula; \mathtt{Interior\ angle\ =\ \frac{( n-2) \ \times 180}{n}} Where; n = number of sides of polygon ## Sum of all interior angle formula for polygon The measure of sum of all interior angle is given by following formula; \mathtt{Sum\ =\ ( n\ -\ 2) \ \times \ 180} Where; n = number of sides of polygon ## Calculating interior angles of regular polygons I hope you understood the above mentioned concept. Let us now calculate the interior angle values of different polygons. ### Interior Angles of Triangle It’s a polygon with 3 sides. Given below is triangle ABC with interior angles ∠A, ∠B & ∠C #### Calculating interior angle of Triangle The measure of interior angle is given by following calculation; \mathtt{Interior\ angle\ =\ \frac{( n-2) \ \times 180}{n}}\\\ \\ \mathtt{Interior\ angle\ =\ \frac{( 3-2) \ \times 180}{3}}\\\ \\ \mathtt{Interior\ angle\ =\ \frac{1\ \times 180}{3}}\\\ \\ \mathtt{Interior\ angle\ =\ 60\ degree} Hence, each angle of regular triangle measures 60 degree. #### Sum of all angle of Triangle The sum of all interior angle of triangle is calculated as; \mathtt{Sum\ =\ ( n\ -\ 2) \ \times \ 180}\\\ \\ \mathtt{Sum\ =\ ( 3\ -\ 2) \ \times \ 180}\\\ \\ \mathtt{Sum\ =\ 180\ } It is a polygon with 4 sides. Given below is the regular quadrilateral with interior angles ∠A, ∠B, ∠C and ∠D. #### Calculating Interior angles of Quadrilateral \mathtt{Interior\ angle\ =\ \frac{( n-2) \ \times 180}{n}}\\\ \\ \mathtt{Interior\ angle\ =\ \frac{( 4-2) \ \times 180}{4}}\\\ \\ \mathtt{Interior\ angle\ =\ \frac{2\ \times 180}{4}}\\\ \\ \mathtt{Interior\ angle\ =\ 90\ degree} Hence, each angle of regular quadrilateral measures 90 degree. #### Sum of all Interior angle of Quadrilateral \mathtt{Sum\ =\ ( n\ -\ 2) \ \times \ 180}\\\ \\ \mathtt{Sum\ =\ ( 4\ -\ 2) \ \times \ 180}\\\ \\ \mathtt{Sum\ =\ 360\ degree} The sum of all interior angle of quadrilateral measures 360 degree. ### Interior angles of Pentagon It’s a polygon with 5 sides. Given below is the regular polygon with interior angles marked in red color. #### Calculating Interior angle of Pentagon The interior angles of regular pentagon is calculated as; \mathtt{Interior\ angle\ =\ \frac{( n-2) \ \times 180}{n}}\\\ \\ \mathtt{Interior\ angle\ =\ \frac{( 5-2) \ \times 180}{5}}\\\ \\ \mathtt{Interior\ angle\ =\ \frac{3\ \times 180}{5}}\\\ \\ \mathtt{Interior\ angle\ =\ 108\ degree} Hence, each angle of pentagon measures 108 degree. #### Sum of all interior angle of Pentagon \mathtt{Sum\ =\ ( n\ -\ 2) \ \times \ 180}\\\ \\ \mathtt{Sum\ =\ ( 5\ -\ 2) \ \times \ 180}\\\ \\ \mathtt{Sum\ =\ 540\ degree} Hence, sum of all interior angles of pentagon measures 540 degree. ### Interior angle of Hexagon It is a polygon with six sides. Given below is the image of regular hexagon. #### Calculating Interior angle of Hexagon The interior angle of hexagon is calculated as; \mathtt{Interior\ angle\ =\ \frac{( n-2) \ \times 180}{n}}\\\ \\ \mathtt{Interior\ angle\ =\ \frac{( 6-2) \ \times 180}{6}}\\\ \\ \mathtt{Interior\ angle\ =\ \frac{4\ \times 180}{6}}\\\ \\ \mathtt{Interior\ angle\ =\ 120\ degree} Hence, each interior angle of regular polygon measure 120 degree. #### Sum of all interior angle of Hexagon The sum of all interior angles of hexagon is calculated as; \mathtt{Sum\ =\ ( n\ -\ 2) \ \times \ 180}\\\ \\ \mathtt{Sum\ =\ ( 6\ -\ 2) \ \times \ 180}\\\ \\ \mathtt{Sum\ =\ 720\ degree} ### Interior angles of Heptagon It’s a polygon with 7 sides. Given below is the image of regular heptagon. #### Calculating Interior angle of Heptagon The interior angles of heptagon is measured as; \mathtt{Interior\ angle\ =\ \frac{( n-2) \ \times 180}{n}}\\\ \\ \mathtt{Interior\ angle\ =\ \frac{( 7-2) \ \times 180}{7}}\\\ \\ \mathtt{Interior\ angle\ =\ \frac{5\ \times 180}{7}}\\\ \\ \mathtt{Interior\ angle\ =\ 128.5\ degree} Hence, each angle of regular heptagon measures 128.5 degree. #### Sum of all interior angles of Heptagon The sum of all interior angles of heptagon is calculated as; \mathtt{Sum\ =\ ( n\ -\ 2) \ \times \ 180}\\\ \\ \mathtt{Sum\ =\ ( 7\ -\ 2) \ \times \ 180}\\\ \\ \mathtt{Sum\ =\ 600\ degree} ### Interior angles of Octagon It’s a polygon with 8 sides. Given below is the image of regular octagon. #### Calculating Interior angle of Heptagon The interior angle of octagon is calculated as; \mathtt{Interior\ angle\ =\ \frac{( n-2) \ \times 180}{n}}\\\ \\ \mathtt{Interior\ angle\ =\ \frac{( 8-2) \ \times 180}{8}}\\\ \\ \mathtt{Interior\ angle\ =\ \frac{6\ \times 180}{8}}\\\ \\ \mathtt{Interior\ angle\ =\ 135\ degree} Hence, each angle of regular octagon measures 135 degree. #### Sum of all interior angles of Heptagon The sum of interior angles of octagon is calculated as; \mathtt{Sum\ =\ ( n\ -\ 2) \ \times \ 180}\\\ \\ \mathtt{Sum\ =\ ( 8\ -\ 2) \ \times \ 180}\\\ \\ \mathtt{Sum\ =\ 1080\ degree} ### Interior angles of Nonagon It’s a polygon with 9 sides. Given below is the image of regular nonagon. #### Calculating Interior angle of Nonagon The interior angle of Nonagon is calculated as; \mathtt{Interior\ angle\ =\ \frac{( n-2) \ \times 180}{n}}\\\ \\ \mathtt{Interior\ angle\ =\ \frac{( 9-2) \ \times 180}{9}}\\\ \\ \mathtt{Interior\ angle\ =\ \frac{7\ \times 180}{9}}\\\ \\ \mathtt{Interior\ angle\ =\ 140\ degree} Hence, each interior angle measures 140 degree. #### Sum of all interior angles of Nonagon Sum of all interior angles of Nonagon is calculated as; \mathtt{Sum\ =\ ( n\ -\ 2) \ \times \ 180}\\\ \\ \mathtt{Sum\ =\ ( 9\ -\ 2) \ \times \ 180}\\\ \\ \mathtt{Sum\ =\ 1260\ degree} ### Interior angles of Decagon It’s a polygon with 10 sides. Given below is the image of regular decagon with angled marked in red color. #### Calculating interior angles of Decagon The interior angle of regular decagon is calculated as; \mathtt{Interior\ angle\ =\ \frac{( n-2) \ \times 180}{n}}\\\ \\ \mathtt{Interior\ angle\ =\ \frac{( 10-2) \ \times 180}{10}}\\\ \\ \mathtt{Interior\ angle\ =\ \frac{8\ \times 180}{10}}\\\ \\ \mathtt{Interior\ angle\ =\ 144\ degree} Hence. each angle measures 144 degree. #### Sum of angles of Decagon Sum of all interior angles is calculated as; \mathtt{Sum\ =\ ( n\ -\ 2) \ \times \ 180}\\\ \\ \mathtt{Sum\ =\ ( 10\ -\ 2) \ \times \ 180}\\\ \\ \mathtt{Sum\ =\ 1440\ degree}	2022-06-25 00:55:22	{"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": 1.0000100135803223, "perplexity": 11341.783557880566}, "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-2022-27/segments/1656103033925.2/warc/CC-MAIN-20220625004242-20220625034242-00385.warc.gz"}
https://www.physicsforums.com/threads/absolute-values.739407/	# Absolute Values 1. Feb 20, 2014 ### TheRedDevil18 1. The problem statement, all variables and given/known data Solve the given inequality by interpreting it as a statement about distances in the real line: \|x-3\| < 2\|x\| 2. Relevant equations 3. The attempt at a solution I have no clue what to do here and I do not understand the answer in the textbook Goes something like this.............. x^2 - 6x + 9.....................Have no idea how they got that = (x-3)^2 ...............and so forth 2. Feb 20, 2014 ### shortydeb There are two possibilities for x: either x is greater than or equal to 3, or x is less than 3. If x ≥ 3, what does the inequality look like (i.e. without the absolute value)? 3. Feb 20, 2014 ### LCKurtz They used $\|a\|<\|b\| \leftrightarrow a^2<b^2$.	2017-10-19 00:41:23	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4117341935634613, "perplexity": 894.2926916232118}, "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/1508187823168.74/warc/CC-MAIN-20171018233539-20171019013539-00111.warc.gz"}
https://www.omnimaga.org/math-and-science/a-'new'-compression-format-(subject-to-changes)/	### Author Topic: A "new" compression format [subject to changes] (Read 6474 times) 0 Members and 1 Guest are viewing this topic. #### harold • Posts: 226 • Rating: +41/-3 ##### A "new" compression format [subject to changes] « on: November 09, 2013, 03:17:32 pm » I was trying to write a Deflate decompressor, and that's perfectly doable, but occasionally annoying. The "new" format I'm suggesting is basically Deflate with some changes. Some to make it less annoying, others because "why not". The goal is mostly "improved decompression speed". The main structure is the same. Matches in a sliding window, with literals and match-lengths (and End of Block marker) in the same alphabet, and Huffman coding. But there are changes. The main change is that the Huffman codes are packed in dwords (specifically to help decoding with a 32bit "window"), starting at the most significant bit (helps with decoding). Those dwords are then saved with little-endian byte-order (to help x86 - ARM can work this more easily than x86 could in the reverse situation). They're aligned to their natural boundary, so there may be a little padding after the header (the point is to read in dwords, aligning them makes that more efficient on some architectures and only costs 3 bytes at worst). The Huffman codes are, of course, canonical, but a different kind of canonical than in Deflate: long codes must begin with zeroes. IE, using the rule "shorter codes have a numerically (if filled with zeros to the right) higher value than longer codes". This ensures that no more than 9 rightmost bits can ever differ from zero, which keeps your decoding tables small without trickery. The maximum length for a Huffman code is 31, up from 15. (that usually won't help, I'm throwing that in "because why not") Furthermore, the match-length symbols start at 256, with EndOfBlock assigned the code 288. This makes it slightly more convenient to index a table directly with the match-symbol. All 32 match-length symbols are used, following the same pattern of "extra bits" as in Deflate, but extended. The last 4 match-length-symbols have 6 "extra bits". The distance codes work like in Deflate64 (which is like Deflate, but with codes 30 and 31 being valid and getting 14 "extra bits"). The suggested way to decode this is to take two dwords, shift them left (with the bits of the second dword appearing in the lsb of the first dword) by some amount, count the leading zeroes (you can OR with 1 and use BSR since the maximum length of one symbol is 31 bits anyway), shift right by some amount depending on the number of leading zeroes, add some offset depending on the number of leading zeroes, then index a table with that. Alternatively, take a (possibly unaligned) qword, rotate it by 32 bits, then shift left (normally), count leading zeroes, etc.. The header is changed to this: if the first byte is 0, end of stream otherwise, the first dword (little-endian) has some bitfields: bit 0(lsb): 1 (constant) (to ensure the first byte is not zero) bit 1: 0 if the block is compressed, 1 is the block is stored uncompressed bit 2-31: length of this block if block is stored: just the bytes, raw. if block is compressed: dword (little endian): length of this block when decompressed uint16 (little endian): is_present, bit[n] (numbered from lsb up) indicates that the range of symbols [n * 16 .. n * 16 + 15] is part of the alphabet (1) or not (0) uint16 (little endian): is_default, bit[n] (see above) indicates that the range (see above) has its own 16bit mask (0) or is completely present (1) For every range that needs a mask (ie is present and not default), an uint16 (little endian) that indicates for every symbol whether it's part of the alphabet (1) or not(0) Two more masks for the match-lengths (these two masks are always present) Then, some bytes to store the code lengths: The lowest 5 bits indicate the code length (a length of 0 may be coded - enables you to drop a subrange mask and save a byte sometimes), the upper 3 bits are a repeat-count. If the repeat-count is 0, the repeat count is 8 + next_byte. [up to 3 bytes padding to align the next part] None of those changes are really spectacular, but IMO they all contribute a little bit to make the format less annoying. The part that I'm least sure about is the header. I'm also not sure whether allowing longer match distances is a mistake or not. Actually I'm not sure whether the whole deal for encoding matches is the right way to go or not. I couldn't come up with anything better, but it's a pain in the behind to parse. For most lengths n, parsing the match takes so much time that decoding n literals would have been faster. As always, suggestions are welcome (that's why I'm posting this in the first place). Especially if you've written a Deflate decompressor before (otherwise the changes might seem random). Check the haroldbot thread for the supported commands and syntax. You can use haroldbot from this website. #### SpiroH • Posts: 726 • Rating: +153/-23 ##### Re: A "new" compression format [subject to changes] « Reply #1 on: November 09, 2013, 04:14:13 pm » This sounds interesting although a bit too technical for many of the users. As you seem to be mostly interested in performance, i would suggest you carry out some benchmarking and then try to 'sell it' afterwards. Good luck with it. #### harold • Posts: 226 • Rating: +41/-3 ##### Re: A "new" compression format [subject to changes] « Reply #2 on: November 09, 2013, 04:23:29 pm » I'll benchmark it of course, but it'll be hard to filter out the contribution of the format. I already have an implementation mostly done. Check the haroldbot thread for the supported commands and syntax. You can use haroldbot from this website. #### harold • Posts: 226 • Rating: +41/-3 ##### Re: A "new" compression format [subject to changes] « Reply #3 on: November 12, 2013, 06:20:58 am » Ok, here are the first benchmark results. For the file alice29.txt from the Canterbury Corpus, using only Huffman compression and no matches (those don't quite work yet), the decompression time on my machine is (approximately and rounded upwards) 1.2 miliseconds. That's purely the decompression time, ie not counting file IO and such. The uncompressed file size is 152,089 bytes, the compressed file size is 87,784 bytes (it would be a lot smaller with matches, about 55kb, but they don't really work yet). That gives a output-throughput of 120MB/s. For reference, 7zip decompression speed (as measured by its built-in benchmarker) is about 50MB/s for a single core on my machine (limiting to a single core because my code is currently single-threaded, but that's not a limitation due to the format). I expect it to slow down when I enable matches, though. Long matches will be fast (as fast as memcpy, essentially), but the more common short matches have a lot of overhead. It isn't the most relevant comparison of course, LZMA doesn't even use Huffman compression at all. But it was the easiest comparison to make, and it's something. « Last Edit: November 12, 2013, 06:27:32 am by harold » Check the haroldbot thread for the supported commands and syntax. You can use haroldbot from this website. #### DJ Omnimaga • Former TI programmer • CoT Emeritus • LV15 Omnimagician (Next: --) • Posts: 55918 • Rating: +3152/-232 • CodeWalrus founder & retired Omnimaga founder ##### Re: A "new" compression format [subject to changes] « Reply #4 on: November 12, 2013, 11:54:07 am » I wonder if such compression would be fast enough for calcs? #### harold • Posts: 226 • Rating: +41/-3 ##### Re: A "new" compression format [subject to changes] « Reply #5 on: November 12, 2013, 12:03:06 pm » Depends. On ARM calcs, no problem. But decoding the Huffman codes will probably suck on the z80 calcs. Check the haroldbot thread for the supported commands and syntax. You can use haroldbot from this website. #### Streetwalrus • LV12 Extreme Poster (Next: 5000) • Posts: 3821 • Rating: +80/-8 ##### Re: A "new" compression format [subject to changes] « Reply #6 on: November 12, 2013, 03:37:12 pm » It sounds really cool. I hope you can finish it and it becomes widely used. #### thepenguin77 • z80 Assembly Master • LV10 31337 u53r (Next: 2000) • Posts: 1591 • Rating: +823/-5 • The game in my avatar is bit.ly/p0zPWu ##### Re: A "new" compression format [subject to changes] « Reply #7 on: November 12, 2013, 03:45:21 pm » My fourVid used deflate to store the videos. That's how I was getting like 30% compressed video sizes. Some quick math tells me that it was decompressing at least 30 KB/s on the z80s. (Though it might be much higher, that's just the lower bound.) What I find interesting about your method is that you are kind of abusing Huffman codes in a way that makes them no longer Huffman codes. I thought the idea was the huffman codes are shorter for the more common values, but yours all end up taking the same amount of space. This would of course greatly speed up decompression, but, it's not really huffman anymore Wouldn't you be able to get better ratios if instead of using huffman, you simply made your lookup tables bigger and used all of the bits in your huffman fields? zStart v1.3.013 9-20-2013 All of my utilities TI-Connect Help You can build a statue out of either 1'x1' blocks or 12'x12' blocks. The 1'x1' blocks will take a lot longer, but the final product is worth it. -Runer112 #### harold • Posts: 226 • Rating: +41/-3 ##### Re: A "new" compression format [subject to changes] « Reply #8 on: November 12, 2013, 03:48:17 pm » No they're Huffman codes, with longer ones and shorter ones.. the thing about the "9 rightmost bits" doesn't change that. That's just a property that those Huffman codes automatically have due to their construction and the size of the alphabet. « Last Edit: November 12, 2013, 03:48:42 pm by harold » Check the haroldbot thread for the supported commands and syntax. You can use haroldbot from this website. #### thepenguin77 • z80 Assembly Master • LV10 31337 u53r (Next: 2000) • Posts: 1591 • Rating: +823/-5 • The game in my avatar is bit.ly/p0zPWu ##### Re: A "new" compression format [subject to changes] « Reply #9 on: November 12, 2013, 03:54:01 pm » Oh, alright. I guess I misunderstood. Your algorithm is a lot more complex than what I had originally realized. zStart v1.3.013 9-20-2013 All of my utilities TI-Connect Help You can build a statue out of either 1'x1' blocks or 12'x12' blocks. The 1'x1' blocks will take a lot longer, but the final product is worth it. -Runer112 #### harold • Posts: 226 • Rating: +41/-3 ##### Re: A "new" compression format [subject to changes] « Reply #10 on: November 12, 2013, 03:56:02 pm » I guess I'll do a quick example: suppose you have the data 4, 5, 5, 4, 0, 1, 2, 3, 4, 5, 5, EOS The codelengths and codes assigned to those symbols will be 0: 3 001 1: 3 010 2: 4 0000 3: 4 0001 4: 2 10 5: 2 11 EOS: 3 011 So the message would be 10 11 11 10 001 010 0000 0001 10 11 11 011 That would be packed as 0xBE2806F6 or, as bytes, F6 06 28 BE (and a header of course but that one's simple) « Last Edit: November 12, 2013, 03:56:25 pm by harold » Check the haroldbot thread for the supported commands and syntax. You can use haroldbot from this website. #### harold • Posts: 226 • Rating: +41/-3 ##### Re: A "new" compression format [subject to changes] « Reply #11 on: November 23, 2013, 04:54:38 am » Alright, new benchmark results. With sliding-window back-matches enabled, the compressed file (still with the same input, alice29.txt from the Canterbury Corpus, which is 152,089 bytes of ascii text) is now 56,052 bytes (this is not optimal, I haven't implemented things like lazy matches (yet?) in the compressor). That's about 55kb, with a high quality Deflate compressor you can get 51kb - but like I said, no lazy matches. More importantly, it takes about 0.5 miliseconds to decompress it (it varies a bit between 0.450 and 0.520, more often on the low side, but let's just round it up to 0.500). That's 290MB/s of output-throughput, or if you'd rather measure input-throughput, 106MB/s (that's not a common measure afaik, but there you go). I really expected it to get slower, since the histogram of match lengths showed the vast majority of them to be tiny, and tiny matches have a lot of overhead per byte. Will edit with code soon (it is at this point just test code though, not a usable product. Various paths are hardcoded. And the decompressor will only run on Haswell CPUs) « Last Edit: November 23, 2013, 04:59:16 am by harold » Check the haroldbot thread for the supported commands and syntax. You can use haroldbot from this website. #### harold • Posts: 226 • Rating: +41/-3 ##### Re: A "new" compression format [subject to changes] « Reply #12 on: November 25, 2013, 01:26:56 pm » Here's a version that works on older CPUs as well, probably a bit more useful It's slightly slower, about 4 to 5%, which means I can still round it off to 0.5 miliseconds on the test file that I've been using so far. The difference is real and detectable though, not just noise. Various paths and buffer sizes are still hardcoded, but it shouldn't be hard to get it to work. The only changes are in the files decoder.asm and altdeflate_infl_ref.h Check the haroldbot thread for the supported commands and syntax. You can use haroldbot from this website. #### Eiyeron • Urist McEiyolobster • LV10 31337 u53r (Next: 2000) • Posts: 1430 • Rating: +130/-10 • (-_(//)); ##### Re: A "new" compression format [subject to changes] « Reply #13 on: November 25, 2013, 01:54:21 pm » Now make it work on calc! :p Is the decompress or light enough for calc? If yes, will it give better results than PuCrunch?	2022-01-20 08:55:28	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4183894693851471, "perplexity": 5432.349592366059}, "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/1642320301730.31/warc/CC-MAIN-20220120065949-20220120095949-00599.warc.gz"}
