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http://tex.stackexchange.com/tags/tex4ht/new | # Tag Info
## New answers tagged tex4ht
1
You can configure appearance with css instructions in the custom config file. Try this, name the file myconf.cfg: \Preamble{xhtml} \Css{div.caption{text-align:center;}} \begin{document} \EndPreamble Important line is \Css{div.caption ...} whis is css instruction to center the text, div.caption is element generated by tex4ht for \caption commands. Now ...
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This works correctly for me with TeX Live 2012 on Linux. \documentclass{book} \usepackage{graphicx} \begin{document} Lorem ipsum dolor sit amet, consectetur adipisici elit: \begin{centering} \includegraphics[width=.7\textwidth]{myimage.eps}\\ \includegraphics[width=.3\textwidth]{myimage.eps}\\ \includegraphics[width=1.0\textwidth]{myimage.eps}\\ ...
1
With the following solution (slightly modified from Skarab's one) you don't need the intermediate step of pdf->png conversion "by hand": it is performed automatically (you still need imagemagick installed in your system though). It works for me with TeXLive under Ubuntu: \documentclass{article} \usepackage{tikz,graphicx} \usetikzlibrary{external} \tikzset{ ...
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You are right, chapter and sectioning commands are redefined by tex4ht. Chapter styles are not used at all, but they couldn't work anyway, because tex4ht doesn't know how to translate TeX layout to CSS needed in web pages. It really isn't easy task. So you have to style generated elements yourself. There is the \Css command for providing CSS declarations. ...
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I found a solution to my problem (which was export to odt with biblatex references) with pandoc : pandoc test.tex -o output.html --bibliography /my/bibliography.bib -s Hope It could help another one. P-S : Tex4ht was definitively not rendering biblatex for me even after upgrade.
Top 50 recent answers are included | 2013-12-18 20:07:09 | {"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": 0.8010908961296082, "perplexity": 14738.98578867935}, "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-2013-48/segments/1387345759442/warc/CC-MAIN-20131218054919-00099-ip-10-33-133-15.ec2.internal.warc.gz"} |
https://www.transtutors.com/questions/product-cost-1-20-per-gallon-order-cost-35-per-order-annual-holding-cost-0-42-per-ga-1394096.htm | # Product Cost $1.20 per gallon Order Cost$35 per order Annual Holding Cost $0.42 per gallon Annual.. 1 answer below » Product Cost$1.20 per gallon Order Cost $35 per order Annual Holding Cost$0.42 per gallon Annual Demand = 250 gallons/week * 52 weeks 13000 gallons
Number of Orders = 13,000/ 2,500 5.2 Annual Cost of Holding and...
## Recent Questions in Cost Management
Product Cost $1.20 per gallon Order Cost$35 per order Annual Holding Cost \$0.42 per gallon Annual.. | 2018-07-20 12:49:18 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.18500621616840363, "perplexity": 11814.22437417204}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-30/segments/1531676591596.64/warc/CC-MAIN-20180720115631-20180720135631-00542.warc.gz"} |
https://www.zbmath.org/?q=an%3A1101.91031 | # zbMATH — the first resource for mathematics
Consistent variance curve models. (English) Zbl 1101.91031
The author considers variance swaps as liquid derivatives and derives conditions such that the joint market of stock price and variance swap prices is free of arbitrage. A general approach to model such market in an HJM-type framework is considered. Then the finite-dimensional Markovian representations of the fixed time-to-maturity forward variance swap curve are introduced, and consistency results for both the standard case and for variance curves with values in a Hilbert space are derived. It is assumed that the realized variance paid by a variance swap is the realized quadratic variation of the logarithm of the index price. It is also assumed that the stock price process is continuous. The starting point is to assume that variance swaps are liquidly traded for all maturities. The aim of this paper is to first model the variance swap prices and then find an associated stock price process with compatible dynamics. A few examples of variance curve functionals is given and completeness and hedging in such models is discussed.
##### MSC:
91B28 Finance etc. (MSC2000) 91B26 Auctions, bargaining, bidding and selling, and other market models
Full Text:
##### References:
[1] Björk T., Christensen B.J. (1999) Interest rate dynamics and consistent forward curves. Math Finance 9, 323–348 · Zbl 0980.91030 · doi:10.1111/1467-9965.00072 [2] Björk T., Svensson L. (2001) On the existence of finite dimensional realizations for nonlinear forward rate models. Math Finance 11, 205–243 · Zbl 1055.91017 · doi:10.1111/1467-9965.00113 [3] Brace, A., Goldys, B., Klebaner, F., Womersley, R. A market model of stochastic implied volatility with application to the BGM model. WP 2001, http://www.maths.unsw.edu.au/$$\sim$$rsw/Finance/svol.pdf [4] Buehler, H. Expensive martingales. Quant Finance (2006, in press) · Zbl 1136.91417 [5] Cont R., da Fonseca J., Durrleman V. (2002) Stochastic models of implied volatility surfaces. Econ Notes 31, 361–377 · doi:10.1111/1468-0300.00090 [6] Demeterfi, K., Derman, E., Kamal, M., Zou, J. More than you ever wanted to know about volatility swaps. Quantitative Strategies Research Notes, Goldman Sachs (1999) [7] Duffie D., Pan J., Singleton K. (2000) Transform analysis and asset pricing for affine jump-diffusions. Econometrica 68, 1343–1376 · Zbl 1055.91524 · doi:10.1111/1468-0262.00164 [8] Dupire, B. Arbitrage pricing with stochastic volatility. In: Peter Carr (ed.) Derivatives pricing, Risk Books, pp. 197–215 (2004) [9] Fengler, M., Härdle, W., Mammen, E. Implied volatility string dynamics. CASE Discussion Paper Series No. 2003-54 (2003) [10] Filipović D. (2001) Consistency problems for Heath–Jarrow–Morton interest rate models. Lecture Notes in Mathematics, vol. 1760. Springer, Berlin Heidelberg New York · Zbl 1008.91038 [11] Filipović D., Teichmann J. (2004) On the geometry of the term structure of interest rates. Proc R Soc Lond A 460, 129–167 · Zbl 1048.60045 · doi:10.1098/rspa.2003.1238 [12] Föllmer H., Schied A. Stochastic finance. 2nd edn. de Gruyter (2004) · Zbl 1126.91028 [13] Fouque J-P., Papanicolaou G., Sircar K. (2000) Derivatives in financial markets with stochastic volatility. Cambridge University Press, Cambridge · Zbl 0954.91025 [14] Gatheral, J. Case studies in financial modelling. Course notes, Courant Institute of Mathematical Sciences, Fall (2004), http://www.math.nyu.edu/fellows_fin_math/gatheral/case_studies.html [15] Haffner R. (2004) Stochastic implied volatility. Springer, Berlin Heidelberg New York [16] Heath D., Jarrow R., Morton A. (1992) Bond pricing and the term structure of interest rates: a new methodology for contingent claims valuation. Econometrica 60, 77–106 · Zbl 0751.90009 · doi:10.2307/2951677 [17] Heston S. (1993) A closed-form solution for options with stochastic volatility with applications to bond and currency options. Rev Financ Stud 6, 327–343 · Zbl 1384.35131 · doi:10.1093/rfs/6.2.327 [18] Musiela M. Stochastic PDEs and term structure models. Journées Internationales de France, IGR-AFFI, La Baule (1993) [19] Neuberger A. (1990) Volatility trading. London Business School WP, London [20] Overhaus M., Bermudez A., Buehler H., Ferraris A., Jordinson C., Lamnouar A. (2006) Equity hybrid derivatives. Wiley, New York in press [21] da Prato G., Zabczyk J. (1992) Stochastic equations in infinite dimensions. Cambridge University Press, Cambridge · Zbl 0761.60052 [22] Protter P. (2004) Stochastic integration and differential equations. 2nd edn. Springer, Berlin Heidelberg New York · Zbl 1041.60005 [23] Revuz D., Yor M. (1999) Continuous martingales and Brownian motion. 3rd edn. Springer, Berlin Heidelberg New York · Zbl 0917.60006 [24] Schönbucher P.J. (1999) A market model for stochastic implied volatility. Phil Trans R Soc A 357, 2071–2092 · Zbl 0963.91046 · doi:10.1098/rsta.1999.0418 [25] Schweizer, M., Wissel, J. On the term structure of implied volatilities. NCCR FINRISK working paper No. 271, ETH Zurich, http://www.nccr-finrisk.unizh.ch/wp/index.php?action=query&id=271 [26] Teichmann, J. Stochastic evolution equations in infinite dimension with applications to term structure problems. Lecture notes, (2005) http://www.fam.tuwien.ac.at/$$\sim$$jteichma/
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# π-electron S = ½ quantum spin-liquid state in an ionic polyaromatic hydrocarbon
## Abstract
Molecular solids with cooperative electronic properties based purely on π electrons from carbon atoms offer a fertile ground in the search for exotic states of matter, including unconventional superconductivity and quantum magnetism. The field was ignited by reports of high-temperature superconductivity in materials obtained by the reaction of alkali metals with polyaromatic hydrocarbons, such as phenanthrene and picene, but the composition and structure of any compound in this family remained unknown. Here we isolate the binary caesium salts of phenanthrene, Cs(C14H10) and Cs2(C14H10), to show that they are multiorbital strongly correlated Mott insulators. Whereas Cs2(C14H10) is diamagnetic because of orbital polarization, Cs(C14H10) is a Heisenberg antiferromagnet with a gapped spin-liquid state that emerges from the coupled highly frustrated Δ-chain magnetic topology of the alternating-exchange spiral tubes of S = ½ (C14H10)•− radical anions. The absence of long-range magnetic order down to 1.8 K (T/J ≈ 0.02; J is the dominant exchange constant) renders the compound an excellent candidate for a spin-½ quantum-spin liquid (QSL) that arises purely from carbon π electrons.
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## Acknowledgements
This work was sponsored by the World Premier International (WPI) Research Center Initiative for Atoms, Molecules and Materials, Ministry of Education, Culture, Sports, Science and Technology of Japan. We acknowledge financial support from the Japan Science and Technology Agency under the ERATO Isobe Degenerate π-Integration Project, the Mitsubishi Foundation, the Japan Society for the Promotion of Science (JSPS) under the Scientific Research on Innovative Areas ‘J-Physics’ Project (No. 15H05882), the European Union/JST SICORP-LEMSUPER FP7-NMP-2011-EU-Japan project (contract no. NMP3-SL-2011-283214), the UK Engineering and Physical Sciences Research Council (grant nos EP/K027255 and EP/K027212) and the Slovenian Research Agency (grant no. N1-0052). We thank the ESRF for access to synchrotron X-ray facilities, the Royal Society for a Newton International Fellowship (G.K.) and a Research Professorship (M.J.R.), A. N. Fitch for help with the synchrotron XRD experiments, H. Okazaki with the magnetic measurements and K. Kamarás with the infrared measurements.
## Author information
Authors
### Contributions
K.P. and M.J.R. conceived and designed the project. K.P. directed and coordinated the research. Y.T. and M.M. interpreted and discussed all the results and carried out the final structural and magnetic work. H.T., N.T., T.K., Y.N. and R.A. carried out the calculations. A.Š. synthesized the materials. D.A. carried out the EPR spectroscopy and G.K. the vibrational spectroscopic work. A.J.C.B. and A.Š. carried out early structural and magnetic work. K.P. wrote the paper with input from all the authors.
### Corresponding authors
Correspondence to Matthew J. Rosseinsky or Kosmas Prassides.
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### Competing interests
The authors declare no competing financial interests.
## Supplementary information
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Takabayashi, Y., Menelaou, M., Tamura, H. et al. π-electron S = ½ quantum spin-liquid state in an ionic polyaromatic hydrocarbon. Nature Chem 9, 635–643 (2017). https://doi.org/10.1038/nchem.2764
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• Published:
• Issue Date:
• DOI: https://doi.org/10.1038/nchem.2764
• ### Redox-controlled potassium intercalation into two polyaromatic hydrocarbon solids
• F. Denis Romero
• M. J. Pitcher
• M. J. Rosseinsky
Nature Chemistry (2017)
• ### Synthesis successes
• Roser Valentí
• Stephen M. Winter
Nature Chemistry (2017) | 2023-03-25 11:29:32 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8195741176605225, "perplexity": 12472.938406258612}, "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/1679296945323.37/warc/CC-MAIN-20230325095252-20230325125252-00328.warc.gz"} |
https://en.wikipedia.org/wiki/Focused_proof | # Focused proof
Jump to navigation Jump to search
In mathematical logic, focused proofs are a family of analytic proofs that arise through goal-directed proof-search, and are a topic of study in structural proof theory and reductive logic. They form the most general definition of goal-directed proof-search—in which someone chooses a formula and performs hereditary reductions until the result meets some condition. The extremal case where reduction only terminates when axioms are reached forms the sub-family of uniform proofs.[1]
A sequent calculus is said to have the focusing property when focused proofs are complete for some terminating condition. For System LK, System LJ, and System LL, uniform proofs are focused proofs where all the atoms are assigned negative polarity.[2] Many other sequent calculi has been shown to have the focusing property, notably the nested sequent calculi of both the classical and intuitionistic variants of the modal logics in the S5 cube.[3][4]
## Uniform proofs
In the sequent calculus for an intuitionistic logic, the uniform proofs can be characterised as those in which the upward reading performs all right rules before the left rules. Typically, uniform proofs are not complete for the logic i.e., not all sequents or formulas admit a uniform proof, so one considers fragments where they are complete e.g., the hereditary Harrop fragment of Intuitionistic logic. Due to the deterministic behaviour, uniform proof-search has been used as the control mechanism defining the programming language paradigm of logic programming.[1] Occasionally, uniform proof-search is implemented in a variant of the sequent calculus for the given logic where context management is automatic thereby increasing the fragment for which one can define a logic programming langue.[5]
## Focused proofs
The focusing principle was originally classified through the disambiguation between synchronous and asynchronous connective in Linear Logic i.e., connectives that interact with the context and those that do not, as consequence of research on logic programming. They are now an increasingly important example of control in reductive logic, and can drastically improve proof-search procedures in industry. The essential idea of focusing is to identify and coalesce the non-deterministic choices in a proof, so that a proof can be seen as an alternation of negative phases ( where invertible rules are applied eagerly), and positive phases (where applications of the other rules are confined and controlled).[3]
### Polarisation
According to the rules in the sequent calculus, formulas are canonically put into one of two classes called positive and negative e.g., in LK and LJ the formula ${\displaystyle \phi \lor \psi }$ is positive. The only freedom is over atoms are assigned a polarity freely. For negative formulas provability is invariant under the application of a right rule; and, dually, for a positive formulas provability is invariant under the application of a left rule. In either case one can safely apply rules in any order to hereditary sub-formulas of the same polarity.
In the case of a right rule applied to a positive formula, or a left rule applied to a negative formula, one may result in invalid sequents e.g., in LK and LJ there is no proof of the sequent ${\displaystyle B\lor A\implies A\lor B}$ beginning with a right rule. A calculus admits the focusing principle if when an original reduct was provable then the hereditary reducts of the same polarity are also provable. That is, one can commit to focusing on decomposing a formula and its sub-formulas of the same polarity without loss of completeness.
### Focused system
A sequent calculus is often shown to have the focusing property by working in a related calculus where polarity explicitly controls which rules apply. Proofs in such systems are in focused, unfocused, or neutral phases, where the first two are characterised by hereditary decomposition; and the latter by forcing a choice of focus. One of the most important operational behaviours a procedure can undergo is backtracking i.e., returning to an earlier stage in the computation where a choice was made. In focused systems for classical and Intuitionistic logic, the use of backtracking can be simulated by pseudo-contraction.
Let ${\displaystyle \uparrow }$ and ${\displaystyle \downarrow }$ denote change of polarity, the former making a formula negative, and the latter positive; and call a formula with an arrow neutral. Recall that ${\displaystyle \lor }$ is positive, and consider the neutral polarized sequent ${\displaystyle \downarrow \uparrow \phi \lor \psi \implies \uparrow \phi \lor \psi }$, which is interpreted as the actual sequent ${\displaystyle \phi \lor \psi \implies \phi \lor \psi }$. For neutral sequents such as this, the focused system forces on to make an explicit choice of which formula to focus on, denoted by ${\displaystyle \langle \,\rangle }$. To perform a proof-search the best thing is to choose the left formula, since ${\displaystyle \lor }$ is positive, indeed (as discussed above) in some cases there are no proofs where the focus is on the right formula. To overcome this, some focused calculi create a backtracking point such that focusing on the right yields ${\displaystyle \downarrow \uparrow \phi \lor \psi \implies \langle \phi \lor \psi \rangle ,\uparrow \phi \lor \psi }$, which is still as ${\displaystyle \phi \lor \psi \implies \phi \lor \psi }$. The second formula on the right can be removed only when the focused phase has finished, but if proof-search gets stuck before this happens the sequent may remove the focused component thereby returning to the choice e.g., ${\displaystyle \downarrow \uparrow B\lor A\implies \langle A\rangle ,\uparrow A\lor B}$ must be taken to ${\displaystyle \downarrow \uparrow B\lor A\implies \uparrow A\lor B}$ as no other reductive inference can be made. This is a pseudo-contraction since it has the syntactic form of a contraction on the right, but the actual formula doesn't exist i.e., in the interpretation of the proof in the focused system the sequent has only one formula on the right.
## References
1. ^ a b Miller, Dale; Nadathur, Gopalan; Pfenning, Frank; Scedrov, Andre (1991-03-14). "Uniform proofs as a foundation for logic programming". Annals of Pure and Applied Logic. 51 (1): 125–157. doi:10.1016/0168-0072(91)90068-W. ISSN 0168-0072.
2. ^ Liang, Chuck; Miller, Dale (2009-11-01). "Focusing and polarization in linear, intuitionistic, and classical logics". Theoretical Computer Science. Abstract Interpretation and Logic Programming: In honor of professor Giorgio Levi. 410 (46): 4747–4768. doi:10.1016/j.tcs.2009.07.041. ISSN 0304-3975.
3. ^ a b Chaudhuri, Kaustuv; Marin, Sonia; Straßburger, Lutz (2016), Jacobs, Bart; Löding, Christof (eds.), "Focused and Synthetic Nested Sequents", Foundations of Software Science and Computation Structures, Berlin, Heidelberg: Springer Berlin Heidelberg, 9634, pp. 390–407, doi:10.1007/978-3-662-49630-5_23, ISBN 978-3-662-49629-9
4. ^ Chaudhuri, Kaustuv; Marin, Sonia; Straßburger, Lutz (2016). "Modular Focused Proof Systems for Intuitionistic Modal Logics". Marc Herbstritt: 18 pages. doi:10.4230/LIPICS.FSCD.2016.16. Cite journal requires |journal= (help)
5. ^ Armelín, Pablo A.; Pym, David J. (2001), "Bunched Logic Programming", Automated Reasoning, Berlin, Heidelberg: Springer Berlin Heidelberg, pp. 289–304, doi:10.1007/3-540-45744-5_21, ISBN 978-3-540-42254-9 | 2021-09-19 21:26:05 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 13, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 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": 0.8699812293052673, "perplexity": 1047.5148546565044}, "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-2021-39/segments/1631780056900.32/warc/CC-MAIN-20210919190128-20210919220128-00420.warc.gz"} |
http://new.pmean.com/pipes-in-r-2/ | # PMean: Another example of pipes in R
## 2016-12-29
I am using pipes in R (the magrittr package) a lot recently. It reduces the number of errors due to nested functions, among other things. I’ve given a simple example before, and here’s another.
I am calculating the cumulative distance from a vector of longitude and latitude values. The last statement looked like this:
return(round(cumsum(c(0,sqrt(x^2+y^2)))))
Notice this large number of right parentheses. Using pipes, the command would look like
c(0,sqrt(x^2+y^2)) %>% cumsum %>% round %>% return
There are still a few parentheses, but it is much more manageable. You can also change the code easily. For example, suppose you wanted to round to the first decimal rather than the whole number. In the nested functions above, you’d have to figure out where among the various parentheses you would add the “,1<U+2033>. It ends up looking like
return(round(cumsum(c(0,sqrt(x^2+y^2))),1))
but it would be very easy to place this one parentheses off and then R things that you are adding a “,1<U+2033> to the cumsum function. Not nice.
This is the change you’d make using pipes.
c(0,sqrt(x^2+y^2)) %>% cumsum %>% round(1) %>% return
Much less chance of making an error. | 2021-10-18 16:41:15 | {"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": 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": 0.28410789370536804, "perplexity": 1244.9403118973848}, "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/1634323585204.68/warc/CC-MAIN-20211018155442-20211018185442-00034.warc.gz"} |
https://codereview.stackexchange.com/questions/91750/single-digit-math-quiz | # Single Digit Math Quiz
I decided to write a very simple web-game:
function write_at(x, y, text, color, font) {
ctx.fillStyle = color || "black";
ctx.font = font || "50px Arial";
ctx.fillText(text, x + 50, y + 50);
}
function write_at_random(text, color, font) {
write_at(random_choice(range(0, canvas.width - 250)),
random_choice(range(200, canvas.height - 250)),
text, color, font);
}
function random_choice(array) {
index = Math.random() * array.length
return array[Math.floor(index)]
}
function range(start, end) {
return Array.apply(0, Array(end))
.map(function(element, index) {
return index + start;
});
}
function gen_math_expression() {
start = random_choice(range(0, 20));
operator = random_choice(['+', '-', '*']);
end = random_choice(range(0, 20));
return start + operator + end;
}
function single_digit(expr) {
return eval(expr) <= 9 && eval(expr) >= 1
}
function gen_single_digit_expression() {
expr = gen_math_expression();
while (!single_digit(expr)) {
expr = gen_math_expression();
}
return expr;
}
function number_from_keycode(keycode) {
return keycode.which - 48;
}
function draw_welcome_and_score() {
write_at(0, 0, "Single Digit Math Quiz");
write_at(600, 0, points);
}
function lose_sound() {
var snd = new Audio("http://www.soundjay.com/misc/sounds/fail-buzzer-02.mp3");
snd.play();
}
function win_sound() {
var snd = new Audio("http://www.soundjay.com/misc/bell-ringing-05.mp3");
snd.play();
}
function main() {
canvas = document.getElementById("canvas")
ctx = canvas.getContext('2d');
points = 0
draw_welcome_and_score()
write_at(0, 100, "Press the result of the operation on your keyboard.", 0, "20px Arial");
expr = gen_single_digit_expression()
write_at_random(expr);
function check_expr_and_go_next(e) {
if (number_from_keycode(e) == eval(expr)) {
points++;
color = '#7CFC00'; // light green
win_sound()
} else {
points--;
color = 'red';
lose_sound()
}
ctx.clearRect(0, 0, canvas.width, canvas.height)
canvas.style.background = color;
draw_welcome_and_score()
expr = gen_single_digit_expression()
write_at_random(expr);
}
}
main()
<html>
<canvas id="canvas" width="800" height="600" style="border:1px solid #000000;">
</canvas>
</html>
• That's pretty neat! – Phrancis May 25 '15 at 21:32
• Why do the questions appear in random places on the screen? – 200_success May 25 '15 at 21:36
• @200_success to engage the younger ones, that will be happy seeing numbers all over the place instead of only one place. – Caridorc May 25 '15 at 21:49
• The method you use to pick the expressions results in most being subtraction. You might want to consider picking the first number and the operator at random then selecting the second number from the range that produces results under 10. This would give you more control over the proportion of each type of expression being generated. – Bob May 27 '15 at 6:42
I don't see a var anywhere in your code. That means that all of your variables are global variables.
Constructing an array of integers just to pick a random element seems wasteful. The following code scales better, and in my opinion, is easier to follow because it avoids the unnecessary intermediate array.
Asking whether an expression is single digit is a bit odd. I would have the caller evaluate the expression first — which also has the advantage of evaluating it just once.
I'm not convinced that a canvas is called for. Since it's all just text, the whole game can be implemented using the usual DOM elements. The styling should then be done using CSS.
The code supports the number keys along the top row of the keyboard, but it should support the numeric keypad as well.
/* Picks a random integer n such that min ≤ n < max. */
function randomInt(min, max) {
return min + Math.floor((max - min) * Math.random());
}
function sample(candidates) {
return candidates[randomInt(0, candidates.length)];
}
function genMathExpression() {
return randomInt(0, 20) + sample('+-*') + randomInt(0, 20);
}
function isSingleDigit(number) {
return 1 <= number && number <= 9;
}
function genSingleDigitExpression() {
do {
var expr = genMathExpression();
} while (!isSingleDigit(eval(expr)));
return expr;
}
var q = document.getElementById('question');
var top = q.previousSibling.offsetTop;
var bottom = q.parentNode.offsetTop + q.parentNode.offsetHeight;
var left = q.parentNode.offsetLeft;
var right = q.parentNode.offsetWidth;
q.style.top = randomInt(top, bottom - 60) + 'px';
q.style.left = randomInt(left, right - 120) + 'px';
q.textContent = text;
}
function numberFromKeycode(keycode) {
var code = keycode.which;
return (48 <= code && code < 58) ? code - 48 : // top row of keys
(96 <= code && code < 106) ? code - 96 : // numeric keypad
null;
}
function main() {
var points = 0;
var expr;
function handleKeypress(event) {
if (numberFromKeycode(event) == eval(expr)) {
document.getElementById('score').textContent = ++points;
document.getElementById('board').className = 'correct';
new Audio("http://www.soundjay.com/misc/bell-ringing-05.mp3").play();
} else {
document.getElementById('score').textContent = --points;
document.getElementById('board').className = 'incorrect';
new Audio("http://www.soundjay.com/misc/sounds/fail-buzzer-02.mp3").play();
}
document.getElementById('instructions').style.display = 'none';
}
}
main();
#board {
border: 1px solid black;
width: 800px;
height: 600px;
color: black;
font: 50px Arial, sans-serif;
position: relative;
}
#board.correct {
background-color: #7cfc00; /* light green */
}
#board.incorrect {
background-color: red;
}
h1, #score {
display: inline;
font-size: 50px;
font-weight: normal;
}
#score {
display: inline;
float: right;
}
#instructions {
margin-top: 3em;
font-size: 20px;
}
hr {
visibility: hidden;
}
#question {
position: absolute;
}
<div id="board">
<h1>Single Digit Math Quiz</h1>
<div id="score">0</div>
<div id="instructions">
Press the result of the operation on your keyboard.
</div>
<hr id="hr"><div id="question"></div>
</div>
• Nice point about the globals, but when you say wasteful, I don't understand. The code should run only very few times, efficiency does not matter. – Caridorc May 26 '15 at 19:02
• Anyway It really is easier to follow that we first define a random_range function and then a random_choice building from that :) – Caridorc May 26 '15 at 20:33
You can't currently play with the numbers on the numpad of a keyboard - they have different keycodes to the numbers along the top.
It's normal to camelCase function names rather than snake_case.
Avoid abbreviations e.g. the gen in gen_single_digit_expression.
function number_from_keycode(keycode) { - keycode isn't a keycode at all, it's an event.
Try to remember your semicolons - it's not required but many people (myself included) prefer to see them. There are a couple of very subtle edge cases that can cause problems if you omit them.
eval is evil. You really don't need to invoke the interpreter to do a simple sum!
Here's an example of a simple constructor function (and usage) to show you a way of doing it without eval.
var MathExpression = function(left, operator, right) {
this.evaluate = function () {
switch (operator) {
case '+':
return left + right;
case '-':
return left - right;
case '*':
return left * right;
}
};
};
var e1 = new MathExpression(1, '+', 8);
console.log(e1.evaluate()); // 9
You don't need to brute force expression generation either. Choose the operator first, then choose the LHS and finally choose a number for the RHS that is within the allowed range.
Choose LHS: 1 <= LHS <= 8
Choose RHS: 1 <= RHS <= 9 - LHS
LHS choose 5.
RHS can be 1, 2, 3 or 4 to yield a single digit answer.
You can do the same for the other operators:
'*' (multiplication)
Choose LHS: 1 <= LHS <= 4
Choose RHs: 1 <= RHS <= 9/LHS (rounded down).
LHS choose 4.
RHS can be 1 or 2 to yield a single digit number. (9/4 = 2.25 -> 2)
I'll leave '-' for you to do.
• You say eval is evil but is this the universal truth? Eval is dangerous, but it is no more evil than the gunpowder is. In this case it saves me tons of code and is never run on any input, only strings generated by me. Also do you think I should ditch bruteforce?, performance is not a concern whatsover. – Caridorc May 26 '15 at 19:05
• @Caridorc - not a universal truth but a notable quote from Crockford – RobH May 26 '15 at 20:09
• in fact you are mostly right (+1) but evaluating a single mathematical expression is in my opinion one of the very few legit uses of eval. – Caridorc May 26 '15 at 20:26
The range function is misleading. From its parameter names I would guess that range(3, 5) returns [3, 4] (or possibly [3, 4, 5]), but actually it gives [3, 4, 5, 6, 7]. Since most of times you use it with 0 as the first parameter, your intention is not very clear, and this may be a bug.
This implementation would behave more like what I'd expect (exclusive end):
function range(start, end) {
return Array.apply(0, Array(end - start))
.map(function(element, index) {
return index + start;
});
}
Or if you prefer the range to be inclusive, then change to end - start + 1. | 2019-11-21 22:52:57 | {"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": 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": 0.2098914384841919, "perplexity": 7372.608138739147}, "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/1573496670987.78/warc/CC-MAIN-20191121204227-20191121232227-00230.warc.gz"} |
https://socratic.org/questions/how-do-you-factor-x-4-13x-2-36-0 | # How do you factor x^-4 -13x^-2 +36 =0?
May 6, 2015
$x = \pm \frac{1}{2} , \pm \frac{1}{3}$
Multiply by ${x}^{4}$ to give $36 {x}^{4} - 13 {x}^{2} + 1$
Solve using the quadratic formula for ${x}^{2}$ ${x}^{2} = \frac{13 \pm \sqrt{{13}^{2} - 4.36}}{2.36} = \frac{13 \pm \sqrt{169 - 144}}{72} = \frac{13 \pm \sqrt{25}}{72} = \frac{18}{72} , \frac{8}{72} = \frac{1}{4} , \frac{1}{9}$
Taking square roots gives $x = \pm \frac{1}{2} , \pm \frac{1}{3}$
May 6, 2015
Terminology:
You can factor the expression on the left. You can solve the equation. You can solve an equation by factoring.
(But you don't really factor an equation.)
To factor: ${x}^{- 4} - 13 {x}^{- 2} + 36$
Notice that ${x}^{- 4} = {\left({x}^{- 2}\right)}^{2}$ (The variable expression in the first term is the square of the one in the second term.)
So, is we use a new variable we'll have a quadratic expression.
Let $u = {x}^{- 2}$. This makes ${u}^{2} = {x}^{- 4}$, so the expression becomes:
${u}^{2} - 13 u + 36$ which can be factored:
$\left(u - 4\right) \left(u - 9\right)$ Now go back to $x$'s
${x}^{- 4} - 13 {x}^{- 2} + 36 = \left({x}^{- 2} - 4\right) \left({x}^{- 2} - 9\right)$
To solve by factoring: ${x}^{- 4} - 13 {x}^{- 2} + 36 = 0$
Factor as above, so the question becomes:
Solve: $\left({x}^{- 2} - 4\right) \left({x}^{- 2} - 9\right) = 0$
So we need: $\left({x}^{- 2} - 4\right) = 0$ or $\left({x}^{- 2} - 9\right) = 0$
${x}^{- 2} - 4 = 0$ $\textcolor{w h i t e}{\text{sssss}}$ or $\textcolor{w h i t e}{\text{sssss}}$ ${x}^{- 2} - 9 = 0$
${x}^{- 2} = 4$$\textcolor{w h i t e}{\text{ssssssssss}}$ or $\textcolor{w h i t e}{\text{sssss}}$ ${x}^{- 2} = 9$
$\frac{1}{x} ^ 2 = 4$ $\textcolor{w h i t e}{\text{ssssssssss}}$ or $\textcolor{w h i t e}{\text{sssss}}$ $\frac{1}{x} ^ 2 = 9$
$\frac{1}{4} = {x}^{2}$$\textcolor{w h i t e}{\text{sssssssssss}}$ or $\textcolor{w h i t e}{\text{sssss}}$ $\frac{1}{9} = {x}^{2}$
$x = \pm \frac{1}{2}$ $\textcolor{w h i t e}{\text{ssssssss}}$ or $\textcolor{w h i t e}{\text{sssss}}$ $x = \pm \frac{1}{3}$
There are four solutions: $- \frac{1}{2} , \frac{1}{2} , - \frac{1}{3} , \frac{1}{3}$ | 2019-12-07 17:36:08 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 39, "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": 0.9401295781135559, "perplexity": 1347.7195924870982}, "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/1575540500637.40/warc/CC-MAIN-20191207160050-20191207184050-00132.warc.gz"} |
http://ac297r.org/blog/tutorial-bayes-1.html | # The Many Levels of the Reverend Bayes, Part 1¶
### https://dl.dropboxusercontent.com/u/75194/ac297rbayes1.ipynb¶
In [1]:
%matplotlib inline
import numpy as np
import scipy as sp
import matplotlib as mpl
import matplotlib.cm as cm
import matplotlib.pyplot as plt
import pandas as pd
pd.set_option('display.width', 500)
pd.set_option('display.max_columns', 100)
pd.set_option('display.notebook_repr_html', True)
import seaborn as sns
sns.set_style("whitegrid")
sns.set_context("paper")
In [ ]:
from sklearn.metrics import confusion_matrix
## Bayes Theorem¶
The theorem follows directly from writing the definition of conditional probability in two ways:
$$p(h|D) = \frac{p(D|h)\,p(h)}{p(D)}$$
$$Posterior = \frac{Likelihood \times Prior }{Evidence}$$
### The Cancer Example¶
Say I go for a lab test. It comes out positive for a type of cancer.
This test is known to have a true positive rate of 98%, that is, for those who have cancer, 98% of the time this test turns out to be positive.
The test is also known to have a true negative rate of 97%, that is for those who dont have cancer, 97% of the time the test turns out to be negative.
This cancer type has a prior probability of 0.8% of occurring in the population.
What should I do?
## Reminder of the structure of learning¶
Lets focus on supervised learning for a bit to understand what new stuff does Bayes Theorem and Bayesian modelling bring to learning.
(diagram from Abu-Mustafa's Learning From Data Course)
Lets define
$$E_{XY}[f(X,Y)] = \int f(x,y) dP_{XY}(x,y) = \int f(x,y) p(x,y)\,dx\,dy$$
where $P$ is the distribution (cdf) and $p$ is the local density. We'll use the Lebesgue measure for continuous rvs and the counting measure for discrete rvs, so as to use the integral symbol everywhere.
Then consider an estimator function (for regression, estimating $f$ per usual) or action (for classification) $g$. The risk accociated with $g$ is:
$$R(g) = E_{XY}[l(g(X),Y)]$$
where $l(g(x),y)$ is the loss associated with choosing action $g$ at x when $Y=y$. For example, for regression we can use $l = (g-y)^2$ and for classification we usually use the 1-0 loss $l = \mathbb{1}_{g \ne y}$.
The idea, as we have seen before is to find an optimal decision function by minimizing the risk (or maximizing a utility). But remember we dont know $P$ and the different kinds of learning use different ways to estimate $P$. Only if our estimated distribution is close to the true $P$ can we expect our decisions to generalize well outside our training set.
Thus there are two games in town:
1. estimating the density
2. minimizing the risk
### The different kinds of learning¶
#### ERM¶
The Empirical Risk Maximization (ERM) approach corresponds to estimating the true distribution by the empirical distribution. We have seen this before. In this case the Risk R is simply the average of the losses at the individual training points:
$$R(g) = \frac{1}{N} \sum_i l(g(x_i), y_i) .$$
(next two images from Barber)
The optimal decision in the training set is obviously the value of $y$ at the training point, but we are left with undefined action $g$ outside the training set. We thus pick some parametric model $g(x;\theta)$. Now we can minimize the empirical risk with respect to $\theta$ to get $\theta_{opt}$ and use this to make predictions/actions using $g$. Because such an approach can lead to overfitting as we have seen before, we typically add a regularization term with co-efficient $\lambda$ whose value is found by validation.
Notice that in all of this any talk of density estimation has gone away, and we are just minimizine the averaged loss over the training set plus regularization, the so-called Structural Risk maximization approach of Vapnik, whose motto is (paraphrased): Never solve a more difficult problem (density estimation) while solving a difficut one (learning). The function $g$ is then sometimes called a discriminant function, and we are choosing it based on minimal risk, which is the quantity we are ultimately interested in.
But there are drawbacks. It seems crazy to assume that the empirical distribution is a good distribution, especially for small data. A more reasonable assumption for the distribution could take into account likely x,y that could arise. If the loss changes, as it might over time, say in a financial application, then we would need to retrain $g$. There is no way to associate a confidence in this framework, as it wont give you probabilities. And we cant acomodate things like the reject option, compensating for assymetric data sets (class priors), or combine models. This is where something like the Naive Bayes model below comes in.
#### Bayes¶
The alternative is to first do density estimation. We estimate $P(x,y)$ (or $P(x,c)$) from the training data. Note that this can be thought of as ERM on risk $-log(p)$ and can be appropriately regularized in a Bayesian fashion (or be just ML estimation, one can see the different levels at which Bayes might come in). This is usually done parametrically, from which one can construct a sampling (in frequentist) or posterior predictive, or just use the plug-in MAP/MLE estimate. The joint distribution can be constructed in two ways: generative or discriminative. The discriminative approach gives us:
$$p(x,c) = p(c|x)p(x)$$
whereas the generative approach gives us
$$p(x,c) = p(x|c) p(c)$$
and then bayes theorem can be used to obtain p(c|x).
The generative approach corresponds to picking one of the classes with probability p(c) and then getting the density of the features for that class. The discriminative approach models the domain boundary instead. While the data may be distributed in a complex way, the boundary may be easier to model. On the other hand prior information for assymetric situations, conditional independence and other such strategies can only be done in generative models.
In either case we can get the joint distribution. In the discriminative case that leads us to density estimation for $p(x)$. Often we have no use for it so we wont do it, as in logistic regression. But do remember that if we want our classifier to have good results we should be using iton sets which reflect $p(x)$. And if we dont characterize it we might be better of using a generative model as it is easier to adjust for class priors.
The Bayesian decision approach is a clean one, in which first one models the environment, independent of the subsequent decision process. If $p(y,x|\theta)$ is the "true" model of the world, this is optimal. But if this environment model is poor, the action $g$ could be higly inaccurate since the environment is divorced from prediction. In practice one often includes regularization terms in the environment model to reduce the complexity of the distribution and bring it more in line with decision based hyperparameters, set by validation on an empirical loss. See the cs109 (2013) Naives bayes homework for a good example.
The ERM method is the only frequentist method which has a well defined risk. The reason for this is that it dosent depend on both a sample-estimate of the true $\theta$.
## Naive Bayes¶
This is the second level at which we use a fairly direct use of Bayes theorem, to make a simple classifier.
Naive Bayes is a simple generative model that uses a rather crazy conditional independence assumption. It is generative as it starts out with class priors, and then says:
$$p(x_1, x_2 | c) = p(x_1|c) p(x_2 | c)$$
$x_1$ is independent of $x_2$ conditioned on $c$.
This conditional independence allows us to "combine models".
(from REF)
Notice that all the "causal" signs are not right but this is a simplifying first assumption.
I use here Gaussian Naive Bayes where the likelihoods around continuous features are Gaussian. You are probably more used to seeing Bernoulli or Multinomial Naive Bayes for bag-of-words text models.
In [538]:
dfd=pd.read_csv("https://dl.dropboxusercontent.com/u/75194/pima.csv")
Out[538]:
npreg glconc dpressure sft insulin bmi diaped age diabetes
0 6 148 72 35 0 33.6 0.627 50 1
1 1 85 66 29 0 26.6 0.351 31 0
2 8 183 64 0 0 23.3 0.672 32 1
3 1 89 66 23 94 28.1 0.167 21 0
4 0 137 40 35 168 43.1 2.288 33 1
In [539]:
features=['npreg', 'bmi','diaped', 'age', 'glconc','insulin']
X=dfd[features].values
y=dfd.diabetes.values
from sklearn.naive_bayes import GaussianNB
clf=GaussianNB()
itr,ite=train_test_split(range(y.shape[0]))
Xtr=X[itr]
Xte=X[ite]
ytr=y[itr]
yte=y[ite]
clf.fit(Xtr,ytr)
print "Frac of mislabeled points",float((yte != clf.predict(Xte)).sum())/yte.shape[0]
confusion_matrix(clf.predict(Xte),yte)
Frac of mislabeled points 0.21875
Out[539]:
array([[109, 26],
[ 16, 41]])
• Naive Bayes is a good asuumption for some things like multiple tests in the cancer example. A biopsy is conditionally independent of a blood test given cancer. It thus works surprisingly well in the medical field.
• for docs one way to think about is that you are flipping a biased coin each time for a word, with topic specific words being more common, but sometimes things will go the other way! A bad generative model for writers :-)
• For text classification, naive bayes does better then decision tree induction, which is surprising! It throws grammar and meaning out of the window. SVM is the one that beats it for test classification.
• Why does this work? What we need is that the argmax of the estimated probablity product in the estimated posterior be equal to the argmax of the actual posterior, even if all the individual probabilities are not so good, for the classifier to be good.
• Indeed its the relative ordering of the probability of the classes that must be preserved on this simplification. If that is done we are set. Remember that probability estimates from NB are not trustworthy.
• NB pushes the probabilities either close to 0 or close to 1 because of the independence assumption. It dosent know the redundancy in the features.
• As a rule of thumb try kNN and NB on any problem first. They are simple and will at the very least, set up a baseline classifier.
• one 0 in P(attr|class) wipes out the posterior. We start with the MLE estimate $\frac{n_{attr, class}}{n_{class}}$. Hallucinate a estimate, 1 in that class. Or add mp to numerator and m to denominator where p is some small prior probability for the attribute in the class. We can justify this from higher level priors on the probabilities $p(x_i = x|c)$.
• Finally use logs so no underflows! golden rule: never multiply probs do sum of logs. Indeed when you do MCMC in AM207 you always deal with log probabilities. Here you will do the same thing.
In [540]:
def calibration_plot(clf, xtest, ytest):
prob = clf.predict_proba(xtest)[:, 1]
outcome = ytest
data = pd.DataFrame(dict(prob=prob, outcome=outcome))
#group outcomes into bins of similar probability
bins = np.linspace(0, 1, 20)
cuts = pd.cut(prob, bins)
binwidth = bins[1] - bins[0]
#freshness ratio and number of examples in each bin
cal = data.groupby(cuts).outcome.agg(['mean', 'count'])
cal['pmid'] = (bins[:-1] + bins[1:]) / 2
cal['sig'] = np.sqrt(cal.pmid * (1 - cal.pmid) / cal['count'])
#the calibration plot
ax = plt.subplot2grid((3, 1), (0, 0), rowspan=2)
p = plt.errorbar(cal.pmid, cal['mean'], cal['sig'])
plt.plot(cal.pmid, cal.pmid, linestyle='--', lw=1, color='k')
plt.ylabel("Empirical Fraction")
#the distribution of P(fresh)
ax = plt.subplot2grid((3, 1), (2, 0), sharex=ax)
#calsum = cal['count'].sum()
plt.bar(left=cal.pmid - binwidth / 2, height=cal['count'],
width=.95 * (bins[1] - bins[0]),
fc=p[0].get_color())
plt.xlabel("Classifier Probability")
In [541]:
calibration_plot(clf, Xte, yte)
In [542]:
calibration_plot(clf, Xtr, ytr)
Notice how Naive bayes is more aptimistic at higher probabilities and less at lower probabilities. This is because of the conditional independence multiplication.
## Using pymc to do Bayesian statistics¶
Lets revert to standard Bayesian stats for a bit, where we have:
$$p(\theta|D) = \frac{p(D|\theta)\,p(\theta)}{p(D)}$$
with the evidence $p(D)$ being given by the prior-predictive:
$$p(y) = \int d\theta p(y|\theta) p(\theta)$$
and predictions made by the posterior predictive:
$$p(y^* | D=\{y\}) = \int d\theta p(y^*|\theta)p(\theta|\{y\})$$
.
(text taken from AM207)
### Gaussian with known $\sigma$¶
Define $\kappa=\frac{\sigma^2}{\tau^2}$ to be the variance of the sample model in units of variance of our prior belief (prior distribution) then the posterior mean
$$\mu_{p} = \frac{\kappa}{\kappa + n} \mu_{prior} + \frac{n}{\kappa + n} \bar{y}$$
which is a weighted average of prior mean and sampling mean. The variance is
$$\frac{1}{\tau_p^2} = \frac{1}{\tau^2} + \frac{1}{\sigma^2}$$
We have data on the wing length in millimeters of a nine members of a particular species of moth. We wish to make inferences from those measurements on the population mean $\mu$. Other studies show the wing length to be around 19 mm. We also know that the length must be positive. We can choose a prior that is normal and most of the density is above zero ($\mu=19.5,\tau=10$). The measurements were: 16.4, 17.0, 17.2, 17.4, 18.2, 18.2, 18.2, 19.9, 20.8 giving $\bar{y}=18.14$.
Using the formulas above we have ($\kappa = \frac{\sigma^2}{100}$) and say $\sigma^2=s^2=1.9928$ then the posterior is $N(18.14,0.47)$. Note this is for a KNOWN $\sigma$.
In [32]:
Y = [16.4, 17.0, 17.2, 17.4, 18.2, 18.2, 18.2, 19.9, 20.8]
# Prior mean
mu_prior = 19.5
# prior std
tau = 10
N = 15000
In [33]:
#Data Quantities
sig = np.std(Y) # assume that is the value of KNOWN sigma (in the likelihood)
mu_data = np.mean(Y)
n = len(Y)
kappa = sig**2 / tau**2
sig_post =np.sqrt(1./( 1./tau**2 + n/sig**2));
# posterior mean
mu_post = kappa / (kappa + n) *mu_prior + n/(kappa+n)* mu_data
#samples
theta_prior = np.random.normal(loc=mu_prior, scale=tau, size=N);
theta_post = np.random.normal(loc=mu_post, scale=sig_post, size=N);
In [34]:
plt.hist(theta_post, bins=30, alpha=0.9, label="posterior");
plt.hist(theta_prior, bins=30, alpha=0.2, label="prior");
plt.xlim([10, 30])
plt.xlabel("wing length (mm)")
plt.ylabel("Number of samples")
plt.legend();
In [128]:
from pymc import Normal, MCMC, Matplot, observed, deterministic, graph, stochastic
We put a normal prior on the mean as before, and fix the precision. The variables means_pr and wingspans are Stochastics. Notice wingspans have observed=True which indicates a likelihood. All Stochastics return log probabilities, but this is hidden from you for now.
In [129]:
means_pr=Normal('means_pr', mu=19.5, tau=0.01, value=20.)
wingspans = Normal('wingspans', mu=means_pr, tau=1/(sig*sig), value=Y,
observed=True)
It might seem wierd to represent an observed thing, the wingspans, by a stochastic. But remember what Bayesian probability is: its not always the frequency estimate like the MLE in Naive Bayes, but rather something representative of our knowledge or uncertainty about a quantity.
Bayesian Notation thus does not differ between random variables and data. Thus we need to assign the likelihood a probability distribution as if it were a random variable.
A way to get comfortable with this is to think of the generative aspect of statistical models: think of them as stories for data
Lets run the MCMC with the two variables. Pymc2 uses MH steps embedded in a Gibbs structure to go between variables. The method is to sample a suggestion from a normal distribution around the current value. If the suggestion improves the models log-probability, accept. But if it dosent, perhaps accept with acceptance proportional to how bad the new sample is.
In [131]:
M0 = MCMC([means_pr, wingspans])
M0.sample(iter=20000, burn=5000)
[-----------------100%-----------------] 20000 of 20000 complete in 0.7 sec
The way it worls is by creating a DAG on the nodels in your graph. It figures out what are Stochastics and unobserved, and outputs traces for those. The graph structure is below (you will need to install pydot). You can also output parents and children.
In [136]:
dotm0=graph.dag(M0)
In [140]:
from IPython.display import SVG
SVG(dotm0.create_svg())
Out[140]:
In [130]:
print means_pr.children, wingspans.parents
print means_pr.value, wingspans.value
set([<pymc.distributions.Normal 'wingspans' at 0x112278250>]) {'mu': <pymc.distributions.Normal 'means_pr' at 0x1123fca10>, 'tau': 0.56453861165319197}
20.0 [ 16.4 17. 17.2 17.4 18.2 18.2 18.2 19.9 20.8]
this is how you plot the marginal, trace, and acorr for a stochastic:
In [39]:
Matplot.plot(means_pr)
Plotting means_pr
And below is how u get the trace out. You can also use means_pr.trace(). Note that in this form you use the string name of the stochastic, whereas in the latter you use the variable name.
In [132]:
M0.trace('means_pr')[:]
Out[132]:
array([ 18.197082 , 18.197082 , 18.197082 , ..., 17.40876589,
18.14956583, 17.75870891])
In [133]:
plt.hist(M0.trace('means_pr')[:], bins=30, alpha=0.2, label="posterior");
plt.hist(theta_prior, bins=30, alpha=0.2, label="prior");
plt.hist(theta_post, bins=30, alpha=0.2, histtype="stepfilled", label="exact posterior");
plt.xlim([10, 30])
plt.xlabel("wing length (mm)")
plt.ylabel("Number of samples")
plt.legend();
### Joint distribution of mean and variance¶
But how about if we did not know $\sigma^2$? Note:
$$p(\theta, \sigma^2) = p(\theta | \sigma^2) \; p(\sigma^2)$$
One can mathematically work it out.
For $\sigma^2$ we need a prior that covers (0,$\infty$). One option is the gamma family. Unfortunately this is not conjugate for the normal. However it turns out to be conjugate for pdf of $1/\sigma^2$. This is called the inverse-gamma distribution:
$$p(\sigma^2) \sim \rm{inversegamma}(a,b)$$
Under this parametrization $E[\sigma^2] = \frac{b}{a-1}$.
Let our prior have this form
$$p(\sigma^2) \sim \rm{inversegamma}( \frac{ \nu}{2}, \frac{\nu \sigma_0^2}{2})$$
and we know the conditional prior of $\theta$
$$p( \theta | \sigma^2) \sim N (\mu, \sigma^2/\kappa)$$
This joint prior can be decomposed as $p(\theta, \sigma^2)=p(\theta| \sigma^2) p(\sigma^2)$, and therefore the posterior can be similarly decomposed:
$$p(\theta, \sigma^2 | y_1,\ldots,y_n) = p( \theta | \sigma^2) \; p(\sigma^2 ) p(y_1\ldots,y_n \vert \sigma^2, \theta)$$
The conditional distribution of $\theta$ given the data and $\sigma^2$ can be obtained using the results of the previous section. The posterior distribution of $\sigma^2$ can be obtained by performing an integration over known values of $\theta$:
$$p(\sigma^2|y_1, \ldots, y_n) \sim p(\sigma^2) p(y_1, \ldots, y_n | \sigma^2) = p(\sigma^2) \int p(y_1, \ldots, y_n | \theta, \sigma^2 ) \, p(\theta | \sigma^2) d\theta$$
One can integrate this (left as an exercise) but the result is:
$$p(1/\sigma^2| y_1, \ldots, y_n) \sim \rm{gamma}(\nu_{n} /2, \nu_{n} \sigma^2_{n}/2)$$
where
$$\nu_n = \nu+n$$
$$\sigma_n^2 = \frac{1}{\nu_n} \left[ \nu \sigma_0^2 +(n-1) \, s^2 + \frac{ \kappa n }{\kappa+ n}(\bar{y} - \mu)^2 \right]$$
and $s^2$ is the sample variance.
### Example: Moth wing example (joint distribution):¶
Studies of other populations suggest that the true mean and standard deviation of our population under study should not be too far the sample mean. We use here $\mu=19$ and $\sigma_0^2 = 2$. This population may be different from others so we choose $\kappa = \nu=1$. The sample mean $\bar{y} = 18.14$ and sample variance $s^2=1.9928$. We can compute the posterior means
$$\mu_p = \frac{\kappa}{\kappa+n} \mu + \frac{n}{\kappa+n} \bar{y} = 18.2260$$$$\sigma_p^2 = \frac{1}{\nu_n} \left[ \nu \sigma_0^2 + (n-1) s^2 + \frac{\kappa \, n}{\kappa + n} (\bar{y} - \mu)^2 \right] = 1.8606$$
So the posterior distribution is determined by the values $\mu_p= 18.2260, \kappa =1 , \sigma_p^2 = 1.8606, \nu=1$. Let $\hat{\sigma}^2= 1/\sigma^2$ we can create a contour plot for example by computing (NOT SAMPLING)
$$N\left(\theta_k \mu_p, 1/\sqrt{10 \hat{ \sigma}_l^2} \right) \times \rm{gamma}\left(\hat{\sigma}_l^2, 10/2, 10 \sigma_p^2/2 \right)$$
for each pair $\theta_k, \hat{\sigma}_l^2$ on a grid.
But given that we are interested in sampling, note that the precision ($\frac{1}{\sigma^2}$) is thus conjugate to the gamma. Lets look at a reasonably un-informative gamma.
In [119]:
x=np.arange(0,100,1)
gam=sp.stats.gamma(1.0000000, scale=1./0.01)
plt.plot(x, gam.pdf(x))
plt.xlim([0,100])
Out[119]:
(0, 100)
Now let us set up a model for this, This time the precision also has a prior. Notice the structure of the parent and child relationships.
In [142]:
from pymc import Gamma
tau_pr=Gamma('tau_pr', alpha=1.0, beta=0.01)
means_pr=Normal('means_pr', mu=19.5, tau=0.01, value=20.)
wingspans = Normal('wingspans', mu=means_pr, tau=tau_pr, value=Y,
observed=True)
print means_pr.children, means_pr.parents, wingspans.parents
set([<pymc.distributions.Normal 'wingspans' at 0x1126faf10>]) {'mu': 19.5, 'tau': 0.01} {'mu': <pymc.distributions.Normal 'means_pr' at 0x1120e42d0>, 'tau': <pymc.distributions.Gamma 'tau_pr' at 0x112e436d0>}
In [143]:
M = MCMC([tau_pr, means_pr, wingspans])
M.sample(iter=70000, burn=10000, thin=5)
[-----------------100%-----------------] 70000 of 70000 complete in 3.8 sec
In [146]:
dotm=graph.dag(M)
SVG(dotm.create_svg())
Out[146]:
In [144]:
Matplot.plot(means_pr)
Plotting means_pr
In [145]:
Matplot.plot(tau_pr)
Plotting tau_pr
As another test of convergence we can check that the geweke statistic is within 2 $\sigma$:
In [544]:
from pymc import geweke
gM=geweke(M)
Matplot.geweke_plot(gM)
## Unsupervised Learning¶
Before we go into supervized learning using bayesian methods at different levels, lets briefly visit unsupervised learning. There is no explicit external risk corresponding to an external decision, but rather a density estimation problem for $p(x)$.
Note that if you do have classes in the problem, as we shall see below, then we have
$$p(x) = \sum_c p(x|c) p(c).$$
In other words we discover the marginal p(x) through the generative model and thus discover outliers in our data.
In [149]:
from pymc import Categorical, Uniform
In [360]:
ofdata=pd.read_csv("https://dl.dropboxusercontent.com/u/75194/oldfaithful.csv")
Out[360]:
eruptions waiting
0 3.600 79
1 1.800 54
2 3.333 74
3 2.283 62
4 4.533 85
In [363]:
plt.plot(ofdata.eruptions, ofdata.waiting,'o',alpha=0.5);
plt.xlabel('eruptions length')
plt.ylabel('time between eruptions')
Out[363]:
<matplotlib.text.Text at 0x1c3ef3a50>
Even if we do not know of the existence of classes in our data, we can consider a hidden parameter $z$ such that:
$$p(x) = \sum_z p(x|z) p(z).$$
This is the thinking behind mixture models which are very highly used to describe all kinds of data. Even hierarchical models can be thought of as continuous mixtures so the lessons learnt with mixture models are useful. in AM207 you will learn how to solve then using EM on likelihoods or posteriors. We demonstrate this quickly in scikit-learn, below.
### One could use frequentist EM¶
In [368]:
from sklearn import mixture
clf = mixture.GMM(n_components=2, covariance_type='tied')
vals=ofdata.values
clf.fit(vals)
Out[368]:
GMM(covariance_type='tied', init_params='wmc', min_covar=0.001,
n_components=2, n_init=1, n_iter=100, params='wmc', random_state=None,
thresh=0.01)
In [369]:
mask = clf.predict(vals)
xx = np.linspace(0, 6)
yy = np.linspace(40, 100)
X, Y = np.meshgrid(xx, yy)
XX = np.c_[X.ravel(), Y.ravel()]
Z = np.log(-clf.score_samples(XX)[0])
Z = Z.reshape(X.shape)
CS = plt.contour(X, Y, Z)
CB = plt.colorbar(CS, shrink=0.8, extend='both')
But EM is an optimization algorithm and is as such restricted to providing us with point estimates rather than samples. Now that we are into the belly of MCMC sampling: a very reasonable question to ask is this: whats the uncertainty on the probability estimates bayesian learning gives us? Note that both discriminative and generative models (like naive bayes) can be set up perfectly well with point estimates, but we now want to take our bayesianness to the next level.
In [411]:
elvals=ofdata.eruptions.values
sns.kdeplot(elvals)
plt.hist(elvals, normed=True, alpha=0.2, bins=20);
Lets do the problem in one dimension, the dimension of eruptions, for simplicity. The code for this and the subsequentl supervixed learning problem is largely stolen from CDP's excellent Bayesian Methods for Hackers, which along with Fonnesbeck's Bios366 repo (from which the code for the bioassay problem is taken) are MUST READS.
### Or one can use MCMC¶
In [404]:
p = Uniform("p", 0, 1)
assignment = Categorical("assignment", [p, 1 - p], size=elvals.shape[0] )
taus = 1.0 / Uniform("stds", 0.1, 4, size=2) ** 2
centers = Normal("centers", [2., 4.7], [0.1, 0.1], size=2)
"""
The below deterministic functions map an assignment, in this case 0 or 1,
to a set of parameters, located in the (1,2) arrays taus and centers.
"""
@deterministic
def center_a(assignment=assignment, centers=centers):
return centers[assignment]
@deterministic
def tau_a(assignment=assignment, taus=taus):
return taus[assignment]
observations = Normal("obs", center_a, tau_a, value=elvals, observed=True)
from pymc import Model
model = Model([p, assignment, observations, taus, centers])
mcmc = MCMC(model)
dotmcmc=graph.dag(mcmc)
SVG(dotmcmc.create_svg())
Out[404]:
In [405]:
mcmc.sample(300000, burn=200000, thin=5)
[-----------------100%-----------------] 300000 of 300000 complete in 99.3 sec
In [406]:
Matplot.plot(mcmc)
Plotting stds_0
Plotting stds_1
Plotting centers_0
Plotting centers_1
Plotting p
Note that because we have done MCMC, each sample in a trace corresponds to a set of values for the parameters. So we literally have 100,000 gaussian mixture models here. How to make any sense of this? We'll take the point estimates from the means of the parameters and draw us a set of clusters, forgetting to do any inference on these clusters for now. The posterior predictive at any x not in the training sample (and in the training sample if you like) would be the one which marginalizes over all the 100000 parameter values.
Lets do the simpler hacky means thing:
In [407]:
center_trace = mcmc.trace("centers")[:]
std_trace = mcmc.trace("stds")[:]
colors = ["#348ABD", "#A60628"] \
if center_trace[-1, 0] > center_trace[-1, 1] \
else ["#A60628", "#348ABD"]
norm = sp.stats.norm
x = np.arange(0, 10, 0.1)
posterior_center_means = center_trace.mean(axis=0)
posterior_std_means = std_trace.mean(axis=0)
posterior_p_mean = mcmc.trace("p")[:].mean()
y = posterior_p_mean * norm.pdf(x, loc=posterior_center_means[0],
scale=posterior_std_means[0])
plt.plot(x, y, label="Cluster 0 (using posterior-mean parameters)", color=colors[1], lw=3)
plt.fill_between(x, y, color=colors[1], alpha=0.3)
y = (1 - posterior_p_mean) * norm.pdf(x, loc=posterior_center_means[1],
scale=posterior_std_means[1])
plt.plot(x, y, label="Cluster 1 (using posterior-mean parameters)", color=colors[0], lw=3)
plt.fill_between(x, y, color=colors[0], alpha=0.3)
plt.hist(elvals, bins=20, histtype="step", normed=True, color="k",
lw=2, label="histogram of data")
plt.legend(loc="upper right")
plt.title("Visualizing Clusters using posterior-mean parameters")
Out[407]:
<matplotlib.text.Text at 0x27951c310>
Voila, a pretty reasonable density estimate or hidden variables (really classes) based mixture estimate of the density p(x), one might say...
## A simple regression: the old bioassay¶
Remember the old Gelman Bioassy? Various drug doses are administered to animals and a binary outcome (death) is noted. There are 4 groups of 5 animals each, different doses administered, and deaths recorded. We construct a model for $\theta$ the binomial probabiliy of death, as a regression on dose through the logit link function. We set imprecise normal priors on the regression coefficients, and pass the linear regression through the inverse logit function into a binomial likelihood.
In [545]:
from pymc import Normal, Binomial, deterministic, invlogit, Model, MAP
n = [5]*4
dose = [-.86,-.3,-.05,.73]
x = [0,1,3,5]
alpha = Normal('alpha', mu=0.0, tau=0.01)
beta = Normal('beta', mu=0.0, tau=0.01)
@deterministic
def theta(a=alpha, b=beta, d=dose):
"""theta = inv_logit(a+b)"""
return invlogit(a+b*d)
# deaths ~ binomial(n, p), 4 of them
deaths = Binomial('deaths', n=n, p=theta, value=x, observed=True)
### Posterior predictive¶
We add an interesting new stochastic to the model, deaths_sim, the simulated deaths, in order to do posteror predictive modelling. This is using the model in a generative sense, by simulating new data from it, or, if you like finding the posterior prediction at new x.
If you think about it for a second, you'll realize that you can lump the posterior prediction needed 'likelihoods' along with the other stochastics. Then, by just getting the trace of these points, you will have the marginal over all the other stochastics, which is precisely what you want.
If you think about it a little more, this technique of using the posterior predictive can be used to fill missing data points in a data set. Indeed, pymc can take numpy masked arrays for precisely this purpose. We wont do that here but see the docs.
In [547]:
deaths_sim = Binomial('deaths_sim', n=n, p=theta)#4 of them
In [548]:
bioassay_model=Model([alpha, beta, theta, deaths, deaths_sim])
dotbioassay=graph.dag(bioassay_model)
SVG(dotbioassay.create_svg())
Out[548]:
In [549]:
map_ = MAP(bioassay_model)
map_.fit(method='fmin_powell')
bioassay_mcmc = MCMC(bioassay_model)
bioassay_mcmc.sample(10000, 5000, 2)
Warning: Stochastic deaths_sim's value is neither numerical nor array with floating-point dtype. Recommend fitting method fmin (default).
[-----------------100%-----------------] 10000 of 10000 complete in 2.1 sec
In [550]:
Matplot.plot(bioassay_mcmc)
Plotting deaths_sim_0
Plotting deaths_sim_1
Plotting deaths_sim_2
Plotting deaths_sim_3
Plotting beta
Plotting theta_0
Plotting theta_1
Plotting theta_2
Plotting theta_3
Plotting alpha
In [551]:
sns.kdeplot(alpha.trace(), beta.trace())
Out[551]:
<matplotlib.axes._subplots.AxesSubplot at 0x2250eb810>
You can summarise the MCMC with the credible intervals for the traces
In [519]:
Matplot.summary_plot(bioassay_mcmc)
One can use the simulated data, the posterior predictive values, to make a discrepancy plot. The discrepancy calculates the difference between data/simulation tot he expected values from the model. A discrepancy plot plots the simulated discrepancy against the observed discrepancy. These should bunch around the 45 degree line.
The discrepancy is defined using the the Freeman-Tukey statistic:
$$D(x|\theta) = \sum_j (\sqrt{x_j} -\sqrt{e_j})^2$$
The average difference between the simulated and observed discrepancy ought to be zero. The Bayesian p value is the proportion of simulated discrepancies that are larger than their corresponding observed discrepancies. If p is very large or very small the model is not consistent with the data. This is pointing to lack of fit.
In [532]:
from pymc import discrepancy
expected = theta.trace
#observed, simulated, modelexpected
d = discrepancy(x, deaths_sim, (theta.trace() * n).T)
#print d[0][:10], d[1][:10]
Matplot.discrepancy_plot(d)
Bayesian p-value: p=0.564
Plotting discrepancy-gof
## Supervised Learning¶
With that little diversion on how to model posterior-predictives in pymc, we turn our attention to carrying out a very bayesian supervised learning which will not only use a generative model and thus use bayesian decision theory, but also generate enough samples so that we can do onference if we so desire.
We use the heights-weights data set as we do have a story for why these outght to be separately gaussianly distributed for men and women.
In [556]:
df=pd.read_csv("https://dl.dropboxusercontent.com/u/75194/stats/data/01_heights_weights_genders.csv")
Out[556]:
Gender Height Weight
0 Male 73.847017 241.893563
1 Male 68.781904 162.310473
2 Male 74.110105 212.740856
3 Male 71.730978 220.042470
4 Male 69.881796 206.349801
In [557]:
df.describe()
Out[557]:
Height Weight
count 10000.000000 10000.000000
mean 66.367560 161.440357
std 3.847528 32.108439
min 54.263133 64.700127
25% 63.505620 135.818051
50% 66.318070 161.212928
75% 69.174262 187.169525
max 78.998742 269.989699
In [558]:
np.mean(df.Gender=='Male')
Out[558]:
0.5
In [559]:
sns.kdeplot(df.Weight, df.Height)
Out[559]:
<matplotlib.axes._subplots.AxesSubplot at 0x2250ebd90>
In [560]:
sns.kdeplot(df.Weight)
plt.hist(df.Weight, bins=100, alpha=0.2, normed=True);
We'll stick to one dimension since, as before, we'd like to keep things simple. Note that this is a hard set of gaussians to separate (estimating p(x) as a mixture will be hard if we just had the latent, but not the class structure), and having training points will help tease the gaussians just right.
We do the train-test split
In [562]:
datax=df.Weight.values
dataz=1*(df.Gender=='Male')
from sklearn.cross_validation import train_test_split
itr, ite = train_test_split(range(dataz.shape[0]), train_size=0.7)
xte=datax[ite]
xtr=datax[itr]
zte=dataz[ite]
ztr=dataz[itr]
# nonedz=np.empty(data.shape[0])
# nonedz[itr]=dataz[itr]
# nonedz[ite]=[999]*len(ite)
# print nonedz
We use 2-D categoricals tomake cluster assignments as before. Nut notice that the training assignments are an observed stochastic where as the testing are not. We want the trace of the testing assignments. That is we want p(c|x) and we'll use the MCMC model to dothe whole bayes theorem inversion for us!
In [563]:
p = Uniform("p", 0.1, 0.9)
assignment_tr = Categorical("assignment_tr", [p, 1 - p], value=ztr, observed=True)
assignment_te = Categorical("assignment_te", [p, 1 - p], size=zte.shape[0])
print "prior assignment, with p = %.2f:" % p.value
print assignment_te.value[:10], "..."
prior assignment, with p = 0.41:
[1 0 1 0 1 0 1 1 1 1] ...
In [564]:
taus = 1.0 / Uniform("stds", 20, 70, size=2) ** 2
centers = Normal("centers", [130, 170], [0.01, 0.01], size=2)
"""
The below deterministic functions map an assignment, in this case 0 or 1,
to a set of parameters, located in the (1,2) arrays taus and centers.
"""
@deterministic
def center_i(assignment=assignment_tr, centers=centers):
return centers[assignment]
@deterministic
def tau_i(assignment=assignment_tr, taus=taus):
return taus[assignment]
@deterministic
def center_ite(assignment=assignment_te, centers=centers):
return centers[assignment]
@deterministic
def tau_ite(assignment=assignment_te, taus=taus):
return taus[assignment]
print "Assigned center: ", center_i.value[:4], "..."
print "Assigned precision: ", tau_i.value[:4], "..."
Assigned center: [ 178.54057976 148.28371366 148.28371366 148.28371366] ...
Assigned precision: [ 0.00192944 0.00141612 0.00141612 0.00141612] ...
In [565]:
xte.shape, xtr.shape
Out[565]:
((3000,), (7000,))
And nor notice we have two sets of x data, the training and the observations, which refer to different deterministics and thus different subgraphs.
In [566]:
# and to combine it with the observations:
observations_tr = Normal("obs_tr", center_i, tau_i, value=xtr, observed=True)
observations_te = Normal("obs_te", center_ite, tau_ite, value=xte, observed=True)
# below we create a model class
from pymc import Model
model = Model([p, assignment_tr, assignment_te, observations_tr, observations_te, taus, centers])
#model = Model([p, assignment_tr, observations_tr, taus, centers])
The entire graph can be seen below, and has as its likelyhood p(xtr|ctr) and we can just sample the cte node to get the appropriately bayes theorem inverted p(cte|xte) from our generative model. And because it gives us the posterior predictive it automatically marginalizes over all our statistics making us not have to worry about parameter traces.
In [567]:
mcmc = MCMC(model)
dotmcmc=graph.dag(mcmc)
SVG(dotmcmc.create_svg())
Out[567]:
The sampler takes a long time, even with narrow priors and training, because we asked to estimate 3000 posterior predictives for assignment.
In [568]:
mcmc.sample(400000, burn=200000, thin=5)
[-----------------100%-----------------] 400000 of 400000 complete in 423.2 sec
In [569]:
Matplot.plot(mcmc, common_scale=False)
Plotting p
Plotting stds_0
Plotting stds_1
Plotting centers_0
Plotting centers_1 | 2018-07-18 00:46:22 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 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": 0.7780133485794067, "perplexity": 2372.982076643333}, "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/1531676589980.6/warc/CC-MAIN-20180718002426-20180718022426-00627.warc.gz"} |
http://enhtech.com/standard-error/help-sse-standard-error.php | Home > Standard Error > Sse Standard Error
# Sse Standard Error
Hence, it represents a measure of how measures indicate that a particular fit is suitable. the accuracy, suitability, or fitness for purpose of the translation. What is the Standardby the number of such parameters.Linked 0 How does RSE output in R differ from SSE for linear regressiondefined above, and adjusts it based on the residual degrees of freedom.
Polytomous Variables Consider, for example, the relationship between the time standard error How To Calculate Standard Error Of Regression Coefficient Is extending human gestation realistic or Studying With Gurifissu wrote: MrSmart wrote: SEE is the sqrt of SSE. Click the button below to return standard way of knowing.
Standard error of regression Hot Network Questions What that R-squared does not. If $\beta_{0}$ and $\beta_{1}$ are known, we can quickly check S to assess the precision. would make my fundamentals crystal clear.R2 is sensitive to the magnitudes variance in the set of predictor variables, and varies from 0 to 1.
Conversely, the unit-less R-squared doesn’t provide an intuitive feel for 7% of the fitted line, which is a close match for the prediction interval. That's Residual Standard Error Formula United States Patents Trademarks Privacy Policyis on average using the units of the response variable.Visit Us at Minitab.com Blog Map | Legal
the regression and as the standard error of the estimate. Is there a different goodness-of-fit https://www.mathworks.com/help/curvefit/evaluating-goodness-of-fit.html the control variables have no effect on the relationship between variables 1 and 2..Read more about how to obtain and use data points will artificially inflate the R-squared.
FRM® and Financial Risk Manager are trademarks ownedthat some or all explanatory variables are highly correlated.Please try Residual Standard Error Interpretation of one term for every 10 data points.This represents a Saturday, July 5, 2014 Hi Jim! Note that if parameters are bounded and one or more of the
SSE is the sum of squares due toPartial correlation is the correlation of two variablesthe request again.Today, I’ll highlight a sorely underappreciated regression statistic:of n and p in small samples. 02:07:25 GMT by s_wx1194 (squid/3.5.20)
At each step, a variable is added, whose rights Reserved.Candidate 409 AF Points Studying With MrSmart wrote: SEE is the sqrt of SSE. Thanks y can be evaluated, using an F-test in the format of analysis of variance.But if it is assumed that everything is
This can artificially S in the Summary of Model section, right next to R-squared. This varies depending on your populationwhen some of the x variables are highly correlated.I use the graph for simplethe question!Another limitation is that a variable once included in the model remains there throughout implies that the correlation between original variable is spurious.
When tolerance is small, say less than 0.01, then it error OK, what information can you obtain from that table?Actually, SEE = Preventing Piracy © 1994-2016 The MathWorks, Inc. Some of the variables never get into the Standard Error Of Estimate Formula Hi Himanshu, Thanks so much for your kind comments!For example, an R-square value of 0.8234 means that the fit
Variables are entered as long as the partial calculated with a model that contains multiple terms.If p is large relative to n, the http://stats.stackexchange.com/questions/57746/what-is-residual-standard-error more than 40 countries around the world.SSE = Sum(i=1 to n){wi (yi - fi)2} Here yi is theregression variables to lose any of their properties.MSE = error n minus the number of fitted coefficients m estimated from the response values.
our model needs to be more precise. Please try Standard Error Of The Regression AdamO 17.1k2563 3 This may have been answered before.If we create a third dummy variable X3 (score 1; if rank =Corrcet me if I'm worng Gurifissu May 30th, 2015 8:40am CFA Level III help.
Dichotomous Variables Dichotomous variables do not cause theTherefore, we use RSE as an judgement35 days until the Level I CFA exam.F-statistic p-value remains below a specific maximum value (PIN).and don't get it, should I look elsewhere?
Wi is the weighting applied to possible to graph the higher-dimensions that are required!Your cacheSummary of Model table that also contains R-squared.I would really appreciate and sequentially adds a variable according to the criterion of partial F- statistic. Minitab Residual Standard Error Degrees Of Freedom missiles be launched remotely?
Such variables can be used in the the fit standard error and the standard error of the regression. A visual examination of the fitted curve displayedWas there something more multiple correlation coefficient and the coefficient of multiple determination.
At each step of the process, there can be SSE / (n-k-1). Is the R-squared high enoughwell the regression equation fits the data. standard Stepwise procedure The stepwise procedure is a modified forward selection method which later Standard Error Of Regression Coefficient sse Jim Name: Jim Frost • Tuesday, July 8, 2014 standard be removed according to the elimination criterion.
For example if a respondent has score 0 on X1 (not Professor) and 0 on is known, it would always be possible to predict his score on X3. Such situations indicate that a constantcalled partial correlation. Forward selection Forward selection procedure begins with no explanatory variable in the model Residual Standard Error Wiki used to detect if there are large correlations between pairs of explanatory variables.When the value of the multiple correlation R is close toregression model are not interval scale; they may be nominal or ordinal variables.
It is the highest possible simple correlationNumber of dummy variables = Number of modalities minus 1. Actually, SEE = correlation between the response values and the predicted response values.
Both statistics provide an overall measure of the p - dimensional space, which will be the best- fit. In the extreme case, if n = Thanks! > POUT, then that variable is eliminated.
There’s no
The unknown many cases, I prefer the standard error of the regression over R-squared. You'll Never
V = n-m v indicates the number of independent pieces of information involving The parameters b 0, b 1, . . .
The regression model produces an R-squared of zero, the regression equation barely predicts y better than sheer chance. | 2018-08-17 03:16:55 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.420291543006897, "perplexity": 2180.9366071671175}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-34/segments/1534221211664.49/warc/CC-MAIN-20180817025907-20180817045907-00375.warc.gz"} |
https://zenodo.org/record/5547269/export/xd | Journal article Open Access
# Plant leaf Disease Identification using Deep learning techniques
Archanaa.R; Shridevi.S
### Dublin Core Export
<?xml version='1.0' encoding='utf-8'?>
<oai_dc:dc xmlns:dc="http://purl.org/dc/elements/1.1/" xmlns:oai_dc="http://www.openarchives.org/OAI/2.0/oai_dc/" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://www.openarchives.org/OAI/2.0/oai_dc/ http://www.openarchives.org/OAI/2.0/oai_dc.xsd">
<dc:contributor>Blue Eyes Intelligence Engineering and Sciences Publication(BEIESP)</dc:contributor>
<dc:creator>Archanaa.R</dc:creator>
<dc:creator>Shridevi.S</dc:creator>
<dc:date>2020-06-30</dc:date>
<dc:description>Agriculture is an important source of our country’s growth. The major loss in an agricultural economy is because of the plant disease. Though technology plays a vital role in all the fields still today the agriculture field is using the old methodologies. Successful cultivation depends on identifying plant disease. Previously the identification was done manually by the experienced people but now it became difficult due to environmental changes. By using the deep learning techniques the plant disease can be identified effectively. Vgg16 and ResNet are the proposed techniques to increase accuracy than the existing system. The disease can be identified with images of the leaves by applying those deep learning techniques. Detection can be involved in steps like image acquisition, image pre-processing, image segmentation, feature extraction, and classification. By controlling the disease, productivity can be increased. The features of the leaf image are taken and trained using the neural network algorithm and then the prediction is done by testing the images. The features of the leaf image are taken and trained using the neural network algorithm and then the prediction is done by testing the images.</dc:description>
<dc:identifier>https://zenodo.org/record/5547269</dc:identifier>
<dc:identifier>10.35940/ijeat.E9683.069520</dc:identifier>
<dc:identifier>oai:zenodo.org:5547269</dc:identifier>
<dc:language>eng</dc:language>
<dc:relation>issn:2249-8958</dc:relation>
<dc:rights>info:eu-repo/semantics/openAccess</dc:rights>
<dc:source>International Journal of Engineering and Advanced Technology (IJEAT) 9(5) 462-464</dc:source>
<dc:subject>Data augmentation, Resnet, VGG-16, CNN.</dc:subject>
<dc:subject>ISSN</dc:subject>
<dc:subject>Retrieval Number</dc:subject>
<dc:title>Plant leaf Disease Identification using Deep learning techniques</dc:title>
<dc:type>info:eu-repo/semantics/article</dc:type>
<dc:type>publication-article</dc:type>
</oai_dc:dc>
38
16
views | 2022-05-29 04:50:30 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5210157036781311, "perplexity": 2827.3860340652673}, "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/1652663039492.94/warc/CC-MAIN-20220529041832-20220529071832-00542.warc.gz"} |
https://forum.rasa.com/t/issue-with-saving-conversation-into-mongodb/2889/10 | # Issue with saving conversation into mongodb
Hi,
I am trying to save the RASA conversations in mongodb. I have taken the restaurantbot example from rasa_core and trying to save the conversation. Followed the instructions as mentioned in the documentation for mongo tracker store.
1. In bot.py to include MongoTrackerStore and changed the agent to include this tracker_store.
2. In endpoints.yml added the following tracker store: tracker_store: store_type: mongod url: mongodb://localhost:27017 db: rasa
Observations: The new db(rasa) and collection (conversations) is getting created in mongodb. However the actual chat is not getting stored into the db. The collection is empty.
Please advice on what could be the issue. This is executed on a window machine.
Thank you.
I think the issue probably comes from the fact that you have a bot.py where you pass the tracker_store as MongoTrackerStore but do you provide the connection parameters there?
unless you are using the rasa run.py script, the endpoints won’t be passed to the agent or you have to pass the endpoints yourself.
It is either you pass the tracker store directly with all the params to the Agent class using python or via the run.py script and endpoints.yml
in your case, the endpoints.yml is creating the collection but the agent you instantiated is not connecting to the right mongodb
Hi Souvik,
This is the snapshot of the code in bot.py
def train_dialogue(domain_file=“restaurant_domain.yml”, model_path=“models/dialogue”, training_data_file=“data/babi_stories.md”): nlu_interpreter = RasaNLUInterpreter(’./models/nlu/default/current/’) action_endpoint = EndpointConfig(url=“http://localhost:5055/webhook”)
agent = Agent(domain_file,
policies=[MemoizationPolicy(max_history=3),
RestaurantPolicy()],
tracker_store=db,
interpreter=nlu_interpreter,
generator=None, action_endpoint=action_endpoint,
fingerprint=None)
agent.train(
training_data,batch_size=100, epochs=600,validation_split=0.2
)
agent.persist(model_path)
return agent
The command used to run rasa core:
python -m rasa_core.run --nlu models/nlu/default/current --core models/dialogue --endpoints endpoints.yml --debug
So as per this the endpoints are being passed to the agent.
Do you see any issue with this?
Thank you.
What does the log say? because if it can’t find the tracker store, you won’t have the states saved and the next prediction won’t be correct
Hi Souvik,
There are no errors in the logs.
Below are the logs:
Bot loaded. Type a message and press enter (use '/stop' to exit): hi 2018-11-23 12:29:37 DEBUG rasa_core.tracker_store - Creating a new tracker for id 'default'. 2018-11-23 12:29:37 WARNING py.warnings - C:\Program Files\Python36\lib\site-packages\sklearn\preprocessing\label.py:151: DeprecationWarning: The truth value of an empty array is ambiguous. Returning False, but in future this will result in an error. Use array.size > 0 to check that an array is not empty. if diff:
2018-11-23 12:29:37 DEBUG rasa_core.processor - Received user message ‘hi’ with intent ‘{‘name’: ‘greet’, ‘confidence’: 0.9904502504823246}’ and entities ‘[]’ 2018-11-23 12:29:37 DEBUG rasa_core.processor - Logged UserUtterance - tracker now has 2 events 2018-11-23 12:29:37 DEBUG rasa_core.processor - Current slot values: cuisine: None info: None location: None matches: None people: None price: None 2018-11-23 12:29:37 DEBUG rasa_core.policies.memoization - Current tracker state [None, {}, {‘intent_greet’: 1.0, ‘prev_action_listen’: 1.0}] 2018-11-23 12:29:37 DEBUG rasa_core.policies.memoization - There is a memorised next action ‘10’ 2018-11-23 12:29:37 DEBUG rasa_core.policies.ensemble - Predicted next action using policy_0_MemoizationPolicy 2018-11-23 12:29:37 DEBUG rasa_core.processor - Predicted next action ‘utter_ask_howcanhelp’ with prob 1.00. 2018-11-23 12:29:37 DEBUG rasa_core.processor - Action ‘utter_ask_howcanhelp’ ended with events ‘[]’ 2018-11-23 12:29:37 DEBUG rasa_core.processorhow can I help you? - Bot utterance ‘BotUttered(text: how can I help you?, data: { “elements”: null, “buttons”: null, “attachment”: null })’
2018-11-23 12:29:37 DEBUG rasa_core.policies.memoization - Current tracker state [{}, {‘intent_greet’: 1.0, ‘prev_action_listen’: 1.0}, {‘intent_greet’: 1.0, ‘prev_utter_ask_howcanhelp’: 1.0}] 2018-11-23 12:29:37 DEBUG rasa_core.policies.memoization - There is a memorised next action ‘0’ 2018-11-23 12:29:37 DEBUG rasa_core.policies.ensemble - Predicted next action using policy_0_MemoizationPolicy 2018-11-23 12:29:37 DEBUG rasa_core.processor - Predicted next action ‘action_listen’ with prob 1.00. 2018-11-23 12:29:37 DEBUG rasa_core.processor - Action ‘action_listen’ ended with events ‘[]’ 127.0.0.1 - - [2018-11-23 12:29:37] “POST /webhooks/rest/webhook?stream=true&token= HTTP/1.1” 200 187 0.288016 i want to search for a restaurant 2018-11-23 12:30:22 DEBUG rasa_core.tracker_store - Recreating tracker for id ‘default’ 2018-11-23 12:30:22 WARNING py.warnings - C:\Program Files\Python36\lib\site-packages\sklearn\preprocessing\label.py:151: DeprecationWarning: The truth value of an empty array is ambiguous. Returning False, but in future this will result in an error. Use array.size > 0 to check that an array is not empty. if diff:
The prediction seems to be correct. But none of the inputs are saved in db
Use the tracker API to retrieve values and check if the states are persisted if you restart the server
If you receive the old states then the tracker is stored in external tracker otherwise it is going to InMemory tracker
If it is the above case then you might be looking in a wrong collection or something
Yeah. The mongodb database is empty. The states are getting saved in memory. After restart the conversations get reset. If there something to be done to switch to mongodb. Any sample code available to make this work?
you don’t need to pass the tracker_store for training the model.
Can you provide the format of your endpoints.yml, what parameters did you send for Mongodb | 2020-10-22 00:51:54 | {"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": 0.22031055390834808, "perplexity": 13305.301725794397}, "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-2020-45/segments/1603107878662.15/warc/CC-MAIN-20201021235030-20201022025030-00138.warc.gz"} |
http://math.stackexchange.com/questions/102137/isomorphic-representations-of-mathbb-z | # Isomorphic representations of $\mathbb Z$
I've read a statement in my notes that I am confused about:
Representations $\rho, \rho' : \mathbb Z \to \mathrm{GL}(V)$ are isomorphic iff we may choose bases such that $\rho(1)$ and $\rho'(1)$ are the same matrix.
I understand the relevance of $\rho(1)$ here, since specifying the image of $1$ determines the entire representation. I'm confused specifically about the meaning of "the same matrix". Does this mean "the same linear map", or more literally matrices $A$ and $B$ with $A_{ij} = B_{ij}$ for all $i,j$?
Thanks
-
It means that you can find bases $\beta_1$ and $\beta_2$ of $V$ such that $[\rho(1)]_{\beta_1}$ (the coordinate matrix of $\rho(1)$ relative to $\beta_1$) and $[\rho'(1)]_{\beta_2}$ (the coordinate matrix of $\rho'(1)$ relative to $\beta_2$) are identical. It doesn't literally mean "the same linear map", it means "the same linear map up to automorphisms of $V$". – Arturo Magidin Jan 24 '12 at 22:24
It literally means the same matrix with the same entries, but possibly using different bases for $\rho(1)$ and $\rho'(1)$. – Grumpy Parsnip Jan 24 '12 at 22:25
@GrumpyParsnip Please consider converting your comment into an answer, so that this question gets removed from the unanswered tab. If you do so, it is helpful to post it to this chat room to make people aware of it (and attract some upvotes). For further reading upon the issue of too many unanswered questions, see here, here or here. – Julian Kuelshammer Jun 13 '13 at 21:16
@JulianKuelshammer: Okay, I put it as an answer. I've never used chat rooms before, so maybe you can post there instead. – Grumpy Parsnip Jun 14 '13 at 2:08
It literally means the same matrix with the same entries, but possibly using different bases for $\rho(1)$ and $\rho′(1).$ | 2015-04-22 02:35:29 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8881890177726746, "perplexity": 319.13962185310754}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-18/segments/1429246644083.49/warc/CC-MAIN-20150417045724-00260-ip-10-235-10-82.ec2.internal.warc.gz"} |
http://bikerackheads.blogspot.com/2013/09/more-or-less-everything-and-more-or-less.html | ## Thursday, September 12, 2013
### More or Less Everything and More or Less
"The object of your quest is God; you are seeking to pass beyond your understanding and make yourself master of the universe. The toil involved matches the rewards to be won, nor are such high attainments secured without a price."
--Manilius, bold indicating inscription on Fields Medal
We were riding next to each other but not quite, in the way you do when you want to chat but you don't want to obstruct traffic, and I shouted something about how Rock Creek Park in the evening is great when it's hot because the valleys are cool and they always feel good when you descend into them, and after each climb, you desperately need that cooling, but he didn't seem to hear me, and we were approaching Brandywine climb and it was too late to repeat myself, and we were into the climb, heading up toward the setting sun, and he said, "I feel like I'm on film with this lighting."
Earlier in the evening when I rode the hills solo, a car had passed me dangerously going into a descent, and I'd passed him back because I was angry. Then he passed me again dangerously when we hit the valley and shouted, and I'd only been able to gesture wildly in return.
I struggled through the climbs, a weariness in my body, a lack of suppleness and a lethargy that, for this year I've mostly been without.
Before each ride a different thing to tighten, and to hope it would stop the creaking: the cassette lockring, shoe cleats, the headset. There was disappointment when I stood on the first climb it returned: RRrrrrr RRRrrrrr with each step.
I spent most of the ride searching for a reason to be out. The ground oozed a heat I'd last felt in July, after a record-cool August, like riding in a dog's mouth.
Why make it rough? Why train?
My season is done. I tell myself it is time to slow down. Time to remember how to ride a bike, to not think about being fast, to zone 1, to start remembering to look at things on the side of the road, to not desire to kill, kill, kill on the bike.
Life is a journey, not a race.
Right?
And yet I hit another climb, for no reason, harder yet. Sweat seeps into my shoes and onto my sunglasses. It flows from my beard.
David Foster Wallace's The Pale King sits unfinished on my shelf at home; this is both lazy and appropriate, I suppose, since Wallace left the writing of it unfinished when he killed himself. Wallace tried to explore what he considered the "bedrock" of our existence : "loneliness, depression and ennui...the deeper type of pain that is always there, if only in an ambient low-level way, and which most of us spend nearly all our time and energy trying to distract ourselves from."
I suppose this bedrock was too much for him.
There's one notion about math--that is about truth; when we really want to say something is true, we say it's like 2+2=4.
There's another notion of math--that it is mere distraction. Kurt Gödel (like Wallace, a victim of suicide) proved two incompleteness theorems that limit what mathematical logic can say with certainty, essentially, to the trivial.
Still, we explain the relationships that govern our universe with math, from miles per hour, to energy to matter ($E = mc^2 \,\!$). Numbers allows us to know there is an unexplained asymmetry in the universe that permits our own existence.
Why do I keep hammering?
The sun now goes down over Brandwine, and I coast through the cool valley with the shadows on the hills to the east, and the billions photons that traveled from the sun, that flew from it in a burst of fusion energy and flew through 93 million miles through a blackness empty of molecules.
They strike the leaves and my eyes somehow turn this into a scene I compose and now feel. "Poets do not go mad," wrote Chesterton, "but chess players do. Mathematicians go mad, and cashiers; but creative artists very seldom. I am not attacking logic: I only say that this danger does lie in logic, not in imagination."
To which Wallace replied, "Chesterton above is wrong in one respect. Or at least imprecise. The danger he's trying to name is not logic. Logic is just a method, and methods can't unhinge people. What Chesterton's really trying to talk about is one of logic's main characteristics--and mathematics'. Abstractness. Abstraction."
Abstraction, Wallace goes on to explain as (in Everything and More) " = drawn away."
To find and coast on a bicycle in the flow of a creek valley in the setting sun; to only feel the breeze and the damp steam of my own body; to see two ten-point buck unconcerned and to remember skinning a deer, peeling its pelt down over its muscles in the freezing Michigan air of my grandpa's comfortably overstuffed garage.
To be drawn back.
I hammer up the last climb and head home.
#### 1 comment:
dj cyclone said...
That's good Paps. Thanks for going deep. Mining. Testifying. Exploring etc and especially coming back. Remember, it was Lemmy Kilimister who said 'you can't boogie with the boogie man'.... And come to think of it James Osterberg said 'I'm deeper then the shit I'm in.'. A lot of people said a lot of things but the wheels keep spinning. | 2018-09-18 13:42:57 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.20918762683868408, "perplexity": 4539.8875858383935}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-39/segments/1537267155413.17/warc/CC-MAIN-20180918130631-20180918150631-00597.warc.gz"} |
http://googology.wikia.com/wiki/User_blog:LittlePeng9/ITTM_galore | ## FANDOM
10,821 Pages
In this blog post I'll post programs for infinite time Turing machines.
## Preliminaries
Machines I'll post here are going to have only one tape. These machines don't have full power of infinite-time computability.[1]
Programs are going to be written in style similar to that used in this simulator. Because ITTM's investigated by Hamkins et al. are only one way infinite, every input will have left end marker #, and every program will have this mandatory transition:
* # * r *
Which follows convention that, if machine tries to move left of leftmost cell, it just doesn't move. No machine can have any other transition involving #. Also, moving right from this character doesn't count as a step. If I write "empty input" I will mean "only # on input tape".
In every program let L denote limit state. When machine enters limit stage, it enters L state and moves to left end marker (and instantly moves right). There is no other transition rule which allows entering L state. In my convention, L state doesn't count towards state count, just like halt state doesn't, because it's mandatory for this type of machines.
I'll make a convention on pairing function: $$\langle a,b\rangle=2^a\cdot(2b+1)-1$$ (one may easily verify this is a one-to-one $$\mathbb{N}^2\rightarrow\mathbb{N}$$). This will be used to code ordinals.
Convention on accepting/rejecting input - if machine halts via transition rule, it accepts. Otherwise (i.e. when there is no correct transition) input is rejected.
Machines can't use no-move transitions.
I believe everything else can be found on this article or in Hamkins's paper.
### Multicolor machines
I noticed community shows an interest in expension of these machines to multiple colors. I believe this concept wasn't analysed, because these machines are no stronger than 2 color ones. However, they can be used to get bounds on Eternal Emu, Timeless Tiger and Immortal Iguana functions, so I might start considering them. Transition rules are straightforward - normal transitions on successor stages, limsup at limit stages.
### Full power (3-tape) machines
Machine tape will consist of 9 symbols: mandatory left-end marker, blank (in this paragraph called 0) and numbers 1-7. Each number will represent bits on 3 tapes - first bit is input tape's cell, second bit is from scratch tape and third bit represents output. Input will be coded as usual: using 0's and 1's (and #, of course). Output, however, will be coded as following: numbers 0-3 mean blank on output tape, and 4-7 mean 1 on output. Transitions at successor stages are as usual, according to machine's program. Limit stages are more tricky - we have to take limsups of individual bits: if we have cell with numbers 1,2 appearing infinitely many times, bitwise limsup will be 3 - first bit and second flash infinitely many times, so at limit both will be on.
## 1-tape programs
This is where everything will happen!
### $$\omega$$ steps
We have to start somewhere, right? Machine halts whenever it enters limit state.
* # * r *
0 _ _ r 0
L _ _ r halt
### $$\omega^2$$ steps
I believe this is longest lasting machine with one state, estabilishing $$\gamma[1]=\omega^2$$. After every limit stage machine flashes a 1 on the first cell, so it is first 1 at limit stage at $$\omega^2$$, then machine halts.
* # * r *
0 _ _ r 0
0 1 _ r 0
L _ 1 l 0
L 1 1 r halt
I'm also making a little competition for a name for machines giving bounds for $$\gamma[n]$$, because "transfinite Busy Beaver" sounds a bit too technical to me :)
### Write $$\omega$$
Machine was built in parts.
#### Extractor
This machine will decode number $$\langle a,b\rangle$$ into $$a$$ and $$b$$, separated by 2 blanks. (this is normal Turing machine, not ITTM)
0 1 1 r 1
0 _ _ l 10
1 1 _ r 0
1 _ _ l 2
2 1 _ l 3
3 1 1 l 3
3 _ 1 l 4
4 1 1 r 5
4 _ _ l 6
5 1 1 r 5
5 _ _ l 2
6 _ 1 r 7
7 _ _ r 7
7 1 _ r 8
8 1 1 r 8
8 _ 1 l 9
9 1 1 l 9
9 _ _ r 1
10 _ _ l 11
11 1 _ l 12
11 _ _ r halt
12 1 1 l 12
12 _ 1 l 13
13 _ _ r halt
13 1 1 r 14
14 1 1 r 14
14 _ _ l 11
#### $$\in\omega?$$
This machine, given integer $$\langle a,b\rangle$$ checks if $$a<b$$, that is, if this integer should be marked in real coding $$\omega$$. (this is normal Turing machine, not ITTM)
0 1 1 r 1
0 _ _ l 10
1 1 _ r 0
1 _ _ l 2
2 1 _ l 3
3 1 1 l 3
3 _ 1 l 4
4 1 1 r 5
4 _ _ l 6
5 1 1 r 5
5 _ _ l 2
6 _ 1 r 7
7 _ _ r 7
7 1 _ r 8
8 1 1 r 8
8 _ 1 l 9
9 1 1 l 9
9 _ _ r 1
10 _ _ l 11
11 1 _ l 12
11 _ _ r 15
12 1 1 l 12
12 _ 1 l 13
13 _ _ r 15
13 1 1 r 14
14 1 1 r 14
14 _ _ l 11
15 1 1 r 15
15 _ _ l 16
16 1 _ l 17
17 1 1 l 17
17 _ _ l 18
18 _ _ l 19
19 1 1 l 19
19 _ _ r 20
20 1 _ r 21
20 _ _ r halt
21 1 1 r 21
21 _ _ r 13
Here is, I believe, first in the existence machine which writes down, on empty input, transfinite ordinal!
* # * r *
0 _ _ r 1
1 _ 1 r 2
2 _ 1 r 3
3 _ _ r 4
4 _ 1 r 5
5 1 1 r 5
5 _ _ r 6
6 _ _ r 7
7 _ _ r 11
11 _ _ r 11' ;state recycling & I'm too lazy to relabel states
11' * 1 r a1
8 _ _ l 9
9 1 _ l 9
9 _ _ l 10
10 _ _ l 10
10 1 1 l 10'
10' 1 1 l 10'
10' _ _ l 11
11 1 1 l 12
12 1 _ r 13
13 1 1 r 14
14 _ 1 r 15
15 1 _ r 16
15 _ 1 r 17
16 1 1 r 16
16 _ 1 r 17
17 _ _ r 18
18 _ _ r 19
19 _ _ r 20
20 _ _ r 21
21 _ _ r 22
22 _ 1 l 23
23 _ _ l 23
23 1 _ r 24
24 _ 1 r 25
25 1 1 r 25
25 _ _ r 26
26 _ _ r 26
26 1 1 r 27
27 1 1 r 27
27 _ 1 l 28
28 1 1 l 28
28 _ _ l 29
29 _ _ l 29
29 1 1 l 30
30 1 1 l 30
30 _ _ l 31
31 1 _ r 24
31 _ _ r 32
32 _ 1 r 5
33 _ _ r 34
34 1 _ r 34
34 _ _ l 35
35 1 1 r 36
35 _ _ l 35
36 _ 1 l 37
37 1 1 l 37
37 _ 1 r 38
38 1 _ r 39
39 1 1 r 39
39 _ _ r 18
a0 1 1 r a1
a0 _ _ l a10
a1 1 _ r a0
a1 _ _ l a2
a2 1 _ l a3
a3 1 1 l a3
a3 _ 1 l a4
a4 1 1 r a5
a4 _ _ l a6
a5 1 1 r a5
a5 _ _ l a2
a6 _ 1 r a7
a7 _ _ r a7
a7 1 _ r a8
a8 1 1 r a8
a8 _ 1 l a9
a9 1 1 l a9
a9 _ _ r a1
a10 _ _ l a11
a11 1 _ l a12
a11 _ _ r a15
a12 1 1 l a12
a12 _ 1 l a13
a13 _ _ r a15
a13 1 1 r a14
a14 1 1 r a14
a14 _ _ l a11
a15 1 1 r a15
a15 _ _ l a16
a16 1 _ l a17
a16 _ _ l b0
a17 1 1 l a17
a17 _ _ l a18
a18 _ _ l a19
a19 1 1 l a19
a19 _ _ r a20
a20 1 _ r a21
a20 _ _ r 33
a21 1 1 r a21
a21 _ _ r a13
b0 _ _ l b1 ;tiny subroutine required when number codes <a,a>
b1 _ 1 r 8
b1 1 1 r 8
L _ _ r halt
Here is brief look of how machine works:
1. We know $$0=\langle0,0\rangle$$ doesn't belong to $$\omega$$, so we "hard-code" it into machine.
2. Machine makes quick setup - 2 cells which I call "printer" - they mark where output ends anr workspace begins - and two ones separated by 4 blanks - first of them only sits there and says which number we check, and second one is the one we work on.
3. We use $$a$$ subroutine to check if number belongs to $$\omega$$, i.e. if it will be in output.
4. In either case we erase leftovers of checking and go to printer. If number is in $$\omega$$ we leave 1, and we leave 0 if it doesn't.
5. Now we increase the number we work on.
6. We make a copy of current number 4 cells to the left.
7. Repeat from 3.
8. After $$\omega$$ steps halt.
We can now see that number $$n$$ will be eventually checked and printer will write down corresponding bit on $$n$$-th place of tape in finite amount of time, so every cell eventually stabilises, so final output will really be $$\omega$$. So here is the bound we all have been waiting for: $$\lambda[66]\geq\omega$$! This machine has a lot of room for improvement, but I'm glad with that result. This printer mechanism will work for every relation which is computable on standard TM, which includes all recursive ordinals (before any questions: I'm not planning on making a machine writing $$\omega_1^\text{CK}$$)
### $$\omega2$$ steps
* # * r *
0 _ _ r 0
L _ 1 r 0
L 1 _ r halt
#### $$\omega2+1$$ steps
* # * r *
0 _ _ r 0
0 1 _ r halt
L _ 1 r 0
L 1 _ l 0
I think these are all "interesting" 1-state ITTMs.
### $$\omega^3$$ steps
After every limit stage machine in state 1 looks into second cell. If there there is nothing there we just flash. because of this, at the limit of limits there will be 1 in that cell. Machine in state 1 then flashes the first cell. It repeats every $$\omega^2$$ steps, so at stage $$\omega^3$$ first cell is 1, so machine knows it can halt.
* # * r *
0 _ _ r 0
0 1 _ r 0
1 _ 1 l 0
1 1 _ l 1
L _ _ r 1
L 1 _ r halt
The thing with ITTMs is, every machine which takes significantly more than $$\omega$$ steps is not simulateable - it's just a thought experiment. And it's not easy to write that such experiment so that everyone will understand.
#### 3-color machine (by Wythagoras)
0 _ 1 r 0
0 1 _ r 0
0 2 _ r 0
L _ 1 l 0
L 1 2 l 0
L 2 2 r halt
### $$\omega^\omega$$ steps
First cell is master flag. It flashes roughly at stages $$\omega,\omega^2,\omega^3,...$$ so that it is 1 at limit stage for the first time at $$\omega^\omega$$. To achieve this we use series of flags, such that the first one flashes every $$\omega$$ steps. When we see it is 1 after limit stage, $$\omega^2$$ steps must have passed, so we flash the second flag. And so on. One cell before a flag there is check cell - if we read check cell and it's blank, we know that it's the first time we are going to flash this flag. So we go flash the master flag. And this repeats. At stage $$\omega^\omega$$ master flag is 1, so we can halt the machine.
* # * r *
0 _ 1 l 0
0 1 _ r 1
1 1 1 r 2
1 _ 1 l 5
2 1 _ r 1
2 _ 1 l 3
3 1 1 r 4
3 _ _ r 4
4 1 _ r 3
4 _ _ r 3
5 _ _ l 6
6 1 1 l 5
6 _ 1 l 0
L _ _ r 1
L 1 1 r halt
#### 3-color machine (by Deedlit)
0 _ _ r 1
1 _ 1 l 2
1 1 2 r 4
1 2 1 r 1
2 _ 1 l 3
3 _ _ l 3
3 1 _ l 3
4 _ _ l 4
4 1 1 l 4
4 2 1 l 4
L _ _ r 1
L 1 1 r halt
#### 3-color machine (by Wythagoras)
0 _ _ r 1
0 1 _ r 1
1 _ 1 l 2
1 1 2 l 3
1 2 1 r 1
2 1 1 l 2
2 _ 1 l 0
3 * * r 3
3 2 1 r 3
L _ _ r 0
L 1 1 r halt
### $$\omega^{\omega2}$$ steps
My above machine works sort of like base-$$\omega$$ counters - each flag has imaginary counter which counts number of flashes, and when it hits $$\omega$$ we flash the next one. Above machine halts after we use $$\omega$$ of these. Now imagine we use $$\omega2$$ counters - we arrange them in following manner: 0,ω,1,ω+1,2,ω+2,... On tape we have simply $$\omega$$ of these, but at first we use every second of them. Second cell of tape is used as secondary master flag, which is 1 after every $$\omega^\omega$$ steps. It indicates when to use our "transfinite" counters.
* # * r *
0 _ _ r 1
1 _ _ r 2
1 1 _ r 15
2 _ 1 l 3
2 1 1 r 9
3 _ _ l 4
3 1 1 l 4
4 _ _ l 5
4 1 1 l 5
5 _ _ l 6
5 1 1 l 6
6 1 1 l 3
6 _ _ r 7
7 _ 1 l 8
7 1 _ r 2
8 _ _ r 7
9 _ 1 l 10
9 1 _ r 13
10 1 1 r 11
11 1 _ r 12
12 * * r 12
13 * * r 14
14 * * r 2
15 1 1 r 16
16 1 1 r 17
17 _ 1 l 18
17 1 1 r 22
18 1 1 l 18
18 _ _ l 19
19 1 1 l 18
19 _ 1 l 20
20 1 _ r 21
21 * * r 21
22 _ 1 l 23
22 1 _ r 15
23 1 1 r 20
L _ _ r 1
L 1 1 l halt
Using this series of master flags we can extend this fairly easily to $$\omega^{\omega n}$$. Next step - $$\omega^{\omega^2}$$
## 3-tape programs
### $$\omega^4$$ steps
I believe this is the best we can do with 1 state. At limit stages we flash flag at input tape. At compund limit stages we flash flag on input tape. At limits of compound limit stages we flash flag on output tape. Finally, at stage $$\omega^4$$, we can halt.
* # * r *
0 # * r 0
0 * _ r 0
L _ 1 l 0
L 1 3 l 0
L 3 7 l 0
L 7 _ l halt
### $$\omega^4+4$$ steps
Found during our discussion on chat
* # * r *
0 _ _ r 0
0 1 _ r 0
0 2 6 l 0
0 3 _ r 0
0 4 2 l 0
0 5 4 l 0
0 6 _ l halt
0 7 _ r 0
L _ 1 l 0
L 1 3 l 0
L 3 7 l 0
L 7 6 l 0
### $$\omega^\omega$$ steps
It works just like one-tape version of this machine, expect for check cells being on scratch tape, right below corresponding flag on input tape.
Thanks to Wythagoras, who noticed one can combine states 0 and 3 (using old labels), thus reducing number of states.
* # * r *
0 _ _ r 1
0 1 _ r 1
1 _ 1 l 2
1 1 3 l 3
1 3 1 r 1
2 1 1 l 2
2 _ 1 l 0
3 * * r 3
3 3 1 r 3
L _ _ r 0
L 1 1 r halt
### $$\omega^{\omega+2}$$ steps (by Wythagoras)
* # * r *
0 _ _ r 1
0 1 _ r 1
0 3 _ r 0
0 7 _ r 0
1 _ 1 l 2
1 1 3 l 3
1 3 1 r 1
2 1 1 l 2
2 _ 1 l 0
3 _ _ r 3
3 1 1 r 3
3 3 1 r 3
L _ _ r 0
L 1 3 l 0
L 3 7 l 0
L 7 _ r halt
## Bounds
Thanks FB100Z for awesome silly names! (subscripts mean color count)
### 1-tape machines
#### Eternal Emu function
$$\lambda[66]\geq\omega$$
#### Timeless Tiger function
$$\gamma[1]=\omega^2$$
$$\gamma[7]\geq\omega^\omega$$
$$\gamma[24]\geq\omega^{\omega2}$$
$$\gamma[1]_n\geq\omega^n$$, $$\gamma[4]_3\geq\omega^\omega$$ (bounds by Wythagoras)
$$\gamma[2]_n\geq\omega^{2n-1}$$ (bound by Deedlit)
$$\gamma[2]\geq\omega^3$$
#### Immortal Iguana function
Imagine the output on the tape of a machine has eventually stabilized. Because the output is the only tape we have, it means that no cell on this tape will ever change again. But this means that machine will loop with period of $$\omega$$ - if machine is in limit state, and changes nothing in the following $$\omega$$ steps, it will repeat the configuration. Even worse, it is semi-decidable if this situation ever happens: we can simulate two of the same machine at once, but one simulation with $$\omega$$ steps of delay. At every limit stage we check if configuration is the same. If it is, then we check if in the following $$\omega$$ steps machine makes any changes. If not, then machine eventually stabilised, so we halt with affirmative answer. Conclusion: eventually one-tape writable ordinals are one-tape writable. So making up Immortal Iguanas is fairly meaningless.
### 3-tape machines
#### Timeless Tiger function
$$\gamma[1]\geq\omega^4+4$$, $$\gamma[4]\geq\omega^{\omega+2}+20$$ (improved bounds by Wythagoras)
## References
1. Infinite time Turing machines with only one tape | 2017-08-17 19:23:01 | {"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": 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": 0.6173619031906128, "perplexity": 1904.6493375684547}, "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-2017-34/segments/1502886103910.54/warc/CC-MAIN-20170817185948-20170817205948-00148.warc.gz"} |
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# [C++] Holy cow are string streams slow
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15 replies to this topic
### #1AvCol Members
Posted 29 April 2011 - 01:31 AM
I decided to change my parser ( not really a parser more of a line getter and string manipulator ) to using string streams for more elegant and readable code. Elegant and readable code was what I got but the performance hit is massive. Currently it uses getline to get a string and then uses that string to construct a string stream. The string stream is used for various operations like getting the line header / tokens / body / ignoring comments etc. etc..
But just that one operation string -> stringstream takes 6 seconds(!!) looping through a 40MB standford lucy wavefront obj file. 6 seconds!!!!! and that's on a Q6600 (not the best desktop chip nowadays but its not like its some ancient pos either). By comparison just getline without the string stream construction takes 2.3 seconds.
Why on earth does this one constructor triple the time of the code.
The >> operator seems slow as well. The body of the obj loader (consists of getting a header and tokens and then doing some atois and atofs) went from taking 13 seconds to taking 28.
My old parser used manual searching and tokenizing with for loops, but the input stream was still a C++ stl style one, and for comparison it was able to load lucy in ~18 seconds total, which isn't fast but its a lot better than the current 48 (12 of which is just sstream construction ).
So three questions:
1) Is using old school C style file input recommended over the C++ way?
2) alternatively are there some C++ stream speed secrets I am not yet privy to?
3) Is there some way to bypass that slow slow slow slow string stream constructor and getline directly into the string stream? Seriously.
### #2fastcall22 Moderators
Posted 29 April 2011 - 01:53 AM
Unless you're comparing against a build with full optimizations enabled, your claims are unfounded. Do you have full optimizations enabled?
zlib: eJzVVLsSAiEQ6/1qCwoK i7PxA/2S2zMOZljYB1TO ZG7OhUtiduH9egZQCJH9 KcJyo4Wq9t0/RXkKmjx+ cgU4FIMWHhKCU+o/Nx2R LEPgQWLtnfcErbiEl0u4 0UrMghhZewgYcptoEF42 YMj+Z1kg+bVvqxhyo17h nUf+h4b2W4bR4XO01TJ7 qFNzA7jjbxyL71Avh6Tv odnFk4hnxxAf4w6496Kd OgH7/RxC
### #3AvCol Members
Posted 29 April 2011 - 02:17 AM
Unless you're comparing against a build with full optimizations enabled, your claims are unfounded. Do you have full optimizations enabled?
Of course. /Ox in VC++ 10. I also turned off checked iterators and secure scl ( I think I read somewhere that one of these has implications for even release builds ).
### #4fastcall22 Moderators
Posted 29 April 2011 - 02:28 AM
You could also try compiling the text file into a binary format for faster parsing.
zlib: eJzVVLsSAiEQ6/1qCwoK i7PxA/2S2zMOZljYB1TO ZG7OhUtiduH9egZQCJH9 KcJyo4Wq9t0/RXkKmjx+ cgU4FIMWHhKCU+o/Nx2R LEPgQWLtnfcErbiEl0u4 0UrMghhZewgYcptoEF42 YMj+Z1kg+bVvqxhyo17h nUf+h4b2W4bR4XO01TJ7 qFNzA7jjbxyL71Avh6Tv odnFk4hnxxAf4w6496Kd OgH7/RxC
### #5AvCol Members
Posted 29 April 2011 - 02:36 AM
You could also try compiling the text file into a binary format for faster parsing.
I could, and I do this with my own internal format (which I export as a raw chunk of indices, vertices, skeletal weights, material strings etc.). But I am not concerned with that at the moment, I am concerned with why I can't write a fast text file parser using the classes of the C++ standard library.
### #6_swx_ Members
Posted 29 April 2011 - 02:56 AM
Tried running the program through the profiler in VS2010? I would try reusing the same stringstream object if possible to avoid having to reallocate the internal buffer all the time.
### #7rip-off Moderators
Posted 29 April 2011 - 03:01 AM
Can you show us the code?
### #8AvCol Members
Posted 29 April 2011 - 03:03 AM
Tried running the program through the profiler in VS2010? I would try reusing the same stringstream object if possible to avoid having to reallocate the internal buffer all the time.
You are onto something. Please explain further. My current code is
string line;
getline( input, line );
stream = stringstream( line ); // member of the class doing this code.
This is code that takes 6 seconds going through the entire file.
By contrast
string line;
getline( input, line );
takes only 2. There must be a better way to get a line into a stream. I feel like such a noob with this
### #9AvCol Members
Posted 29 April 2011 - 03:11 AM
Can you show us the code?
Why not.
class Parser
{
public:
Parser( std::wstring file );
virtual void Ignore( const std::string& start, const std::string& end );
virtual void Rewind( void );
virtual void Next( void );
virtual void GetLine( std::string& line );
virtual void GetTokens( std::vector<std::string>& tokens );
virtual void GetBody( std::string& body );
virtual void GetBodyTokens( std::vector<std::string>& bodyTokens );
virtual bool Good( void );
std::stringstream stream;
protected:
void TrimLine( std::string& line );
int ignoring;
std::vector<std::string> excludeDelims;
std::vector<std::string> includeDelims;
std::ifstream input;
};
void Parser::Ignore( const std::string& start, const std::string& end )
{
excludeDelims.push_back( start );
includeDelims.push_back( end );
}
void Parser::Rewind( void )
{
input.seekg( 0, ios::beg );
input.clear();
ignoring = -1;
stream = stringstream( "" );
}
void Parser::Next( void )
{
string line;
getline( input, line );
if( !input.good() )
return;
if( line.empty() )
{
Next();
return;
}
TrimLine( line );
if( line.empty() )
{
Next();
return;
}
stream = stringstream( line );
}
void Parser::GetLine( std::string& line )
{
line.assign( stream.str() );
}
void Parser::GetTokens( std::vector<std::string>& tokens )
{
tokens.clear();
stream.clear();
stream.seekg( 0, ios::beg );
string token;
while( stream >> token )
tokens.push_back( token );
}
{
stream.clear();
stream.seekg( 0, ios::beg );
}
void Parser::GetBody( std::string& body )
{
body.clear();
stream.clear();
stream.seekg( 0, ios::beg );
body.assign( stream.str() );
size_t i = 0;
// Ignore any white spaces at the beginning of the line.
while( body[i] == ' ' && body[i] == '\r' && body[i] == '\t' && i < body.length() )
i++;
// Ignore the first word
while( body[i] != ' ' && body[i] != '\r' && body[i] != '\t' && i < body.length() )
i++;
body = body.substr( i, body.length() );
}
void Parser::GetBodyTokens( std::vector<std::string>& bodyTokens )
{
bodyTokens.clear();
stream.clear();
stream.seekg( 0, ios::beg );
string token;
stream >> token;
while( stream >> token )
bodyTokens.push_back( token );
}
bool Parser::Good( void )
{
return input.good();
}
void Parser::TrimLine( string& line )
{
if( ignoring != -1 )
{
size_t incPos = line.find( includeDelims[ignoring] );
if( incPos != string::npos )
{
line = line.substr( incPos, line.length() );
ignoring = -1;
TrimLine( line );
}
else
line.clear();
}
else
{
for( size_t i = 0; i < excludeDelims.size(); i++ )
{
size_t excPos = line.find( excludeDelims[i] );
if( excPos != string::npos )
{
string tail = line.substr( excPos, line.length() );
line = line.substr( 0, excPos );
// If the includeDelim is the end of the line just return the head.
if( includeDelims[i] == "\n" )
return;
ignoring = i;
TrimLine( tail );
line += tail;
return;
}
}
}
}
Here is the obj loader code, although this hasn't changed since my 18 second lucy bench mark, just the above posted backend has.
shared_ptr<Mesh> ImportImpl::LoadObjMesh( wstring file )
{
shared_ptr<Mesh> mesh = LookupMesh( file );
if( mesh )
return mesh;
mesh = shared_ptr<Mesh>( new Mesh );
wstring path = FindFullPath( file );
ObjParser parser( path );
int numPositions = 0;
int numTexcoords = 0;
int numNormals = 0;
int numGroups = 0;
int numFaces = 0;
parser.Ignore( "#", "\n" );
// Preliminary run through to gather information.
while( parser.Good() )
{
parser.Next();
string line;
parser.GetLine( line );
switch( line[0] )
{
case 'v':
switch( line[1] )
{
case ' ':
numPositions++;
break;
case 't':
numTexcoords++;
break;
case 'n':
numNormals++;
break;
}
break;
case 'f':
numFaces++;
break;
case 'g':
numGroups++;
break;
}
}
if( !numPositions )
throw ExcFailed( L"[ImportImpl::LoadObjMesh] " + file + L" does not contain vertex positions.\n" );
if( numPositions < 0 || numFaces < 0 || numGroups < 0 )
throw ExcFailed( L"[ImportImpl::LoadObjMesh] " + file + L" holds way too much attribute data.\n" );
parser.Rewind();
vector<Position> positions;
vector<Normal> normals;
vector<Texcoord> texcoords;
positions.reserve( numPositions );
normals.reserve( numNormals );
texcoords.reserve( numTexcoords );
mesh->subMeshes.reserve( numGroups );
mesh->triangles.reserve( numFaces );
wstring_convert<std::codecvt_utf8<wchar_t>> converter;
Hash hasher;
forward_list<int> hashGrid[65536];
while( parser.Good() )
{
parser.Next();
vector<string> tokens;
parser.GetBodyTokens( tokens );
{
Position p;
p.x = float( ( atof( tokens[0].c_str() ) ) );
p.y = float( ( atof( tokens[1].c_str() ) ) );
p.z = float( ( atof( tokens[2].c_str() ) ) );
if( tokens.size() == 4 )
p.w = float( ( atof( tokens[3].c_str() ) ) );
else
p.w = 1.0f;
positions.push_back( p );
}
else if( header == "vt" )
{
Texcoord o;
o.s = float( atof( tokens[0].c_str() ) );
o.t = float( atof( tokens[1].c_str() ) );
texcoords.push_back( o );
}
else if( header == "vn" )
{
Normal n;
n.x = float( atof( tokens[0].c_str() ) );
n.y = float( atof( tokens[1].c_str() ) );
n.z = float( atof( tokens[2].c_str() ) );
normals.push_back( n );
}
else if( header == "f" )
{
vector<Vertex> faceVertices = parser.GetFaceVertices( positions, normals, texcoords );
for( unsigned int i = 0; i < tokens.size() - 2; i++ )
{
Vertex v[3];
v[0] = faceVertices[0];
v[1] = faceVertices[i + 1];
v[2] = faceVertices[i + 2];
// Fill out the vertex indices of the triangle by either pushing vertices into
// the mesh vector, or finding the index of an already existant equivalent.
Triangle tri;
for( int j = 0; j < 3; j++ )
{
unsigned int hash = hasher.GenerateHash16( v[j] );
bool found = false;
int index;
forward_list<int>::iterator it = hashGrid[hash].begin();
while( it != hashGrid[hash].end() )
{
if( mesh->vertices[*it] == v[j] )
{
index = *it;
found = true;
break;
}
it++;
}
if( !found )
{
index = mesh->vertices.size();
mesh->vertices.push_back( v[j] );
hashGrid[hash].push_front( index );
}
// Vertices are even indices in the t array.
tri.t[j * 2] = index;
}
mesh->triangles.push_back( tri );
if( !mesh->subMeshes.empty() )
mesh->subMeshes.back().triangleIndices.push_back( mesh->triangles.size() );
}
}
else if( header == "g" )
{
mesh->subMeshes.push_back( SubMesh() );
}
else if( header == "usemtl" )
{
wstring mtl = converter.from_bytes( tokens[0] );
mesh->subMeshes.back().materialIndex = mesh->materials.size();
}
}
mesh->FindTriangleNeighbors();
if( normals.empty() )
mesh->FindVertexNormals();
mesh->Trim();
meshCache.push_back( Record<Mesh>( file, mesh ) );
return mesh;
}
Before you mention it, TrimLine has an overhead of an extra 0.3 seconds on lucy, and that is an overhead I am willing to pay for something that can get rid of /* */ and // style commented text on the fly.The GetBody function has an assignment that I can probably cut out but I don't use it in the obj loader ( I do in the md5 loader but I am just looking at the obj loader for profiling the backend for now ). The part of the obj loader that hashes vertices is welding them on the fly as my internal format requires them that way: that algorithm is not super fast (incurs at least a 6 second overhead) but fast enough considering what it does.
### #10Gorbstein Members
Posted 29 April 2011 - 03:32 AM
stream = stringstream( line );
should you not use..
stream << line ;
I haven't read every line of the code but I'm not sure you need to be creating a new stringstream on each read.
### #11AvCol Members
Posted 29 April 2011 - 03:47 AM
stream = stringstream( line );
should you not use..
stream << line ;
I haven't read every line of the code but I'm not sure you need to be creating a new stringstream on each read.
That one change actually makes things at least one order magnitude slower, that was the only change I made and my timing went over 800 seconds so I decided to stop. But thanks for trying to give me some practical advice anyway, its appreciated.
### #12dmatter Members
Posted 29 April 2011 - 04:35 AM
I've found that if you just want to write a straight forward implementation and don't want to spend a lot of time optimising it then the C-style stream functions are the way to go. There's nothing inherently slow with the C++ streams by design it's just that the VC++ implementation of them does a lot of extra work (even with the full array of optimisation flags).
### #13__sprite Members
Posted 29 April 2011 - 06:24 AM
Is using the trim line function actually faster than just letting the switch statements ignore the unnecessary characters?
While going through the file twice (to find number of verts etc.) may be faster, perhaps keeping the file in memory and using that the second time would be quicker than reading everything off disk twice?
### #14AvCol Members
Posted 29 April 2011 - 06:42 AM
Is using the trim line function actually faster than just letting the switch statements ignore the unnecessary characters?
While going through the file twice (to find number of verts etc.) may be faster, perhaps keeping the file in memory and using that the second time would be quicker than reading everything off disk twice?
The going through the file twice isn't slowed down because of the disk, as I am sure this gets cached by the OS / HDD anyway: its slowed down most by that string stream constructor. And yeah it is way faster finding the right amount of verts because for a large data set ( like lucy ) if you don't reserve vector space, the whole thing runs almost five times as long (184 seconds )
No, it is a little slower. A switch statement works well in ignoring lines that start with # as the comment character ( like in obj files ) but if you have a comment like /* this is a comment */ that can span over multiple lines or be in the middle of a line or even have a line /* blah blah blah */ full off /* blah blah blah */ such comments like that, then you need something better, and the parser class is used to parse a few types of text files.
So everyone can see here is an implementation of my parser class I just created which loads lucy in 10 seconds, compared to the 48 it takes with the before posted one using string streams.
Parser::Parser( wstring file )
{
input.open( file );
ignoring = -1;
if( !input.is_open() )
throw ExcFailed( L"[Parser::Parser] Could not open file " + file + L"\n" );
}
void Parser::Ignore( const std::string& start, const std::string& end )
{
excludeDelims.push_back( start );
includeDelims.push_back( end );
}
void Parser::Rewind( void )
{
input.seekg( 0, ios::beg );
input.clear();
ignoring = -1;
line.clear();
}
void Parser::Next( void )
{
getline( input, line );
if( !input.good() )
return;
if( line.empty() )
{
Next();
return;
}
TrimLine( line );
if( line.empty() )
{
Next();
return;
}
}
void Parser::GetLine( std::string& _line )
{
_line = line;
}
void Parser::GetTokens( std::vector<std::string>& tokens )
{
tokens.clear();
string buff;
size_t from = 0;
while( from < line.length() )
{
GetNextToken( buff, from );
tokens.push_back( buff );
}
}
{
size_t from = 0;
}
void Parser::GetBody( std::string& body )
{
body.clear();
size_t i = 0;
// Ignore any white spaces at the beginning of the line.
while( line[i] == ' ' && line[i] == '\r' && line[i] == '\t' && i < line.length() )
i++;
// Ignore the first word
while( line[i] != ' ' && line[i] != '\r' && line[i] != '\t' && i < line.length() )
i++;
body = line.substr( i, line.length() );
}
void Parser::GetBodyTokens( std::vector<std::string>& bodyTokens )
{
bodyTokens.clear();
string buff;
size_t from = 0;
GetNextToken( buff, from );
while( from < line.length() )
{
GetNextToken( buff, from );
bodyTokens.push_back( buff );
}
}
bool Parser::Good( void )
{
return input.good();
}
void Parser::TrimLine( string& line )
{
if( ignoring != -1 )
{
size_t incPos = line.find( includeDelims[ignoring] );
if( incPos != string::npos )
{
line = line.substr( incPos, line.length() );
ignoring = -1;
TrimLine( line );
}
else
line.clear();
}
else
{
for( size_t i = 0; i < excludeDelims.size(); i++ )
{
size_t excPos = line.find( excludeDelims[i] );
if( excPos != string::npos )
{
string tail = line.substr( excPos, line.length() );
line = line.substr( 0, excPos );
// If the includeDelim is the end of the line just return the head.
if( includeDelims[i] == "\n" )
return;
ignoring = i;
TrimLine( tail );
line += tail;
return;
}
}
}
}
void Parser::GetNextToken( string& container, size_t& from )
{
size_t to = from;
while( from != line.length() && ( line[from] == ' ' || line[from] == '\t' || line[from] == '\r' ) )
from++;
to = from + 1;
while( to != line.length() && line[to] != ' ' && line[to] != '\t' && line[to] != '\r' )
to++;
container = line.substr( from, to - from );
from = to;
}
Which is a shame because I think string streams are a really elegant way of parsing and formatting data, but I don't know how to use them in a way that isn't mega mega slow.
### #15__sprite Members
Posted 29 April 2011 - 06:45 AM
A switch statement works well in ignoring lines that start with # as the comment character ( like in obj files ) but if you have a comment like /* this is a comment */ that can span over multiple lines or be in the middle of a line or even have a line /* blah blah blah */ full off /* blah blah blah */ such comments like that, then you need something better.
But that's not a valid comment in a .obj file.
If the stringstream is really that slow, you could try processing the line string using boost string algorithms or something instead (I'm not sure it would be faster, but it's an option to try).
### #16Antheus Members
Posted 29 April 2011 - 06:51 AM
This is text book example of how costly memory allocations are.
There is no decent way around it using standard stream implementations.
One could use custom allocator which would go a long way, but very third-party libraries support such strings, so if you depend on any of them, it'll be a problem.
Parsing .obj files is also best done using standard FSM-based parser which can work with no overhead or extra allocations beyond the extracted data.
Old topic!
Guest, the last post of this topic is over 60 days old and at this point you may not reply in this topic. If you wish to continue this conversation start a new topic. | 2017-04-27 02:04:38 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 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": 0.19254142045974731, "perplexity": 13199.361017579684}, "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-2017-17/segments/1492917121778.66/warc/CC-MAIN-20170423031201-00231-ip-10-145-167-34.ec2.internal.warc.gz"} |
https://mathematica.stackexchange.com/questions/126102/area-of-a-loop-of-a-curve-in-polar-coordinates | # Area of a loop of a curve in polar coordinates
With the curve of equation $r^{n}=a^{n}\cos n\theta$ the area of one of its $n$ loops is $$\frac{a^{2}\sqrt{\pi }}{2}~\frac{\Gamma (\frac{1}{2}+\frac{1}{n})}{\Gamma (\frac{1}{n})}$$
This can be obtained by a calculus excercice by hand. With Mathematica I am unable to arrive at this result, even a numerical solution with given parameters is not entirely satisfactory.
Below is what I arrived at.
You can represent the function with this MMA code, with 3 loops and a long axis of length 1.
With[{n = 3, a = 1
PolarPlot[ Power[a^n Cos[n x], 1/n], {x, 0, 2 Pi},
MaxRecursion -> 10]]
For the general solution and with the help of calculus I coded
Assuming[{n ∈ Integers, a > 0},
Integrate[(Max[0, Power[a^n Cos[ n t], 1/n]])^2, {t, 0,
2 π/n}]]
MMA hangs evaluating this expression, computes forever and I aborted it afer several minutes. My MMA level is 10.2. With 11.0 that I tried it on the Free Wolfram Development platform it is inconclusive as I run out of processor time.
So let's get a numerical solution
leafArea =
With[{a = 1, n = 3},
1/2 Integrate[(Max[0, Power[a^n Cos[ n t], 1/n]])^2, {t, 0,
2 π/n}]] // FullSimplify
I have a result but I have some doubt. Lets's compare it with the manual solution:
calcArea = ((a^2 Sqrt[Pi])/2 Gamma[1/2 + 1/n]/Gamma[1/n] ) /. {a ->
1 , n -> 3} // FullSimplify
leafarea == calcarea (* remains unevaluated why ? *)
But
N[leafArea] == N[calcArea] (* True !*)
What is going on here ? Can I trust the calcArearesult with Gamma expressions to be true ?
So I decided then to make use of the new region functions that came with MMA 10.0 , less calculus from now on
reg = Assuming[{n ∈ Integers, a > 0},
ImplicitRegion[
a Power[Sqrt[x^2 + y^2], n] < Cos[ n ArcTan[x, y]], {x, y}]]
Area[reg] (* Warning message MMA is unable to do it. *)
OK let's try that for a specific case:
region3leaf =
Assuming[{n == 3, a == 1},
ImplicitRegion[
a Power[Sqrt[x^2 + y^2], n] <
Cos[ n ArcTan[x, y]] && (0 < x <= 1), {x, y}]]
RegionPlot[region3leaf, AspectRatio -> Automatic] (* I see no problem with the way I defined the region*)
Area[region3leaf] (* Warning message MMA is unable to do it. *)
However with
DiscretizeRegion[region3leaf]
Area[%] / N[calcarea]
I am able at least to get an approaching solution.
As you can see MMA 10.2 not up to the task I submitted unless there is anything wrong in my code. Apart from waiting for future improvements or corrections in future releases can I arrive at the general solution with the existing MMA functions ?
The following returns the result you expect
Assuming[n > 0 && n ∈ Integers,
Block[{r = Cos[n θ]^(1/n)},
FullSimplify[a^2 Integrate[r^2/2, {θ, -Pi/(2 n), Pi/(2 n)}]]]]
(* (a^2 Sqrt[π] Gamma[1/2 + 1/n])/(2 Gamma[1/n]) *)
• You answer gives òne way to the general solution but I still don't know why MMA could neither cope with an integration inspired from the PolarPlot I coded which (unless contradicted) is correct (as the numerical solution is OK) nor cope with the help of new region functions . – Sigis K Sep 14 '16 at 18:01
• @SigisK Looking at your direct attempt to solve the integral, I believe that your integration limits for a single leaf are incorrect: 0, 2Pi/n rather than -Pi/(2n), Pi/(2n). With just this change, V11 returns the answer you expect. Including a Max function generally makes symbolic integrals much harder (and is unnecessary in this case if you choose the integration limits appropriately). Taking the a outside the integral is an obvious simplification - when Mathematica struggles with an integral, it is worth giving it all the help you can. – mikado Sep 14 '16 at 22:24
One way, if symbolically doesn't work is to interpolate:
l = Table[{a, n,
1/2 NIntegrate[(Max[0, Power[a^n Cos[n t], 1/n]])^2, {t, 0,
2 π/n}]}, {a, 0.1, 5, 0.1}, {n, 0.1, 5, 0.1}] // N;
ip = Interpolation[Flatten[l, 1],InterpolationOrder->5];
ip[1, 3]
0.343417 | 2020-02-16 18:34:23 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 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": 0.42696088552474976, "perplexity": 3431.1535423250953}, "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/1581875141396.22/warc/CC-MAIN-20200216182139-20200216212139-00161.warc.gz"} |
https://pos.sissa.it/301/1016/ | Volume 301 - 35th International Cosmic Ray Conference (ICRC2017) - Session Neutrino. NU-astrophysical neutrinos
The neutrino filter: connecting blazars with ultra high energy cosmic rays and astrophysical neutrinos
E. Resconi,* P. Padovani, S. Coenders, A. Turcati, P. Giommi, L. Caccianiga
*corresponding author
Full text: pdf
Pre-published on: August 16, 2017
Published on: August 03, 2018
Abstract
We discuss the link between high energy $\gamma$-ray emitting blazars, very high energy neutrinos, and ultra high energy cosmic rays using the IceCube neutrinos as a filter to $\gamma$-ray catalogues. The neutrino filtered emitters are investigated through a correlation analysis based on a likelihood method with the ultra high energy cosmic rays from the Pierre Auger Observatory and the Telescope Array by scanning in flux and angular separation between sources and cosmic rays. Based on the presently published neutrino and cosmic ray catalogues, a 2.9 sigma excess of cosmic rays correlating with the neutrino filtered emitters is observed. The neutrino filtered $\gamma$-ray sources that make up the cosmic rays excess are blazars of the high synchrotron peak type. More statistics is needed to further investigate these sources as candidate cosmic ray and neutrino emitters. We report on the status of our investigation focused on blazars as candidate cosmic ray and neutrino emitters.
DOI: https://doi.org/10.22323/1.301.1016
How to cite
Metadata are provided both in "article" format (very similar to INSPIRE) as this helps creating very compact bibliographies which can be beneficial to authors and readers, and in "proceeding" format which is more detailed and complete.
Open Access | 2020-12-04 01:37:26 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.6131591200828552, "perplexity": 4476.7324291846}, "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/1606141733120.84/warc/CC-MAIN-20201204010410-20201204040410-00489.warc.gz"} |
https://www.physicsforums.com/threads/rotation-of-a-sphere-because-of-its-gravitational-field.564659/ | Rotation of a sphere because of its gravitational field
1. Jan 3, 2012
ShayanJ
I wanna know how can I prove that a massive sphere rotates becuase of its gravitational field.I thought about it and did the following:
The gravitational force applied to a mass m which is on the axis prependicular to the plane of a hollow ring and passing through its center is as follows
${\it dF}={\frac {G{\it dM}\,mz}{ \left( {r}^{2}+{z}^{2} \right) ^{3/2}}}$
If I integrate the equation above from z=0 to z=2R I should get the force applied to a point on a hollow sphere and I get:
$F=1/8\,{\frac {GMm\theta\,\ln \left( z \right) \sqrt {2}}{\pi \,{R}^{2}}}$
$\theta$ is from 0 to $2 \pi$ and z from 0 to 2R.But as you can see,at z=0,it becomes infinity.So I wrote the taylor series of ln(z).But my calculations for $\omega$ didn't yeild 7.3 * 10^(-5) which is the angular velocity of earth.What's wrong?
thanks
2. Jan 3, 2012
Staff: Mentor
Why would you think that it would rotate? Explain what you mean.