https://solvedlib.com/we-can-now-use-the-standard-normal-distribution,71602	# We can now use the Standard Normal Distribution Table to find the probability P(-0.25 sz s... ###### Question: We can now use the Standard Normal Distribution Table to find the probability P(-0.25 sz s 1). 0.05 0.06 0.07 0.08 0.09 -0.2 0.4013 0.3974 0.3936 0.3897 0.3859 0.00 0.01 0.02 0.03 0.04 Using these 1.0 0.8413 0.8438 0.8461 0.8485 0.8531 The table entry for z = -0.25 is 0.00 and the table entry for z = 1 is values to calculate the probability gives the following result. PC-0.25 sz s 1) P(Z < 1) - P(Z 5 -0.25) 10. [6/9 Points] DETAILS PREVIOUS ANSWERS BBUNDERSTAT12 6.3. This question has several parts that must be completed sequentially. If you skip a pan Tutorial Exercise Assume that x has a normal distribution with the specified mean and standard d probability, P(5 S XS 15); 1 = 7; 0 = 8 Step 1 First, to find areas and probabilities for a random variable x that follows a nom and standard deviation 0 = 8, we must convert the x values to z values. x- Use the formula z = to convert the given x values to z values. 0 Find the z value for x = 5 with mean l = 7 and standard deviation 0 = 8. X-H 0 Z = 5 5 7 #### Similar Solved Questions ##### Complete the reaction, add hydrogen atoms and charges to the appropriate atoms Single bond O ZO... Complete the reaction, add hydrogen atoms and charges to the appropriate atoms Single bond O ZO NaOH CH3 -CH2-C-N-CH CH, + H TH H Theat heat... ##### 1 (15 points in total) Assume that we have & random sample {(Yi, Xi) i =1, N} of Size N from the joint distribution of the two random variables Y and X. Also assume that the first and second moments of this distribution are finite. Answer the following questions:a) (5 points) The population covariance between two random variables is defined by cov (X,Y) = E [(X - E(X)) * (Y _ E(Y))] Show that this expression can be rewritten asCOU (X,Y) = E(X Y) - E(X) E(Y) (5 points) Write down the Metho 1 (15 points in total) Assume that we have & random sample {(Yi, Xi) i =1, N} of Size N from the joint distribution of the two random variables Y and X. Also assume that the first and second moments of this distribution are finite. Answer the following questions: a) (5 points) The population cov... ##### [z2 1 1 3 Mh NF { { Uh [ 1 { { H 2 6 1 5 I J V 1 W 1 5 1 1 W L 1 1 1 0 \| 9 I { 1 [z2 1 1 3 Mh NF { { Uh [ 1 { { H 2 6 1 5 I J V 1 W 1 5 1 1 W L 1 1 1 0 \| 9 I { 1... ##### 3a) Is the sequence 4n sin(Zin) convergent? If so what does it converge to? [1pts]3b). Determine the following limit: If the limit does not exist write DNE and explain why: [2pts] lim sin (2tx) 740 3a) Is the sequence 4n sin(Zin) convergent? 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Find the x-value representing the absolute minimum value of... f(x) = x3 - 9x2 + 24x; (0,5), and find the y-value of the minimum value.... ##### Denote the bator nerantnnd turbinc bludaedcdHeionstrlbutijn;Axi0iathoquistmadeldeenuanVenry [MEis a lagitimae pdt7mvnFunutOrobahilin tha /Most ?007 Less than 20072007 (Rojra Yiur ansefaur decima placen\|SupposeMees(naRasl 20](c) Whetthopmnbabili;athatBotncanJeeuttun^139)7 (Round ,qur %iunier tour de0Mi 34025 \|(di Givnexpression for Ax $4}.At<4pjut7 denote the bator nerant nnd turbinc bluda edcd Heion strlbutijn; Axi0i athoquist madel deenuan Venry [ME is a lagitimae pdt 7mvn Funut Orobahilin tha / Most ?007 Less than 2007 2007 (Rojra Yiur anse faur decima placen\| Suppose Mees(na Rasl 20] (c) Whet thopmnbabili;athat Botncan Jeeuttun^ 139)7 (Ro... 5 answers ##### Consider the following reaction: Nz(g) + 3H2(g) 2NH3 (g) Kc = 1.45x 10-5 Equilibrium pressure (atm) 0.432 0.928 What is the pressure of NH3 at equilibrium? 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A) 5.00 x 10 6 atm B) 2.41 x 10-3 atm 2.24 * 10-3 8 atm 1.71 X 10-2 atm 2.14 x 102 atm... 5 answers ##### Question3616O1, find the first derivative, 9' (c)Given the function g(1 (c)Notice thal g= 0 whenthat IS; 9' ( =1) = 0local minimum Or local maximum at F50 we will use the secondNow we want to know whether there derivatie test Fld tie second dervatie g"()Evaluate 0 ( 9""'( =I)Dnsed on the sign O thts number, does tlts menn the graph o gr) concave UR Or cOncave_(lwn al$ (Answet ether Up Or dowm watch youI spellingi] AII the graph of g(T) concaveElased on the concavit Question 361 6O1, find the first derivative, 9' (c) Given the function g(1 (c) Notice thal g = 0 when that IS; 9' ( =1) = 0 local minimum Or local maximum at F 50 we will use the second Now we want to know whether there derivatie test Fld tie second dervatie g "() Evaluate 0 ( 9"... ##### Question 201 ptsA scale measures the weight of a Skg object submerged in water to be 10 Newtons: If the density of water is 1000 kg/m? what is the density of the object?900 kg/m^31250 kg/m^31120 kg/m^31190 kg/m^3 Question 20 1 pts A scale measures the weight of a Skg object submerged in water to be 10 Newtons: If the density of water is 1000 kg/m? what is the density of the object? 900 kg/m^3 1250 kg/m^3 1120 kg/m^3 1190 kg/m^3... ##### (14 points) Let F (y8,22) and let C be the arc â‚¬ = 1-y? from (0,~1) to (0,1) followed by the line segment from (0,1) to (0, ~1). Evaluate IcF dr. (14 points) Let F (y8,22) and let C be the arc â‚¬ = 1-y? from (0,~1) to (0,1) followed by the line segment from (0,1) to (0, ~1). Evaluate IcF dr.... ##### Assume a company needs to produce 60 units of product A and 30 units of product... Assume a company needs to produce 60 units of product A and 30 units of product Beach day. The firm operates using Lean practices. Also assume the firm has reduced setup times to negligible values. Which of the following would be the best strategy for the company to follow to produce these units? Se... ##### Question Let V â‚¬ R. Define the addition and multiplication byX+y=x cx = r Examine the following axioms, state whether it satisfy Or nota) (u+v)+w=u+(v+w)b) clu+v)= (cu)+ (cv)Question 2Given matrix AFindBasis of column space of matrix A b) Rank (A) Question Let V â‚¬ R. Define the addition and multiplication by X+y=x cx = r Examine the following axioms, state whether it satisfy Or not a) (u+v)+w=u+(v+w) b) clu+v)= (cu)+ (cv) Question 2 Given matrix A Find Basis of column space of matrix A b) Rank (A)... ##### Round all answers to 3 decimal places.If no answer exists, enter DNE for all answers.Assume âˆ A is opposite side a, âˆ B is oppositeside b, and âˆ C is opposite side c.âˆ A=111âˆ˜,a=33 ,b=11. Round all answers to 3 decimal places. If no answer exists, enter DNE for all answers. Assume âˆ A is opposite side a, âˆ B is opposite side b, and âˆ C is opposite side c. âˆ A=111âˆ˜,a=33 ,b=11.... ##### QUESTION 3 2.5 points If traveler's checks were $1000 higher and saving deposits were$500 higher.... QUESTION 3 2.5 points If traveler's checks were $1000 higher and saving deposits were$500 higher. MI would be a $1.000 high and M2 would be$500 higher b. $500 higher aad M2 would be$1,500 higher. c.$1,000 higher and M2 would be$1,500 higher d. M2 and MI would be S1,500 higher. QUESTION 4 2.5... ##### A Carnot engine removes $1.20 \times 103 \mathrm{~J} 1.20 \times 10^{3} \mathrm{~J}$ of heat from a hightemperature source and dumps $6.00 \times 102 \mathrm{~J} 6.00 \times 10^{2} \mathrm{~J}$ to the atmosphere at $20.0^{\circ} \mathrm{C}$. (a) What is the efficiency of the engine? (b) What is the temperature of the hot reservoir? A Carnot engine removes $1.20 \times 103 \mathrm{~J} 1.20 \times 10^{3} \mathrm{~J}$ of heat from a hightemperature source and dumps $6.00 \times 102 \mathrm{~J} 6.00 \times 10^{2} \mathrm{~J}$ to the atmosphere at $20.0^{\circ} \mathrm{C}$. (a) What is the efficiency of the engine? (b) What is the ...	2022-07-03 18:13:51	{"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": 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": 0.6104843020439148, "perplexity": 2391.0809083287877}, "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/1656104248623.69/warc/CC-MAIN-20220703164826-20220703194826-00789.warc.gz"}
https://www.mersenneforum.org/showthread.php?s=3a90f619f89b9baa3e60cbcbc61793ee&t=21544&page=13	mersenneforum.org Your help wanted - Let's buy GIMPS a KNL development system! Register FAQ Search Today's Posts Mark Forums Read 2016-09-16, 16:38 #133 bsquared "Ben" Feb 2007 7·13·41 Posts Quote: Originally Posted by xathor Traditional hyperthreading is out of order and only used when needed. KNL hyperthreading is round robin in order. KNC is round robin in order. KNL allows back to back instructions from a single thread. "Thread count requirement reduced to ~70 (one thread per core) from 120 (2 threads per core)". Etc. Unless the KNL guides from Colfax are completely false. must register, but otherwise free: colfaxresearch.com/knl-webinar look at slide 26. The hot chips paper/presentation is also a good read: http://www.hotchips.org/wp-content/u...dani-Intel.pdf http://ieeexplore.ieee.org/document/...number=7453080 Last fiddled with by bsquared on 2016-09-16 at 16:43 Reason: links 2016-09-16, 17:16 #134 xathor Sep 2016 19 Posts Quote: Originally Posted by science_man_88 http://www.mersenne.org/various/math...rial_factoring Forgive me if I am a little bit ignorant to this, but what command is he wanting me to run exactly? I have no problem doing it, I just want to make sure I get his request correct. If there are any MPI capable builds, I can scale this up quite high for testing. I also have a UV2000 available if you guys are interested in seeing scaling up to 256 physical cores with OpenMP. Last fiddled with by xathor on 2016-09-16 at 17:18 2016-09-16, 17:58 #135 henryzz Just call me Henry "David" Sep 2007 Liverpool (GMT/BST) 37·163 Posts Quote: Originally Posted by xathor Forgive me if I am a little bit ignorant to this, but what command is he wanting me to run exactly? I have no problem doing it, I just want to make sure I get his request correct. If there are any MPI capable builds, I can scale this up quite high for testing. I also have a UV2000 available if you guys are interested in seeing scaling up to 256 physical cores with OpenMP. Trial division under mprime to see how it scales with threads. 2016-09-16, 18:23 #136 jasonp Tribal Bullet Oct 2004 32×5×79 Posts Quote: Originally Posted by xathor If there are any MPI capable builds, I can scale this up quite high for testing. I also have a UV2000 available if you guys are interested in seeing scaling up to 256 physical cores with OpenMP. AFAIK there are no Mersenne testing programs that use MPI. Building one and tuning it for many MPI processes, each with ideally a few threads, would be very challenging. But potentially it can buy a lot of performance because it forces programmers to be explicit about what is shared between MPI instances, and that's really critical on highly-NUMA systems like this one. For reference, I started the MPI port of Msieve back in 2010; because the multithreading scheme for the linear algebra was not that efficient back then the speedup was immediately better than multithreading alone, even when limited to a single (admittedly very nice) box. But it took a year or two before using MPI and multithreading together ran faster than either one alone on a single box. fivemack had to use black magic arguments to pin MPI processes to the correct physical nodes, but with that in place his old 48-core AMD system absolutely flew. 2016-09-16, 21:21 #137 ewmayer 2ω=0 Sep 2002 República de California 5·2,351 Posts Quote: Originally Posted by xathor I'm not sure what you guys mean by trial factoring. Here is a Haswell (dual E5-2670v3 24c AVX2) for comparison: 1000 iterations of M77597293 with FFT length 4194304 = 4096 K Res64: 5F87421FA9DD8F1F. AvgMaxErr = 0.292735259. MaxErr = 0.343750000. Program: E14.1 ... Clocks = 00:00:09.605 Here is a Ivy-Bridge (dual E5-2670v2 20c AVX) for comparison: 1000 iterations of M77597293 with FFT length 4194304 = 4096 K Res64: 5F87421FA9DD8F1F. AvgMaxErr = 0.249028471. MaxErr = 0.312500000. Program: E14.1 ... Clocks = 00:00:08.712 How many threads are those with? Are you using the -nthread flag to control threadcount for those (an on your KNL)? Without that flag, Mlucas will use as many threads as virtual cores it detects on the system. (This seems to be OS-dependent - on my debian-running Haswell quad with HT enabled at boot that number is 4, on my dual-core Broadwell NUC under Ubuntu it is again 4, i.e. 2x the number of physical cores on the latter system.) 2016-09-16, 22:27 #138 Prime95 P90 years forever! Aug 2002 Yeehaw, FL 815010 Posts Quote: Originally Posted by xathor I'm not sure what you guys mean by trial factoring. mprime/prime95 can also do trial factoring, but graphics cards are better suited to that job. mprime/prime95 will have difficulty using a vast number of threads. As currently implemented, only one thread can do the sieving -- 63 TFing threads would easily outpace the one sieving thread. I'd suggest something like 8 workers using 16 threads or 16 workers using 8 threads. I do suspect hyperthreading will be beneficial. 2016-09-17, 00:49 #139 ewmayer ∂2ω=0 Sep 2002 República de California 5·2,351 Posts First-look timings for a build of my dev-branch Mlucas code (no major performance differences over current release, but has enhanced thread affinity-setting options. I need 8 cores to get a timing close to the 23 ms/iter xathor noted in post #123, without any mention of threadcount. Build using gcc 5.1, default core-affinity [n threads ==> affinity set to cores 0:n-1], per-iter times in ms: Code: #thr iter(ms) 1 197.72 2 103.86 4 48.26 8 27.48 16 13.71 32 9.85 64 9.95 Summary: o Base 1-thread timing is dismal ... about the same as my aged Core2; o Scaling quite good up to 32 threads, plateaus there. Here is output of 'numactl -H' ... David, am I right in surmising that to mean "no NUMA clustering"? available: 1 nodes (0) node 0 cpus: 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 181 182 183 184 185 186 187 188 189 190 191 192 193 194 195 196 197 198 199 200 201 202 203 204 205 206 207 208 209 210 211 212 213 214 215 216 217 218 219 220 221 222 223 224 225 226 227 228 229 230 231 232 233 234 235 236 237 238 239 240 241 242 243 244 245 246 247 248 249 250 251 252 253 254 255 node 0 size: 98178 MB node 0 free: 94363 MB node distances: node 0 0: 10 Time to play with non-default thread affinity-setting, e.g. every-other-core and every-fourth-core. My new coreset-summary messaging at least is a definite improvement over the current release's verbose per-thread messaging: Set affinity for the following 64 cores: 0.1.2.3.4.5.6.7.8.9.10.11.12.13.14.15.16.17.18.19.20.21.22.23.24.25.26.27.28.29.30.31.32.33.34.35.36.37.38.39.40.41.42.43.44.45.46.47.48.49.50.51.52.53.54.55.56.57.58.59.60.61.62.63. 2016-09-17, 00:55 #140 airsquirrels "David" Jul 2015 Ohio 10000001012 Posts Quote: Originally Posted by ewmayer ... Here is output of 'numactl -H' ... David, am I right in surmising that to mean "no NUMA clustering"?... That's correct, currently the system is configured with MCDRAM as: Mode: Cache In theory that means the cores should be hitting MCDRAM for everything unless you happen to exceed a 16GB working set. After we've had some time to do benchmarking we will likely want to set this back to Flat and work with the Clustering modes. At that point "numactl -m 0 ./Mlucas" can be used to force DDR4 only and -m 1 can be used to force MCDRAM. In the other clustering modes we would need to be pretty explicit about which cores are being used to maximize locality to the MCDRAM bank. http://colfaxresearch.com/knl-numa/ for more details. 2016-09-17, 01:02 #141 kladner "Kieren" Jul 2011 In My Own Galaxy! 1015810 Posts Both KNL threads are incredibly interesting and informative. I feel privileged to be able to audit them. All of the main participants are bloody amazing! Thanks! 2016-09-17, 01:17 #142 airsquirrels "David" Jul 2015 Ohio 11×47 Posts PSA: Those who PM'd or emailed me with an SSH public key should now have accounts active. If you wanted an account and do not have one, please PM or email me an SSH public key and I will get that setup. 2016-09-17, 01:34 #143 airsquirrels "David" Jul 2015 Ohio 11×47 Posts Quote: Originally Posted by kladner Both KNL threads are incredibly interesting and informative. I feel privileged to be able to audit them. All of the main participants are bloody amazing! Thanks! Glad we can provide some entertainment! Here is another fun tidbit. Based on the sample output of numactl -H when configured in subNuma clustering modes, latency within a quadrant is always 10 whether MCDRAM or DDR4 is used. However cross-quadrant latency is odd - DDR4 is always only 21 between quadrants, but MCDRAM has a latency of 41 between nodes. That suggests that there is a tradeoff between bandwidth and latency when scaling up beyond 16 threads, and that the DDR4 is going to be a bit quicker for sparse access unless we are saturating the 90Gb/s bandwidth... Finally, the L2 cache for each tile pair is available via the cache grid to other tiles with latency 10 (same quadrant) or 21. That suggests that careful cache management could keep 32MB on die ahead of the HBM. Last fiddled with by airsquirrels on 2016-09-17 at 01:39 Similar Threads Thread Thread Starter Forum Replies Last Post Jean Penné Software 39 2012-04-27 12:33 Jean Penné Software 6 2011-04-28 06:21 Surge Hardware 5 2010-12-09 04:07 Unregistered Hardware 6 2005-07-04 04:27 Uncwilly Software 46 2004-02-05 09:38 All times are UTC. The time now is 08:16. Fri Jan 27 08:16:25 UTC 2023 up 162 days, 5:44, 0 users, load averages: 1.20, 1.20, 1.11	2023-01-27 08:16:25	{"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": 0.2737070620059967, "perplexity": 1763.6434164947862}, "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-2023-06/segments/1674764494974.98/warc/CC-MAIN-20230127065356-20230127095356-00511.warc.gz"}
https://www.cnjacu.com/fertility-type-quiz/	Phone (732) 285-4184 Email info@cNJacu.com 1 Bethany Rd, Ste 83, Hazlet, NJ 07730 This self assessment questionnaire is a collaborative effort of Sami S.David, MD and Jill Blakeway, LAc. It’s a part of their book “Making Babies” ## What Is Your Fertility Type? Not every approach to improving fertility will work for every person, no matter what most fertility clinics seem to think. Not should it: we are all individuals and should be considered as such when we receive any kind of health care. The operative question isn’t “What works?” It’s “What works FOR ME?” ### Find out what could be obstructing your fertility and where to focus your attention to conceive and have a healthy baby This is a very detailed questionnaire that digs into your symptoms, analyzes them from different perspectives and gives you a final result. Are you ready? LET’S GO!!! * BBT stands for Basal Body Temperature. Email: ## COVID-19 PROTOCOL Central New Jersey Acupuncture & Wellness Clinic will not be treating any known, or suspected, cases of Covid-19 in the office. If you are sick, please wait for 2 weeks until you are symptom free before making an appointment. If you have been exposed to someone with Covid-19 in the past 14 days, we will not be able to accept you as a patient at this time. If you have traveled outside of the US in the past 30 days we will not be able to accept you as a patient at this time. Following are our new procedures: 1. You will be pre-screened via phone prior to your appointment. 2. When you arrive at the office, please wait in your car and call the office. You will be told when it is safe to come in. 3. Upon entering the office, you will be required to hand sanitize. We will do a quick temperature check, if you are over 100.4F, you will be asked to leave and call your doctor. Temperature checks will be done at each visit. 4. All patients are required to wear a face covering of some kind the entire time you are in the office, for the duration of the treatment. We have masks available if you need one. 5. You will be taken to the treatment room directly. No waiting in the waiting room. 6. Follow up appointments can be scheduled prior to leaving the office. You may also email info@cNJacu.com to set up your next appointment Cleaning: 1. Tables are wiped down between treatments. 2. High tough surfaces- door handles, light switches, call buttons, etc will be sanitized between patients 3. Providers will wear masks and gloves while cleaning 4. Alcohol based hand sanitizers are available in all rooms. What we need from you: 1. Please do not bring anyone with you to your appointment. 2. Please DO NOT bring unnecessary items into the office with you, i.e., purses, bags, water bottles, iPads, etc.. The fewer items you have the less likely we will have any issues with contamination. 3. Please DO NOT come to your appointment if you have, or have had, any cold or flu symptoms in the past 14 days, have traveled recently, or have any of the symptoms that are associated with Covid-19. This applies if you have a household member with any symptoms. We know that there are many silent carriers, but our goal is to minimize any exposure of known causes. Instead, call us to reschedule. If you are uncomfortable coming to the office for any reason, we understand. If any of us tests positive for Covid-19, the office will be closed for 14 days, disinfected, and all patients who have been in 14 days prior will be notified. ## Book an Appointment ### NEW PATIENT OFFER New Patients Enjoy A $40 Initial Consultation Normally$240. Save \$200. Call Us 24/7 To Schedule Your Consultation: (732) 285-4184 Or use this form and we will contact you	2021-04-17 10:26:11	{"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": 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": 0.1955774575471878, "perplexity": 2925.749817452727}, "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-2021-17/segments/1618038119532.50/warc/CC-MAIN-20210417102129-20210417132129-00331.warc.gz"}
https://www.gradesaver.com/textbooks/math/precalculus/precalculus-6th-edition-blitzer/chapter-2-test-page-434/25	## Precalculus (6th Edition) Blitzer Domain: $(-\infty,\infty)$ See graph. Step 1. The domain requirement is that the denominator can not be zero, which gives $x^2+3\ne0$. Thus the domain can be written as: $(-\infty,\infty)$ Step 2. As $f(-x)=f(x)$, the function is even, and we can not identify any vertical asymptotes. Step 3. As $x\to\pm\infty, y\to4$, we find a horizontal asymptote as $y=4$ Step 4. See graph.	2023-03-25 22:43:44	{"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": 0.9909368753433228, "perplexity": 275.0318459958947}, "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/1679296945376.29/warc/CC-MAIN-20230325222822-20230326012822-00298.warc.gz"}
https://gamedev.stackexchange.com/questions/156836/a-trigger-collider-in-unity-doesnt-allow-the-player-to-pass-through	# A trigger collider in Unity doesn't allow the player to pass through It's my understanding that checking "Is Trigger" on a collider should allow rigid bodies to pass through without collision prevention, but my trigger isn't doing that consistently. The player cannot pass through, but other objects can. The trigger is a cube primitive with "mesh rendered" unchecked so that the cube won't be visible. The cube has a box collider with "is trigger" checked, and also a script. The script currently doesn't do anything except output from Debug.Log in OnTriggerEnter() as a test. Primitives are able to pass through. The test is just a basic Unity sphere with all the default settings, including a sphere collider. The only change was that I added a RigidBody. Imported models also pass through. I have a simple .blend model with a rigid body and box collider. The player does not pass through. The main difference between the player's object and the other imported model is that the player is using a custom mesh collider which is different than the player's rendered mesh. The rendered mesh is the game object and the low-poly collision mesh is a child of that object. I've tried several variations, including switching the player from a mesh collider to a box collider, and also moving the box collider to the main mesh rather than a child mesh, but the results are the same. The player stops moving as soon as he hits the location of the trigger's box collider. Any ideas are appreciated. Thanks in advance. Here's a screenshot of the scene layout, with the invisible trigger collider selected. Also pictured is the hierarchy showing that the trigger collider ("Jump Pad Trigger") has no children. On the right is displayed every component that the trigger object has. I've also moved the collider from its original location several times to prove that it's not another object in the scene that the player is colliding with. • Indeed, a rigidbody can avoid collisions, but not triggers, if a collider's hasTrigger is activated (true), and if they are set to collide on the collision's matrix. It would be good if you shared your scene layout, maybe there is another object on the same position where the collider is collisionable, or maybe the trigger mesh has a child object that is collisionable. – LifGwaethrakindo Mar 25 '18 at 20:22 • @LinkWindcrafter: Thanks, I've updated the answer. – Nightmare Games Mar 26 '18 at 1:21 • Ordinarily a body should pass right through a trigger, as you say. So if could be a script on the player or on the trigger is reacting to the trigger overlap and preventing movement. Can you show us the contents of this JumpPad script, and any movement scripts being used on your player? It's also possible that it's coincidence, and the player is getting snagged on the edge of some non-trigger collider that just happens to sit near the trigger boundary. – DMGregory Mar 26 '18 at 1:28 • @DMGregory: Yeah, it was the car script. Thanks again! – Nightmare Games Mar 26 '18 at 5:04 It was in the car's movement code. Even though trigger colliders can be passed through, raycasting into them apparently still fires off the moving object's collision handlers (depending on your physics settings). Good to know. I figured it out when I tried driving backward through the trigger and it worked fine. I had a bit of old code in my vehicle's movement script to mitigate some weird physics behaviors during head-on collisions only. void OnCollisionEnter () { check_forward_collision (); } void OnCollisionStay () { check_forward_collision (); } void check_forward_collision () { bool collision_front = false; Vector3 ray_start = transform.position + new Vector3 (0f, .5f, 0f); collision_front = Physics.Raycast (ray_start, transform.forward, 3f); if (collision_front && forward_speed > .3f) { forward_speed = .3f; // <-- THIS LINE } } When the player would continue pushing forward against a wall with the button held down, the car would sometimes climb the side of the wall or drive underneath it. This function would raycast forward and reduced the player's speed if the hit was directly in front. This old code was apparently getting fired off by the trigger collider. I don't need it any longer, since I fixed those physics problems a different way.	2020-07-10 22:58:18	{"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": 0.20395629107952118, "perplexity": 2608.711756201413}, "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-2020-29/segments/1593655912255.54/warc/CC-MAIN-20200710210528-20200711000528-00309.warc.gz"}
https://publications.hse.ru/en/articles/226471505	• A • A • A • ABC • ABC • ABC • А • А • А • А • А Regular version of the site ## Topological conjugacy of gradient-like flows on surfaces Динамические системы. 2018. Vol. 8(36). No. 1. P. 15-21. The class of C^1-smooth gradient-like flows (Morse flows) on closed surface is the subclass of the Morse-Smale flows class, which are rough. Their non-wandering set consists of a finite number of hyperbolic fixed points and a finite number of hyperbolic limit cycles, and they does not have trajectories connecting saddle points. It is well known that the topological equivalence class of a Morse- Smale flow on a surface can be described combinatorially, for example, by the directed Peixoto graph, or by the Oshemkov-Sharko molecule. However, the description of the class of the topological conjugacy of such a system already requires the introduction of continuous invariants (moduli), corresponding to the periods of limit cycles at least. Thus, one class of the equivalence contains continuum classes of the topological conjugacy. Gradient-like flows are Morse-Smale flows without limit cycles. In this paper we prove that gradient-like flows on a closed surface are topologically conjugate iff they are topologically equivalent.	2021-09-27 23:47:00	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8366749286651611, "perplexity": 1949.9000913209509}, "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/1631780058552.54/warc/CC-MAIN-20210927211955-20210928001955-00437.warc.gz"}
https://www.ucale.com/trigonometrical-ratios-of-submultiple-of-an-angle/	Ucale Trigonometrical Ratios of Submultiple of an Angle An angle of the form A/n where n is an integer is called submultiple angle of A 1. $\displaystyle \left\| \sin { \frac { A }{ 2 } } +\cos { \frac { A }{ 2 } } \right\|$=$\displaystyle\sqrt { 1+\sin { A } }$ or$\displaystyle \sin { \frac { A }{ 2 } } +\cos { \frac { A }{ 2 } }$=$\displaystyle \pm \sqrt { 1+\sin { A } }$ $\displaystyle \left[ +ve, if\quad 2n\pi -\frac { \pi }{ 4 } \le \frac { A }{ 2 } \le 2n\pi +\frac { 3 }{ 4 } \\ otherwise,-ve\right]$ 2. $\displaystyle \left\| \sin { \frac { A }{ 2 } } -\cos { \frac { A }{ 2 } } \right\|$=$\displaystyle \sqrt { 1-\sin { A } }$ or $\displaystyle \sin { \frac { A }{ 2 } } -\cos { \frac { A }{ 2 } }$=$\displaystyle \pm \sqrt { 1-\sin { A } }$ $\displaystyle \left[ +ve,if\quad 2n\pi +\frac { \pi }{ 4 } \le \frac { A }{ 2 } \le 2n\pi +\frac { 5\pi }{ 4 } \\ otherwise,-ve\right]$ 3. $\displaystyle \tan { \frac { A }{ 2 } }$=$\displaystyle \frac { \pm \sqrt { { tan }^{ 2 }A+1 } -1 }{ \tan { A } }$ February 22, 2019 Which class you are presently in? Choose an option. You can change your option at any time. You will be solving questions and growing your critical thinking skills.	2023-02-03 16:46:39	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 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": 0.5373931527137756, "perplexity": 4568.188325834429}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764500058.1/warc/CC-MAIN-20230203154140-20230203184140-00713.warc.gz"}
https://yufeizhao.com/a34/fa19/	# 18.A34 Mathematical Problem Solving (Putnam Seminar) Fall 2019, MIT (Link to the most current version of the course) Class meetings: Mondays and Wednesdays 1–2pm in 4-149 Instructor: Yufei Zhao (see link for contact info) Please include “18.A34” in the subject line of your emails ## Course description 18.A34 is a first-year undergraduate seminar on mathematical problem solving. One of the goals of the seminar is to prepare students for the Putnam Mathematical Competition in December. Each week, one meeting will be a lecture (often by a guest speaker) on some topic/technique, and the other meeting will be student presentations of homework problems, where there will be emphasis on developing good classroom presentation skills. This seminar is most suitable for students with some previous exposure to Mathematical Olympiads. William Lowell Putnam Mathematics Competition: The Putnam Competition is an annual mathematics contest for undergraduates in the USA and Canada. This year it will be held Saturday, December 7, 2019. • All students officially registered in the class are required to participate in the Putnam competition. MIT students should sign up here. Grading. Based on homework and in-class presentations. • Homework will be graded on correctness and presentation. Illegible or extremely sloppy write-ups are unacceptable. • Students are expected to regularly present solutions in class. Class presentations will be critiqued on correctness and clarity. • Class attendance is required. Please notify me in advance if you cannot make it to class. Too many unexcused absences is cause for concern. Listeners. MIT students who are not officially enrolled in the subject are welcome to sit in and participate, but should not hand in homework. Student Support Services (S3) and Student Disability Services ## Schedule Subject to change. See homework policy below Lecture date Topic Lecturer HW due & presentations Supplementary set W 9/4 Probability Evan Chen M 9/9 #1 W 9/11 Independence and uniformity Yufei Zhao M 9/16 #2 W 9/18 Abstract algebra Zilin Jiang M 9/23 #3 W 9/25 Inequalities Mehtaab Sawhney M 9/30 #4 W 10/2 Analysis Ashwin Sah M 10/7 #5 W 10/9 Generating functions (notes) Benjamin Gunby W 10/16 (HW due & new lec.) #6 W 10/16 Congruences and divisibility Junyao Peng M 10/21 #7 W 10/23 Sums and integrals Shengtong Zhang M 10/28 #8 W 10/30 Recurrences Hung-Hsun Yu M 11/4 #9 W 11/6 Combinatorial configurations Yuan Yao W 11/13 #10 M 11/18 (extra presentations) W 11/20 Linear algebra Allen Liu M 11/25 #11 W 11/27 (extra presentations) M 12/2: Students present solutions to Putnam 2018 exam W 12/4: Students present solutions to Putnam 2017 exam Saturday 12/7: Putnam Competition M 12/9: Students present solutions to part A of Putnam 2019 exam W 12/11: Students present solutions to part B of Putnam 2019 exam ## Homework Each problem set contains a (sometimes long) list of problems. You are encouraged to work on as many as you like, but only hand in your six best solutions (please do not submit more than six), at least four from the topics problem set based on the lecture. Do not hand in supplementary problems rated strictly less than [2]; these are too easy. For multi-part problems, you may decide what counts as “one solution”, as long as it is reasonable (i.e., not too trivial). Homework submissions are due at the end of class. Please type or write legibly. They will be graded similarly to the Putnam competition. (Non-registered students should not hand in solutions.) Sources. It is required to write, at the top of each assignment, a list of collaborators and sources (people, books, websites, etc.). If no additional sources are consulted, you must write Sources consulted: none. Failure to do this will result in an automatic 10% deduction. Do not look up solutions to homework problems online (or offline). Collaborations. You are strongly encouraged to work on the problems on your own, though reasonable collaboration is permitted. Everyone must write their solutions individually and acknowledge their collaborators as noted above.	2021-04-12 00:49:35	{"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": 0.33117640018463135, "perplexity": 5707.359321927788}, "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-2021-17/segments/1618038065903.7/warc/CC-MAIN-20210411233715-20210412023715-00285.warc.gz"}
https://aviation.stackexchange.com/questions/12015/are-the-implications-of-weight-and-balance-for-rotary-wing-aircraft-similar-to-f	# Are the implications of weight and balance for rotary wing aircraft similar to fixed wing aircraft? Original discussion here. How does weight and balance for rotary wing aircraft work? Is lateral CG more or less of an issue, or is it comparable to fore/aft CG since the lifting surface is rotating around the aircraft? What are the major differences compared to fixed wing, if any? I don't understand your statement since the lifting surface is rotating around the aircraft - not sure where you are going with that. In general, the principles are the same as for fixed wing. If the CofG limits are exceeded, it is possible to run out of control authority to the extent that the aircraft becomes uncontrollable. Many helicopters, I believe any with a tail rotor, have different left/right CofG limits. The tail rotor opposes the rotation of the fuselage due to torque from the rotor. Since it does this by producing thrust to the side, there is a "translating tendency" which would cause the helicopter to drift in the direction in which the TR is generating lift. To counter this, the cyclic neutral position is rigged to the left (CCW rotor) or right (CW rotor) which results in the skid on that side hanging low. Clearly, there is now less possible movement in the cyclic to that side and therefore, a narrower lateral CofG. For example, in a helicopter with a counter-clockwise rotating rotor, the fuselage wants to rotate clockwise. The tail rotor will produce a thrust to the left to counter this and a long lever couple is formed with the main rotor which, if not corrected, would cause the helicopter to drift to the right. With a perfectly balanced helicopter, in a no wind steady hover, you want the cyclic to be in the centre with no force required from the pilot. To stop the drift to the right, the cyclic is moved to the left. To prevent the need for the pilot to do this, the cyclic is rigged so that its neutral position is to the left. The main rotor thrust vector is now titled to the left which introduces a horizontal component of thrust acting to the left. This also causes the left skid to hang low. This is why in most CCW rotor helicopters, the captain sits on the right and in some of these helicopters, solo flight from the left is prohibited. Since the cyclic is already left of centre, there is less cyclic authority to the left than to the right. Therefore, the right limit of lateral CofG is less than the left limit. This could not be achieved by moving the CofG since this would level the fuselage but would not oppose the thrust. I will never forget the sensation of my first solo lift into the hover after the talking ballast (my instructor) got out and left me to it. I lifted with the control positions where I was used to them. Because there was 170lbs less weight in the front left corner, this meant that I started to pitch up and move back. I was surprised by how much forward and left I had to move the cyclic to counter it. It felt very odd for a few minutes. • i would think that because a helicopter can hover, therefore lateral cg plays a much bigger role than longitudinal compared to fixed wing, all flight vectors being equal. in general during forward flight, i expect the effects of w&b on rotary wing to approximate those of fixed wing. – Erich Jan 26 '15 at 4:54 • to counteract "translating tendency", instead of trimming the cyclic left or right, would moving the lateral cg left or right also do the trick? – Erich Jan 26 '15 at 4:56 • I edited my answer. – Simon Jan 26 '15 at 5:06 • @Federico edit his post if it's that important to you. – Erich Jan 26 '15 at 11:46 • for the record, @Simon, i don't think I have ever used full cyclic deflection in the R-22, and I don't think I would ever want to. – rbp Jan 26 '15 at 14:45	2020-07-10 00:04:04	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.5990121364593506, "perplexity": 1035.0661628226842}, "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-2020-29/segments/1593655902377.71/warc/CC-MAIN-20200709224746-20200710014746-00353.warc.gz"}
https://tex.stackexchange.com/questions/456254/prooftrees-sty-why-are-line-numbers-overlapping-and-how-to-avoid	prooftrees.sty - why are line numbers overlapping and how to avoid? Why are the "6" and "7" overlapping here and how can I avoid it? \documentclass[border=10pt]{standalone} \usepackage{prooftrees} \begin{document} \begin{prooftree}{} [A [A [A [A [A [A [A] ] ] ] ] ] [A [A [A [A] [A] ] ] ] ] \end{prooftree} \end{document} I'm aware of the "move by" option, however in a large, complex tree this adds a lot of sometimes quite long single branches which doesn't look very good imo. It seems to me that when a branching rule is applied, the branches should either increase/decrease in size so that this doesn't happen. Edit - something like this is what I would like: • Welcome to TeX.SE! Good to see you've provided a minimal working example, that is very helpful. It is however not fully clear (to me) how you would want the output to look, and how automated this should be. Do you want the branching step (7) below the last single row (8 in this case). i.e., extending the edges? Or do you want it to be positioned between 6 and 7 as it is now, with extra space within the single branch? Or maybe something like pushing the 7 a bit to the left and keep the other numbers as they are? – Marijn Oct 22 '18 at 15:19 • Did you try it with \begin{prooftree}{single branches=true} ? Is that output acceptable? – Thruston Oct 22 '18 at 15:24 • Pages 22-23 of the manual (try texdoc prooftrees) suggests that automatically fixing issues like this is not always possible... – Thruston Oct 22 '18 at 15:38 • @Marijn Thank you for your reply. I would either like the branching step (7) extended like you said or preferably I would like it aligned with either 6 or 8. I know the latter would produce the "merge delimiter" / "Merging conflicting justifications" warning, but I don't really mind having multiple justifications on the same line. The problem is I don't know how to tell prooftrees to align it with either 6 or 8. – Ben Loewe Oct 22 '18 at 22:33 • @Thruston Thank you for your reply. I have tried constructing trees with \begin{prooftree}{single branches=true} however this doesn't necessarily fix it. Well that might explain a lot as I have fiddled around with the "move by" along with the "single branches" option quite a bit, and still struggled to to avoid these issues. I can do it, it just ends up adding lots of long single branches, which I don't really want. – Ben Loewe Oct 22 '18 at 22:50 prooftrees should not generate this output by default. That is, I consider it a bug. However, it is unlikely to generate precisely the output you want by default either. That I do not consider a bug. With current prooftrees, I suggest \documentclass[border=10pt]{standalone} \usepackage{prooftrees} \begin{document} \begin{prooftree}{} [A [A [A [A [A [A [A] ] ] ] ] ] [A [A [A [A, move by=1] [A, move by=1] ] ] ] ] \end{prooftree} \end{document} Alternatively, \begin{prooftree}{} [A [A [A [A [A, move by=1, edge+={draw=none}% if you don't want an edge [A [A] ] ] ] ] ] [A [A [A [A] [A] ] ] ] ] \end{prooftree} A (just invented) development version, however, avoids overlapping in the default case: \begin{prooftree}{} [A [A [A [A [A [A [A] ] ] ] ] ] [A [A [A [A] [A] ] ] ] ] \end{prooftree} This will make it into the next version of prooftrees if (a) anyone seems interested and (b) it doesn't cause worse effects than those it avoids and which I cannot figure out how to prevent. • I didn't answer the 'why?' The answer is that when proof line numbers are assigned, the disjunctions on the right have a smaller y than the non-disjunction on the left, because the one on the left has l set to \baselineskip, whereas the ones on the right don't. This avoids overly leggy trees. So the disjunctions get a new line number, because y is less than it is for the non-disjunction, but the l on the right is not twice \baselineskip, so the next number on the left gets another number which overlaps with it. The disjunction should be on line 5. – cfr Nov 11 '18 at 3:57 • Thank you for this solution. Not sure if you've released the updated version or not, but if you have I'll definitely give it a try, otherwise the second solution looks quite elegant too. – Ben Loewe Nov 23 '18 at 3:23 • @BenLoewe I haven't published it yet. If you are prepared to try it, send me an email and I'll send you a copy to test. That way, you can drop it in your working directory and just delete it if things go wrong. And you can let me know if it works OK or you find problems. – cfr Nov 23 '18 at 6:02 • Yeah I'll give it a try, I'm a LaTeX newbie so I don't how well I'll be able to diagnose any problems if they arise. – Ben Loewe Nov 23 '18 at 6:31 • where can I find your email? – Ben Loewe Nov 23 '18 at 7:19	2019-07-16 23:33:33	{"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": 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": 0.6132921576499939, "perplexity": 778.5914444228538}, "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-2019-30/segments/1563195524972.66/warc/CC-MAIN-20190716221441-20190717003441-00384.warc.gz"}
https://converter.ninja/volume/imperial-quarts-to-milliliters/71-imperialquart-to-ml/	# 71 imperial quarts in milliliters ## Conversion 71 imperial quarts is equivalent to 80693.0975 milliliters.[1] ## Conversion formula How to convert 71 imperial quarts to milliliters? We know (by definition) that: $1\mathrm{imperialquart}\approx 1136.5225\mathrm{ml}$ We can set up a proportion to solve for the number of milliliters. $1 ⁢ imperialquart 71 ⁢ imperialquart ≈ 1136.5225 ⁢ ml x ⁢ ml$ Now, we cross multiply to solve for our unknown $x$: $x\mathrm{ml}\approx \frac{71\mathrm{imperialquart}}{1\mathrm{imperialquart}}*1136.5225\mathrm{ml}\to x\mathrm{ml}\approx 80693.0975\mathrm{ml}$ Conclusion: $71 ⁢ imperialquart ≈ 80693.0975 ⁢ ml$ ## Conversion in the opposite direction The inverse of the conversion factor is that 1 milliliter is equal to 1.23926337069909e-05 times 71 imperial quarts. It can also be expressed as: 71 imperial quarts is equal to $\frac{1}{\mathrm{1.23926337069909e-05}}$ milliliters. ## Approximation An approximate numerical result would be: seventy-one imperial quarts is about eighty thousand, six hundred and ninety-three point one zero milliliters, or alternatively, a milliliter is about zero times seventy-one imperial quarts. ## Footnotes [1] The precision is 15 significant digits (fourteen digits to the right of the decimal point). Results may contain small errors due to the use of floating point arithmetic.	2022-06-30 23:37:11	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 6, "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": 0.898032546043396, "perplexity": 3545.330265587872}, "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/1656103915196.47/warc/CC-MAIN-20220630213820-20220701003820-00686.warc.gz"}
https://yutsumura.com/given-the-characteristic-polynomial-find-the-rank-of-the-matrix/	# Given the Characteristic Polynomial, Find the Rank of the Matrix ## Problem 484 Let $A$ be a square matrix and its characteristic polynomial is given by $p(t)=(t-1)^3(t-2)^2(t-3)^4(t-4).$ Find the rank of $A$. (The Ohio State University, Linear Algebra Final Exam Problem) ## Solution. Note that the degree of the characteristic polynomial $p(t)$ is the size of the matrix $A$. Since the degree of $p(t)$ is $3+2+4+1=10$, the size of the matrix $A$ is $10\times 10$. From the characteristic polynomial, we see that the eigenvalues of $A$ are $1,2,3,4$. In particular, $0$ is not an eigenvalue of $A$. Hence the null space of $A$ is zero dimensional, that is, the nullity of $A$ is $0$. By the rank-nullity theorem, we have $\text{rank of A} +\text{ nullity of A}=10.$ Hence the rank of $A$ is $10$. ## Final Exam Problems and Solution. (Linear Algebra Math 2568 at the Ohio State University) This problem is one of the final exam problems of Linear Algebra course at the Ohio State University (Math 2568). The other problems can be found from the links below. ### 8 Responses 1. 06/28/2017 […] Given the Characteristic Polynomial, Find the Rank of the Matrix […] 2. 06/28/2017 […] Given the Characteristic Polynomial, Find the Rank of the Matrix […] 3. 07/12/2017 […] Given the Characteristic Polynomial, Find the Rank of the Matrix […] 4. 08/02/2017 […] Given the Characteristic Polynomial, Find the Rank of the Matrix […] 5. 08/12/2017 […] Given the Characteristic Polynomial, Find the Rank of the Matrix […] 6. 08/17/2017 […] Given the Characteristic Polynomial, Find the Rank of the Matrix […] 7. 10/05/2017 […] Given the Characteristic Polynomial, Find the Rank of the Matrix […] 8. 10/15/2017 […] Given the Characteristic Polynomial, Find the Rank of the Matrix […] This site uses Akismet to reduce spam. Learn how your comment data is processed. ##### Diagonalize the 3 by 3 Matrix Whose Entries are All One Diagonalize the matrix $A=\begin{bmatrix} 1 & 1 & 1 \\ 1 &1 &1 \\ 1 & 1 & 1 \end{bmatrix}.$... Close	2019-07-20 22:48:07	{"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": 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": 0.9097553491592407, "perplexity": 315.3229442032358}, "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/1563195526714.15/warc/CC-MAIN-20190720214645-20190721000645-00267.warc.gz"}
https://puzzling.stackexchange.com/questions/36934/queen-paralyzed-king-vs-king	Queen & “paralyzed” King vs King In this puzzle the standard chess rules apply on standard 8x8 board, white has King & Queen and black has his King, the only exception to normal rules is that white is not allowed to move his king. White wins when he mates black, black wins if white can't mate in any finite amount of moves. It is a well known fact that white can mate if he is allowed to move both his king and queen pretty easily. The challenge is to do it only moving the queen, the white king remains paralyzed in his starting position. The initial position I propose for this question is: White king starts at c3. White queen and black king are setup somewhere (in a legal position) and it is white to move. Can white force mate? For which starting positions? Edit: Black can't "touch" the white king by moving next to it, it's an illegal move, as in normal chess. Think of this puzzle as a situation in which white has to mate normally but doesn't want to move the white king. Edit: I'm having confusions between Row, rank file. Before answering you, please clarify me this. In matrix notation i normally call them rows (horizontal lines) and columns (vertical lines). Please answer me if you are using with this convention: column=file. Row=rank • Would the white king be allowed to attack the black king if it got in range? – Areeb Jul 3 '16 at 18:35 • @Areeb You are not allowed to put your own King in check, assuming that White can still attack if this happens. – Klyzx Jul 3 '16 at 19:26 • @CaelanO'Toole that's what I was confused about, since white can't move it is really check? – Areeb Jul 3 '16 at 19:33 • Well, black doesn't win by killing the white king, so, maybe. – Klyzx Jul 3 '16 at 19:36 • @Areeb I understand from what place that doubt came from. No, the intended puzzle you can't put your own black king in check, and yes, that's check because as in the post, standard chess rules apply. – Santropedro Jul 3 '16 at 21:07 I believe White can force mate from every legal starting position. Unfortunately, I believe this because my crappy home-grown Python program thinks it has found mates for white from every position in at most 46 half-moves -- but the program might consist entirely of bugs. (Earlier I said "30 moves" which was ambiguous between full-moves and half-moves and wrong either way. I've fixed a bug. The conclusion is the same.) Here is my crappy Python program: # position is encoded as follows: WQ x,y in bits 0..2, 3..5; # BK x,y in bits 6..8, 9..11; BTM flag in bit 12. moves = [[] for i in range(8193)] # possible successors states = [None for i in range(8193)] # (Wwin?, halfmoves to end) states[8192] = (0,0) # immediate draw (pseudo-position for when WQ captured) # compute moves available in each position and note immediate # win/draw results for qx in range(8): for qy in range(8): for kx in range(8): for ky in range(8): for btm in (0,1): p = qx \| (qy<<3) \| (kx<<6) \| (ky<<9) \| (btm<<12) m = [] if btm: for dx in (-1,0,+1): for dy in (-1,0,+1): if dx==dy==0: continue kx2,ky2 = kx+dx,ky+dy if not (0<=kx2<8 and 0<=ky2<8): continue # off the board if max(abs(kx2-2),abs(ky2-2))<=1: continue # into check by WK if kx2==qx or ky2==qy or abs(kx2-qx)==abs(ky2-qy): # into check by WQ if kx2==qx and ky2==qy and max(abs(qx-2),abs(qy-2))>1: m.append(8192) # capture WQ else: continue m.append(qx \| (qy<<3) \| (kx2<<6) \| (ky2<<9) \| (0<<12)) moves[p] = m if not m: # no moves: checkmate or stalemate if kx==qx or ky==qy or abs(kx-qx)==abs(ky-qy): states[p] = (1,0) # immediate win else: states[p] = (0,0) # immediate draw else: # wtm m = [] for dx in (-1,0,+1): for dy in (-1,0,+1): if dx==dy==0: continue for n in range(1,9): qx2,qy2 = qx+ndx,qy+ndy if not (0<=qx2<8 and 0<=qy2<8): break # off board if qx2==2 and qy2==2: break # collision with WK # no need to check for collision with BK because # WTM positions where this can happen are illegal m.append(qx2 \| (qy2<<3) \| (kx<<6) \| (ky<<9) \| (1<<12)) moves[p] = m def descr(p): return "WQ%s%d BK%s%d %stm" % ("abcdefgh"[p&7], ((p>>3)&7)+1, "abcdefgh"[(p>>6)&7], ((p>>9)&7)+1, "wb"[(p>>12)&1]) # now very inefficiently keep looking for positions whose win/draw status # is resolvable changed = True while changed: changed = False for i in range(8192): if not moves[i]: continue btm = (i>>12)&1 wtm = 1-btm seen = [0,0,0] for j in moves[i]: if states[j]==None: seen[2]=1 else: seen[states[j][0]]=1 if seen[wtm]: # W has a move to (1,---) so this is a W win # or B has a move to (0,---) so this is a draw minr = min(states[j][1] for j in moves[i] if states[j] is not None and states[j][0]==wtm) if states[i] is None or states[i][1]>minr+1: states[i] = (wtm,minr+1) changed = True print descr(i), states[i] elif not seen[2]: # all W moves lead