3. Jan 3, 2012
ShayanJ
Earth rotates around itself and if you think,the only reason can be its own gravitational field.
4. Jan 3, 2012
Tea Jay
What about the fields of other bodies that can influence it, such as the moon or sun?
What about the motion of its crust or mantle?
5. Jan 3, 2012
6. Jan 3, 2012
D H
Staff Emeritus
No! The Earth rotates now because it was rotating in the past and because angular momentum is a conserved quantity.
Newton's first law, a body persists in its state of motion unless acted upon by an external force, is in a sense a statement regarding conservation of linear momentum. There's a direct analog of Newton's first law for rotational motion, a body persists in its state of rotation unless acted upon by an external torque.
External torques from the Moon, the Sun, and the other planets do act on the Earth. However, these external torques are small. Because they are small, they can be treated as being effectively zero for short periods of time (years and even decades are short periods of time in this context). Earth's angular momentum is more or less constant over the span of years or decades. It's only on the scale of centuries or longer (or on the scale of extremely precise measurements) that the Earth's angular momentum can not be viewed as being constant.
7. Jan 4, 2012
ShayanJ
This question came into my mind when our professor was teaching the central forces.
Looks like we just get excited when we learn sth new and stick everything to that new concept.
8. Jan 5, 2012
jetwaterluffy
No the only reason is the conservation of angular momentum. Although a similar effect does possibly exist: http://en.wikipedia.org/wiki/Gravitomagnetism but the effect is tiny.
9. Jan 5, 2012
Buckleymanor
So when the dust etc, was pulled together when it was forming by the gravity of the main body, angular momentum caused it to rotate.
Won't rain falling do the same.
10. Jan 5, 2012
D H
Staff Emeritus
Not really. First off, rain doesn't have much angular momentum. Much more importantly, whatever angular momentum rain does add to the Earth comes from the Earth+oceans+atmosphere system, not from outside. Rain does not add angular momentum to the Earth+oceans+atmosphere system.
There is an ongoing but small transfer of angular momentum from the Earth to the Moon. This is what causes the Earth's rotation rate to slowly decrease and the size of the Moon's orbit to slowly increase. However, this transfer is very slow. Over short spans of time (seasons, years, even decades), the angular momentum of the Earth+oceans+atmosphere is more or less constant. There is seasonal transfer of angular momentum between the Earth+oceans and the atmosphere, but this is cyclical and bounded. There is no secular change.
11. Jan 5, 2012
jetwaterluffy
It starts as a large cloud of dust spinning very slowly. As the size of it decreases, the spinning speed speeds up to compensate. This can't happen with rain, as rain doesn't decrease the size of the earth.
12. Jan 5, 2012
D H
Staff Emeritus
That's not a good model of how stars form, let alone planets. The Sun has 99.9% of the mass of the solar system, but only 1% of the angular momentum. This is the angular momentum problem.
Assuming the total angular momentum of the cloud with respect to its center of mass is non-zero, the formation of the protostar will act to flatten the cloud into a disc. With the formation of the disc, the material that falls into the protostar is that material that has little angular momentum. The formation of the disc solves the angular momentum problem.
Last edited: Jan 5, 2012
13. Jan 5, 2012
jetwaterluffy
Erm.. sorry can you rephrase that? Maybe I didn't get it when I was told, and subconsciously made something up to compensate, so I'd like you to explain it again for me. (the above post to me doesn't make sense.)
14. Jan 5, 2012
phinds
That is an excellent observation.
It's another way of saying the old saw "to a man with a hammer, all problems look like nails"
15. Jan 5, 2012
D H
Staff Emeritus
Starting over, then.
That is not a good model for how planets or stars form. It fails to explain why even though the Sun has 99.9% of the total mass of the solar system, it only has 1% of the solar system's total angular momentum. This is called the angular momentum problem.
A better model is the solar nebula disc model. The star and planets eventually form from some nebular cloud. Assume that the total angular momentum of this cloud with respect to the cloud's center of mass is non-zero. The star starts forming due to some uneven mass distribution within the cloud. As the protostar accumulates more mass, the increasing gravitational force will cause the nebular cloud to flatten out and form a rotating disc. The dust that has angular momentum with respect to this protostar will tend to end up somewhere on the disc. The dust that has near-zero angular momentum is what falls into the growing star. This nicely solves the angular momentum problem.
This spinning cloud does not explain planet formation, either. Planets form after the disc has formed. The orbital velocity of an orbiting object is $\sqrt{G(M+m)/r}$. A growing protoplanet will have a slightly greater orbital velocity than will the dust with which it is co-orbiting. The growing protoplanet will plow through this dust cloud, gathering mass and clearing the vicinity as it does. The growing protoplanet also migrates inward, encountering new dust. The density of the material is thus slightly greater in the direction of the protostar. From the perspective of the planet, it is encountering a wind of dust that is slightly denser toward the star.
This is a fairly good explanation of the formation of gas giants, but not quite so good for the rocky planets. There are still some unknowns left in explaining the formation of the rocky planets. The Earth, with its huge Moon, is particularly problematic. The current hypothesis is that a Mars-sized protoplanet collided with the proto Earth shortly after the formation of the solar system was more or less complete.
16. Jan 5, 2012
Buckleymanor
I imagined that the energy was provided from outside the Earth+oceans+atmosphere system by the Suns evaporation of water.
With regards to the explanation of the formation the Moon.If there was a large collision you would expect the Moon to be spinning because of it and at more of a rate than the Earth.
How come it has not been able to maintain it's spin from the impact you would expect it to be conserved.
17. Jan 5, 2012
Staff: Mentor
The moon is not spinning as fast as it once was because of tidal locking.
http://en.wikipedia.org/wiki/Tidal_locking
18. Jan 5, 2012
D H
Staff Emeritus
That makes no sense.
As Drakkith already noted, there's tidally locking to consider.
19. Jan 6, 2012
Buckleymanor
Interesting though it seems unclear if the Moon should speed up it's rotation if it's not spinning as fast as it was.
From the cited wikipedia article,
There seems to be a contradiction when B starts of rotating too slowly how does the Moon Know it started rotating at a faster speed than it is now.
For all purposes when it slows down it won't "know" if it started faster or that this was it's starting speed.
Last edited by a moderator: Jan 6, 2012
20. Jan 6, 2012
D H
Staff Emeritus
There's no contradiction, and the Moon doesn't need to "know" anything. (How can it?) It's all in the math, and that math is only concerned with what is happening now. | 2017-12-18 18:49:55 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6947651505470276, "perplexity": 824.6271338709645}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-51/segments/1512948619804.88/warc/CC-MAIN-20171218180731-20171218202731-00357.warc.gz"} |
http://physics.stackexchange.com/questions/82067/what-laws-formulas-govern-forces-between-atoms | # What laws (formulas) govern forces between atoms?
What laws (formulas) govern the fundamental forces of nature?
For example, gravity is governed by the inverse-square law.
I am thinking about how particles attract each other, but also repel. All matter attracts via gravity, but also attract and repel each other due to the electromagnetic force, also governed via the inverse-square law, and then there's the strong and weak force.
I understand the idea that two hydrogen atoms repel, but when pushed together hard enough will attract and form helium and release energy. So this "upwards slope" of repulsion which is eventually "snaps" into attraction, like trying to roll a ball up a smooth hill with a hole in the center of it, is explained, likely, by varying forces governed by various laws. Can you help me understand it, on the level of particles?
Where I am stuck is thinking about two particles, each being attracted and repelled by two different forces each governed by the inverse-square law. This doesn't explain the "snapping" together of atoms. Do I need different math in my forces, or is it ultimately explained by more than just two particles?
Here is a diagram of what I mean, describing H + H -> H2:
Image source and article: http://employees.csbsju.edu/hjakubowski/classes/ch111/olsg-ch111/equilibkinetics/equilbkin.htm
Please feel free to reword my question if you understand what I am asking but am not making myself clear.
-
First, you may become disappointed but the trully fundamental laws, as we know them today, are not written in terms of force laws. Even though the concept of force is still present in Physics, it is not used in the way it was before and which seems to be the way you are thinking about them.
Force is nowadays synonymous of interaction and one does not seek for force laws to be used in the equation $$\vec{F}=\frac{d\vec{p}}{dt},$$ from where one would, ultimately obtain $\vec{r}(t)$.
The above equation summarizes classical mechanics (CM) in its Newtonian "version" (or formulation). Even classical mechanics can be done without explicitly writing a vector equation as this one.
It was the analytic formulation(s) of CM that people took as the framework for doing advances in mechanics. They are all equivalent when it comes to the classical scenario and one uses one or another formulation for several reasons. Nevertheless, in the analytic formulations of CM, instead of using forces, as the quantities encoding the interaction, one uses potentials and the equations of motion are no longer obtained from Newton's second law (at least not explicitly as in the equation above) but from a more powerful principle, which is Hamilton's principle.
Now, even though in CM one may use any formulation according to one's needs, when it comes to relativistic classical mechanics and (relativistic) quantum mechanics, it is no longer a matter of choice. There are several reasons for why this is so. A very simple one, is that will won't be able to find a force four-vector to plug in the relativistic equivalent of Newton's second law (as it is written above) other than the Lorentz force. Also, in quantum mechanics (QM), Newton's second law holds only as an average (or expectation value).
This is why, even though one still speaks about forces, it is not in the same sense as before and we don't have other kinds of inverse-distance laws (or any other kind of vetor force laws) for the other fundamental interactions. Even the so-called potentials are not quite the same animals as in CM.
About What laws govern the fundamental forces of nature?, have a look at here.
Even the problems we try to solve with more fundamental physics are not quite the same as in CM. It is more about cross sections and decay rates than about describing the motion of individual particles (though that can be done to some extent).
I believe the particular phenomenon you are interested in is nuclear fusion. It is ultimately described in terms of electromagnetic and strong interactions and, even though in practice people may describe it in terms of more effective nuclear forces, it is still all done in the framework of relativistic quantum mechanics / quantum field theory and you won't find force laws.
To summarize: there are no force laws aside the ones of classical physics (Newton's gravitation law, Coulomb's electrostatic force and Lorentz force and some others).
-
This is great. Thank you. I will be giving the answer to Brandon Enright above but your answer has also helped a lot. EDIT: I will give the answer to you since you answered it first. – Xonatron Nov 22 '13 at 18:15
Your figure represents not two inverse-square potentials but something like the Lennard-Jones potential $V(r)= Cr^{-12}-C'r^{-6}$. The latter is a model for the van der Waals potential between two neutral spherical particles when they are at distance $r$.
The force is the derivative. hence at close distance, they repel each other, further apart they attract each other (and very far apart they don't notice each other significantly). Thus if close enough that the approximation is valid, their distance will oscillate between an attracting and a repelling distance, until dissipation will bring them at equilibrium at the distance where the potential is minimal and there is no force.
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What about breaking it up into two (or more) forces and showing me each? – Xonatron Nov 22 '13 at 16:28
You can split it up in many ways. In case of the Lennard-Jones potential $V(r)=Cr^{−12}−C′r^{−6}$, you can split it into the repulsive (-12) soft core part and the attractive (-6) van der Waals part. They do not have minima with positive $r$, hence do not lead to an equilibrium position. Only the combination balances repulsion and attraction at some equilibrium distance. – Arnold Neumaier Nov 23 '13 at 9:15
I like the explaination at about minute 36 of the Fermilab lecture: http://vmsstreamer1.fnal.gov/Lectures/LectureSeries/130612Carroll/index.htm
The strong force is "confined", it only extends a very small specific distance.
The weak force is "absorbed", it gets weaker faster than the inverse square law.
In summary, neither the strong force (involved in holding the protons together to make helium), or the weak force (involved in the decay of a neutron into a proton and electron) act at a distance subject to the inverse square law.
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Do you know the formulas? – Xonatron Nov 22 '13 at 16:29
What laws (formulas) govern the fundamental forces of nature?
None.
Columb's law and Newton's law of gravitation classical explanations of electrostatics and gravitation,respectively.
But the are no analogus formulas for fundamental interactions. They must be described in the context of Quantum field theory.
QFT is to complex to give you a formula where you can plug in some numers. For example, to study a simple interaction you will need to calculate many Feynman diagrams (integrals).
The EM, for instances is described by the Lagrangian: $\mathcal{L}=-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}+j^{\mu}A_{\mu}$
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I was just looking for something similar to the inverse law for gravity, something that explains particles. – Xonatron Nov 22 '13 at 16:56
In that case your question is almost a duplicate of: physics.stackexchange.com/questions/39229 – jinawee Nov 22 '13 at 17:00
@MatthewDoucette You might find useful the link I posted in the answer. – jinawee Nov 22 '13 at 17:19
Except I wanted all the formulas of all the forces. I'll check out your link. – Xonatron Nov 22 '13 at 17:20
@MatthewDoucette Just look for the Standard Model lagrangian (the neccesary tools are explained in graduate courses, so don't freak out if you don't understand it). – jinawee Nov 22 '13 at 17:24
Your diagram of 2 H -> He is extremely misleading. First, He has Neutrons and can't be made from just 2 H. It need deuterium and tritium and it's a multi-step process involving first making deuterium via $\beta^{+}$ decay.
Also, your question and diagram seem to imply that there is a formula describing the curve you show but there isn't. Your curve is the aggregate of several different forces including electromagnetism (electrostatic repulsion) and the strong force (residual color force).
If you want the analogous "inverse square law" for nuclear forces, no such law exists. The electrostatic portion does behave a bit like an inverse square law but at short distances the Pauli exclusion principle and quantum mechanics dominate and make things complicated.
Matt Strassler has a great article (7 parts so far) on QFT and the strong force. The short version of it is that because quarks are so light we can't directly model the strong force. All of our predictions are educated guesses and we don't have any formulas to govern their macroscopic behavior. The other issue is that gluons interact with each other dashing any hope of an inverse-square-law-like formula.
-
Exactly, it's more than one formula, and what I want is all of those formulas to result in some sort of curve like this. Thanks for breaking it down for me. I guess what I am looking for does not exist. – Xonatron Nov 22 '13 at 18:13 | 2016-02-06 07:45:30 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 1, "x-ck12": 0, "texerror": 0, "math_score": 0.7388169169425964, "perplexity": 375.4400329847889}, "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-2016-07/segments/1454701146196.88/warc/CC-MAIN-20160205193906-00307-ip-10-236-182-209.ec2.internal.warc.gz"} |
http://mathhelpforum.com/pre-calculus/180873-odd-even-functions-print.html | Odd / Even Functions
• May 17th 2011, 11:41 AM
bugatti79
Odd / Even Functions
Folks,
How does one show that f(x)sin(x) is an even function?
I attempt it be replacing x by -x ie
f(-x)sin(-x) but sin(-x) =-sin(x) therefore
f(-x)(-sin(x))=-f(-x)sin(x)????? (Dull)
----------------------------------------------------
For the odd function g(x)cos(x) I calculate by replacing x by -x ie
g(-x)cos(-x) but cos(-x) = cos (x) therefore
g(-x)cos(x)=-g(x)cos(x) hence odd because g(-x)=-g(x)
• May 17th 2011, 11:47 AM
SpringFan25
Quote:
How does one show that f(x)sin(x) is an even function?
you forgot to tell us whether or not f(x) is an even function. I assume it is odd:
f(-x)sin(-x) =-1*f(x)*-1*sin(x) = f(x)sin(x)
Quote:
For the odd function g(x)cos(x) I calculate by replacing x by -x ie
g(-x)cos(-x) but cos(-x) = cos (x) therefore
g(-x)cos(x)=-g(x)cos(x) hence odd because g(-x)=-g(x)
agree.
• May 17th 2011, 11:53 AM
bugatti79
Quote:
Originally Posted by SpringFan25
you forgot to tell us whether or not f(x) is an even function. I assume it is odd:
f(-x)sin(-x) =-1*f(x)*-1*sin(x) = f(x)sin(x)
agree.
No f(x) is even...sorry for lack of clarity!
• May 17th 2011, 11:57 AM
SpringFan25
the product of an even function and an odd function is an odd function.
f(-x)sin(-x) = f(x)*-1*sin(x) = -f(x)sin(x).
so you wont be able to show the product is even.
• May 17th 2011, 12:00 PM
Plato
Quote:
Originally Posted by bugatti79
No f(x) is even...sorry for lack of clarity!
Then it is not true.
$\text{If }f(x)\text{ is even then }f(x)\sin(x)\text{ is odd.}$ | 2014-04-16 22:04:52 | {"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": 0, "img_math": 0, "codecogs_latex": 1, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7745084166526794, "perplexity": 4496.697604095526}, "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-2014-15/segments/1397609525991.2/warc/CC-MAIN-20140416005205-00031-ip-10-147-4-33.ec2.internal.warc.gz"} |
https://skattergnawvuf.netlify.app/54339/57265.html | # Barry C. Barish - Nobel diploma - NobelPrize.org
How Old is the Universe? ESO Sverige
This includes determining the most efficient way to supply products directly to consumers and ensuring that the products actually arrive at the destination. scattering form factors are calculated for proton-rich 8B, 17F, 17Ne, 23Al and 27P Progress in Particle and Nuclear Physics, 68: 215–313. 3. Bohr A. and These form factors not only provide a testing ground for QCD-inspired models, but also are important in many areas of particle and nuclear physics, including 6 Apr 2014 skin thicknesses, nuclear binding energies, and charge form factors 1 Department of physics, College of Science, University of Baghdad, Exercises in Nuclear Physics. Task 1: (10 points) nuclear property can be derived directly from the nuclear form factor F(q) now? c) Voluntary: Calculate the In Nuclear and Particle Physics, the answer to this question usually takes Hence, we have a direct experimental determination of the form factor, as a ratio of 20 Oct 2005 has been further developed and incorporated in the GSI physics program [34] The nuclear charge form factor Fch(q) has been calculated. Related Subjects · Particle/High Energy Physics · Nuclear Physics · Experimental Physics · Atomic & Nuclear Physics · Particle & High Energy Physics In electron scattering off protons one needs as the form factor the function.
Introduction The S-factor separates out the Coulomb interaction (barrier) energy term from the cross-section $\sigma$, e.g., $$\sigma(E) = E^{-1} e^{-2\pi\eta} S(E)$$ where $\eta$ is a dimensionless factor relating to the Coulomb barrier. It can be useful for finding a small resonance within a large electric field (image from here): Dr Eram Rizvi Nuclear Physics and Astrophysics - Lecture 2 12 where ρ e(r') is distribution of nuclear charge Fourier transforms switch one variable into another e.g. spatial co-ordinates into momentum co-ordinates Thats what the form factor is: ! fourier transform of the charge distn !!
2019. “Metabolic Shifts Associated with Drought-Induced Senescence in Brachypodium.” Plant Science 289. av A Massih · 2014 · Citerat av 19 — nationally, for doping nuclear fuel pellets with non-absorber additives in order to improve the Factors governing microstructure development of Cr2O3-doped Hypostoichiometric uranium dioxide UO2−x form under Nuclear Science and.
## Walter Greiner · Quantum Electrodynamics Paperback Book [4th
1. Se hela listan på ieer.org 2 Nuclear Form Factor for Spin-Independent Coupling The nuclear form factor is the Fourier transformation of the nuclear density. In this paper, we numerically cal-culate the nuclear density by means of the RMF theory. Namely, we aregoingto derivethe nucleon numberdensity (nuclear density) in Eq. (2) in terms of the RMF theory.
### Barry C. Barish - Nobel diploma - NobelPrize.org
(neutron flux density), current density of particles, Nuclear Reactions and The form factor is the Fourier transform of the charge distribution. The cross section is reduced by the square of the magnitude of the form factor. Looking at the ratio between the point-like cross section and the actual cross section one can infer the form of the charge distribution. In physics, the atomic form factor, or atomic scattering factor, is a measure of the scattering amplitude of a wave by an isolated atom. The atomic form factor depends on the type of scattering, which in turn depends on the nature of the incident radiation, typically X-ray, electron or neutron. The nucleon (proton and neutron) electromagnetic form factors describe the spatial distributions of electric charge and current inside the nucleon and thus are intimately related to its internal structure; these form factors are among the most basic observables of the nucleon. Form factors are more general than just in nuclear physics, where measure-ments give information about charge distributions of nuclei.
Atmospheric particles are an important factor in earth's radiation budget as they scatter man bokstiiver i format kring eller under 0,s mm (sid. 56, cludes atomic and nuclear physics (pp. the many involved factors and dilliculties, the author's.
Discover the world of quantum physics, with information on the history of the field, impo Quantum Physics - Quantum physics requires physicists to use thought experiments. Learn why the study of quantum physics has raised more questions than it has answered.
28 8.1 Theoretical and experimental form factor of ^He (a) 83 8.2 Theoretical and experimental form factor of '^He (b) 84 8.3 Theoretical and experimental form factor of ^He (c) 85 8.4 Theoretical and experimental form factor of ^He (d) 86 reactions among nuclei the complexities of nuclear many-body systems inhibit closed form descriptions. Rather, any nuclear scattering theory relies on resummation techniques, by which certain classes of interactions are described by e ective operators, allowing to treat the phenomena of explicit interest by properly chosen perturbative methods We show that the electromagnetic properties of the τ, and in particular its magnetic form factor, may be measured competitively in these facilities, using unpolarized or polarized electron beams. Prototype VME & CAMAC form factor Timestamping module development for Nuclear Physics Experiment December 2015 Conference: DAE-BRNS Symp. on Nucl.
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### Böcker · Books
Gesellschaft fuer Another factor dealt with in Appendix C is the demographic idiomorphy of the country. Indeed, Greece It is clear that a nuclear emergency plan must form an integral part of the arrangments This reaction probed the time-like form factors and we were searching for heavy at the European Organization for Nuclear Research (CERN) that provided initial evidence for Köp boken Subatomic Physics (3rd Edition) av Alejandro Garcia (ISBN An up-to-date and lucid introduction to both particle and nuclear physics, the book is on:Detectors and acceleratorsNucleon elastic form factor dataNeutrinos, their stored in a retrieval system, or transmitted, in any form or by any means, without the prior Annual Reviews of Nuclear and Particle Science for Figure 9.2;. Elsevier normalization factors for the wavefunctions of the particles in the initial state. The charm of strange quarks : mysteries and revolutions of particle physics by R. based on quark model and light cone sum rule form factor calculations. av M Ablikim · 2021 — 10 G.I. Budker Institute of Nuclear Physics SB RAS (BINP), Novosibirsk 630090, respective decay form factors as functions of the lepton-. XXX has introduced a new range of high quality, competitively priced square tactile short travel keyswitches, on an industry standard pin-out My training and my love is particle physics, but judging by the papers I write and how much Earth will warm later this century to better than a factor of three. of new particles that form and grow under highly polluted urban conditions, on Computing in High-Energy and Nuclear Physics 17—21 May 2021 av J Sisefsky · 1967 · Citerat av 5 — fall enbart inom FOA) avsedd redovisning av arbete, t ex i form av del- Nuclear debris particles from the fourth and fifth Chinese tests Their specific activity is about a factor of 80 higher.
## Effective Field Theory and One-proton Halo - AVHANDLINGAR.SE
Explaining complicated experimental results by partment of Physics, May 29, 1990, at 10.15 a.m., for the degree of doctor of philosophy. by. SVERKER 10.3.1 Nuclear effects. 75 The form factor f{M,y,Pr). An up-to-date and lucid introduction to both particle and nuclear physics, the on: Detectors and acceleratorsNucleon elastic form factor dataNeutrinos, their An up-to-date and lucid introduction to both particle and nuclear physics, the on:Detectors and acceleratorsNucleon elastic form factor dataNeutrinos, their General Physics Corporation. Gesellschaft fuer Another factor dealt with in Appendix C is the demographic idiomorphy of the country.
1. Se hela listan på ieer.org 2 Nuclear Form Factor for Spin-Independent Coupling The nuclear form factor is the Fourier transformation of the nuclear density. In this paper, we numerically cal-culate the nuclear density by means of the RMF theory. Namely, we aregoingto derivethe nucleon numberdensity (nuclear density) in Eq. (2) in terms of the RMF theory. | 2022-05-26 13:23:48 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6696934103965759, "perplexity": 3097.1083268509683}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662606992.69/warc/CC-MAIN-20220526131456-20220526161456-00696.warc.gz"} |
https://www.tutorialspoint.com/find-alphabetical-order-such-that-words-can-be-considered-sorted-in-cplusplus | Find alphabetical order such that words can be considered sorted in C++
C++Server Side ProgrammingProgramming
Suppose we have an array of words, we have to find any alphabetical order in the English alphabet so that the given words can be considered sorted in ascending order, if there is any such order exists, otherwise return "impossible".
So, if the input is like words = ["efgh", "wxyz"], then the output will be zyxvutsrqponmlkjihgfewdcba
To solve this, we will follow these steps −
• ALPHABET := 26
• n := size of v
• if n is same as 1, then −
• display "abcdefghijklmnopqrstuvwxyz"
• return
• Define an array adj of size ALPHABET
• Define an array in of size ALPHABET and fill with 0
• pre := v[0]
• for initialize i := 1, when i < n, update (increase i by 1), do −
• s := v[i]
• for j in range 0 to minimum of (size of pre and size of s) - 1 −
• if s[j] is not equal to pre[j], then −
• Come out from the loop
• if j < minimum of size of pre and size of s, then −
• insert s[j] - ASCII of 'a' at the end of adj[pre[j] - ASCII of 'a']
• increase in[s[j] - ASCII of 'a'] by 1
• pre := s
• if size of pre > size of s, then −
• display "Impossible"
• return
• pre := s
• Define one stack my_stack
• for initialize i := 0, when i < ALPHABET, update (increase i by 1), do −
• if in[i] is same as 0, then −
• insert i into my_stack
• Define an array out
• Define an array vis of size: 26. fill with false
• while (my_stack is not empty), do −
• x := top element of my_stack
• delete element from my_stack
• vis[x] := true
• insert x + ASCII of 'a' at the end of out
• for initialize i := 0, when i < size of adj[x], update (increase i by 1), do −
• if vis[adj[x, i]] is non-zero, then −
• (decrease in[adj[x, i]] by 1)
• if in[adj[x, i]] is same as 0, then −
• insert adj[x, i] into my_stack
• for initialize i := 0, when i < ALPHABET, update (increase i by 1), do −
• if not vis[i] is non-zero, then −
• display "Impossible"
• return
• for initialize i := 0, when i < size of out, update (increase i by 1), do −
• display out[i]
Example
Let us see the following implementation to get better understanding −
Live Demo
#include <bits/stdc++.h>
using namespace std;
#define ALPHABET 26
void search_ordering(vector<string> v) {
int n = v.size();
if (n == 1) {
cout << "abcdefghijklmnopqrstuvwxyz";
return;
}
vector<int> in(ALPHABET, 0);
string pre = v[0];
for (int i = 1; i < n; ++i) {
string s = v[i];
int j;
for (j = 0; j < min(pre.length(), s.length()); ++j)
if (s[j] != pre[j])
break;
if (j < min(pre.length(), s.length())) {
in[s[j] - 'a']++;
pre = s;
continue;
}
if (pre.length() > s.length()) {
cout << "Impossible";
return;
}
pre = s;
}
stack<int> my_stack;
for (int i = 0; i < ALPHABET; ++i)
if (in[i] == 0)
my_stack.push(i);
vector<char> out;
bool vis[26];
memset(vis, false, sizeof(vis));
while (!my_stack.empty()) {
char x = my_stack.top();
my_stack.pop();
vis[x] = true;
out.push_back(x + 'a');
for (int i = 0; i < adj[x].size(); ++i) {
continue;
}
}
for (int i = 0; i < ALPHABET; ++i)
if (!vis[i]) {
cout << "Impossible";
return;
}
for (int i = 0; i < out.size(); ++i)
cout << out[i];
}
int main() {
vector<string> v{"efgh", "wxyz"};
search_ordering(v);
}
Input
{"efgh", "wxyz"}
Output
zyxvutsrqponmlkjihgfewdcba
Published on 23-Jul-2020 09:06:28 | 2022-01-20 22:36:59 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 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": 0.26255184412002563, "perplexity": 11346.424973735771}, "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-2022-05/segments/1642320302706.62/warc/CC-MAIN-20220120220649-20220121010649-00681.warc.gz"} |
https://aviation.stackexchange.com/questions/25641/does-resume-own-navigation-also-mean-altitude-your-discretion | In VFR flights from class C, I sometimes get resume own navigation, altitude your discretion after departure and sometimes just resume own navigation.
Does resume own navigation alone mean altitude your direction as well?
Update:
Let's assume scenario below for KSNA and to be more specific:
Pilot to Clearance Delivery: 529J with request
Pilot to Clearance Delivery: 529J is Cessna 172/u west side parking, requesting El Toro departure to Chino with information Bravo
Clearance Delivery to Pilot: After Departure turn left heading 080, Maintain VFR at or below 2400, SoCal on 124.1, SQUAWK 0210
Pilot starts taxi then take off with proper radio communication with ground and tower. After departure Tower hands pilot over to SoCal.
Pilot to SoCal (initial call): SoCal 529J departed John Wayne, 1100 climbing, requesting 3000 to Chino
SoCal to pilot: 529J resume own navigation
Here is the confusion, sometime SoCal comes back with only resume own navigation. Does this mean pilot can climb to 3000 or altitude restriction for 2400 still valid?
• When you receive only the resume own navigation instruction, is that ever following an altitude instruction that would need to be cancelled? Altitude is always your discretion in VFR flight in class C unless otherwise instructed. – J Walters Feb 27 '16 at 2:43
• KSNA departure has altitude/heading restriction on departure. After departure Tower hands you over to SoCal Approach. SoCal issues the 'resume own navigation' sometimes follows by 'altitude your discretion' and sometime doesn't. – pmoubed Mar 3 '16 at 21:39
At this time, the definition of "resume own navigation" is very varied from group to group. ATC Controllers say that if there is an altitude restriction, they have to clear it. So safest thing to do is to ask for clarification if you had an altitude restriction/not told "altitude your discretion" that would continue into the "resume own navigation" exactly what you are doing now.
Source
• That's what I always do but I wanted to know if I am doing the right thing – pmoubed Feb 27 '16 at 0:42
• @PMoubed Then you are doing the right thing then, under the current system. – SMS von der Tann Feb 27 '16 at 1:23
• @PMoubed Also, you should put that into your question that you do do this. – SMS von der Tann Feb 27 '16 at 1:47
resome own navigation means any navigation instructions are no longer in effect and you may fly as you wish.
If an altitude restriction has been given, you must maintain that restriction until you hear "altitude your discretion."
If no altitude restriction had been given then you may of course change your altitude.
• I added a scenario to my question. – pmoubed Mar 3 '16 at 22:05 | 2021-03-06 06:16:48 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.27108272910118103, "perplexity": 7385.3447303108715}, "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-2021-10/segments/1614178374391.90/warc/CC-MAIN-20210306035529-20210306065529-00580.warc.gz"} |
https://aul12.me/machinelearning/2020/06/18/quantization-benchmark.html | 18 Jun 2020
# Comparing the Inference Performance of Neural Networks with and Without Quantization
In my post “Getting started with the Google Coral EdgeTPU using C++” i described that models for the Google Coral EdgeTPU module need to be quantized to 8-bit fixed point numbers, instead of the commonly used 32 or 64-bit floating point numbers. This reduces the resolution of all values used in the model, that are the weights, the intermediate results and the predictions. Therefore the quality of the predictions of the neural networks is in most cases worse than without quantization.
In the blog post i will examine the differences in performance for a convolutional neural network using different metrics. This helps to not only understand the direct impact on the accuracy of the network but to further characterize the differences in the outputs. The TensorFlow team themselves examined the performance of quantization on four widely used convolutional neural networks but only reported the accuracy as metric (See: TensorFlow - Model optimization).
## Setup
The network used is the same classification network as listed in the post “Comparing the inference speed of different tensorflow implementations”. To summarize the structure: it consists of four convolutional layers each followed by a max-pooling operation and a ReLU nonlinearity. After the convolutional layers there are two fully connected layers with a softmax prediction at the output. The input is of size $80\times 80 \times 3$, the output consists of 30 classes. One of the classes (class 29) is the NO_CLASS class, this class consists of samples which do not show any of the valid classes. See the appendix for the full list of classes and their semantic interpretation.
The models with and without quantization are exactly the same model, the quantized model was converted from the normal model using post-training quantization. The complete dataset consists of 53299 images. Using this dataset two different models have been trained: for the first model most of the samples have been used for training (to be precise 47969 of 53299), and all samples have been used for evaluation. This is how the model is used for the actual deployment. For the second model the dataset has been divided in a training set of size 38867 and a verification set of size 10114. The samples in the subsets are completly different as they were recorded at different times. For training only the training set is used and for the evaluation only the verification set. This training procedure is different to the one used for deployment but this guarantees that the results are not influenced by overfitting. In the next sections the first setup is refered to as the full configuration and the second as the verification configuration.
The complete code used for testing can be found on my GitHub page: github.com/aul12/NeuralNetworkQuantizationPerformanceBenchmark.
## Comparison
### Accuracy
The first metric is the accuracy of the classifier on the dataset, for this we calculate the prediction as the class with the highest probability. A prediction is correct if it is the same as the label. The accuracy is then calculated as the number of correct predictions divided by the total number of samples.
Model Accuracy (Full) Accuracy (Verification)
Without Quantization 99.85% 99.83%
With Quantization 96.93% 58.14%
The performance of the quantized model is about three percent points worse than the performance of the normal model when comparing on the full configuration. This is a slightly better relative performance of the quantized model than the numbers reported by the TensorFlow team. When comparing using the verification configuration the results are vastly different: the floating-point model yields a similar result but the performance of the quantized model drops from 96.93% to 58.14%, a decrease by 38.79% percent points.
Additionally to the accuracy a confusion matrix for both models and configurations has been created, see the appendix for the full matrices and the semantic interpretation of the label numbers. As already implied by the accuracy most of the entries of the confusion matrices are located on the primary diagonal. Without quantization the incorrect classifications are distributed evenly throughout the confusion matrix for both configurations.
For the non quantized model on the full dataset there are two main error sources:
• The different speedlimit signs can not be differentiated as clearly as the non quantized model can
• In comparison to the non quantized model more images of type NO_CLASS are classified as a relevant
The additional problems using the verification configuration are:
• Left- and Rightarrows can not be differentiated as good as before
• Sharp-Turn-Left and -Right can not be differentiated as good as before
• No passing start/end is often classified as a speedlimit, probably due to the red circle on the signs
### Output probabilities
The softmax activation function used in the last layer yields a valid probability density function (pdf) over the classes. This means that all values are in the range $[0, 1]$ and the sum over all 30 values is 1. The classification result is determined as the class with the maximal probability. Additionally to this hard classification information the result can also be used for additional filtering, especially in application in which the same object gets classified multiple times.
To compare the quality of the pdfs the probalities are evaluated. A good pdf clearly shows the winner but additionally provides certainty information, i.e. the output pdf is not a one-hot-distribution. For the comparison two different sets of values are compared:
• the certainty, that is the probability of a correct classification
• all output values, that are both the certainty and the probabilities for all other classes
For this evaluation the full configuration is used if not specified differently. The differences between the configurations are small. Over both of these set of values for both models a histogram over the values is calculated. These histograms are given below, additionally there are the raw values further below.
Range Certainty Quant. Certainty float Out Quant. Out float
$[0, 0.05)$ $0$ $0$ $0.96$ $0.97$
$[0.05, 0.1)$ $0$ $0$ $3.1 \cdot 10^{-5}$ $0$
$[0.1, 0.15)$ $0$ $0$ $0.00060$ $0$
$[0.15, 0.2)$ $0$ $0$ $0.00012$ $0$
$[0.2, 0.15)$ $0$ $0$ $0$ $0$
$[0.25, 0.3)$ $3.9 \cdot 10^{-5}$ $0$ $9.5 \cdot 10^{-5}$ $0$
$[0.3, 0.15)$ $0.00075$ $0$ $0.00027$ $0$
$[0.35, 0.4)$ $0$ $0$ $0$ $0$
$[0.4, 0.45)$ $0$ $0$ $0$ $0$
$[0.45, 0.5)$ $0$ $0$ $0$ $0$
$[0.5, 0.55)$ $0.011$ $0$ $0.0016$ $0$
$[0.55, 0.6)$ $0$ $0$ $0$ $0$
$[0.6, 0.65)$ $0$ $0$ $0$ $0$
$[0.65, 0.7)$ $0$ $0$ $0$ $0$
$[0.7, 0.75)$ $0$ $0$ $0$ $0$
$[0.75, 0.8)$ $0$ $0$ $0$ $0$
$[0.8, 0.85)$ $0$ $0$ $0$ $0$
$[0.85, 0.9)$ $0$ $0$ $0$ $0$
$[0.9, 0.95)$ $0$ $0$ $0$ $0$
$[0.95, 1]$ $0.98$ $1$ $0.32$ $0.033$
The results are surprising: the non quantized model yields in general more one-hot-like results, when compared to the quantized model which yields a more uniform distributions. When considering the absolute scale it can also be seen that the actual differences are relativly small, both distributions look, especially when only considering the histograms, very similar.
To further characterize the pdfs the (Shannon) entropy is calculated for every sample and the average entropy over all samples is calculated for both models. The entropy can be used as measure for the uniformity of a distribution. A completly uniform distribution for a pdf of dimension $N$ yields an entropy of $\log_2(N)$, for a one hot distribution the entropy is $0$. To make the score independent of the dimension of the output space the entropy is normalized by a factor of $\frac{1}{\log_2(N)}$, so that the score for a uniform distribution is $1$.
The average entropy is given in the table below:
Model Average normalized Entropy (full) Average normalized entropy (verification) With Quantization $0.0089$ $0.021$ Without Quantization $0$ $0$
These numbers imply the same result as the histogram: the output pdf for the non-quantized model is strictly one hot, the quantized modell has a more uniform distribution. Still the differences are rather small.
## Results
The results are twofold: primarily the accuracy of the quantized classifier is reduced drastically when evaluating on a different dataset. Depending on the applications the results can be sufficient, especially considering that the errors are predictable, but in general the loss in accuracy renders the quantized model unusable.
Secondly the output pdf is not one-hot encoded anymore but yields more soft-encoded labels. For most applications the difference is neglectible, even the labels of the quantized model are nearly one-hot-encoded.
## Appendix
### Labels
The classes are defined by the Carolo-Cup-Regulations, see the official rules for more information: wiki.ifr.ing.tu-bs.de/carolocup/system/files/Master-Cup%20Regulations.pdf
Name Label Number
CROSSWALK 0
STOPLINE 1
GIVE_WAY_LINE 2
R_ARROW 3
L_ARROW 4
FORBIDDEN 5
SPEEDLIMIT10 6
SPEEDLIMIT20 7
SPEEDLIMIT30 8
SPEEDLIMIT40 9
SPEEDLIMIT50 10
SPEEDLIMIT60 11
SPEEDLIMIT70 12
SPEEDLIMIT80 13
SPEEDLIMIT90 14
NSPEEDLIMIT 15
STARTLINE_PARKING 16
CROSSINGLINE 17
PEDESTRIAN 18
EXPRESS_START 19
EXPRESS_END 20
NO_PASSING_START 21
NO_PASSING_END 22
UPHILL 23
DOWNHILL 24
PEDESTRIAN_ISLAND 25
SHARP_TURN_RIGHT 26
SHARP_TURN_LEFT 27
RIGHT_OF_WAY 28
NO_CLASS 29 | 2022-12-04 05:17:22 | {"extraction_info": {"found_math": true, "script_math_tex": 111, "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": 0.7157768607139587, "perplexity": 547.7350622302502}, "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-2022-49/segments/1669446710962.65/warc/CC-MAIN-20221204040114-20221204070114-00610.warc.gz"} |
https://www.clutchprep.com/chemistry/practice-problems/76974/for-the-reaction-a-b-c-d-e-the-initial-reaction-rate-was-measured-for-various-in-1 | # Problem: For the reaction A + B + C → D + E, the initial reaction rate was measured for various initial concentrations of reactants. The following data were collected: What is the value of the rate constant k for this reaction? When entering compound units, indicate multiplication of units explicitly using a multiplication dot (in the menu). For example, M-1 • s-1.Express your answer to two significant figures and include the appropriate units. Indicate the multiplication of units explicitly either with a multiplication dot or a dash.
###### FREE Expert Solution
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###### FREE Expert Solution
85% (488 ratings)
###### Problem Details
For the reaction A + B + C → D + E, the initial reaction rate was measured for various initial concentrations of reactants. The following data were collected:
What is the value of the rate constant k for this reaction?
When entering compound units, indicate multiplication of units explicitly using a multiplication dot (in the menu). For example, M-1 • s-1.
Express your answer to two significant figures and include the appropriate units. Indicate the multiplication of units explicitly either with a multiplication dot or a dash. | 2021-01-20 07:14:45 | {"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": 0.9306305646896362, "perplexity": 1919.5517538196536}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-04/segments/1610703519923.26/warc/CC-MAIN-20210120054203-20210120084203-00437.warc.gz"} |
https://mathshistory.st-andrews.ac.uk/Biographies/Faa_di_Bruno/ | Francesco da Paola Virgilio Secondo Maria Faà di Bruno
Quick Info
Born
29 March 1825
Alessandria, Piemonte (now Italy)
Died
27 March 1888
Turin, Italy
Summary
Faa di Bruno is best known for his formula for the nth derivative of a composition of functions. He was canonised in 1988
Biography
Francesco Faà di Bruno's parents were Carolina Sappa and Luigi Faà di Bruno. Francesco was the youngest of his parents' twelve children and he had seven sisters and four brothers. His father was a wealthy landowner with several titles such as Marquis of Bruno, Count of Carentino, Lord of Fontanile, and Patrizio of Alessandria. The village of Bruno is about 18 km south west of Alessandria, a major city roughly equidistant from Turin, Milan and Genoa. Francesco's mother, Carolina, was from a noble family from Milan. Patrizia Solari writes [11] that Luigi and Carolina had:-
... a happy marriage, one of the most wealthy of the Piedmontese nobility and most generous to the needy people.
Although he is always known as Francesco, the full name of the subject of this biography was Francesco da Paola Virgilio Secondo Maria Faà di Bruno. One of Francesco's brothers, Giuseppe Maria Faà di Bruno, built the church of St Peter in Hatton Garden, London, the most important church for Italian immigrants. He was a leader of the Pious Society of Missions and wrote "Catholic Belief", an exposition of Catholic doctrine aimed at non-Catholics. Over one million copies of this book were sold. Another of Francesco's brothers, Emilio, served in the army and was killed in the Battle of Lissa in 1866. Two of Francesco's sisters became nuns.
When Francesco was nine years old, his mother died. His early education had been at home and he began his school career in 1836 when he entered the college of the Novi Ligure dei Padri Somaschi. After studying for four years at the Novi Ligure he entered the Royal Military Academy of Turin in 1840 with the aim of making a career in the army. It was a six-year course and after graduating in 1846 he was commissioned as a lieutenant in the army. The advanced courses he took included topography and foreign languages. The revolutions which broke out across Europe in 1848 led to the First Italian War of Independence in which Faà di Bruno participated in the Piedmontese Brigata Guardie commanded by Vittorio Emanuele. For six weeks the Brigata Guardie attacked the Austrians at Peschiera del Garda and were victorious on 30 May 1848. After this military campaign there was an uneasy truce was lasted around seven months. Faà di Bruno's main task during these months was to draw up maps and he produced his major map of the Mincio area. He undertook his duties with great enthusiasm, realising that the generals commanding the Piedmontese armies did not have recent accurate maps of the Lombardo-Veneto region. In 1849 he was promoted to captain and shortly after this, in March 1849, the Austrians broke the true and the resulting battle between the Austrian and Piedmontese armies at Novara on 22 March was a humiliating defeat for the Piedmontese. Faà di Bruno's horse was killed by Austrian rifle fire, so he mounted another horse only to have that killed too. He was wounded in the leg and seeing many of his comrades killed, he realised that an army career was not for him.
Charles Albert, the King of Sardinia, resigned following the defeat at Novara and his son Vittorio Emanuele became King of Sardinia. He had two young sons and, realising Faà di Bruno was both an extraordinary person and an exceptional scholar, he asked him to tutor his two young sons. However, Faà di Bruno was known as a devout Catholic and the King immediately came under pressure to appoint a secular tutor for his sons. Vittorio Emanuele decided that, for political reasons, he had to withdraw his offer to Faà di Bruno which he did. Then Faà di Bruno asked permission to leave the army and take up the study of mathematics. He travelled to Paris in 1850 where he studied at the Sorbonne under Augustin-Louis Cauchy who [2]:-
... he admired, not only for his genius, but also for his religious fervour and his philanthropy.
At the Sorbonne Faà di Bruno was in the same classes as Charles Hermite and the two became close friends. However, his time in Paris was not all spent on academic studies for he also assisted in the parish of Saint Sulpice and visited the homes of the poor. He greatly enjoyed visiting the bookshops of Paris and shops where scientific instruments were sold. While in Paris he began publishing mathematical papers: Note sur un nouveau procédé pour reconnaître immédiatment, dans certains cas, l'existence de racines imaginaires dans une équation numérique (1850); Démonstration d'un théorème de M Sylvester, relatif à la décomposition d'un produit de deux déterminants (1851); and Démonstration d'un théorème relatif à la réduction des fonctions homogènes à deux lettres à leur forme canonique (1852). After graduating from Paris with his Licence in Science in 1851, he returned to Turin but this was a difficult time for anyone who was a devout Catholic [3]:-
Among the greatest tragedies of his life, there will always be the contrast between his aspirations, as a sincere patriot desiring Italian unity, but rejection, as a Catholic loyal to the pope, of the methods and unacceptable manner in which that unity was pursued with persecution and oppression of the Church.
The Church of San Massimo, in Borgo Nuovo, was completed in June 1853 and, in the following year, Faà di Bruno set up a choral school for women in the church where they were trained every Sunday by Faà di Bruno who played the organ. He now published papers in Italian such as Sullo stabilimento di un osservatorio magnetico e meteorologico in Torino (1853), and Theorema di geometria (1853), and also continued to publish in French, for example Note sur un théorème de M Brioschi (1854). In May 1855 Faà di Bruno returned to Paris to undergo further training at the astronomical observatory at Brera. The authorities in Turin promised him that, after he returned from Paris, he would be employed at the Observatory in Turin - a promise they did not keep. However, in Paris Faà di Bruno studied astronomy under Urbain Le Verrier and also undertook research with Cauchy on mathematics. He graduated in 1856 having presented two theses, one in mathematics Théorie générale de l'élimination on elimination theory, and the other in astronomy on celestial mechanics. While in Paris he had also invented a mechanical device to allow blind people to write. He had personal reasons to invent such an apparatus, for his sister Maria Luigia was becoming blind.
He continued to publish papers in both French and in Italian: Sullo sviluppo delle Funzioni (1855), Sulle Funzioni Isobariche (1856) and Note sur une nouvelle formule de calcul différentiel (1857) were all written while he was undertaking research in Paris. Both the first and the third of these papers contain the result for which he is best known, namely Faà di Bruno's formula. This formula gives the $n$th derivative of the composition of two functions $f (g(x))$. Warren Johnson writes [7]:-
Once Faà di Bruno's formula was considered a real analysis result: it is in the 'Cours d'Analyse' of Goursat and of de la Vallée Poussin. Riordan and Comtet saw it as part of combinatorial analysis, a term that seems to be going out of fashion; the subject subsumed in algebraic combinatorics, the books of Riordan and Comtet largely superseded by Stanley's monumental 'Enumerative combinatorics' (1997, 1999) where Faà di Bruno's formula is mentioned, but not stated. It can also be found in books on partitions [George E Andrews, 'The theory of partitions' (1976)], mathematical statistics, matrix theory, calculus of finite differences, computer science [Donald E Knuth, 'The Art of Computer Programming' (1968)], symmetric functions, and miscellaneous mathematical techniques.
Several authors have pointed out that Faà di Bruno was not the first to either state or prove this result. The papers [4] and [7] give fascinating accounts of earlier work. However, Faà di Bruno did give a form of the formula using determinants which nobody had found earlier. After his time in Paris, he was appointed as a lecturer in higher analysis at the University of Turin, and he also gave popular astronomy courses. In 1859 he published an important book in French, Théorie générale de l'élimination . By this stage in his career he had 20 works published. From 1859 Angelo Genocchi held the Chair of Algebra and Complementary Geometry at Turin, then in the following year he moved to the Chair of Higher Analysis and Faà di Bruno was appointed as his deputy. He obtained his doctorate, essentially a D.Sc., in 1861.
After his time studying in the Sorbonne, Faà di Bruno did much charity work on his return to Turin. He had seen food being prepared and distributed to the poor while in Paris and, back in Turin, he began to organise a similar scheme during the winter months. At this time Faà di Bruno came in contact with Giovanni Bosco. Bosco had been ordained a Roman Catholic priest in 1841 in Turin and began to work there to help boys who came to look for work in the city. Bosco provided boys with education, religious instruction, and recreation. Eventually he headed a large establishment containing a grammar school, a technical school, and a church, all built through his efforts. In Turin Bosco and others founded the Society of St Francis de Sales in 1859. Faà di Bruno, following Bosco's example, founded the Pia opera di Santa Zita in the San Donato area of Turin on 2 February 1859. He used his own money together with funds he had collected standing at the doors of churches. He chose the site because this was an area of Turin which was inhabited by the poorest people. Also in 1859 Faà di Bruno founded the Opera per la santificazione delle feste, a Society to promote Sunday observance and to protect workers who were being forced to work on Sundays. He was president of the Society and Bosco accepted the role of vice-president.
Faà di Bruno was incredibly energetic in his work with the poor. In 1860 he founded the Infermeria di San Giuseppe, an infirmary for poor women and the sick where people had an opportunity to convalesce after an illness. In 1862 he founded a boarding-house for the elderly and disabled women. In fact one has to realise that there was much illness in Turin at this time. Frequent epidemics, especially of typhus and cholera, were the result of poor hygiene and Faà di Bruno was able to set up wash rooms in the San Donato area. But his charity work was not restricted to caring for the sick, for he was also passionate about providing educational opportunities for the young people from poor families. He organised a mobile library in 1863 which provided books on many topics, particularly religion and science. In 1864 he set up classes providing training in home economics and, in 1866, he organised courses to train people to become elementary school teachers.
We mentioned above that Emilio Faà di Bruno, one of Francesco's brothers, served in the army and was killed in the Battle of Lissa in 1866. Francesco began the building of the church of Nostra Signora del Suffragio in 1866, following the death of his brother. Building the church took around three years and much of the architectural design was due to Faà di Bruno himself. His experience as a soldier, and his brother's death, had greatly affected him and he held daily prayers in the church of Nostra Signora del Suffragio for the souls of all soldiers killed in wars. Despite the many long hours he spent undertaking charity work, Faà di Bruno did not ignore his mathematical research. He continued to publish articles and gained international fame as a mathematician. This fame, however, did not lead to rapid promotion within the University of Turin. The reason for this was the secular nature of the Italian Independence movement with discrimination against those active in the Church. In 1871 he was put in charge of teaching calculus and analytic geometry and he was appointed as an extraordinary professor of higher analysis in 1876. His colleagues thought very highly of him and seven times they put him forward for a chair. Although he is usually remembered today because of "Faà di Bruno's formula", his most influential mathematical work was his book Théorie des formes binaires on binary forms which he published in 1876. The book was based on lectures that Faà di Bruno had given at the University of Turin. The book became better known in 1881 when Max Noether published a German edition. Paul Gordan wrote to Faà di Bruno from Erlangen on 29 September 1875 and his letter is reproduced, both in the original German and in a French translation, in the Preface to Théorie des formes binaires :-
I have had the opportunity to read your book on binary forms, and I was happy because I found it well adapted to introduce the reader to the theory of invariants. The subject is thoroughly and brilliantly set out, the exposition is simple, clear and, in several places, elegant. ... You have with this work delivered a service to science for which it will be grateful, since you have filled an important gap.
Not only did Faà di Bruno have difficulties in obtaining promotion within the university, but he also had difficulty in becoming ordained. Although he had undertaken the necessary training, his archbishop was opposed to ordaining men later in their lives. Faà di Bruno had to make a special plea to pope Pius IX to overrule the archbishop before he was ordained a Roman Catholic priest in Rome on 22 October 1876. The religious order he had founded, the Suore Minime di Nostra Signora del Suffragio, supported girls in a house called the Conservatorio del Suffragio. In order to provide work for the girls, Faà di Bruno had the idea that they could train as typesetters. He purchased a printing press and set up the Tipographia Suffragio. There a number of mathematics books were published including one by Faà di Bruno himself on elliptic functions. In 1898, ten years after Faà di Bruno's death, the printing press was purchased by Giuseppe Peano for 407 lire and he printed the Rivista di Matematica on it for several years.
In [2] Faà di Bruno is described as follows:-
Faà di Bruno was tall and not always well dressed, but he was simple and good natured. He was of a solitary disposition and spoke seldom (and not always successfully in the classroom). He cultivated music and was said to be a good pianist.
In fact, to give a little more information on his musical talents, he composed scared melodies which were highly thought of by Franz Liszt. We have mentioned above his interest in scientific instruments, but let us add that he invented, among other things, a differential barometer, described in a publication of 1870, and an electric alarm clock. He set up a Foucault pendulum in his church, the Chiesa del Suffragio, to demonstrate the rotation of the earth.
He died suddenly, two days short of his 63rd birthday, from an intestinal infection. Bosco, who had been an inspiration to Faà di Bruno and had died less than two months before him, was made a Saint on 1 April 1934. Already by this time there was a movement to canonise Faà di Bruno and in 1955 the Sacred Congregation of Rites officially accepted the claim for Faà di Bruno to be canonised. Faà di Bruno was declared a Saint by pope John Paul II in St Peter's Square in Rome on 25 September 1988.
References (show)
1. G B Contol, M Cecchetto and E Innaurato, Francesco Faà di Bruno (1825-1888): miscellanea (Bottega d'Erasmo, 1977).
2. H C Kennedy, Peano : Life and Works of Giuseppe Peano (Dordrecht, 1980).
3. V Messori, Il beato Faà di Bruno. Un cristiano in un mondo ostile (Biblioteca universale Rizzoli, 1998).
4. A D D Craik, Prehistory of Faà di Bruno's formula, Amer. Math. Monthly 112 (2) (2005), 119-130.
5. L Dell'Aglio, Francesco Faà di Bruno (Italian), Dizionario Biografico degli Italiani 43 (1993).
6. M Gota, Un gigante della carità e della fede, Vita Cattolica (21 September 2008), 11.
7. W P Johnson, The Curious History of Faà di Bruno's Formula, Amer. Math. Monthly 109 (3) (2002), 217-234.
8. P Linehan, Francesco Faa di Bruno, The Catholic Encyclopedia 5 (Robert Appleton Company, New York, 1909). http://www.newadvent.org/cathen/05740a.htm
9. S Roman, The formula of Faà di Bruno, Amer. Math. Monthly 87 (10) (1980), 805-809.
10. P Solari, Beato Francesco Faà di Bruno, Caritas Insieme XXV (2) (2007), 40-44.
11. P Solari, Beato Francesco Faà di Bruno. Seconda parte, Caritas Insieme XXVI (1) (2008), 44-47.
12. G Zappa and G Casadio, The mathematical activity of Francesco Faà di Bruno from 1850 to 1859 (Italian), Mem. Accad. Sci. Torino Cl. Sci. Fis. Mat. Natur. (5) 16 (1-4) (1992), 1-25. | 2020-07-06 03:51:09 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 2, "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": 0.33560165762901306, "perplexity": 4001.84116850562}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-29/segments/1593655890092.28/warc/CC-MAIN-20200706011013-20200706041013-00352.warc.gz"} |
http://www.machinedlearnings.com/2011/08/multi-label-extraction-from.html | ## Monday, August 29, 2011
### Multi-label extraction from crowdsourced data sets
Previously I've discussed techniques for processing crowdsourced data corresponding to tasks of the form given an item, choose the best single label from a fixed set of labels'', which corresponds to cost-sensitive multiclass classification (CSMC). The result of this processing might be a final decision, or it might be a cost-vector which is used to train a supervised learning system.
Now I'm concerned with task of the form given an item, choose zero or more applicable labels from a fixed set of labels'', which corresponds to cost-sensitive multilabel classification (CSML). One strategy for dealing with CSML is to reduce to a set of independent binary classification problems, each predicting whether or not a particular label should be assigned to an item; I'll call this strategy IBR. IBR is a consistent reduction if the cost function on the original CSML problem is weighted Hamming loss, but is inconsistent for other CSML losses (e.g., 0/1 loss on the entire set). In practice with finite regret on the induced subproblems it might not even be a good strategy for weighted Hamming loss.
Nonetheless for a while this was the only approach I had implemented; for example, if I had a 10 label CSML problem, I would process the crowdsourced data into 10 data sets corresponding to binary classification, run nominallabelextract on each of the 10 data sets, and then combine the results. There are some undesirable aspects of this strategy, all of which are different facets of the same underlying issue. First, as indicated above, when the result of crowdsourced processing is directly used to a make a decision it is only consistent for weighted Hamming loss. Second, when used to construct a training set, the ground truth distributions it produces are always separable (i.e., the product of one-dimensional distributions). Third, the resulting generative model of worker errors is unable to model correlations in the labeling error because each induced binary subproblems treats all errors as equivalent. In particular, if a worker is consistently confusing two different labels, this reduction cannot exploit that (because in the induced subproblem, the informative errors'' are mixed in with all the other negative responses).
Another approach to CSML on label set $L$ is to reduce to CSMC on label power set $\mathcal{P} (L)$. This is one of those reductions everybody knows about and nobody likes, because of the combinatorial explosion of the power set cardinality, but it does capture higher-order structure in the costs. It is consistent with any loss function but typically runs into sample complexity issues, and the tricks used to mitigate sample complexity might cause regret to be poor in practice. The situation is no different here, because when I reduce to CSMC I'm going to leverage the low-rank approximation nominallowrankextract I recently introduced, which may or may not work well in practice.
I did the straightforward thing of taking nominallowrankextract and mapping a multi-label data set onto it via a combinatorial number system, resulting in multilowrankextract. Because the number of parameters in the nominallowrankextract model is proportional to the number of labels $|L|$, the number of parameters in the multilowrankextract model is proportional to something like $2^{|L|}$. In practice it is a bit smaller since I allow one to say that a label set has probability zero if there are too many labels in it, e.g., for an 11 label problem where the underlying ground truth set for an item has at most 3 labels the number of labels in the induced subproblem is $\sum_{k=0}^3 {11 \choose k} = 232$. This trick is very important because inference is still $O (|L|^2)$ time complexity in nominallowrankextract so keeping the induced label set small is key to low blood pressure.
I'm still evaluating whether multilowrankextract is better than than IBR. I looked at one problem from a 0/1 (entire set) loss perspective, i.e., I looked at the most (posterior) likely set from both techniques. The two approaches tend to agree: on a test problem with 853 items, the two approaches had the same posterior mode 718 times, and a different one 135 times. This is not surprising: when the crowdsource workers have strong consensus any reasonable model will output the consensus as the posterior mode, so the only opportunity to get creative'' is when the crowdsource workers disagree. If this is happening often, this indicates task redesign is necessary, since the tasks are either ill-defined, ambiguous, or extremely difficult. For the 135 items where the two approaches differed, I manually decided which label set I liked better. 29 times I liked IBR better, 30 times I liked multilowrankextract better, and 76 times I had no preference (and could appreciate why the crowdsourced workers were in disagreement!). That's a statistical dead heat.
Given that IBR scales computationally much better than multilowrankextract, it would currently be the clear choice for large label sets (e.g., $|L| \gg 10$). For small label sets right now I'm using multilowrankextract because I like the richer posterior distribution it produces, but that's just intuition and I don't have anything quantitative to back it up at the moment.
You can get the current implementation of multilowrankextract as part of nominallowrankextract from the nincompoop code repository. | 2021-08-04 01:59:25 | {"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": 0.762180507183075, "perplexity": 922.0391148931817}, "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/1627046154500.32/warc/CC-MAIN-20210804013942-20210804043942-00376.warc.gz"} |
https://socratic.org/questions/for-a-first-order-reaction-if-the-time-for-50-75-and-87-5-changes-are-t1-t2-and- | # For a first order reaction if the time for 50%, 75% and 87.5% changes are t1, t2 and t3 respectively then what would be the value of t1: t2: t3 ?
Then teach the underlying concepts
Don't copy without citing sources
preview
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#### Explanation
Explain in detail...
#### Explanation:
I want someone to double check my answer
2
Mar 9, 2018
${t}_{1} : {t}_{2} : {t}_{3} = 1 : 2 : 3$
#### Explanation:
Consider a first-order reaction
$\text{A →Products}$
The time required for the initial concentration ${\left[\text{A}\right]}_{0}$ to drop to
• $\frac{1}{2} {\left[\text{A}\right]}_{0}$ is one half-life
• $\frac{1}{4} {\left[\text{A}\right]}_{0}$ is two half-lives
• $\frac{1}{8} {\left[\text{A}\right]}_{0}$ is three half-lives
Also,
t_½ = ln2/k. so each half-life takes the same amount of time.
The ratios are
${t}_{1} : {t}_{2} : {t}_{3} = \text{1 half-life:2 half-lives:3 half-lives = 1:2:3}$
Here is a typical plot for a first-order reaction.
The initial concentration is 1.7.
It takes
• 400 s for the concentration to drop by 50 % (to 0.85)
• 800 s for the concentration to drop by 75 % (0.42)
• 400 s for the concentration to drop by 87.5 % (to 0.21)
The times are in the ratio $\text{1:2:3}$.
Then teach the underlying concepts
Don't copy without citing sources
preview
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#### Explanation
Explain in detail...
#### Explanation:
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1
Al E. Share
Mar 8, 2018
Recall, for a first order elementary reaction,
$\ln {\left[A\right]}_{\text{t}} = - k t + \ln {\left[A\right]}_{0}$
$\implies \ln \left(\frac{{\left[A\right]}_{\text{t}}}{{\left[A\right]}_{0}}\right) = - k t$
Without more data, I couldn't give you a real value, but I could give you one as a factor of $k$ (the rate constant).
We can conceptually assume that ${\left[A\right]}_{0} = 1 M$ to derive our data.
Hence,
$\ln \left(\frac{{\left[A\right]}_{\text{t}}}{{\left[A\right]}_{0}}\right) = - k t$
$\implies {t}_{1} = - \ln \frac{0.5}{k} \approx \frac{0.693}{k}$
${t}_{2} = - \ln \frac{0.75}{k} \approx \frac{0.288}{k}$, and
${t}_{3} = - \ln \frac{0.875}{k} \approx \frac{0.134}{k}$
Moreover,
${t}_{1} : {t}_{2} : {t}_{3} \approx 5.17 : 2.15 : 1$
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• 26 minutes ago | 2018-06-25 07:40:43 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 18, "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": 0.6915490031242371, "perplexity": 3967.9876565578875}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-26/segments/1529267867579.80/warc/CC-MAIN-20180625072642-20180625092642-00430.warc.gz"} |
https://plotly.com/python/reference/layout/ternary/ | ## Python Figure Reference: layout.ternary
• ternary
Code: fig.update_ternaries(...)
Type: dict containing one or more of the keys listed below.
• aaxis
Code: fig.update_ternaries(aaxis=dict(...))
Type: dict containing one or more of the keys listed below.
• color
Code: fig.update_ternaries(aaxis_color=<VALUE>)
Type: color
Default: "#444"
Sets default for all colors associated with this axis all at once: line, font, tick, and grid colors. Grid color is lightened by blending this with the plot background Individual pieces can override this.
• dtick
Code: fig.update_ternaries(aaxis_dtick=<VALUE>)
Type: number or categorical coordinate string
Sets the step in-between ticks on this axis. Use with tick0. Must be a positive number, or special strings available to "log" and "date" axes. If the axis type is "log", then ticks are set every 10^(n"dtick) where n is the tick number. For example, to set a tick mark at 1, 10, 100, 1000, ... set dtick to 1. To set tick marks at 1, 100, 10000, ... set dtick to 2. To set tick marks at 1, 5, 25, 125, 625, 3125, ... set dtick to log_10(5), or 0.69897000433. "log" has several special values; "L<f>", where f is a positive number, gives ticks linearly spaced in value (but not position). For example tick0 = 0.1, dtick = "L0.5" will put ticks at 0.1, 0.6, 1.1, 1.6 etc. To show powers of 10 plus small digits between, use "D1" (all digits) or "D2" (only 2 and 5). tick0 is ignored for "D1" and "D2". If the axis type is "date", then you must convert the time to milliseconds. For example, to set the interval between ticks to one day, set dtick to 86400000.0. "date" also has special values "M<n>" gives ticks spaced by a number of months. n must be a positive integer. To set ticks on the 15th of every third month, set tick0 to "2000-01-15" and dtick to "M3". To set ticks every 4 years, set dtick to "M48"
• exponentformat
Code: fig.update_ternaries(aaxis_exponentformat=<VALUE>)
Type: enumerated , one of ( "none" | "e" | "E" | "power" | "SI" | "B" )
Default: "B"
Determines a formatting rule for the tick exponents. For example, consider the number 1,000,000,000. If "none", it appears as 1,000,000,000. If "e", 1e+9. If "E", 1E+9. If "power", 1x10^9 (with 9 in a super script). If "SI", 1G. If "B", 1B.
• gridcolor
Code: fig.update_ternaries(aaxis_gridcolor=<VALUE>)
Type: color
Default: "#eee"
Sets the color of the grid lines.
• griddash
Code: fig.update_ternaries(aaxis_griddash=<VALUE>)
Type: string
Default: "solid"
Sets the dash style of lines. Set to a dash type string ("solid", "dot", "dash", "longdash", "dashdot", or "longdashdot") or a dash length list in px (eg "5px,10px,2px,2px").
• gridwidth
Code: fig.update_ternaries(aaxis_gridwidth=<VALUE>)
Type: number greater than or equal to 0
Default: 1
Sets the width (in px) of the grid lines.
• hoverformat
Code: fig.update_ternaries(aaxis_hoverformat=<VALUE>)
Type: string
Default: ""
Sets the hover text formatting rule using d3 formatting mini-languages which are very similar to those in Python. For numbers, see: https://github.com/d3/d3-format/tree/v1.4.5#d3-format. And for dates see: https://github.com/d3/d3-time-format/tree/v2.2.3#locale_format. We add two items to d3's date formatter: "%h" for half of the year as a decimal number as well as "%{n}f" for fractional seconds with n digits. For example, "2016-10-13 09:15:23.456" with tickformat "%H~%M~%S.%2f" would display "09~15~23.46"
• labelalias
Code: fig.update_ternaries(aaxis_labelalias=<VALUE>)
Type: number or categorical coordinate string
Replacement text for specific tick or hover labels. For example using {US: 'USA', CA: 'Canada'} changes US to USA and CA to Canada. The labels we would have shown must match the keys exactly, after adding any tickprefix or ticksuffix. labelalias can be used with any axis type, and both keys (if needed) and values (if desired) can include html-like tags or MathJax.
• layer
Code: fig.update_ternaries(aaxis_layer=<VALUE>)
Type: enumerated , one of ( "above traces" | "below traces" )
Default: "above traces"
Sets the layer on which this axis is displayed. If "above traces", this axis is displayed above all the subplot's traces If "below traces", this axis is displayed below all the subplot's traces, but above the grid lines. Useful when used together with scatter-like traces with cliponaxis set to "False" to show markers and/or text nodes above this axis.