to (0,---) so this is a draw # or all B moves lead to (1,---) so this is a W win maxr = max(states[j][1] for j in moves[i] if states[j] is not None and states[j][0]==btm) if states[i] is None or states[i][1]<maxr+1: states[i] = (btm,maxr+1) changed = True print descr(i), states[i] def bestfrom(p): if states[p] is None: return None btm = (p>>12)&1 win = states[p][0] remoteness = states[p][1] if remoteness <= 0: return None return [m for m in moves[p] if states[m] is not None and states[m][0]==win and states[m][1]==remoteness-1][0] Code to display an example longest White win (it happens to start with a position with Black to move): p = 3+86+647+512*7+4096 # or whatever while p is not None: print descr(p) p = bestfrom(p) which, translating from the horrible notation this gives you, goes like this. Start with WQ at d7 and BK at h8, and Black to move. Then: 1. ... Kg8; 2. Qc6 Kg7; 3. Qa8 Kf7; 4. Qg2 Kf6; 5. Qg8 Kf5; 6. Qf7 Kg5; 7. Qe6 Kf4; 8. Qd5 Kg4; 9. Qe5 Kf3; 10. Qd4 Kg3; 11. Qe4 Kh3; 12. Qf4 Kg2; 13. Qe3 Kh1; 14. Qe2 Kg1; 15. Qe4 Kh2; 16. Qf3 Kg1; 17. Qh3 Kf2; 18. Qg4 Ke3; 19. Qf5 Ke2; 20. Qf4 Ke1; 21. Qd2 Kf1; 22. Qh2 Ke1; 23. Qg2 Kd1; 24. Qf1# (This seems superficially plausible to me but I haven't attempted to check carefully that neither player can improve on it.) Code to list all positions with WTM that are not white wins (there are none, so this prints nothing): for p in range(4096): # 4096 not 8192 because others are BTM if states[p] is not None and states[p][0]: continue print p, states[p], moves[p], descr(p) • Could I review your program? @Gareth – Areeb Jul 4 '16 at 0:25 • If you just take a setup with the white queen on a1 and the black king in the center and play the game it found, you should be able to convince yourself whether the technique is sound. I doubt it will vary if that position is solvable. Clearly if you can force the king to the 1 row or a column you have a win. – Ross Millikan Jul 4 '16 at 1:29 • You say "clearly" but consider the position with BK a8 and WQ b5 and white to move; it may be a win for white (and my program thinks it is) but it's not immediately obvious how to do it. Unfortunately I am at work, my program is at home, and I haven't yet looked at any of the nontrivial output of the program to see (1) whether it makes sense or (2) how it wins any particular nontrivial-looking position. – Gareth McCaughan Jul 4 '16 at 10:08 • @Areeb, my program has now had one bug fixed and is in the answer above. – Gareth McCaughan Jul 7 '16 at 22:19 • I viewed your longest mating secuence and it's indeed identical after 16 Qf3 to the algorithm I thought. The first part up to 14 Qe2 is easy, but the move 15 Qe4! Shocked me, it gaves a winning system faster than what i conceived, both systems reach the position Q on f3 and king h2(black to move). I can;t review your program for lack of knoledge in programming. So I will accept because the sample secuence it's really cool and indeed works. I appreciate your efforts! – Santropedro Jul 9 '16 at 5:17 White can force mate from any position. As an example, put the white queen on a1 and the black king on h8, as far away from mate as possible. White plays Qa8+ and black must move to the seventh rank. Say he moves Kg7. White can move Qe8. If black moves to the sixth rank, white can move the queen to the seventh. If black moves to h7, white can move to f8. Black will then be forced to move down the h column and white can follow down the f column. In any case, he can force the black king to the first rank. When he gets black to the first rank, he needs to leave space for black to move between g1 and h1. With black trapped to g1/h1, white can move to e2, then to g4 and (if necessary) h4 to get on the far side of the king. More of the same will force the king back to the first rank, where the queen can chase it to b1 or c1 and deliver mate. • Not clear at all after "with black trapped". – Santropedro Jul 4 '16 at 1:56 • @Santropedro: I added more detail in the answer so it would be under the spoiler. – Ross Millikan Jul 4 '16 at 2:44 • So it's clear that if you get black to move to the a-file or 1-rank you win. And you can move him along without too much problem. The only remaining question is if he can run interference on your king. Eg can he get himself to an awkward position on d5 and e4? I doubt it, but your proof seems incomplete without it. – Dr Xorile Jul 4 '16 at 3:36 • If he is on d5 I can move to b6. Then he has to move to d4 or e5 or something easier. In either case I can keep restricting his freedom, chasing him toward h1. – Ross Millikan Jul 4 '16 at 4:05 • I don't understand your solution after taking black king to h1. While pushing the king to the left he goes up, over the white king, and if you want it to go to the a file, you have to accept he will achieve getting to a8-a7 and you are on the same situation as before (simmetrically equivalent) so i don;t see progress. One way to show me your strategy easier will be if you play a game at lichess (free, very easyto register) or any site with me and we setup the position, and play, and I see your algorithm working, please consider this, since there is a point I think you are overlooking. – Santropedro Jul 4 '16 at 17:53	2019-04-19 11:05:35	{"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": 0.2346714735031128, "perplexity": 5296.011746971962}, "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-2019-18/segments/1555578527566.44/warc/CC-MAIN-20190419101239-20190419123239-00483.warc.gz"}
https://spinw.org/swplot_line	### Syntax hLine = swplot.line(rStart, rEnd) hLine = swplot.line(r,[]) hLine = swplot.line(rStart, rEnd, lineStyle, lineWidth, multiPatch) ### Description hLine = swplot.line(rStart, rEnd) creates disconnected line segments between multiple rStart(:,i) rEnd(:,i) pairs of 3D coordinates. The lines are shown as patch faces. hLine = swplot.line(r,[]) creates connected line segments between the consicutive points r(:,i). hPatch = swplot.line(handle, ...) adds the generated patch object to a given axis if handle is an axis handle or adds the lines to an existing patch object, if the given handle points to a patch object. ### Input Arguments handle Handle of an axis or triangulated patch object. In case of patch object, the constructed faces will be added to the existing object. rStart Coordinate(s) of the starting point, either a 3 element vector or a matrix with dimensions of [3\times n_{lineSegment}] to plot multiple line segments. rEnd Coordinate(s) of the end point, either a 3 element vector or a matrix with dimensions of $[3\times n_{lineSegment}]$ to plot multiple line segments. r Matrix with dimensions of $[3\times n_{obj}\times n_{lineSegment}]$. The function will plot $n_{obj}$ number of disconnected curves. The $i$th curve will follow the x=r(1,i,:), y=r(2,i,:), z=r(3,i,:) (parameteric) segmented curve. lineStyle Line style, default value is '-' for continuous line. Any other Matlab line style string is accepted: '--'\|':'\|'-.'\|'none'. lineWidth Line width in pt, default value is 0.5. mPatch If true, a separate patch object will be created per line segment. Default is false, a single patch object will store all line segments.	2022-06-26 23:58:33	{"extraction_info": {"found_math": true, "script_math_tex": 4, "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": 0.21003711223602295, "perplexity": 4527.245296126786}, "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/1656103322581.16/warc/CC-MAIN-20220626222503-20220627012503-00634.warc.gz"}
http://math.stackexchange.com/questions/603/what-transformations-of-the-plane-are-geometrically-constructable-compass-str	# What transformations of the plane are geometrically constructable (compass & straight edge)? Congruence transformations (isometries) and similarity transformations (isometries + dilations) should be constructable. What about other affine transformations? Other conformal mappings? edit: by constructable, I mean given the defining information for the transformation in a geometric way (e.g. a dilation requires a center and a ratio, so the given could be a point and two segments), can you construct the image of a point under the transformation from its preimage? - edit (2010-07-26): The question is much more involved than I'd originally thought. As implied in the question, I knew that congruence and similarity transformations are constructible. Immediately below this section is my original answer, which only demonstrates congruence transformations and was intended more to give an idea of what an answer might look like (since, at the time, there was another answer that was not particularly helpful). In the last section of this answer is my justification that all affine transformations of the plane are constructible. In re-reading that now, I realize that I'd assumed the ability to construct a point, say $P'$, on a line, say $\overleftrightarrow{RP}$, such that $\frac{RP'}{RP}$ is equal to some known ratio. This is equivalent to being able to construct the dilation of $P$ by the known ratio about center $R$. I've added the construction of such a dilation below the congruence transformation section. edit (2012-01-28): A conversation with some colleagues reminded me about this problem and in starting to ask them about it, I realized I'd completely missed that all Möbius transformations are constructible. Since any Möbius transformation can be expressed as a composition of translation, reflection, inversion, dilation, rotation, and translation (I think there's a typical decomposition that's roughly in that order, hence my listing translation twice). The only one of these that I have not yet shown is constructible is inversion, so I have appended that construction. As a partial answer, here are constructions of the basic congruence transformations, assuming basic construction techniques like constructing a line parallel or perpendicular to a given line through a given point and angle-copying: 1. reflection Given a point $P$ and a line $\ell$, construct the line perpendicular to $\ell$ through $P$, and construct the circle centered at the intersection of this new line and $\ell$ and passing through $P$. The image of $P$ under a reflection over the line $\ell$ is the point of intersection of the circle and the new line (the one not at $P$). 2. translation Given a point $P$ and a vector $\overrightarrow{AB}$ (from $A$ to $B$), construct the line through $A$ and $P$, the line through $B$ parallel to line $\overleftrightarrow{PA}$, and the line through $P$ parallel to $\overrightarrow{AB}$. The image of $P$ under translation by vector $\overrightarrow{AB}$ is the intersection of the two constructed parallels. 3. rotation Given a point $P$, a center of rotation $R$, and an $\angle ABC$ (from $A$ to $C$), construct the line through $P$ and $R$, copy $\angle ABC$ such that the copy of $B$ coincides with $R$ and the copy of $A$ is on ray $\overrightarrow{RP}$ and let the copy of $C$ be called $D$, construct the circle with center at $R$ and passing through $P$. The image of $P$ under rotation by $\angle ABC$ about point $R$ is the intersection of the circle with ray $\overrightarrow{RD}$. dilation Given a point $P$, a center of dilation $R$, and a ratio $\frac{AC}{AB}$ (where point $C$ lies on ray $\overrightarrow{AB}$), translate $B$ to $B'$ and $C$ to $C'$ by the translation that takes $A$ to $R$, construct line $\overleftrightarrow{B'P}$, construct the line through $C'$ parallel to $\overleftrightarrow{B'P}$, and construct ray $\overrightarrow{RP}$. The image of $P$ under a dilation about $R$ by a factor of $\frac{AC}{AB}$ is the intersection of ray $\overrightarrow{RP}$ and the line through $C'$ parallel to $\overleftrightarrow{B'P}$. All affine transformations are constructible. Per MathWorld and Wikipedia, an affine transformation of the plane is a transformation of the plane that preserves collinearity and preserves ratios of distances on any given line. First, to show that affine transformations preserve parallelism, suppose that two lines $\overleftrightarrow{MN}$ and $\overleftrightarrow{PQ}$ are parallel, and that their images, lines $\overleftrightarrow{M'N'}$ and $\overleftrightarrow{P'Q'}$, intersect at $X'$, the image of $X$. Since affine transformations preserve collinearity, $X$ must be on $\overleftrightarrow{MN}$ and on $\overleftrightarrow{PQ}$, which is a contradiction, so $\overleftrightarrow{M'N'}$ and $\overleftrightarrow{P'Q'}$ cannot intersect. Thus, affine transformations preserve parallelism. An affine transformation is determined by a $\triangle ABC$ and its image, $\triangle A'B'C'$ (per MathWorld; Wikipedia talks about defining an affine transformation by a parallelogram and its image, which is equivalent since affine transformations preserve parallelism). Given point $P$ and triangles $\triangle ABC$ and $\triangle A'B'C'$, construct line $\ell_1$ through $P$ parallel to $\overline{AB}$ and line $\ell_2$ through $P$ and parallel to $\overline{AC}$, call the intersection of $\ell_1$ with $\overline{AC}$ $I_1$ and call the intersection of $\ell_2$ with $\overline{AB}$ $I_2$, extend $\overline{A'B'}$ past $B'$ to a point $I'_2$ such that $\frac{AB}{AI_2}=\frac{A'B'}{A'I'_2}$, extend $\overline{A'C'}$ past $C'$ to a point $I'_1$ such that $\frac{AC}{AI_1}=\frac{A'C'}{A'I'_1}$, construct line $\ell'_1$ through $I'_1$ parallel to $\overline{A'B'}$ and line $\ell'_2$ through $I'_2$ parallel to $\overline{A'C'}$. The image of $P$ under the affine transformation mapping $\triangle ABC$ onto $\triangle A'B'C'$ is the intersection of lines $\ell'_1$ and $\ell'_2$. inversion Given a point $P$ and a circle centered at $O$, construct ray $\overrightarrow{OP}$, let $X$ be the point of intersection of $\overrightarrow{OP}$ with the circle. The image of $P$ under an inversion through the circle is the image of $X$ under a dilation by $\frac{OX}{OP}$ centered at $O$. - Couldn't all transformation which send each point (x,y) to another point (x',y') which can be computed from the first one by performing only the four operations and extraction of square root? - I'm pretty sure that's a subset of all conformal mappings, but without a bit clearer definition of what can be used with +, -, , and /, and sqrt(), I'm not sure about constructability. Could you give some examples of what you have in mind (and perhaps some examples of what doesn't fit your description)? – Isaac Jul 26 '10 at 6:06 All Möbius transformations (i.e., translations, rotations, dilations, and transversions) can certainly be constructed since they are just compositions of inversions (which are of course generalized reflections) by the Cartan-Dieudonne theorem. Indeed, these are just the transformations that relate all the possible geometric primitives (in this case generalized circles) to each other. However, the above claim that all affine transformations can be constructed appears misleading to me. All triangles are equivalent modulo an affine transformation, but it is well known that only a dense subset of all real-valued angles is constructible with a compass and straightedge. For instance, no triangle with a 20° angle can be constructed, so no existing triangle can be transformed into one with a 20° angle. If one is presented with two arbitrary triangles as above, then they can be used to define a particular affine transformation, but neither of the triangles may be themselves constructible and so a generic affine transformation is not constructible from generalized circles with a compass and straightedge. - The bit about affine transformations may be misleading, but I don't think it's incorrect—to construct the affine transformation taking one particular triangle onto a second particular triangle that has a 20° angle requires being given that second triangle, so not constructing the 20° angle from a simpler set of given objects. – Isaac Jan 28 '13 at 21:59 Right, I didn't mean to suggest that it's wrong. I just read the question as "What transformations can be constructed from scratch with a compass and straightedge?", which is more restrictive than your actual question. I suppose it would also be possible to combine a generic affine transformation with an inversion to create a generic projective transformation as well? – Andrew Shevchuk Jan 28 '13 at 22:15 # Projective Transformations Projective transformations of the (projective) plane $\mathbb{RP}^2$ can be constructed using only straightedge. Defining input are four preimage points and their corresponding images. Four points $A$ through $D$ in general position, together with their image points $A'$ through $D'$ again in general position, uniquely define a projective transformation. For every point $E$, the line $AE$ intersects $BC$ in a given point, and $AD$ intersects $BC$ in another point. Together with the points $B$ and $C$ themselves, this gives four points on the line $AB$. By a similar construction, four points on $AD$ can be constructed, using the connections $BE$ and $BC$. The cross ratio of four points remains the same under projective transformations. On the image side, the lines $B'C'$ and $A'D'$ already have both their endpoints and their intersection already defined. So the fourth point which has the same cross ratio is uniquely defined. Using at most two pserspectivities, one can transfer the cross ratio from the preimage to the image line. Doing so for both lines gives two points, which connected to $A$ resp. $B$ will intersect in the image point $E'$. # Computations You can choose a projective basis of four points, corresponding to the following homogenous coordinates: \begin{align} A&=\begin{pmatrix}1\\0\\0\end{pmatrix} & B&=\begin{pmatrix}0\\1\\0\end{pmatrix} & C&=\begin{pmatrix}0\\0\\1\end{pmatrix} & D&=\begin{pmatrix}1\\1\\1\end{pmatrix} \end{align*} In common embeddings of the projective plane, two of these would be “at infinity”, in which case the computations outlined below would involve things like parallel lines, for which you'd probably need some form of compass. But you can also choose a projectively transformed situation (or a different embedding, which is a different view on the same thing) so that all four points are finite, and you can do most of the stuff I'll outline below using straightedge alone. You can obtain lines serving as coordinate axes from the points given above. You can also project any point in the plane onto these axes. There are constructions given by von Staudt which perform additions and multiplications on a projective line. For addition you need points for “zero” and “infinity”, which along the $x$ axis would be $C$ and $A$. For multiplication you also need a point at position “one”, which would be the intersection of $AC$ with $BD$. So given a certain projective reference frame (the four points $A$ through $D$), you can do the following: • construct points on a line corresponding to original coordinates • construct points on that same line corresponding to arbitrary rational numbers • constructively perform elementary arithmetic operations like addition, subtraction, multiplication and division • use these operations to compute/construct new coordinates • turn these coordinates back into a point in the plane Unless you hit the special case of parallel lines resp. points at infinity, all of the above can be done using just a straightedge. So many things you could compute from the coordinates can in fact be constructed. I guess this is what @mau meant in his answer. Since you asked about compass and straightedge, you won't have to worry about the special case of parallel lines either, and you can add taking squeare roots to your sets of primitive operations. -	2016-05-27 04:44:27	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.895370364189148, "perplexity": 197.6156327456247}, "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/1464049276537.37/warc/CC-MAIN-20160524002116-00218-ip-10-185-217-139.ec2.internal.warc.gz"}
http://lists.w3.org/Archives/Public/xsl-editors/1999AprJun/0072.html	# Re: XSL-FO: Using an SGML parser to validate an XSL document Date: Tue, 22 Jun 1999 07:42:06 -0700 Message-Id: <3.0.1.32.19990622074206.0075ac34@mail-321> The DTD was not intended for use in validation, but rather to reflect the general orgainzation of the FO-tree. There are a number of violations of the XML syntax for DTDs. In addition, it does not reflect where a property may be specified, rather it describes where a property is used/consumed. This also means it cant be used for validation. There has been significant confusion over the use of the DTD-like description in this manner, we will most-likely remove it from the next draft and seek another method for describing this information. ---Stephen Deach At 08:30 1999-06-22 GMT, you wrote: >Hi, > >Sorry to bother you, but you are listed as the editor on the xsl page on >www.w3.org and I am having a difficult time trying to validate an XSL >document. I took the DTD fragments from http://www.w3.org/TR/WD-xsl/ and >http://www.w3.org/TR/1999/WD-xslt-19990421, pasted them together into a file >I called stylesheet.dtd, put in the doctype declaration into my file, and >tried to validate the document with sp version 1.3. > >Here are some of the errors I found (if you want I can send you log file): > >fo:block is defined like this: > ><!ELEMENT fo:block ( > fo:first-line-marker?, > ( > #PCDATA > \| %inlines; > \| %block-level; > )+ > ) >> > >But that is not valid, since #PCDATA must be the first element in a mixed >content model. (rule 51 in section 3.2.2 of the XML specification). > >Then the fo:table-header, fo:table-footer, and fo:table-body are defined as >a group (which is not allowed in XML). > >Am I missunderstanding the DTD and the error messages I get? > > Jeffrey M\kern-.05em\raise.5ex\hbox{\b c}\kern-.05emArthur > a.k.a. Jeffrey McArthur ATLIS Publishing Services > http://members.home.net/jeffmcarthur/ > > ---------------------------------------------------------------------------- ------- Stephen Deach \| Sr Computer Scientist 408-536-6521 (office) \| Adobe Systems Inc. 408-537-4214 (fax) \| Mail Stop E15-420	2014-08-20 19:26:50	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.588101863861084, "perplexity": 8572.460028069023}, "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/1408500811913.46/warc/CC-MAIN-20140820021331-00277-ip-10-180-136-8.ec2.internal.warc.gz"}
http://www.math.uni.wroc.pl/seminar-item/2676	# Some twisting around the Cantor space Seminarium: Topologia Osoba referująca: A twisted sum of Banach spaces $X$ and $Y$ is another space $Z$ containing $Y$ as a subspace such that $Z/Y = X$. In this talk we study the behaviour of twisted sums in which $X$ is a $C(K)$-space, using techniques from Topology and Functional Analysis. Most of the action will take place in the space of continuous functions on the Cantor space.	2020-07-13 17:39:32	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.9343836307525635, "perplexity": 128.0060413321325}, "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-2020-29/segments/1593657146247.90/warc/CC-MAIN-20200713162746-20200713192746-00451.warc.gz"}
https://www.anubih.ba/Journals/vol.15,no-1,y19/contentvol15no1y19.htm	SARAJEVO JOURNAL OF MATHEMATICS ISSN 1840-0655 (Printed edition) ISSN 2233-1964 (Online edition) ## Contents of Vol. 15, No. 1 In Memoriam Prof. Dr. Harry I. Miller (1939 - 2018) 3 Arthur W. Apter Tall, Strong, and Strongly Compact Cardinals 7 Mohamed Bezziou, Zoubir Dahmani, and Amina Khameli On Some Double Weighted Fractional Integral Inequalities 23 Steven G. From, and Silvestru S. Dragomir Some New Refinements of Jensen's Discrete Inequality 37 Sami Rebhi Equivalence of K-Functionals and Modulus of Smoothness Generated by The $q$- Rubin Operator 49 Ömer Kişi $\mathcal{I}_{2}$-Asymptotically Lacunary Statistical Equivalence of Weight $g$ of Double Sequences of Sets 57 Belkacem Chaouchi, and Marko Kostić On the Study of an Initial Value Problem for a Second Order Differential Equation Set in a Singular Cylindrical Domain 67 Gabdolla Akishev On Orders of Approximation Functions of Generalized Mixed Smoothness in Lorentz Spaces 81 George A. Anastassiou Multivariate Approximation with Rates by Perturbed Kantorovich-Shilkret Neural Network Operators 97 Emre Taş, and Tuğba Yurdakadim Variational Approximation for Modified Meyer-König and Zeller Operators 113 ### Contact: Tel: + 387 33 560 741 +387 33 560 700 Fax: + 387 33 560 703 E-mail: sarajevojmath@anubih.ba	2021-11-28 05:56:43	{"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": 0.5722262859344482, "perplexity": 14443.097378172575}, "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-2021-49/segments/1637964358469.34/warc/CC-MAIN-20211128043743-20211128073743-00219.warc.gz"}
http://neoxfiles.com/absolute-error/absolute-error-and-relative-error-formula.php	Home > Absolute Error > Absolute Error And Relative Error Formula # Absolute Error And Relative Error Formula ## Contents The absolute error is 1 mm. The approximation error is the gap between the curves, and it increases for x values further from 0. Question2: If the approximate value of $\pi$ is 3.14. Your last reading for the dog's mass M, with absolute error included, is Which measurement is more precise? this contact form The precision is said to be the same as the smallest fractional or decimal division on the scale of the measuring instrument. About Today Living Healthy Chemistry You might also enjoy: Health Tip of the Day Recipe of the Day Sign up There was an error. Know your tools! Imperfect equipment used either for measurement or studies, such as very small, precise measurements or burners that provide uneven heat.[6] Method 2 Calculating Relative Error 1 Divide the Absolute Error by pop over to these guys ## How To Calculate The Relative Error c.) the percentage error in the measured length of the field Answer: a.) The absolute error in the length of the field is 8 feet. The smaller the unit, or fraction of a unit, on the measuring device, the more precisely the device can measure. For example if you know a length is 0.428 m ± 0.002 m, the 0.002 m is an absolute error. Not only have you made a more accurate determination of the value, you also have a set of data that will allow you to estimate the uncertainty in your measurement. Zellmer Chem 102 February 9, 1999 Home Numbers Algebra Geometry Data Measure Puzzles Games Dictionary Worksheets Show Ads Hide AdsAbout Ads Errors in Measurement Error? Tolerance intervals: Error in measurement may be represented by a tolerance interval (margin of error). Unlike absolute error where the error decides how much the measured value deviates from the true value the relative error is expressed as a percentage ratio of absolute error to the Mean Absolute Error Formula In the example if the estimated error is 0.02 m you would report a result of 0.43 ± 0.02 m, not 0.428 ± 0.02 m. Solution: Given: The measured value of metal ball xo = 3.97 cm The true value of ball x = 4 cm Absolute error $\Delta$ x = True value - Measured Difference Between Relative And Absolute Error Avoid the error called "parallax" -- always take readings by looking straight down (or ahead) at the measuring device. Relative Precision Error Relative Standard Deviation (RSD) Coefficient of Variation (CV) Example: The CV of 53.15 %Cl, 53.56 %Cl, and 53.11 %Cl is (0.249 %Cl/53.27 %Cl)x100% = 0.47% relative uncertainty. check over here Chemistry Chemistry 101 - Introduction to Chemistry Chemistry Tests and Quizzes Chemistry Demonstrations, Chemistry Experiments, Chemistry Labs & Chemistry Projects Periodic Table and the Elements Chemistry Disciplines - Chemical Engineering and A measuring instrument shows the length to be 508 feet. Absolute Error Formula Excel Finally, let us see what the convention is for reporting relative error. So we use the maximum possible error. Another word for this variation - or uncertainty in measurement - is "error." This "error" is not the same as a "mistake." It does not mean that you got the wrong ## Difference Between Relative And Absolute Error Another possibility is that the quantity being measured also depends on an uncontrolled variable. (The temperature of the object for example). http://www.owlnet.rice.edu/~labgroup/pdf/Error_analysis.htm if then In this and the following expressions, and are the absolute random errors in x and y and is the propagated uncertainty in z. How To Calculate The Relative Error The percent of error is found by multiplying the relative error by 100%. Absolute Error Formula Chemistry The accepted value for her experiment was 34 grams. To record this measurement as either 0.4 or 0.42819667 would imply that you only know it to 0.1 m in the first case or to 0.00000001 m in the second. http://neoxfiles.com/absolute-error/absolute-error-chemistry-formula.php Babbage [S & E web pages] No measurement of a physical quantity can be entirely accurate. Yes No Can you please put wikiHow on the whitelist for your ad blocker? You'll need to calculate both types of error in science, so it's good to understand the difference between them and how to calculate them.Absolute ErrorAbsolute error is a measure of how Absolute Error Formula Physics However, there should be a way to compare the precision of different measurements. b.) the relative error in the measured length of the field. This means that your percent error would be about 17%. http://neoxfiles.com/absolute-error/absolute-error-formula.php Clearly, if the errors in the inputs are random, they will cancel each other at least some of the time. When reporting relative errors it is usual to multiply the fractional error by 100 and report it as a percentage. Absolute Deviation Formula Example: Alex measured the field to the nearest meter, and got a width of 6 m and a length of 8 m. Looking at the measuring device from a left or right angle will give an incorrect value. 3. ## For example, when an absolute error in a temperature measurement given in Celsius is 1° and the true value is 2°C, the relative error is 0.5 and the percent error is they could both be the smallest possible measure, or both the largest. Well, now we can make a direct comparison. Greatest Possible Error: Because no measurement is exact, measurements are always made to the "nearest something", whether it is stated or not. Percent Error Formula The result is the relative error. The errors in a, b and c are assumed to be negligible in the following formulae. You can compare your own results to get Absolute Error, which measures how far off you were from the expected results. You measure the book and find it to be 75 mm. his comment is here Privacy policy About Wikipedia Disclaimers Contact Wikipedia Developers Cookie statement Mobile view PHYSICS LABORATORY TUTORIAL Contents > 1. > 2. > 3. Assume you made the following five measurements of a length: Length (mm) Deviation from the mean 22.8 0.0 23.1 0.3 22.7 0.1 The same measurement in centimeters would be 42.8 cm and still be a three significant figure number. Note that absolute error is reported in the same units as the measurement.Alternatively, you may have a known or calculated value and you want to use absolute error to express how Thus, relative error is useful for comparing the precision of different measurements.	2018-05-27 17:41:04	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.556488037109375, "perplexity": 723.6992215384427}, "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-2018-22/segments/1526794869732.36/warc/CC-MAIN-20180527170428-20180527190428-00116.warc.gz"}
https://math.stackexchange.com/questions/2067036/is-it-possible-that-each-element-of-galois-group-preserve-a-root	# Is it possible that each element of Galois group preserve a root? Let $f(x) \in \mathbb{Q}[x]$ be an irreducible polynomial. Denote by $K$ the splitting field for $f(x)$. Let $G$ be Galois group of $K$ over $\mathbb{Q}$. Let $\alpha_1, \dots, \alpha_n \in K$ be all the roots of $f(x) = \Pi (x - \alpha_i)$. Question: Does there exist such an $f(x)$ that for each $g \in G$ there exist $\alpha_i$ such that $g( \alpha_i ) = \alpha_i$? Edited: I am interested in the case $\deg f \geq 2$. • For a trivial extension – yes – johnnycrab Dec 21 '16 at 10:14 ## 2 Answers The action of the Galois group of an irreducible polynomial is transitive and every transitive permutation group on a finite set has an element that fixes no point.	2019-06-20 21:25:43	{"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": 0.9787865281105042, "perplexity": 214.86648414833192}, "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-2019-26/segments/1560627999273.79/warc/CC-MAIN-20190620210153-20190620232153-00273.warc.gz"}
http://costruzionipallotta.it/knlt/inverse-sinc-matlab.html	The corresponding inverse Fourier transform is a wavelet pattern. 1 inverse Matrix einer 2x2-Matrix; Gleichungssystem lösen. The effort you put into asking a question is often matched by the quality of our answers. 10:05 ADSP , MATLAB PROGRAMS. The GCV-FFT package is a set of Matlab functions for the computation of analog convolutions of the form. But all the communication (data transmission and reception) is happening in time domain. It is considered the ideal time domain filter, just as the sinc is the ideal frequency domain filter. In the MATLAB, a vector refers to a one dimensional (1×N or N×1) matrix, commonly referred to as an array in other programming languages. The observed spectrum is as shown below. This effectively interpolates between each pixel with a sinc shaped basis function, and is commonly used to up-scale low resolution medical imaging data. 19) and continuous time (Eq. From the above spectrum plot, the sinc shaped filter does not result in perfectly bandlimited spectrum from from Hz to Hz. The wizard contains 8 different pages: Goal page, Name and Type page, Variables and Parameters page, Function Body page, Parameter Initialization Code page, Bounds and General Linear Constrains page, Derived Parameters page and Script Before or. Liste des fonctions de traitement du signal sous Matlab. Inverse 2D FT 2. The Data Types and Code Generation panes are not available for blocks in the DSP System Toolbox™ Filter Designs library. Windowed-sinc filters are used to separate one band of frequencies from another. The observed spectrum is as shown below. Both results carry the sign of x and are floats. Thanks for the link, I saw it weeks ago though. For another example, the fft of [0 1 0 1] is [2 0 -2 0]. Inverse Fourier Transform 7 •Use of Matlab symbolics tool to compute the Fourier transform analytically. Instead, it's the period from 0 to. El término "sinc" es una contracción del nombre latino completo de la función sinus cardinalis (seno cardinal). 3 Fourier Transform Example: To find in frequency domain, /2 22 2 22 /2 2 sin( ) sin( ) sinc a fa fa jj jft a h Xf he dt e e jf hfa fa ha ffa ha fa ππ π π π π ππ − − −. 17 DFT and linear convolution. From Inverse Problems to Software Design. Interpolate F(r,q) to Fˆ(u,v) (polar to rectangular coordinates – e. This is essentially the inverse of function frexp(). And similarly for each of the inverse trigonometric functions. Write and solve the differential equation for a first-order RC circuit. As the sinc filter has infinite impulse response in both positive and negative time directions, it must be approximated for real-world (non-abstract) applications; a windowed sinc filter is often used instead. ) Verify that it works correctly by comparing the results of your function with the Matlab command conv. This is a full (but evolving) list of each project (directory) with links. tan(2x) = 2 tan(x) / (1. 익명함수-----1. Similarly, the analytic inverse Fourier transform is given by g x y G f f j f x f y df df,(,)exp2X YXYXYπ. 0000 We can only use the method described earlier, multiplying by the inverse of the coefficient matrix to obtain a solution, if the coefficient matrix is square. Y = asind(X) returns the inverse sine (sin-1) of the elements of X in degrees. [1] [2] [3] A szakirodalomban az elnevezések nem egységesek, különösen angol nyelvű könyvekben sinc néven hivatkoznak mind a normált, mind a nem. For noise, this. 303 Linear Partial Diﬀerential Equations Matthew J. MATLAB界面简介下面我们主要对MATLAB基本界面进行介绍。. However, before you can implement your filter within an FPGA, you must generate a set of filter coefficients using one of several software tools, such as Octave, MATLAB, or even Excel. 2 Inverse Laplace Transform. As the sinc filter has infinite impulse response in both positive and negative time directions, it must be approximated for real-world (non-abstract) applications; a windowed sinc filter is often used instead. m performs monotonic Steffen interpolation 38) stineman. The complex plane has a real axis (in place of the x-axis) and an imaginary axis (in place of the y-axis). Learn more about inverse sinc MATLAB. fft2 — two-dimension fast Fourier transform; fftshift — rearranges the fft output, moving the zero frequency to the center of the spectrum; hilb — FIR approximation to a Hilbert transform filter. Since the CIC has a sinc-like response, it can be compensated for the droop with a lowpass filter that has an inverse-sinc response in the passband. 6 MATLAB Laboratory Experiment. Properties of Sinc matrices-- This is joint work that was completed by Iyad Abu-Jeib from SUNY at Fredonia and myself. 303 Linear Partial Diﬀerential Equations Matthew J. Within the sampling cones of Fourier space, it can be shown that the amplitude of the “sinc” function increases as the object thickness is reduced (Fig. Create a spatial filter to get the horizontal edge of the image; Create a spatial filter to get the vertical edge of the image (read the MATLAB documentation of fspecial). cos - trigonometric cosine 3. Sine calculator. eat 1 sa- 3. Note : Do not write matlab code. Opérations sur les chaînes La fonction strcmp. Поэтому ответ на запрос. For example, the FT of a square is a 2D sinc function, and the FT of a circle is an airy function (like a Bessel function). Ergänzungen: 06A. Select angle type of degrees (°) or radians (rad) in the combo box. Examine why solving a linear system by inverting the matrix using inv(A)b is inferior to solving it directly using the backslash operator, x = A\b. 1 with the filter coefficients of FIR1. Also, the Sinc function approaches zero as x goes towards infinity, with the envelope of sinc(x) tapering off as 1/x. Instead, it's the period from 0 to. Due to product-to-sum identity (5) the sinc function (4) can be rewritten. En raison de limitations techniques, la typographie souhaitable du titre, « Calcul avec les nombres complexes : Module et argument Calcul avec les nombres complexes/Module et argument », n'a pu être restituée correctement ci-dessus. Different choices of definitions can be specified using the option FourierParameters. MATLAB ajoute automatiquement le nombre de blancs nécessaires à la chaîne la plus courte. sinc π (x) : Sinus cardinal normalisé En mathématiques , la fonction sinus cardinal est une fonction spéciale définie à partir de la fonction trigonométrique sinus apparaissant fréquemment dans des problèmes de physique ondulatoire. Inverse Cosine Calculator You can enter input as either a decimal or as the adjacent over the hypotenuse There are 2 different ways that you can enter input into our arc cos calculator. This MATLAB function constructs an inverse sinc highpass filter specification object D, applying default values for the default specification 'Fst,Fp,Ast,Ap'. You can see that the output from MATLAB is one period of the DTFT, but it's not the period normally plotted, which is from to. 2, Myint-U & Debnath §2. Home / ADSP / MATLAB PROGRAMS / MATLAB program for plotting the power spectral densities of NRZ unipolar, NRZ polar, NRZ bipolar and Manchester polar line coding schemes. IFFTN N-dimensional inverse discrete Fourier transform. modf (x) ¶ Return the fractional and integer parts of x. Skills required from Chapt. The designing of the Chebyshev and Windowed-Sinc filters depends on a mathematical technique called as the Z-transform. найдет множество согласованных ответов. IFFT Inverse discrete Fourier transform. The following Matlab project contains the source code and Matlab examples used for modified interpft to do aliased sinc interpolation. MATLAB Central contributions by Kobi. Also, the Sinc function approaches zero as x goes towards infinity, with the envelope of sinc(x) tapering off as 1/x. Yaroslavsky, Signal sinc-interpolation: a fast computer algorithm, Bioimaging, 4, p. Thanks for the link, I saw it weeks ago though. A sinc function is an even function with unity area. The output was collapsed across directions, to provide an overall estimate of the correlation of nearby points in the image. The main feature of Chebyshev filter is their speed, normally faster than the windowed-sinc. The rectangular function is an idealized low-pass filter, and the sinc function is the non-causal impulse response of such a filter. , t = ± 1, ± 2, …), but at time t = 0, it reaches its maximum of 1. Matlab function return \ Enter a brief summary of what you are selling. you could use Matlab functions interp2 or griddata) 3. 1 Matlab Desktop Command-Window (immer vorhanden) Eingabe der Kommandos nach dem Matlab-Prompt. MATLAB2016a的帮助系统. Windowed-sinc filters are used to separate one band of frequencies from another. Plotting of Discrete and Continuous signal. The jinc function is analogous to sinc, but using a Bessel function in place of sine: jinc(x) = J 1 (x)/x. matlab\elmat - Elementary matrices and matrix manipulation. 1 The Z Transform and Its Properties. The function accepts both real and complex inputs. 35) sincdint. Itturns outthatsuch 2Npointconvolutionof evenlyextended signal with zero padded discrete sinc-function is reduced to inverse discrete cosine. Beenden von Matlab durch Eingabe von quit, exit oder Strg+Q. 1 Lineares Gleichungssystem, Gaußsches Eliminationsverfahren, Cramer-Regel, inverse Matrix 26:22 06A. Important observations that apply to test patterns in general include: (a) the nulls in the sinc2(f) envelope occur at integer multiples of the data rate; (b) spectral lines are evenly spaced at an interval that is the inverse of the pattern length; and (c) the. Sine calculator. ISBN 9780123948120, 9780123948434. One "quick and dirty" way to interpolate a small image to a larger size is to Fourier transform it, pad the Fourier transform with zeros, and then take the inverse transform. Also called Astop. In this work, sinc functions are utilized as basic functions to introduce a new sinc basis Method of Moment (SMOM) for solving the planar D-bar equation. MATLAB Central contributions by Kobi. INVERSE HYPERBOLIC FUNCTIONS. Since the CIC has a sinc-like response, it can be compensated for the droop with a lowpass filter that has an inverse-sinc response in the passband. MATLAB (matrix laboratory) е нумеричко анализаторска околина и програмски јазик. This is a very desirable property in a pulse, as it helps to avoid intersymbol interference, a major cause of degradation in digital transmission. Figure 2 shows how the sinc 1, sinc 3 and sinc 5 filters respond to a unit step input. The output was collapsed across directions, to provide an overall estimate of the correlation of nearby points in the image. Certain MATLAB functions are essentially used on scalars, but operate element-wise when applied to a matrix (or vector). According to the inverse relations: y = arcsin x implies sin y = x. Create a spatial filter to get the horizontal edge of the image; Create a spatial filter to get the vertical edge of the image (read the MATLAB documentation of fspecial). Code to add this calci to your website Just copy and paste the below code to your webpage where you want to display this calculator. So the inverse of csc is arccsc etc. We can use built-in functions very similar to Matlab’s, such as meshgrid, plot3d, surf and others or we can use a more native function such as param3d (as always, you can type ‘help’ on your Scilab command window to see a comprehensive list of available functions). >> TestEgalite=strcmp('MATLAB', 'matlab') TestEgalite = 0 >> TestEgalite=strcmp('MATLAB', 'MATLAB') TestEgalite = 1. 1 Erste Schritte mit Matlab Starten von Matlab mit Doppelclick der linken Maustaste (cll) auf das Matlab Symbol oder die Datei matlab. The information we’ve reviewed so far details what windowed-sinc filters are, the importance of windowing and how to generate filters of different topologies. Using MATLAB, plot Bode diagrams for the closed-loop system shown in Figure 8-94 for K = 1, K = 10, and K = 20. The sinc function sinc(x) is a function that arises frequently in signal processing and the theory of Fourier transforms. sinc(x) 51 squeeze(a) 51 std(m) 51 sum(m) 51 svd(m) 51 trapz(y,x=None) 51 tri(N, M=N, k=0, typecode=None) 51 tril(m,k=0) 51 triu(m,k=0) 51 The multiarray object 52 Typecodes 52 Indexing in and out, slicing 53 Ellipses 54 NewAxis 54 Set-indexing and Broadcasting 54 Axis specifications 55 Textual representations of arrays 55 Comparisons 57. One "quick and dirty" way to interpolate a small image to a larger size is to Fourier transform it, pad the Fourier transform with zeros, and then take the inverse transform. sin(2x) = 2 sin x cos x. A Sinc is defined in the following equation using the sine function. sinc(a) The unnormalized cardinal sine of a: asin(a) The inverse sine of a: acos(a) The inverse cosine of a: atan(a) The inverse tangent of a: acot(a) The inverse cotangent of a: asec(a) The inverse secant of a: acsc(a) The inverse cosecant of a: atan2(a, b) The inverse tangent of a divided by b: sinh(a) The hyperbolic sine of a: cosh(a) The. Ainatpuglia. 1 with the filter coefficients of FIR1. The real part of a complex number is obtained by real(x) and the imaginary part by imag(x). To explain the MATLAB output we're looking at, let me show a DTFT magnitude plot that shows three periods instead of just one. and Implementing FIR Filters 1. Opérations sur les chaînes La fonction strcmp. is the triangular function 13 Dual of rule 12. t 3 2s2 p 6. In some sense it is akin to the derivative of the Heaviside unit step. I have to compute Fourier Transform and Inverse Fourier Transform for a signal and plot its graphs (magnitude and phase). you could use Matlab functions interp2 or griddata) 3. XFOURIER Graphics demo of Fourier series expansion. 실제로는 stop-band attenuation이 대략 -30dB가 안되는것을 볼 수 있고, pass-band에서 큰 ripple이 있는것 (위 그림에서 대략 0. to know 1) uv t V and 1) uv t U an inverse easily. This source was analysed considering the available studies of the area, and a conceptual model of the volcanic-hydrothermal system was designed. The book covers the 108 key topics including the thorough analysis of some discrepancies between the Fourier transform, the Laplace transform, the discrete-time Fourier transform, and the z transform , for example, of u(t)and u(n). As the sinc filter has infinite impulse response in both positive and negative time directions, it must be approximated for real-world (non-abstract) applications; a windowed sinc filter is often used instead. F-1 is the inverse Fourier transform. Scilab 3D Plots In this brief article we’re going to describe how to create 3D Plots in Scilab. ) Verify that it works correctly by comparing the results of your function with the Matlab command conv. 45 Normalized Frequency위치)또한 볼 수 있다. The first figure shows the normalized magnitude response of the ramp filter and the second figure shows the reconstructed phantom using the filter. The GCV-FFT package is a set of Matlab functions for the computation of analog convolutions of the form. 3 Fourier Transform Example: To find in frequency domain, /2 22 2 22 /2 2 sin( ) sin( ) sinc a fa fa jj jft a h Xf he dt e e jf hfa fa ha ffa ha fa ππ π π π π ππ − − −. The 2 main functions for plotting are. 2, Myint-U & Debnath §2. Problem 2: How to plot a Sin Function in MATLAB? The function of y(x)= Sin(x) for 0 To know the details about any Matlab command, you can simply click on that command in the editor and press F1. tan - trigonometric tangent 4. Inverse Sinc Matlab - ainatpuglia. The closed-loop transfer function of the system is given by C(s) -- - K R(s) s(s + l)(s + 5) + K - K s3 + 6s' + 5s + K. INSTDFFT Inverse non-standard 1-D fast Fourier transform. 3/2 1/w 0 w g(t) 0 d(t) (a) (b) Figure 3. Labels: Matlab: Ideal_Low_Pass_Filter. Создаден е од MathWorks MATLAB овозможува лесна манипулација со матрици, исцртување на функции, визуализација на податоци, имплементација на алгоритами. The Excel EXP function returns the result of the constant e raised to the power of a number. Sinc (sinx/x) signal contains uniformly distributed amplitude lines in its frequency spectrum. Labels: Matlab: Ideal_Low_Pass_Filter. MATLAB news, code tips and tricks, questions, and discussion! We are here to help, but won't do your homework or help you pirate software. MatLab's autocorrelation function was used to estimate orientation change across the image. $$h(t)=sinc \left( \frac{t}{T} \right )-sinc \left( \frac{t-2T}{T} \right)$$ Decoding Process: 5) The receiver consists of a modified duobinary decoder and a postcoder (inverse of precoder). , when the image is blurred by a known lowpass filter, it is possible to recover the image by inverse filtering or generalized inverse filtering. 0b 10-Jan-94. When an image is scaled down to a lower size, the inverse question is what will be the color of the remaining pixels. 