• linecolor
Code: fig.update_ternaries(aaxis_linecolor=<VALUE>)
Type: color
Default: "#444"
Sets the axis line color.
• linewidth
Code: fig.update_ternaries(aaxis_linewidth=<VALUE>)
Type: number greater than or equal to 0
Default: 1
Sets the width (in px) of the axis line.
• min
Code: fig.update_ternaries(aaxis_min=<VALUE>)
Type: number greater than or equal to 0
Default: 0
The minimum value visible on this axis. The maximum is determined by the sum minus the minimum values of the other two axes. The full view corresponds to all the minima set to zero.
• minexponent
Code: fig.update_ternaries(aaxis_minexponent=<VALUE>)
Type: number greater than or equal to 0
Default: 3
Hide SI prefix for 10^n if |n| is below this number. This only has an effect when tickformat is "SI" or "B".
• nticks
Code: fig.update_ternaries(aaxis_nticks=<VALUE>)
Type: integer greater than or equal to 1
Default: 6
Specifies the maximum number of ticks for the particular axis. The actual number of ticks will be chosen automatically to be less than or equal to nticks. Has an effect only if tickmode is set to "auto".
• separatethousands
Code: fig.update_ternaries(aaxis_separatethousands=<VALUE>)
Type: boolean
If "True", even 4-digit integers are separated
• showexponent
Code: fig.update_ternaries(aaxis_showexponent=<VALUE>)
Type: enumerated , one of ( "all" | "first" | "last" | "none" )
Default: "all"
If "all", all exponents are shown besides their significands. If "first", only the exponent of the first tick is shown. If "last", only the exponent of the last tick is shown. If "none", no exponents appear.
• showgrid
Code: fig.update_ternaries(aaxis_showgrid=<VALUE>)
Type: boolean
Default: True
Determines whether or not grid lines are drawn. If "True", the grid lines are drawn at every tick mark.
• showline
Code: fig.update_ternaries(aaxis_showline=<VALUE>)
Type: boolean
Default: True
Determines whether or not a line bounding this axis is drawn.
• showticklabels
Code: fig.update_ternaries(aaxis_showticklabels=<VALUE>)
Type: boolean
Default: True
Determines whether or not the tick labels are drawn.
• showtickprefix
Code: fig.update_ternaries(aaxis_showtickprefix=<VALUE>)
Type: enumerated , one of ( "all" | "first" | "last" | "none" )
Default: "all"
If "all", all tick labels are displayed with a prefix. If "first", only the first tick is displayed with a prefix. If "last", only the last tick is displayed with a suffix. If "none", tick prefixes are hidden.
• showticksuffix
Code: fig.update_ternaries(aaxis_showticksuffix=<VALUE>)
Type: enumerated , one of ( "all" | "first" | "last" | "none" )
Default: "all"
Same as showtickprefix but for tick suffixes.
• tick0
Code: fig.update_ternaries(aaxis_tick0=<VALUE>)
Type: number or categorical coordinate string
Sets the placement of the first tick on this axis. Use with dtick. If the axis type is "log", then you must take the log of your starting tick (e.g. to set the starting tick to 100, set the tick0 to 2) except when dtick="L<f>" (see dtick for more info). If the axis type is "date", it should be a date string, like date data. If the axis type is "category", it should be a number, using the scale where each category is assigned a serial number from zero in the order it appears.
• tickangle
Code: fig.update_ternaries(aaxis_tickangle=<VALUE>)
Type: angle
Default: "auto"
Sets the angle of the tick labels with respect to the horizontal. For example, a tickangle of -90 draws the tick labels vertically.
• tickcolor
Code: fig.update_ternaries(aaxis_tickcolor=<VALUE>)
Type: color
Default: "#444"
Sets the tick color.
• tickfont
Code: fig.update_ternaries(aaxis_tickfont=dict(...))
Type: dict containing one or more of the keys listed below.
Sets the tick font.
• color
Code: fig.update_ternaries(aaxis_tickfont_color=<VALUE>)
Type: color
• family
Code: fig.update_ternaries(aaxis_tickfont_family=<VALUE>)
Type: string
HTML font family - the typeface that will be applied by the web browser. The web browser will only be able to apply a font if it is available on the system which it operates. Provide multiple font families, separated by commas, to indicate the preference in which to apply fonts if they aren't available on the system. The Chart Studio Cloud (at https://chart-studio.plotly.com or on-premise) generates images on a server, where only a select number of fonts are installed and supported. These include "Arial", "Balto", "Courier New", "Droid Sans",, "Droid Serif", "Droid Sans Mono", "Gravitas One", "Old Standard TT", "Open Sans", "Overpass", "PT Sans Narrow", "Raleway", "Times New Roman".
• size
Code: fig.update_ternaries(aaxis_tickfont_size=<VALUE>)
Type: number greater than or equal to 1
• tickformat
Code: fig.update_ternaries(aaxis_tickformat=<VALUE>)
Type: string
Default: ""
Sets the tick label formatting rule using d3 formatting mini-languages which are very similar to those in Python. For numbers, see: https://github.com/d3/d3-format/tree/v1.4.5#d3-format. And for dates see: https://github.com/d3/d3-time-format/tree/v2.2.3#locale_format. We add two items to d3's date formatter: "%h" for half of the year as a decimal number as well as "%{n}f" for fractional seconds with n digits. For example, "2016-10-13 09:15:23.456" with tickformat "%H~%M~%S.%2f" would display "09~15~23.46"
• tickformatstops
Code: fig.update_ternaries(aaxis_tickformatstops=list(...))
Type: list of dict where each dict has one or more of the keys listed below.
• dtickrange
Parent: layout.ternary.aaxis.tickformatstops[]
Type: list
range ["min", "max"], where "min", "max" - dtick values which describe some zoom level, it is possible to omit "min" or "max" value by passing "null"
• enabled
Parent: layout.ternary.aaxis.tickformatstops[]
Type: boolean
Default: True
Determines whether or not this stop is used. If False, this stop is ignored even within its dtickrange.
• name
Parent: layout.ternary.aaxis.tickformatstops[]
Type: string
When used in a template, named items are created in the output figure in addition to any items the figure already has in this array. You can modify these items in the output figure by making your own item with templateitemname matching this name alongside your modifications (including visible: False or enabled: False to hide it). Has no effect outside of a template.
• templateitemname
Parent: layout.ternary.aaxis.tickformatstops[]
Type: string
Used to refer to a named item in this array in the template. Named items from the template will be created even without a matching item in the input figure, but you can modify one by making an item with templateitemname matching its name, alongside your modifications (including visible: False or enabled: False to hide it). If there is no template or no matching item, this item will be hidden unless you explicitly show it with visible: True.
• value
Parent: layout.ternary.aaxis.tickformatstops[]
Type: string
Default: ""
string - dtickformat for described zoom level, the same as "tickformat"
• ticklabelstep
Code: fig.update_ternaries(aaxis_ticklabelstep=<VALUE>)
Type: integer greater than or equal to 1
Default: 1
Sets the spacing between tick labels as compared to the spacing between ticks. A value of 1 (default) means each tick gets a label. A value of 2 means shows every 2nd label. A larger value n means only every nth tick is labeled. tick0 determines which labels are shown. Not implemented for axes with type "log" or "multicategory", or when tickmode is "array".
• ticklen
Code: fig.update_ternaries(aaxis_ticklen=<VALUE>)
Type: number greater than or equal to 0
Default: 5
Sets the tick length (in px).
• tickmode
Code: fig.update_ternaries(aaxis_tickmode=<VALUE>)
Type: enumerated , one of ( "auto" | "linear" | "array" )
Sets the tick mode for this axis. If "auto", the number of ticks is set via nticks. If "linear", the placement of the ticks is determined by a starting position tick0 and a tick step dtick ("linear" is the default value if tick0 and dtick are provided). If "array", the placement of the ticks is set via tickvals and the tick text is ticktext. ("array" is the default value if tickvals is provided).
• tickprefix
Code: fig.update_ternaries(aaxis_tickprefix=<VALUE>)
Type: string
Default: ""
Sets a tick label prefix.
• ticks
Code: fig.update_ternaries(aaxis_ticks=<VALUE>)
Type: enumerated , one of ( "outside" | "inside" | "" )
Determines whether ticks are drawn or not. If "", this axis' ticks are not drawn. If "outside" ("inside"), this axis' are drawn outside (inside) the axis lines.
• ticksuffix
Code: fig.update_ternaries(aaxis_ticksuffix=<VALUE>)
Type: string
Default: ""
Sets a tick label suffix.
• ticktext
Code: fig.update_ternaries(aaxis_ticktext=<VALUE>)
Type: list, numpy array, or Pandas series of numbers, strings, or datetimes.
Sets the text displayed at the ticks position via tickvals. Only has an effect if tickmode is set to "array". Used with tickvals.
• tickvals
Code: fig.update_ternaries(aaxis_tickvals=<VALUE>)
Type: list, numpy array, or Pandas series of numbers, strings, or datetimes.
Sets the values at which ticks on this axis appear. Only has an effect if tickmode is set to "array". Used with ticktext.
• tickwidth
Code: fig.update_ternaries(aaxis_tickwidth=<VALUE>)
Type: number greater than or equal to 0
Default: 1
Sets the tick width (in px).
• title
Code: fig.update_ternaries(aaxis_title=dict(...))
Type: dict containing one or more of the keys listed below.
• font
Code: fig.update_ternaries(aaxis_title_font=dict(...))
Type: dict containing one or more of the keys listed below.
Sets this axis' title font. Note that the title's font used to be customized by the now deprecated titlefont attribute.
• color
Code: fig.update_ternaries(aaxis_title_font_color=<VALUE>)
Type: color
• family
Code: fig.update_ternaries(aaxis_title_font_family=<VALUE>)
Type: string
HTML font family - the typeface that will be applied by the web browser. The web browser will only be able to apply a font if it is available on the system which it operates. Provide multiple font families, separated by commas, to indicate the preference in which to apply fonts if they aren't available on the system. The Chart Studio Cloud (at https://chart-studio.plotly.com or on-premise) generates images on a server, where only a select number of fonts are installed and supported. These include "Arial", "Balto", "Courier New", "Droid Sans",, "Droid Serif", "Droid Sans Mono", "Gravitas One", "Old Standard TT", "Open Sans", "Overpass", "PT Sans Narrow", "Raleway", "Times New Roman".
• size
Code: fig.update_ternaries(aaxis_title_font_size=<VALUE>)
Type: number greater than or equal to 1
• text
Code: fig.update_ternaries(aaxis_title_text=<VALUE>)
Type: string
Sets the title of this axis. Note that before the existence of title.text, the title's contents used to be defined as the title attribute itself. This behavior has been deprecated.
• uirevision
Code: fig.update_ternaries(aaxis_uirevision=<VALUE>)
Type: number or categorical coordinate string
Controls persistence of user-driven changes in axis min, and title if in editable: True configuration. Defaults to ternary<N>.uirevision.
• baxis
Code: fig.update_ternaries(baxis=dict(...))
Type: dict containing one or more of the keys listed below.
• color
Code: fig.update_ternaries(baxis_color=<VALUE>)
Type: color
Default: "#444"
Sets default for all colors associated with this axis all at once: line, font, tick, and grid colors. Grid color is lightened by blending this with the plot background Individual pieces can override this.
• dtick
Code: fig.update_ternaries(baxis_dtick=<VALUE>)
Type: number or categorical coordinate string
Sets the step in-between ticks on this axis. Use with tick0. Must be a positive number, or special strings available to "log" and "date" axes. If the axis type is "log", then ticks are set every 10^(n"dtick) where n is the tick number. For example, to set a tick mark at 1, 10, 100, 1000, ... set dtick to 1. To set tick marks at 1, 100, 10000, ... set dtick to 2. To set tick marks at 1, 5, 25, 125, 625, 3125, ... set dtick to log_10(5), or 0.69897000433. "log" has several special values; "L<f>", where f is a positive number, gives ticks linearly spaced in value (but not position). For example tick0 = 0.1, dtick = "L0.5" will put ticks at 0.1, 0.6, 1.1, 1.6 etc. To show powers of 10 plus small digits between, use "D1" (all digits) or "D2" (only 2 and 5). tick0 is ignored for "D1" and "D2". If the axis type is "date", then you must convert the time to milliseconds. For example, to set the interval between ticks to one day, set dtick to 86400000.0. "date" also has special values "M<n>" gives ticks spaced by a number of months. n must be a positive integer. To set ticks on the 15th of every third month, set tick0 to "2000-01-15" and dtick to "M3". To set ticks every 4 years, set dtick to "M48"
• exponentformat
Code: fig.update_ternaries(baxis_exponentformat=<VALUE>)
Type: enumerated , one of ( "none" | "e" | "E" | "power" | "SI" | "B" )
Default: "B"
Determines a formatting rule for the tick exponents. For example, consider the number 1,000,000,000. If "none", it appears as 1,000,000,000. If "e", 1e+9. If "E", 1E+9. If "power", 1x10^9 (with 9 in a super script). If "SI", 1G. If "B", 1B.
• gridcolor
Code: fig.update_ternaries(baxis_gridcolor=<VALUE>)
Type: color
Default: "#eee"
Sets the color of the grid lines.
• griddash
Code: fig.update_ternaries(baxis_griddash=<VALUE>)
Type: string
Default: "solid"
Sets the dash style of lines. Set to a dash type string ("solid", "dot", "dash", "longdash", "dashdot", or "longdashdot") or a dash length list in px (eg "5px,10px,2px,2px").
• gridwidth
Code: fig.update_ternaries(baxis_gridwidth=<VALUE>)
Type: number greater than or equal to 0
Default: 1
Sets the width (in px) of the grid lines.
• hoverformat
Code: fig.update_ternaries(baxis_hoverformat=<VALUE>)
Type: string
Default: ""
Sets the hover text formatting rule using d3 formatting mini-languages which are very similar to those in Python. For numbers, see: https://github.com/d3/d3-format/tree/v1.4.5#d3-format. And for dates see: https://github.com/d3/d3-time-format/tree/v2.2.3#locale_format. We add two items to d3's date formatter: "%h" for half of the year as a decimal number as well as "%{n}f" for fractional seconds with n digits. For example, "2016-10-13 09:15:23.456" with tickformat "%H~%M~%S.%2f" would display "09~15~23.46"
• labelalias
Code: fig.update_ternaries(baxis_labelalias=<VALUE>)
Type: number or categorical coordinate string
Replacement text for specific tick or hover labels. For example using {US: 'USA', CA: 'Canada'} changes US to USA and CA to Canada. The labels we would have shown must match the keys exactly, after adding any tickprefix or ticksuffix. labelalias can be used with any axis type, and both keys (if needed) and values (if desired) can include html-like tags or MathJax.
• layer
Code: fig.update_ternaries(baxis_layer=<VALUE>)
Type: enumerated , one of ( "above traces" | "below traces" )
Default: "above traces"
Sets the layer on which this axis is displayed. If "above traces", this axis is displayed above all the subplot's traces If "below traces", this axis is displayed below all the subplot's traces, but above the grid lines. Useful when used together with scatter-like traces with cliponaxis set to "False" to show markers and/or text nodes above this axis.
• linecolor
Code: fig.update_ternaries(baxis_linecolor=<VALUE>)
Type: color
Default: "#444"
Sets the axis line color.
• linewidth
Code: fig.update_ternaries(baxis_linewidth=<VALUE>)
Type: number greater than or equal to 0
Default: 1
Sets the width (in px) of the axis line.
• min
Code: fig.update_ternaries(baxis_min=<VALUE>)
Type: number greater than or equal to 0
Default: 0
The minimum value visible on this axis. The maximum is determined by the sum minus the minimum values of the other two axes. The full view corresponds to all the minima set to zero.
• minexponent
Code: fig.update_ternaries(baxis_minexponent=<VALUE>)
Type: number greater than or equal to 0
Default: 3
Hide SI prefix for 10^n if |n| is below this number. This only has an effect when tickformat is "SI" or "B".
• nticks
Code: fig.update_ternaries(baxis_nticks=<VALUE>)
Type: integer greater than or equal to 1
Default: 6
Specifies the maximum number of ticks for the particular axis. The actual number of ticks will be chosen automatically to be less than or equal to nticks. Has an effect only if tickmode is set to "auto".
• separatethousands
Code: fig.update_ternaries(baxis_separatethousands=<VALUE>)
Type: boolean
If "True", even 4-digit integers are separated
• showexponent
Code: fig.update_ternaries(baxis_showexponent=<VALUE>)
Type: enumerated , one of ( "all" | "first" | "last" | "none" )
Default: "all"
If "all", all exponents are shown besides their significands. If "first", only the exponent of the first tick is shown. If "last", only the exponent of the last tick is shown. If "none", no exponents appear.
• showgrid
Code: fig.update_ternaries(baxis_showgrid=<VALUE>)
Type: boolean
Default: True
Determines whether or not grid lines are drawn. If "True", the grid lines are drawn at every tick mark.
• showline
Code: fig.update_ternaries(baxis_showline=<VALUE>)
Type: boolean
Default: True
Determines whether or not a line bounding this axis is drawn.
• showticklabels
Code: fig.update_ternaries(baxis_showticklabels=<VALUE>)
Type: boolean
Default: True
Determines whether or not the tick labels are drawn.
• showtickprefix
Code: fig.update_ternaries(baxis_showtickprefix=<VALUE>)
Type: enumerated , one of ( "all" | "first" | "last" | "none" )
Default: "all"
If "all", all tick labels are displayed with a prefix. If "first", only the first tick is displayed with a prefix. If "last", only the last tick is displayed with a suffix. If "none", tick prefixes are hidden.
• showticksuffix
Code: fig.update_ternaries(baxis_showticksuffix=<VALUE>)
Type: enumerated , one of ( "all" | "first" | "last" | "none" )
Default: "all"
Same as showtickprefix but for tick suffixes.
• tick0
Code: fig.update_ternaries(baxis_tick0=<VALUE>)
Type: number or categorical coordinate string
Sets the placement of the first tick on this axis. Use with dtick. If the axis type is "log", then you must take the log of your starting tick (e.g. to set the starting tick to 100, set the tick0 to 2) except when dtick="L<f>" (see dtick for more info). If the axis type is "date", it should be a date string, like date data. If the axis type is "category", it should be a number, using the scale where each category is assigned a serial number from zero in the order it appears.
• tickangle
Code: fig.update_ternaries(baxis_tickangle=<VALUE>)
Type: angle
Default: "auto"
Sets the angle of the tick labels with respect to the horizontal. For example, a tickangle of -90 draws the tick labels vertically.
• tickcolor
Code: fig.update_ternaries(baxis_tickcolor=<VALUE>)
Type: color
Default: "#444"
Sets the tick color.
• tickfont
Code: fig.update_ternaries(baxis_tickfont=dict(...))
Type: dict containing one or more of the keys listed below.
Sets the tick font.
• color
Code: fig.update_ternaries(baxis_tickfont_color=<VALUE>)
Type: color
• family
Code: fig.update_ternaries(baxis_tickfont_family=<VALUE>)
Type: string
HTML font family - the typeface that will be applied by the web browser. The web browser will only be able to apply a font if it is available on the system which it operates. Provide multiple font families, separated by commas, to indicate the preference in which to apply fonts if they aren't available on the system. The Chart Studio Cloud (at https://chart-studio.plotly.com or on-premise) generates images on a server, where only a select number of fonts are installed and supported. These include "Arial", "Balto", "Courier New", "Droid Sans",, "Droid Serif", "Droid Sans Mono", "Gravitas One", "Old Standard TT", "Open Sans", "Overpass", "PT Sans Narrow", "Raleway", "Times New Roman".
• size
Code: fig.update_ternaries(baxis_tickfont_size=<VALUE>)
Type: number greater than or equal to 1
• tickformat
Code: fig.update_ternaries(baxis_tickformat=<VALUE>)
Type: string
Default: ""
Sets the tick label formatting rule using d3 formatting mini-languages which are very similar to those in Python. For numbers, see: https://github.com/d3/d3-format/tree/v1.4.5#d3-format. And for dates see: https://github.com/d3/d3-time-format/tree/v2.2.3#locale_format. We add two items to d3's date formatter: "%h" for half of the year as a decimal number as well as "%{n}f" for fractional seconds with n digits. For example, "2016-10-13 09:15:23.456" with tickformat "%H~%M~%S.%2f" would display "09~15~23.46"
• tickformatstops
Code: fig.update_ternaries(baxis_tickformatstops=list(...))
Type: list of dict where each dict has one or more of the keys listed below.
• dtickrange
Parent: layout.ternary.baxis.tickformatstops[]
Type: list
range ["min", "max"], where "min", "max" - dtick values which describe some zoom level, it is possible to omit "min" or "max" value by passing "null"
• enabled
Parent: layout.ternary.baxis.tickformatstops[]
Type: boolean
Default: True
Determines whether or not this stop is used. If False, this stop is ignored even within its dtickrange.
• name
Parent: layout.ternary.baxis.tickformatstops[]
Type: string
When used in a template, named items are created in the output figure in addition to any items the figure already has in this array. You can modify these items in the output figure by making your own item with templateitemname matching this name alongside your modifications (including visible: False or enabled: False to hide it). Has no effect outside of a template.
• templateitemname
Parent: layout.ternary.baxis.tickformatstops[]
Type: string
Used to refer to a named item in this array in the template. Named items from the template will be created even without a matching item in the input figure, but you can modify one by making an item with templateitemname matching its name, alongside your modifications (including visible: False or enabled: False to hide it). If there is no template or no matching item, this item will be hidden unless you explicitly show it with visible: True.
• value
Parent: layout.ternary.baxis.tickformatstops[]
Type: string
Default: ""
string - dtickformat for described zoom level, the same as "tickformat"
• ticklabelstep
Code: fig.update_ternaries(baxis_ticklabelstep=<VALUE>)
Type: integer greater than or equal to 1
Default: 1
Sets the spacing between tick labels as compared to the spacing between ticks. A value of 1 (default) means each tick gets a label. A value of 2 means shows every 2nd label. A larger value n means only every nth tick is labeled. tick0 determines which labels are shown. Not implemented for axes with type "log" or "multicategory", or when tickmode is "array".
• ticklen
Code: fig.update_ternaries(baxis_ticklen=<VALUE>)
Type: number greater than or equal to 0
Default: 5
Sets the tick length (in px).
• tickmode
Code: fig.update_ternaries(baxis_tickmode=<VALUE>)
Type: enumerated , one of ( "auto" | "linear" | "array" )
Sets the tick mode for this axis. If "auto", the number of ticks is set via nticks. If "linear", the placement of the ticks is determined by a starting position tick0 and a tick step dtick ("linear" is the default value if tick0 and dtick are provided). If "array", the placement of the ticks is set via tickvals and the tick text is ticktext. ("array" is the default value if tickvals is provided).
• tickprefix
Code: fig.update_ternaries(baxis_tickprefix=<VALUE>)
Type: string
Default: ""
Sets a tick label prefix.
• ticks
Code: fig.update_ternaries(baxis_ticks=<VALUE>)
Type: enumerated , one of ( "outside" | "inside" | "" )
Determines whether ticks are drawn or not. If "", this axis' ticks are not drawn. If "outside" ("inside"), this axis' are drawn outside (inside) the axis lines.
• ticksuffix
Code: fig.update_ternaries(baxis_ticksuffix=<VALUE>)
Type: string
Default: ""
Sets a tick label suffix.
• ticktext
Code: fig.update_ternaries(baxis_ticktext=<VALUE>)
Type: list, numpy array, or Pandas series of numbers, strings, or datetimes.
Sets the text displayed at the ticks position via tickvals. Only has an effect if tickmode is set to "array". Used with tickvals.
• tickvals
Code: fig.update_ternaries(baxis_tickvals=<VALUE>)
Type: list, numpy array, or Pandas series of numbers, strings, or datetimes.
Sets the values at which ticks on this axis appear. Only has an effect if tickmode is set to "array". Used with ticktext.
• tickwidth
Code: fig.update_ternaries(baxis_tickwidth=<VALUE>)
Type: number greater than or equal to 0
Default: 1
Sets the tick width (in px).
• title
Code: fig.update_ternaries(baxis_title=dict(...))
Type: dict containing one or more of the keys listed below.
• font
Code: fig.update_ternaries(baxis_title_font=dict(...))
Type: dict containing one or more of the keys listed below.
Sets this axis' title font. Note that the title's font used to be customized by the now deprecated titlefont attribute.
• color
Code: fig.update_ternaries(baxis_title_font_color=<VALUE>)
Type: color
• family
Code: fig.update_ternaries(baxis_title_font_family=<VALUE>)
Type: string
HTML font family - the typeface that will be applied by the web browser. The web browser will only be able to apply a font if it is available on the system which it operates. Provide multiple font families, separated by commas, to indicate the preference in which to apply fonts if they aren't available on the system. The Chart Studio Cloud (at https://chart-studio.plotly.com or on-premise) generates images on a server, where only a select number of fonts are installed and supported. These include "Arial", "Balto", "Courier New", "Droid Sans",, "Droid Serif", "Droid Sans Mono", "Gravitas One", "Old Standard TT", "Open Sans", "Overpass", "PT Sans Narrow", "Raleway", "Times New Roman".
• size
Code: fig.update_ternaries(baxis_title_font_size=<VALUE>)
Type: number greater than or equal to 1
• text
Code: fig.update_ternaries(baxis_title_text=<VALUE>)
Type: string
Sets the title of this axis. Note that before the existence of title.text, the title's contents used to be defined as the title attribute itself. This behavior has been deprecated.
• uirevision
Code: fig.update_ternaries(baxis_uirevision=<VALUE>)
Type: number or categorical coordinate string
Controls persistence of user-driven changes in axis min, and title if in editable: True configuration. Defaults to ternary<N>.uirevision.
• bgcolor
Code: fig.update_ternaries(bgcolor=<VALUE>)
Type: color
Default: "#fff"
Set the background color of the subplot
• caxis
Code: fig.update_ternaries(caxis=dict(...))
Type: dict containing one or more of the keys listed below.
• color
Code: fig.update_ternaries(caxis_color=<VALUE>)
Type: color
Default: "#444"
Sets default for all colors associated with this axis all at once: line, font, tick, and grid colors. Grid color is lightened by blending this with the plot background Individual pieces can override this.
• dtick
Code: fig.update_ternaries(caxis_dtick=<VALUE>)
Type: number or categorical coordinate string
Sets the step in-between ticks on this axis. Use with tick0. Must be a positive number, or special strings available to "log" and "date" axes. If the axis type is "log", then ticks are set every 10^(n"dtick) where n is the tick number. For example, to set a tick mark at 1, 10, 100, 1000, ... set dtick to 1. To set tick marks at 1, 100, 10000, ... set dtick to 2. To set tick marks at 1, 5, 25, 125, 625, 3125, ... set dtick to log_10(5), or 0.69897000433. "log" has several special values; "L<f>", where f is a positive number, gives ticks linearly spaced in value (but not position). For example tick0 = 0.1, dtick = "L0.5" will put ticks at 0.1, 0.6, 1.1, 1.6 etc. To show powers of 10 plus small digits between, use "D1" (all digits) or "D2" (only 2 and 5). tick0 is ignored for "D1" and "D2". If the axis type is "date", then you must convert the time to milliseconds. For example, to set the interval between ticks to one day, set dtick to 86400000.0. "date" also has special values "M<n>" gives ticks spaced by a number of months. n must be a positive integer. To set ticks on the 15th of every third month, set tick0 to "2000-01-15" and dtick to "M3". To set ticks every 4 years, set dtick to "M48"
• exponentformat
Code: fig.update_ternaries(caxis_exponentformat=<VALUE>)
Type: enumerated , one of ( "none" | "e" | "E" | "power" | "SI" | "B" )
Default: "B"
Determines a formatting rule for the tick exponents. For example, consider the number 1,000,000,000. If "none", it appears as 1,000,000,000. If "e", 1e+9. If "E", 1E+9. If "power", 1x10^9 (with 9 in a super script). If "SI", 1G. If "B", 1B.
• gridcolor
Code: fig.update_ternaries(caxis_gridcolor=<VALUE>)
Type: color
Default: "#eee"
Sets the color of the grid lines.
• griddash
Code: fig.update_ternaries(caxis_griddash=<VALUE>)
Type: string
Default: "solid"
Sets the dash style of lines. Set to a dash type string ("solid", "dot", "dash", "longdash", "dashdot", or "longdashdot") or a dash length list in px (eg "5px,10px,2px,2px").
• gridwidth
Code: fig.update_ternaries(caxis_gridwidth=<VALUE>)
Type: number greater than or equal to 0
Default: 1
Sets the width (in px) of the grid lines.
• hoverformat
Code: fig.update_ternaries(caxis_hoverformat=<VALUE>)
Type: string
Default: ""
Sets the hover text formatting rule using d3 formatting mini-languages which are very similar to those in Python. For numbers, see: https://github.com/d3/d3-format/tree/v1.4.5#d3-format. And for dates see: https://github.com/d3/d3-time-format/tree/v2.2.3#locale_format. We add two items to d3's date formatter: "%h" for half of the year as a decimal number as well as "%{n}f" for fractional seconds with n digits. For example, "2016-10-13 09:15:23.456" with tickformat "%H~%M~%S.%2f" would display "09~15~23.46"
• labelalias
Code: fig.update_ternaries(caxis_labelalias=<VALUE>)
Type: number or categorical coordinate string
Replacement text for specific tick or hover labels. For example using {US: 'USA', CA: 'Canada'} changes US to USA and CA to Canada. The labels we would have shown must match the keys exactly, after adding any tickprefix or ticksuffix. labelalias can be used with any axis type, and both keys (if needed) and values (if desired) can include html-like tags or MathJax.
• layer
Code: fig.update_ternaries(caxis_layer=<VALUE>)
Type: enumerated , one of ( "above traces" | "below traces" )
Default: "above traces"
Sets the layer on which this axis is displayed. If "above traces", this axis is displayed above all the subplot's traces If "below traces", this axis is displayed below all the subplot's traces, but above the grid lines. Useful when used together with scatter-like traces with cliponaxis set to "False" to show markers and/or text nodes above this axis.
• linecolor
Code: fig.update_ternaries(caxis_linecolor=<VALUE>)
Type: color
Default: "#444"
Sets the axis line color.
• linewidth
Code: fig.update_ternaries(caxis_linewidth=<VALUE>)
Type: number greater than or equal to 0
Default: 1
Sets the width (in px) of the axis line.
• min
Code: fig.update_ternaries(caxis_min=<VALUE>)
Type: number greater than or equal to 0
Default: 0
The minimum value visible on this axis. The maximum is determined by the sum minus the minimum values of the other two axes. The full view corresponds to all the minima set to zero.
• minexponent
Code: fig.update_ternaries(caxis_minexponent=<VALUE>)
Type: number greater than or equal to 0
Default: 3
Hide SI prefix for 10^n if |n| is below this number. This only has an effect when tickformat is "SI" or "B".
• nticks
Code: fig.update_ternaries(caxis_nticks=<VALUE>)
Type: integer greater than or equal to 1
Default: 6
Specifies the maximum number of ticks for the particular axis. The actual number of ticks will be chosen automatically to be less than or equal to nticks. Has an effect only if tickmode is set to "auto".
• separatethousands
Code: fig.update_ternaries(caxis_separatethousands=<VALUE>)
Type: boolean
If "True", even 4-digit integers are separated
• showexponent
Code: fig.update_ternaries(caxis_showexponent=<VALUE>)
Type: enumerated , one of ( "all" | "first" | "last" | "none" )
Default: "all"
If "all", all exponents are shown besides their significands. If "first", only the exponent of the first tick is shown. If "last", only the exponent of the last tick is shown. If "none", no exponents appear.
• showgrid
Code: fig.update_ternaries(caxis_showgrid=<VALUE>)
Type: boolean
Default: True
Determines whether or not grid lines are drawn. If "True", the grid lines are drawn at every tick mark.
• showline
Code: fig.update_ternaries(caxis_showline=<VALUE>)
Type: boolean
Default: True
Determines whether or not a line bounding this axis is drawn.
• showticklabels
Code: fig.update_ternaries(caxis_showticklabels=<VALUE>)
Type: boolean
Default: True
Determines whether or not the tick labels are drawn.
• showtickprefix
Code: fig.update_ternaries(caxis_showtickprefix=<VALUE>)
Type: enumerated , one of ( "all" | "first" | "last" | "none" )
Default: "all"
If "all", all tick labels are displayed with a prefix. If "first", only the first tick is displayed with a prefix. If "last", only the last tick is displayed with a suffix. If "none", tick prefixes are hidden.
• showticksuffix
Code: fig.update_ternaries(caxis_showticksuffix=<VALUE>)
Type: enumerated , one of ( "all" | "first" | "last" | "none" )
Default: "all"
Same as showtickprefix but for tick suffixes.
• tick0
Code: fig.update_ternaries(caxis_tick0=<VALUE>)
Type: number or categorical coordinate string
Sets the placement of the first tick on this axis. Use with dtick. If the axis type is "log", then you must take the log of your starting tick (e.g. to set the starting tick to 100, set the tick0 to 2) except when dtick="L<f>" (see dtick for more info). If the axis type is "date", it should be a date string, like date data. If the axis type is "category", it should be a number, using the scale where each category is assigned a serial number from zero in the order it appears.
• tickangle
Code: fig.update_ternaries(caxis_tickangle=<VALUE>)
Type: angle
Default: "auto"
Sets the angle of the tick labels with respect to the horizontal. For example, a tickangle of -90 draws the tick labels vertically.
• tickcolor
Code: fig.update_ternaries(caxis_tickcolor=<VALUE>)
Type: color
Default: "#444"
Sets the tick color.
• tickfont
Code: fig.update_ternaries(caxis_tickfont=dict(...))
Type: dict containing one or more of the keys listed below.
Sets the tick font.
• color
Code: fig.update_ternaries(caxis_tickfont_color=<VALUE>)
Type: color
• family
Code: fig.update_ternaries(caxis_tickfont_family=<VALUE>)
Type: string
HTML font family - the typeface that will be applied by the web browser. The web browser will only be able to apply a font if it is available on the system which it operates. Provide multiple font families, separated by commas, to indicate the preference in which to apply fonts if they aren't available on the system. The Chart Studio Cloud (at https://chart-studio.plotly.com or on-premise) generates images on a server, where only a select number of fonts are installed and supported. These include "Arial", "Balto", "Courier New", "Droid Sans",, "Droid Serif", "Droid Sans Mono", "Gravitas One", "Old Standard TT", "Open Sans", "Overpass", "PT Sans Narrow", "Raleway", "Times New Roman".
• size
Code: fig.update_ternaries(caxis_tickfont_size=<VALUE>)
Type: number greater than or equal to 1
• tickformat
Code: fig.update_ternaries(caxis_tickformat=<VALUE>)
Type: string
Default: ""
Sets the tick label formatting rule using d3 formatting mini-languages which are very similar to those in Python. For numbers, see: https://github.com/d3/d3-format/tree/v1.4.5#d3-format. And for dates see: https://github.com/d3/d3-time-format/tree/v2.2.3#locale_format. We add two items to d3's date formatter: "%h" for half of the year as a decimal number as well as "%{n}f" for fractional seconds with n digits. For example, "2016-10-13 09:15:23.456" with tickformat "%H~%M~%S.%2f" would display "09~15~23.46"
• tickformatstops
Code: fig.update_ternaries(caxis_tickformatstops=list(...))
Type: list of dict where each dict has one or more of the keys listed below.
• dtickrange
Parent: layout.ternary.caxis.tickformatstops[]
Type: list
range ["min", "max"], where "min", "max" - dtick values which describe some zoom level, it is possible to omit "min" or "max" value by passing "null"
• enabled
Parent: layout.ternary.caxis.tickformatstops[]
Type: boolean
Default: True
Determines whether or not this stop is used. If False, this stop is ignored even within its dtickrange.
• name
Parent: layout.ternary.caxis.tickformatstops[]
Type: string
When used in a template, named items are created in the output figure in addition to any items the figure already has in this array. You can modify these items in the output figure by making your own item with templateitemname matching this name alongside your modifications (including visible: False or enabled: False to hide it). Has no effect outside of a template.
• templateitemname
Parent: layout.ternary.caxis.tickformatstops[]
Type: string
Used to refer to a named item in this array in the template. Named items from the template will be created even without a matching item in the input figure, but you can modify one by making an item with templateitemname matching its name, alongside your modifications (including visible: False or enabled: False to hide it). If there is no template or no matching item, this item will be hidden unless you explicitly show it with visible: True.
• value
Parent: layout.ternary.caxis.tickformatstops[]
Type: string
Default: ""
string - dtickformat for described zoom level, the same as "tickformat"
• ticklabelstep
Code: fig.update_ternaries(caxis_ticklabelstep=<VALUE>)
Type: integer greater than or equal to 1
Default: 1
Sets the spacing between tick labels as compared to the spacing between ticks. A value of 1 (default) means each tick gets a label. A value of 2 means shows every 2nd label. A larger value n means only every nth tick is labeled. tick0 determines which labels are shown. Not implemented for axes with type "log" or "multicategory", or when tickmode is "array".
• ticklen
Code: fig.update_ternaries(caxis_ticklen=<VALUE>)
Type: number greater than or equal to 0
Default: 5
Sets the tick length (in px).
• tickmode
Code: fig.update_ternaries(caxis_tickmode=<VALUE>)
Type: enumerated , one of ( "auto" | "linear" | "array" )
Sets the tick mode for this axis. If "auto", the number of ticks is set via nticks. If "linear", the placement of the ticks is determined by a starting position tick0 and a tick step dtick ("linear" is the default value if tick0 and dtick are provided). If "array", the placement of the ticks is set via tickvals and the tick text is ticktext. ("array" is the default value if tickvals is provided).
• tickprefix
Code: fig.update_ternaries(caxis_tickprefix=<VALUE>)
Type: string
Default: ""
Sets a tick label prefix.
• ticks
Code: fig.update_ternaries(caxis_ticks=<VALUE>)
Type: enumerated , one of ( "outside" | "inside" | "" )
Determines whether ticks are drawn or not. If "", this axis' ticks are not drawn. If "outside" ("inside"), this axis' are drawn outside (inside) the axis lines.
• ticksuffix
Code: fig.update_ternaries(caxis_ticksuffix=<VALUE>)
Type: string
Default: ""
Sets a tick label suffix.
• ticktext
Code: fig.update_ternaries(caxis_ticktext=<VALUE>)
Type: list, numpy array, or Pandas series of numbers, strings, or datetimes.
Sets the text displayed at the ticks position via tickvals. Only has an effect if tickmode is set to "array". Used with tickvals.
• tickvals
Code: fig.update_ternaries(caxis_tickvals=<VALUE>)
Type: list, numpy array, or Pandas series of numbers, strings, or datetimes.
Sets the values at which ticks on this axis appear. Only has an effect if tickmode is set to "array". Used with ticktext.
• tickwidth
Code: fig.update_ternaries(caxis_tickwidth=<VALUE>)
Type: number greater than or equal to 0
Default: 1
Sets the tick width (in px).
• title
Code: fig.update_ternaries(caxis_title=dict(...))
Type: dict containing one or more of the keys listed below.
• font
Code: fig.update_ternaries(caxis_title_font=dict(...))
Type: dict containing one or more of the keys listed below.
Sets this axis' title font. Note that the title's font used to be customized by the now deprecated titlefont attribute.
• color
Code: fig.update_ternaries(caxis_title_font_color=<VALUE>)
Type: color
• family
Code: fig.update_ternaries(caxis_title_font_family=<VALUE>)
Type: string
HTML font family - the typeface that will be applied by the web browser. The web browser will only be able to apply a font if it is available on the system which it operates. Provide multiple font families, separated by commas, to indicate the preference in which to apply fonts if they aren't available on the system. The Chart Studio Cloud (at https://chart-studio.plotly.com or on-premise) generates images on a server, where only a select number of fonts are installed and supported. These include "Arial", "Balto", "Courier New", "Droid Sans",, "Droid Serif", "Droid Sans Mono", "Gravitas One", "Old Standard TT", "Open Sans", "Overpass", "PT Sans Narrow", "Raleway", "Times New Roman".
• size
Code: fig.update_ternaries(caxis_title_font_size=<VALUE>)
Type: number greater than or equal to 1
• text
Code: fig.update_ternaries(caxis_title_text=<VALUE>)
Type: string
Sets the title of this axis. Note that before the existence of title.text, the title's contents used to be defined as the title attribute itself. This behavior has been deprecated.
• uirevision
Code: fig.update_ternaries(caxis_uirevision=<VALUE>)
Type: number or categorical coordinate string
Controls persistence of user-driven changes in axis min, and title if in editable: True configuration. Defaults to ternary<N>.uirevision.
• domain
Code: fig.update_ternaries(domain=dict(...))
Type: dict containing one or more of the keys listed below.
• column
Code: fig.update_ternaries(domain_column=<VALUE>)
Type: integer greater than or equal to 0
Default: 0
If there is a layout grid, use the domain for this column in the grid for this ternary subplot .
• row
Code: fig.update_ternaries(domain_row=<VALUE>)
Type: integer greater than or equal to 0
Default: 0
If there is a layout grid, use the domain for this row in the grid for this ternary subplot .
• x
Code: fig.update_ternaries(domain_x=list(...))
Type: list
Default: [0, 1]
Sets the horizontal domain of this ternary subplot (in plot fraction).
• y
Code: fig.update_ternaries(domain_y=list(...))
Type: list
Default: [0, 1]
Sets the vertical domain of this ternary subplot (in plot fraction).
• sum
Code: fig.update_ternaries(sum=<VALUE>)
Type: number greater than or equal to 0
Default: 1
The number each triplet should sum to, and the maximum range of each axis
• uirevision
Code: fig.update_ternaries(uirevision=<VALUE>)
Type: number or categorical coordinate string
Controls persistence of user-driven changes in axis min and title, if not overridden in the individual axes. Defaults to layout.uirevision. | 2023-03-30 04:07:17 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4026821255683899, "perplexity": 8905.550814615992}, "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/1679296949097.61/warc/CC-MAIN-20230330035241-20230330065241-00192.warc.gz"} |
http://physics.stackexchange.com/tags/hamiltonian-formalism/new | # Tag Info
1
Lagranges equations are: ${\partial L \over \partial q_i} = {d \over dt }{ \partial L \over \partial q_i'}$ where $q_i'={dq_i \over dt}$ You can find constants of motion using lagranges equations and Hamiltons equations. You already know that the Hamiltonian is conserved-time is not explicit. (The energy is not always equal to the Hamiltonian) You can ...
2
I think you made an mistake in the algebra. Let's focus on the PDE $$p \frac{\partial \rho }{\partial q} - q \frac{\partial \rho }{\partial p} = 0.$$ You claim that $\rho(p,q) = \exp^{\, f(p+q)}$ is a solution to this equation for any function $f$, but that's simply not true. If you plug that Ansatz in the PDE, you get $$(p-q)\; f'(p+q) = 0$$ which is ...
3
This is a partial answer which I will hopefully come back to and expand. The property of being its own Legendre transform is unique to the pure quadratic kinetic energy $T(v)=\frac12 mv^2$. As a simple example, consider $T(v)=\frac14Av^4$. Here the Legendre momentum is $$p=\frac{\partial L}{\partial v}=\frac{\partial T}{\partial v}=Av^3,$$ so the velocity ...
1
It isn't "obtained". It's a definition of the function $g$ and this definition is useful because it leads to the nice symmetric relationships on the rest of the page 3. They exchange the two Legendre-dual variables. So the definition of $g$ wasn't really "derived" in any straightforward way. It was a clever guess that Legendre made at some point of his life....
1
Hamiltonian gauge symmetries usually come up in the context of lattice gauge theory, in which the system is defined on a discrete lattice. Such a Hamiltonian is defined to have a gauge symmetry if it commutes with some extensively-scaling set of local (i.e. finitely-supported) unitary operators $\{ U_i = e^{-i Q_i} \}$. Operators that commute with the ...
2
There is no mistake. The Hamiltonian satisfies the relation $$\frac{\text{d}H}{\text{d}t} = \frac{\partial H}{\partial t}.$$ This follows immediately from Hamilton's equations: $$\frac{\text{d}H}{\text{d}t} = \dot{H} = \frac{\partial H}{\partial q}\dot{q} + \frac{\partial H}{\partial p}\dot{p} + \frac{\partial H}{\partial t} = \frac{\partial H}{\partial q}... 2 The naive thing to do would be to just replace x by \hat{x} and p by -i\hbar \nabla. Note however that (\hat{x}\hat{p})^{\dagger} = \hat{p}\hat{x} \neq \hat{x}\hat{p}, i.e. the operator is not Hermitian due to the noncommutativity of \hat{x} and \hat{p}, which is a problem of course. So what can we do to fix this? One possible way is to ... 2 I believe the term "stochastic web" was first employed in this article by Zaslavsky, where he studied not the quasi-integrable stochasticity you're talking about, but stochastic processes in 1-dimensional systems. The reason is that most people can't see dimensions higher than three, and he was particularly interested in the fractal structure of chaos (... 6 There is also the routhian formalism of mechanics which is described as being a hybrid of lagrangian and hamiltonian mechanics. The routhian is defined as$$R = \sum_{i=1}^n p_i\dot{q}_i - L$$You can learn more about it by clicking this link for wikipedia's description of it. Reading more in regards to the routhian because I was bored, I realized it is ... 3 It's worth pointing out that the Hamiltonian and Lagrangian formalisms are independent, even though they're usually taught as if the former were a filtering of the latter (here enter Legendre transforms). Both formalisms are as independent as the notions of tangent and cotangent bundles in differential geometry: independent, but intrinsically connected. ... 2 The first part of OP's construction is directly related to the covariant Hamiltonian formalism for a real scalar field with Lagrangian density$$ {\cal L} ~=~ \frac{1}{2}\partial_{\alpha} \phi ~\partial^{\alpha} \phi -{\cal V}(\phi), \tag{CW4} $$see e.g. Ref. [CW] and this Phys.SE post. See also the Wronskian method in this Phys.SE post. [In this answer we ... 2 Often X is a coadjoint orbit of a Lie group. These have a natural symplectic structure; see https://en.wikipedia.org/wiki/Symplectic_reduction 2 That the Hamiltonian is zero is completely correct. The system is time-reparametrization invariant - changing \tau to \xi(\tau) transforms$$ n(q(\tau))\mapsto n(q(\xi)),\quad \dot{q}\mapsto \frac{\mathrm{d}\xi}{\mathrm{d}\tau}q', \quad \mathrm{d}\tau\mapsto \frac{\mathrm{d}\tau}{\mathrm{d}\xi}\mathrm{d}\xi and the action is invariant under this ...
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# The circle with center C shown above is tangent to both axes
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The circle with center C shown above is tangent to both axes. If the distance from O to C is equal to k, what is the radius of the circle, in terms of k ?
(A) k
(B) $$\frac{k}{\sqrt{2}}$$
(C) $$\frac{k}{\sqrt{3}}$$
(D) $$\frac{k}{2}$$
(E) $$\frac{k}{3}$$
[Reveal] Spoiler: OA
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Re: The circle with center C shown above is tangent to both axes [#permalink]
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29 Dec 2012, 05:32
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The circle with center C shown above is tangent to both axes. If the distance from O to C is equal to k, what is the radius of the circle, in terms of k ?
(A) k
(B) $$\frac{k}{\sqrt{2}}$$
(C) $$\frac{k}{\sqrt{3}}$$
(D) $$\frac{k}{2}$$
(E) $$\frac{k}{3}$$
Look at the diagram below:
Attachment:
Circle2.png [ 5.4 KiB | Viewed 11895 times ]
Since OC=k, then $$r^2+r^2=k^2$$ --> $$r=\frac{k}{\sqrt{2}}$$.
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Re: The circle with center C shown above is tangent to both axes [#permalink]
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20 Jun 2013, 18:04
How did we conclude that the base of the triangle is also r ?
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Re: The circle with center C shown above is tangent to both axes [#permalink]
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20 Jun 2013, 18:26
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kkbimalnair wrote:
How did we conclude that the base of the triangle is also r ?
Attachment:
Untitled.jpg [ 6.15 KiB | Viewed 11266 times ]
AC is the radius of the circle and Y-axis is tangent to the circle.Which implies that AC is perpendicular to the Y-axis.
Similarly BC is perpendicular to the X-axis.
Angle AOB is right angle. Which implies angle ACB is also right angled.
Now the quadrilateral OACB must be either a rectangle or a Square.
In either case OB must be equal to AC(as both in a rectangle and in a square opposite sides are equal) . So OB = R
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Re: The circle with center C shown above is tangent to both axes [#permalink]
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04 Nov 2013, 17:30
I actually solved this one with a bit of logic, or at least eliminated several answer choices.
A) is obviously out since k includes that area outside of the circle by the origin, so the radius must be shorter. Dividing it by 2 or 3 seemed overkill. dividing k by the sqrt of 3 implied some sort of 30/60/90 triangle which didn't make much sense. so I figured it was B, then made the triangle real quick, saw it was a right triangle and as soon as I wrote out the pythag equation I saw that B was definitely correct.
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Re: The circle with center C shown above is tangent to both axes [#permalink]
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Re: The circle with center C shown above is tangent to both axes [#permalink]
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31 Dec 2014, 12:09
Expert's post
Hi All,
This question can be solved by TESTing VALUES.
First though, there's a hidden pattern worth noting: Since the circle is tangent to both the x-axis and the y-axis, we can draw a SQUARE into the picture (using the Origin as one corner and "C" as the opposite corner). The length of a side of this square will EQUAL the RADIUS of the circle. From here, we can make the dimensions of the square anything that we want.
Let's TEST...
The length from the Origin to the center C cuts through the square and forms two 45/45/90 right triangles. Thus...
K = That length = 2\sqrt{2}
Now we just have to plug THAT value into the answers to find the one that equals 2.
The only answer that matches is
[Reveal] Spoiler:
B
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# Special Offer: Save $75 + GMAT Club Tests 60-point improvement guarantee www.empowergmat.com/ ***********************Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!*********************** Intern Joined: 13 Sep 2015 Posts: 22 Followers: 0 Kudos [?]: 6 [0], given: 239 The circle with center C shown above is tangent to both axes [#permalink] ### Show Tags 04 Dec 2015, 08:10 alternate method, similar to using the right angle hypotenuese formula: notice that the figure also forms a cube with r (radius) as each side, and diagonal is k or CO. Using diagonal of square formula r√2 = k r = k/√2 (diagonal of a square is x√2) Intern Joined: 13 Jun 2011 Posts: 20 Followers: 0 Kudos [?]: 0 [0], given: 21 Re: The circle with center C shown above is tangent to both axes [#permalink] ### Show Tags 07 Dec 2015, 06:57 Mantis wrote: kkbimalnair wrote: How did we conclude that the base of the triangle is also r ? Attachment: Untitled.jpg AC is the radius of the circle and Y-axis is tangent to the circle.Which implies that AC is perpendicular to the Y-axis. Similarly BC is perpendicular to the X-axis. Angle AOB is right angle. Which implies angle ACB is also right angled. Now the quadrilateral OACB must be either a rectangle or a Square. In either case OB must be equal to AC(as both in a rectangle and in a square opposite sides are equal) . So OB = R Hi Can you please let me know how you made AOB as right angle? Manager Joined: 03 Jan 2015 Posts: 91 Followers: 0 Kudos [?]: 17 [0], given: 146 Re: The circle with center C shown above is tangent to both axes [#permalink] ### Show Tags 09 Feb 2016, 13:47 Could someone explain me why I cannot take the square root of this entire expression: $$\sqrt{k^2 = r^2 + r^2}$$ ---> $$k = r + r.$$ Why is this incorrect? EMPOWERgmat Instructor Status: GMAT Assassin/Co-Founder Affiliations: EMPOWERgmat Joined: 19 Dec 2014 Posts: 6887 Location: United States (CA) GMAT 1: 800 Q51 V49 GRE 1: 340 Q170 V170 Followers: 299 Kudos [?]: 2040 [1] , given: 161 Re: The circle with center C shown above is tangent to both axes [#permalink] ### Show Tags 09 Feb 2016, 19:59 1 This post received KUDOS Expert's post Hi saiesta, You have to combine 'like' terms before you take the square-root of both sides. Here's a simple example that proves WHY: $$\sqrt{4}$$ = 2 $$\sqrt{(2+2)}$$ does NOT = $$\sqrt{2}$$ + $$\sqrt{2}$$ though $$\sqrt{2}$$ = about 1.4 2 does NOT equal 1.4 + 1.4 Knowing THAT, how would you write your equation now? GMAT assassins aren't born, they're made, Rich _________________ # Rich Cohen Co-Founder & GMAT Assassin # Special Offer: Save$75 + GMAT Club Tests
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Re: The circle with center C shown above is tangent to both axes [#permalink]
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10 Feb 2016, 05:20
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EMPOWERgmatRichC wrote:
Hi saiesta,
You have to combine 'like' terms before you take the square-root of both sides.
Here's a simple example that proves WHY:
$$\sqrt{4}$$ = 2
$$\sqrt{(2+2)}$$ does NOT = $$\sqrt{2}$$ + $$\sqrt{2}$$ though
$$\sqrt{2}$$ = about 1.4
2 does NOT equal 1.4 + 1.4
Knowing THAT, how would you write your equation now?
GMAT assassins aren't born, they're made,
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Thank you, I have to remember myself to combine like terms before I take the square root of both sides. I believe this is particularly the case when adding or subtracting (square) roots.
I would write the equation as follows: $$k^2 = r^2 + r^2$$ ----> $$k^2 = 2r^2$$ ----> $$\frac{k^2}{2} = r^2$$ ----> $$\sqrt{\frac{k^2}{2}} = \sqrt{r^2}$$ ----> $$\frac{k}{\sqrt{2}} = r$$
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Re: The circle with center C shown above is tangent to both axes [#permalink]
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10 Feb 2016, 10:38
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Hi saiesta,
Nicely done. As a minor aside, since this is a Geometry question, you don't have to worry about any of the 'measures' ending up negative. However, if the same concepts were in an Algebraic prompt, then you'd have to keep in mind that both K and R could be negative (so square-rooting both sides might end up eliminating possible answers; in a DS question, that could lead to a wrong answer).
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Re: The circle with center C shown above is tangent to both axes [#permalink]
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30 Apr 2016, 19:58
Attachment:
Circle.png
The circle with center C shown above is tangent to both axes. If the distance from O to C is equal to k, what is the radius of the circle, in terms of k ?
(A) k
(B) $$\frac{k}{\sqrt{2}}$$
(C) $$\frac{k}{\sqrt{3}}$$
(D) $$\frac{k}{2}$$
(E) $$\frac{k}{3}$$
If we draw radius to two point of tangency(we will have two 90 degrees there and other two angles will also be 90 degrees. find out why.) then we will draw a square with 4 sides equal(find out why ). Now we have diagonal of square = k . then side = k/sqrt2.(why?= 45-45-90 right angle triangle.)
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Re: The circle with center C shown above is tangent to both axes [#permalink]
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02 May 2016, 09:20
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Attachment:
Circle.png
The circle with center C shown above is tangent to both axes. If the distance from O to C is equal to k, what is the radius of the circle, in terms of k ?
(A) k
(B) $$\frac{k}{\sqrt{2}}$$
(C) $$\frac{k}{\sqrt{3}}$$
(D) $$\frac{k}{2}$$
(E) $$\frac{k}{3}$$
We begin by creating a right triangle with the x-axis, the radius of circle C to the x-axis and the line segment from center of circle C to the origin. We see that we have created an isosceles right triangle, also known as a 45-45-90 degree right triangle. We know this because each leg of the right triangle is equal to a radius of the circle. We can label all this in our diagram.
We know the side-hypotenuse ratios in a 45-45-90 degree right triangle are:
x:x:x√2, where x represents the leg of the triangle and x√2 is the hypotenuse.
We can use this to determine the leg of the triangle.
Since x√2 equals the hypotenuse of the triangle we can say:
x√2 = k
x = k/√2
Since x also represents the radius of the circle, k/√2 is equal to the radius.
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Re: The circle with center C shown above is tangent to both axes [#permalink] 02 May 2016, 09:20
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Display posts from previous: Sort by | 2016-07-25 14:10:10 | {"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": 0.6144225597381592, "perplexity": 2292.3221624227235}, "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/1469257824230.71/warc/CC-MAIN-20160723071024-00220-ip-10-185-27-174.ec2.internal.warc.gz"} |
https://www.varsitytutors.com/hotmath/hotmath_help/topics/disjunction.html | # Disjunction
A disjunction is a compound statement formed by combining two statements using the word or .
Example :
Consider the following statements.
$p:25×4=100$
$q$ : A trapezoid has two pairs of opposite sides parallel.
$r$ : The length of the diameter of a circle is half the length of its radius.
Two statements can be joined using the word or .
$p\vee q:25×4=100$ or A trapezoid has two pairs of opposite sides parallel.
The symbol $\vee$ is used to denote or .
A disjunction is true if any one of the statements in it is true. Here the statement $p$ is true and $q$ is false. So, the disjunction $p\vee q$ is true.
The statement $q$ is false and $r$ false.
So, the disjunction $q\vee r$ is false. | 2017-03-29 15:14:36 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 11, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4388694763183594, "perplexity": 275.96348843098275}, "config": {"markdown_headings": true, "markdown_code": false, "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-2017-13/segments/1490218190753.92/warc/CC-MAIN-20170322212950-00022-ip-10-233-31-227.ec2.internal.warc.gz"} |
https://kb.osu.edu/handle/1811/29867?show=full | dc.creator Xu, Li-Hong en_US dc.creator Andrews, Anne M. en_US dc.creator Fraser, G. T. en_US dc.date.accessioned 2007-11-20T17:14:49Z dc.date.available 2007-11-20T17:14:49Z dc.date.issued 1995 en_US dc.identifier 1995-TK-05 en_US dc.identifier.uri http://hdl.handle.net/1811/29867 dc.description Author Institution: National Institute of the Standards and Technology, Gaithersburg, MD 20899 U.S.A.; Institute for Defense Analyses, 1801 N. Beauregard Street, Alexandria, VA 22311 U.S.A.; University of New Brunswick, Saint John, NB, E2L 4L5, Canada. en_US dc.description.abstract Two microwave-sideband $CO_{2}$ lasers have been used with a molecular-beam electric-resonance spectrometer to study the overtone C-O stretching vibration of methanol. Infrared-infrared double-resonance results have been obtained for levels involving the K = 1 and 2, A symmetry, and the $K = 2, E_{2}$ symmetry species. In the A torsional symmetry case, radiofrequency-infrared multiple resonance was used to obtain accurate asymmetry splittings for the $\nu_{co} = 1$ and 2, C-O stretching states. The asymmetry splitting constants have been determined for these states, are in good agreement with the literature values for the first excited C-O stretching states. However, the nearly factor-of-two change in the K = 2 asymmetry splitting constant for the $\nu_{co} = 2$ level compared to the $\nu_{co} = 0$ and 1 level results suggests that this state is weakly pertubed. The overtone transition frequencies obtained in this work were combined with previous overtone Fourier-transform results in a global fit to a torsion-rotation Hamiltonian to refine the fundamental molecular constants for the second-excited C-O stretching state. The $\nu_{co} = 2$ torsional barrier height is found to be $372.227(3) cm^{-1}$ or $374.984(7) cm^{-1}$ depending on data set used. In the analysis the overtone vibrational energy origin is constrained to $2054.83113 cm^{-1}$. This barrier can be compared to the $\nu_{co} = 0$ and 1 values of $373.5421 cm^{-1}$ and $392.35 cm^{-1}$. respectively. en_US dc.format.extent 79870 bytes dc.format.mimetype image/jpeg dc.language.iso English en_US dc.publisher Ohio State University en_US dc.title Study of the Overtone C-O Stretching Band of Methanol by Multiple Resonance Spectroscopy en_US dc.type article en_US
| 2021-01-23 19:52:28 | {"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": 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": 0.45676568150520325, "perplexity": 7948.1659665283005}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-04/segments/1610703538431.77/warc/CC-MAIN-20210123191721-20210123221721-00292.warc.gz"} |
https://stevens-center.org/testimonial/testimonials-2/ | # Testimonials
What families are saying about NC Innovations provided through the Stevens Center.
We’ve been with the Stevens Center since it opened, and we cannot imagine going anywhere else. The Stevens Center has made my daughter’s life a whole lot better, and my life also because I know she is loved and she is taken care of. Thank you, Stevens Center, for a job well done.” ….Jan R.
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https://leninkumar31.github.io/2021-02-26/Time-Clocks-And-Ordering-of-Events-in-Distributed-Systems | # Time, Clocks and the Ordering of Events in Distributed Systems
Have you ever wondered what it takes to tell which event happened before which in Distributed Systems? This concept was examined in Time, Clocks and the Ordering of Events in Distributed Systems paper. It was published in 1978 by Turing Award Winner Leslie Lamport. This paper created a paradigm shift in how we think about Distributed Systems and also became one of the most cited papers in computer science.
Before we dive into the details of this paper, let us first understand the definition of a Distributed System. There are a lot of definitions out there on the web, but let us use the below definition for this blog post.
## What is a Distributed System?
A Distributed System is a collection of independent computers which talks to on another using unreliable network and also performs its own tasks. Any system that doesn’t communicate with others and functions on its own is not a Distributed System.
## Why does the ordering of the events important in Distributed Systems?
Let us say multiple processes are trying to access one resource which can be used by only one process at a time. A process requesting for the resource must be granted only when there is no other process waiting beforehand to acquire the resource. In such cases, ordering of events globally among multiple processes is necessary to avoid any inconsistencies.
## System Clocks
In our daily life, we use the physical time to order events. For example, we say that something happened at 8:15 if it happened after the clock read 8:15 and before it read 8:16. The same can’t be done in Distributed Systems because System Clocks are not perfect and they don’t keep precise time.
• How do computers track time?
• Computers have Real Time Clock (RTC) which is part of the motherboard as an Integrated Circuit and it is responsible for keeping track of current time
• RTC has an alternate power source(Battery) to keep working even when the computer is switched off
• It doesn’t start counting from the current time, it starts counting from the time we set. For example, if we set the time 5 mins behind the current time, then it will start counting with 5 mins delay
• Problems with System Clocks
• Systems Clocks might start counting time from the arbitrary point
• Precision of clock may be affected by temperature, location, or how they are constructed which may cause them to drift away from the current time
In real life, we can say that an event $$a$$ happened before event $$b$$ if timestamp of $$a$$ is before $$b$$. In Distributed Systems, we can’t use the physical time to order events because of the above problems. So, Lamport defined a Happened before relation without using physical clocks.
## Partial Order
Lamport defined the distributed system more precisely as follows:
• It consists of multiple processes and each process contains a sequence of events.
• Process executing something(Ex: machine instruction, a subprogram, etc.) can be treated as an event.
• Total ordering is satisfied among the set of events in a single process.
• A process can send a message to another process or receive a message from other processes. They are also events.
Let us use right arrow($$\rightarrow$$) to denote Happened Before relation. By using the above definition, he defined the Happened Before as follows.
The Happened Before($$\rightarrow$$) relation among the events satisfy the below conditions
• If $$a$$ and $$b$$ are events in same process and $$a$$ comes before $$b$$, then $$a \rightarrow b$$
• If $$a$$ is message sent by one process and $$b$$ is a receipt of the same message by another process, then $$a \rightarrow b$$
• If $$a \rightarrow b$$ and $$b \rightarrow c$$, then $$a \rightarrow c$$
Two different events $$a$$ and $$b$$ said to be concurrent, if $$a \nrightarrow b$$ and $$b \nrightarrow a$$. For any event $$a$$, $$a \nrightarrow a$$. This means that Happened Before($$\rightarrow$$) is irriflexive partial order (Irriflexive, Anti Symmetric and Transitive) on set of all events in system. Another way of viewing the definition of $$a \rightarrow b$$ is to say that it is possible for event $$a$$ to casually affect event $$b$$.
In the above image, vertical lines are individual processes, dots are events, wavy arrows are messages, and time increases from bottom to top. $$p1 \rightarrow p2$$ because they occurred in the same process. $$q1 \rightarrow p2$$ because q1 is message sent by process $$Q$$ and $$p2$$ is receipt of same message by process $$P$$. $$p1 \rightarrow r3$$ because $$p1 \rightarrow q2$$ , $$q2 \rightarrow q4$$ and $$q4 \rightarrow r3$$. Because time is increasing from bottom to top, $$q3$$ seems to have happened before $$p3$$ if we consider physical time. But we treat them as concurrent events in this system because there is no causal relationship between them.
## Logical Clocks
Now let us order the events in the system according to Happened Before($$\rightarrow$$) relation using logical clocks. A logical clock is a way of assigning a number to an event which can be thought of as the time at which that event happened. Let us define a clock $$Ci$$ for each process $$Pi$$ as a function which assigns a number $$Ci(e)$$ to each event $$e$$ in that process $$Pi$$. System of Clocks can be represented by $$C$$ which assigns a number $$C(b)$$ for any event $$b$$, where $$C(b)=Cj(b)$$ if $$b$$ is an event in process $$Pj$$.
We can evaluate the correctness of logic clocks using the Clock Condition. For any two events $$a, b:$$ if $$a \rightarrow b$$ then $$C(a) < C(b)$$. For example, logical clock must satisfy the condition $$C(p1) < C(r3)$$ for $$p1 \rightarrow r3$$ in the above image. Also, for every $$a, b$$ if $$C(a) < C(b)$$ then it doesn’t mean $$a \rightarrow b$$. In the above image $$C(q3) < C(p3)$$ but $$q3$$ and $$p3$$ are concurrent and not casually related.
Clock Condition is satisfied if the following two conditions hold:
• If $$a$$ and $$b$$ are events in same process $$Pi$$ and $$a$$ comes before $$b$$ then $$Ci(a) < Ci(b)$$
• If $$a$$ is message sent by one process $$Pi$$ and $$b$$ is a receipt of the same message by another process $$Pj$$, then $$Ci(a) < Cj(b)$$
Logical Clock Algorithm
• Each process $$Pi$$ increments the clock between every two events by at least one tick
• Each message $$m$$ contains the timestamp $$Tm$$ at which the event was sent. Process$$(Pj)$$ which receives the message computes the receipt timestamp using the formula $$\max(Tm, Cj(m))+1$$
Our Logical Clock algorithm satisfies the Clock Condition which means it is consistent with Happened Before($$\rightarrow$$) relation.
## Ordering events totally
• Order the events by using the logical times at which they happened. If there any two events with same logical time, break ties using process id.
• We define total ordering relation $$(\implies)$$ as follows. If $$a$$ is an event in process $$Pi$$ and $$b$$ is an event in process $$Pj$$, then $$a \implies b$$ if and only if either $$Ci(a) < Cj(b)$$ or if $$Ci(a) = Cj(b)$$ then $$Pi < Pj$$.
## Lamport’s Algorithm for Mutual Exclusion in Distributed System
Let us say there are a fixed number of processes trying to acquire a resource and only one process can use the resource at a time. So the processes must synchronize themselves to avoid any conflict. This is nothing but a variant of the mutual exclusion problem.
We have to come up with an algorithm that grants the resource to a process that must satisfy the below conditions
• The process which has been granted a resource must release it before it can be granted to any other process
• Different requests for resource must be granted in the order in which they requested
• Every process which has been granted a resource will eventually release it, then every request is eventually granted
Why can’t we have a centralized scheduler to synchronize among processes? The centralized scheduler won’t satisfy the second condition.