9 Fourier Transform Properties Solutions to Recommended Problems S9. You would already know that IFFT (Inverse Fast Fourier Transform) is the tool to convert a frequency domain data into a time domain data. The figures below illustrate the effect of using a simple ramp filter. Trigonometric sine calculator. matlab\elfun - Elementary math functions. To find the derivatives of f, g and h in Matlab using the syms function, here is how the code will look like syms x f = cos(8x) g = sin(5x)exp(x) h =(2x^2+1)/(3x) diff(f) diff(g) diff(h) Which returns the following ( You can decide to run one diff at a time, to prevent the confusion of having all answers displayed all at the same time ). This effectively interpolates between each pixel with a sinc shaped basis function, and is commonly used to up-scale low resolution medical imaging data. In this work, sinc functions are utilized as basic functions to introduce a new sinc basis Method of Moment (SMOM) for solving the planar D-bar equation. The band pass filter is a combination of low pass and high pass filters. According to the inverse relations: y = arcsin x implies sin y = x. Convert hex color to or from matlab rgb vector. local function. Trigonometric sine calculator. The inverse hyperbolic functions are multiple-valued and as in the case of inverse trigonometric functions we restrict ourselves to principal values for which they can be considered as single-valued. %Ap —amount rippleallowed defaultunits). This MATLAB function returns the Inverse Sine (sin-1) of the elements of X in radians. The information we’ve reviewed so far details what windowed-sinc filters are, the importance of windowing and how to generate filters of different topologies. I will have to read and learn how functions in MATLAB work and then do what I need. perfect discrete sinc-interpolation, the N-point discrete sinc-function must be padded with N zeros to the full 2N length, and DFT convolution should be carried out with2Npointsignals. - Seite Matlab Skript 2 - 2 Matlab Grundlagen 2. Opérations sur les chaînes La fonction strcmp. Figure 2 shows how the sinc 1, sinc 3 and sinc 5 filters respond to a unit step input. Table of Laplace Transforms f(t) = L-1 {Fs( )} F(s) = L{ ft( )} f(t) = L-1 {Fs( )} F(s) = L{ ft( )} 1. 执行该命令可以查询有关于FUN函数的使用信息。. Create a spatial filter to get the horizontal edge of the image; Create a spatial filter to get the vertical edge of the image (read the MATLAB documentation of fspecial). 1 Matlab Desktop Command-Window (immer vorhanden) Eingabe der Kommandos nach dem Matlab-Prompt. The second filter of the conventional DDC compensates for the passband droop caused by the CIC. The problem is I'm not familiar with the 'function' command in MATLAB, I just wanted to have a simple function like u(t-1) so I can shift the function by varying 't', convolute it with another signal, etc. The corresponding inverse Fourier transform is a wavelet pattern. An example MATLAB code is: a. The Inverse Fourier Transform The Fourier Transform takes us from f(t) to F(ω). 5 From Laplace to the z-Transform. (That’s one of the reasons sinc comes up so often in Fourier analysis, as Woodward observed. There are three parameters that define a rectangular pulse: its height , width in seconds, and center. This is a full (but evolving) list of each project (directory) with links. x4 (t) is sinc(t) contracted by a factor of 10. Matlab also has numerous libraries geared toward scientific and commercial applications including: solving systems of ODEs, signal processing, wavelets, equation solving, linear and nonlinear optimization, neural networks, image processing, interpolation, polynomials, data analysis, Fourier transforms, elementary and special mathematical functions, and digital audio. For noise, this. The firmware for the MMDVM (Multi-Mode Digital Voice Modem) - g4klx/MMDVM. Scilab 3D Plots In this brief article we’re going to describe how to create 3D Plots in Scilab. Thanks for the link, I saw it weeks ago though. How to do this in Matlab? As I know Matlab provides built in function fft which computes DFT and probably it is possible to convert results from DFT to DTFT. To explain the MATLAB output we're looking at, let me show a DTFT magnitude plot that shows three periods instead of just one. sawtooth - Sawtooth function. MATLAB ajoute automatiquement le nombre de blancs nécessaires à la chaîne la plus courte. Labels: Matlab: Ideal_Low_Pass_Filter. For math, science, nutrition, history. One "quick and dirty" way to interpolate a small image to a larger size is to Fourier transform it, pad the Fourier transform with zeros, and then take the inverse transform. Matlab function return \ Enter a brief summary of what you are selling. It's a complicated set of integration by parts, and then factoring the complex exponential such that it can be rewritten as the sine function, and so on. This book is a self-teaching reference focused on visualization of signals and systems with MATLAB. 26 한컴오피스 (hwp) \| 3페이지 \| 가격 500원. psinc is the 'periodic sinc function. So we write this as the inner product between sinc of t minus n, this is the basis vector number n, and x of t are signal because of the symmetry of the sinc we can flip the sign of the argument of the sinc, and we obtain that this inner product is once again the convolution between a sinc and our bandlimited signal. tan(2x) = 2 tan(x) / (1. The input is a 10-bit video signal and the output is scaled to accommodate the gain of the inverse sinc response. Generalizing The Inverse Fft Off Unit Circle Scientific Reports Fourier Transforms Matlab Simulink Sinc Function An Overview Sciencedirect Topics. Similarly we define the other inverse hyperbolic functions. The sinc function computes the mathematical sinc function for an input vector or matrix. MATLAB news, code tips and tricks, questions, and discussion! We are here to help, but won't do your homework or help you pirate software. 2 Inverse of the Z Transform. FFT Zero Padding. The sinc function has a value of 1 where x is zero, and a value of. I'd like to call your attention especially to the dots ("") at the left and right on the cosine and sinc function plots. Canny edge detection goes a bit further by removing speckle noise with a low pass filter first, then applying a Sobel filter, and then doing non-maximum suppression to pick out the best pixel for DA: 92 PA: 80 MOZ Rank: 75 Up or Down: Up. Similarly, the analytic inverse Fourier transform is given by g x y G f f j f x f y df df,(,)exp2X YXYXYπ. Online documentation. In using MATLAB to plot. See Inverse Sinc Filter Design — Main Pane for more information about the parameters of this block. Here is how you can relate between the time and frequency domains. In this work, sinc functions are utilized as basic functions to introduce a new sinc basis Method of Moment (SMOM) for solving the planar D-bar equation. Mathematical function and library in MATLAB: This is the cluster of functions like cosine, sum, sine, and complex arithmetic along with the more functions like eigenvalues, matrix inverse, Fourier transforms, Bessel functions and fast. The wizard contains 8 different pages: Goal page, Name and Type page, Variables and Parameters page, Function Body page, Parameter Initialization Code page, Bounds and General Linear Constrains page, Derived Parameters page and Script Before or. Note that the order of the sinc filter matches the number of samples it takes to settle to the input. C numerics library. Trigonometric sine calculator. They are very stable, produce few surprises, and can be pushed to incredible performance levels. , t = ± 1, ± 2, …), but at time t = 0, it reaches its maximum of 1. m performs piecewise discrete sinc interpolation 36) sincint. Se representa mediante el mismo símbolo que el productorio, siendo la función una fracción: la parte de arriba (t) representa en función de que variable esta , la parte de abajo (T) representa la extensión de la función, que irá desde –T/2 a T/2. m: 232 mcentral/winsinc: Applies a windowed sinc filter. Notice the left term in the last line above is n s times the inverse Fourier transform of B [k] ^ f 0 where B is the. In Topic 19 of Trigonometry, we introduced the inverse trigonometric functions. Header declares a set of functions to compute common mathematical operations and transformations: Functions Trigonometric functions cos Compute cosine (function ). XFOURIER Graphics demo of Fourier series expansion. The Fourier transform of the box function, the function box(x) that is 1 on the interval [-1/2, 1/2] and zero everywhere else, is sinc(π ω). The Excel EXP function returns the result of the constant e raised to the power of a number. for all other elements of x. Matlab – Loop types There may be a situation when you need to execute a block of code several times. For noise, this. The inverse hyperbolic functions are multiple-valued and as in the case of inverse trigonometric functions we restrict ourselves to principal values for which they can be considered as single-valued. g(x) = f ∗ h(x),that is, convolutions between an analog signal f(x) (defind by a set of discrete samples c[k]) and an analog convolution kernel h(x), defined by its analog frequency response ĥ(ν). The inverse cosecant function - arccsc. They are very stable, produce few surprises, and can be pushed to incredible performance levels. The rectangular function is an idealized low-pass filter, and the sinc function is the non-causal impulse response of such a filter. 10:05 ADSP , MATLAB PROGRAMS. This is essentially the inverse of function frexp(). The high'DC' components of the rect function lies in the origin of the image plot and on the fourier transform plot, those DC components should coincide with the center of the plot. inverse of the sampling rate, centred if necessary. abs - Magnitude. Les noms des constantes n'est pas réservé, c'est-à-dire qu'il est possible de définir des variables de même nom. Supported with. Wiener Filtering. This analytic expression corresponds to the continuous inverse Fourier transform of a rectangular pulse of width 2π and height 1: sinc t = 1 2 π ∫ − π π e j ω t d ω. The argument of Sinc is assumed to be in radians. If x[n]is a signal whose length exceeds N, e. Due to product-to-sum identity (5) the sinc function (4) can be rewritten. of sinc and the scaling property of the F. So we write this as the inner product between sinc of t minus n, this is the basis vector number n, and x of t are signal because of the symmetry of the sinc we can flip the sign of the argument of the sinc, and we obtain that this inner product is once again the convolution between a sinc and our bandlimited signal. To explain the MATLAB output we're looking at, let me show a DTFT magnitude plot that shows three periods instead of just one. global function ( primary function) MATLAB의 가장 일반적인 함수 정의 방법입니다. IFFTN N-dimensional inverse discrete Fourier transform. The sinc function is the continuous inverse Fourier transform of the rectangular pulse of width 2pi and height 1. Create a random matrix A of order 500 that is constructed so that its condition number, cond(A), is 1e10, and its norm, norm(A), is 1. Results from inverse modelling indicate that a source located beneath the volcano edifice at a mean depth of 4 km is producing a volume change of approximately 0. Part_1_The_Z_Transform click here for the slides. 0b 10-Jan-94. The Fourier transform of the box function, the function box(x) that is 1 on the interval [-1/2, 1/2] and zero everywhere else, is sinc(π ω). But a Gaussian is a special case where it's so smooth, that the FT of a Gaussian is just another Gaussian , so you won't see any periodicity or oscillations at all. Um die Matlab-Hausaufgaben durchführen zu können, gibt es mehrere Möglichkeiten: Der Matlab-spezifische Teil der Hausaufgaben wird an einem PC im Rechenzentrum durchgeführt. X Y For the Fourier transform to be realizable in a mathematical sense, g(x,y). sometimes we will use the DFT, in which case we should interpret the inverse DFT as follows x[n] = 1 N NX 1 k=0 X[k]e\|2Nˇ kn: This is indeed a N-periodic expression. Recall the definition of sinc function: As a result, we have:. The first thing is to try is the command hold off before the new plot command. In this context, MATLAB should have rejected the above case on the ground that [AT1 A1 ] is singular and so its inverse does not exist. Suggestion: Google the F. However, before you can implement your filter within an FPGA, you must generate a set of filter coefficients using one of several software tools, such as Octave, MATLAB, or even Excel. Plot the sinc function Plots: sin(x)/x Ck sincu:= , u ∈ℝ. Posted by Shannon Hilbert in Digital Signal Processing on 4-22-13. Part_1_The_Z_Transform click here for the slides. Now, you can go through and do that math yourself if you want. Inverse Fourier Transform 7 •Use of Matlab symbolics tool to compute the Fourier transform analytically. To find the derivatives of f, g and h in Matlab using the syms function, here is how the code will look like syms x f = cos(8x) g = sin(5x)exp(x) h =(2x^2+1)/(3x) diff(f) diff(g) diff(h) Which returns the following ( You can decide to run one diff at a time, to prevent the confusion of having all answers displayed all at the same time ). This effectively interpolates between each pixel with a sinc shaped basis function, and is commonly used to up-scale low resolution medical imaging data. In Matlab complex numbers can be created using x = 3 - 2i or x = complex(3, -2). as an incomplete cosine expansion of the sinc function. Also, the Sinc function approaches zero as x goes towards infinity, with the envelope of sinc(x) tapering off as 1/x. This MATLAB function returns the Inverse Sine (sin-1) of the elements of X in radians. It is a series of Dirac delta functions in the frequency domain, and is an even function, meaning symmetrical about the origin. m performs rational Thiele interpolation 40) trigint. To simplify the presentation, consider the case of an 1D image f(x) that we want to magnify by a factor of 2 to create the new image g(x). matlab\elfun - Elementary math functions. m performs piecewise discrete sinc interpolation 36) sincint. matlab의 freqz함수를 이용해서 저 filter의 magnitude/phase response를 구해보면 다음과 같다. Se representa mediante el mismo símbolo que el productorio, siendo la función una fracción: la parte de arriba (t) representa en función de que variable esta , la parte de abajo (T) representa la extensión de la función, que irá desde –T/2 a T/2. tnn-12,=1,2,3,K. Print Book & E-Book. 26 한컴오피스 (hwp) \| 3페이지 \| 가격 500원. Table of Laplace Transformations. The Sinc Function The function, crops up again and again in Fourier analysis. Behind all that complicated mathematics, there is a simple logic. 6 Response to arbitrary dynamic excitation 148 6. MATLAB code for the given mathematical function: Here is a simple code in MATLAB, to draw the graph for the given equation. The closest to the origin components are at f0, the fundamental. How about going back? Recall our formula for the Fourier Series of f(t) : Now transform the sums to integrals from –∞to ∞, and again replace F m with F(ω). If you are a newbie in this field, have a look at our MATLAB tutorials to get familiar with it. 12 s i n c 2 ( a t ) {\displaystyle \mathrm {sinc} ^{2}(at)\,}. MATLAB界面简介下面我们主要对MATLAB基本界面进行介绍。. The effect of the multiplication is the appearance of ℱ−1of 𝟏−𝐵,𝐵𝜉 in a convolution, which is the sinc function. Bottom: Inverse sinc response, f = 0 to f s /2. Backprojection-Filtering Backprojection means that we “smear” the projection data back across the object space. Euler (pronounced "oiler'') was born in Basel in 1707 and died in 1783, following a life of stunningly prolific mathematical work. So we have the formula for the inverse Fourier Transform, the integral over the entire frequency axis of the Fourier Transform of the sinc times the Fourier Transform of the sinc. Using MATLAB, plot Bode diagrams for the closed-loop system shown in Figure 8-94 for K = 1, K = 10, and K = 20. help inverse. The inverse Fourier transform of a function is by default defined as. 303 Linear Partial Diﬀerential Equations Matthew J. Opérations sur les chaînes La fonction strcmp. INVERSE HYPERBOLIC FUNCTIONS. I will have to read and learn how functions in MATLAB work and then do what I need. Programming with MATLAB WP240 (v1. Filter analysis and implementation. So the inverse of csc is arccsc etc. >> TestEgalite=strcmp('MATLAB', 'matlab') TestEgalite = 0 >> TestEgalite=strcmp('MATLAB', 'MATLAB') TestEgalite = 1. 6th order Low-pass Chebyshev-2 filter with a cut-off frequency of 3 MHz when the signal is sampled at 10 MHz. m performs monotonic Steffen interpolation 38) stineman. One "quick and dirty" way to interpolate a small image to a larger size is to Fourier transform it, pad the Fourier transform with zeros, and then take the inverse transform. For example, the FT of a square is a 2D sinc function, and the FT of a circle is an airy function (like a Bessel function). Поэтому ответ на запрос. Itturns outthatsuch 2Npointconvolutionof evenlyextended signal with zero padded discrete sinc-function is reduced to inverse discrete cosine. 1 inverse Matrix einer 2x2-Matrix; Gleichungssystem lösen. Different choices of definitions can be specified using the option FourierParameters. Demostraciones: Pulso rectangular []. Windowed-sinc filters are used to separate one band of frequencies from another. DFTMTX Discrete Fourier transform matrix. The input x must be a real array with elements between 0 and 1, inclusive. The EXP function is the inverse of the LN (natural logarithm) function. Sinc-based methods have been applied to diverse scientiﬁc and engineering problems comprising heat conduction [9], [10], population growth [11], inverse problems. , when the image is blurred by a known lowpass filter, it is possible to recover the image by inverse filtering or generalized inverse filtering. 1 sin (𝑥) 𝑥 (6) which has the following normalisation [6]. Always sample by a power of 2 4. tnn,=1,2,3,K 1! n n s + 4. A few days ago, I was trying to do the convolution between a Sinc function and a Gaussian function. Convert hex color to or from matlab rgb vector. Matlab function return \ Enter a brief summary of what you are selling. 与大家常用的 Fortran 和 C 等高级语言相比，MATLAB 的语法规则更简单，更贴近人的思维方式，被称之为“草稿纸式的语言” 。 MATLAB 软件主要由主包、仿真系统（simulink）和工具箱（toolbox）三大部分组成。 2. Also called Astop. freqz(b,a) plt. When an image is scaled down to a lower size, the inverse question is what will be the color of the remaining pixels. The dots are there to remind you that these functions have infinite extent. Important observations that apply to test patterns in general include: (a) the nulls in the sinc2(f) envelope occur at integer multiples of the data rate; (b) spectral lines are evenly spaced at an interval that is the inverse of the pattern length; and (c) the. In order to calculate sin(x) on the calculator: Enter the input angle. Interpolate F(r,q) to Fˆ(u,v) (polar to rectangular coordinates – e. Preventing MATLAB from putting a new plot on top of an old one. This is a full (but evolving) list of each project (directory) with links. If MATLAB ignores hold off (which it does occasionally), give the command. DFTMTX Discrete Fourier transform matrix. The backprojection operator for a single projection looks like this: ∫ ∞ −∞. MATLAB에서도 아주 간단하게 함수를 정의 할 수 있더군요!! MATLAB에서 함수를 정의하는 방법 3가지를 소개합니다!-----1. Code to add this calci to your website Just copy and paste the below code to your webpage where you want to display this calculator. Instead, it's the period from 0 to. Suggestion: Google the F. Yaroslavsky, Signal sinc-interpolation: a fast computer algorithm, Bioimaging, 4, p. m performs piecewise discrete sinc interpolation 36) sincint. (This is a MATLAB function. The rectangular pulse and the normalized sinc function 11 Dual of rule 10. 19) and continuous time (Eq. Например, MATLAB не имеет функции с именем inverse. Inverse Fourier Transform (10-8) 1 2 g f t e dt it Fourier Transform (10-9) There are a lot of notable things about these relations. The first statement in a function is executed first, followed by the second, and so on. NSTDFFT Non-standard 1-D fast Fourier. Remembering the fact that we introduced a factor of i. Then, since cos(0) = 1, we can apply the Squeeze Theorem to show that the sinc function approaches one as x goes to zero. 1 sin (𝑥) 𝑥 (6) which has the following normalisation [6]. This can be seen by looking at ContourPlot :. Title: Microsoft PowerPoint - Lecture 10 - Fourier Transform. Problem 2: How to plot a Sin Function in MATLAB? The function of y(x)= Sin(x) for 0 To know the details about any Matlab command, you can simply click on that command in the editor and press F1. The Inverse Fourier Transform The Fourier Transform takes us from f(t) to F(ω). sawtooth - Sawtooth function. DFTMTX Discrete Fourier transform matrix. ) fftfilt: FFT-based FIR filtering using the overlap-add method. Figures 10d and 10e show that nearby locations in natural scenes and noise tend to be of similar orientation. The windowed-sinc filter that is described in this article is an example of a Finite Impulse Response (FIR) filter. When an image is scaled down to a lower size, the inverse question is what will be the color of the remaining pixels. Octave-Forge is a collection of packages providing extra functionality for GNU Octave. Woodwardtól származik 1953-ból. It is recommended the final value of an even function be dropped when performing DFTs (cyclic nature) Rules of DFTs. See full list on tutorialspoint. 2 mit Cramer-Regel 3x3-Matrix invertieren 10:43 06A. Matlab – Loop types There may be a situation when you need to execute a block of code several times. The GCV-FFT package is a set of Matlab functions for the computation of analog convolutions of the form. Windowed-sinc filters are used to separate one band of frequencies from another. Smith III, W3K Publishing, 2011, ISBN 978-0-9745607-3-1. is the triangular function 13 Dual of rule 12. Sinc Filter. Interpolate F(r,q) to Fˆ(u,v) (polar to rectangular coordinates – e. The function accepts both real and complex inputs. This analytic expression corresponds to the continuous inverse Fourier transform of a rectangular pulse of width 2π and height 1: sinc t = 1 2 π ∫ − π π e j ω t d ω. 2251e-308 et realmax = 1. 执行该命令可以查询有关于FUN函数的使用信息。. Windowed-sinc filters are used to separate one band of frequencies from another. A Sinc is defined in the following equation using the sine function. Title: Microsoft PowerPoint - Lecture 10 - Fourier Transform. 1 sin (𝑥) 𝑥 (6) which has the following normalisation [6]. {yn } is presented in Figure 2. Se representa mediante el mismo símbolo que el productorio, siendo la función una fracción: la parte de arriba (t) representa en función de que variable esta , la parte de abajo (T) representa la extensión de la función, que irá desde –T/2 a T/2. Backprojection-Filtering Backprojection means that we “smear” the projection data back across the object space. matlab의 freqz함수를 이용해서 저 filter의 magnitude/phase response를 구해보면 다음과 같다. An example MATLAB code is: a. Always sample by a power of 2 4. This is a general feature of Fourier transform, i. FFT Zero Padding. Matlab Libraries. Matlab help file explains the usage and other details about the commands like fft,sin and so on. FIR0_2x = zeros(1,2length(FIR0)); FIRO_2x(1:2:end) = FIR0; 2. These files will work under Octave and, with a little tweaking, Matlab (Octave allows multiple functions to be defined in a single script file, so these have to be separated out into their own files for Matlab. You would already know that IFFT (Inverse Fast Fourier Transform) is the tool to convert a frequency domain data into a time domain data. Windowed-sinc filters are used to separate one band of frequencies from another. First, the reconstruction in SampleMania is computed not in the frequency domain with a low-pass filter, but in the time domain via convolution with a sinc (this is done in order to avoid the additional time that would be required to compute an inverse Fourier transform). A Sinc is defined in the following equation using the sine function. The syntax for chebfuns is almost exactly the same as the usual MATLAB syntax for vectors, with the familiar MATLAB commands for vectors overloaded in natural ways. In general, statements are executed sequentially. The inverse Fourier transform of a function is by default defined as. Always sample by a power of 2 4. FFT Zero Padding. IFFT Inverse discrete Fourier transform. The Discrete Two Dimensional Fourier Transform in Polar Coordinates A thesis submitted to the Faculty of Engineering in partial fulfillment of the requirements for the. Integration of sinx in python. and , we utilize the command sinc so that there is no need to separately handle the “0/0” cases due to the sine functions. KY - White Leghorn Pullets). Euler (pronounced "oiler'') was born in Basel in 1707 and died in 1783, following a life of stunningly prolific mathematical work. Here I’m going to show you how signals can be generated in MATLAB. And similarly for each of the inverse trigonometric functions. Posted by Shannon Hilbert in Digital Signal Processing on 4-22-13. 