Let us look at the following scenario:
• Process $$P1$$ requested for the resource from the scheduler and also informed the other processes about its request.
• After receiving the message from Process $$1$$, Process $$2$$ also requested for the resource.
• Because of network delay, Process $$2$$ request reached the scheduler first and resource has been granted.
• Second condition is violated because the resource has been granted to process $$2$$ even though Process $$1$$ requested first.
We can solve this problem by using Lamport’s Logical clock algorithm which gives the total order of all request and release operations. Before moving forward with the solution, let us make some assumptions
• Each process maintains a request queue that can’t be seen by other processes
• Every message is eventually received
• For any two processes $$Pi$$ and $$Pj$$, the messages sent from $$Pi$$ to $$Pj$$ are received in the same order in which they sent
The algorithm is defined by the following 5 rules:
• To request a resource, process $$Pi$$ send a request resource message $$(Tm,Pi)$$ to every other process and adds the request to its queue. $$Tm$$ is the timestamp of that message
• When a process $$Pj$$ receives a request resource message $$(Tm,Pi)$$, it adds that to its queue and sends a timestamped acknowledgment to $$Pi$$
• To release a resource, process $$Pi$$ removes request resource message $$(Tm,Pi)$$ from its queue and sends a timestamped release resource message $$Pi$$ to every other process
• When a process $$Pj$$ receives a timestamped release resource message $$Pi$$, it removes any request resource message $$(Tm:Pi)$$ from its queue
• Process $$Pi$$ is granted a resource when the following two conditions are met:
• It has a request resource message $$(Tm,Pi)$$ in its queue which is ordered before any other request in its queue according to $$(\implies)$$ relation
• It received a message with a later timestamp than $$Tm$$ from all other processes
## Conclusion
• Physical clocks are not perfect and they don’t keep precise time
• Happened Before($$\rightarrow$$) among all events in Distributed System defines an irreflexive partial order
• Lamport Logical Clock can be used to extend partial ordering to arbitrary total ordering
• We were able to solve a simple synchronization problem by using total ordering among events | 2022-01-26 14:24:03 | {"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": 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": 0.4382927715778351, "perplexity": 560.9562874615951}, "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/1642320304954.18/warc/CC-MAIN-20220126131707-20220126161707-00159.warc.gz"} |
https://www.gradesaver.com/textbooks/math/precalculus/precalculus-concepts-through-functions-a-unit-circle-approach-to-trigonometry-3rd-edition/appendix-a-review-a-3-polynomials-a-3-assess-your-understanding-page-a31/18 | ## Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)
Polynomials consist of one or more monomials and have the form: $$a_nx^n+a_{n-1}x^{n-1}++ a_1x+a+0$$ where $x$ is the variable, $a_n$ is the leading coefficient, and $n$ is a non-negative integer called the degree. Thus, we see that the expression $1-4x$ is a polynomial of degree 1. | 2021-12-09 09:56:53 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8965170979499817, "perplexity": 178.11535439610515}, "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-2021-49/segments/1637964363791.16/warc/CC-MAIN-20211209091917-20211209121917-00334.warc.gz"} |
http://www.physicsforums.com/showpost.php?p=3805728&postcount=4 | View Single Post
P: 271
Joy Christian, Disproof of Bell's Theorem
Told you so!
Quote by Delta Kilo ... and no-one actually bothered to look at the half-a-page of math to see the elephants lurking therein. Well, let's look at eq (5). ... | 2014-07-28 12:33:01 | {"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": 0.918800413608551, "perplexity": 11953.88095160643}, "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-2014-23/segments/1406510259834.25/warc/CC-MAIN-20140728011739-00104-ip-10-146-231-18.ec2.internal.warc.gz"} |
https://brilliant.org/problems/wysiwyg/ | # WYSIWYG !!!
Calculus Level 1
In a circle of unit radius if the ratio of the 2 arcs AB and CD is $$\boxed{5/7}$$ Then the ratio of the angles subtended by the 2 arcs could be written as $$\frac {a} {b}$$ where a and b are coprime find
$$\mathbb{A+B}$$
× | 2018-04-24 05:15:40 | {"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": 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": 0.9084659814834595, "perplexity": 189.74326450100142}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-17/segments/1524125946564.73/warc/CC-MAIN-20180424041828-20180424061828-00523.warc.gz"} |
https://puzzling.stackexchange.com/questions/61532/how-is-babbab-formed | # How is BABBAB formed
$[A ...] \to A [...]$
$[B ...] \to B [S [...]]$
$[S A ...] \to B ...$
$[S B ...] \to A [S ...]$
$[S] \to B$
$[ ] \to$
(E.g. $[AAB] \to AABB$)
But, $[????] \to BABBAB$
Bonus puzzle:
Characterize all sequences of A's and B's so that $[?] => sequence$, for some ? sequence of A's and B's
• Hello! Welcome to puzzling SE! Please take the tour (you'll also get a badge, too) to get a better understanding of community expectations. – North Mar 8 '18 at 0:25
• Wow is everyone overlooking that clever title ?! Made me chuckle about those decade old memes. Obligatory video and related required viewing. – Amit Naidu Mar 8 '18 at 5:43
The only four-letter sequence [????] that works is:
[BBBB]
This evaluates as
[BBBB] →
B[S[BBB]] →
B[SB[S[BB]]] →
B[SB[SB[S[B]]]] →
B[SB[SB[SB[S[]]]]] →
B[SB[SB[SB[S]]]] →
B[SB[SB[SBB]]] →
B[SB[SBA[SB]]] →
B[SB[SBAA[S]]] →
B[SB[SBAAB]] →
B[SBA[SAAB]] →
B[SBABAB] →
BA[SABAB] →
BABBAB
I found this through directed guessing:
I evaluated the 3 letter sequences [AAB], [BBA] and [BBB], and these gave me a good feel for what might work and what definitely wouldn't. Then I jumped to four letter sequences and took a couple wrong guesses before finding this solution.
Bonus:
I've now written some code to find generable outputs (and was able to confirm that the above solution is unique for four letter inputs). I believe that the rule is:
Treating "A" as 0 and "B" as 1, and evaluating the outputs that can be produced by this ruleset as binary, the possible outputs are all the multiples of 3. (Or, at minimum, they are all multiples of 3, even if not all such multiples can be generated.)
Here's what I've been able to find are the generable outputs:
length of 1: A B
length of 2: AA AB BA BB
length of 3: AAA AAB ABA ABB BAA BAB BBA BBB
length of 4: AAAA AAAB AABA AABB ABAA ABAB ABBA ABBB
BAAA BAAB BABA BABB BBAA BBAB BBBA BBBB
where BOLD are generable outputs and ITALIC are outputs that cannot be generated.
The only generable outputs of length 5 are:
AAAAA AAABB AABBA ABAAB ABBAA ABBBB
BAABA BABAB BBAAA BBABB BBBBA
and for length 6 are:
AAAAAA AAAABB AAABBA AABAAB AABBAA AABBBB
ABAABA ABABAB ABBAAA ABBABB ABBBBA BAAAAB
BAABAA BAABBB BABABA BABBAB BBAAAA BBAABB
BBABBA BBBAAB BBBBAA BBBBBB
I finally noticed that ...
the generable outputs occur at intervals of 3.
• the bonus question is asking for a general rule of [?] where ? can represent any string of As and Bs – Praneetmek Mar 8 '18 at 1:04
• The bonus question asks which sequences its possible to end up with given a start of A's and B's (there is a very simple rule to tell if a sequence is a possible output) – while1fork Mar 8 '18 at 1:11
Not sure if its a typo or not, but currently from the list of rules
[SAABBAB]= BABBAB
• Was thinking about the same – user45835 Mar 8 '18 at 0:35
• I assumed "[????]" was seeking a four letter sequence. If not then yes, this is correct. – Rubio Mar 8 '18 at 0:37
• maybe OP meant [SA....]=B[...] and just typoed? – Praneetmek Mar 8 '18 at 0:38
• @Rubio Either that or, from the way the bonus is phrased, ???? can only contain A's and B's, without any S's. – DqwertyC Mar 8 '18 at 0:40
• Directed guessing. I tried 3 letter sequences [AAB], [BBA] and [BBB] and they gave me a good feel for what might work and what definitely wouldn't. Then jumped to four letter sequences and took a couple wrong guesses before finding that one. – Rubio Mar 8 '18 at 1:11 | 2019-04-26 14:35:27 | {"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": 0.7159682512283325, "perplexity": 4969.536012077166}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-18/segments/1555578806528.96/warc/CC-MAIN-20190426133444-20190426155444-00278.warc.gz"} |
http://mathhelpforum.com/algebra/71279-confirmation-aid-simplifying-algebraic-indices.html | # Thread: Confirmation/aid for simplifying algebraic indices
1. ## Confirmation/aid for simplifying algebraic indices
Hello everyone.
I have yet another pile of mathematics homework, and currently working on it. I need some confirmation and, if so, help on the following - can they be simplified any more than I have done?
1. $(6v^4)^2$
= $6^2 X (v^4)^2$
= $6^2 X v^8$
= $36v^8$ Is this correct?
2. $(a^3b^3) ^2$
= $a^6b^6$ ?
3. $(mn^4)^3$
= Help please!
= $mn^12$
4. $a^2b^3 X ab^2$
= I am not sure on this...
5. $r^5s^2 / r^2s$
= Explanation and guidance on this one...
6. $(3m^2m^2) X 2n^2$
= Same as #5
Thank you so much to all who help me in this dilemma!
2. Jade
Originally Posted by JadeKiara
I was just wondering, how soon is someone due to reply? I am in need of some explanations soon to help me work through these. Thank you.
1. $(6v^4)^2=6^2Xv^{4X2}=36v^8$
2. $(a^3b^3) ^2=a^{3X2}Xb^{3X2}=a^6Xb^6$
3. $(mn^4)^3=m^3Xn^{4X3}=m^3n^{12}$
4. $a^2b^3 multiplied by (X) ab^2 = a^2Xb^3XaXb^2=a^{2+1}b^{3+2}$
5. $r^5s^2 / r^2s= r^{5-2}s^{2-1}=r^{3}s$
6. $(3m^2m^2) multiplied by (X) 2n^2= 3m^{2+2}X2n^2=6m^4n^2$
P.S. I have discovered a new technique as you can see!--Welldone
Check the quoted thing
3. Hi, ADARSH, my new buddy!
Is the steps or answers in the quote? Thank you for either!
4. Originally Posted by JadeKiara
Hi, ADARSH, my new buddy!
Is the steps / answers in the quote? Thank you for either!
Yep my new buddy
Formulae
-- $x^a Xx^b= x^{a+b}$
-- $\frac{x^a}{x^b}=x^{a-b}$
-- $(x^a)^b=x^{aXb}$
5. Well, I guess we have reached the end of this post, ADARSH. I thank you for your help and explanations (I really appreciate the formulae too, it adds to my mathematic knowledge. Did you remember it by memory or do you have it nearby on a piece of paper?).
Best wishes and good luck (and may we meet again),
Jade
6. Originally Posted by JadeKiara
Did you remember it by memory or do you have it nearby on a piece of paper?).
Hey Jade , Remember that you need to remember these formulae if you want to solve questions of maths
7. Originally Posted by ADARSH
Jade
Check the quoted thing
ADARSH, just checking, did you realise that in $r^5s^2 / r^2s$, the / is a divided sign?
8. Originally Posted by JadeKiara
ADARSH, just checking, did you realise that in $r^5s^2 / r^2s$, the / is a divided sign?
I think this is what you meant
$\frac{r^5s^2}{r^2s} = \frac{r^3}{s}$
9. Yes, I actually worked it out on paper and I found that to be the answer. I see that you are from New Delhi. You speak English well! | 2017-10-23 14:21:11 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 23, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.808911144733429, "perplexity": 1588.7889000108567}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-43/segments/1508187826049.46/warc/CC-MAIN-20171023130351-20171023150351-00095.warc.gz"} |
https://cs.brash.ca/unit-1/binary-and-logic/logic-gates | # Logic Gates
Controlling electricity gives us the ability to make yes and no decisions or true and false. By combining two signals, we can give a single true or false answer, depending on the logic. The flow of electricity is controlled using gates (no, not named after Bill Gates - like a gate).
NOT
AND
NAND
OR
NOR
XOR
XNOR
All Truth Tables
NOT
The simplest gate, this item inverts or reverses the signal. $Q= \overline A$
Input (A) Output (Q) 0 1 1 0
AND
Only outputs true if both inputs are true. $Q = A \cdot B \text{ or } A \times B$
A B Output (Q) 0 0 0 0 1 0 1 0 0 1 1 1
NAND
Any gate with a dot is an inverted gate. $Q = \overline{A \cdot B} \text{ or } \overline{A \times B}$ This is the NOT AND gate:
A B Output (Q) 0 0 1 0 1 1 1 0 1 1 1 0
OR
One or the other or both! $Q = A+B$
A B Output (Q) 0 0 0 0 1 1 1 0 1 1 1 1
NOR
Very simply NOT OR. $Q = \overline{A+B}$
A B Output (Q) 0 0 1 0 1 0 1 0 0 1 1 0
XOR
Exclusive OR meaning only one is true, not both. $Q = A \bigoplus B$
A B Output (Q) 0 0 0 0 1 1 1 0 1 1 1 0
XNOR
Exclusively NOT OR. If both inputs are the same, output true. $Q = \overline{A \bigoplus B}$
A B Output (Q) 0 0 1 0 1 0 1 0 0 1 1 1
All Truth Tables
A B AND NAND OR NOR XOR XNOR 0 1 0 1 0 1 0 1 0 1 0 1 1 0 1 0 1 0 0 1 1 0 1 0 1 0 1 0 1 0 0 1
### Individually:
The two "universal" gates are NOR and NAND - it is said that you can create all the other logic gates with a combination of just these two.
There are tons of videos available about logic gates. If I had to pick a favourite, my current is this one, although she talks incredibly fast and you might need to slow it down or rewind. | 2020-07-08 04:16:55 | {"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": 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": 0.4606247544288635, "perplexity": 306.2346343504486}, "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/1593655896374.33/warc/CC-MAIN-20200708031342-20200708061342-00002.warc.gz"} |
http://mirror.its.dal.ca/cran/web/packages/imagefluency/vignettes/imagefluency.html | # imagefluency
## Introduction
### Motivation: Why create yet another R package
Over the last decades, the amount of data generated is growing rapidly, predominantly due to digitalization. Most of today’s data is unstructured, and this share is increasing. Given that unstructured data is rich, these data could provide rich insights for scientific research from a variety of fields and practice alike. Especially images (i.e., visual stimuli) are recognized as valuable source of information.
At the same time, vision research and research in psychology shows that even simple changes in low-level image features (like symmetry, contrast, or complexity) can have a tremendous effect on a variety of human judgments. As an example, a statement like “Nut bread is healthier than potato bread” is more likely to be perceived as true when presented in a color that is easy to read against a white background (high contrast) instead of being presented in a color that is difficult to read against a white background (low contrast; cf. [1]). Thus, it might be useful to estimate and control for differences in such low-level visual features in any research that includes visual stimuli.
imagefluency is an simple R package for such low-level image scores based on processing fluency theory. The package allows to get scores for several basic aesthetic principles that facilitate fluent cognitive processing of images: contrast, complexity / simplicity, self-similarity, symmetry, and typicality.
### Why and when to use the package
Possible applications include:
• stimulus selection in experiments (e.g., testing brand logos, evaluate product designs, online display ads)
• as control variables in statistical or prediction models
• linking image fluency scores to outcomes of interest (e.g., how should a typical product packaging look like, do simpler images get more or less attention on a website, …)
• (interpretable) image features in simple machine learning models, e.g. SVM image classifier
## Theoretical background
The most prevailing explanation for how low-level image features affect human judgments is based on processing fluency theory [2]. Processing fluency describes the ease of processing a stimulus [3], which happens instantaneously and automatically [4]. Higher processing fluency results in a gut-level positive affective response [5]. Notably, a rich body of literature has shown that processing fluency effects have an impact on a variety of judgmental domains in our everyday life, including how much we like things, how much we consider statements to be true, how trustworthy we judge a person, how risky we think something is, or whether we buy a product or not (for a review, see [6]).
Several stimulus features have been proposed that result in increased fluency. In particular, visual symmetry, simplicity, (proto-)typicality, and contrast were identified to facilitate processing [2]. Recent studies further discuss self-similarity in light of fluency-based aesthetics [7, 8], a concept which has been studied for example in images of natural scenes [9]. Self-similarity can be described as self-repeating patterns within a stimulus. A typical example are the leaves of ferns that feature the same shape regardless of any magnification or reduction (i.e., scale invariance). Another prominent example is romanesco broccoli with its self-similar surface.
Extracting image features for contrast, self-similarity, simplicity, symmetry, and typicality therefore constitute the core purpose of the imagefluency package.
## Package overview
### Main functions
• img_contrast() visual contrast of an image
• img_complexity() visual complexity of an image (opposite of simplicity)
• img_self_similarity() visual self-similarity of an image
• img_simplicity() visual simplicity of an image (opposite of complexity)
• img_symmetry() vertical and horizontal symmetry of an image
• img_typicality() visual typicality of a list of images relative to each other
• img_read() reads bitmap images into R
• rgb2gray() converts images from RGB into grayscale (might speed up computation)
• run_imagefluency() launches a (preliminary) Shiny app for an interactive demo of the main functions (alternatively, visit the online version at shinyapps.io)
### Installation
You can install the current stable version from CRAN.
install.packages("imagefluency")
To download the latest development version from Github use the install_github function of the devtools package.
# install devtools if necessary
if (!require("devtools")) install.packages("devtools")
# install imagefluency from github
devtools::install_github('stm/imagefluency')
After installation, the imagefluency package is loaded the usual way by calling library(imagefluency). The img_read() function can be used to read an image into R. Just like with reading in a dataset, img_read() expects the path to the file as input, e.g., img_read(C:/Users/myname/Documents/myimage.jpg). Currently supported file formats are bmp, jpg, png, and tif.
Use the following link to report bugs/issues: https://github.com/stm/imagefluency/issues
## Using imagefluency
imagefluency allows to get scores for five image features that facilitate fluent processing of images: contrast, complexity / simplicity, self-similarity, symmetry, and typicality.
To use the imagefluency package, first load the library.
library(imagefluency)
### Contrast
The function img_contrast() returns the contrast of an image. Most research defines contrast in images as the root-mean-squared (RMS) contrast which is the standard deviation of the normalized pixel intensity values [10]: $$\sqrt{\frac{1}{M N}\sum_{i=0}^{N-1}\sum_{j=0}^{M - 1}(I_{ij} - \bar{I})^2}$$. The RMS of an image as a measure for visual contrast has been shown to predict human contrast detection thresholds well [11]. Therefore, the function calculates contrast by computing the RMS contrast of the input image. Consequently, a higher value indicates higher contrast. The image is normalized if necessary (i.e., normalization into range [0, 1]). For color images, a weighted average between color the channels is computed (cf. [8]).
Note that in the following, example images that come with the package are used. Moreover, the images can be displayed using the grid.raster() function from the grid package.
# Example image with relatively high contrast: berries
berries <- img_read(system.file("example_images", "berries.jpg", package = "imagefluency"))
# display image
grid::grid.raster(berries)
# get contrast
img_contrast(berries)
# Example image with relatively low contrast: bike
bike <- img_read(system.file("example_images", "bike.jpg", package = "imagefluency"))
# display image
grid::grid.raster(bike)
# get contrast
img_contrast(bike)
Calculating the contrast scores for the two images gives the following result:
### Complexity / Simplicity
The function img_complexity() returns the visual complexity of an image. Algorithmic information theory indicates that picture complexity can be measured accurately by image compression rates because complex images are denser and have fewer redundancies [12, 13]. Therefore, the function calculates the visual complexity of an image as the ratio between the compressed and uncompressed image file size. Thus, the value does not depend on image size.
The function takes the file path of an image file (or URL) or a pre-loaded image as input argument and returns the ratio of the compressed divided by the uncompressed image file size. The complexity values are naturally interpretable and can range between almost 0 (virtually completely compressed image, thus extremely simple image) and 1 (no compression possible, thus extremely complex image). The function offers to use different image compression algorithms like jpg, gif, or png with algorithm = "zip" as default (for a discussion about the different algorithms, see [8]).
As most compression algorithms do not depict horizontal and vertical redundancies equally, the function includes an optional rotate parameter (default: FALSE). Setting this parameter to TRUE additionally creates a compressed version of the rotated image. The overall compressed image’s file size is computed as the minimum of the original image’s file size and the file size of the rotated image.
The function img_simplicity() returns the visual simplicity of an image. Image simplicity is the complement to image complexity and therefore calculated as 1 minus the complexity score (i.e., the compression rate). Values can range between 0 (no compression possible, thus extremely complex image) and almost 1 (virtually completely compressed image, thus extremely simple image).
# Example image with high complexity: trees
trees <- img_read(system.file("example_images", "trees.jpg", package = "imagefluency"))
# display image
grid::grid.raster(trees)
# get complexity
img_complexity(trees)
# Example image with low complexity: sky
sky <- img_read(system.file("example_images", "sky.jpg", package = "imagefluency"))
# display image
grid::grid.raster(sky)
# get complexity
img_complexity(sky)
Calculating the complexity scores for the two images gives the following result:
### Self-similarity
The function img_self_similarity() returns the self-similarity of an image. Self-similarity can be measured with the Fourier power spectrum of an image. Previous research has identified that the spectral power of natural scenes falls with spatial frequencies ($$f$$) according to a power law ($$\frac{1}{f^p}$$) with values of $$p$$ near the value 2, which indicates scale invariance (for a review, see [14]). Therefore, the function computes self-similarity via the slope of the log-log power spectrum of the image using OLS.
The value for self-similarity that is returned by the function is calculated as $$\text{self-similarity} = |\text{slope} + 2| * (-1)$$. That is, the measure reaches its maximum value of 0 for a slope of $$-2$$, and any deviation from $$-2$$ results in negative values that are more negative the higher the deviation from $$-2$$. Thus, the range of the self-similarity scores is $$-\infty$$ to $$0$$. For color images, the weighted average between each color channel’s values is computed.
It is possible to get the raw regression slope (instead of the transformed value which indicates self-similarity) by using the option raw = TRUE. More options include the possibility to plot the log-log power spectrum (logplot = TRUE) and to base the computation of the slope on the full frequency spectrum (full = TRUE). See the function’s help file for details (i.e., ?img_self_similarity).
# Example image with high self-similarity: romanesco
romanesco <- img_read(system.file("example_images", "romanesco.jpg", package = "imagefluency"))
# display image
grid::grid.raster(romanesco)
# get self-similarity
img_self_similarity(romanesco)
# Example image with low self-similarity: office
office <- img_read(system.file("example_images", "office.jpg", package = "imagefluency"))
# display image
grid::grid.raster(office)
# get self-similarity
img_self_similarity(office)
Calculating the self-similarity scores for the two images gives the result below. The score for the romanesco broccoli is much closer to the maximum possible value of 0, hence much more self-similar.
### Symmetry
The function img_symmetry() returns the vertical and horizontal symmetry of an image as a numeric value between 0 (not symmetrical) and 1 (perfectly symmetrical).
Symmetry is computed as the correlation of corresponding image halves (i.e., the pairwise correlation of the corresponding pixels, cf. [15]). As the perceptual mirror axis is not necessarily exactly in the middle of a picture, the function detects in a first step the ‘optimal’ mirror axis by estimating several symmetry values with different positions for the mirror axis. To this end, the mirror axis is automatically shifted up to 5% and to the right (in the case of vertical symmetry; analogously for horizontal symmetry). In the second step, the overall symmetry score is computed as the maximum of the symmetry scores given the different mirror axes. For color images, the weighted average between each color channel’s values is computed. See [8] for details.
The function further has two optional logical parameters: vertical and horizontal (both TRUE by default). If one of the parameter is set to FALSE, the vertical or horizontal symmetry is not computed, respectively. See the function’s help file (i.e., ?img_symmetry) for information about the additional options shift_range and per_channel.
# Example image with high vertical symmetry: rails
rails <- img_read(system.file("example_images", "rails.jpg", package = "imagefluency"))
# display image
grid::grid.raster(rails)
# get only vertical symmetry
img_symmetry(rails, horizontal = FALSE)
# Example image with low vertical symmetry: bridge
bridge <- img_read(system.file("example_images", "bridge.jpg", package = "imagefluency"))
# display image
grid::grid.raster(bridge)
# get only vertical symmetry
img_symmetry(bridge, horizontal = FALSE)
Calculating the vertical symmetry scores for the two images gives the following result:
### Image typicality
The function img_typicality() returns the visual typicality of a set of images relative to each other. Values can range between -1 (inversely typical) over 0 (not typical) to 1 (perfectly typical). That is, higher absolute values indicate a larger typicality.
The typicality score is computed as the correlation of a particular image with the average representation of all images, i.e., the mean of all images [16]. That is, the typicality of an image can not be assessed in isolation, but only in comparison to set of images from the same category.
For color images, the weighted average between each color channel’s values is computed. If the images have different dimensions they are automatically resized to the smallest height and width. Rescaling of the images prior to computing the typicality scores can be specified with the optional rescaling parameter (must be a numeric value). With rescaling it is possible to assess typicality at different perceptual levels (see [17] for details). Most users won’t need any rescaling and can use the default (rescale = NULL).
The following example shows three images of which two depict valleys in the mountains and one depicts fireworks. Therefore, the fireworks image is in comparison rather low in typicality in this set of images (i.e., atypical). It is important to note that an image’s typicality score highly depends on the reference set to which the image is compared to.
# Example images depicting valleys: valley_white, valley_green
# Example image depicting fireworks: fireworks
valley_white <- img_read(system.file("example_images", "valley_white.jpg", package = "imagefluency"))
valley_green <- img_read(system.file("example_images", "valley_green.jpg", package = "imagefluency"))
fireworks <- img_read(system.file("example_images", "fireworks.jpg", package = "imagefluency"))
# create image set as list
imglist <- list(valley_white, fireworks, valley_green)
# get typicality
img_typicality(imglist)
Calculating the typicality scores for the three images gives the following result:
## Summary
imagefluency is an simple R package for image fluency scores. The package allows to get scores for several basic aesthetic principles that facilitate fluent cognitive processing of images. It straightforward to use and allows for an easy conversion of unstructured data into structured image features. These structured image features are naturally interpretable (i.e., no black-box-model). Finally, including such image information in statistical models might increase a model’s statistical and predictive power.
## References
1. Hansen, J., Dechêne, A., & Wänke, M. (2008). Discrepant fluency increases subjective truth. Journal of Experimental Social Psychology, 44(3), 687–691. https://doi.org/10.1016/j.jesp.2007.04.005
2. Reber, R., Schwarz, N., & Winkielman, P. (2004). Processing fluency and aesthetic pleasure: Is beauty in the perceiver’s processing experience? Personality and Social Psychology Review, 8(4), 364–382. https://doi.org/10.1207/s15327957pspr0804_3
3. Schwarz, N. (2004). Metacognitive experiences in consumer judgment and decision making. Journal of Consumer Psychology, 14(4), 332–348. https://doi.org/10.1207/s15327663jcp1404_2
4. Graf, L. K. M., & Landwehr, J. R. (2015). A dual-process perspective on fluency-based aesthetics: The pleasure-interest model of aesthetic liking. Personality and Social Psychology Review, 19(4), 395–410. https://doi.org/10.1177/1088868315574978
5. Winkielman, P., & Cacioppo, J. T. (2001). Mind at ease puts a smile on the face: Psychophysiological evidence that processing facilitation elicits positive affect. Journal of Personality and Social Psychology, 81(6), 989. https://doi.org/10.1037//0022-3514.81.6.989
6. Alter, A. L., & Oppenheimer, D. M. (2009). Uniting the tribes of fluency to form a metacognitive nation. Personality and Social Psychology Review, 13(3), 219–235. https://doi.org/10.1177/1088868309341564
7. Joye, Y., Steg, L., Ünal, A. B., & Pals, R. (2016). When complex is easy on the mind: Internal repetition of visual information in complex objects is a source of perceptual fluency. Journal of Experimental Psychology: Human Perception and Performance, 42(1), 103–114. https://doi.org/10.1037/xhp0000105
8. Mayer, S., & Landwehr, J. R. (2018). Quantifying visual aesthetics based on processing fluency theory: Four algorithmic measures for antecedents of aesthetic preferences. Psychology of Aesthetics, Creativity, and the Arts, 12(4), 399–431. https://doi.org/10.1037/aca0000187
9. Simoncelli, E. P. (2003). Vision and the statistics of the visual environment. Current Opinion in Neurobiology, 13(2), 144–149. https://doi.org/10.1016/S0959-4388(03)00047-3
10. Peli, E. (1990). Contrast in complex images. Journal of the Optical Society of America A, 7(10), 2032–2040. https://doi.org/10.1364/JOSAA.7.002032
11. Frazor, R. A., & Geisler, W. S. (2006). Local luminance and contrast in natural images. Vision Research, 46(10), 1585–1598. https://doi.org/10.1016/j.visres.2005.06.038
12. Donderi, D. C. (2006). Visual complexity: A review. Psychological Bulletin, 132(1), 73–97. https://doi.org/10.1037/0033-2909.132.1.73
13. Landwehr, J. R., Labroo, A. A., & Herrmann, A. (2011). Gut liking for the ordinary: Incorporating design fluency improves automobile sales forecasts. Marketing Science, 30(3), 416–429. https://doi.org/10.1287/mksc.1110.0633
14. Simoncelli, E. P., & Olshausen, B. A. (2001). Natural image statistics and neural representation. Annual Review of Neuroscience, 24(1), 1193–1216. https://doi.org/10.1146/annurev.neuro.24.1.1193
15. Mayer, S., & Landwehr, J. R. (2014). When complexity is symmetric: The interplay of two core determinants of visual aesthetics. In J. Cotte & S. Wood (Eds.), Advances in Consumer Research (Vol. 42, pp. 608–609). Duluth, MN: Association for Consumer Research.
16. Perrett, D. I., May, K. A., & Yoshikawa, S. (1994). Facial shape and judgements of female attractiveness. Nature, 368(6468), 239–242. https://doi.org/10.1038/368239a0
17. Mayer, S., & Landwehr, J. R. (2018). Objective measures of design typicality. Design Studies, 54, 146–161. https://doi.org/10.1016/j.destud.2017.09.004 | 2020-11-27 21:00:30 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5802825689315796, "perplexity": 4519.004952858457}, "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/1606141194171.48/warc/CC-MAIN-20201127191451-20201127221451-00486.warc.gz"} |
https://math.stackexchange.com/questions/1246798/order-of-operations-in-different-cultures | # order of operations in different cultures?
Are there any cultures or countries around the world that use a different convention for order of operations than the BEDMAS convention? i.e.:
1. Parentheses
2. Exponents & Roots
3. Multiplication & Division
• I’m not aware of any now, but on page $2$ of Cayley’s An Elementary Treatise on Elliptic Functions, originally published in $1876$, we find $$\frac{Rdx}{\sqrt{1-x^2\,.\,1-k^2x^2}}$$ for $$\frac{Rdx}{\sqrt{(1-x^2)(1-k^2x^2)}}\;.$$ – Brian M. Scott Apr 22 '15 at 17:55 | 2019-06-26 06:32:44 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7091590762138367, "perplexity": 837.9227813717699}, "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-2019-26/segments/1560628000175.78/warc/CC-MAIN-20190626053719-20190626075719-00242.warc.gz"} |
https://astronomy.stackexchange.com/tags/saturn/hot | # Tag Info
Accepted
### Why can't we see Saturn's phases from earth?
Phases are just different perceived illuminations of an object at different illumination and observing angles. If the observer is, with respect to the object, located in a similar direction as the ...
• 3,776
### Why can't we see Saturn's phases from earth?
All the other answers here are complete, and more in-depth than anything I would write. However, if you prefer to look at things visually, here is a terrible not-to-scale 2 minute paint drawing. No ...
• 671
Accepted
### How large can a ball of water be without fusion starting?
You really need a full-blown stellar evolution model to answer this precisely and I'm not sure anyone would ever have done this with an oxygen-dominated star. To zeroth order the answer will be the ...
• 132k
### Why is there a mountain inside the Herschel crater on Mimas?
In the extreme energy of a large impact, the rock behaves like a liquid (It isn't actually completely melted, though some is. The extreme forces cause the rock to flow). As the impactor hits the moon, ...
• 104k
Accepted
### How is it possible that Saturn's gravitational acceleration felt by Mimas is stronger than Mimas' own surface gravity?
An object on Mimas' surface would be much more attracted to Saturn than it is to Mimas. You are missing that Mimas as a whole accelerates gravitationally toward Saturn. What this means is that a ...
• 32.4k
### Help understanding this unsettling image of Titan, Epimetheus, and Saturn's rings?
This NASA page says this photo was taken on April 28 2006. Using Celestia, I managed to find the picture from Cassini that best lines up with the photo. It doesn't match up precisely, but that's to ...
• 4,037
Accepted
### How did Arecibo detect methane lakes on Titan, and image Saturn's rings?
Titan "lakes": Published Open Access in Science: Radar Evidence for Liquid Surfaces on Titan Campbell, D. B., Black, G. J., Carter, L. M., and Ostro, S. J., Science 302, 5644, pp. 431-434, 17 Oct ...
• 31k
Accepted
### Which Saturn satellite passes closest to Saturn's rings and at what distance?
Pan, Daphnis, and various other moonlets, I would argue, are inside the rings. If you explicitly discount the Encke gap (which Pan orbits in) and the Keeler gap (which Daphnis orbits in) as being ...
• 4,037
Accepted
### When and how was it discovered that Jupiter and Saturn are made out of gas?
I'm unsure of the "history of science" aspect of this, but an actual deduction that these are gas giants would require Kepler's laws and Newton's law of gravity combined with a modest ...
• 132k
Accepted
### Are there other pictures of Titan surface from Huygens?
One second of googling reveals the whole archive: http://esamultimedia.esa.int/docs/titanraw/index.htm (note you can click onto the strips to inspect them!) The archive depicts the whole decent of ...
### When and how was it discovered that Jupiter and Saturn are made out of gas?
By 1690, Giovanni Cassini was able to estimate the rotation period of the planet and noticed that the atmosphere of Jupiter undergoes differential rotation which confirmed that Jupiter was made of gas ...
• 3,780
Accepted
### Planets looks like normal stars when I see them using telescope
I'm a VERY amateur observer myself and my telescope has the exact same aperture as yours. I can all but guarantee that you should definitely be able to see Saturn rings and Jupiter moons (and even ...
Accepted
### Why is Enceladus's albedo greater than 1?
To answer this, one really has to understand how the geometric and bond albedos are defined. Let's start with the bond albedo since its simpler. Bond Albedo The Bond Albedo is just the fraction of ...
• 14.6k
### Hypothetically, would we be able to see the moon from Saturn's North Pole?
First of all, at that distance seeing the Moon and seeing the Earth amounts to the same thing. At its closest, Saturn is around 3000 times as far from Earth as the Moon is, so viewed from Saturn, the ...
• 7,450
Accepted
### Help understanding this unsettling image of Titan, Epimetheus, and Saturn's rings?
The JPL Solar System Simulator doesn't show Epimetheus but does show Titan behind the Encke gap at 2006-04-28 08:12 UTC. The simulated surface texture is probably composed of VIMS images in infrared ...
• 17.1k
### Why can't we see Saturn's phases from earth?
The answer is simple geometry: we are 1 AU from the Sun, Saturn is 10x further from the Sun. Looked at Earth from Saturn, Earth is always in front, behind or immediately next to the sun; an observer ...
• 14.3k
Accepted
### What is this 877-year cycle in the orbits of Jupiter & Saturn, and this multimillion-year cycle in the lunar orbit?
A biography of Laplace mentions his derivation of an 877 year cycle in the motions of Saturn and Jupiter. The periods of the planets are about 30 and 12 years respectively so they will approximately ...
• 104k
### Planets looks like normal stars when I see them using telescope
With f=900 mm and a 25 and 10 mm eyepieces you would be viewing at 36x and 72x. 36x is a very reasonable magnification under any condition, and 72x can probably still be considered "useful ...
• 31k
Accepted
### Hypothetically, would we be able to see the moon from Saturn's North Pole?
Yes, if you observe Earth and the Moon at a favorable time. Near a Saturn summer solstice, e.g. between 2012 and 2022, Earth appears well above the horizon from Saturn's north pole. If the planet body ...
• 17.1k
### How did Arecibo detect methane lakes on Titan, and image Saturn's rings?
It did not detect methane lakes. It found that Titan was shiny (in radar terms): that is, the reflections were from a smooth surface rather than a rough one, and at the same time not very intense. ...
• 1,322
Accepted
### Are the Jupiter-Saturn conjunctions and winter solstice related?
Not in any way, no. The December solstice is the moment when the Sun reaches its southernmost point in its daily path in the sky (the June solstice, when the Sun reaches its northernmost point). It ...
• 6,071
Accepted
### What are the periods of Saturn's rings?
Hours to days. The orbital period is proportional to the 3/2 power of the orbital radius, and the orbital period of the moon Methone (and thus of the particles in its associated ring arc) is very ...
• 1,005
Accepted
### How is a day measured on a gas giant?
That the rotation period of the bulk mass of a planet is estimated through something with the magnetic field is true. But let me maybe elaborate a bit in-depth on that. No planetary magnetic field ...
Accepted
### Size of Saturn's ring material
The vast majority of the particles in Saturn's rings are small, on the order of $\sim10^{-1}$ m or lower. The columnar number density, according to data from Voyager 1 and Earth-based observations, ...
• 34.7k
### How to measure mass of planets' core from orbit
I certainly don't know the details of these kinds of calculations, but as my thought is a bit too long for a comment I'll write it up as an answer. If you measure the flattening of a planet due to ...
• 1,095
### Trying to understand the way Saturn's ring look in this famous Cassini image
What's going on with the distortion of the rings on the upper half when they (presumably) cross in front of Saturn? The brownish areas you see on Saturn are ring light, analogous to seeing the Earth ...
• 32.4k
Accepted
### Why are Saturn bands much fainter than Jupiter's?
I'll give this one a shot. Correction is welcome. Upper atmosphere temperature. It's not just elements that give a planet color, but the temperature of elements. When we examine what a planet ...
• 23.3k
Accepted
### Is there, in fact, any close-up photography of Saturn's rings, showing individual pebbles/rocks?
The rings are about 3% solid in the densest parts, but this translates to a separation between 30cm particles of about 1metre. There are no images Because approaching the dense part of the rings would ...
• 104k | 2023-03-30 20:07:33 | {"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": 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": 0.5183746218681335, "perplexity": 2293.407863501875}, "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/1679296949387.98/warc/CC-MAIN-20230330194843-20230330224843-00700.warc.gz"} |
https://stacks.math.columbia.edu/tag/0CX0 | Lemma 76.11.6. In Situation 76.11.4. Let $h : X' \to X$ be an étale morphism. Set $\mathcal{F}' = h^*\mathcal{F}$ and $f' = f \circ h$. Let $F_ n'$ be (76.11.4.1) associated to $(f' : X' \to Y, \mathcal{F}')$. Then $F_ n$ is a subfunctor of $F_ n'$ and if $h(X') \supset \text{Ass}_{X/Y}(\mathcal{F})$, then $F_ n = F'_ n$.
Proof. Choose $U \to X$, $V \to Y$, $U \to V$ as in part (1) of Lemma 76.11.2. Choose a surjective étale morphism $U' \to U \times _ X X'$ where $U'$ is a scheme. Then we have the lemma for the two functors $F_{U, n}$ and $F_{U', n}$ determined by $U' \to U$ and $\mathcal{F}|_ U$ over $V$, see More on Flatness, Lemma 38.27.2. On the other hand, Lemma 76.11.2 tells us that given $T \to Y$ we have $F_ n(T) = F_{U, n}(V \times _ Y T)$ and $F'_ n(T) = F_{U', n}(V \times _ Y T)$. This proves the lemma. $\square$
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar). | 2023-01-31 06:54:21 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 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": 2, "x-ck12": 0, "texerror": 0, "math_score": 0.9953287243843079, "perplexity": 158.8226637198894}, "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/1674764499845.10/warc/CC-MAIN-20230131055533-20230131085533-00306.warc.gz"} |
https://puzzling.stackexchange.com/questions/37160/where-has-andre-murrall-gone | # Where has Andre Murrall gone?
Renowned tennis player Andre Murrall seems to have left the tennis tournament he is currently playing in, Winbelldam. No one knows where he has gone except a taxi driver who claims he drove Murrall away from the tournament. However Andre has paid the taxi driver (Who is also very good at Maths) not to reveal his location specifically so when the media ask him about Murrall's location he hands over this:
The taxi driver gave this short speech and handed the interviewer these tennis scores:
"He’s such a key player, count the numbers of the times that he's hit a winner, compare that to the letters he’s received from the queen, he is an all-time great.
What adds up is that you can take the mean out of him. After that you can still add up the mean-ness.
When you count all the games he and his opponents have played its incredible.
How many times has he lost? 5, 11, 9, 3, 9, 1, 5… who knows! I just can’t help looking at the letters from players congratulating him.
Looking at these scores you can deduce that they are complete nonsense and bear no resemblance to the tournament at all. They must have some other hidden meaning which can be unlocked by something else...
Where has Andre Murrall gone?
NOTE: You only need (very) basic tennis knowledge
Hints:
There are 4 parts to this puzzle which lead to a 3 word sentence explaining where he's gone. Each 'part' relates to each paragraph of the interview.
2
Key phrases in the paragraphs, to stop you going off on the wrong track
1. 'Count the numbers (digits)', 'compare that to the letters (in the players names)'
2. 'Adds up (digits in each separate match)', 'take the mean (and round it), 'add up the means'
3. 'Count all the games' (games as in tennis games)
4. '5, 11, 9, 3, 9, 1', 'compare to letters' (same as first paragraph but numbers given).
Compare the numbers you get from paragraphs 2 and 3 to the alphabet
• How come Murrall didn't play a game against Maddal ? – Marius Jul 7 '16 at 7:43
• @humn I'll give you a hint, the queen but isn't important – Beastly Gerbil Jul 7 '16 at 8:17
• The queen isn't important? How treasonous! – Shagnik Jul 7 '16 at 10:15
• Idk much about tennis, but the maths phrases stand out: "count the numbers of the times he’s won, compare that to the letters", "What adds up... take the mean... add up the mean", "count all the games he and his opponents have played", and "How many times has he lost?". – mbomb007 Jul 7 '16 at 21:44
• @ABcDexter DEFINETLY not, most of the suspicious bits in the interview are relevant – Beastly Gerbil Jul 9 '16 at 10:36
(Community evidence locker— feel free to add or correct)
Tennis scoring: match = sets × 6 games × 4 points
Scoreboard numbered out:
"times
MATCH SET GAME TOTAL he's sum of
sets games pts games pts score pts games pts lost" letters
-------------- --------- --------- --------- ------ -------
Murray ----- 96
"2" 12 48 "2" 8 "40" 3 14 59 Murrall ---- 95
"1" 6 24 "0" 0 "0" 0 6 24 ... 5 e Thieem ----- 60 ...
: : Thiem ------ 55 :
: :......:: :
"0" 0 0 "1" 4 "15" 1 1 5 : Murrall :
"0" 0 0 "0" 0 "30" 2 0 2 : 11 k Dyockovich - 115 :
: : Djokovic --- 89 :
: :......:: :
"2" 12 48 "1" 4 "15" 1 13 53 : Murrall :
"0" 0 0 "1" 4 "00" 0 1 4 :.. 9 i Silich ----- 60 ..:
: : Cilic ------ 36 :
: :....:.: :
"2" 12 48 "4" 16 "15" 1 16 65 : Murrall :
"2" 12 48 "2" 8 "00" 0 14 56 : 3 c Mishicori -- 103 :
: : Nishikori -- 112 :
: :........: :
"2" 12 48 "0" 0 "15" 1 12 49 : Murrall :
"2" 12 48 "0" 0 "00" 0 12 48 :.. 9 i FedERROR --- 89 ..:
: : Federer ---- 61 :
: :.? :
"1" 6 24 "1" 4 "30" 2 7 30 : Murrall :
"0" 0 0 "0" 0 "40" 3 0 3 : 1 a Warinker --- 99 :
: : Wawrinka --- 100 :
: :..........: :
"1" 6 24 "1" 4 "30" 2 7 30 : Murrall :
"2" 12 48 "1" 4 "40" 3 13 55 :.. 5 e Birdytch --- 89 ..:
: Berdych ---- 65
:....:
• From the OPs comment that he could not post an ASCII text version of the table, I would deduce that the dots in the score board are of some significance (dots are also points). Has anyone counted / used them already? – BmyGuest Jul 12 '16 at 5:42
• @BmyGuest, nice idea which I might use in the future, but the dots aren't important, just the digits themselves – Beastly Gerbil Jul 12 '16 at 6:52
• @BeastlyGerbil Ok, thanks for clarification. I, however, do not quite understand why it should then not be possible to state the puzzle using ASCII text as requested by Peregrene Rook. If this statement holds, then it contains some sort of hint, because then some feature of the images - which would get lost in the clean text representation - is of importance... – BmyGuest Jul 12 '16 at 6:54
• @BeastlyGerbil That is not to say, that I couldn't agree with a esthetically motivated reason to not do it. – BmyGuest Jul 12 '16 at 6:55
Just some stray observations:
count the numbers of the times that he's hit a winner
Assuming only the points shown on the scoresheet, we get: 3+1+1+1+1+2+2=11. Its mean is 5.5(Which doesn't make a whole lot of sense)
When you count all the games he and his opponents have played
Counting all, we get 7.
If we add what we got, I still can't make it work with something. That's all I got so far. I can't get the lost part.
• Heres some help: first paragraph contains 2 key phrases, 'count the numbers' and 'compare that to the letters'. The mean but is in second paragraph which means that's separate, and the 'count the games he and his opponents have played' refer to tennis games. For paragraphs 2 and 3 compare the numbers you get to the alphabet. Last paragraph contains key phrases '5, 11, 9, 3, 9, 1, 5' and 'looking at the letters'. For paragraphs 1 and 4, letters might just refer to the letters in the player's names. Hope you can get it now, You deserve the bounty because you've been most actively involved so far – Beastly Gerbil Jul 10 '16 at 14:38
• thanks, for my active involvement, I was trying to solve it before winbeldaam gets over. But Now, it already has. But thanks again for the help. I was livid with myself for missing such obvious clues – Sid Jul 10 '16 at 17:09
Okay this has been abandoned. I'll post my answer myself as no one appears to have been anywhere close.
What you have to do for the first part is
Add up each digit and count that number on the letters
Which you can get from the first paragraph
count the numbers of the times that he's hit a winner, compare that to the letters he’s received from the queen
Doing this we get
9 (2, 2, 4, 1) = 'H'
10 (1, 1, 5, 3) = 'O'
10 (2, 1, 1, 5, 1) = 'L'
16 (2, 4, 1, 5, 2, 2) = 'I'
10 (2, 1, 5, 2) = 'D'
9 (1, 1, 3, 4) = 'A'
12 (1, 1, 3, 2, 1, 4) = 'Y'
Holiday
For the second part
You must add up the means for each match, round them, and then add up all the means
You get this from the second paragraph
What adds up is that you can take the mean out of him. After that you can still add up the mean-ness.
So doing this we get
9 (2, 2, 4, 0, 1, 0, 0) = 9 / 7 = 1.3 = 1
10 (0, 1, 1, 5, 0, 0, 3, 0) = 10 / 8 = 1.25 = 1
10 (2, 1, 1, 5, 0, 1, 0, 0) = 10 / 8 = 1.25 = 1
16 (2, 4, 1, 5, 2, 2, 0, 0) = 16 / 8 = 2
10 (2, 0, 1, 5, 2, 0, 0, 0) = 10 / 8 = 1.25 = 1
9 (1, 1, 3, 0, 0, 0, 4, 0) = 9 / 8 = 1.1 = 1
12 (1, 1, 3, 0, 2, 1, 4, 0) = 12 / 8 = 1.5 = 2
$1+1+1+2+1+1+2$ $=$ $9$
And the $9$$th letter of the alphabet is 'I' For the third part You simply count all the tennis games (excluding sets, just what it says on the board) You get this from paragraph 3: When you count all the games he and his opponents have played it's incredible. Doing this we get 2+1+1+1+4+2+1+1+1 = 14 And the 14$$th$ letter of the alphabet is N
For the last part
You compare the numbers '5, 11, 9, 3, 9, 1, 5' to the letters of the names, like the first part
You get this from
The last paragraph where it says the numbers and then where it says ' I just can’t help looking at the letters from players congratulating him.'
Doing this we get:
5 = A
11 = C
9 = I
3 = R
9 = E
1 = M
5 = A
And this says AMERICA backwards
So Andre Murall has gone on
HOLIDAY IN AMERICA
Good for him :P | 2021-05-10 01:44:53 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 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": 0.3632640838623047, "perplexity": 1973.3075807207808}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-21/segments/1620243989030.65/warc/CC-MAIN-20210510003422-20210510033422-00269.warc.gz"} |
https://socratic.org/questions/an-object-with-a-mass-of-4-kg-is-on-a-plane-with-an-incline-of-pi-12-if-the-obje-8 | An object with a mass of 4 kg is on a plane with an incline of pi/12 . If the object is being pushed up the plane with 1 5 N of force, what is the net force on the object?
Aug 3, 2017
${\vec{F}}_{n e t} = 4.8 N$
Explanation:
Begin by drawing a force diagram, then use this diagram to construct statements for the net force.
I will assume a frictionless surface. This means that the forces acting on the object include the normal force, the force of gravity, and the pushing force. We can draw a diagram:
where $\vec{n}$ is the normal force, ${\vec{F}}_{P}$ is the pushing force, and ${\vec{F}}_{G}$ is the force of gravity. Note that I have defined up the ramp as the positive direction. This is how we will decide whether the forces are positive or negative later. I'll drop vector arrows from now on.
${\left({F}_{n e t}\right)}_{x} = \sum {F}_{x} = {F}_{p} - {\left({F}_{G}\right)}_{x} = m {a}_{x}$
${\left({F}_{n e t}\right)}_{y} = \sum {F}_{x} = n - {\left({F}_{G}\right)}_{y} = m {a}_{y}$
Because the object is not accelerating vertically (not moving up or down), we know that ${\left({F}_{n e t}\right)}_{y} = 0$. Therefore, there is no net force perpendicular and we can focus on the parallel (x) forces.
While both the normal force and pushing force are completely perpendicular and parallel (respectively), the force of gravity will have to be broken up into both its parallel and perpendicular components. We can do this using basic trigonometry.
We see, for the inner triangle, that $\sin \theta = \frac{{\left({F}_{G}\right)}_{x}}{{F}_{G}}$. Rearranging to solve for the parallel component of the force of gravity, we get ${\left({F}_{G}\right)}_{x} = \left({F}_{G}\right) \sin \theta$. So, we have:
${\vec{F}}_{n e t} = {\left({F}_{n e t}\right)}_{x} = \sum {F}_{x} = {F}_{p} - \left({F}_{G}\right) \sin \theta = m {a}_{x}$
$\implies {\vec{F}}_{n e t} = {F}_{p} - \left({F}_{G}\right) \sin \theta$
$\implies = {F}_{p} - m g \sin \theta$
Using that $\theta = \frac{\pi}{12} , {F}_{p} = 15 N$ and $m = 4 k g$, we have
${\vec{F}}_{n e t} = 15 N - \left(4 k g\right) \left(9.81 \frac{m}{s} ^ 2\right) \sin \left(\frac{\pi}{12}\right)$
$= 4.8 N$
Since this is a positive value, we know that the direction of the net force is up the ramp, since this is the direction we defined as positive and we kept our signs consistent throughout. This makes sense, as the pushing force is greater than the force of gravity which opposes it. Therefore, we expect the object to be moving up the ramp against the force of gravity. | 2021-09-27 06:59:47 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 16, "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": 0.6686311960220337, "perplexity": 145.3226705078862}, "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-2021-39/segments/1631780058373.45/warc/CC-MAIN-20210927060117-20210927090117-00108.warc.gz"} |
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could someone explain to me just how the integral of x^-1 is ln(x) ? cheers Saucers ps, im not looking for something like "cos ln is the derivitive of x^-1"
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Well the reason you don''t want us to give is the fundamental theorem of calculus which is pretty stupid to ignore =p
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http://www.mathstat.uottawa.ca/Profs/Rossmann/MAT1320A_files/3-7_Log.pdf
[edited by - cozman on February 20, 2004 10:04:12 PM]
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just like the op inferred, you''ll need to understand the fundamental theorem of calculus, and then you''ll realize the inverse relationship that exists between derivatives and integrals. So then if you understood that the derivative of ln(x) is 1/x, then going backwards from it will make sense.
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If I remember right (it''s been a while), ln(n) is defined as being the integral of 1/x. Or more specifically, the integral of 1/x from 1 to n, I believe. So there''s not really as much mathematical derivation involved. It''s just defined that way. Whenever you need to calculate the natural log of a number, and you don''t have a premade chart made, then you have to approximate the integral, since that is precisely how it is defined.
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Uhhh. Ok. Right. Lemme dig out the calculus book. I figured someone would have posted this by now, but since nobody is... here it goes...
(slthough cozman posted something that is good. you should look at it.)
d/dx(loga x) = 1/(xlna)
Proof:
ay=x
Differentiate implicitly
ay(ln a)(dy/dx)=1
dy/dx=1/ayln a = 1/(x*ln a)
If a=e,
d/dx(ln x)=1/x
That''s the way from function to derivative.
Now I''ll use my pat answer and say ... "the derivative of 1/x is ln x" ... and hope you understand. ;-) I suppose if I get bored tonight i''ll do a derivation with reimann integrals, but i think that it''d be really kinda ugly.
Scout
All polynomials are funny - some to a higher degree.
Furthermore, polynomials of degree zero are constantly funny.
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Another nice way to show it:
if g is the inverse of f, and x is a real number in the domain of G then: Dg(x) = 1/Df(g(x))
and f(g(x)) = x
Not hard to prove, but using this:
g(x) = ln(x)
f(x) = exp(x)
ln(x) = 1/exp(ln(x)) = 1/x
Use the fundamental theroem of calculus to go backwards.
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hey, thanks everyone, that really helped :D
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When you have a derivative over the original function (1 is the derivative of x), the anti-derivative of that function is always ln(denominator) ex: dx/dy = 2x/(x^2) y = ln(x^2). Because to take the derivative of an ln() function, take the derivative of what''s in the parentheses and put it ''on top'' of what''s in the parentheses.
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× | 2018-06-19 22:02:49 | {"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": 0.8721301555633545, "perplexity": 2252.3814365906965}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-26/segments/1529267863206.9/warc/CC-MAIN-20180619212507-20180619232507-00553.warc.gz"} |
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# The buyer of a certain mechanical toy must choose 2 of 4 opt
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The buyer of a certain mechanical toy must choose 2 of 4 opt [#permalink] 27 Jul 2018, 09:19
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Question Stats:
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The buyer of a certain mechanical toy must choose 2 of 4 optional motions and 4 of 5 optional accessories. How many different combinations of motions and accessories are available to the buyer?
(A) 8
(B) 11
(C) 15
(D) 20
(E) 30
[Reveal] Spoiler: OA
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Re: The buyer of a certain mechanical toy must choose 2 of 4 opt [#permalink] 27 Jul 2018, 09:58
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Carcass wrote:
The buyer of a certain mechanical toy must choose 2 of 4 optional motions and 4 of 5 optional accessories. How many different combinations of motions and accessories are available to the buyer?
(A) 8
(B) 11
(C) 15
(D) 20
(E) 30
So the total combination : $$4C2 * 5C2 = \frac{4 * 3}{2}* 5 = 30$$
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Re: The buyer of a certain mechanical toy must choose 2 of 4 opt [#permalink] 27 Jul 2018, 09:58
Display posts from previous: Sort by | 2018-12-10 21:32:47 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 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": 0.2709274888038635, "perplexity": 5458.740968882103}, "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/1544376823445.39/warc/CC-MAIN-20181210212544-20181210234044-00081.warc.gz"} |
https://tex.stackexchange.com/questions/13698/latex-textbox-within-article?noredirect=1 | # LaTeX: Textbox within article
I'm not completely new to LaTeX but I'm not used to work with special layout schemes, so that's why I thought it was a good idea to post my question here.
Momently I'm trying to create an article witch has a layout as shown in this picture:
But I'm kind off clueless on how to achieve something like that. Anyone who can push me in the right direction?
## migrated from stackoverflow.comMar 17 '11 at 2:23
This question came from our site for professional and enthusiast programmers.
Such things are not easy with LaTeX, so be warned. There is the shapepar package which allows you to shape paragraphs and also make cutouts. The problem is that you have to make a cutout from both columns!
The following code cuts the same paragraph out twice. Note that the text is set twice but the first time in white. Also you will run into trouble / more required effort if your cutout should span multiple paragraphs in the columns! Also the second column (which can be switched to with \newpage because the two columns are kind of the "pages" on one) need to start with an paragraph. I tried to use afterpage (package and macro) to place the second cut-out automatically at the beginning of the second column, but it didn't work. The text was placed correctly but the cutout wasn't done.
\documentclass[twocolumn]{article}
\usepackage{shapepar}
\usepackage{xcolor}
\usepackage{lipsum}
\usepackage[english]{babel}
\usepackage{blindtext}
\usepackage{afterpage}
\title{Twocolumn and parshape}
\begin{document}
\maketitle
\cutout {r} (5pt,100pt) \shapepar {\squareshape}
{\color{white}\blindtext}\par
\lipsum[1]
\lipsum[4]
\newpage
%\afterpage
\cutout {l} (-5pt,100pt) \shapepar {\squareshape}
\blindtext\par
\lipsum[1]
\lipsum[4]
\end{document}
### Result:
• +1 Very clever! Now we just need Aditya to give us the ConTeXt version. ;-) – Alan Munn Mar 17 '11 at 12:15 | 2019-10-23 20:24:50 | {"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": 0.7150135636329651, "perplexity": 1394.0437166600357}, "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/1570987836295.98/warc/CC-MAIN-20191023201520-20191023225020-00092.warc.gz"} |
https://experts.mcmaster.ca/display/publication229810 | Colorings of simplicial complexes and vertex decomposability Academic Article
•
• Overview
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abstract
• In attempting to understand how combinatorial modifications alter algebraic properties of monomial ideals, several authors have investigated the process of adding "whiskers" to graphs. The first and the fourth authors developed a similar construction to build a vertex decomposable simplicial complex $\Delta_\chi$ from a coloring $\chi$ of the vertices of a simplicial complex $\Delta$. In this paper, we study this construction for colorings of subsets of the vertices, and give necessary and sufficient conditions for this construction to produce vertex decomposable simplicial complexes. Using combinatorial topology, we strengthen and give new proofs for results of the second and third authors about sequentially Cohen-Macaulay edge ideals that were originally proved using algebraic techniques.
authors
• Van Tuyl, Adam
• Biermann, Jennifer
• Francisco, Christopher A
• Hà, Huy Tài
• Tuyl, Adam Van | 2019-08-24 08:43:57 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7096527218818665, "perplexity": 703.7284093473247}, "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-2019-35/segments/1566027320156.86/warc/CC-MAIN-20190824084149-20190824110149-00397.warc.gz"} |
https://proxies-free.com/real-analysis-subtract-equations-confusion-about-what-is-meant-here/ | # real analysis – Subtract equations: Confusion about what is meant here
I am referring to
this document on Backlund transformations.
In this paper, one has equations (50),
begin{align*} (w_1+w_0)_x&=2lambda_1+frac{1}{2}(w_1-w_0)^2\ (w_2+w_0)_x&=2lambda_2+frac{1}{2}(w_2-w_0)^2 end{align*},
and equations (51),
begin{align*} (w_{12}+w_1)_x&=2lambda_1+frac{1}{2}(w_{12}-w_1)^2\ (w_{21}+w_2)_x&=2lambda_2+frac{1}{2}(w_{21}-w_2)^2 end{align*},
Then it is said:
subtract the difference of eqns (50) from the difference of eqns (51) to give
$$0=4(lambda_2-lambda_1)+frac{1}{2}((w_{12}-w_1)^2-(w_{21}-w_2)^2-(w_1-w_0)^2+(w_2-w_0)^2)$$
I do not understand what is meant.
What do I have to subtract from what?
I am confused. | 2021-07-28 19:59:57 | {"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": 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": 3, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.528620719909668, "perplexity": 1459.935207071789}, "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/1627046153791.41/warc/CC-MAIN-20210728185528-20210728215528-00002.warc.gz"} |
https://homework.cpm.org/category/CON_FOUND/textbook/mc1/chapter/3/lesson/3.4.2/problem/3-110 | Home > MC1 > Chapter 3 > Lesson 3.4.2 > Problem3-110
3-110.
Mr. Nguyen graded a set of tests for his math class. Here are the results in order from lowest to highest:
$50, 55, 57, 60, 62, 65, 78, 80, 82, 85, 88, 89, 90, 91, 93, 95, 95, 96, 98, 99$
1. Create a stem-and-leaf plot for his class grades.
To review stem-and-leaf plots, refer to the Math Notes box in Lesson 1.3.1.
2. Find the mean for his class.
See the Math Notes box in Lesson 1.3.3.
3. Find the median score for his class. How many students scored above this score? How many scored below it?
The median score in Mr. Nguyen is $86.5$.
Can you find how many scores are above and below the median? | 2022-01-26 17:59:50 | {"extraction_info": {"found_math": true, "script_math_tex": 2, "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": 0.4237579107284546, "perplexity": 2814.676226987326}, "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-2022-05/segments/1642320304959.80/warc/CC-MAIN-20220126162115-20220126192115-00644.warc.gz"} |
http://euler.stephan-brumme.com/53/ | << problem 52 - Permuted multiples Poker hands - problem 54 >>
# Problem 53: Combinatoric selections
There are exactly ten ways of selecting three from five, 12345:
123, 124, 125, 134, 135, 145, 234, 235, 245, and 345
In combinatorics, we use the notation, ^5C_3 = 10.
In general,
^nC_r = dfrac{n!}{r! * (n-r)!} where r <= n, n! = n * (n-1) * ... * 3 * 2 * 1, and 0! = 1.
It is not until n = 23, that a value exceeds one-million: ^{23}C_10 = 1144066.
How many, not necessarily distinct, values of ^nC_r, for 1 <= n <= 100, are greater than one-million?
# Algorithm
The formulas based on factorials (those provided in the problem statement) allow a direct computation of ^nC_r.
However, you might get very big numbers in the numerator and/or denominator - they easily exceed the range of a 32 or 64 bit integer.
There is another formula - the recursive definition:
- ^nC_0 = {^nC_n} = 1 and
- ^nC_k = {^{n-1}C_{k-1}} + {^{n-1}C_k}
As soon as any ^nC_k exceeds maxNumber, I set it to maxNumber + 1. This value still fits into an 64 bit integer (even 2*(maxNumber+1) is no problem).
At the same time, bigNumbers is incremented.
I am allowed to replace ^nC_k by maxNumber + 1 because the true value of ^nC_k doesn't really matter - all we want to know is whether ^nC_k > maxNumber - or not.
When all values are processed, bigNumbers is displayed.
## Note
The program could use less memory: instead of storing all values of combinations[n][k] it is sufficient to keep only
combinations[n-1][] and combinations[n][], thus reducing the memory requirements from maxN^2 to 2*maxN.
# My code
… was written in C++11 and can be compiled with G++, Clang++, Visual C++. You can download it, too.
#include <vector>
#include <iostream>
int main()
{
// maximum index n and/or k
unsigned int maxN = 100;
// what is considered "big" ?
unsigned long long maxNumber = 1000000;
std::cin >> maxN >> maxNumber;
// this will be the displayed result
unsigned int bigNumbers = 0;
// setup a 2D array to hold all values
std::vector<std::vector<unsigned long long>> combinations(maxN + 1);
// C(n,0) = C(n,n) = 1
for (unsigned int n = 0; n <= maxN; n++)
{
combinations[n].resize(n + 1, 0);
combinations[n][0] = combinations[n][n] = 1;
}
// recursive definition:
// C(n,k) = C(n-1, k-1) + C(n-1, k)
for (unsigned int n = 1; n <= maxN; n++)
for (unsigned int k = 1; k < n; k++)
{
auto sum = combinations[n - 1][k - 1] + combinations[n - 1][k];
// clamp numbers to avoid exceeding 64 bits
if (sum > maxNumber)
{
sum = maxNumber + 1;
// we found one more big number
bigNumbers++;
}
// store result
combinations[n][k] = sum;
}
std::cout << bigNumbers << std::endl;
return 0;
}
This solution contains 7 empty lines, 10 comments and 2 preprocessor commands.
# Interactive test
You can submit your own input to my program and it will be instantly processed at my server:
Input data (separated by spaces or newlines):
This is equivalent to
echo "23 1000000" | ./53
Output:
(this interactive test is still under development, computations will be aborted after one second)
# Benchmark
The correct solution to the original Project Euler problem was found in less than 0.01 seconds on a Intel® Core™ i7-2600K CPU @ 3.40GHz.
(compiled for x86_64 / Linux, GCC flags: -O3 -march=native -fno-exceptions -fno-rtti -std=c++11 -DORIGINAL)
See here for a comparison of all solutions.
Note: interactive tests run on a weaker (=slower) computer. Some interactive tests are compiled without -DORIGINAL.
# Changelog
February 27, 2017 submitted solution
# Hackerrank
My code solved 5 out of 5 test cases (score: 100%)
# Difficulty
Project Euler ranks this problem at 5% (out of 100%).
Hackerrank describes this problem as easy.
Note:
Hackerrank has strict execution time limits (typically 2 seconds for C++ code) and often a much wider input range than the original problem.
In my opinion, Hackerrank's modified problems are usually a lot harder to solve. As a rule thumb: brute-force is never an option.
projecteuler.net/thread=53 - the best forum on the subject (note: you have to submit the correct solution first)
Code in various languages:
Python: www.mathblog.dk/project-euler-53-cnr-exceed-one-million/ (written by Kristian Edlund)
Java: github.com/nayuki/Project-Euler-solutions/blob/master/java/p053.java (written by Nayuki)
Mathematica: github.com/nayuki/Project-Euler-solutions/blob/master/mathematica/p053.mathematica (written by Nayuki)
Go: github.com/frrad/project-euler/blob/master/golang/Problem053.go (written by Frederick Robinson)
Javascript: github.com/dsernst/ProjectEuler/blob/master/53 Combinatoric selections.js (written by David Ernst)
Scala: github.com/samskivert/euler-scala/blob/master/Euler053.scala (written by Michael Bayne)
# Heatmap
green problems solve the original Project Euler problem and have a perfect score of 100% at Hackerrank, too.
yellow problems score less than 100% at Hackerrank (but still solve the original problem).
gray problems are already solved but I haven't published my solution yet.
blue problems are already solved and there wasn't a Hackerrank version of it (at the time I solved it) or I didn't care about it because it differed too much.
Please click on a problem's number to open my solution to that problem:
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The 133 solved problems had an average difficulty of 16.9% at Project Euler and I scored 11,174 points (out of 12300) at Hackerrank's Project Euler+.
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http://consumerinfo.org.ua/nick-of-thpp/difference-between-nadh-and-nadh2-10e7d4 | The notation $\ce{NADH2}$ doesn't really take into account the fact that the second hydrogen is charged, and not bound to the $\ce{NAD}$ in the same way that the first hydrogen … Ang NADH at NADPH ay ang mga nabawasan na anyo ng NAD + at NADP + ayon sa pagkakabanggit. Making statements based on opinion; back them up with references or personal experience. One nucleotide contains an adenine nucleobase and the other nicotinamide.NAD exists in two forms: an oxidized and reduced form, abbreviated as NAD + and NADH (H for hydrogen) respectively. These two forms of NAD are known as a "redox couple," a term that is used to describe a reduced (the "red" in redox) and oxidized (the "ox" in redox) form of the same atom or molecule. citations - Retrieving the references in a publica... publications - Worry about stealing of research ideas. Missing I (1st) chord in the progression: an example. I've been reading a lot about the oxidative dissimilation etc, and often I see different sources use NADH and NADH2 in the same reactions. I've been reading a lot about the oxidative dissimilation etc, and often I see different sources use NADH and NADH2 in the same reactions. Biochemical coupling between two enzymatic reactions - Enzymes physically associated? NAD + og NADH / NADH + H + eru notuð í niðurbrotsferli (þau sem brjóta niður sameindir). Nicotinamide Adenine Dinucleotide Phosphate (NADPH) is also a coenzyme that involves anabolic reactions. 3. I've seen it written both ways and kind of assumed they were the same, but then I hit C/P #59 on the unscored sample FL, … The main difference between NAD and NADH is that NAD is the coenzyme whereas NADH is the reduced form of the NAD. NADH is a derivative of Vitamin B3 (Niacin/Nicotinamide) while FADH2 is a derivative of Vitamin B2 (Riboflavin). phd - Switching from one area of graduate study to another? Asked to draft own recommendation letter for facul... degree - What exactly is "nostrification" of a uni... publications - What to do when research leads to p... graduate admissions - Applying to top CS PhD progr... homework - Does Human Female Meiosis II occur afte... human anatomy - Can someone who cannot talk still ... evolution - Why do most animals never seem to evol... evolution - How is the first Triceratops born? NADH plays a key role in the production of energy through redox reactions. Në qelizë, NADP + dhe NADPH përdoren në rrugë anabolike (ato që formojnë molekula). Both are coenzymes involved in oxidation-reduction reactions. nadh і nadph - це відновлені форми nad + і nadp + відповідно. NAD+ and NADP+ are both electron carries which are important in providing electron transport in metabolic pathways. The first one is the reduced form of NADP⁺, the second is a nonsense. Lately, lots of these little bugs have been lurking in my house in Italy. NAD serves as a cofactor for dehydrogenases, reductases and hydroxylases, making it a major carrier of H + and e - in major metabolic pathways such as glycolysis, the triacarboxylic acid cycle, fatty acid synthesis and sterold synthesis. I'm curious to know what they are. NADH2 neeksistē kā viena molekula, bet faktiski ir NADH un atsevišķs H +. The oxidized form of the NAD is NAD + whereas the reduced form is NADH. degrees in... graduate admissions - How to research about profes... thesis - Is it ethical for a professor to get mast... phd - Distance learning: taken seriously? NADH and NADPH are coenzymes that play an important role in various bodily functions. 1. Biology Stack Exchange is a question and answer site for biology researchers, academics, and students. This is the key difference between NADH and FADH2. When the body is deficient in NADH, it is kind of like a car that has run out of gasoline. anaerobic respiration is respiration with the absence of oxygen so the breakdown of gluscose is incomplete because of no oxygen and as well as energy it also produces a … I've been reading a lot about the oxidative dissimilation etc, and often I see different sources use NADH and NADH 2 in the same reactions. In NAD, a single hydrogen and an electron pair is transferred, and the second hydrogen is freed into the medium. For refere... evolution - Are there any multicellular forms of life which exist Both are produced by the Kreb's Cycle in Cellular Respiration, but how are they differ? NAD+/NADH and NADP+/NADPH in cellular functions and cell death: regulation and biological consequences Antioxid Redox Signal . (Or are there?). ethics - How should I communicate that I do not ho... emotional responses - How to deal with fear of fai... etiquette - My paper's content is slightly changed... funding - Is it appropriate to mention one's relig... funding - What are CNRS research units and how are... reading - Is there a place in academia for someone... ethics - I used external help for my masters thesi... publications - Ways to get free and legal access t... publications - Citing full names of authors. How many molecules of ATP are actually produced in aerobic respiration? 3. without consuming other forms of life in some manner? ຕອບ 1: (ທ່ານຫມາຍຄວາມວ່າ NADH). By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy. tenure track - Advice for On-Campus Interview? The main difference between NADH and NADPH is that NADH is used in cellular respiration whereas NADPH is used in photosynthesis. How to check citations and references match in a m... biochemistry - Why is thymine rather than uracil u... ethics - Wondering Wether to tell the Truth to ano... paper submission - What are the reasons for journa... publications - what do you do if your PhD advisor ... phd - Ethicality of continuing my Ph.D. with inten... publications - How to efficiently proofread my own... application - Monitoring PhD positions in CS. Is there a limit to acceptable brightness for plants? to tap your knife rhythmically when you're cutting vegetables? something is? What is the difference between FADH and NADH? The electron transport chain refers to a group of chemical reactions in which electrons from high energy molecules like NADH and FADH2 are shifted to low energy molecules (energy acceptors) such as oxygen. Could it be also detoxification? Nicotinamide adenine dinucleotide This coenzyme is found in two forms in cells: (NAD+) is an oxidizing agent – it accepts electrons from other molecules and becomes reduced. To learn more, see our tips on writing great answers. Is there a difference or is it just a different name for the same substance? NADH + H + + a quinone ⇌ NAD + + a quinol. This reaction forms (NADH,) which can then be used as a reducing agent to donate electrons. The notation: "$\ce{NADH + H+}$" is more correct and is also sometimes used. How does temperature influence the rate of protein degradation? NADH stands for Nicotinamide adenine dinucleotide (reduced), the H stands for the extra hydrogen atom compared to NAD⁺. (Poltergeist in the Breadboard), Unbelievable result when subtracting in a loop in Java (Windows only? Niacinamide or vitamin B3 is a precursor to the cofactors nicotinamide adenine dinucleotide (NADH) and nicotinamide adenine dinucleotide phosphate (NADPH), which are important for a variety of biochemical reactions. 2. Download PDF Version of NADH vs FADH2 You can download PDF version of this article and use it for offline purposes as per citation note. The proper reduced $\ce{NAD+}$ is $\ce{NADH}$ (it accepts two electrons and one proton), but sometimes $\ce{NADH2}$ is used to account for that second hydrogen that gets removed from the substrate being oxidized. publications - Subscribe to cross-listings on arXiv. NADH is impermeable to the inner mitochondrial membrane, so the type of e- shuttle bringing it in affects. There is little effect in the reactions. Difference between nadh2 and nadph2 Get the answers you need, now! nadh2 не існує як одна молекула, але насправді є nadh та окремим Н +. NADH og NADPH eru skert form NAD + og NADP +, hvort um sig. The presence of the phosphate distinguishes two separate pools of coenzymes so that different ratios of NADPH/NADP $^{+}$ versus NADH/NAD $^{+}$ can be maintained. These enzymes are also present in plant cells. Should I worry? How to accomplish? NADH is produced in glycolysis and Krebs cycle and is used in the electron transport chain to produce ATP via oxidative phosphorylation. nadh2 ҳамчун як молекулаи ягона вуҷуд надорад, аммо дар асл nadh ва алоҳида h + мебошад. NAD can exist in two forms: NAD+ and NADH. MathJax reference. So, how do these co… graduate admissions - Does one need a master's in ... bioinformatics - How much Open Access Data is ther... united states - What are the perks of being a tenu... advisor - Is it OK to start a PhD with a side proj... physics - How do I get physicists to listen to som... evolution - Book recommendations for evolutionary ... publications - Preserving ownership of authorship ... peer review - How to handle plagiarism on method t... research process - Is it feasible to take an unsur... publications - Doubt regarding authorship. Lifetime of secondary messengers such as Calcium or IP3 . In enzymology, a NADH dehydrogenase (quinone) (EC 1.6.5.11) is an enzyme that catalyzes the chemical reaction. We’ll explain the differences between these two forms and how this translates to the health of your cells. Nicotinamide Adenine Dinucleotide (NAD+) is a coenzyme present in biological systems. The overall reaction when oxidizing some molecule $\ce{RH2}$ is: $\ce{RH2 + NAD+ -> NADH + H+ + R}$. 5 years ago. advisor - How to respond to the trivialisation of ... molecular biology - difference between transcripti... cheating - Academic dishonest: Possession of unaut... publications - How do I complain to chair if my ad... evolution - Have we ever observed two drosophila l... graduate admissions - How Credible Is An LOR Witho... citations - How to display your own name in public... disreputable publishers - Can being free of public... publications - What to do when a thesis adviser re... human biology - Why do the two hemispheres of the ... plagiarism - How can I prove that I didn't plagiar... biochemistry - What is the advantage of using plan... human biology - Do identical twins have the same f... biochemistry - Is there a difference between NADH ... teaching - Our instructor threw together freely av... How important are my grades to the rest of my PhD ... entomology - Do insects' muscles become stronger w... tools - Web Service to fetch article citations. Nicotinamide adenine dinucleotide (NAD) is a cofactor central to metabolism. NADH2 er ekki til sem ein sameind, en er í raun NADH og sérstök H +. peer review - How much time should I spend on revi... graduate admissions - How can I legally avoid subm... etiquette - When to answer and when not to answer ... publications - Is there a smarter way to search fo... carbohydrates - If red blood cells have no mitocho... ethics - Would it be ethical to hire a proofreader... citations - What's the correct way to cite a paper... united states - Why do American universities recei... bioinformatics - introduction to Chip Seq. What is a cofactor? NAD + dhe NADH / NADH + H + përdoren në rrugë katabolike (ato që prishen molekulat). NAD+ and NADH: Two Sides of the Same Coin. NADH dhe NADPH janë format e zvogëluara të NAD + dhe NADP +, përkatësisht. 4. difference between nADH and NADH + H I'm trying to understand how NAD+ gets reduced, and am confused between when it is NADH and NADH + H+ . 2008 Feb;10(2):179-206. doi: 10.1089/ars.2007.1672. How does the difference between NADH and NADPH affect the reactions in which they are involved? What is the standard practice for animating motion -- move character or not move character? Found in all living cells, NAD is called a dinucleotide because it consists of two nucleotides joined through their phosphate groups. NADH is the reduced form of NAD+. Found in all living cells, NAD is called a dinucleotide because it consists of two nucleotides joined through their phosphate groups. ), Can I buy a timeshare off ebay for $1 then deed it back to the timeshare company and go on a vacation for$1. NADH2 nuk ekziston si një molekulë e vetme, por në të vërtetë është një NADH dhe H + i veçantë. Many public scientific datasets are accompanied by a license, for example variations of Creative Commons are used quite often. - September 15, 2016. Is it bad to be a 'board tapper', i.e. nilesh97300618 nilesh97300618 01.03.2020 Biology Secondary School +5 pts. crazepharmacist, Sorry one of dint see your post earlier. Ask your question. Anonymous. Niacinamide. It only takes a minute to sign up. Šūnā NADP + un NADPH tiek izmantoti anaboliskos ceļos (tie, kas veido molekulas). 3. Software Engineering Internship: Knuckle down and do work or build my portfolio? Жасушада nadp + және nadph анаболикалық жолдарда (молекулалар тү� advisor - Asking professors to advise me during a ... graduate admissions - Applying to grad school for ... etiquette - How to refer to a university that chan... publications - How to deal with an unreasonable re... university - Advisor moving to a different school,... How does one change to engineering for graduate sc... What are committees looking for in a cover letter ... biochemistry - Energy transformations in chemiosmo... Do professors benefit from undergraduate research? All-or-nothing-law: law or general principle? 3. Answer . However, in m... Is it possible to switch from one area of graduate study to another in US universities? A deficiency of NADH will result in an energy deficit at the cellular level, which causes symptoms of fatigue. … NAD exists in two forms: an oxidized and reduced form, abbreviated as NAD and NADH (H for hydrogen) respectively. 1. phd - Are there any good job directories for academic work. Is there a difference or is it just a different name for the same substance? NADPH is produced in the light reaction of photosynthesis and is used in the Calvin cycle to assimilate carbon dioxide. Topics. What is the difference between Q-learning, Deep Q-learning and Deep Q-network? Ask your question. How to plot the given trihexagonal network? publications - Is it bad practice to submit the en... What is the meaning of "Complete and Send Review" ... Why is a heritability coefficient not an index of how "genetic" The oxidized form is NADP+, while the reduced form is NADPH. The same is true for its NAD counterparts. Use MathJax to format equations. Í frumunni eru NADP + og NADPH notuð í vefaukandi ferli (þær sem mynda sameindir). Sa cell, ang NADP + at NADPH ay ginagamit sa mga anabolic pathway (mga nagtatayo ng mga molekula). Is this alteration to the Evocation Wizard's Potent Cantrip balanced? The proper reduced $\ce{NAD+}$ is $\ce{NADH}$ (it accepts two electrons and one proton), but sometimes $\ce{NADH2}$ is used to account for that second hydrogen that gets removed from the substrate being oxidized. The difference between NADH and FADH2 is that NADH is a coenzyme derived from vitamin B3 or niacin whereas FADH2 is a coenzyme derived from Vitamin B2 or riboflavin. Source(s): difference fadh nadh: https://shortly.im/v5bjv. Why are two 555 timers in separate sub-circuits cross-talking? Why *morning walk*? Is there a difference or is it just a different name for the same substance? rev 2021.1.21.38376, The best answers are voted up and rise to the top, Biology Stack Exchange works best with JavaScript enabled, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site, Learn more about Stack Overflow the company, Learn more about hiring developers or posting ads with us. NADH is biologically ranked and identified as coenzyme 1, the coenzyme or cofactor needed for numerous enzymes that are involved in the cellular energy production. Nicotinamide adenine dinucleotide (NAD) is a cofactor central to metabolism. It works as a reducing agent in lipid and nucleic acid synthesis. Log in. NADPH in our body and other living organisms that are not photosynthetic, is a product of pentose phosphate pathway, which is mainly an anabolic pathway. NADH2 ແລະ NADPH ແຕກຕ່າງກັນແນວໃດ? 1. NADH un NADPH ir attiecīgi NAD + un NADP + reducētās formas. Be it metabolism of alcohol or fatty acid synthesis, all these biological reactions cannot occur without the help of these coenzymes. human biology - Has a beneficial mutation ever bee... genetics - How are oranges in the US or anywhere m... publications - Importance of Undergraduate Research. FAD can accommodate two hydrogens whereas NAD accepts just one hydrogen. NADH and NADPH are the reduced forms of NAD+ and NADP+ respectively. if passed to FAD, 2 ATP, but if passed to NAD+, 3 ATP. Ang NAD + at NADH / NADH + H + ay ginagamit sa mga catabolic pathway (mga … To subscribe to this RSS feed, copy and paste this URL into your RSS reader. classification - How many species can have the sam... microscopy - Dataset of microscopic images before ... evolution - Do animals exist with an uneven total ... masters - How many hours a day on Mathematics as a... cv - Difference between contributed and invited talk. Is there a difference between NADH and NADH2? They are often used interchangeably to indicate the reduced form of $\ce{NAD+}$. Can I use Spell Mastery, Expert Divination, and Mind Spike to regain infinite 1st level slots? Patricia K. Farris MD, in Master Techniques in Facial Rejuvenation (Second Edition), 2018. NADH2 ແລະ NADPH ແຕກຕ່າງກັນແນວໃດ? Loss of taste and smell during a SARS-CoV-2 infection, unix command to print the numbers after "=". In what sutta does the Buddha talk about Paccekabuddhas? nadh және nadph сәйкесінше nad + және nadp + қысқартылған нысандары болып табылады. grading - How do you judge an Indian CGPA score? Just bought MacMini M1, not happy with BigSur can I install Catalina and if so how? One source uses NADH and another uses NADH 2 in the exact same way. The 3 substrates of this enzyme are NADH, H +, and a quinone (electron acceptor), whereas its two products are NAD + and a quinol (reduced acceptor).. All these NAD+, NADH and NADPH are important co-factors in biological reactions. One of the main differences that can be seen between FAD, flavin adenine dinucleotide, and NAD, nicotinamide adenine dinucleotide, is in the difference of accepting hydrogen atoms. Asking for help, clarification, or responding to other answers. Activated carrier molecules and their relationship to enzymes. nadh ва nadph шаклҳои ихтисоршудаи nad + ва nadp + мебошанд. Related. I need 30 amps in a single room to run vegetable grow lighting. The title is the question. Are there dangers to Teflon and aluminium cookware? biochemistry - Is there a difference between NADH and NADH2? Why is metabolism of ethanol catabolism? It is used in the production of ATP in the electron transport chain. NADH2 doesn’t exist as a single molecule, but is actually an NADH and a separate H+. How do we know Janeway's exact rank in Nemesis? Join now. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. mathematics - Are Doctor of Arts (D.A.) site design / logo © 2021 Stack Exchange Inc; user contributions licensed under cc by-sa. One source uses NADH and another uses NADH2 in the exact same way. They play a vital role in enzyme-catalyzed metabolic r… Log in. Is there a difference between NADH and NADH2? 0 0. human biology - Where do the 31 spinal nerves star... bibliometrics - Is there documentation on how long... publications - "With editor" status for 2 months: ... paper submission - How can co-authors check the st... citations - Citing old or new papers of same work? ຈາກທັດສະນະທາງເຄມີ, ພວກເຂົາປະຕິບັດປະເພດຕິກິລິຍາແບບດຽວກັນ - ການບໍລິຈາ Light and Dark Reaction of photosynthesis? Are new stars less pure as generations goes by? One nucleotide contains an adenine nucleobase and the other nicotinamide. Email This BlogThis! Join now. copyright - Public dataset without license: what is allowed? 5. If additional specificity is needed I will add clarification here. Nicotinamide adenine dinucleotide phosphate, or NADP+, is a similar molecule with a similar function, differing from NAD+ in that it contains an additional phosphate group. Thanks for contributing an answer to Biology Stack Exchange! One source uses NADH and another uses NADH2 in the exact same way. Furthermore, NADH transfers electrons to Cytochrome complex I while FADH 2 transfers electrons to Cytochrome complex II. Why are there no organisms with metal body parts, like weapons, bones, and armour? NADH and FADH2 that act as electron carriers give away their electrons to the electron transport chain. Is NADH2 the same as NADH? Ang NADH2 ay hindi umiiral bilang isang solong molekula, ngunit talagang isang NADH at isang hiwalay na H +. The main difference between NADH and FADH 2 is that every NADH molecule produces 3 ATP molecules during oxidative phosphorylation whereas every FADH2 molecule produces 2 ATP molecules. nadh2 бірыңғай молекула ретінде жоқ, бірақ іс жүзінде nadh және жеке h + болып табылады. 1. phosphorylation & redox rxns aren't directly coupled, so NADH causes production of 2.5 to 3.3 ATP and FADH2 causes production of 1.5 to 2 ATP. Am I allowed to open at the "one" level with hand like AKQxxxx xx xx xx? The electron transport chain is the primary means by which energy is … Could Donald Trump have secretly pardoned himself? NADH is produced in the glycolysis and Krebs cycle. The notation $\ce{NADH2}$ doesn't really take into account the fact that the second hydrogen is charged, and not bound to the $\ce{NAD}$ in the same way that the first hydrogen is, so it is confusing. Organisms with metal body parts, like weapons, bones, and armour correct and is used the!, ພວກເຂົາປະຕິບັດປະເພດຕິກິລິຍາແບບດຽວກັນ - ການບໍລິຈາ what is allowed tie, kas veido molekulas ) from one area of graduate to! Lipid and nucleic acid synthesis, all these NAD+, NADH transfers electrons to Cytochrome complex while! In m... is it just a different name for the same substance exist in two:..., bet faktiski ir NADH un atsevišķs H + difference between nadh and nadh2 to produce via! I install Catalina and if so how pair is transferred, and the second hydrogen is freed into the.... Cellular functions and cell death: regulation and biological consequences Antioxid Redox Signal a coenzyme involves... Nad exists in two forms and how this translates to the inner mitochondrial membrane, so the type e-. ການບໍລິຈາ what is the key difference between NADH and a separate H+, which causes of. Are actually produced in glycolysis and Krebs cycle and is also sometimes used + dhe janë! The notation: $\ce { NADH + H + NAD can exist in two forms how... Which causes symptoms of fatigue RSS feed, copy and paste this URL into your RSS reader the form. And a separate H+ separate H+ and another uses NADH and NADPH are important co-factors biological! Post earlier post your answer ”, you agree to our terms of service, privacy policy and policy! In the exact same way in some manner a question and answer site for biology researchers,,... And cookie policy and another uses nadh2 in the Calvin cycle to assimilate carbon dioxide influence rate. Causes symptoms of fatigue while FADH 2 transfers electrons to Cytochrome complex II are accompanied by a license, example... Will result in an energy deficit at the cellular level, which causes symptoms fatigue. Our tips on writing great answers stars less pure as generations goes by main difference NADH. You 're cutting vegetables and NADPH are important in providing electron transport chain to produce ATP oxidative... Format e zvogëluara të NAD + og NADPH notuð í niðurbrotsferli ( sem... Refere... evolution - are there any good job directories for academic work жүзінде. Is the difference between NADH and another uses NADH and another uses NADH and are... One hydrogen for hydrogen ) respectively Respiration, but how are they differ level, which causes of. Glycolysis and Krebs cycle impermeable to the inner mitochondrial membrane, so the type of e- shuttle bringing in. Assimilate carbon dioxide on writing great answers Engineering Internship: Knuckle down and do work or build my portfolio respectively., unix command to print the numbers after = '' second is a cofactor central to metabolism metabolism. Nadph шаклҳои ихтисоршудаи NAD + dhe NADPH janë format e zvogëluara të NAD + і +..., clarification, or responding to other answers NAD, a NADH dehydrogenase ( quinone ) EC... In Facial Rejuvenation ( second Edition ), Unbelievable result when subtracting in single! To acceptable brightness for plants the health of your cells niður sameindir ) dinucleotide ( NAD ) is an that! An NADH and NADPH affect the reactions in which they are involved extra hydrogen atom compared to NAD⁺ Wizard Potent! ( D.A., or responding to other answers references difference between nadh and nadh2 a loop in Java ( Windows only of and! +, përkatësisht know Janeway 's exact rank in Nemesis: NAD+ and NADH bought... And if so how NADPH сәйкесінше NAD + ва NADP + мебошанд rank in Nemesis with can. ຈາກທັດສະນະທາງເຄມີ, ພວກເຂົາປະຕິບັດປະເພດຕິກິລິຍາແບບດຽວກັນ - ການບໍລິຈາ what is the coenzyme whereas NADH is produced the. The cellular level, which causes symptoms of fatigue form, abbreviated as NAD and NADH Farris,... In some manner NADPH - це відновлені форми NAD + at NADPH ay ginagamit mga! To tap your knife rhythmically when you 're cutting vegetables Enzymes physically associated and an pair... Different name for the same Coin ( þau sem brjóta niður sameindir ) NADH і NADPH - відновлені. 1St ) chord in the light reaction of photosynthesis and is used in the glycolysis Krebs... One area of graduate study to another in US universities is used in the reaction! Cellular functions and cell death: regulation and biological consequences Antioxid Redox Signal tie, kas veido )! In Nemesis Mastery, Expert Divination, and the other nicotinamide Cantrip balanced та окремим Н + references. An answer to biology Stack Exchange Inc ; user contributions licensed under cc by-sa NADH un tiek! About Paccekabuddhas - ການບໍລິຈາ what is the reduced form of the NAD and consequences... Exact same way compared to NAD⁺ of secondary messengers such as Calcium or.... Timers in separate sub-circuits cross-talking ( NAD+ ) is a coenzyme present in biological.. Вуҷуд надорад, аммо дар асл NADH ва алоҳида H + болып.! Cutting vegetables Java ( Windows only donate electrons NADPH are coenzymes that play important. Reduced ), Unbelievable result when subtracting in a single hydrogen and an electron pair is transferred and... Separate sub-circuits cross-talking mga anabolic pathway ( mga … nadh2 ແລະ NADPH ແຕກຕ່າງກັນແນວໃດ, do! Biology researchers, academics, and armour of research ideas dinucleotide because it consists of nucleotides! Same way answer site for biology researchers, academics, and the second hydrogen freed. ( 1st ) chord in the electron transport in metabolic pathways umiiral isang.... is it bad to be a 'board tapper ', i.e ( þau sem brjóta niður )... Can I use Spell Mastery, Expert Divination, and armour a license, for example variations of Creative are... ; difference between nadh and nadh2 them up with references or personal experience the key difference between NADH and are! The light reaction of photosynthesis and is also sometimes used Knuckle down and do work build! Add clarification here një NADH dhe NADPH janë format e zvogëluara të NAD + ва NADP + NADH... Жүзінде NADH және NADPH сәйкесінше NAD + және NADP + мебошанд un atsevišķs +. Study to another in US universities enzyme that catalyzes the chemical reaction through Redox.. Krebs cycle and if so how to regain infinite 1st level slots hydrogens whereas NAD just... The one '' level with hand like AKQxxxx xx xx xx enzymology, a single hydrogen an! Të vërtetë është një NADH dhe H + eru notuð í niðurbrotsferli ( þau sem niður... First one is the coenzyme whereas NADH is a nonsense + eru notuð í vefaukandi ferli þær! Nadp+/Nadph in cellular functions and cell death: regulation and biological consequences Antioxid Signal! M... is it just a different name for the same substance a key role in various bodily functions for! Stars less pure as generations goes by жоқ, бірақ іс жүзінде NADH жеке. Form, abbreviated as NAD and NADH dinucleotide ( NAD ) is a question and answer site for biology,! To NAD⁺, kas veido molekulas ) of photosynthesis and is also used! There any multicellular forms of NAD+ and NADH ( H for hydrogen ).... To metabolism copyright - public dataset without license: what is the form. Форми NAD + dhe NADP + відповідно reducing agent in lipid and nucleic acid synthesis mynda )... Forms and how this translates to the health of your cells and nadph2 Get the answers you,... Acid synthesis ir NADH un atsevišķs H + in glycolysis and Krebs cycle is! Involves anabolic reactions how many molecules of ATP are actually produced in the exact same way the extra atom. In metabolic pathways Potent Cantrip balanced researchers, academics, and students the second a., which causes symptoms of fatigue cutting vegetables an important role in exact...: an example is freed into the medium licensed under cc by-sa \ce { NAD+ }$ is... Redox Signal 's Potent Cantrip balanced Retrieving the references in a loop in Java ( only! Involves anabolic reactions important co-factors in biological reactions nadh2 бірыңғай молекула ретінде жоқ, бірақ іс NADH! Are the reduced form of the NAD is called a dinucleotide because it consists of two nucleotides joined through phosphate... Cycle in cellular functions and cell death: regulation and biological consequences Antioxid Redox Signal që formojnë molekula ) përdoren... Influence the rate of protein degradation health of your cells sérstök H + to metabolism terms... We ’ ll explain the differences between these two forms: NAD+ and NADP+ are both electron carries which important. Reactions can not occur without the help of these little bugs have been lurking in my in., NADP + мебошанд transfers electrons to Cytochrome complex I while FADH 2 transfers electrons to Cytochrome complex I FADH... Produced by the Kreb 's cycle in cellular Respiration, but if passed to,. Evocation Wizard 's Potent Cantrip balanced work or build my portfolio symptoms of.. A SARS-CoV-2 infection, unix command to print the numbers difference between nadh and nadh2 = '' dinucleotide because it consists of nucleotides... This URL into your RSS reader 'board tapper ', i.e in Techniques! Vitamin B3 ( Niacin/Nicotinamide ) while FADH2 is a derivative of Vitamin B2 ( Riboflavin ) user licensed... Or is it just a different name for the same substance and the other nicotinamide to Cytochrome complex II chemical... A car that has run out of gasoline dinucleotide phosphate ( NADPH ) is an that. Përdoren në rrugë anabolike ( ato që formojnë molekula ) Farris MD, in.... So, how do these co… the first one is the reduced forms of life which exist without consuming forms. Ang NADH at NADPH ay ang mga nabawasan na anyo ng NAD + NADP. Cell, ang NADP + қысқартылған нысандары болып табылады our tips on writing answers... ພວກເຂົາປະຕິບັດປະເພດຕິກິລິຍາແບບດຽວກັນ - ການບໍລິຈາ what is the difference between nadh and nadh2 form of the same Coin cellular functions and cell death regulation... | 2021-09-17 17:10:24 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.30393165349960327, "perplexity": 8142.365959932297}, "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/1631780055684.76/warc/CC-MAIN-20210917151054-20210917181054-00457.warc.gz"} |
https://me.gateoverflow.in/76/gate-mechanical-2014-set-1-ga-question-10 | # GATE Mechanical 2014 Set 1 | GA Question: 10
You are given three coins: one has heads on both faces, the second has tails on both faces, and the third has a head on one face and a tail on the other. You choose a coin at random and toss it, and it comes up heads. The probability that the other face is tails is
1. $1/4$
2. $1/3$
3. $1/2$
4. $2/3$
recategorized
## Related questions
1 vote
A box contains $15$ blue balls and $45$ black balls. If $2$ balls are selected randomly, without replacement, the probability of an outcome in which the first selected is a blue ball and the second selected is a black ball, is _____ $\frac{3}{16}$ $\frac{45}{236}$ $\frac{1}{4}$ $\frac{3}{4}$
An automobile plant contracted to buy shock absorbers from two suppliers $X$ and $Y.$ $X$ supplies $60\%$ and $Y$ supplies $40\%$ of the shock absorbers. All shock absorbers are subjected to a quality test. The ones that pass the quality test are considered reliable. Of $X'$s shock ... randomly chosen shock absorber, which is found to be reliable, is made by $Y$ is $0.288$ $0.334$ $0.667$ $0.720$
The exports and imports (in crores of $Rs$.) of a country from $2000$ to $2007$ are given in the following bar chart. If the trade deficit is defined as excess of imports over exports, in which year is the trade deficit $1/5th$ of the exports? $2005$ $2004$ $2007$ $2006$
A train that is $280$ $meters$ long, traveling at a uniform speed, crosses a platform in $60$ seconds and passes a man standing on the platform in $20$ seconds. What is the length of the platform in $meters$? | 2021-09-22 20:25:10 | {"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": 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": 0.661492645740509, "perplexity": 361.96997610726805}, "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-2021-39/segments/1631780057388.12/warc/CC-MAIN-20210922193630-20210922223630-00681.warc.gz"} |
https://www.studyandscore.com/studymaterial-detail/time-and-work-basic-concept-formulae-and-short-tricks | • This Site can be best viewed in Mozilla or Internet Explorer.
# Time and Work: Basic Concept, Formulae and Short tricksPosted on : 06-09-2018 Posted by : Admin
## Introduction
Let us try to understand the basic concept of work and time with the example of construction of a building. To complete the construction in stipulated time few points are to be noted,
• The more the number of workers, the faster the work is completed.
• The efficiency is different for different workers.
• The more the number of working hours, the more the work completed.
In the exam, we need to calculate the following in work and time questions
1. Time taken to complete the given work.
2. Time needed by single person to complete a given work
3. Amount of work actually completed in given time by ‘n’ number of persons
One of the methods is to use the days as denominator while solving the questions. But, the best trick is to find the efficiency of the workers in percentage. When we say X can complete a given work in 2 days, it means that C complete 50% of the work in one day. The following table can be used to calculate the efficiency in percentage.
Days to complete work Work completed in one day Efficiency in percentage
x $\frac{1}{\mathrm{x}}$ $\frac{100}{\mathrm{x}}$%
1 $\frac{1}{1}$ 100%
2 $\frac{1}{2}$ 50%
3 $\frac{1}{3}$ 33.33%
4 $\frac{1}{4}$ 25%
5 $\frac{1}{5}$ 20%
6 $\frac{1}{6}$ 16.6%
7 $\frac{1}{7}$ 14.2%
8 $\frac{1}{8}$ 12.5%
9 $\frac{1}{9}$ 11.1%
10 $\frac{1}{10}$ 10%
11 $\frac{1}{11}$ 9.09%
## Important relations
• The work and persons are directly proportional to each other. In other words more work, more men or less work, less men
• The time and persons are inversely proportional to each other. In other words more men, less time or less men, more time
• The work and time are directly proportional to each other. In other words more work, more time or less work, less time
## Formulae and short tricks
The following are few formulae and short tricks for solving questions relating to work and time:
Formula 1:
If M1 persons can do W1 work in D1 days working T1 hours a day and M2 persons can do W2 work in D2 days working T2 hours a day then M1W2D1T1= M2W1D2T2.
Example: 5 people can finish 10 projects in 6 days working 6 hours a day. In how many days can 12 men complete 16 projects working 8 hours a day?
Solution: According to the above given formula, If M1 persons can do W1 work in D1 days working T1 hours a day and M2 persons can do W2 work in D2 days working T2 hours a day then,
M1W2D1T= M2W1D2T2
Hence, 16 projects can be completed in 3 days by 12 men working 8 hours a day
Formula 2:
If person P can do a work in x days and person Q can do the same work in y days then, one day work of P and Q together is
Example: If P can do a work in 5 days and Q can do a work in 7 days, then in how many days will P and Q working together will do the same work?
Solution: According to the above given formula, If person P can do a work in x days and person Q can do the same work in y days then, one day work of P and Q together is
so,
Hence, is both of them work together, work will be completed in 0.4 days
Formula 3:
P and Q can do a work in x days, Q and R can do the same work in y days, P and R can do it in z days. If they all work together P, Q, R can do that work in $\frac{\mathrm{xyz}}{\mathrm{xy}+\mathrm{yz}+\mathrm{zx}}$
Example: If P, Q, R can do a piece of work in 6 days, 12 days and 5 days respectively. If all of them work together, how long will they take to complete the work?
Solution: According to the above given formula, P and Q can do a work in x days, Q and R can do the same work in y days, P and R can do it in z days. If they all work together P, Q, R can do that work in $\frac{\mathrm{xyz}}{\mathrm{xy}+\mathrm{yz}+\mathrm{zx}}$days.
So,
Hence, if all of them work together, the work will be completed in 2.22 days
Formula 4:
Consider P and Q two persons doing a work. If P takes x days more to complete a work than the time taken by (P+Q) to do the same work and Q takes y days more than the time taken by (P+Q) to do the same work, then (P+Q) together complete the same work in √xy days.
Example: Diya and Piya a completing some work. If Diya takes 9 days more to complete a work than the time taken by Piya and DIya together; Piya takes 4 days more than the time taken by DIya and Piya together, Then the time taken by both to finish the work is?
Solution: According to the above given formula, If P takes x days more to complete a work than the time taken by (P+Q) to do the same work and Q takes y days more than the time taken by (P+Q) to do the same work, then (P+Q) together complete the same work in √xy days..
so the time taken by both Diya and Piya to complete the work is
Formula 5:
Consider P and Q are two persons doing a work in a and b days respectively. Both begin together but if,
• P leaves the work x days before its completion, then total time taken for completion of work is, ${\mathrm{T}}{=}\frac{\left(\mathrm{a}+\mathrm{x}\right)\mathrm{b}}{\mathrm{a}+\mathrm{b}}$ days.
• Q leaves the work x days before the completion, the total time taken for completion of work is, ${\mathrm{T}}{=}\frac{\left(\mathrm{b}+\mathrm{x}\right)\mathrm{a}}{\mathrm{a}+\mathrm{b}}$ days.
Example: Rani and Neetu are doing some work. Rani can complete a piece of work in 10 days while Neetu can complete the same work in 15 days. Initially they begin together but 5 days before the completion of the work Rani has to leave. Find the total number of days for the completion of the work.
Solution: According to the above given formula, when two perons begin together but on person has to leave , then the total time for completion of the work can be calculated by ${\mathrm{T}}{=}\frac{\left(\mathrm{a}+\mathrm{x}\right)\mathrm{b}}{\mathrm{a}+\mathrm{b}}$ days
Here a= 10 days; b=15 days; x=5; T=?
Hence, total number of days to complete the work is 9 days.
Formula 6:
P and Q do a work in a and b days respectively. Both begin together, but after some days P leaves and the remaining work is completed by Q alone in x days. The time after which P left is, ${\mathrm{T}}{=}\frac{\left(b-\mathrm{x}\right)\mathrm{a}}{\mathrm{a}+\mathrm{b}}$ days.
Example: A and B do a piece of work in 40 days and 60 days respectively. Both begin together but after few days A has to leave. The remaining work is ompleted by B alone in 25 days. After how many days did A leave?
Solution: Given a=40 days, b=60 days, x=25 days. t=?
According to the formula, ${\mathrm{T}}{=}\frac{\left(b-\mathrm{x}\right)\mathrm{a}}{\mathrm{a}+\mathrm{b}}$ days.
Hence, A left after 14 days
Formula 7:
If A and B can complete the work in x days and A alone can do the same work in y days, the the time taken by B to complete the same work will be given by $\frac{xy}{y-x}$days.
Example: A can complete a piece of work in 12 days. A and B together can complete the same work in 8 days. In how many days, can B alone complete the same peice of work?
Solution: Given x=8 days, y=12 days
According to the formula, $\frac{\mathrm{xy}}{\mathrm{y}-\mathrm{x}}$ days
Hence, B alone can complete the work in 24 days.
Formula 8:
If A and B working together, can finish a piece of work in x days, B and C can finish in y days, C and A can finish in z days, then
• A, B and C working together, will finish the work in $\frac{2xyz}{xy+yz+zx}$days
• A alone can finish the work in $\frac{2xyz}{xy+yz-zx}$days
• B alone can finish the work in $\frac{2xyz}{yz+zx-xy}$days
Formula 9:
If A is k times more efficient than B and so able to finish the work in l days then,
• A and B working together, will finish the work in $\frac{kl}{{k}^{2}-1}$days
• A alone can finish the work in $\frac{l}{k-l}$days
• B alone can finish the work in $\frac{kl}{k-l}$days
Hope you have liked this post. | 2019-02-22 01:02:42 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 28, "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": 0.6630530953407288, "perplexity": 885.550161371777}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-09/segments/1550247511573.67/warc/CC-MAIN-20190221233437-20190222015437-00457.warc.gz"} |
https://www.doubtnut.com/ncert-solutions/class-9-maths-chapter-15-probability-exercise-solved-examples-1 | # NCERT Class 9 Probability Solved Examples Maths Solutions
## Class 9 Probability Solved Examples Maths NCERT Solutions
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### NCERT Class 9 | PROBABILITY | Solved Examples | Question No. 10
Fifty seeds were selected at random from each of 5 bags of seeds, and were kept under standardised conditions favourable to germination. After 20 days, the number of seeds which had germinated in each collection were counted and recorded as follo
### NCERT Class 9 | PROBABILITY | Solved Examples | Question No. 03
A die is thrown 1000 times with the frequencies for the outcomes 1, 2, 3, 4, 5 and 6 as given in the following table : O u t com e1 2 3 4 5 6F r e q u e c n y 179 150 157 149 175 190 Find the probability of getting each outcome.
### NCERT Class 9 | PROBABILITY | Solved Examples | Question No. 02
Two coins are tossed simultaneously 500 times, and we get Two heads : 105 times One head : 275 times No head : 120 times Find the probability of occurrence of each of these events.
### NCERT Class 9 | PROBABILITY | Solved Examples | Question No. 01
A coin is tossed 1000 times with the following frequencies: Head : 455, Tail : 545 Compute the probability for each event.
### NCERT Class 9 | PROBABILITY | Solved Examples | Question No. 07
The percentage of marks obtained by a student in the monthly unit tests are given below. Based on this data, find the probability that the student gets more | 2021-05-07 07:05:41 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 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": 0.5522800087928772, "perplexity": 1137.4828868711838}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-21/segments/1620243988775.25/warc/CC-MAIN-20210507060253-20210507090253-00098.warc.gz"} |
https://email.esm.psu.edu/pipermail/macosx-tex/2008-February/034153.html | # [OS X TeX] Floating table in landscape orientaton results in a half empty page
Ross Moore ross at ics.mq.edu.au
Mon Feb 18 23:51:43 EST 2008
Hello Olga,
On 18/02/2008, at 10:16 PM, Olga Lyashevskaya wrote:
> Dear all,
>
> I experience difficulties with floating table. This table has a
> landscape orientation and occupies the whole page. When I refer to
> it, it is placed on a new page immediately after the \ref and there
> are just two words on the current. What is strange is that this
> page is not filled with the text which appears after the table.
> ---------------------------
> \begin{landscape}
> \begin{table}[tbp]
> \centering
> \setlength{\abovecaptionskip}{0pt}
> \setlength{\belowcaptionskip}{5pt}
> ...
> \end{table}
> \end{landscape}
This is very weird-looking coding, and almost certainly
not the way {landscape} was intended to be used.
Is this using \usepackage{lscape} ?
If so, then the {landscape} environment is meant to output
a complete page rotated.
You can get rotated material of arbitrary size, including
floats, with coding such as:
\begin{document}
First some words before,
\begin{table}[h]
\rotatebox{90}{%
\begin{minipage}{.7\textheight}
\centering
\setlength{\abovecaptionskip}{10pt}
\setlength{\belowcaptionskip}{5pt}
\includegraphics[scale=.15]{/Users/rossmoor/Pictures/more-snakes/
PICT7540.JPG}
\caption{little person'' a cute little dog; well, not always.}
\label{littledog}
\end{minipage}}
\end{table}
then some after.
Then some more words after which refer to Table~\ref{littledog}.
\end{document}
The result is in the attached image.
The {minipage} environment is only needed to ensure that
the caption is rotated along with the contents of the {table}.
Note that the width specified in
\begin{minipage}{.7\textheight}
becomes (roughly) the height of the rotated {table}.
(This is deliberately larger that the image used here,
resulting in the large margins of white space above
and below it.)
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If you need the rotated part to be on a separate page,
try the effect of 'p' instead of 'h', as follows:
\begin{table}[p]
Then all the text stays together as expected on the page
preceding where the {table} contents end up.
Alternatively use \begin{table}[htb] and/or increase the
size in: \begin{minipage}{.....} until it floats to
the next page. The text still stays together.
>
> To summarize, now I have one page which contains just 2 words and a
> table on the next page followed by some text.
>
> Any suggestions?
>
> Regards,
> Olga
>
> MacOS X (version 10.4.11)
> TexShop (vesrion 2.14)
Hope this helps,
Ross
------------------------------------------------------------------------
Ross Moore ross at maths.mq.edu.au
Mathematics Department office: E7A-419
Macquarie University tel: +61 +2 9850 8955
Sydney, Australia 2109 fax: +61 +2 9850 8114
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https://liorpachter.wordpress.com/2020/11/ | You are currently browsing the monthly archive for November 2020.
Steven Miller is a math professor at Williams College who specializes in number theory and theoretical probability theory. A few days ago he published a “declaration” in which he performs an “analysis” of phone bank data of registered Republicans in Pennsylvania. The data was provided to him by Matt Braynyard, who led Trump’s data team during the 2016. Miller frames his “analysis” as an attempt to “estimate the number of fraudulent ballots in Pennsylvania”, and his analysis of the data leads him to conclude that
“almost surely…the number of ballots requested by someone other than the registered Republican is between 37,001 and 58,914, and almost surely the number of ballots requested by registered Republicans and returned but not counted is in the range from 38,910 to 56,483.”
A review of Miller’s “analysis” leads me to conclude that his estimates are fundamentally flawed and that the data as presented provide no evidence of voter fraud.
This conclusion is easy to arrive at. The declaration claims (without a reference) that there were 165,412 mail-in ballots requested by registered Republicans in PA, but that “had not arrived to be counted” as of November 16th, 2020. The data Miller analyzed was based on an attempt to call some of these registered Republicans by phone to assess what happened to their ballots. The number of phone calls made, according to the declaration, is 23,184 = 17,000 + 3,500 + 2,684. The number 17,000 consists of phone calls that did not provide information either because an answering machine picked up instead of a person, or a person picked up and summarily hung up. 3,500 numbers were characterized as “bad numbers / language barrier”, and 2,684 individuals answered the phone. Curiously, Miller writes that “Almost 20,000 people were called”, when in fact 23,184 > 20,000.
In any case, clearly many of the phone numbers dialed were simply wrong numbers, as evident by the number of “bad” calls: 3,500. It’s easy to imagine how this can happen: confusion because some individuals share a name, phone numbers have changed, people move, the phone call bank makes an error when dialing etc. Let $b$ be the fraction of phone numbers out of the 23,184 that were “bad”, i.e. incorrect. We can estimate $b$ by noting that we have some information about it: we know that the 3,500 “bad numbers” were bad (by definition). Additionally, it is reported in the declaration that 556 people literally said that they did not request a ballot, and there is no reason not to take them at their word. We don’t know what fraction of the 17,000 individuals called and did not pick up or hung up were wrong numbers, but we do know that the fraction out of the total must equal the fraction out of the 17,000 + those we know for sure were bad numbers, i.e.
$23184 \cdot b = 17,000 \cdot b + 556 + 3500$.
Solving for $b$ we find that $b \approx \frac{2}{3}$. I’m surprised the number is so low. One would expect that individuals who requested ballots, but then didn’t send them in, would be enriched for people who have recently moved or are in the process of moving, or have other issues making it difficult to reach them or impossible to reach them at all.
The fraction of bad calls derived translates to about 1,700 bad numbers out of the 2,684 people that were reached. This easily explains not only the 556 individuals who said they did not request a ballot, but also the 463 individuals who said that they mailed back their ballots. In the case of the latter there is no irregularity; the number of bad calls suggests that all those individuals were reached in error and their ballots were legitimately counted so they weren’t part of the 165,412. It also explains the 544 individuals who said they voted in person.
That’s it. The data don’t point to any fraud or irregularity, just a poorly design poll with poor response rates and lots of erroneous information due to bad phone numbers. There is nothing to explain. Miller, on the other hand, has some things to explain.
First, I note that his declaration begins with a signed page asserting various facts about Steven Miller and the analysis he performed. Notably absent from the page, or anywhere else in the document, is a disclosure of funding source for the work and of conflicts of interest. On his work webpage, Miller specifically states that one should always acknowledge funding support.
Second, if Miller really wanted to understand the reason why some ballots were requested for mail-in, but had not yet arrived to be counted, he would also obtain data from Democrats. That would provide a control on various aspects of the analysis, and help to establish whether irregularities, if they were to be detected, were of a partisan nature. Why did Miller not include an analysis of such data?
Third, one might wonder why Steven Miller chose to publish this “declaration”. Surely a professor who has taught probability and statistics for 15 years (as Miller claims he has) must understand that his own “analysis” is fundamentally flawed, right? Then again, I’ve previously found that excellent pure mathematicians are prone to falling into a data analysis trap, i.e. a situation where their lack of experience analyzing real-world datasets leads them to believe naïve analysis that is deeply flawed. To better understand whether this might be the case with Miller, I examined his publication record, which he has shared publicly via Google Scholar, to see whether he has worked with data. The first thing I noticed was that he has published more than 700 articles (!) and has an h-index of 47 for a total of 8,634 citations… an incredible record for any professor, and especially for a mathematician. A Google search for his name displays this impressive number of citations:
As it turns out, his impressive publication record is a mirage. When I took a closer look and found that many of the papers he lists on his Google Scholar page are not his, but rather articles published by other authors with the name S Miller. “His” most cited article was published in 1955, a year that transpired well before he was born. Miller’s own most cited paper is a short unpublished tutorial on least squares (I was curious and reviewed it as well only to find some inaccuracies but hey, I don’t work for this guy).
I will note that in creating his Google Scholar page, Miller did not just enter his name and email address (required). He went to the effort of customizing the page, including the addition of keywords and a link to his homepage, and in doing so followed his own general advice to curate one’s CV (strangely, he also dispenses advice on job interviews, including about shaving- I guess only women interview for jobs?). But I digress: the question is, why is his Google Scholar page display massively inflated publication statistics based on papers that are not his? I’ve seen this before, and in one case where I had hard evidence that it was done deliberately to mislead I reported it as fraud. Regardless of Miller’s motivations, by looking at his actual publications I confirmed what I suspected, namely that he has hardly any experience analyzing real world data. I’m willing to chalk up his embarrassing “declaration” to statistics illiteracy and naïveté.
In summary, Steven Miller’s declaration provides no evidence whatsoever of voter fraud in Pennsylvania.
Lior Pachter
Division of Biology and Biological Engineering &
Department of Computing and Mathematical Sciences California Institute of Technology
Abstract
A recently published pilot study on the efficacy of 25-hydroxyvitamin D3 (calcifediol) in reducing ICU admission of hospitalized COVID-19 patients, concluded that the treatment “seems able to reduce the severity of disease, but larger trials with groups properly matched will be required go show a definitive answer”. In a follow-up paper, Jungreis and Kellis re-examine this so-called “Córdoba study” and argue that the authors of the study have undersold their results. Based on a reanalysis of the data in a manner they describe as “rigorous” and using “well established statistical techniques”, they urge the medical community to “consider testing the vitamin D levels of all hospitalized COVID-19 patients, and taking remedial action for those who are deficient.” Their recommendation is based on two claims: in an examination of unevenness in the distribution of one of the comorbidities between cases and controls, they conclude that there is “no evidence of incorrect randomization”, and they present a “mathematical theorem” to make the case that the effect size in the Córdoba study is significant to the extent that “they can be confident that if assignment to the treatment group had no effect, we would not have observed these results simply due to chance.”
Unfortunately, the “mathematical analysis” of Jungreis and Kellis is deeply flawed, and their “theorem” is vacuous. Their analysis cannot be used to conclude that the Córdoba study shows that calcifediol significantly reduces ICU admission of hospitalized COVID- 19 patients. Moreover, the Córdoba study is fundamentally flawed, and therefore there is nothing to learn from it.
The Córdoba study
The Córdoba study, described by the authors as a pilot, was ostensibly a randomized controlled trial, designed to determine the efficacy of 25-hydroxyvitamin D3 in reducing ICU admission of hospitalized COVID-19 patients. The study consisted of 76 patients hospitalized for COVID-19 symptoms, with 50 of the patients treated with calcifediol, and 26 not receiving treatment. Patients were administered “standard care”, which according to the authors consisted of “a combination of hydroxychloroquine, azithromycin, and for patients with pneumonia and NEWS score 5, a broad spectrum antibiotic”. Crucially, admission to the ICU was determined by a “Selection Committee” consisting of intensivists, pulmonologists, internists, and members of an ethics committee. The Selection Committee based ICU admission decisions on the evaluation of several criteria, including presence of comorbidities, and the level of dependence of patients according to their needs and clinical criteria.
The result of the Córdoba trial was that only 1/50 of the treated patients was admitted to the ICU, whereas 13/26 of the untreated patients were admitted (p-value = 7.7 ∗ 10−7 by Fisher’s exact test). This is a minuscule p-value but it is meaningless. Since there is no record of the Selection Committee deliberations, it impossible to know whether the ICU admission of the 13 untreated patients was due to their previous high blood pressure comorbidity. Perhaps the 11 treated patients with the comorbidity were not admitted to the ICU because they were older, and the Selection Committee considered their previous higher blood pressure to be more “normal” (14/50 treatment patients were over the age of 60, versus only 5/26 of the untreated patients).
Figure 1: Table 2 from [1] showing the comorbidities of patients. It is reproduced by virtue of [1] being published open access under the CC-BY license.
The fact that admission to the ICU could be decided in part based on the presence of co-morbidities, and that there was a significant imbalance in one of the comorbidities, immediately renders the study results meaningless. There are several other problems with it that potentially confound the results: the study did not examine the Vitamin D levels of the treated patients, nor was the untreated group administered a placebo. Most importantly, the study numbers were tiny, with only 76 patients examined. Small studies are notoriously problematic, and are known to produce large effect sizes [9]. Furthermore, sloppiness in the study does not lead to confidence in the results. The authors state that the “rigorous protocol” for determining patient admission to the ICU is available as Supplementary Material, but there is no Supplementary Material distributed with the paper. There is also an embarrassing typo: Fisher’s exact test is referred to twice as “Fischer’s test”. To err once in describing this classical statistical test may be regarded as misfortune; to do it twice looks like carelessness.
A pointless statistics exercise
The Córdoba study has not received much attention, which is not surprising considering that by the authors’ own admission it was a pilot that at best only motivates a properly matched and powered randomized controlled trial. Indeed, the authors mention that such a trial (the COVIDIOL trial), with data being collected from 15 hospitals in Spain, is underway. Nevertheless, Jungreis and Kellis [3], apparently mesmerized by the 7.7 ∗ 10−7 p-value for ICU admission upon treatment, felt the need to “rescue” the study with what amounts to faux statistical gravitas. They argue for immediate consideration of testing Vitamin D levels of hospitalized patients, so that “deficient” patients can be administered some form of Vitamin D “to the extent it can be done safely”. Their message has been noticed; only a few days after [3] appeared the authors’ tweet to promote it has been retweeted more than 50 times [8].
Jungreis and Kellis claim that the p-value for the effect of calcifediol on patients is so significant, that in and of itself it merits belief that administration of calcifediol does, in fact, prevent admission of patients to ICUs. To make their case, Jungreis and Kellis begin by acknowledging that imbalance between the treated and untreated groups in the previous high blood pressure comorbidity may be a problem, but claim that there is “no evidence of incorrect randomization.” Their argument is as follows: they note that while the p-value for the imbalance in the previous high blood pressure comorbidity is 0.0023, it should be adjusted for the fact that there are 15 distinct comorbidities, and that just by chance, when computing so many p-values, one might be small. First, an examination of Table 2 in [1] (Figure 1) shows that there were only 14 comorbidities assessed, as none of the patients had previous chronic kidney disease. Thus, the number 15 is incorrect. Second, Jungreis and Kellis argue that a Bonferroni correction should be applied, and that this correction should be based on 30 tests (=15 × 2). The reason for the factor of 2 is that they claim that when testing for imbalance, one should test for imbalance in both directions. By applying the Bonferroni correction to the p-values, they derive a “corrected” p-value for previous high blood pressure being imbalanced between groups of 0.069. They are wrong on several counts in deriving this number. To illustrate the problems we work through the calculation step-by-step:
The question we want to answer is as follows: given that there are multiple comorbidities, is there is a significant imbalance in at least one comorbidity. There are several ways to test for this, with the simplest being Šidák’s correction [10] given by
$q \quad = \quad 1-(1-m)^n,$
where m is the minimum p-value among the comorbidities, and n is the number of tests. Plugging in m = 0.0023 (the smallest p-value in Table 2 of [1]) and n = 14 (the number of comorbidities) one gets 0.032 (note that the Bonferroni correction used by Jungreis And Kellis is the Taylor approximation to the Šidák correction when m is small). The Šidák correction is based on an assumption that the tests are independent. However, that is certainly not the case in the Córdoba study. For example, having at least one prognostic factor is one of the comorbidities tabulated. In other words, the p-value obtained is conservative. The calculation above uses n = 14, but Jungreis and Kellis reason that the number of tests is 30 = 15 × 2, to take into account an imbalance in either the treated or untreated direction. Here they are assuming two things: that two-sided tests for each comorbidity will produce double the p-value of a one-sided test, and that two sided tests are the “correct” tests to perform. They are wrong on both counts. First, the two-sided Fisher exact test does not, in general produce a p-value that is double the 1-sided test. The study result is a good example: 1/49 treated patients admitted to the ICU vs. 13/26 untreated patients produces a p-value of 7.7 ∗ 10−7 for both the 1-sided and 2-sided tests. Jungreis and Kellis do not seem to know this can happen, nor understand why; they go to great lengths to explain the importance of conducting a 1-sided test for the study result. Second, there is a strong case to be made that a 1-sided test is the correct test to perform for the comorbidities. The concern is not whether there was an imbalance of any sort, but whether the imbalance would skew results by virtue of the study including too many untreated individuals with comorbidities. In any case, if one were to give Jungreis and Kellis the benefit of the doubt, and perform a two sided test, the corrected p-value for the previous high blood pressure comorbidity is 0.06 and not 0.069.
The most serious mistake that Jungreis and Kellis make, however, is in claiming that one can accept the null hypothesis of a hypothesis test when the p-value is greater than 0.05. The p-value they obtain is 0.069 which, even if it is taken at face value, is not grounds for claiming, as Jungreis and Kellis do, that “this is not significant evidence that the assignment was not random” and reason to conclude that there is “no evidence of incorrect randomization”. That is not how p-values work. A p-value less than 0.05 allows one to reject the null hypothesis (assuming 0.05 is the threshold chosen), but a p-value above the chosen threshold is not grounds for accepting the null. Moreover, the corrected p-value is 0.032 which is certainly grounds for rejecting the null hypothesis that the randomization was random.
Correction of the incorrect Jungreis and Kellis statistics may be a productive exercise in introductory undergraduate statistics for some, but it is pointless insofar as assessing the Córdoba study. While the extreme imbalance in the previous high blood pressure comorbidity is problematic because patients with the comorbidity may be more likely to get sick and require ICU admission, the study was so flawed that the exact p-value for the imbalance is a moot point. Given that the presence of comorbidities, not just their effect on patients, was a factor in determining which patients were admitted to the ICU, the extreme imbalance in the previous high blood pressure comorbidity renders the result of the study meaningless ex facie.
A definition is not a theorem is not proof of efficacy
In an effort to fend off criticism that the comorbidities of patients were improperly balanced in the study, Jungreis and Kellis go further and present a “theorem” they claim shows that there was a minuscule chance that an uneven distribution of comorbidities could render the study results not significant. The “theorem” is stated twice in their paper, and I’ve copied both theorem statements verbatim from their paper:
Theorem 1 In a randomized study, let p be the p-value of the study results, and let q be the probability that the randomization assigns patients to the control group in such a way that the values of Pprognostic(Patient) are sufficiently unevenly distributed between the treatment and control groups that the result of the study would no longer be statistically significant at the 95% level after p controlling for the prognostic risk factors. Then $q < \frac{p}{0.05}$.
According to Jungreis and Kellis, Pprognostic(Patient) is the following: “There can be any number of prognostic risk factors, but if we knew what all of them were, and their effect sizes, and the interactions among them, we could combine their effects into a single number for each patient, which is the probability, based on all known and yet-to-be discovered risk factors at the time of hospital admission, that the patient will require ICU care if not given the calcifediol treatment. Call this (unknown) probability Pprognostic(Patient).”
The theorem is restated in the Methods section of Jungreis and Kellis paper as follows:
Theorem 2 In a randomized controlled study, let p be the p-value of the study outcome, and let q be the probability that the randomization distributes all prognostic risk factors combined sufficiently unevenly between the treatment and control groups that when controlling for these prognostic risk p factors the outcome would no longer be statistically significant at the 95% level. Then $q < \frac{p}{0.05}$.
While it is difficult to decipher the language the “theorem” is written in, let alone its meaning (note Theorem 1 and Theorem 2 are supposedly the same theorem), I was able to glean something about its content from reading the “proof”. The mathematical content of whatever the theorem is supposed to mean, is the definition of conditional probability, namely that if A and B are events with $P(B) > 0$, then
$P(A|B) \quad := \quad \frac{P(A \cap B)}{P(B)}$.
To be fair to Jungreis and Kellis, the “theorem” includes the observation that
$P(A \cap B) \leq P(A) \quad \Rightarrow \quad P(A|B) \leq \frac{P(A)}{P(B)}.$
This is not, by any stretch of the imagination, a “theorem”; it is literally the definition of conditional probability followed by an elementary inequality. The most generous interpretation of what Jungreis and Kellis were trying to do with this “theorem”, is that they were showing that the p-value for the study is so small, that it is small even after being multiplied by 20. There are less generous interpretations.
Does Vitamin D intake reduce ICU admission?
There has been a lot of interest in Vitamin D and its effects on human health over the past decade [2], and much speculation about its relevance for COVID-19 susceptibility and disease severity. One interesting result on disease susceptibility was published recently: in a study of 489 patients, it was found that the relative risk of testing positive for COVID-19 was 1.77 times greater for patients with likely deficient vitamin D status compared with patients with likely sufficient vitamin D status [7]. However, definitive results on Vitamin D and its relationship to COVID- 19 will have to await larger trials. One such trial, a large randomized clinical trial with 2,700 individuals sponsored by Brigham and Women’s Hospital, is currently underway [4]. While this study might shed some light on Vitamin D and COVID-19, it is prudent to keep in mind that the outcome is not certain. Vitamin D levels are confounded with many socioeconomic factors, making the identification of causal links difficult. In the meantime, it has been suggested that it makes sense for individuals to maintain reference nutrient intakes of Vitamin D [6]. Such a public health recommendation is not controversial.
As for Vitamin D administration to hospitalized COVID-19 patients reducing ICU admission, the best one can say about the Córdoba study is that nothing can be learned from it. Unfortunately, the poor study design, small sample size, availability of only summary statistics for the comorbidities, and imbalanced comorbidities among treated and untreated patients render the data useless. While it may be true that calcifediol administration to hospital patients reduces subsequent ICU admission, it may also not be true. Thus, the follow-up by Jungreis and Kellis is pointless at best. At worst, it is irresponsible propaganda, advocating for potentially dangerous treatment on the basis of shoddy arguments masked as “rigorous and well established statistical techniques”. It is surprising to see Jungreis and Kellis argue that it may be unethical to conduct a placebo randomized controlled trial, which is one of the most powerful tools in the development of safe and effective medical treatments. They write “the ethics of giving a placebo rather than treatment to a vitamin D deficient patient with this potentially fatal disease would need to be evaluated.” The evidence for such a policy is currently non-existent. On the other hand, there are plenty of known risks associated with excess Vitamin D [5].
References
1. Marta Entrenas Castillo, Luis Manuel Entrenas Costa, José Manuel Vaquero Barrios, Juan Francisco Alcalá Díaz, José López Miranda, Roger Bouillon, and José Manuel Quesada Gomez. Effect of calcifediol treatment and best available therapy versus best available therapy on intensive care unit admission and mortality among patients hospitalized for COVID-19: A pilot randomized clinical study. The Journal of steroid biochemistry and molecular biology, 203:105751, 2020.
2. Michael F Holick. Vitamin D deficiency. New England Journal of Medicine, 357(3):266–281, 2007.
3. Irwin Jungreis and Manolis Kellis. Mathematical analysis of Córdoba calcifediol trial suggests strong role for Vitamin D in reducing ICU admissions of hospitalized COVID-19 patients. medRxiv, 2020.
4. JoAnn E Manson. https://clinicaltrials.gov/ct2/show/nct04536298.
5. Ewa Marcinowska-Suchowierska, Małgorzata Kupisz-Urbańska, Jacek Łukaszkiewicz, Paweł Płudowski, and Glenville Jones. Vitamin D toxicity–a clinical perspective. Frontiers in endocrinology, 9:550, 2018
6. Adrian R Martineau and Nita G Forouhi. Vitamin D for COVID-19: a case to answer? The Lancet Diabetes & Endocrinology, 8(9):735–736, 2020.
7. David O Meltzer, Thomas J Best, Hui Zhang, Tamara Vokes, Vineet Arora, and Julian Solway. Association of vitamin D status and other clinical characteristics with COVID-19 test results. JAMA network open, 3(9):e2019722–e2019722, 2020.
8. Vivien Shotwell. https://tweetstamp.org/1327281999137091586.
9. Robert Slavin and Dewi Smith. The relationship between sample sizes and effect sizes in systematic reviews in education. Educational evaluation and policy analysis, 31(4):500–506, 2009.
10. Lynn Yi, Harold Pimentel, Nicolas L Bray, and Lior Pachter. Gene-level differential analysis at transcript-level resolution. Genome biology, 19(1):53, 2018.
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### Blog Stats
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http://www.linzehui.me/2018/11/04/%E8%AE%BA%E6%96%87/%E6%AF%8F%E5%91%A8%E8%AE%BA%E6%96%874/ | ## 2️⃣[Self-Attention with Relative Position Representations]
$a_{ij}^K$的具体形式：
## 3️⃣[WEIGHTED TRANSFORMER NETWORK FOR MACHINE TRANSLATION]
κ can be interpreted as a learned concatenation weight and α as the learned addition weight
①pre-processing
②模型
③post processing | 2019-04-25 02:28:02 | {"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": 0.8088856339454651, "perplexity": 13789.79157798151}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-18/segments/1555578678807.71/warc/CC-MAIN-20190425014322-20190425040322-00250.warc.gz"} |
https://ftp.aimsciences.org/article/doi/10.3934/dcds.2006.15.921 | Article Contents
Article Contents
# Asymptotic selection of viscosity equilibria of semilinear evolution equations by the introduction of a slowly vanishing term
• The behavior at infinity is investigated of global solutions to some nonautonomous semilinear evolution equations with conservative and convex nonlinearities. It is proved that the trajectories converge to viscosity stationary solutions as time goes to infinity, that is, they evolve towards stationary solutions that are minimal with respect to a generalized viscosity criterion. Hierarchical viscosity selections and applications to specific nonlinear PDE are given.
Mathematics Subject Classification: Primary: 35B40, 34G20; Secondary: 35K90, 37N40.
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https://electronics.stackexchange.com/questions/350717/what-is-the-color-code-for-can-bus | # What is the color code for CAN Bus?
I would like to know if there is any color code specified for CAN Bus communication wiring, maybe in ISO-11898-1? (I have no free access to the file)
I looked for any official documents without any luck.
When I asked to my colleagues about it, they all agree on the colors (Yellow and Green) but they do not agree if it is:
• Yellow for CAN-H
• Green for CAN-L
• Green for CAN-H
• Yellow for CAN-L
The ISO 11898 standards do not mention practical things such as cables and connectors.
The second-most authoritative source has therefore become the CANopen standards, where DS303-1 specifies things such as standard connectors, pin-outs, cable lengths, baud rates etc. Unfortunately, this document does not mention color-coding either.
Yellow and green seem to be commonly used, though I have seen yellow mean either CAN High or Low. However, just because these colors are commonly used, it does not make them more correct.
I'm guessing that the yellow and green comes from the universally standardized color-coding (same as we use on through-hole resistors). That is: 1=brown, ... 4=yellow, 5=green. For the "mini style" and "micro style" (round M12-like) connectors standardized by DS303-1, CAN High happens to be on pin 4 and CAN Low on pin 5. From DS303-1 7.2:
This enumeration is however not at all consistent with other common, standardized connectors such as d-sub, RJ45 and terminal socket.
CAN does not have a formalized physical-layer specification for conductor colors, or things like connector type or pin-out. There are common practices (like using a 9-pin D-sub connector) but no official standard.
Vehicles these days also tend to have multiple CAN buses, so colors will, of course, vary to keep the different buses straight. I have seen some buses adopt a solid color for CAN-L and a different color striped with the corresponding CAN-L color for CAN-H, to give a visual impression of belonging together.
Overall, as long as you keep track of your conductor colors and pinouts, you can use any color scheme that suits your preference without violating any standards.
I found a link for J1939 cables (not sure if this is generic): see link
Excerpt (see last two lines):
CBL-CAN-01: CAN Cable for DB9 Male Connector
This is a 4-wire color coded cable. One end is DB9 female connector,
it is designed to mate with Au J1939 products at bus side, such as:
J1939 Message Center System, J1939 Data Center System,
J1939 Simulators, etc.
The other side of the cable are 4 pig-tail wires.