3YF2 Fourier Series – Solutions 4 4. Interp_4x = conv(FIR0_2x,FIR1). Demostraciones: Pulso rectangular []. abs - Magnitude. The GCV-FFT package is a set of Matlab functions for the computation of analog convolutions of the form. 45 Normalized Frequency위치)또한 볼 수 있다. tan(x y) = (tan x tan y) / (1 tan x tan y). This MATLAB function constructs an inverse sinc lowpass filter specification object d, applying default values for the default specification, 'Fp,Fst,Ap,Ast'. Code to add this calci to your website Just copy and paste the below code to your webpage where you want to display this calculator. To explain the MATLAB output we're looking at, let me show a DTFT magnitude plot that shows three periods instead of just one. [How to cite this work] [Order a printed hardcopy] [Comment on this page via email] Spectral Audio Signal Processing'', by Julius O. By the end of this course you should be able to perform Convolution with matlab, perform Discrete Fourier Transform (DFT) with matlab, perform Inverse Discrete Fourier Transform (IDFT) with matlab , design and develop Finite Impulse Response (FIR) filters with matlab, design and develop Infinite Impulse Response (IIR) filters with matlab. 6 MATLAB Laboratory Experiment. Windowed-sinc filters are used to separate one band of frequencies from another. Problem 2: How to plot a Sin Function in MATLAB? The function of y(x)= Sin(x) for 0 To know the details about any Matlab command, you can simply click on that command in the editor and press F1. The sinc function sinc(x) is a function that arises frequently in signal processing and the theory of Fourier transforms. Compare 2 chaînes de caractères données en arguments. Diese Definition zeigt, dass man von der Delta-Distribution nicht wie von einer gewöhnlichen Funktion denken sollte, die ausschließlich für x = 0. The following Matlab project contains the source code and Matlab examples used for modified interpft to do aliased sinc interpolation. 0b 10-Jan-94. Always sample by a power of 2 4. But, because the roundoff errors make a very small number appear to be a zero or make a real zero appear to be a very small number (as will be mentioned in Remark 2. Sampling and replication duality Sketch replication in the opposite domain for one and two dimensions Restoration using a sinc interpolation filter Units of 2D sampling – cycles/ mm Discrete Fourier Transform Relationship between number of points sampled, sampling rate, and length of sampling (N= 2BL) in terms of the frequency domain. eat 1 sa- 3. Analogue signal pre-processing was done on simple amplifier circuit designated for ECG signal measurement. Here I’m going to show you how signals can be generated in MATLAB. ) conv2: Two-dimensional convolution. 2) is its own Fourier transform. Liste des fonctions de traitement du signal sous Matlab. INSTDFFT Inverse non-standard 1-D fast Fourier transform. userpath: list the directories where MATLAB looks for m-files. 실제로는 stop-band attenuation이 대략 -30dB가 안되는것을 볼 수 있고, pass-band에서 큰 ripple이 있는것 (위 그림에서 대략 0. 14 Shows that the Gaussian function exp( - a. The designing of the Chebyshev and Windowed-Sinc filters depends on a mathematical technique called as the Z-transform. Interp_4x = conv(FIR0_2x,FIR1). Different choices of definitions can be specified using the option FourierParameters. Sinc can be evaluated to arbitrary numerical precision. - Seite Matlab Skript 2 - 2 Matlab Grundlagen 2. 225-231, 1996 Discrete sinc interpolation versus other. A sinc function is an even function with unity area. 익명함수-----1. m performs piecewise sinc interpolation 37) steffen. Paul Cu Slides courtesy of John Pauly (Stanford) Princeton University Fall 2011-12 Cu (Lecture 3) ELE 301: Signals and Systems Fall 2011-12 1 / 55. MATLAB ajoute automatiquement le nombre de blancs nécessaires à la chaîne la plus courte. For math, science, nutrition, history. ) Verify that it works correctly by comparing the results of your function with the Matlab command conv. Write and interpret MATLAB code for calculating and plotting signals in continuous and discrete time. Supported with. The sinc function has a value of 1 where x is zero, and a value of. how to make inverse fourrier transform in the correct way? function there is a difference between sinc(x) and sin(x)/x. m performs piecewise sinc interpolation 37) steffen. MATLAB界面简介下面我们主要对MATLAB基本界面进行介绍。. Behind all that complicated mathematics, there is a simple logic. The inverse Fourier transform of the output signal is. Which of the following is the Analysis equation of Fourier Transform?. Interp_4x = conv(FIR0_2x,FIR1). So we have the formula for the inverse Fourier Transform, the integral over the entire frequency axis of the Fourier Transform of the sinc times the Fourier Transform of the sinc. These exceptional frequency domain characteristics are obtained at the expense of poor performance in the time domain, including excessive ripple and overshoot in the. The sinc function is the continuous inverse Fourier transform of the rectangular pulse of width 2pi and height 1. The inverse filtering is a restoration technique for deconvolution, i. Yaroslavsky, Signal sinc-interpolation: a fast computer algorithm, Bioimaging, 4, p. >> TestEgalite=strcmp('MATLAB', 'matlab') TestEgalite = 0 >> TestEgalite=strcmp('MATLAB', 'MATLAB') TestEgalite = 1. A kardinális szinusz elnevezés Philip M. The real part of a complex number is obtained by real(x) and the imaginary part by imag(x). help inverse. global function ( primary function) MATLAB의 가장 일반적인 함수 정의 방법입니다. If False, values of x can be in any order and they are sorted first. %Ap —amount rippleallowed defaultunits). Interpolate F(ρ,θ) to Fˆ(u,v)(polar to rectangular coordinates – e. From these relations and the properties of exponential multiplication you can painlessly prove all sorts of trigonometric identities that were immensely painful to. Posted by Shannon Hilbert in Digital Signal Processing on 4-22-13. If we have oversampling (supp(𝑓) strictly in (−𝐵,𝐵)), we can choose 𝜒 smooth, hence 𝜒 will be in the Schwartz class unlike the sinc function. The wizard contains 8 different pages: Goal page, Name and Type page, Variables and Parameters page, Function Body page, Parameter Initialization Code page, Bounds and General Linear Constrains page, Derived Parameters page and Script Before or. Recall that the Fourier transform of a sine plate consists of two lines modulated by a “sinc” function along the direction perpendicular to the pitch axis. Also, the Sinc function approaches zero as x goes towards infinity, with the envelope of sinc(x) tapering off as 1/x. Preventing MATLAB from putting a new plot on top of an old one. Filter analysis and implementation. Include your state for easier searchability. Write a Matlab function that uses the DFT (fft) to compute the linear convolution of two sequences that are not necessarily of the same length. Woodwardtól származik 1953-ból. m performs monotonic Steffen interpolation 38) stineman. I have to compute Fourier Transform and Inverse Fourier Transform for a signal and plot its graphs (magnitude and phase). Here I’m going to show you how signals can be generated in MATLAB. You can see that the output from MATLAB is one period of the DTFT, but it's not the period normally plotted, which is from to. DFTMTX Discrete Fourier transform matrix. Important observations that apply to test patterns in general include: (a) the nulls in the sinc2(f) envelope occur at integer multiples of the data rate; (b) spectral lines are evenly spaced at an interval that is the inverse of the pattern length; and (c) the. The backprojection operator for a single projection looks like this: ∫ ∞ −∞ b. From Inverse Problems to Software Design. XFOURIER Graphics demo of Fourier series expansion. Sinc-based methods have been applied to diverse scientiﬁc and engineering problems comprising heat conduction [9], [10], population growth [11], inverse problems. Upload Computers & electronics Software matlab. Backprojection-Filtering Backprojection means that we “smear” the projection data back across the object space. As the sinc filter has infinite impulse response in both positive and negative time directions, it must be approximated for real-world (non-abstract) applications; a windowed sinc filter is often used instead. Use the known F. Remembering the fact that we introduced a factor of i. This page will generally introduce how to use the Fitting Function Builder, a friendly and easily used wizard, to create user-defined fitting functions. Re 1 z Im 1 z Figure 1. The following Matlab project contains the source code and Matlab examples used for modified interpft to do aliased sinc interpolation. Different choices of definitions can be specified using the option FourierParameters. Matlab - Generalità Cosa è Matlab L'ambiente a riga di comando General Purpose Commands demo help info lookfor path type what which clear disp length load pack save sxze who whos Cd delete diary dir getenv urn x Managing Commands and Functions Run demos. Extended Capabilities C/C++ Code Generation Generate C and C++ code using MATLAB® Coder™. 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https://masuday.github.io/fortran_tutorial/control.html	# Flow control February 2020 Back to index.html. # Flow control As previously noted, a Fortran program sequentially executes the code from the beginning through the end. By default, the program does not skip any single statement and does not go back to the previous statements. We are going to see some control statements which change the flow or repeat the same statements many times. ## Conditional ### The if statement #### Simple example You often want to change the flow by the condition. A typical example is to test the input value; you can print an error message if the values are invalid. To make branches by condition, Fortran provides the if statement, as seen in the following example. program cond implicit none real :: x if(x<0.0) then print ,'The value should be 0 or positive.' stop else print ,'square root=',sqrt(x) end if end program cond Note that I put an indent to the if statement to make the structure clearer (and it is a good practice). The stop keyword stops the program immediately. You can see how to use the if statement with this example. • The if statement starts with if (condition) then and ends with end if. The condition is a logical expression which returns .true. or .false.. • The if statement has 2 blocks: the first one between if and else, and the second one between else and end if. Each block is exclusively executed by the condition in the if statement. • If the condition is true, the program proceeds the first block and exits the statement (jumps to the next line of end if). • If the condition is false, the program jumps to the second block and exit the statement. The above if statement has the following structure. if(condition 1) then {block} else {the other block} end if The if statement has some flexible structure. The following code runs in the same way as above (only the kernel of the program is shown). ! Alternative 1 if(x>=0.0) then print ,'square root=',sqrt(x) else print ,'The value should be 0 or positive.' stop end if ! ---------------------------------------------- ! Alternative 2 if(x<0.0) then print ,'The value should be 0 or positive.' stop end if print ,'square root=',sqrt(x) The first alternative has the opposite condition as the original one. The second alternative has no else block; the program stops if the conditional meets, so it does not need the else block. #### General form of if The if statement has the following general form. if(condition 1) then {block 1} else if(condition 2) then {block 2} else if(condition 3) then {block 3} ... else {the other block} end if In the above case, the program can be interpreted as follows. • The program first tests Condition 1, and if true, it executes Block 1 only and exits the statement. • If Condition 1 is not true, then it tests condition 2, and if true, it executes Block 2 and exits the statement. • If Condition 2 is not true, then it tests condition 3, and if true, it executes Block 3 and exits the statement. • You can put more else if statements between the first if and else. • You can optionally put else at the bottom of the statement. The Else block will be executed if all the conditions do not meet. When there is no else block, if all conditions do not meet, the program does nothing and exits the statement. By the way, Fortran accepts endif instead of end if (and elseif vs. else if). Also, spaces are allowed to be between if and ( i.e. if (condition) then. This tutorial uses end if and else if, but it is my preference. You can use your favorite style. #### One-line if When you have only 1 block, i.e. if(...) then ... end if and the block has only 1 statement, you can write the code in 1 line. program cond implicit none real :: val if(val==0.0) stop print ,val end program cond #### Logical expressions You can combine several conditions with .and. and .or.. Recall the rule to make logical expressions. The following program precisely finds the error of input values and gives the user messages about what is wrong. It accepts two positive numbers from the keyboard and calculates a value. program cond implicit none real :: x,y if(x<0 .and. y<=0) then print ,'error: x<0 and y<=0' else if(x<0) then print ,'error: x<0' else if(y<=0) then print ,'error: y<=0' else print ,'sqrt(x)log(y)=',sqrt(x)log(y) end if print ,'done.' end program cond #### Nested if statements The if statements can be nested, which means you can put it in another if statement. You can cascade a conditional test to another. The following example is an alternative version of the previous program. program cond implicit none real :: x,y if(x<0) then print ,'error: x<0' else if(y<=0) then print ,'error: y<=0' else print ,'sqrt(x)log(y)=',sqrt(x)log(y) end if end if print ,'done.' end program cond In this case, the program does not detect the case where both x and y are invalid at the same time. The original code looks flat and checks all possible cases on the same line. The above code shows a clear intent that checks x first and y second. I have to admit that the above example is artificial to show the nested-if. The code is equivalent to the following program with a single if-elseif-else block. if(x<0) then print ,'error: x<0' else if(y<=0) then print ,'error: y<=0' else print ,'sqrt(x)log(y)=',sqrt(x)log(y) end if #### Summary • The if statement starts with if(condition) then and ends with end if. • The condition has a logical expression resulting in a single .true. or .false.. • You can optionally insert else if(condition) as many as possible and one else into the statement. • If the condition meets, the program executes the immediately beneath block and exits the current if statement (moves to the next line of end if). • If the condition does not meet, the program jumps to the next else if and evaluates the condition and decide whether the program takes the block or not. • If all the conditions are false and there is the else block, the program executes it. • The if statement can be nested. #### Exercises • Read 1 integer (code) and 1 real numbers (v) from the keyboard. If the code is 1, compute the square root of v. If the code is 2, compute the sine of v. If the code is the other values, print an error message and stop the program. • Read an integer and show if it is an even or odd number. ### The select statement #### Simple example There is a convenient statement (select) to make branches conditioned by a single value. The select statement is useful to compare between a scalar with a literal (or, a range of literals). Here is a piece of code that shows a typical case where select works well. integer :: x if(x==1) then print ,'case 1' else if(x==2 .or. x==3) then print ,'case 2 or 3' else print ,'failed' end if With the select statement, the code becomes readable and straightforward as follows. integer :: x select case(x) case(1) print ,'case 1' case(2,3) print ,'case 2 or 3' case default print ,'failed' end select The select statement has the following structure. • The statement starts with select case(x) and end select where x is a single variable or a literal (i.e., scalar). • The statement has 1 or more case(y) where y is a literal, a range, or a list of them. y should not be a variable. For example, case(1), case(1,2,5), case(2:3), case(2:3,5) are all valid. • The program starts comparing x with y from the first case block to the last. If x equals to y, the program performs this block, and exit the entire select statement without moving toward the next block. • If x does not match any y, the program executes the optional case default block. If you omit this block, the program does not do anything. Here is a general form of the select statement. select case(scalar) case(literal list 1) {block 1} case(literal list 2) {block 2} ... case default {the other block} end select #### Exercises 1. Confirm that the select statement can accept characters. Declare a character variable (say, character(len=10)::s), and make the branches by the variable (select case(s) or select case(s(1:2))). ## Iteration ### The do statement #### Small example There is a case where you have to repeat the same operation many times. Or, you need to repeat slightly different but almost the same operation by round. The simplest way is to write all the repeated operations as a flat program, but it should be in trouble in its readability and redundancy. Fortran provides a systematic way to repeat the same (or similar) operation with the do statement. For example, consider printing Hello, world! 3 times. It can be written in Fortran as follows. program loop implicit none integer :: i do i=1,3 print ,"Hello, world!" end do end program loop Hello, world! Hello, world! Hello, world! You may need some explanation. • The operations (block) that you want to repeat are placed between do and end do. • The do keyword accompanies with an integer variable, here, i. • The variable i is automatically initialized to 1 (as specified in the do statement) before the loop starts. • In the 1st iteration, i is compared with the last value 3 as i<=3. It is true, and the program executes the block. • When the program reaches end do, it jumps back to do, and increment i by 1. In this case, i becomes 2. • In the 2nd iteration, i is compared with the last value 3 as i<=3. It is true, and the program executes the block. • When the program reaches end do, it jumps back to do, and increment i by 1. In this case, i becomes 3. • In the 3rd iteration, i is compared with the last value 3 as i<=3. It is true, and the program executes the block. • When the program reaches end do, it jumps back to do, and increment i by 1. In this case, i becomes 4. • In the 4th iteration, i is compared with the last value 3 as i<=3. It is false, and the program jumps out of end do. The key to understanding the do loop is as follows. 1. The do statement needs an integer variable (any name; i in this case) as a counter. 2. The do statement has the following form: do variable = starting value, final value. 3. Before the loop, the variable is initialized to the staring value. 4. Whenever the program reaches do, the variable is compared to the final value. If the variable is smaller than or equals to the final value, the focus moves to the next line; Otherwise, the program exits the current loop. 5. Whenever the program reaches end do, the focus goes back to the corresponding do, then the variable is incremented by 1 (and going back to Step 5). Here is a general form of do. do counter_variable = initial_value, last_value {block} end do In this usage, do needs a counter variable. This behavior is like to count up numbers by hand. The counter variable is checked only at the beginning of loops, and it is incremented when going back from end do to do. By the way, this program also works when i=1,3 is replaced with i=0,2 or even i=-5,-3 (please consider why). Non-1 starting value looks unusual but can be useful in future use. #### Use of the counter variable The counter is an integer variable, so the program refers to the value of this variable in each iteration (but can not rewrite the variable because the statement protects it). In the above case (i=1:3), the counter i is 1 in the 1st iteration, 2 in the 2nd, and 3 in the last iteration. After the loop, i becomes 4. The previous example can be slightly changed to show the current iteration number. program loop implicit none integer :: i do i=1,3 print ,"iteration",i,": Hello, world!" end do end program loop iteration 1: Hello, world! iteration 2: Hello, world! iteration 3: Hello, world! This variable can also be used in a numerical formula. The following example computes the multiple of 3. program loop implicit none integer :: i do i=1,9 print ,i," x 3 = ",i3 end do end program loop 1 x 3 = 3 2 x 3 = 6 3 x 3 = 9 4 x 3 = 12 5 x 3 = 15 6 x 3 = 18 7 x 3 = 21 8 x 3 = 24 9 x 3 = 27 Any other variables can be used inside the loop. The following program calculates the factorial of an arbitrary integer read from the keyboard. program loop implicit none integer :: i,n,fact fact = 1 do i=1,n fact = facti end do print ,n,"! = ",fact end program loop # when putting 5: 5 ! = 120 Note that the variable fact for storing the result must be initialized by 1 before the loop. This program does not work if you put a large number (it overflows). The variable is changed to double precision to fix this issue (see exercises). The counter variable should be an integer. If you need a real number that incrementally changes, you have to calculate the value from the counter. The following program calculates the square root of a real number ranged from 0 to 1 by 0.1. program loop implicit none integer :: i real :: v do i=0,10 v = real(i)/10 print ,sqrt(v) end do end program loop #### Break and skip the loop You may break (exit) a loop when a condition is true. The keyword exit immediately exits the do statement by jumping to the next line of end do. It may happen when you repeatedly read a value from the keyboard and stop reading when the input is invalid. program loop implicit none integer :: i real :: val ! read value 10 times but exit the loop if the input is negative. do i=1,10 if(val < 0) exit print ,sqrt(val) end do end program loop If you want to skip the rest of the statements in the loop and start the next iteration, you can use the cycle keyword, which immediately moves the focus to the beginning of do. In the above code, if you replace exit by cycle, it jumps back to the do statement and starts the next round. You should see if this program reads the value precisely 10 times. program loop implicit none integer :: i real :: val ! read value 10 times but skip the round if the input is negative. do i=1,10 if(val < 0) cycle print ,sqrt(val) end do end program loop #### Infinite loops If you don’t know how many iterations are needed, do continues the loop forever. Such a structure is called infinite loops, and it never ends unless the program meets exit in the loop. Making infinite loops, you just drop the iterator and use do alone. program loop implicit none real :: val ! read a value but exit the loop if the input is negative. do if(val < 0) exit print ,sqrt(val) end do end program loop Here is a general from of the infinite loops. The code block should have the exit statement. do {block with exit} end do Again, this unconditional do statement creates the infinite loops, and the program may never stop unless the program reaches exit. The compiler can not detect the infinite loops as a failure. If you run into this issue, hit the Control key (Ctrl) and the C key simultaneously (hit C while pushing Ctrl). #### Writing a loop The loop in a programming language has a unique syntax that is not necessarily clear to the beginners. If you know that you have to write a loop but do not know how to write it, please consider the following method. First, you can think of the result of the loops. For example, you want to get the following output on the screen. number = 3 number = 4 number = 5 Then, you look for a common thing across iterations. • Each output should be from the print statement, and you can find the common message in the output, number =. • The common statement over the loops is print ,"number =". • You can put a variable to stuff that changes by iteration: print ,"number = ",x. Finally, you enclose the common code by the do statement. do print ,"number = ",x end do If the variable systematically changes by iteration, it can be a counter variable. In this case, x ranges from 3 to 5, and here is the code. do x=3,5 print ,"number = ",x end do Or, you may think of the following code. x = 3 do print ,"number = ",x x = x + 1 if(x>=6) exit end do The first loop is more transparent for any people than the latter, while the latter code is very straightforward (but it may look more complicated for some people). Although the first code is preferred, we would need a loop like the other code. There are many cases to generate a sequence in Fortran. The do statement is a typical mechanism to generate the sequence, and it is a solution for the purpose. #### Practical use of loops The loop has 2 different viewpoints in practice. 