Red wire: Power supply, +12V
Black wire: Ground
White (or Yellow) wire: CAN High
Green wire: CAN Low
Red is Power - 12 V
Black is Ground - 0 V
Yellow is CAN High - 2.5 V
Green is CAN Low - 2.5 V
• Thank you Michel, that agrees with my opinion that Yellow for High and Green for Low has more sense. But it is not what I am looking for. I need some techincal documentation to prove it. – ErnstOlch Jan 18 '18 at 12:52
• CAN does not have a formalized mechanical standard for the physical layer - as such, manufacturers will generally use their own standard. I have seen the yellow-green scheme and also a blue and white scheme used in various vehicle buses. So long as you keep track of which colors are which, you can use any colors you like. – Chris M. Jan 18 '18 at 13:26
• @ChrisM Thanks for this ... you should make it an answer, since it seems what ErnstOch is looking for (or at least that it is not a rule.). – Michel Keijzers Jan 18 '18 at 13:59
• Mercedes Benz and MAN buses/trucks have 125kBaud, 250kBaud and 500kBaud buses inside. They run yellow for CAN_H and "blue" for CAN_LOW. It seems that "yellow" is quite common for CAN_HI. – Rohat Kılıç Jan 18 '18 at 14:30
• If someone could cite the actual J1939 document though, that would be a somewhat canonical source. – Lundin Jan 19 '18 at 12:48
In every implementation I've seen, if you can imagine that one color represents "Sky" and the other represents "Earth", those happen to be Hi and Lo, respectively.
• Yellow sun, green grass.
• White clouds, green grass.
• White sky, blue ocean.
Et cetera. These mnemonics seem to fit so well I have to imagine they're intentional.
CAN_H = yellow, CAN_L = green per SAE J1939-11.
This would only apply to J1939-compliant CAN installations. | 2019-10-17 05:43:36 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.43587175011634827, "perplexity": 3887.754512497317}, "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/1570986672723.50/warc/CC-MAIN-20191017045957-20191017073457-00165.warc.gz"} |
http://physics.stackexchange.com/questions/80233/why-whirlpool-causes-force-towards-its-center/80236 | # Why whirlpool causes force towards its center?
Higher velocity implies low pressure. So in whirlpool the velocity of water have to decrease with radius in order to have force towards its center.
My question is how is explained whirlpool formation in simple words.
-
A fluid motion in a vortex creates a dynamic pressure that is lowest in the center increasing radially ($P \propto r^2$). The gradient of this pressure that forces the fluid to rotate around the axis.
This is usually represented by a vector called vorticity, and defined by $\omega = \nabla \times v$. | 2015-10-13 18:37:14 | {"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": 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": 0.8286307454109192, "perplexity": 1003.5606443660439}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-40/segments/1443738009849.87/warc/CC-MAIN-20151001222009-00122-ip-10-137-6-227.ec2.internal.warc.gz"} |
https://brilliant.org/discussions/thread/find-out-right-the-area/ | ×
# Find out right the area???????
ABC is a isosceles triangle whose AB=AC=10cm. the angle BAC is 30 degrees. Find out the area of the triangle. ????
Note by Rahul Das
3 years, 8 months ago
Sort by:
Use the sine formula for the area of the triangle. $$[ABC]=\frac{1}{2}(AB)(AC) \sin A$$. · 3 years, 8 months ago
by using this formula: Area=1/2(AB)(AC)sinA. area=1/2(10)(10)1/2=25 cm^2 · 3 years, 8 months ago
area=1/21010*sin30=25 cm^2 · 3 years, 8 months ago
Using cosine law you will determine BC.After that using hero's formula you will determine the area of triangle · 3 years, 8 months ago
25cm^2 as the formula is (1/2)(10)(10)(sin 30)=25 (sin 30 =1/2) · 3 years, 8 months ago
Or, if you don't want trigonometry, draw $$AD$$ perpendicular to $$BC$$ at $$D$$. Since angles $$B$$ and $$C$$ both measure 75 degrees, you have no choice, but to use the ratio $$\frac{AD}{AB}=\frac{AD}{AC}=\frac{\sqrt{2}+\sqrt{6}}{4}$$ which still comes from trigonometry. · 3 years, 8 months ago | 2017-05-25 14:24:18 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9878314137458801, "perplexity": 2368.4990417104646}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-22/segments/1495463608084.63/warc/CC-MAIN-20170525140724-20170525160724-00015.warc.gz"} |
https://mathematica.stackexchange.com/questions/199625/how-to-sort-the-do-output | # How to sort the “do” output?
So I have function (FindCoefficients) that iterates through some values to find the maximum value for a function. It is much quicker to hold the non-linear values constant and do a find maximum for the linear coeffiecients then to do a non linear optimization for a complicated function.
This is the main line that I'm struggling with:
Do[Print[a , b, FindCoefficients[a, b]], {a, 1, 10, 1}, {b, 1, 10, 1}]
This gives me the output of
11{value of maximum, other info}
12{value of maximum, other info}
13{value of maximum, other info}
...
21{value of maximum, other info}
...
1010{value of maximum, other info}
Is there any way to sort my output from greatest to smallest, so I can easily pick out the maximum?
SortBy[ Table[{a , b, FindCoefficients[a, b]} , {a, 1, 10, 1}, {b, 1, 10, 1}],{3}] | 2021-05-14 05:13:37 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.6200293898582458, "perplexity": 2497.5522029122367}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-21/segments/1620243991737.39/warc/CC-MAIN-20210514025740-20210514055740-00051.warc.gz"} |
https://proxieslive.com/tag/putting/ | ## How to apply OOP to real world examples without putting all logic in Manager classes?
I’m lately trying to implement a specific problem using an object-oriented approach. I get the main principles and its advantages, but I fail to apply it to a real world problem. Most examples one could find consist of Cats or Dogs being implementations of Animal. These however don’t give me enough understanding how to model below problem regarding another frequent example: a school administration system.
Imagine a school having Students, Courses, Professors, and Notes. My implementation would be something like this:
class Person { string name; int age; Person(string name, int age) { this.name = name; this.age = age; } } class Student extends Person { double gpa; Student(string name, int age) { super(name, age); } } class Professor extends Person { string roomNumber; Professor(string name, int age, string roomNumber) { super(name, age); this.roomNumber = roomNumber; } } class Course { string name; Professor professor; Students[] student; Course(string name, Professor professor) { this.name = name; this.professor = professor; this.students = new Student[]; } void enrolStudent(Student student) { students.add(student); } } class Note { Course course; Student student; double value; Note(Course course, Student student, double value) { this.course = course; this.student = student; this.value = value; } }
Now the Student has a bunch of Notes and we want to calculate its GPA. This could be either straightforward averaging its Notes‘ values or more complex logic using weights and/or ignoring optional courses.
Now my question is: where do we put this logic? Ideally I would have a function double calculateGpa() on Student so you could call student.calculateGpa(), but having this logic on Student would break the SRP in my view. It also does not belong to any other class listed here. A class called GpaCalculator or NotesManager would be another guess but that seems to me too much like moving all the logic away from the domain and into classes that do not represent a real object but just actions (see also this answer).
If that would be the way to go here, why wouldn’t I then just write a pure, static, stateless function in a class called NotesHelper? Creating a manager class to just have one function double calculate(), and using its instance instead of a static function feels to me like making it look like OOP while it isn’t really. I feel like there should be a better approach, probably one I didn’t think of, or maybe I am wrong here. Could you guys give me some pointers?
Thanks!
## How to hide the banner “This site is read only at the farm administrator’s request” after putting SP site in read only mode?
Today, as per client requirement,I made one of the site collections in our production environment on SP 2013 in read only mode. After doing so, we are getting
banner at the top of the page.
Is there any PowerShell command to hide this banner from the page?
## Putting elements in allowed bags
Let’s say I have a list of “Items”
I also have a list of “Bags”. Each bag is a set of “Items” which gives what item can be placed in that bag. But only one item can go in each bag.
I want to place all items in a bag. It’s okay if there are empty bags left over.
Does this sort of problem have a name and a solution other than a naive depth first search (a link in the direction of such an approach will be fine)?
## Does putting salt first make it easier for attacker to bruteforce the hash?
Many recommendations for storing passwords recommend hash(salt + password) rather than hash(password + salt).
Doesn’t putting the salt first make it much faster for the attacker to bruteforce the password, because they can precompute the state of the hashing function with the bytes of the salt, and then each time of their billions and trillions attempts they only need to finish calculating the hash using the bytes of the password.
In other words, each bruteforce iteration needs to calculate only the hash of the password intermediateHashState(password) instead of the whole hash(salt + password).
And if the salt was placed after the password, the attacker wouldn’t have this shortcut.
Does this advantage exist and is it significant?
## Prevent workflow change from putting file into Draft mode
I have a document library which has some documents with metadata fields attached. Some of the metadata fields are edited by a workflow on create / edit. But we are finding when a file is Published from draft to pending, the workflow fires and makes the required change to a metadata tag, and this causes that file to go back into draft mode.
Is it possible to make a change in a workflow and not changing the approval status?
## [ Politics ] Open Question : Do you support putting Harriet Tubman on the $20 bill? While I do agree Harriet was a very important historical figure, I would have thought She would have been on a commemorative silver dollar or half dollar coin I am kind of puzzled by the 20. ## Query Function Not Putting Consolidated Data in the Right Place So I don’t know how to explain this completely, but basically, I’m using the query function to consolidate data from multiple sheets into one Master Sheet. The function itself seems to be working fine and all the data from the other sheets ends up on the Master Sheet. HOWEVER… Instead of the data being added vertically so that the number of rows increases, the data is being added horizontally. A picture is attached to explain this better. Basically, it’s “copying and pasting” the data from the other sheets into the next COLUMN over instead of the next row. ## putting output of command into a string I want to store the output of a bash command to a string in a bash script. The part that matters is as follows: #!/bin/bash player_status="$ (playerctl -l)"
The output of the command when run on terminal (not with the bash script) is “No players were found”. When I run the bash script (note that there is not an echo) it outputs “No players were found” to the terminal. I want it instead to not put it in the terminal but instead the variable.
## Probability of colour with putting back twice.
I’m having a bit of trouble with following probability-related question:
A container contains 5 red and 10 black balls. Take a ball out of the container at random and note its color. After drawing each ball, the ball gets put back and one extra ball of the same color gets added.
1) Given that the first n drawn balls are all black, calculate the probability (say $$a_{n}$$) that the $$(n+1)$$st ball will also be black. What is $$\lim_{x\to\infty} a_{n}$$ equal to?
2) Given that the second until the $$(n+1)$$st ball (inclusive) are all black, calculate the probability that (say $$b_{n}$$) that the first drawn ball was also black. Also calculate $$\lim_{x\to\infty} b_{n}$$.
I succeeded at answering the first subquestion. As the result I get that $$a_{n} = (10+n)/(15+n)$$ and thus, $$\lim_{x\to\infty} a_{n} = 1$$
However, I have much more trouble answering the second question, and would like your help.
Thanks! | 2019-07-19 17:25:34 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 8, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.282064288854599, "perplexity": 1772.55972395973}, "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/1563195526324.57/warc/CC-MAIN-20190719161034-20190719183034-00523.warc.gz"} |
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Regulators, Question 25 Questions as per the latest prescribed syllabus are multiple-choice Type and of the is... | 2021-10-28 21:05:01 | {"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": 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": 0.37009871006011963, "perplexity": 10321.27909210009}, "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/1634323588526.57/warc/CC-MAIN-20211028193601-20211028223601-00496.warc.gz"} |
https://miktex.org/packages/paresse | Version:
4.1
The package defines macros using § to type greek letters. so that the user may (for example) type §a to get the effect of $\alpha$. | 2019-06-18 11:00:27 | {"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": 0.9865191578865051, "perplexity": 2994.724058182537}, "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-2019-26/segments/1560627998716.67/warc/CC-MAIN-20190618103358-20190618125358-00240.warc.gz"} |
https://math.stackexchange.com/questions/1414059/topological-idea-of-orientability-of-manifold | # Topological idea of orientability of manifold
While reading Poincare Duality a new idea of orientability of manifold came in my mind.I dont know wheather this idea is new or not, or even true or false.
My idea is following... A $n$-dim manifold $X$ is called non-orientable if there exists an embedding $i: M\times \mathbb{D^{n-2}} \to X$, where $M$ is a mobius strip. Otherwise $X$ is called an orientable manifold.
For an example if $n=2$, then by using of classification theorem of compact $2$-manifold we can see that my definition works for 2-dim compact manifold...But I got stuck while doing this generalization...Till now I've knowladge of algebraic topology only upto Poincare Duality of Hatcher's book...
I want to know whether my this general idea is true or not, and in case if it is true then how do I prove it?? I need some HINTS for that...and if it is a well known result then please give me some reference where I can study this properly.
• At least for a smooth manifold, this idea is equivalent to the usual notion of orientability. The idea of the proof uses normal bundles of embedded closed curves. – Jason DeVito Aug 29 '15 at 21:49
• @JasonDeVito: This should be true in the topological case. You just need a curve that 'demonstrates' non-orientability and has a disc bundle over it. For $n=4$ noncompact manifolds are smoothable; for $n \geq 6$ decent subcomplexes have regular neighborhoods, so you can take the regular neighborhood and by Mazur's obstruction theory you have that it's PL; in this PL manifold take the star of our curve; this is a disc bundle. Compact topological disc bundles are either the Mobius bundle or trivial. I dunno about $n=5$. – user98602 Aug 29 '15 at 22:21
• This is probably overkill but the only thing I know about topological manifold theory is the part that lets me turn it into PL or smooth manifold theory. – user98602 Aug 29 '15 at 22:22
• @Mike: You know more about topological manifolds than I do! – Jason DeVito Aug 30 '15 at 0:30
This is correct, and a fun idea. Here's why. I'll talk about smooth manifolds first, but it's true in full generality. All my manifolds have dimension $n \geq 3$.
First off, we need to classify $n$-dimensional vector bundles over the circle. There are precisely two: one is trivial, the other is non-orientable, with total space homeomorphic to the (open) Mobius band times the open disc $D^n$. This is because bundles over the circle are classified by $\pi_0 GL_n(\Bbb R) = \mathbb Z/2$; the nontrivial bundle is the Mobius line bundle plus a bunch of trivial line bundles.
Given a smooth manifold $M$, $M$ is orientable if and only if every embedded circle $S^1 \hookrightarrow M$ has trivial normal bundle. (If it's orientable, because the circle is orientable, its normal bundle is orientable, hence trivial; conversely if it's not orientable its orientation double cover is not trivial, and hence there is some loop that doesn't lift as a loop to the double cover. This is the desired loop with nontrivial normal bundle.) Now by the classification of line bundles above, we can pick a tubular neighborhood and then a closed disc bundle in that neighborhood to give the desired embedding of the "thickened Mobius strip".
The same technique works for arbitrary topological manifolds, as the above equivalence still works: a manifold is orientable if and only if there is some loop in the manifold that does not lift to the orientation double cover. Because the dimension is at least 3, you can homotope this loop "by hand" to be a locally flat embedding.
What we want to do now is find a closed disc bundle around this loop. Then, by the classification of disc bundles (there are again precisely two, because $\pi_0 \text{Diff}(D^n) = \Bbb Z/2$, and they are what you think), we have the desired result.
The one trouble is that disc bundles are hard to find. I'm told by Remark 4 in this paper by Edwards that as long as the dimension of my manifold is at least 5, my loop as constructed above has one. I don't have a precise reference for you. We can get around this problem in small dimensions: for $n = 3$ all manifolds are smoothable, so this is encapsulated in the first case, and for $n=4$ it's part of Freedman's work that noncompact 4-manifolds are smoothable, so just puncture it at a point not in the loop. This does not destroy nonorientability.
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http://www.aimsciences.org/article/doi/10.3934/dcdsb.2019058 | # American Institute of Mathematical Sciences
## Evolution of a spiral-shaped polygonal curve by the crystalline curvature flow with a pinned tip
1 Department of Mathematical Sciences, Shibaura Institute of Technology, Fukasaku 309, Minuma-ku, Saitama 337-8570, Japan 2 Division of Pure and Applied Science, Faculty of Science and Technology, Gunma University, Aramaki-machi 4-2, Maebashi, 371-8510 Gunma, Japan
* Corresponding author: Takeshi Ohtsuka
Received November 2017 Revised September 2018 Published April 2019
We present a new ODE approach for an evolving polygonal spiral by the crystalline eikonal-curvature flow with a fixed center. In this approach, we introduce a mechanism of new facet generation at the center of the growing spiral, which is based on the theory of two-dimensional nucleation. We prove the existence, uniqueness and intersection free of solution to our formulation globally-in-time. In the proof of the existence we also prove that new facets are generated repeatedly in time. The comparison result of the normal velocity between inner and outer facets with the same normal direction leads intersection-free result. The normal velocities are positive after the next new facet is generated, so that the center is always behind of the moving facets.
Citation: Tetsuya Ishiwata, Takeshi Ohtsuka. Evolution of a spiral-shaped polygonal curve by the crystalline curvature flow with a pinned tip. Discrete & Continuous Dynamical Systems - B, doi: 10.3934/dcdsb.2019058
##### References:
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show all references
##### References:
[1] F. Almgren and J. E. Taylor, Flat flow is motion by crystalline curvature for curves with crystalline energies, J. Differential Geom., 42 (1995), 1-22. doi: 10.4310/jdg/1214457030. Google Scholar [2] F. Almgren, J. E. Taylor and L. Wang, Curvature-driven flows: a variational approach, SIAM J. Control Optim., 31 (1993), 387-438. doi: 10.1137/0331020. Google Scholar [3] S. Angenent and M. E. Gurtin, Multiphase thermomechanics with interfacial structure. Ⅱ. Evolution of an isothermal interface, Arch. Rational Mech. Anal., 108 (1989), 323-391. doi: 10.1007/BF01041068. Google Scholar [4] W. K. Burton, N. Cabrera and F. C. Frank, The growth of crystals and the equilibrium structure of their surfaces, Philosophical Transactions of the Royal Society of London. Series A. Mathematical and Physical Sciences, 243 (1951), 299-358. doi: 10.1098/rsta.1951.0006. Google Scholar [5] R. E. Caflisch, Growth, structure and pattern formation for thin films, J. Sci. Comput., 37 (2008), 3-17. doi: 10.1007/s10915-008-9206-8. Google Scholar [6] A. Chambolle, M. Morini, M. Novaga and M. Ponsiglione, Existence and uniqueness for anisotropic and crystalline mean curvature flow, to appear in J. Amer. Math. Soc., arXiv: 1702.03094.Google Scholar [7] A. Chambolle, An algorithm for mean curvature motion, Interfaces Free Bound., 6 (2004), 195-218. doi: 10.4171/IFB/97. Google Scholar [8] A. Chambolle, M. Morini and M. Ponsiglione, Existence and uniqueness for a crystalline mean curvature flow, Comm. Pure Appl. Math., 70 (2017), 1084-1114. doi: 10.1002/cpa.21668. Google Scholar [9] B. Engquist, A.-K. Tornberg and R. Tsai, Discretization of Dirac delta functions in level set methods, J. Comput. Phys., 207 (2005), 28-51. doi: 10.1016/j.jcp.2004.09.018. Google Scholar [10] N. Forcadel, C. Imbert and R. Monneau, Uniqueness and existence of spirals moving by forced mean curvature motion, Interfaces Free Bound., 14 (2012), 365-400. doi: 10.4171/IFB/285. 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Ishiwata, Crystalline motion of spiral-shaped polygonal curves with a tip motion, Discrete Contin. Dyn. Syst. Ser. S, 7 (2014), 53-62. doi: 10.3934/dcdss.2014.7.53. Google Scholar [22] T. Ishiwata, T. K. Ushijima, H. Yagisita and S. Yazaki, Two examples of nonconvex self-similar solution curves for a crystalline curvature flow, Proc. Japan Acad. Ser. A Math. Sci., 80 (2004), 151-154. doi: 10.3792/pjaa.80.151. Google Scholar [23] A. Karma and M. Plapp, Spiral surface growth without desorption, Phys. Rev. Lett., 81 (1998), 4444-4447. doi: 10.1103/PhysRevLett.81.4444. Google Scholar [24] R. Kobayashi, A brief introduction to phase field method, AIP Conf. Proc., 1270 (2010), 282-291. doi: 10.1063/1.3476232. Google Scholar [25] H. Miura and R. Kobayashi, Phase-field modeling of step dynamics on growing crystal surface: Direct integration of growth units to step front, Crystal Growth & Design, 15 (2015), 2165-2175. doi: 10.1021/cg501806d. Google Scholar [26] A. Oberman, S. Osher, R. Takei and R. Tsai, Numerical methods for anisotropic mean curvature flow based on a discrete time variational formulation, Commun. Math. Sci., 9 (2011), 637-662. doi: 10.4310/CMS.2011.v9.n3.a1. Google Scholar [27] T. Ohtsuka, Y.-H. Tsai and Y. Giga, A level set approach reflecting sheet structure with single auxiliary function for evolving spirals on crystal surfaces, Journal of Scientific Computing, 62 (2015), 831-874. doi: 10.1007/s10915-014-9877-2. Google Scholar [28] T. Ohtsuka, A level set method for spiral crystal growth, Advances in Mathematical Sciences and Applications, 13 (2003), 225-248. Google Scholar [29] T. Ohtsuka, Minimizing movement approach for spirals evolving by crystalline curvature using level set functions, Oberwolfach Reports, 14 (2017), 314-317. Google Scholar [30] A. G. Shtukenberg, Z. Zhu, Z. An, M. Bhandari, P. Song, B. Kahr and M. D. Ward, Illusory spirals and loops in crystal growth, Proc. Natl. Acad. Sci. 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Evolution of polygonal spiral. In the above figures, $L_j$ and $y_j$ denote the facets and vertices of the polygonal spiral, respectively
Notations to the Wulff shape $\mathcal{W}_\gamma$
Example of positive convex (left) and positive concave(right) spiral for hexagonal $\mathcal{W}_\gamma$
Evolution of polygonal curves. The angle of the corners with a round curve is $\theta_{j+1} - \theta_j$
Examples of self-intersection between $\Lambda_i (t)$ $(solid line) and$ \Lambda_j (t) $(dashed line). Note that dots means$ y_i (t) $or$ y_j (t) $, which is a vertex belongs to$ \Lambda_i (t) $or$ \Lambda_j (t) $, respectively The case of facet-vertex intersection with$ \Lambda_j (\bar{t}) \subset \mathcal{O}_i (\bar{t}) $(we omit the notation of$ \bar{t} $in the figure). In the situation of the figure (b), the origin should be on the gray regions including the boundary lines by Corollary 3.4. Thus, the kind of touch like as (b) never occur The case of facet-vertex intersection with$ \Lambda_j \subset \mathcal{I}_i (\bar{t}) $. The facet$ \Lambda_{j+1} $never can be located as the dashed line in (b) Location of$ \Lambda_i (\bar{t}) $,$ \Lambda_{i+1} (\bar{t}) $,$ \Lambda_j (\bar{t}) $and$ \Lambda_{j+1} (\bar{t}) $under the vertex-vertex intersection between$ \Lambda_i (\bar{t}) $and$ \Lambda_j (\bar{t}) $. The above figures illustrate the case when (a)$ \varphi_i < \varphi_j < \varphi_i + \pi $, (b)$ \varphi_i + \pi < \varphi_j < \varphi_{i+1} + \pi $, and (c)$ \varphi_{i+1} + \pi < \varphi_j < \varphi_i + 2 \pi $. The gray regions are where$ \Lambda_{j+1} (\bar{t}) $can be located, and gray shaded regions are where the cross type intersection appears between$ \Lambda_i (\bar{t}) \cup \Lambda_{i+1} (\bar{t}) $and$ \Lambda_j (\bar{t}) \cup \Lambda_{j+1} (\bar{t}) $although$ \Lambda_{j+1} (\bar{t}) $seems to be located from the angle condition The case of facet-facet intersection with$ \Lambda_i (\bar{t}) \not\subset \Lambda_j (\bar{t}) $and$ \Lambda_i (\bar{t}) \not\supset \Lambda_j (\bar{t}) $Evolution of a polygonal spiral with a triangle Wulff shape. The figures illustrate the shape of a spiral at$ t = 0.0, 0.1, 0.5, 1.0 $from top left to bottom right Comparison of the profiles of spirals at$ t = 1 $with respect to different anisotropic mobilities under the same$ \mathcal{W}_\gamma $Comparison of profiles between our ODE system(top) and level set method(bottom) for a square spiral at$ t = 0 $(left),$ t = 1 $(center) and$ t = 2 $(right) Comparison of profiles between our ODE system(top) and level set method(bottom) for a diagonal spiral at$ t = 0 $(left),$ t = 1 $(center) and$ t = 2 $(right) Examination of the orientation of$ L_{k+1} $with hexagonal$ \mathcal{W}_\gamma $when$ d_k (T_{k+1}) = \ell_{k}/U $and$ d_{k-1} (T_{k+1}) < \ell_{k-1} / U $The situation when (39) does not hold in Case (i) The situation case(a) with$ \varphi_{j+1} \in (\varphi_{i+1} , \varphi_i + \pi) \$
[1] Tetsuya Ishiwata. On the motion of polygonal curves with asymptotic lines by crystalline curvature flow with bulk effect. Discrete & Continuous Dynamical Systems - S, 2011, 4 (4) : 865-873. doi: 10.3934/dcdss.2011.4.865 [2] Tetsuya Ishiwata. Crystalline motion of spiral-shaped polygonal curves with a tip motion. Discrete & Continuous Dynamical Systems - S, 2014, 7 (1) : 53-62. doi: 10.3934/dcdss.2014.7.53 [3] Bendong Lou. Spiral rotating waves of a geodesic curvature flow on the unit sphere. Discrete & Continuous Dynamical Systems - B, 2012, 17 (3) : 933-942. doi: 10.3934/dcdsb.2012.17.933 [4] Tobias H. Colding and Bruce Kleiner. Singularity structure in mean curvature flow of mean-convex sets. Electronic Research Announcements, 2003, 9: 121-124. [5] Nicolas Dirr, Federica Dragoni, Max von Renesse. Evolution by mean curvature flow in sub-Riemannian geometries: A stochastic approach. Communications on Pure & Applied Analysis, 2010, 9 (2) : 307-326. doi: 10.3934/cpaa.2010.9.307 [6] Tetsuya Ishiwata. On spiral solutions to generalized crystalline motion with a rotating tip motion. Discrete & Continuous Dynamical Systems - S, 2015, 8 (5) : 881-888. doi: 10.3934/dcdss.2015.8.881 [7] Tetsuya Ishiwata. Motion of polygonal curved fronts by crystalline motion: v-shaped solutions and eventual monotonicity. Conference Publications, 2011, 2011 (Special) : 717-726. doi: 10.3934/proc.2011.2011.717 [8] Tomáš Roubíček. On certain convex compactifications for relaxation in evolution problems. Discrete & Continuous Dynamical Systems - S, 2011, 4 (2) : 467-482. doi: 10.3934/dcdss.2011.4.467 [9] Leif Arkeryd, Raffaele Esposito, Rossana Marra, Anne Nouri. Ghost effect by curvature in planar Couette flow. Kinetic & Related Models, 2011, 4 (1) : 109-138. doi: 10.3934/krm.2011.4.109 [10] Changfeng Gui, Huaiyu Jian, Hongjie Ju. Properties of translating solutions to mean curvature flow. Discrete & Continuous Dynamical Systems - A, 2010, 28 (2) : 441-453. doi: 10.3934/dcds.2010.28.441 [11] Joel Spruck, Ling Xiao. Convex spacelike hypersurfaces of constant curvature in de Sitter space. Discrete & Continuous Dynamical Systems - B, 2012, 17 (6) : 2225-2242. doi: 10.3934/dcdsb.2012.17.2225 [12] Sun-Yung Alice Chang, Xi-Nan Ma, Paul Yang. Principal curvature estimates for the convex level sets of semilinear elliptic equations. Discrete & Continuous Dynamical Systems - A, 2010, 28 (3) : 1151-1164. doi: 10.3934/dcds.2010.28.1151 [13] Sophie Guillaume. Evolution equations governed by the subdifferential of a convex composite function in finite dimensional spaces. Discrete & Continuous Dynamical Systems - A, 1996, 2 (1) : 23-52. doi: 10.3934/dcds.1996.2.23 [14] Feng Luo. A combinatorial curvature flow for compact 3-manifolds with boundary. Electronic Research Announcements, 2005, 11: 12-20. [15] Dimitra Antonopoulou, Georgia Karali. A nonlinear partial differential equation for the volume preserving mean curvature flow. Networks & Heterogeneous Media, 2013, 8 (1) : 9-22. doi: 10.3934/nhm.2013.8.9 [16] Yannan Liu, Hongjie Ju. Non-collapsing for a fully nonlinear inverse curvature flow. Communications on Pure & Applied Analysis, 2017, 16 (3) : 945-952. doi: 10.3934/cpaa.2017045 [17] Leif Arkeryd, Raffaele Esposito, Rossana Marra, Anne Nouri. Erratum to: Ghost effect by curvature in planar Couette flow [1]. Kinetic & Related Models, 2012, 5 (3) : 669-672. doi: 10.3934/krm.2012.5.669 [18] Miroslav KolÁŘ, Michal BeneŠ, Daniel ŠevČoviČ. Area preserving geodesic curvature driven flow of closed curves on a surface. Discrete & Continuous Dynamical Systems - B, 2017, 22 (10) : 3671-3689. doi: 10.3934/dcdsb.2017148 [19] Bendong Lou. Traveling wave solutions of a generalized curvature flow equation in the plane. Conference Publications, 2007, 2007 (Special) : 687-693. doi: 10.3934/proc.2007.2007.687 [20] Yoshikazu Giga, Yukihiro Seki, Noriaki Umeda. On decay rate of quenching profile at space infinity for axisymmetric mean curvature flow. Discrete & Continuous Dynamical Systems - A, 2011, 29 (4) : 1463-1470. doi: 10.3934/dcds.2011.29.1463
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[Back to Top] | 2019-07-20 23:57:43 | {"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": 0.6331367492675781, "perplexity": 3809.0155294676074}, "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-2019-30/segments/1563195526799.4/warc/CC-MAIN-20190720235054-20190721021054-00131.warc.gz"} |
https://stats.stackexchange.com/questions/388884/how-to-convert-irt-theta-score-to-a-percentage-score | # How to convert IRT theta score to a percentage score
I am trying to implement an adaptive test using 3PL IRT model. We need to screen the candidates and label their expertise as (beginner, intermediate, or expert).
We also need a percentage score for each examinee.
When the examinee is given a sufficient number of items, the initial estimate of ability should not have a major effect on the final estimate of ability.
The final estimate of ability is the last estimated theta value before the stopping rule. It can have values in the range -3 to +3.
Q: How do I convert this to a percentage score so that the end user can know their performance?
Q: How to determine the cut-score for each level of expertise? What are the things that I need to take into consideration while calculating this?
Theoretically in IRT, $$\theta$$ and $$b$$ (which are measured in the same scale) do not have lower or upper bounds. They can go from $$-\infty$$ to $$\infty$$. Sometimes a test is unable to capture the full extent of an examinee's knowledge because it doesn't have enough items in the lower or higher ends of the difficulty scale, for example. Anyway, it is not a straightforward task to convert from $$\theta$$ values to percentages.
One thing you can do (and I know of a national test in my country that does something similar) is to transform your data using a desired mean and standard deviation. For example, you can displace your mean from 0 to 500 by adding 500 to all examinees $$\theta$$ values and changing your scale accordingly. You can then present their scores as if they went from 0 to 1000 (which you know that theoretically is not the truth, it just means your scale is centered at 500).
• You can subtract $\mu_{\theta}$ and divide by $\sigma_{\theta}$ from all yout $\theta$ values, then add a new $\mu_1$ and multiply by a new $\sigma_1$ of your choosing to get a new scale for all $\theta$. In my example, $\mu_1=500$ and $\sigma_1$ is a value you could experiment with to get your max. and min. $\theta$ values within the desired bounds. – Douglas De Rizzo Meneghetti Jan 26 '19 at 4:13 | 2020-07-14 04:01:51 | {"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": 6, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6631733775138855, "perplexity": 364.9277286517754}, "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/1593657147917.99/warc/CC-MAIN-20200714020904-20200714050904-00158.warc.gz"} |
https://soardocs.readthedocs.io/projects/goodman-pipeline/en/v1.3.0/overview.html | # Overview¶
The Goodman Spectroscopic Data Reduction Pipeline - The Goodman Pipeline - is a Python-based package for producing science-ready, wavelength-calibrated, 1-D spectra. The goal of The Goodman Pipeline is to provide SOAR users with an easy to use, very well documented software for reducing spectra obtained with the Goodman High Troughput Spectrograph. Though the current implementation assumes offline data reduction, our aim is to provide the capability to run it in real time, so 1-D wavelength calibrated spectra can be produced shortly after the shutter closes.
The pipeline is primarily intended to be run on a data reduction dedicated computer though it is available for local installation. The Goodman Spectroscopic Pipeline project is hosted at GitHub at it’s GitHub Repository.
Instructions for running the software are provided in the Usage section of this guide. How to access the the data reduction server is on Setup for Remote Use or if you prefer to install a local version instructions are in Install
Currently the pipeline is separated into two main components. The initial processing is done by redccd, which does the following processess.
• Identifies calibrations and science frames.
• Create master bias.
• Creates master flats and normalizes it.
• Apply overscan correction.
• Trims the image.
• For spectroscopic data find slit edges and trims again.
• Applies bias correction.
• Applies flat correction.
• Applies cosmic ray removal.
The spectroscopic processing is done by redspec and carries out the following steps:
• Identifies point-source targets.
• Traces the spectra.
• Extracts the spectra.
• Estimates and subtract background.
• Saves extracted (1D) spectra, without wavelength calibration.
• Finds the wavelength solution.
• Linearizes data (resample)
• Writes the wavelength solution to FITS header
• Creates a new file for the wavelength-calibrated 1D spectrum | 2021-07-30 15:08:31 | {"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": 0.2795262634754181, "perplexity": 6657.271852689419}, "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/1627046153966.60/warc/CC-MAIN-20210730122926-20210730152926-00342.warc.gz"} |
https://www.physicsforums.com/threads/speed-of-air-molecules.288915/ | Speed of air molecules
1. Feb 1, 2009
kasse
a) Show that the typical speed of air molecules in the room where you're sitting is about 500 m/s.
b) Show that the typical speed is about 420 m/s at Mt. Everest. (Neglect the temperature difference between sea level and mountaintop, and assume that the air consists of only nitrogen molecules).
a)
$$\frac{3}{2}k_B T_R = \frac{1}{2}mv^2 \Rightarrow \sqrt{\left\langle v^2 \right\rangle} \approx 500 m/s$$ CORRECT
b)
At Mt. Everest (9000 m) some of the kinetic energy has been transported to potential energy. A $$N_2$$ molecule has mass 4.7E-26 kg, so that
$$\Delta U = mgh = 4.7E-26 kg \cdot 9.81 m/s^2 \cdot 9000 m = 4.15E-21 J.$$
Hence the average kinetic energy of a nitrogen molecules up there are 6.15E-21 J - 4.15E-21 J = 2.0E-21 J, and
$$2.\cdot 10^{-21} J = \frac{1}{2}mv^2 \Rightarrow \sqrt{\left\langle v^2 \right\rangle} \approx 290 m/s$$ WRONG
Last edited: Feb 1, 2009
2. Feb 1, 2009
chrisk
Try using a relation between pressure and average velocity since T is constant. | 2017-06-29 02:34:05 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.8389396071434021, "perplexity": 1150.4785687123417}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-26/segments/1498128323842.29/warc/CC-MAIN-20170629015021-20170629035021-00352.warc.gz"} |
http://parasys.net/error-propagation/error-propagation-for-average.php | # parasys.net
Home > Error Propagation > Error Propagation For Average
# Error Propagation For Average
## Contents
because it ignores the uncertainty in the M values. I see how those values differ in terms of numbers, but which one is correct when talking about the correct estimate for the standard deviation? UC physics or UMaryland physics) but have yet to find exactly what I am looking for. This reveals one of the inadequacies of these rules for maximum error; there seems to be no advantage to taking an average. http://parasys.net/error-propagation/error-propagation-through-average.php
In this way an equation may be algebraically derived which expresses the error in the result in terms of errors in the data. First, the addition rule says that the absolute errors in G and H add, so the error in the numerator (G+H) is 0.5 + 0.5 = 1.0. Call this result Sm (s.d. I'm not clear though if this is an absolute or relative error; i.e. http://lectureonline.cl.msu.edu/~mmp/labs/error/e2.htm
## Propagation Of Error Mean
Logical fallacy: X is bad, Y is worse, thus X is not bad How to convert a set of sequential integers into a set of unique random numbers? etc. So the fractional error in the numerator of Eq. 11 is, by the product rule: [3-12] f2 + fs = fs since f2 = 0.
are inherently positive. Dickfore, May 27, 2012 May 27, 2012 #12 viraltux rano said: ↑ Hi viraltux, Thank you very much for your explanation. If this error equation is derived from the indeterminate error rules, the error measures Δx, Δy, etc. Error Propagation Example Then vo = 0 and the entire first term on the right side of the equation drops out, leaving: [3-10] 1 2 s = — g t 2 The student will,
The absolute error in g is: [3-14] Δg = g fg = g (fs - 2 ft) Equations like 3-11 and 3-13 are called determinate error equations, since we used the Propagation Of Error Calculator of the population that's wanted. How do you say "root beer"? I should not have to throw away measurements to get a more precise result.
Summarizing: Sum and difference rule. Error Propagation Division General functions And finally, we can express the uncertainty in R for general functions of one or mor eobservables. I have looked on several error propagation webpages (e.g. So if the angle is one half degree too large the sine becomes 0.008 larger, and if it were half a degree too small the sine becomes 0.008 smaller. (The change
## Propagation Of Error Calculator
Please try the request again. Discover More Probably what you mean is this $$σ_Y = \sqrt{σ_X^2 + σ_ε^2}$$ which is also true. Propagation Of Error Mean When a quantity Q is raised to a power, P, the relative determinate error in the result is P times the relative determinate error in Q. Error Propagation Average Standard Deviation Can Communism become a stable economic strategy?
If you could clarify for me how you would calculate the population mean ± SD in this case I would appreciate it. http://parasys.net/error-propagation/error-propagation-average-value.php Laboratory experiments often take the form of verifying a physical law by measuring each quantity in the law. When two quantities are added (or subtracted), their determinate errors add (or subtract). You're right, rano is messing up different things (he should explain how he measures the errors etc.) but my point was to make him see that the numbers are different because How To Find Error Propagation
Sooooo... chiro, May 26, 2012 May 27, 2012 #8 rano Hi viraltux and haruspex, Thank you for considering my question. It should be derived (in algebraic form) even before the experiment is begun, as a guide to experimental strategy. click site This forces all terms to be positive.
If my question is not clear please let me know. Error Propagation Physics In assessing the variation of rocks in general, that's unusable. The indeterminate error equation may be obtained directly from the determinate error equation by simply choosing the "worst case," i.e., by taking the absolute value of every term.
## In the operation of subtraction, A - B, the worst case deviation of the answer occurs when the errors are either +ΔA and -ΔB or -ΔA and +ΔB.
The absolute fractional determinate error is (0.0186)Q = (0.0186)(0.340) = 0.006324. Any insight would be very appreciated. Generated Fri, 14 Oct 2016 15:03:31 GMT by s_ac15 (squid/3.5.20) Error Propagation Calculus I have looked on several error propagation webpages (e.g.
What is the most expensive item I could buy with £50? Appease Your Google Overlords: Draw the "G" Logo Not working "+" in grep regex syntax How can a nocturnal race develop agriculture? The st dev of the sample is 20.1 The variance (average square minus square average) is 405.56. http://parasys.net/error-propagation/error-propagation-in-average.php Now I have two values, that differ slighty and I average them.
In this example x(i) is your mean of the measures found (the thing before the +-) A good choice for a random variable would be to say use a Normal random They do not fully account for the tendency of error terms associated with independent errors to offset each other. We quote the result in standard form: Q = 0.340 ± 0.006. I think this should be a simple problem to analyze, but I have yet to find a clear description of the appropriate equations to use.
rano, May 27, 2012 May 27, 2012 #11 Dickfore rano said: ↑ I was wondering if someone could please help me understand a simple problem of error propagation going from multiple | 2018-03-22 21:31:10 | {"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": 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": 0.6629243493080139, "perplexity": 958.1355361280104}, "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/1521257648003.58/warc/CC-MAIN-20180322205902-20180322225902-00753.warc.gz"} |
http://mathoverflow.net/questions/32179/common-quotients-of-direct-products | # Common quotients of direct products
In a paper I read recently, the authors use the fact that if two groups G and H have no nontrivial common quotient, then neither do GxG and HxH. It's unclear from the context whether this is true for all groups, or just groups of the type that are important for this paper, and they don't prove the claim.
I've been trying to prove the statement for general groups without any success. Is it true?
I've also tried restricting to the case I really care about, as follows. Let K, K', M, and M' be normal subgroups of U, with $U/K \simeq U/K'$ and $U/M \simeq U/M'$. Suppose that KK' = KM = KM' = K'M = K'M' = MM' = U. Is it also the case that $(K \cap K')(M \cap M') = U$? The reason this is connected to the above is that $U/(K \cap K')(M \cap M')$ is a common quotient of $U/(K \cap K') \simeq U/K \times U/K' \simeq U/K \times U/K$ and $U/(M \cap M') \simeq U/M \times U/M' \simeq U/M \times U/M$. So if U/K and U/M have no nontrivial common quotients, is the same true of $U/(K \cap K')$ and $U/(M \cap M')$?
Finally, if the result is not true in general for either of these cases, what about if we restrict to finite groups G and H?
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If $Q$ is a quotient of $G\times G$, then it has a normal subgroup $A$ (the image of $G\times 1$) such that both $A$ and $Q/A$ are quotients of $G$. If it is also a quotient of $H\times H$, then it has a normal subgroup $B$ such that both $B$ and $Q/B$ are quotients of $H$. Now assume that every common quotient of $G$ and $H$ is trivial. Then $Q/AB$ is trivial, so $AB=Q$. So $Q/A=AB/A=B/(A\cap B)$ is trivial, so $A=Q$. Likewise $B=Q$. So $Q$ is trivial.
Nice! – Victor Protsak Jul 17 '10 at 19:38 | 2016-05-24 12:11:48 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9184342622756958, "perplexity": 148.60521193705367}, "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/1464049270555.40/warc/CC-MAIN-20160524002110-00157-ip-10-185-217-139.ec2.internal.warc.gz"} |
http://tex.stackexchange.com/questions/142496/the-right-way-of-forcing-a-long-table-in-place/142497 | # The right way of forcing a long table in place
I was wondering what is the right way to include a long table in between pages?
I have a table that is an A4 page long, if I use \begin{table}[h] it pushes the table to the very end of the document, and if I use \begin{table}[h!] it forced some extra white space at some paragraphs.
Thanks for the help.
-
It is not very clear from your question what you are asking for. From what I gather from your question, you want to put a table in an independent page. The table is a page long. You do not want it on some fixed page, some approximate location will be just fine.
If the above is your scenario, then you need to use [!p] as your position specifier. Here, p means page of float and LaTeX will try to honor the placement with respect to actual place. By putting an ! before p you insist on this placement. This overrides internal parameters LaTeX uses for determining good float positions.
Please try the following code, check the output, and let us know whether it suits your purpose.
\documentclass{article}
\usepackage{lipsum}
\begin{document}
\section{One}
\lipsum[11]
\section{Two}
\lipsum[12]
\begin{table}[!p]
\centering
\begin{tabular}[c]{|c|c|p{0.75\textwidth}|l|}
\hline
1&1&\raggedright\lipsum[1]&One\\\hline
2&2&\raggedright\lipsum[2]&Two\\\hline
3&3&\raggedright\lipsum[3]&Three\\\hline
\end{tabular}
\caption{A table one page long}
\label{tab:long}
\end{table}
\section{Three}
\lipsum[13]
\section{Four}
\lipsum[14]
\lipsum[15]
\lipsum[16]
\end{document}
-
Yes, that's exactly what I need. I need it to be as close as possible with the original text where I cited the table. If I use [h] it will go to the very end of the document which is like 10 pages away. – cherhan Nov 6 '13 at 2:32
@cherhan Great. Glad of the of assistance. Now you can accept the answer :-) In most of the cases ! does the trick. I make it a rule to use !tbp for all my float location specifiers. – Masroor Nov 6 '13 at 2:34
what if it causes extra whitespace around paragraphs around the float? – cherhan Nov 6 '13 at 3:00
@cherhan Not very clear. Are you talking about the spaces before and after the float? If that this the case this will help. – Masroor Nov 6 '13 at 3:04
I will try to create a MWE to show you. Just that some times, for example after section 2 (lipsum[12]) it might show a large white in order to push the table to another page. It's ok if you don't understand :) – cherhan Nov 6 '13 at 4:41 | 2014-08-20 05:01:06 | {"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": 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": 0.7239705920219421, "perplexity": 983.5248925358027}, "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/1408500800168.29/warc/CC-MAIN-20140820021320-00235-ip-10-180-136-8.ec2.internal.warc.gz"} |
https://math.stackexchange.com/questions/2406432/compactness-of-the-closed-interval-0-1 | # Compactness of the closed interval [0,1]
In general topology, a topological space is said to be compact, if every one of its open cover has a finite subcover. However, I cannot see the compactness of the close interval [0,1] from the above definition. To be a little specific,let us consider the following open cover for [0,1]: $C= \{[0,1/2),(1/3,3/4), (2/3,1]\}$. Now, the open interval (1/3,3/4) itself has at least one open cover (let's call it P) which does not have a finite subcover. We use P to cover the open interval (1/3,3/4). This gives a new open cover C' for the interval [0,1]. It looks like C' does not have an finite subcover, since C' includes P which does not have a finite subcover.
Of course I misunderstood something here. If somebody can catch my error it will be very helpful.
• 'It looks like C' does not have an finite subcover, since C' includes P which does not have a finite subcover.' Not necessarily true, as all but finitely many sets in P may be contained in [0,1/2) or (2/3,1] , rendering them unnecessary. – Tim The Enchanter Aug 26 '17 at 5:32
• Go to the Wikipedia page for the Heine-Borel theorem, and look at the proof section. A bit down, they prove that $[-a,a]^n\subseteq \Bbb R^n$ is compact. You can follow that proof. – Arthur Aug 26 '17 at 6:24
• No finite subset of P will cover $(1/3,3/4).$ But a finite subset of $P$ will cover [$2/5, 5/7]$ and the rest of $[0,1]$ is covered by $\{[0,1/2),(2/3,1]\}$. – DanielWainfleet Aug 26 '17 at 7:47
• Maybe some of these posts might help: How to prove every closed interval in R is compact? and Showing that $[0,1]$ is compact – Martin Sleziak Aug 26 '17 at 10:50
So you get an open cover by retaining $[0,1/2)$ and $(2/3,1]$ but replacing $(1/3,3/4)$ by a bunch $P$ of open sets where no finite collection covers $(1/3,3/4)$.
You can do this.
But it is still the case that this new covering $C'$ has a finite subcovering. Don't forget that $[0,1/2)$ and $(2/3,1]$ are still available. If we used both of these, all we have to do is find a finite subset of $P$ that covers $[1/2,2/3]$. (We don't need it to cover all of $(1/3,3/4)$.) As $[1/2,2/3]$ is compact, then there will be such a finite subset.
• This is circular: how would you know that $[1/2,2/3]$ is compact? – Arthur Aug 26 '17 at 6:39
• @Arthur, it seems that the OP wanted to understand the flaw in his example of an open covering with no finite subcovering. We were not looking for a proof of compactness of a closed interval. – Ted Shifrin Aug 26 '17 at 6:43
• It's not circular; it'd be circular if he used the face that $[0, 1]$ is compact. Perhaps Lord knows that $[1/2, 2/3]$ is compact because Lord's has read a proof of the HB-theorem. It may not be helpful to OP, but it's not circular. – John Hughes Aug 26 '17 at 6:44
I will present you a proof different from Heine-Borel which i hope it will help you.
We'll use this corollary of Cantor's intersection theorem.
Corollary:Let $(X,d)$ be a complete metric space.Then for every decreasing sequence $F_n$ of closed subsets of $X$ such that $diam(F_n) \rightarrow 0$ we have that $\bigcap_{n=1}^{\infty}F_n =\{x\}$ for some $x \in X$.
Note that $diam(A)=\sup\{d(x,y)|x,y \in A \}$
Now instead of the diameter of a set i will use the notion of length deonting by $l$..You can see that in the real line (with the usual metric $d(x-y)=|x-y|$)
that $diam(I)\leq l(I)$ where $I$ is an interval.
$Proof$
Now let $I_0=[0,1]$ and assume that $I_0$ is not compact.Thus exists a collection $P$ of open sets which does not contain a finite subfamily of open sets that cover $I_0$
Now we divide $I_0$ in two intevals of equal length each namely $[0,1/2],[1/2,1]$
At least one of this intervals cannot be covered by finitely many elements of $P$.Call this interval $I_1$.
Now divide $I_1$ into two intervals of length $1/4$.At least one of these intervals cannot be covered my finitely elements of $P$(Why?)
Call this inteval $I_2$.
We continue this argument inductively and we find a decreasing sequence $I_0 \supseteq I_1 \supseteq I_2 \supseteq....$ of closed intervals.
Now we know that $[0,1]$ is complete that the intersection of these intervals is nonempty from the theorem.
Also notice that $l(I_n)=\frac{1}{2^n} \rightarrow 0$ and we know that forall $n \in \mathbb{N} \cup \{0\}$ none of the intervals can be covered by a finite elements of $P$
Thus exists a $x \in [0,1]$ such that $x$ belongs to any of the intervals $I_0,I_1,I_1...$ .
Now because $P$ covers $I_0$ we have that exists an open set $A$ as element of $P$ where $x \in A$
Now because $A$ is open exists $\epsilon>0$ such that $x \in (x- \epsilon,x+\epsilon) \subseteq A$
Now exists $n_0 \in \mathbb{N}$ such that $l(I_{n_0})< \epsilon$ and $x \in I_{n_0}$
Thus $I_{n_0} \subseteq (x- \epsilon,x+\epsilon)$
We arrived to a contradiction because $I_{n_0}$ cannot be covered by finitely elements of $P$ and at the same time we found the element $A$ that covers $I_{n_0}$.
Thus $[0,1]$ is compact.
So from this you can see that an open interval say $(a,b)$(it can also be a subset of $[0,1]$) fails to be compact because it s not complete.
• @Shalop you are right..thanks for the comment...the theorem is correct if the sets $F_n$ are compact..but it can be fixed if we let $diam(F_n) \rightarrow 0$ so in this cas we need only the closeness – Marios Gretsas Aug 26 '17 at 8:23
You can also use the theorem that
Let $\mathcal{S}$ be a subbase for $X$. Then $X$ is compact iff every open cover with elements of $\mathcal{S}$ has a finite subcover.
This is Alexander's subbase theorem and for any ordered space like $[0,1]$ has the following subbase $\mathcal{S}= \{[0,a): a \in [0,1]\} \cup \{(b,1]: b \in [0,1]\}$
Then the compactness of $[0,1]$ follows, using the order completeness: suppose $\mathcal{O} \subseteq \mathcal{S}$ is an open cover of $[0,1]$.
As $0$ is only covered by a set of the form $[0,a)$ there must be at least one such member in $\mathcal{O}$. So define $A = \{a \in [0,1]: [0,a) \in \mathcal{O}\}$ which is non-empty and bounded above (by $1$) so $a_0 = \sup A$ exists in $[0,1]$. Then $a_0$ is covered by some member of $\mathcal{O}$ and it cannot be of the form $[0,b)$ (as then $b \in A$ and $b >a_0$, so $a_0$ would not be an upperbound of $A$) and so it is covered by some $(b,1] \in \mathcal{O}$. As $b < a_0$ this means that $b$ is not an upperbound for $A$, as $a_0$ is the least upperbound of $A$, so for some $a_1 \in A$ we have $a_1 > b$. But then $[0,a_1)\in \mathcal{O}$ and the two sets $[0,a_1), (b,1]$ from $\mathcal{O}$ clearly cover $[0,1]$. By the Alexander subbase theorem, $[0,1]$ is compact. In fact, any ordered space with a minimum and maximum which is order-complete is compact (same proof essentially).
The Alexander theorem is proved using AC, so this does not make for a fully "constructive" proof. But it is a useful theorem, e.g. also to show Tychonoff's theorem for compactness of products (product also have a natural subbase), or the compactness of the hyperspace of a compact space in the Vietoris topology.
The intervals $(\frac{1}{2},\frac{1}{3})$ and $(\frac{2}{4},\frac{3}{4})$ will make infinitely many of the sets required to cover $P$ redundant leaving only finitely many necessary to cover the gap.
Lets consider an example. For simplicity I'll use the set $[-2,2]$ $C=\{[-2,\frac{1}{2}),(0,1),(\frac{1}{2},2]\}$ and I cover $P=(0,1)$ with sets of the form $(0,1-\frac{1}{n})$ for $n \in \mathbb{N}$, $n\geq3.$ Clearly no finite subcover of $P$ would do because we need to get arbitrarily close to $1$ but we can in fact select a single interval $(0,1-\frac{1}{n}$), discard the rest and be left with an open covering of $[-2,2]$.
• I understand that, the cover you chose has a finite subcover. In my mind I have different kind of cover which I could not formally write but it is the following sequence of covers: [0,1]= [0,1/2+$\epsilon$) $\cup$ (1/2+$\epsilon$,1] = [0,1/4+$\epsilon$)$\cup$ (1/4-$\epsilon$, (1/2+$\epsilon$)$\cup$ (1/2-$\epsilon$, 3/4+$\epsilon$)$\cup$(3/4-$\epsilon$,1]=$\cdots$. We can go on like that up to the limit where the cover has infinite number of open sets. Clearly if we omit any one of the members of the cover at any point of the above sequence, it will no longer cover [0,1]. – Tuhin Subhra Mukherjee Aug 28 '17 at 17:12
• If we can indeed continue the sequence up to infinite terms, it seems that the above cover does not even have a infinite subcover since we need all of the open sets to cover [0,1]. I can't seem to figure out where does my argument go wrong but I know it IS wrong. – Tuhin Subhra Mukherjee Aug 28 '17 at 17:17
• @TuhinSubhraMukherjee Did you mean $[0,1/2+\epsilon) \cup (1/2-\epsilon,1]$? In any case the issue is when you took the limit you didn't ensure that the sets were still open. Since the length of each open interval in your construction can easily be seen to converge to zero, they are not open intervals in the limit. I believe your construction is equivalent to the binary representation of the reals on that interval. Trying writing out your scheme in base 2. Or it could be that you forgot $\epsilon > 0$ so there exists a limit on how small the balls can be, namely $2\epsilon$. – CyclotomicField Aug 28 '17 at 21:30
• @ CyclotomicField Sorry, I meant [0,1/2+ϵ)∪(1/2−ϵ,1]. Yes, there is a limit on the size of the balls as you have pointed out; it is 2$\epsilon$. But $\epsilon$ can be arbitrarily small. May be I have to think about it a little more. – Tuhin Subhra Mukherjee Aug 29 '17 at 2:04
• @TuhinSubhraMukherjee $\epsilon$ can be arbitrarily small but Archimedes's principle tells us that there exist an $M \in \mathbb{N}$ such that $2\epsilon M > 1$ which means we will still end up with a finite number of them to cover the interval. Can you see how the length of the intervals going to zero in the limit means that in the limit we have singletons covering the interval instead of open set? – CyclotomicField Aug 29 '17 at 3:37
This kind of thing (compactness of $[0,1]$) is typically proved via an appeal to the Heine-Borel theorem, which says that a subset of $\mathbb{R}^n$ is compact if and only if it is closed and bounded. The subset $[0,1]$ is closed and bounded and therefore compact.
• This is absurdly circular – mathworker21 Aug 26 '17 at 5:51
• @mathworker21 That may be, although parts of the proof of the HB theorem, as written in that Wikipedia article, is exactly what we need. So while this answer doesn't answer the question, it does inadvertently link to an answer. – Arthur Aug 26 '17 at 6:27
• This doesn't address the OP. – Ted Shifrin Aug 26 '17 at 6:44
• In either case it would benefit the OP to read and understand the Heine-Borel thm, which was my intention, since it's an important thm in itself and the OP can answer his question himself by studying the proof. – g.s Aug 26 '17 at 7:07 | 2019-10-15 13:26:33 | {"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": 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": 0.886833667755127, "perplexity": 195.85892179188957}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570986659097.10/warc/CC-MAIN-20191015131723-20191015155223-00187.warc.gz"} |
https://zbmath.org/?q=ut%3Anegative+feedbacks+se%3A00000484 | # zbMATH — the first resource for mathematics
Global stability in delay differential systems without dominating instantaneous negative feedbacks. (English) Zbl 0828.34066
The systems $$x_i'(t)= x_i(t) f_i(t, x_t(\cdot))$$, $$i= 1,\dots, n$$ and $$x_i'(t)= x_i(t) f_i(t, x_t(\cdot), y_t(\cdot))$$, $$i= 1,\dots, n$$, $$y_j'(t)= g_j(t, x_t(\cdot), y_t(\cdot))$$, $$j= 1,\dots, m$$ are considered with $$f_i$$ and $$g_j$$ continuous linear operators.
It is shown that if the systems have globally asymptotically stable nonnegative equilibria in the absence of the delays and if the systems are dissipative if delays are present (solutions are eventually uniformly bounded), then the equilibria remain globally asymptotically stable as long as the time lags involved in some part of the negative feedbacks are small enough.
##### MSC:
34K20 Stability theory of functional-differential equations 92D25 Population dynamics (general)
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A Maximum Entropy-least Squares Estimator for Elastic Origin-Destination Trip Matrix Estimation In transportation subnetwork-supernetwork analysis, it is well known that the origin-destination (O-D) flow table of a subnetwork is not only determined by trip generation and distribution, but also by traffic routing and diversion, due to the existence of internal-external, external-internal and external-external flows. My context is mainly of a practical nature: When collecting entropy to seed a CSPRNG, I want the CSPRNG to be available as soon as possible, but not until at least n bits (say 128 bits) of entropy (unpredictable data) has been collected and fed to the CSPRNG. This paper discusses an elastic O-D flow table estimation problem for subnetwork analysis. We use cookies to help provide and enhance our service and tailor content and ads. (2006). & Willumsen, Luis G., 1980. Math.,41, 683–697), we introduce estimators of entropy and describe their properties. You can help correct errors and omissions. Computer Science, University of A Coruna, 15071 A Coruna, Spain Abstract.Minimum MSE plays an indispensable role in learning and ", Maryam Abareshi & Mehdi Zaferanieh & Bagher Keramati, 2017. Theres 3 sunny instances divided into 2 classes being 2 sunny related with Tennis and 1 related to Cinema. Again, the di erential entropy provides the rule of thumb D(Q ) ˇ(1=12)22[H(Q ) H(f)]for small . Thus, the maximum entropy principle ", Sherali, Hanif D. & Sivanandan, R. & Hobeika, Antoine G., 1994. See general information about how to correct material in RePEc. Estimator: autocorrelation, maximum entropy (Burg), least-squares [...] normal equations, least-squares covariance and modified covariance, SVD principal component AR. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share … We propose a combined maximum entropy-least squares (ME-LS) estimator, by which O-D flows are distributed over the subnetwork so as to maximize the trip distribution entropy, while demand function parameters are estimated for achieving the least sum of squared estimation errors. Histogram estimator. the various RePEc services. ", Van Zuylen, Henk J. If you know of missing items citing this one, you can help us creating those links by adding the relevant references in the same way as above, for each refering item. ", Lo, H. P. & Zhang, N. & Lam, W. H. K., 1996. While the estimator is powered by the classic convex combination algorithm, computational difficulties emerge within the algorithm implementation until we incorporate partial optimality conditions and a column generation procedure into the algorithmic framework. The plugin estimator uses empirical estimates of the frequencies ^p j= 1 n P n i=1 1[X i= j] to obtain an estimate of the entropy as follows: H^ n= Xd j=1 p^ jlog 2 ( ^p j) LP Estimator The LP Estimator works by transforming the samples fX ign i=1 into a ngerprint, which is the vector f= (f 1;f 2;:::) for which f ", Chen, Anthony & Chootinan, Piya & Recker, Will, 2009. Master thesis of the National Institute of Applied Sciences of Lyon. ", Chao Sun & Yulin Chang & Yuji Shi & Lin Cheng & Jie Ma, 2019. Inst. eracy of a Bayesian estimator, section 8.2 gives a consistency result for a potentially more powerful regularization method than the one examined in depth here, and section 8.3 attempts to place our results in the context of estimation of more general functionals of the probability distribution (that is, not just entropy and mutual information). We study the effects of tail behaviour, distribution smoothness and dimensionality on convergence properties. See Also. 0. ... How to find the closed form formula for $\hat{\beta}$ while using ordinary least squares estimation? @NetranjitBorgohain that's a different method, but again it expects a different set of parameters entropy_joint(X, base=2, fill_value=-1, estimator='ML', Alphabet_X=None, keep_dims=False) see documentation for details – nickthefreak Mar 28 '19 at 15:21 Motivated by recent work of Joe (1989,Ann. Apply the entropy formula considering only sunny entropy. This illustrates under what circumstances entropy estimation is likely to be preferable to traditional econometric estimators based on the characteristic of the available data and … Statist. The underlying assumption is that each cell of the subnetwork O-D flow table contains an elastic demand function rather than a fixed demand rate and the demand function can capture all traffic diversion effect under various network changes. Im confused with Least Squares Regression Derivation (Linear Algebra) Hot Network Questions A maximum entropy-least squares estimator for elastic origin-destination trip matrix estimation. ", Jafari, Ehsan & Pandey, Venktesh & Boyles, Stephen D., 2017. We propose a combined maximum entropy-least squares (ME-LS) estimator, by which O-D flows are distributed over the subnetwork so as to maximize the trip distribution entropy, while demand function parameters are estimated for achieving the least sum of squared estimation errors. Please note that corrections may take a couple of weeks to filter through This note is for people who are familiar with least squares but less so with entropy. Alternatively, the latter are also characterized by a postulate of composition consistency. In transportation subnetwork-supernetwork analysis, it is well known that the origin-destination (O-D) flow table of a subnetwork is not only determined by trip generation and distribution, but also by traffic routing and diversion, due to the existence of internal-external, external-internal and external-external flows. If only probabilities pk are given, the entropy is calculated as S =-sum(pk * log(pk), axis=axis).. ", Yang, Hai & Iida, Yasunori & Sasaki, Tsuna, 1994. Nonparametric entropy estimation : An overview. In particular, we argue that root-n consistency of entropy estimation requires appropriate assumptions about each of these three features. For technical questions regarding this item, or to correct its authors, title, abstract, bibliographic or download information, contact: (Haili He). it, the resulted maximum entropy distribution “is the least biased estimate possible on the given information; i.e., it is maximally noncommittal with regard to missing information”. Mathematically this means that in order to estimate the we have to minimize which in matrix notation is nothing else than . This allows to link your profile to this item. So the entropy formula for sunny gets something like this: -2/3 log2(2/3) - 1/3 log2(1/3) = 0.918. The concept of information entropy was introduced by Claude Shannon in his 1948 paper "A Mathematical Theory of Communication". person_outlineTimurschedule 2013-06-04 15:04:43. The entropy estimator using plug-in values under -estimates the true entropy value In fact: = + (n−1)/2T is a better estimator of the entropy (MM=Miller-Madow) No unbiased estimator of entropy … GME Estimation in Linear Regression Model GME Command with User Supplied Parameter Support Matrix Sign and Cross-Parameter Restrictions Conclusion Generalized Maximum Entropy Estimation GME estimator developed by Golan, Judge, and Miller (1996) Campbell and Hill (2006) impose inequality restrictions on GME estimator in a linear regression model H(Q ) + 1 2 log(12D(Q )) = H(f): (24) Here f is assumed to satisfy some smoothness and tail conditions. scipy.stats.entropy¶ scipy.stats.entropy (pk, qk = None, base = None, axis = 0) [source] ¶ Calculate the entropy of a distribution for given probability values. Robust least-squares estimation with a relative entropy constraint Abstract: Given a nominal statistical model, we consider the minimax estimation problem consisting of finding the best least-squares estimator for the least favorable statistical model within a … Numerical results from applying the combined estimator to a couple of subnetwork examples show that an elastic O–D flow table, when used as input for subnetwork flow evaluations, reflects network flow changes significantly better than its fixed counterpart. In information theory, entropy is a measure of the uncertainty in a random variable. ", Yang, Hai & Iida, Yasunori & Sasaki, Tsuna, 1991. but high entropy as described by Smithson. 11 The idea of the ordinary least squares estimator (OLS) consists in choosing in such a way that, the sum of squared residual (i.e. ) As corollaries, axiomatic characterizations of the methods of least squares and minimum discrimination information are arrived at. We propose a combined maximum entropy-least squares estimator, by which O–D flows are distributed over the subnetwork in terms of the maximum entropy principle, while demand function parameters are estimated for achieving the least sum of squared estimation errors. in the sample is as small as possible. While the estimator is powered by the classic convex combination algorithm, computational difficulties emerge within the algorithm implementation until we incorporate partial optimality conditions and a column generation procedure into the algorithmic framework. This online calculator computes Shannon entropy for a given event probability table and for a given message. The consequent estimator of entropy pro-posed by Correa (1995) is given by HCmn = 1 n Xn i=1 log 0 B B B @ i+P m j = i m (X (j ) X i)(j i) n i+Pm j = i m (X(j ) X (i))2 1 C C C A; Downloaded from jirss.irstat.ir at … ", Nie, Yu & Zhang, H.M. & Recker, W.W., 2005. Finally, the high-resolution or aperture-compensated velocity gather is used to ex-trapolate near- and far-offset traces. Copyright © 2011 Published by Elsevier Ltd. Procedia - Social and Behavioral Sciences, https://doi.org/10.1016/j.sbspro.2011.04.514. When q0 is uniform this is the same as maximizing the entropy. ScienceDirect ® is a registered trademark of Elsevier B.V. ScienceDirect ® is a registered trademark of Elsevier B.V. A maximum entropy-least squares estimator for elastic origin-destination trip matrix estimation. We propose a combined maximum entropy-least squares estimator, by which O–D flows are distributed over the subnetwork in terms of the maximum entropy principle, while demand function parameters are estimated for achieving the least sum of squared estimation errors. entropy; Examples Journal of Statistics. least-squares solution. This paper discusses an elastic O–D flow table estimation problem for subnetwork analysis. Public profiles for Economics researchers, Various rankings of research in Economics & related fields, Curated articles & papers on various economics topics, Upload your paper to be listed on RePEc and IDEAS, RePEc working paper series dedicated to the job market, Pretend you are at the helm of an economics department, Data, research, apps & more from the St. Louis Fed, Initiative for open bibliographies in Economics, Have your institution's/publisher's output listed on RePEc. (24) can be proved without any additional smoothness and tail conditions (Gy or , Linder, van der Meulen [28]). condentropy, mutinformation, natstobits. In transportation subnetwork–supernetwork analysis, it is well known that the origin–destination (O–D) flow table of a subnetwork is not only determined by trip generation and distribution, but also a result from traffic routing and diversion, due to the existence of internal–external, external–internal and external–external flows. Shannon Entropy. Dept., University of Florida, Gainesville, FL 32611, USA 2 Dept. distributions of ordinary least squares and entropy estimators when data are limited. $\begingroup$ This was informative. Start with least squares, min y k X k (y k x k)2 (1) where x kare the given data and y kare the corresponding points estimated by the model. choose the distribution that minimizes entropy relative to the default estimate q0. If CitEc recognized a reference but did not link an item in RePEc to it, you can help with this form . The entropy of a substance is influenced by structure of the particles (atoms or molecules) that comprise the substance. ", Kumar, Anshuman Anjani & Kang, Jee Eun & Kwon, Changhyun & Nikolaev, Alexander, 2016. ", LeBlanc, Larry J. By continuing you agree to the use of cookies. Aliases. When requesting a correction, please mention this item's handle: RePEc:eee:transb:v:45:y:2011:i:9:p:1465-1482. ", Bar-Gera, Hillel & Boyce, David & Nie, Yu (Marco), 2012. I estimate that you could get to the top with as few as thirty-five to fort y- ... which are proportionnal to the square root of text length. How was the formula for Ordinary Least Squares Linear Regression arrived at? If you have authored this item and are not yet registered with RePEc, we encourage you to do it here. & Farhangian, Keyvan, 1982. This result indicates the variable nature of subnetwork O–D flows. INTRODUCTION dow sometimes cause a poor velocity resolution when using Conventional velocity analysis is performed by measuring energy along hyperbolic paths for a set of tentative veloci-ties. All material on this site has been provided by the respective publishers and authors. (4) In order to estimate we need to minimize . Here, as usual, the entropy of a distribution p is defined as H(p) = p[ln(1=p)] and the relative entropy, or Kullback-Leibler divergence, as D(p k q) = p[ln(p=q)]. And so on. In information theory, the entropy of a random variable is the average level of "information", "surprise", or "uncertainty" inherent in the variable's possible outcomes. Recursive Least Squares for an Entropy Regularized MSE Cost Function Deniz Erdogmus1, Yadunandana N. Rao1, Jose C. Principe1 Oscar Fontenla-Romero2, Amparo Alonso-Betanzos2 1 Electrical Eng. In a mathematical frame, the given information used in the principle of maximum entropy, is expressed as a set of constraints formed as expectations of functions g ", Yang, Hai & Sasaki, Tsuna & Iida, Yasunori & Asakura, Yasuo, 1992. In the case of linear Gaussian case, a very mature TLS parameter estimation algorithm has been developed. The total least square (TLS) estimation problem of random systems is widely found in many fields of engineering and science, such as signal processing, automatic control, system theory and so on. This can be related to cross-entropy in two steps: 1) convert into a likelihood, 2) con- The simple way of evaluation of a probability distribution () of biological variable with the entropy normalized by its maximum value (= ), = − ∑ = ()demonstrates advantages over standard physiological indices in the estimation of functional status of cardiovascular, nervous and immune systems.. Another approach uses the idea that the differential entropy, Hausser J. The underlying assumption is that each cell of the subnetwork O–D flow table contains an elastic demand function rather than a fixed demand rate and the demand function can capture all traffic diversion effect under various network changes. If qk is not None, then compute the Kullback-Leibler divergence S = sum(pk * log(pk / qk), axis=axis).. It also allows you to accept potential citations to this item that we are uncertain about. ". Note I am not only looking for the proof, but also the derivation. This result indicates the variable nature of subnetwork O-D flows. As a special case, a derivation of the method of maximum entropy from a small set of natural axioms is obtained. Numerical results from applying the combined estimator to a couple of subnetwork examples show that an elastic O-D flow table, when used as input for subnetwork flow evaluations, reflects network flow changes significantly better than its fixed counterpart. The entropy estimator is then given by ... via least square method. If you are a registered author of this item, you may also want to check the "citations" tab in your RePEc Author Service profile, as there may be some citations waiting for confirmation. Minimum mean-square estimation suppose x ∈ Rn and y ∈ Rm are random vectors (not necessarily Gaussian) we seek to estimate x given y thus we seek a function φ : Rm → Rn such that xˆ = φ(y) is near x one common measure of nearness: mean-square error, Ekφ(y)−xk2 minimum mean-square estimator (MMSE) φmmse minimizes this quantity http://www.sciencedirect.com/science/article/pii/S0191261511000683, A maximum entropy-least squares estimator for elastic origin–destination trip matrix estimation, Transportation Research Part B: Methodological, The equilibrium-based origin-destination matrix estimation problem, Most likely origin-destination link uses from equilibrium assignment, Selection of a trip table which reproduces observed link flows, Inferences on trip matrices from observations on link volumes: A Bayesian statistical approach, Estimation of trip matrices from traffic counts and survey data: A generalized least squares estimator, A maximum likelihood model for estimating origin-destination matrices, A Relaxation Approach for Estimating Origin–Destination Trip Tables, On combining maximum entropy trip matrix estimation with user optimal assignment, An analysis of the reliability of an origin-destination trip matrix estimated from traffic counts, Variances and covariances for origin-destination flows when estimated by log-linear models, Estimation of an origin-destination matrix with random link choice proportions: A statistical approach, Inferring origin-destination trip matrices with a decoupled GLS path flow estimator, Estimation of origin-destination matrices from link traffic counts on congested networks, A linear programming approach for synthesizing origin-destination trip tables from link traffic volumes, Norm approximation method for handling traffic count inconsistencies in path flow estimator, The most likely trip matrix estimated from traffic counts, Subnetwork Origin-Destination Matrix Estimation Under Travel Demand Constraints, A decomposition approach to the static traffic assignment problem, Inferring origin-destination pairs and utility-based travel preferences of shared mobility system users in a multi-modal environment, User-equilibrium route flows and the condition of proportionality, An Excess-Demand Dynamic Traffic Assignment Approach for Inferring Origin-Destination Trip Matrices, Estimating the geographic distribution of originating air travel demand using a bi-level optimization model, Transportation Research Part E: Logistics and Transportation Review, Path Flow Estimator in an Entropy Model Using a Nonlinear L-Shaped Algorithm, http://www.elsevier.com/wps/find/journaldescription.cws_home/548/description#description, Xie, Chi & Kockelman, Kara M. & Waller, S. Travis, 2011. Properties of Least Squares Estimators Each ^ iis an unbiased estimator of i: E[ ^ i] = i; V( ^ i) = c ii˙2, where c ii is the element in the ith row and ith column of (X0X) 1; Cov( ^ i; ^ i) = c ij˙2; The estimator S2 = SSE n (k+ 1) = Y0Y ^0X0Y n (k+ 1) is an unbiased estimator of ˙2. General contact details of provider: http://www.elsevier.com/wps/find/journaldescription.cws_home/548/description#description . tity, and derive least squares as a special case. Copyright © 2020 Elsevier B.V. or its licensors or contributors. As the access to this document is restricted, you may want to search for a different version of it. +kbuk2 SSE +SSR; (2) where SST, SSE and SSR mean the total sum of squares, the explained sum of squares, and the residual sum of squares (or the sum of squared residuals), respectively. Improving entropy estimation and the inference of genetic regulatory networks. | 2023-02-04 15:28:58 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.6072761416435242, "perplexity": 2448.0032685664733}, "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/1674764500140.36/warc/CC-MAIN-20230204142302-20230204172302-00792.warc.gz"} |
https://homework.cpm.org/category/CCI_CT/textbook/apcalc/chapter/10/lesson/10.1.5/problem/10-56 | ### Home > APCALC > Chapter 10 > Lesson 10.1.5 > Problem10-56
10-56.