1. Iteration: repeating the same (or similar) operations a finite number of times. 2. Sequence of numbers: generating a series of numbers. The latter is useful when you manipulate the subscript (index) of an array (e.g., matrix and vector that we are going to see later). When you have to change the index or any numbers systematically, you will use the loops. #### Summary • The do statement defines a loop. The statements between do and the corresponding end do are repeated. • The loop can have an integer variable indicating how many times the loop is repeated. The user can refer to the counter as a variable. • When the program reaches end do, the focus goes back to the corresponding do, then the counter is incremented by 1. • The evaluation of the counter occurs at do, the top of the loop. If the counter is greater than the final value, the program exits the loop, i.e., going to the next line of end do. • The exit statement jumps out of the loop. The cycle statement immediately goes back to do, ignoring the remaining statements. • Without the counter, do makes an infinite loop. The exit statement is the only way to exit the loop. • The loop is useful to repeat a task many times or to generate a sequence of numbers. #### Exercises 1. Rewrite the code to compute a factorial with a larger n using a double-precision variable. 2. Read 10 numbers from the keyboard, then show how many positive numbers were given. 3. Write a fizz buzz program. It prints integer numbers from 1 to n on screen, but when the number is divisible by 3 is replaced by the word fizz, any number divisible by 5 by the word buzz, and the numbers divisible by 15 by the word fizz buzz only (without showing any of fizz and buzz). ### Variation in loop In this subsection, we are going to see a variation of the loop, particularly useful to generate a sequence of numbers. #### Non-executable loop You can use variables as the staring and the last values. In such a case, the loop may not be executed. It is a valid program, and it often appears in some algorithms. program loop implicit none integer :: i,a,b ! not executed when a>b do i=a,b print ,i end do end program loop #### Non-1 increment By default, the counter variable is incremented by 1. The user can change the increment to any non-zero integer as the 3rd parameter in the do statement. For example, the following code shows odd numbers up to 10 (actually 9). program loop implicit none integer :: i ! counter incremented by 2 do i=1,10,2 print ,i end do end program loop 1 3 5 7 9 It is equivalent to write the following code. You can use whatever you prefer. do i=1,5 print ,2i-1 end do Using this technique, the user can define a loop where the counter decrements. program loop implicit none integer :: i ! from 5 to 1 by -1 do i=5,1,-1 print ,i end do end program loop 5 4 3 2 1 #### Summary • If the staring value is already larger than the final value in a loop, such a loop will not be executed. • The increment is defined in the 3rd parameter in do. It should be non zero integers. #### Exercises 1. Print the following sequence of integer numbers on the screen (first 12 numbers of the sequence). 1. $2, 4, 6, 8, 10, 12, 14, \cdots$ 2. $0, 1, 3, 6, 10, 15, 21, \cdots$ 3. $1, -1, 1, -1, 1, -1, \cdots$ 4. $1, 2, 0, 4, 5, 0, 7, 8, 0, \cdots$ 5. $1, 2, 4, 5, 7, 8, \cdots$ 2. Compute $4x$ where $x=1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}+\cdots$ for the first 1 million terms. 3. Factor $2099829353$ into two primes (i.e. find primes $x$ and $y$ so that $xy=2099829353$). ### Nested loops You can define nested loops with multiple dos. Here, for simplicity, we just consider double loops with two do statements. #### Double loops As the if statement, do can also be nested. You can see how the nested loops work in the following example. program loop implicit none integer :: i,j do i=1,3 do j=7,9 print ,i," x ",j,"=",ij end do end do end program loop 1 x 7 = 7 1 x 8 = 8 1 x 9 = 9 2 x 7 = 14 2 x 8 = 16 2 x 9 = 18 3 x 7 = 21 3 x 8 = 24 3 x 9 = 27 This program prints some products. When the program enters the first loop do i=1,3, the initial i is set. Then the program enters the second loop do j=7,8. In the second loop, i is preserved, and j is updated in this loop. When the second loop ends, the focus returns to the top of the first loop, then i is incremented, and the second loop is executed again. When The outmost loop ends, the entire loop block is done. #### exit and cycle With multiple loops, exit and cycle are valid only for the current loop. For example, the following code tries to exit the loop when $i\times j = 16$. program loop implicit none integer :: i,j do i=1,3 do j=7,9 if(ij==16) exit print ,i," x ",j,"=",ij end do end do end program loop The exit statement is active when i=2 and j=8, and it just exits the current (the innermost) loop. So, the output looks like this. 1 x 7 = 7 1 x 8 = 8 1 x 9 = 9 2 x 7 = 14 3 x 7 = 21 3 x 8 = 24 3 x 9 = 27 It surely skips i=2 and j=8. But what if you want to exit the entire loop (the outer loop)? The simplest option is stop, but it stops everything. In Fortran, you can specify do, which you particularly exit by using a label. A label is an arbitrary name stuck to do, and the corresponding end do statements. In the following example, the outer do has a label outer with a colon (:), the corresponding end do has the same label, and exit specifies the label. program loop implicit none integer :: i,j outer: do i=1,3 do j=7,9 if(ij==16) exit outer print ,i," x ",j,"=",i*j end do end do outer end program loop 1 x 7 = 7 1 x 8 = 8 1 x 9 = 9 2 x 7 = 14 The same principle applies to cycle. #### Summary • Nested loops can be defined. • The do (and end do) statement can have a label, which is a sequence of arbitrary characters. • The label is used by exit and cycle` to specify the loop. #### Exercises 1. Show all pairs of integers ($a$ and $b$; $0<a,b<20$) to satisfy that $a/b$ is an integer number. 2. Write a program to find triplets of integer numbers ($a$, $b$, and $c$) to satisfy $a^2+b^2=c^2$ where $0<a,b,c<1000$. Just show the first 10 sets of numbers. Back to index.html.	2021-10-18 02:16:36	{"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": 2, "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": 0.2902459502220154, "perplexity": 1581.5907300294025}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323585186.33/warc/CC-MAIN-20211018000838-20211018030838-00236.warc.gz"}
https://core.ac.uk/display/2193856	Location of Repository ## Universality properties of Gelfand-Tsetlin patterns ### Abstract A standard Gelfand-Tsetlin pattern of depth $n$ is a configuration of particles in $\{1,...,n\} \times \R$. For each $r \in \{1,...,n\}$, $\{r\} \times \R$ is referred to as the $r^\text{th}$ level of the pattern. A standard Gelfand-Tsetlin pattern has exactly $r$ particles on each level $r$, and particles on adjacent levels satisfy an interlacing constraint. Probability distributions on the set of Gelfand-Tsetlin patterns of depth $n$ arise naturally as distributions of eigenvalue minor processes of random Hermitian matrices of size $n$. We consider such probability spaces when the distribution of the matrix is unitarily invariant, prove a determinantal structure for a broad subclass, and calculate the correlation kernel. In particular we consider the case where the eigenvalues of the random matrix are fixed. This corresponds to choosing uniformly from the set of Gelfand-Tsetlin patterns whose $n^\text{th}$ level is fixed at the eigenvalues of the matrix. Fixing $q_n \in \{1,...,n\}$, and letting $n \to \infty$ under the assumption that $\frac{q_n}n \to \a \in (0,1)$ and the empirical distribution of the particles on the $n^\text{th}$ level converges weakly, the asymptotic behaviour of particles on level $q_n$ is relevant to free probability theory. Saddle point analysis is used to identify the set in which these particles behave asymptotically like a determinantal random point field with the Sine kernel.Comment: 32 page Topics: Mathematics - Probability, 60B20 Year: 2011 OAI identifier: oai:arXiv.org:1105.1272	2018-12-12 02:04:42	{"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": 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": 0.8474270701408386, "perplexity": 265.2142550612841}, "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-2018-51/segments/1544376823710.44/warc/CC-MAIN-20181212000955-20181212022455-00103.warc.gz"}
https://www.chemicalforums.com/index.php?topic=96717.0	# Chemical Forums • March 20, 2019, 03:59:12 AM • Welcome, Guest Please login or register. Login with username, password and session length • ### Sponsored links Pages: [1] Go Down ### AuthorTopic: Minimizing microbial growth in circulating water baths (Read 2477 times) !function(d,s,id){var js,fjs=d.getElementsByTagName(s)[0];if(!d.getElementById(id)){js=d.createElement(s);js.id=id;js.src="https://platform.twitter.com/widgets.js";fjs.parentNode.insertBefore(js,fjs);}}(document,"script","twitter-wjs"); (function() {var po = document.createElement("script"); po.type = "text/javascript"; po.async = true;po.src = "https://apis.google.com/js/plusone.js";var s = document.getElementsByTagName("script")[0]; s.parentNode.insertBefore(po, s);})(); 0 Members and 1 Guest are viewing this topic. #### Babcock_Hall • Chemist • Sr. Member • Mole Snacks: +234/-16 • Offline • Posts: 3622 ##### Minimizing microbial growth in circulating water baths « on: September 08, 2018, 06:18:30 AM » I just cleaned what looked like algae out of a circulating bath made by Laura. I am not sure how to minimize their return. I may have used a mixture of ethylene glycol and water in the past. This bath is high enough up on a bench that I don't see how a dog could potentially lap up ethylene glycol (which can happen in other contexts). Any suggestions? Logged #### P • Full Member • Mole Snacks: +63/-15 • Offline • Gender: • Posts: 629 • I am what I am ##### Re: Minimizing microbial growth in circulating water baths « Reply #1 on: September 11, 2018, 12:41:52 AM » I use a water cooler which stores water for my calorimeter. I have to flush it out quite regularly to avoid biological build up. We are supposed to add a biocidal liquid supplied by the manufacturer of the calorimeter - I stopped buying it as they put the price of the biocide up 10 fold. It is about £50 for a few ml now! So, I now use some of the biocide that I use in a paint formulation to add to the water to prevent/retard fungal/biological growth in the water cooler. I guess using biocidal liquids are a no go because of the dog? If it is kept out of his reach I suppose it could be OK. Logged Tonight I’m going to party like it’s on sale for $19.99! - Apu Nahasapeemapetilon #### Babcock_Hall • Chemist • Sr. Member • Mole Snacks: +234/-16 • Offline • Posts: 3622 ##### Re: Minimizing microbial growth in circulating water baths « Reply #2 on: September 11, 2018, 03:02:36 AM » There should not be any dogs around, but one always worries about chance events. I am leaning toward using some antifreeze. Logged #### Arkcon • Retired Staff • Sr. Member • Mole Snacks: +533/-146 • Offline • Posts: 7360 ##### Re: Minimizing microbial growth in circulating water baths « Reply #3 on: September 11, 2018, 04:32:40 AM » Basically, the answer is frequent draining, cleaning and drying. There are many commercial products available, and some good ones on your own, solvents and the like. But in the end, the inhibitor gets consumed, dust accumulates, resistant organisms grow, the water evaporates and gets topped of, overflows get drained, then with refilling, the solution gets dilute again. And you have to drain, wash, and dry. Logged Hey, I'm not judging. I just like to shoot straight. I'm a man of science. #### P • Full Member • Mole Snacks: +63/-15 • Offline • Gender: • Posts: 629 • I am what I am ##### Re: Minimizing microbial growth in circulating water baths « Reply #4 on: September 11, 2018, 05:31:39 AM » nano particulate silver or copper particles would probably work also. Logged Tonight I’m going to party like it’s on sale for$19.99! - Apu Nahasapeemapetilon #### wildfyr • Global Moderator • Sr. Member • Mole Snacks: +130/-9 • Offline • Posts: 1151 ##### Re: Minimizing microbial growth in circulating water baths « Reply #5 on: September 11, 2018, 08:22:43 AM » Antifreeze seems best and simplest to me. If this is a lab how could a dog get to it? A bit a bleach works too if its all plastic. Logged #### Babcock_Hall • Chemist • Sr. Member • Mole Snacks: +234/-16 • Offline • Posts: 3622 ##### Re: Minimizing microbial growth in circulating water baths « Reply #6 on: September 11, 2018, 08:32:44 AM » Dogs should not get in, but sometimes the unexpected happens. Fortunately, the bath is on a bench top; therefore, doubt that this is a serious concern. Logged #### P • Full Member • Mole Snacks: +63/-15 • Offline • Gender: • Posts: 629 • I am what I am ##### Re: Minimizing microbial growth in circulating water baths « Reply #7 on: September 11, 2018, 10:19:50 PM » A bit a bleach works too if its all plastic. I'd thought about bleach for my water cooler, but I was unsure how it would get on in the calorimeter so I stuck with the biocide. I guess a recalibration maybe needed due to the change in specific heat capacity of the bleach/water system compared to just water, but other than that I guess it would be fine. The difference might even be negligible idk. Logged Tonight I’m going to party like it’s on sale for \$19.99! - Apu Nahasapeemapetilon Pages: [1] Go Up Mitch Andre Garcia's Chemical Forums 2003-Present. 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http://math.stackexchange.com/questions/250427/3d-objects-in-2d-drawing-how-to-get-the-size-of-a-box-relative-to-a-known-plane?answertab=oldest	# 3D objects in 2D drawing: How to get the size of a box relative to a known plane? Sorry if this is a really badly worded question. Say you have a box of unknown size, and a planar object of a known size (say, a credit card). You arrange these object somehow (probably with the card aligned with two of the boxes axes) and take a picture of them. Knowing the exact size of the card, is there any maths I could use to calculate the exact size of the box? Would I have to move the card and take another picture to get the size along the 3rd axis? - A photo is usually a projection, the conserved quantity in a projection is the cross ratio of four points on a line. So, for calculation purposes, you want to already know that your credit card shares lines with your box. This approach indeed necessitates more than one picture. I will first assume that you know the mid-points of the side of your credit card and that the side of the credit card coincide with two sides of the box (that is, the credit card is places directly in a corner). We will just look at one side of the box and the credit card now, the other calculation is analogous: Now, measure in your photo $a=$ length from corner to mid-point of credit card side, $b=$ length from mid-point of credit card side to end of credit card side, $c=$ length from corner to other corner of unknown box and $d=$end of credit card side to other corner of unknown box. Denote $l$ the length of this side of the credit card and $x$ the unknown length of the side of the box. You have: $$\frac{a}{b}\frac{d}{c}=\frac{x-l}{x}.$$ So, $$x= \frac{lbc}{bc- ad}.$$ How you can recover a mid-point of a credit card side (or more precisely, $a$ and $b$ as above): Two points, their mid-point and the point at infinity of that line have cross-ratio -1. The two parallel sides of the credit card intersect at the point at infinity. Therefore, with $a,b$ as above, $e$ the length of the credit card in the picture and $m$ the distance of the beginning of the credit card side to the intersection of the credit card side with its "parallel" one in the picture, you get: $$\frac a{e-a} \frac {m-e}{m} = -1$$ which permits you to calculate $a$. - This is awesome, thanks :D. Out of curiousity what do you mean by "This can be recovered as a a harmonic point in the photo"?, I don't think there's any way to have the midpoint of the card available so I'll probably need to calculate it. – J.Ashworth Dec 4 '12 at 22:57 Also will this approach work for a photo from any angle? As long as the edge of the card is exactly parallel with the edge of the box? – J.Ashworth Dec 4 '12 at 22:58 @J.Ashworth I have added some details. My calculation if for the case that the edge of the card is placed exactly on the edge of the box. If it is just parallel, you would first have to shift the card in the picture by connecting it to the respective points at infinity (each family of originally parallel line intersects in one point which permits to construct parallel lines easily). – Phira Dec 4 '12 at 23:29 Thanks :), one more question. Would there be any way to get all 3 dimensions of the box from a single picture? Say, if the camera had a "Y" shape reference over it that you had to line the box up with so you knew the exact perspective? – J.Ashworth Dec 5 '12 at 0:08 If the card is perpendicular to the axis (parallel to the sensor plane), so is the near surface of the box, and they are at the same distance from the camera, the sizes are in the same ratio as the images. So if the near surface of the box looks three times as wide and twice as tall as the card, it is. If the box is aligned this way on axis, the rear surface is hidden and you can tell nothing about the depth. If you can see one of the back sides, the ratio of the image sizes of the edge is the inverse of the ratio of distance from the lens. So if the rear edge is $\frac 45$ as tall as the front edge, it is $\frac 54$ as far from the lens. With today's lenses, finding the optical center of the lens is not easy, so the calculation will have some error. -	2014-12-19 07:37:33	{"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": 0.5403041243553162, "perplexity": 342.82824597044674}, "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-2014-52/segments/1418802768285.15/warc/CC-MAIN-20141217075248-00083-ip-10-231-17-201.ec2.internal.warc.gz"}
https://jira.lsstcorp.org/browse/DM-9849?actionOrder=desc	# stack demo failing on master XMLWordPrintable #### Details • Type: Story • Status: Invalid • Resolution: Done • Fix Version/s: None • Component/s: None • Labels: None • Team: SQuaRE #### Description I have seen two build failures in row from master on the jenkins lsst-dev node while running the stack demo. Failed (4238 of 4238 flags do not match) in column flags_negative. astrometry_net_data bin detected-sources_small.txt expected input output_small README Warning: output results not within error tolerance for: lsst_dm_stack_demo-master *** The simple integration demo completed with some statistical deviation in the output comparison. #### Attachments 1. detected-sources_small.txt 1.18 MB 2. expected-detected-sources_small.txt 1.17 MB #### Activity Hide Joshua Hoblitt added a comment - This was indeed a case of running the demo from master against a build of 13.0. Show Joshua Hoblitt added a comment - This was indeed a case of running the demo from master against a build of 13.0 . Hide Tim Jenness added a comment - The attached file is using a False for the flags_negative column which is what you would expect before DM-903 was merged and this failure would be what happens if you build a v13.0 stack and use the master demo. Show Tim Jenness added a comment - The attached file is using a False for the flags_negative column which is what you would expect before DM-903 was merged and this failure would be what happens if you build a v13.0 stack and use the master demo. Hide Tim Jenness added a comment - If every single row is failing because of a change in one string column that usually means a flag became a number that was a boolean or the reverse. Usually seeing the real change in that column makes the problem obvious. Show Tim Jenness added a comment - If every single row is failing because of a change in one string column that usually means a flag became a number that was a boolean or the reverse. Usually seeing the real change in that column makes the problem obvious. #### People Assignee: Joshua Hoblitt Reporter: Joshua Hoblitt Watchers: Joshua Hoblitt, Kian-Tat Lim, Tim Jenness Votes: 0 Vote for this issue Watchers: 3 Start watching this issue #### Dates Created: Updated: Resolved:	2021-01-27 03:56:31	{"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": 0.27075302600860596, "perplexity": 5999.0199124370565}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "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-2021-04/segments/1610704820894.84/warc/CC-MAIN-20210127024104-20210127054104-00635.warc.gz"}
https://en.wikipedia.org/wiki/Line_chart	# Line chart This simple graph shows data over intervals with connected points A line chart or line graph is a type of chart which displays information as a series of data points called 'markers' connected by straight line segments.[1] It is a basic type of chart common in many fields. It is similar to a scatter plot except that the measurement points are ordered (typically by their x-axis value) and joined with straight line segments. A line chart is often used to visualize a trend in data over intervals of time – a time series – thus the line is often drawn chronologically. In these cases they are known as run charts.[2] ## History Some of the earliest known line charts are generally credited to Francis Hauksbee, Nicolaus Samuel Cruquius, Johann Heinrich Lambert and William Playfair.[3] ## Example In the experimental sciences, data collected from experiments are often visualized by a graph. For example, if one were to collect data on the speed of a body at certain points in time, one could visualize the data by a data table such as the following: Graph of Speed Vs Time Elapsed Time (s) Speed (m s−1) 0 0 1 3 2 7 3 12 4 20 5 30 6 45.6 The table "visualization" is a great way of displaying exact values, but can be a poor way to understand the underlying patterns that those values represent. Because of these qualities, the table display is often erroneously conflated with the data itself; whereas it is just another visualization of the data. Understanding the process described by the data in the table is aided by producing a graph or line chart of Speed versus Time. Such a visualisation appears in the figure to the right. Mathematically, if we denote time by the variable ${\displaystyle t}$, and speed by ${\displaystyle v}$, then the function plotted in the graph would be denoted ${\displaystyle v(t)}$ indicating that ${\displaystyle v}$ (the dependent variable) is a function of ${\displaystyle t}$. ## Best-fit A parody line graph published in 1919. Charts often include an overlaid mathematical function depicting the best-fit trend of the scattered data. This layer is referred to as a best-fit layer and the graph containing this layer is often referred to as a line graph. It is simple to construct a "best-fit" layer consisting of a set of line segments connecting adjacent data points; however, such a "best-fit" is usually not an ideal representation of the trend of the underlying scatter data for the following reasons: 1. It is highly improbable that the discontinuities in the slope of the best-fit would correspond exactly with the positions of the measurement values. 2. It is highly unlikely that the experimental error in the data is negligible, yet the curve falls exactly through each of the data points. In either case, the best-fit layer can reveal trends in the data. Further, measurements such as the gradient or the area under the curve can be made visually, leading to more conclusions or results from the data. A true best-fit layer should depict a continuous mathematical function whose parameters are determined by using a suitable error-minimization scheme, which appropriately weights the error in the data values. Such curve fitting functionality is often found in graphing software or spreadsheets. Best-fit curves may vary from simple linear equations to more complex quadratic, polynomial, exponential, and periodic curves.[4]	2016-07-28 19:29:23	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 5, "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": 0.6948485970497131, "perplexity": 621.3798054148158}, "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-2016-30/segments/1469257828313.74/warc/CC-MAIN-20160723071028-00132-ip-10-185-27-174.ec2.internal.warc.gz"}