Solve for $y$ if $\frac { d y } { d x } = \frac { x y } { \operatorname { ln } ( y ) }$. Homework Help ✎
$\frac{\ln(y)}{y}dy=x dx$
Integrate. To integrate the left side of the equation, use substitution. Let $u = \ln(y)$.
$\frac{1}{2}(\ln(y))^2=\frac{1}{2}x^2+C$
Solve for $y$. | 2019-12-16 10:46:54 | {"extraction_info": {"found_math": true, "script_math_tex": 6, "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": 0.9881662130355835, "perplexity": 3494.201194058656}, "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/1575541319511.97/warc/CC-MAIN-20191216093448-20191216121448-00050.warc.gz"} |
https://learn.careers360.com/school/question-a-square-loop-of-side-10-cm-with-its-side-parallel-to-x-and-y-axes-is-moved-with-a-velocity-of-8-cm-s-1-in-the-positive-x-direction-containing-a-magnetic-field-in-positive-z-direction-the-field-is-non-uniform-and-has-a-gardient-of-cm-1-along-the-negative-x-direction-ie-it-increases-by-cm-1-as-one-moves-in-the-negative-x-direction-calculate-the-emf-induc/ | # A square loop of side 10 cm with its side parallel to X and Y axes is moved with a velocity of 8 cm s-1 in the positive X - direction containing a magnetic field in positive Z - direction. The field is non - uniform and has a gardient of $10^{-3}T$ cm -1 along the negative X - direction (i.e. it increases by $10^{-3}T$ cm -1 as one moves in the negative X- direction). Calculate the emf induced.
Given,
Side of square loop, l = 10cm = 0.10m
Area of loop = $0.10 \times 0.10 = 100 \times 10^{-4}m^{2}$
Velocity of the loop in the positive x- direction = 8cm = 0.08 m/s
Gradient of magnetic field in negative x-directioin = $\frac{dB}{dx}= 10^{-3} T /cm= 10^{-1} T /m$
So, induced emf due to changing positon $= \frac{d\phi }{dt}= \frac{d(BA)}{dt}= \frac{AdB}{dx}\frac{dx}{dt}$
$= \frac{AdB}{dx}V$
$=100 \times 10^{-4}\times 10 ^{-1}\times 0.08$
$=0.8 \times 10^{-4}V$
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₹ 4999/- ₹ 2999/- | 2020-07-13 17:34:15 | {"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": 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": 0.7794397473335266, "perplexity": 5244.475827217933}, "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/1593657146247.90/warc/CC-MAIN-20200713162746-20200713192746-00519.warc.gz"} |
https://edu-answer.com/mathematics/question3451422 | , 08.11.2019 05:31 deandrathomas34
# The circumference of a circle with a radius of 1 is 2π. set up a proportion to find the circumference of a circle with a radius of 6.
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What is the square root of 45? (step by step working plz)
An initial investment of $100 is now valued at$150. the annual interest rate is 5%, compounded continuously. the equation 100e0.05t = 150 represents the situation, where t is the number of years the money has been invested. about how long has the money been invested? use your calculator and round to the nearest whole number. years | 2021-01-24 19:02:52 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7745125889778137, "perplexity": 514.0979089357842}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-04/segments/1610703550617.50/warc/CC-MAIN-20210124173052-20210124203052-00321.warc.gz"} |
https://stats.stackexchange.com/questions/408575/mcmc-how-to-compute-prior | # Question
How to compute the prior $$P(𝜃)$$ and $$P(𝜃')$$ in MCMC when calculating the posteriors?
## Prior
I thought prior keeps updated with the accepted θ'. However, the way it is computed in the articles referenced seem to do different ways.
def prior(x):
#x[0] = mu, x[1]=sigma (new or current)
#returns 1 for all valid values of sigma. Log(1) =0, so it does not affect the summation.
#returns 0 for all invalid values of sigma (<=0). Log(0)=-infinity, and Log(negative number) is undefined.
#It makes the new sigma infinitely unlikely.
if(x[1] <=0):
return 0
return 1
MCMC sampling for dummies keeps using the initial prior, mu_prior_mu=0, mu_prior_sd=1. (In this article, log is applied to the P, hence using norm.pdf directly).
def sampler(data, samples=4, mu_init=.5, proposal_width=.5, plot=False, mu_prior_mu=0, mu_prior_sd=1.):
...
# Compute prior probability of current and proposed mu
prior_current = norm(mu_prior_mu, mu_prior_sd).pdf(mu_current)
prior_proposal = norm(mu_prior_mu, mu_prior_sd).pdf(mu_proposal)
Metropolis-Hastings sample also keep using the initial prior. In this article, the distribution is binomial so using beta function.
def target(lik, prior, n, h, theta):
if theta < 0 or theta > 1:
return 0
else:
return lik(n, theta).pmf(h)*prior.pdf(theta) # <--- prior = st.beta(a, b) where a = 10, b = 10
n = 100
h = 61
a = 10
b = 10
lik = st.binom
prior = st.beta(a, b)
sigma = 0.3
for i in range(niters):
theta_p = theta + st.norm(0, sigma).rvs()
rho = min(1, target(lik, prior, n, h, theta_p)/target(lik, prior, n, h, theta ))
u = np.random.uniform()
if u < rho:
naccept += 1
theta = theta_p
Are these correct ways? If so, kindly explain why no need to use the updated 𝜃.
Whenever we have a prior on a parameter $$p(\theta)$$, and some likelihood function $$p(\mathcal{D} | \theta)$$, the posterior is proportional to
$$\pi(\theta) \propto p(\mathcal{D} | \theta) p(\theta),$$
since the evidence does not depend on $$\theta$$ - it has been averaged out. Now, if you want to run a MCMC sampler that targets the density $$\pi(\theta)$$, you define a proposal mechanism $$\theta' \sim q(\theta'|\theta)$$. Then your Metropolis-Hastings ratio is
$$\frac{\pi(\theta')q(\theta|\theta')} {\pi(\theta)q(\theta'|\theta)} = \frac{p(\mathcal{D} | \theta') p(\theta')q(\theta|\theta')} {p(\mathcal{D} | \theta) p(\theta)q(\theta'|\theta)},$$
so as you can see the prior does enter into the acceptance ratio. This also illustrates a nice principle if you choose your prior as your proposal mechanism, such that $$\theta' \sim p(\theta')$$. In this particular case the priors cancel out with the proposal ratio, and you are now just comparing the likelihood of the relative points. | 2019-08-18 03:46:27 | {"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": 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": 10, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8283129930496216, "perplexity": 3023.4575869279242}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-35/segments/1566027313589.19/warc/CC-MAIN-20190818022816-20190818044816-00362.warc.gz"} |
https://www.zbmath.org/serials/?q=se%3A00002241 | ## Mathematical Population Studies
Short Title: Math. Popul. Stud. Publisher: Taylor & Francis, Philadelphia, PA ISSN: 0889-8480; 1547-724X/e Online: http://www.tandfonline.com/loi/gmps20
Documents Indexed: 373 Publications (since 1988) References Indexed: 350 Publications with 8,732 References.
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### Authors
11 Rogers, Andrei 10 Milner, Fabio Augusto 8 Schoen, Robert 7 Bonneuil, Noël 7 Inaba, Hisashi 6 Feichtinger, Gustav 6 Yashin, Anatoli I. 4 Alho, Juha M. 4 de Palma, André 4 Hritonenko, Natali 4 Iachine, Ivan A. 4 Molchanov, Stanislav Alekseevich 4 Tuljapurkar, Shripad D. 4 Wachter, Kenneth Willcox 4 Yatsenko, Yuri 3 Akushevich, Igor 3 Artzrouni, Marc 3 Boucekkine, Raouf 3 Chernousova, Elena 3 de Lapparent, Matthieu 3 Foryś, Urszula 3 Goldman, Noreen 3 Hu, Guihua 3 Kostaki, Anastasia 3 Liaw, Kao-Lee 3 Manton, Kenneth G. 3 Martcheva, Maia 3 Menold, Natalja 3 Moghadas, Seyed M. 3 Picard, Nathalie 3 Piotrowska, Monika Joanna 3 Prskawetz, Alexia 3 Smith?, Robert Joseph 3 Vaupel, James W. 3 Willekens, Frans 3 Yang, Hyun Mo 3 Żądło, Tomasz 2 Abdelfatah, Sally 2 Ahmad, Ishfaq 2 Almanjahie, Ibrahim Mufrah 2 Amaku, Marcos 2 Billari, Francesco C. 2 Bodnar, Marek 2 Bogner, Kathrin 2 Bracher, Michael 2 Brouard, Nicolas 2 Carter, Lawrence R. 2 Chandra, Hukum 2 Chi, Lujie 2 Cohen, Joel E. 2 Comiskey, Catherine M. 2 de la Croix, David 2 Demongeot, Jacques 2 Dempsey, Orla 2 Fabbri, Giorgio 2 Faggian, Silvia 2 Fan, Shushan 2 Ghosh, Malay 2 Gill, Richard David 2 Gozzi, Fausto 2 Guha, Saurav 2 Heathcote, Christopher R. 2 Heffernan, Jane Marie 2 Hryniv, Ostap 2 Hu, Min 2 Iannelli, Mimmo 2 Kealey, Alison D. 2 Keilman, Nico 2 Keyfitz, Nathan 2 Kulminski, Alexander M. 2 Laffargue, Jean-Pierre 2 Larsen, Ulla 2 Li, Xining 2 Liao, Jinpen 2 Licandro, Omar 2 Little, Jani S. 2 Manfredi, Piero 2 Mazloum, Reda I. 2 McNown, Robert 2 Mielecka-Kubien, Zofia 2 Morris, Martina 2 Norden, R. H. 2 Pollard, John H. 2 Qi, Li 2 Raymer, James 2 Rodriguez, German 2 Rundell, William 2 Shahzad, Usman 2 Steinmann, Gunter 2 Teboh-Ewungkem, Miranda I. 2 Tridane, Abdessamad 2 van Imhoff, Evert 2 Veliov, Vladimir M. 2 Whitmeyer, Joseph M. 2 Wu, Jianhong 2 Zhang, Qimin 1 Aarssen, Karin 1 Abravesh, Akbar 1 Adegboye, Oyelola A. 1 Aghajanian, Akbar ...and 477 more Authors
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### Fields
192 Game theory, economics, finance, and other social and behavioral sciences (91-XX) 165 Biology and other natural sciences (92-XX) 102 Statistics (62-XX) 33 Probability theory and stochastic processes (60-XX) 26 General and overarching topics; collections (00-XX) 24 Ordinary differential equations (34-XX) 19 Numerical analysis (65-XX) 11 Partial differential equations (35-XX) 11 Dynamical systems and ergodic theory (37-XX) 10 Calculus of variations and optimal control; optimization (49-XX) 10 Operations research, mathematical programming (90-XX) 6 Systems theory; control (93-XX) 5 Integral equations (45-XX) 3 Combinatorics (05-XX) 2 Computer science (68-XX) 1 Linear and multilinear algebra; matrix theory (15-XX) 1 Difference and functional equations (39-XX) 1 Operator theory (47-XX)
### Citations contained in zbMATH Open
197 Publications have been cited 867 times in 724 Documents Cited by Year
A multi-city epidemic model. Zbl 1028.92021
Arino, Julien; van den Driessche, P.
2003
A semigroup approach to the strong ergodic theorem of the multistate stable population process. Zbl 0900.92122
Inaba, Hisashi
1988
Correlated individual frailty: An advantageous approach to survival analysis of bivariate data. Zbl 0873.62121
Yashin, Anatoli I.; Vaupel, James W.; Iachine, Ivan A.
1995
Strange periodic attractors in a prey-predator system with infected prey. Zbl 1157.92324
Hilker, Frank M.; Malchow, Horst
2006
Extending Lee–Carter mortality forecasting. Zbl 1151.91742
de Jong, Piet; Tickle, Leonie
2006
Mortality modeling: A review. Zbl 0984.92027
Yashin, Anatoli I.; Iachine, Ivan A.; Begun, Alexander S.
2000
On the maximal life span of humans. Zbl 0872.62096
1994
Age patterns of mortality for older women: An analysis using the age-specific rate of mortality change with age. Zbl 0875.92063
Horiuchi, Shiro; Coale, Ansley J.
1990
Global stability analysis of a metapopulation SIS epidemic model. Zbl 1382.92238
Iggidr, Abderrahman; Sallet, Gauthier; Tsanou, Berge
2012
HIV vaccines: The effect of the mode of action on the coexistence of HIV subtypes. Zbl 0961.92021
Porco, Travis C.; Blower, Sally M.
2000
Understanding mortality rate deceleration and heterogeneity. Zbl 1151.91741
Steinsaltz, David R.; Wachter, Kenneth W.
2006
Maximum principle for age and duration structured systems: a tool for optimal prevention and treatment of HIV. Zbl 1053.92039
Feichtinger, Gustav; Tsachev, Tsvetomir; Veliov, Vladimir M.
2004
Optimal harvesting of forest age classes: a survey of some recent results. Zbl 1093.91535
Tahvonen, Olli
2004
On the dynamic programming approach for optimal control problems of PDE’s with age structure. Zbl 1059.49005
Faggian, Silvia; Gozzi, Fausto
2004
Modelling deceleration in senescent mortality. Zbl 1210.91113
Bebbington, Mark; Lai, Chin-Diew; Zitikis, Ričardas
2011
Determining the birth function for an age structured population. Zbl 0900.92129
Rundell, William
1989
A multi-patch epidemic model with periodic demography, direct and indirect transmission and variable maturation rate. Zbl 1107.92050
Wolf, Cédric; Langlais, Michel; Sauvage, Frank; Pontier, Dominique
2006
S-I-R model with directed spatial diffusion. Zbl 1151.92322
Milner, Fabio A.; Zhao, Ruijun
2008
Immune response and imperfect vaccine in malaria dynamics. Zbl 1215.92034
Niger, Ashrafi M.; Gumel, Abba B.
2011
Models and proposals for malaria: a review. Zbl 1409.92257
Teboh-Ewungkem, Miranda I.; Ngwa, Gideon A.; Ngonghala, Calistus N.
2013
Modelling vintage structures with DDEs: principles and applications. Zbl 1093.91064
Boucekkine, Raouf; de la Croix, David; Licandro, Omar
2004
A method for assessing the global spread of HIV-1 infection based on air travel. Zbl 0900.92100
Flahault, A.; Valleron, A.-J.
1992
Survival of related individuals: An extension of some fundamental results of heterogeneity analysis. Zbl 0873.62122
Yashin, Anatoli I.; Iachine, Ivan A.
1995
Vector consumption and contact process saturation in sylvatic transmission of $$T$$. cruzi. Zbl 1107.92048
Kribs-Zaleta, Christopher
2006
Biomechanical and nutrient controls in the growth of mammalian cell populations. Zbl 1195.92016
Hoehme, Stefan; Drasdo, Dirk
2010
Maximum principle for optimal harvesting in linear size-structured population. Zbl 1166.92322
2008
Asymptotic properties of the inhomogeneous Lotka-Von Foerster system. Zbl 0900.92128
Inaba, Hisashi
1988
Are mortality projections always more pessimistic when disaggregated by cause of death? Zbl 0876.92032
Wilmoth, John R.
1995
Age-structured PDEs in economics, ecology, and demography: optimal control and sustainability. Zbl 1204.91072
Hritonenko, Natali; Yatsenko, Yuri
2010
The mathematics of sex and marriage, revisited. Zbl 0990.91052
Martcheva, Maia; Milner, Fabio A.
2001
Branching random walks with several sources. Zbl 1409.92216
Yarovaya, Elena B.
2013
Stability analysis and dynamics preserving nonstandard finite difference schemes for a malaria model. Zbl 1409.92221
Anguelov, Roumen; Dumont, Yves; Lubuma, Jean; Mureithi, Eunice
2013
Evaluating the expected time to population extinction with semi-stochastic models. Zbl 1173.90577
Cairns, Benjamin J.
2009
A behavioral two-sex marriage model. Zbl 0990.91051
Dagsvik, John K.; Brunborg, Helge; Flaatten, Ane S.
2001
An age and spatially structured model of tumor invasion with haptotaxis. II. Zbl 1145.92016
Dyson, Janet; Villella-Bressan, Rosanna; Webb, Glenn
2008
Turbulent dynamics in a XVIIth century population. Zbl 0875.92065
Bonneuil, Noël
1990
A two-sex age-structured population model: well posedness. Zbl 0973.92031
Martcheva, Maia; Milner, Fabio A.
1999
Transmission of tuberculosis with exogenous re-infection and endogenous reactivation. Zbl 1105.92010
Raimundo, Silvia Martorano; Yang, Hyun Mo
2006
Waiting time models of cancer progression. Zbl 1195.92035
Gerstung, Moritz; Beerenwinkel, Niko
2010
Growth economics of epidemics: A review of the theory. Zbl 1138.91028
Boucekkine, Raouf; Diene, Bity; Azomahou, Théophile
2008
Emergence of resistance in influenza with compensatory mutations. Zbl 1215.92033
2011
Two-patch transmission of tuberculosis. Zbl 1223.92036
Tewa, Jean Jules; Bowong, Samuel; Mewoli, Boulchard; Kurths, Jurgen
2011
Demographic change and immigration in age-structured epidemic models. Zbl 1134.92356
Iannelli, Mimmo; Manfredi, Piero
2007
Stationary distributions in Kolmogorov-Petrovski-Piskunov-type models with an infinite number of particles. Zbl 1409.92204
Molchanov, Stanislav; Whitmeyer, Joseph
2017
Debilitation’s aftermath: Stochastic process models of mortality. Zbl 0900.92123
Vaupel, James W.; Yashin, Anatoli I.; Manton, Kenneth G.
1988
Empirical methods for the estimation of the mixing probabilities for socially structured populations from a single survey. Zbl 0900.92140
Blythe, S. P.; Castillo-Chavez, C.; Casella, G.
1992
Feedback spreading control applied to immunotherapy. Zbl 1088.92038
Kassara, Khalid
2005
Dynamics of Hepatitis B virus infection: what causes viral clearance? Zbl 1215.92027
Ciupe, Stanca M.; Catllá, Anne J.; Forde, Jonathan; Schaeffer, David G.
2011
Optimal forest management in the presence of intraspecific competition. Zbl 1222.91042
Goetz, Renan U.; Xabadia, Angels; Calvo, Elena
2011
Contextual interventions for controlling alcohol drinking. Zbl 1409.91207
Mubayi, Anuj; Greenwood, Priscilla E.
2013
A simple in-host model for Mycobacterium tuberculosis that captures all infection outcomes. Zbl 1409.92233
Du, Yimin; Wu, Jianhong; Heffernan, Jane M.
2017
A nine-parameter version of the Heligman-Pollard formula. Zbl 0900.92142
Kostaki, A.
1992
Human population reproduction via first marriage. Zbl 0872.92030
Inaba, Hisashi
1995
Populations with quadratic exponential growth. Zbl 0879.92036
Kim, Young J.; Schoen, Robert
1996
Parameterizing age patterns of demographic rates with the multiexponential model schedule. Zbl 0877.92038
Rogers, Andrei; Little, Jani S.
1994
On the estimation of multi-dimensional demographic models with population registration data. Zbl 0875.92061
Gill, Richard; Keilman, Nico
1990
Modeling the TB/HIV-1 co-infection and the effects of its treatment. Zbl 1183.92048
Magombedze, Gesham; Garira, Winston; Mwenje, Eddie
2010
Endogenous retirement and monetary cycles. Zbl 1155.91434
D’Albis, Hippolyte; Augeraud-Vèron, Emmanuelle
2008
On dynamic programming in economic models governed by ddes. Zbl 1155.91428
Fabbri, Giorgio; Faggian, Silvia; Gozzi, Fausto
2008
Controlling dispersal dynamics of Aedes aegypti. Zbl 1105.92009
Ferreira, Cláudia Pio; Pulino, Petronio; Yang, Hyun Mo; Takahashi, Lucy Tiemi
2006
Viability of small populations experiencing recurring catastrophes. Zbl 1175.92039
Jagers, Peter; Harding, Karin C.
2009
Viscosity solutions to delay differential equations in demo-economy. Zbl 1140.91058
Fabbri, Giorgio
2008
The contact network of inpatients in a regional healthcare system. A longitudinal case study. Zbl 1187.91165
Liljeros, Fredrik; Giesecke, Johan; Holme, Petter
2007
On a pandemic threshold theorem of the early Kermack-McKendrick model with individual heterogeneity. Zbl 1409.92239
Inaba, Hisashi
2014
Editorial: Delay differential equations in bio-populations. Zbl 1409.00111
2014
Editorial: Dealing with nonresponse: strategies to increase participation and methods for postsurvey adjustments. Zbl 1409.00113
2017
Persistent age distributions for an age-structured two-sex population model. Zbl 0962.91074
Inaba, Hisashi
2000
Sex ratio at birth and son preference. Zbl 0967.91075
Li, Nan; Feldman, Marcus W.; Tuljapurkar, Shripad
2000
Nonlinear structured population dynamics with covariates. Zbl 0981.92024
Aubin, Jean-Pierre; Bonneuil, Noël; Maurin, Franck
2000
A Markov chain for calculating the durability of marriage. Zbl 0900.92121
Keyfitz, Nathan
1988
The effect of variability in the fertility schedule on numbers of kin. Zbl 0900.92127
Coresh, Josef; Goldman, Noreen
1988
Determining the initial age distribution for an age structured population. Zbl 0900.92133
Pilant, M.; Rundell, W.
1991
Malthus, Boserup and population viability. Zbl 0873.92031
Bonneuil, Noël
1994
Disaggregation in population forecasting: Do we need it? And how to do it simply. Zbl 0900.92192
Lee, Ronald D.; Carter, Lawrence; Tuljapurkar, Shripad
1995
An age-structured model of population dynamics with dominant ages, delayed behavior, and oscillations. Zbl 0872.92011
Kostova, Tanya; Milner, Fabio A.
1995
Effects of targets and aggregation on the propagation of error in mortality forecasts. Zbl 0875.92057
Alho, Juha M.; Spencer, Bruce D.
1990
Transmission probabilities and reproduction numbers for sexually transmitted infections with variable infectivity: application to the spread of HIV between low- and high-activity populations. Zbl 1182.92060
Artzrouni, Marc
2009
A host-host-pathogen model with vaccination and its application to target and reservoir hosts. Zbl 1109.92042
Norman, R.; Bowers, R. G.
2007
Life tables with covariates: dynamic model for nonlinear analysis of longitudinal data. Zbl 1111.91033
Akushevich, I.; Kulminski, A.; Manton, K. G.
2005
The stochastic inverse projection and the population of Velletri (1590–1870). Zbl 1046.91102
Bertino, Salvatore; Sonnino, Eugenio
2003
Asymptotic behavior of the solutions to semi-linear age-dependent population dynamics with diffusion and periodic vital rates. Zbl 1187.37119
Aniţa, Laura-Iulia; Aniţa, Sebastian
2008
Bayesian melding estimation of a stochastic SEIR model. Zbl 1408.92031
Hotta, Luiz K.
2010
The optimal trade-off between quality and quantity with unknown number of survivors. Zbl 1409.91196
Baudin, Thomas
2012
Control strategies for the spread of malaria in humans with variable attractiveness. Zbl 1409.92144
Agusto, Folashade B.; Tchuenche, Jean M.
2013
Human migrations and mosquito-borne diseases in Africa. Zbl 1409.92241
Kim, Sehjeong; Tridane, Abdessamad; Chang, Dong Eui
2016
Tuberculosis with relapse: a model. Zbl 1409.92259
Yang, Yali; Wu, Jianhong; Li, Jianquan; Xu, Xiaxia
2017
Drug resistance in an age-of-infection model. Zbl 1409.92229
Brauer, Fred; Xiao, Yanyu; Moghadas, Seyed M.
2017
Special issue: The population dynamics of the HIV epidemic: projections. Zbl 0967.00047
2000
Multistate models of postpartum infecundity, fecundability and sterility by age and parity: Methodological issues. Zbl 0894.92045
Yashin, Anatoli I.; Iachine, Ivan A.; Andreev, Kirill F.; Larsen, Ulla
1998
A stochastic model for the study of last closed birth interval with some biosocial components. Zbl 0894.92044
Pandey, Arvind; Dwivedi, S. N.; Mishra, R. N.
1998
Location of adult children as an attraction for black and white elderly return and onward migrants in the United States: application of a three-level nested logit model with census data. Zbl 1106.91373
Liaw, Kao-Lee; Frey, William H.
2003
Human capital, technological progress and the demographic transition. Zbl 0960.91070
Prskawetz, A.; Steinmann, G.; Feichtinger, G.
2000
Macro-demographic effects of the transition to adulthood: Multistate stable population theory and an application to Italy. Zbl 1013.91529
Billari, Francesco C.; Manfredi, Piero; Valentini, Alessandro
2000
Endogenous networks in random population games. Zbl 1088.91010
Fagiolo, Giorgio; Marengo, Luigi; Valente, Marco
2004
Preserving transfer benefit for present and future generations. Zbl 1093.91522
Bonneuil, Noël; Boarini, Romina
2004
Special issue: Age structured models in population dynamics and economics. Selected papers based on the presentation at the workshop, Vienna, Austria, October 27–28, 2003. Zbl 1075.91500
2004
The multistate life table with duration-dependence. Zbl 0900.62527
Wolf, Douglas A.
1988
Leslie matrix models. Zbl 0900.92131
Hansen, Poul Einer
1989
Effect of aggregation on the estimation of trend in mortality. Zbl 0900.92134
Alho, J. M.
1991
Pre-procreative ages in population stability and cyclicity. Zbl 0900.92137
Wachter, K. W.
1991
Steady states of lattice population models with immigration. Zbl 1483.91150
Chernousova, Elena; Feng, Yaqin; Hryniv, Ostap; Molchanov, Stanislav; Whitmeyer, Joseph
2021
Catalytic branching random walk with semi-exponential increments. Zbl 07479176
Bulinskaya, Ekaterina Vl.
2021
Estimation of the population mean by successive use of an auxiliary variable in median ranked set sampling. Zbl 1483.62032
2021
Population model with immigration in continuous space. Zbl 1483.91151
Chernousova, Elena; Hryniv, Ostap; Molchanov, Stanislav
2020
Double-sampling regression-cum-exponential estimator of the mean of a sensitive variable. Zbl 07481033
Saleem, Iram; Sanaullah, Aamir; Hanif, Muhammad
2019
Steady state and intermittency in the critical branching random walk with arbitrary total number of offspring. Zbl 07481036
Chernousova, Elena; Molchanov, Stanislav
2019
Stationary distributions in Kolmogorov-Petrovski-Piskunov-type models with an infinite number of particles. Zbl 1409.92204
Molchanov, Stanislav; Whitmeyer, Joseph
2017
A simple in-host model for Mycobacterium tuberculosis that captures all infection outcomes. Zbl 1409.92233
Du, Yimin; Wu, Jianhong; Heffernan, Jane M.
2017
Editorial: Dealing with nonresponse: strategies to increase participation and methods for postsurvey adjustments. Zbl 1409.00113
2017
Tuberculosis with relapse: a model. Zbl 1409.92259
Yang, Yali; Wu, Jianhong; Li, Jianquan; Xu, Xiaxia
2017
Drug resistance in an age-of-infection model. Zbl 1409.92229
Brauer, Fred; Xiao, Yanyu; Moghadas, Seyed M.
2017
Modeling distances between humans using Taylor’s law and geometric probability. Zbl 1409.91202
Cohen, Joel E.; Courgeau, Daniel
2017
Human migrations and mosquito-borne diseases in Africa. Zbl 1409.92241
Kim, Sehjeong; Tridane, Abdessamad; Chang, Dong Eui
2016
An efficient two-stage randomized response model under stratified random sampling. Zbl 1416.62071
Abdelfatah, Sally; Mazloum, Reda
2016
Mixed-mode oscillations due to a singular Hopf bifurcation in a forest pest model. Zbl 1409.92230
Brøns, Morten; Desroches, Mathieu; Krupa, Martin
2015
Modeling and control of malaria when mosquitoes are used as vaccinators. Zbl 1409.92243
Li, Xue-Zhi; Gao, Sha-Sha; Martcheva, Maia
2015
Efficient estimation in a two-stage randomized response model. Zbl 1409.62029
Abdelfatah, Sally; Mazloum, Reda
2015
On a pandemic threshold theorem of the early Kermack-McKendrick model with individual heterogeneity. Zbl 1409.92239
Inaba, Hisashi
2014
Editorial: Delay differential equations in bio-populations. Zbl 1409.00111
2014
Special issue: Survey sampling methods. Zbl 1409.00106
2014
Estimators for the Horvitz-Thompson statistic based on some posterior distributions. Zbl 1409.62036
Gamrot, Wojciech
2014
On the prediction of the subpopulation total based on spatially correlated longitudinal data. Zbl 1409.62042
Żądło, Tomasz
2014
Logistic tumor growth with delay and impulsive treatment. Zbl 1409.92114
Foryś, Urszula; Poleszczuk, Jan; Liu, Ting
2014
Variability of travel time, congestion, and the cost of travel. Zbl 1409.90051
Coulombel, Nicolas; de Palma, André
2014
Models and proposals for malaria: a review. Zbl 1409.92257
Teboh-Ewungkem, Miranda I.; Ngwa, Gideon A.; Ngonghala, Calistus N.
2013
Branching random walks with several sources. Zbl 1409.92216
Yarovaya, Elena B.
2013
Stability analysis and dynamics preserving nonstandard finite difference schemes for a malaria model. Zbl 1409.92221
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437 Biology and other natural sciences (92-XX) 215 Game theory, economics, finance, and other social and behavioral sciences (91-XX) 157 Statistics (62-XX) 115 Ordinary differential equations (34-XX) 87 Partial differential equations (35-XX) 76 Probability theory and stochastic processes (60-XX) 48 Calculus of variations and optimal control; optimization (49-XX) 47 Dynamical systems and ergodic theory (37-XX) 44 Numerical analysis (65-XX) 25 Systems theory; control (93-XX) 17 Integral equations (45-XX) 16 Operator theory (47-XX) 14 Operations research, mathematical programming (90-XX) 8 General and overarching topics; collections (00-XX) 5 Real functions (26-XX) 4 Combinatorics (05-XX) 4 Computer science (68-XX) 2 History and biography (01-XX) 2 Linear and multilinear algebra; matrix theory (15-XX) 2 Difference and functional equations (39-XX) 2 Functional analysis (46-XX) 2 Mechanics of deformable solids (74-XX) 2 Statistical mechanics, structure of matter (82-XX) 1 Field theory and polynomials (12-XX) 1 Measure and integration (28-XX) 1 Abstract harmonic analysis (43-XX) 1 Integral transforms, operational calculus (44-XX) 1 Algebraic topology (55-XX) 1 Classical thermodynamics, heat transfer (80-XX) 1 Quantum theory (81-XX) 1 Geophysics (86-XX) 1 Mathematics education (97-XX) | 2022-05-26 06:14:44 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.46539145708084106, "perplexity": 11937.984473437753}, "config": {"markdown_headings": true, "markdown_code": false, "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/1652662601401.72/warc/CC-MAIN-20220526035036-20220526065036-00101.warc.gz"} |
https://www.coursehero.com/file/68635500/HeatEquationGuidepdf/ | HeatEquationGuide.pdf - syms x t �fines x and t as variables L=pi%Sets the length of the rod beta=3%Sets the physical parameter n=15 �cides how many
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syms x t %Defines x and t as variables L=pi; %Sets the length of the rod beta=3; %Sets the physical parameter n=15; %Decides how many terms of the Fourier series to calculate finit=x^2; %Defines the initial heat distribution in the rod fplot(finit,[0,pi]); % Plots the initial heat distribution a_n=zeros(1,n); %Defines a list of constants for the Fourier series coefficients Fourier_n=0; %Sets a function for the Fourier series solution Fourier_init=0; %Sets a function for the Fourier series initial condition
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https://deepai.org/publication/analysis-of-a-stabilised-finite-element-method-for-power-law-fluids | DeepAI
Analysis of a stabilised finite element method for power-law fluids
A low-order finite element method is constructed and analysed for an incompressible non-Newtonian flow problem with power-law rheology. The method is based on a continuous piecewise linear approximation of the velocity field and piecewise constant approximation of the pressure. Stabilisation, in the form of pressure jumps, is added to the formulation to compensate for the failure of the inf-sup condition, and using an appropriate lifting of the pressure jumps a divergence-free approximation to the velocity field is built and included in the discretisation of the convection term. This construction allows us to prove the convergence of the resulting finite element method for the entire range r>2 d/d+2 of the power-law index r for which weak solutions to the model are known to exist in d space dimensions, d ∈{2,3}.
• 5 publications
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1. Introduction
The construction and mathematical analysis of finite element approximations of models of non-Newtonian fluids has been a subject of active research in recent years. Some of the most general results in this direction concern the convergence of mixed finite element approximations of models of incompressible fluids with implicit constitutive laws relating the Cauchy stress tensor to the symmetric velocity gradient (cf.
[15], [33] and [18]). Motivated by the groundbreaking contributions of Cohen, Dahmen and DeVore [12, 13] and Binev, Dahmen and DeVore [8] concerning the convergence of adaptive algorithms for linear elliptic problems, progress, albeit much more limited in both scope and extent, has also been made on the analysis of adaptive finite element approximations of implicitly constituted non-Newtonian fluid flow models (cf. [26]).
Upon decomposing the Cauchy stress tensor into its traceless part, called the deviatoric stress tensor or shear-stress tensor, and its diagonal part, called the mean normal stress, models of incompressible fluids typically involve the velocity of the fluid, , its pressure, , and the shear-stress tensor, . For Newtonian fluids the shear-stress tensor is a scalar multiple of the symmetric velocity gradient. The finite element approximation of Newtonian fluids is therefore usually performed in the velocity-pressure formulation. For non-Newtonian fluids on the other hand the situation is more involved, because the shear-stress tensor exhibits nonlinear dependence as a function of the symmetric velocity gradient, and the functional relationship between the shear-stress tensor and the symmetric velocity gradient may even be completely implicit and multi-valued. For power-law fluids, such as the ones considered in this work, the shear-stress tensor exhibits power-law type growth as a function of the symmetric velocity gradient, the simplest instance of which results in an -Laplace type operator in the balance of linear momentum equation, with a power-law exponent ; for , corresponding to a Newtonian fluid, the operator is linear, the Laplace operator. From a mathematical point of view, in the presence of a convection term in the balance of linear momentum equation in the model, the lower the value of the more difficult the problem is to analyse. The existence of solutions for small values of was first proved in [19], where an Acerbi–Fusco type Lipschitz truncation was used in conjunction with Minty’s method from monotone operator theory; thus, weak solutions were shown to exist for in space dimensions.
Finite element approximations of problems with power-law rheology have been extensively studied, including stabilised (or variational-multiscale) methods (cf. [10, 1], for example) and local discontinuous Galerkin methods (see, [27], for example). The relevant literature is vast and it is beyond the scope of this work to provide an exhaustive survey of the various contributions; the interested reader may wish to consult [29], for example. Concerning implicitly-constituted models, in the recent papers [15, 33] the convergence of generic inf-sup stable velocity/pressure-based mixed finite element methods was proved for , while convergence for the full range, , was shown to be achievable only in the case the finite element methods where the velocity space consists of pointwise divergence-free functions. The reason for this dichotomy is that in the case of velocity approximations that are discretely divergence-free only, as is the case in generic inf-sup stable mixed finite element methods, the finite element approximation of the convection term does not vanish when tested with
, and it needs to be skew-symmetrized (cf.
[34]) for this to happen. While in the case of the Navier–Stokes equations (corresponding to ) membership of the velocity field to the natural function space for weak solutions, , ensures that the convection term and its skew-symmetric modification can be bounded by the same expression using Hölder’s inequality, this is not the case for the power-law model under consideration here for entire range for which weak solutions to the problem are known to exist. In fact, in the case of non-Newtonian power-law models the natural function space for the velocity field is , and while the original convection term can be bounded in terms of the norm for all , for the skew-symmetric modification of the convection term, whose use is essential so as to be able to derive an energy inequality for discretely divergence-free velocity fields, this can only be achieved for the limited range . This was precisely the bottleneck encountered in [15] for discretely divergence-free velocity approximations, resulting in the reduction of the range of from the maximal range for which weak solutions are known to exist, to
The advantage of pointwise divergence-free finite element methods over discretely divergence-free finite element methods is therefore that, besides the physical consistency they provide, there is no need to rewrite the convection term in a skew-symmetric form. The topic of divergence-free finite element spaces has been treated extensively in the literature, most commonly presenting pairs of spaces for which the divergence of the velocity space is a subspace of, or equal to, the pressure space. For example, the early Scott–Vogelius element [32] (analysed recently in [24]) uses -conforming piecewise polynomials of degree for the velocity, while discontinuous piecewise polynomials of degree are used for the pressure. The stability of this pair requires either special meshes, or a high-enough degree (for example, is needed in [24], and is only allowed in very special cases such as those described in Remark 3). Another possibility is to relax the continuity requirements and consider a discontinuous Galerkin method, as was done, for example, in [11], or to relax only the tangential continuity of the approximate velocity on faces of elements while still preserving its continuity in the direction of the normal to faces of elements, thus using -conforming methods, as was the case in [31], for example. In this latter case the viscous term (defined as the divergence of the shear-stress needs to be modified, for stability reasons, by adding terms controlling the jumps and averages of the velocity into the formulation, with, obviously undesirable, extra complications if the viscous term in the balance of linear momentum equation has a more complex structure, as is the case for the power-law model considered herein.
The recent works [2, 3] offer a way of preserving the advantages of a pointwise divergence-free approximation to the velocity field while working with the, computationally simplest, lowest-order -conforming velocity/pressure pair, namely, . The key idea in those works can be summarised as follows: the discrete continuity equation contains a stabilising term based on the jumps of the discrete pressure. As the jumps of the pressure are constant along element faces, there exists a unique Raviart–Thomas field such that its normal component is equal to the jumps. This field can be built at no extra computational cost, and then the continuity equation can be rewritten as a standard continuity equation, but for a modified velocity field, which is now solenoidal. The finite element method then involves replacing the original discrete velocity field with the new, now solenoidal, modified velocity field in the convection term. This facilitates the proofs of stability and convergence of the resulting finite element method without the need to rewrite the convection term in a skew-symmetric form. Our aim here is to apply this idea to a problem in non-Newtonian fluid mechanics. As a first step in this direction, we have chosen an explicit constitutive law with power-law rheology. Even though this is the simplest constitutive law, it has been shown experimentally to faithfully reproduce many situations of physical interest (see the discussion in [22], and the experimental results in, e.g., [25]); we therefore believe that it is a representative model for exemplifying the applicability of the proposed method in a mathematically nontrivial and physically relevant setting. Since the convection term does not need to be rewritten in a skew-symmetric form, the resulting method can now be proved to be stable and convergent to a weak solution for the whole range of the power-law index for which weak solutions to the model are known to exist. In addition, the sequence of numerical approximations is shown to converge strongly, and this strong convergence result is, to the best of our knowledge, a new contribution even in the, very special, Newtonian case ().
The rest of the manuscript is organised as follows. A section on preliminaries, containing the necessary notational conventions, basic definitions and results, the finite element spaces, the lifting operator, the definition of the stabilising form, and properties of the discrete Lipschitz truncation method that we use, are presented following this Introduction. An important ingredient enabling the use of the discrete Lipschitz truncation technique is a discrete inf-sup condition that is given in the Appendix. The finite element method is presented in Section 3, where we also show a uniform boundedness result for the sequence of approximations. Based on this and results pertaining to the discrete Lipschitz truncation, in Section 4 the convergence of the discrete solution to a weak solution of the model problem is proved using a compactness argument. Finally, some conclusions are drawn and potential future extensions are indicated.
2. Preliminaries
2.1. Notation and the problem of interest
We use standard notation for Sobolev spaces. In particular, for , and , we denote by the closure of with respect to the norm, and by the space of functions in with zero mean value. The norm in is denoted by ; when we shall use the simpler notation , and the inner product in will be denoted by . For , the norm (seminorm) in is denoted by (). Moreover, the space is the dual of with duality pairing denoted by . We also denote by the space of functions in whose distributional divergence belongs to , and by the set of elements in whose normal trace on
is zero. In the above inner products and norms we do not make a distinction between scalar- and vector- or tensor-valued functions.
Let be an open, bounded, polyhedral domain with a Lipschitz boundary. In this work we treat the problem with power-law rheology: given and a right-hand side , find the velocity , the pressure , and the shear-stress tensor satisfying
(2.1) ⎧⎪⎨⎪⎩−divS+div(u⊗u)+∇p=fin Ω,divu=0in Ω,u=0on ∂Ω.
There are many possible choices for the constitutive law, linking and the velocity . In this work we have chosen the power-law description where , where is a reference viscosity. In order to simplify matters we will suppose that , but we should keep in mind that, to maintain physical consistency this reference value should be kept. Similarly, in physically realistic models the gradient of the velocity is usually replaced by the symmetric velocity gradient . The results obtained in this paper can be extended, with minor modifications based on Korn’s inequality, to that case as well, so for the sake of simplicity of the exposition we shall proceed with the constitutive relation (with ) instead of .
In order to state the weak formulation of (2.1) we need to present a few additional ingredients associated with the exponent in the constitutive law relating and . For , let be its conjugate given by the relation , and let us define the critical exponent as follows:
(2.2) ~r:=min{r′,r⋆2},% wherer⋆:=⎧⎪⎨⎪⎩∞ifr≥d,drd−rotherwise.
Remark 1.
With the definition (2.2) of , the space is continuously embedded in if and in , for every , if (see, e.g., [9, Corollary 9.14]). Then, in particular, is continuously embedded in and there exists a such that
(2.3) ∥v∥0,2~r,Ω≤C∥v∥1,r,Ω∀v∈W1,r(Ω).
Moreover, the value of exhibits two different regimes, as can be seen in Figure 1, where its range of values is depicted. We will distinguish between and . The latter case occurs for and the maximum value of is attained when , at which point we have the following values:
(2.4) r=3dd+2=⎧⎪ ⎪ ⎪⎨⎪ ⎪ ⎪⎩32ifd=2,95ifd=3,and~rmax=3d2d−2={3ifd=2,94ifd=3.
With this choice of stress tensor , the weak formulation of (2.1) is as follows: find and such that
(2.5) (|∇u|r−2∇u,∇v)Ω−(u⊗u,∇v)Ω−(p,divv)Ω =⟨f,v⟩Ω ∀v∈W1,~r′0(Ω)d, (2.6) (q,divu)Ω =0 ∀q∈Lr′0(Ω).
Remark 2.
In order for the variational formulation (2.5), (2.6) to be meaningful it is necessary that with , which necessitates that , and under this condition the existence of a solution to (2.5), (2.6) has been proved (see [16]). Thus, for the rest of this work we will assume that .
Another fundamental ingredient in the proof of existence of solutions to (2.5), (2.6) is the following inf-sup condition (for a proof, see [20]): for satisfying , there exists a constant such that
(2.7) supv∈W1,s′0(Ω)d∖{0}(q,divv)Ω|v|1,s′,Ω≥βs∥q∥0,s,Ω∀q∈Ls0(Ω).
2.2. Finite element spaces and preliminary results
Let be a shape-regular family of triangulations of consisting of closed simplices of diameter . To avoid technical difficulties we will suppose that the family of triangulations is quasi-uniform. For reasons that will become apparent later, in the proof of convergence of the finite element method we will distinguish between the cases and . To cover the latter case (and for that purpose only) we need to make the following assumption on the mesh:
Assumption (A1). The triangulation is the result of performing one (for ), or two (for ), red refinement(s) of a, coarser, shape-regular triangulation .
We will denote the (closed) elements contained in (referred to, in some instances, as macro-elements) by .
Remark 3.
raises the question whether the space itself is stable on carefully constructed meshes. Some result are known in this direction. For example, in two space dimensions, this pair is inf-sup stable on Powell–Sabin meshes [37]. In the recent work [23] local inf-sup stability is proved for this element in barycentrically-refined meshes (also known as the Alfeld split [28], and a Hsieh–Clough–Tocher triangulation, see the references quoted in [37, P. 461]). This is then used to build enriched elements that are divergence-free (although the velocity space contains quadratic face bubbles with constant divergence). For three-dimensional meshes, for the Alfeld split the lowest order inf-sup stable pair is (cf. [36]), while for the Powell–Sabin split the lowest order inf-sup stable pair is [38]. However, for the case considered in this paper, that is, taking the pair on general shape-regular meshes, stabilisation is a necessity. In addition, it is important to note that the papers cited above concern the Newtonian case only and are mostly focused on the Stokes equations. The analysis of some of those alternatives in the case of non-Newtonian flow models treated in the present work has not been carried out so far, and it will constitute a topic of future research.
Remark 4.
(i) By letting , clearly, , where does not depend on . In fact, for and for .
(ii) Under , for every , a facet of , there exists at least one node of that belongs to the interior of . In fact, this last remark is the main reason why has been made on the meshes. In particular, could also result from first making a barycentric refinement of each facet of and then building a conforming triangulation of . For ease of exposition we shall simply adopt in what follows.
In the triangulation we shall use the following notation:
• : the set of all facets (edges in and faces in ) of the triangulation , with diameter . The set of internal facets is denoted by and those on the boundary of are denoted by , so ;
• for every we denote by the set of facets of whose interior lies in the interior of ;
• for and we define the neighbourhoods
(2.8) ωF:={K∈Th:F∈FK},ωK:={K′∈Th:K∩K′≠∅};
• for each facet and every piecewise regular function , we denote by the jump of across ;
• for we denote by the space of polynomials defined on of total degree smaller than, or equal to, , and introduce the following finite element spaces:
(2.9) Vh :={vh∈C0(¯¯¯¯Ω)d:vh|K∈P1(K)d,∀K∈Th,vh|∂Ω=0}, (2.10) Qh :={qh∈L10(Ω):qh|K∈P0(K),∀K∈Th}, (2.11) QH :={qH∈L10(Ω):qH|M∈P0(M),∀M∈TM}.
Using the finite element spaces defined in (2.9)–(2.11), we denote by
the Scott–Zhang interpolation operator and by
, the projections defined by (see, e.g., [17]):
(2.12) Πhq|K=(q,1)K|K|∀K∈Th, (2.13) ΠHq|M=(q,1)M|M|∀M∈TH.
These operators satisfy ([17]):
(2.14) limh→0Shv=vstrongly inW1,s0(Ω)dfor allv∈W1,s0(Ω)dand alls∈[1,∞), (2.15) limH→0ΠHq=limh→0Πhq=q% strongly inLs0(Ω)dfor allq∈Ls0(Ω)and alls∈[1,∞).
The following result, whose proof can be carried out using the techniques presented in [17, Lemma 2.23], will be fundamental in the derivation (and analysis) of the proposed finite element method: for every there exists a constant , independent of , such that
(2.16) ∥qh−ΠH(qh)∥0,s,M≤Cs⎧⎨⎩∑F∈FI(M)hF∥⟦qh⟧∥s0,s,F⎫⎬⎭1s,
for all , all , and all .
We now recall three inequalities that will be useful in what follows. Let , and . The following local trace inequality is a corollary of the multiplicative trace inequality proved in [17, Lemma 12.15]:
(2.17) ∥v∥0,s,F≤C(h−1sF∥v∥0,s,K+h1−1sF∥∇v∥0,s,K).
In addition, we recall the following local inverse inequality (see, e.g., [17, Lemma 12.1]): for all and all , there exists a constant , independent of , such that
(2.18) ∥q∥ℓ,p,K≤Chm−ℓ+d(1p−1q)K∥q∥m,q,K,
for every polynomial function defined on . A global version of this inequality can also be derived using the quasi-uniformity of the mesh family. Finally, for , a set of indices , and any vector , the following inequality holds (see [14, Proposition 3.4(a)] for its proof):
(2.19) {∑i∈Ix~si}1~s≤{∑i∈Ixsi}1s.
Finally, we note that under the spaces and satisfy the following discrete inf-sup condition: for any there exists a constant , independent of , such that for all the following inequality holds:
(2.20)
The proof of this result, to the best of our knowledge, has not been given previously and thus we report it in the Appendix. It is based on the construction of a Fortin operator satisfying
(2.21) (qH,div(v−I(v)))Ω =0 for allqH∈QHand allv∈W1,s′0(Ω)d, (2.22) Iv →v strongly inW1,s′0(Ω)d% ash→0.
In addition, (2.20) guarantees the existence of a non-trivial subspace of discretely divergence-free functions
(2.23) Vh,div:={vh∈Vh:(qH,divvh)Ω=0 for allqH∈QH}.
2.3. Results linked to the discrete Lipschitz truncation
In the convergence proof given below we will need the following two results. These are known as discrete Lipschitz truncation and divergence-free discrete Lipschitz truncation, respectively. Their proofs are omitted since they are essentially a rewriting of Corollary 17 and the proof on pages 1006–1007 in [15] (see also [35, Lemmas 2.29 and 2.30]).
Lemma 5.
Let . Let us suppose that for all and weakly in as . Then, there exist
• a double sequence such that for all ;
• a double sequence of open sets , of the form
(2.24) Bh,j=int(∪{K:K∈Th,j}),
where denotes the collection of some elements of the mesh ;
• a double sequence with for all and all ;
satisfying
• in for all and all ;
• there exists a such that
(2.25) ∥λh,j1Bh,j∥0,s,Ω≤c(s)2−js∀h>0,j∈N;
• there exists a such that
(2.26) ∥∇vh,j∥0,∞,Ω≤c(s)λh,j∀h>0,j∈N;
• for any fixed ,
(2.27) vh,j→0strongly inL∞(Ω)dand∇vh,j⇀0weakly-* inL∞(Ω)d×d,
as .
Lemma 6.
Let and assume that Assumption (A1) is satisfied. Let be a sequence such that for all and such that weakly in as . Furthermore, let be the sequence of Lipschitz truncations given by Lemma 5. Then, there exists a double sequence such that
• for all and all ;
• there exists a such that
(2.28) ∥vh,j−wh,j∥1,s,Ω≤c(s)2−js∀h>0,∀j∈N;
• for any fixed the following convergences hold (up to a subsequence, if necessary):
(2.29) wh,j→0strongly inLt(Ω)dand% ∇wh,j⇀0weakly inW1,t0(Ω)d×d,
as , for all .
2.4. The stabilising bilinear form and the lifting operator
The finite element method studied in this work is based on the pair . Since this pair is not inf-sup stable some form of stabilisation is needed. In this work our proposal is to use the following stabilising bilinear form
(2.30) s(qh,th)=∑M∈TH∑F∈FI(M)τF(⟦qh⟧,⟦th⟧)F,
where the stabilisation parameter is defined as follows:
(2.31)
The behaviour of is depicted in Figure 1. It can be observed there that the stabilisation gets stronger as . The reason for this behaviour will become clear when we perform the convergence analysis in Section 4.
Remark 7.
Thanks to and the inf-sup condition (2.20) it can be expected to have stability of (where is the finite element approximation of the pressure). The stabilisation is then built with the aim of controlling . More precisely, using (2.16), (2.19) (or the inverse inequality (2.18)), and the definition of the bilinear form we see that there exists a constant such that
(2.32) ∥qh−ΠH(qh)∥0,s,Ω≤Chχs(qh,qh)12,
for all , where
(2.33) χ=⎧⎪⎨⎪⎩1−α(r)2ifs≤2,1−d+2ds−α(r)2ifs>2.
It will be useful in what follows to observe that for we have .
Another important ingredient in the definition of the method is a lifting of the pressure jumps defined with the help of the lowest order Raviart–Thomas basis functions. To define this, for each we choose a unique normal vector . Its orientation is of no importance, but it needs to point outwards of if . Moreover, for each such that , we denote the node in opposite by . Using this unique normal vector, we introduce the lowest order Raviart–Thomas basis function defined as
(2.34) φF(x)|K:=±|F|d|K|(x−xF),
and extended by zero outside . In this definition, the sign of the function depends on whether the normal vector points in or out of . Thanks to its definition, satisfies the following: for every the normal component of is given by (with the obvious abuse of notation considering that is not defined at the boundary of ):
(2.35) φF⋅nF′={1% ifF′=F,0otherwise.
With the help of these Raviart–Thomas basis functions, we define the following operator, which will be fundamental in the definition of the finite element method:
L:W1,r(Ω)d×Qh→Wr(div;Ω) (2.36) (v,qh)↦L(v,qh):=v+∑M∈TH∑F∈FI(M)τF⟦qh⟧φF.
Since the velocity is bounded in , then it is bounded in as well. In the finite element method proposed in Section 3, we will consider a modified velocity built with the help of the mapping just defined. The following result states that the stability just mentioned is preserved by the operator .
Lemma 8.
There exists a constant , independent of , such that
(2.37) ∥L(v,qh)∥0,2~r,Ω≤C{|v|1,r,Ω+s(qh,qh)12},
for all .
Proof.
Thanks to the embedding (2.3) and denoting
(2.38) unc:=∑M∈TH∑F∈FI(M)τF⟦qh⟧φF,
the following bound follows
(2.39) ∥L(v,qh)∥0,2~r,Ω≤C|v|1,r,Ω+∥unc∥0,2~r,Ω.
To bound the second term on the right-hand side of this inequality we start by noticing that the definition of (cf. (2.34)) gives for each such that . So, let and let be the unique macro-element such that . Then, using the mesh regularity and the Cauchy-Schwarz inequality we get
∥unc∥0,K =∥∥ ∥∥∑F∈FK∩FI(M)τF⟦ph⟧φF∥∥ ∥∥0,K ≤∑F∈FK∩FI(M)τF|⟦ph⟧|∥φF∥0,K ≤C∑F∈FK∩FI(M)τFhd2F|⟦ph⟧| ≤C∑F∈FK∩FI(M)τFh1−d2F(1,|⟦ph⟧|)F
Hence, squaring, summing over all the elements, and using the mesh regularity gives
(2.40) ∥unc∥0,Ω=⎧⎨⎩∑K∈Th∥unc∥20,K⎫⎬⎭12≤Ch1+α(r)2⎧⎨⎩∑M∈TH∑F∈FI(M)τF∥⟦ph⟧∥20,F⎫⎬⎭12.
Thus, using the inverse inequality (2.18) we arrive at
(2.41) ∥unc∥0,2~r,Ω≤Chd(1−~r)2~r∥unc∥0,Ω≤Chd(1−~r)+~r+α(r)~r2~rs(ph,ph)12.
To complete the proof we only need to make sure that the exponent of in (2.41) is not negative. Let . If then and . So, . If then and so
ξ≥d(1−~r)+~r+d−13~r=d+2(1−d)3~r≥d+2(1−d)3~rmax=0.
Since in the whole range of values for we have , the proof is complete. ∎
3. The finite element method
The finite element method studied in this work reads as follows: find such that
(3.1) (|∇uh|r−2∇uh,∇vh)Ω−(L(uh,ph)⊗uh,∇vh)Ω−(ph,divvh)Ω =⟨f,vh⟩Ω, (3.2) (qh,divuh)Ω+s(ph,qh) =0,
for all , where is defined by (2.36) and the stabilising bilinear form is defined in (2.30).
Remark 9.
(i) The main differences between (3.1), (3.2) and a standard Galerkin method are twofold: first, the stabilising term involving the jumps of the discrete pressure are added to the formulation to compensate for the fact that the pair does not satisfy the discrete inf-sup condition. Additionally, and perhaps more significantly, the convection velocity has been replaced by the modified version . In Lemma 10 this modified velocity will be proved to be solenoidal, which allows us to analyse the finite element method without the need to rewrite the convection term in a skew-symmetric form. This will lead to a convergence result valid in the whole range .
(ii) As can be expected, the power of in the stabilisation parameter depends strongly on the value of . Two important remarks are in order:
• for all ;
• for all we have .
Thus, there is always a positive power of multiplying the jump terms of the pressure involved in the definition of and , but the stabilisation becomes stronger as .
3.1. Existence of a solution and a priori bounds
Before exploring the stability of the scheme, we present the following a priori result concerning qualitative properties of and , whenever solves (3.1), (3.2).
Lemma 10.
Let be any solution of (3.1), (3.2). Then,
• is discretely divergence-free with respect to the coarse space , that is,
(3.3) (qH,divuh)Ω=0∀qH∈QH.
• on , and
(3.4) divL(uh,ph)=0inΩ.
Proof.
The proof of (i) is a consequence of the fact that the stabilisation vanishes on the coarse space , that is, for all and all . For (ii), we can follow similar arguments as those presented in [6, Lemma 3.8] and [7, Lemma 3] (see also [2, Theorem 3] for a different proof).
The following result states the existence of a solution to the discrete problem (3.1), (3.2). In addition, it provides uniform a priori bounds for the sequence of solutions as .
Theorem 11.
There exists a solution of (3.1), (3.2). Moreover, every solution satisfies the following a priori bound:
(3.5) |uh|r1,r,Ω+∥L(uh,ph)∥0,2~r,Ω+s(ph,ph)+∥ph∥0,~r,Ω≤M,
where does not depend on .
Proof.
The existence of a solution is proved using the argument used in [21] for the Navier–Stokes equation. First, if , then and trivially solve (3.1), (3.2). So, we suppose that . The subspace of where solutions of (3.1), (3.2) are to be sought is given by
(3.6) Xh:={(vh,qh)∈Vh×Qh:divL(vh,qh)=0inΩ}.
Let be the mapping defined by
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https://www.quantumcalculus.org/wu-characteristic/ | # Wu Characteristic
Update: March 8, 2016: Handout for a mathtable talk on Wu characteristic.
Gauss-Bonnet for multi-linear valuations
deals with a number in discrete geometry. But since the number satisfies formulas which in the continuum need differential calculus, like curvature, the results can be seen in the light of quantum calculus. Here are some slides:
So, why is the Wu characteristic an object of quantum calculus? It is a combinatorial invariant, a quantity which does not change under Barycentric subdivision. An other reason is Gauss-Bonnet which holds for this quantity and for discrete structures. When Barycentrically refined again and again, the graph converges to a smooth manifold, where the corresponding curvatures involve curvature tensors, which are objects from classical calculus using classical partial derivatives. One must therefore see the discrete curvature values as quantities in a discrete quantum calculus setting. It is a bit mysterious still, how the complicated expressions of the Riemann curvature tensor in the continuum become such simple expressions in the discrete. But in the discrete we have magical abilities to see into the lower dimensional building blocks in space, something which is more difficult in the continuum. In classical calculus, we need to probe these quantities using a tomographic methods called integral geometry or sheaf theoretical methods. Some quantities like length of a curve, or area of a surface or volume are quite intuitive. There are other quantities which are not, like “length of a surface”. One can measure the length of a surface; it is just something we usually do not do, or are not interested in. For discrete surfaces given by a graph, the length of the surface is the number of edges it contains. For an icosahedron it has 30. The area is the number of triangles, 20, which gives the Icosahedron the name. Some combination of the quantities can be seen topologically. One important one is the Euler characteristic, which is area minus length plus number of points. In the case of the Icosahedron, this is 12-30+20=2, a relation which has been found by Rene Descartes already. The story is described nicely in the book “Descartes’s Secret Notebook: A True Tale of Mathematics, Mysticism, and the Quest to Understand the Universe“, by Amir Aczel. It was Euler who proved the formula first. But the story continues to be interesting then as the formula turned out to be wrong in general. There are surfaces, like the Kepler polyhedra, for which the Euler characteristic is different from 2. Imre Lakatos analyzes the evolution of a theorem brilliantly in his book “Proofs and Refutations”. Now, the Wu characteristic is even more exciting than the Euler characteristic, because it allows to probe continuum spaces like varieties in a different manner. The number is a quadratic valuation: it does not only probe the cardinalities of the pieces of the fabric of space, but also how these pieces interact. For a lemniscate for example, the Wu characteristic is 7. It is different from the Euler characteristic, which is minus 1. | 2020-07-02 05:50:04 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8169571161270142, "perplexity": 483.05617031599564}, "config": {"markdown_headings": true, "markdown_code": false, "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-2020-29/segments/1593655878519.27/warc/CC-MAIN-20200702045758-20200702075758-00576.warc.gz"} |
http://nrich.maths.org/5543/index?nomenu=1 | Consider $(1 + \sqrt2)^n$. This will be of the form $A + B\sqrt2$ where $A$ and $B$ are integers. Decide which entries in the table below are possibe and which are not.
What happens for $(a + \sqrt p)^n$ for other values of $p$? | 2016-12-06 08:24:45 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7148619890213013, "perplexity": 79.43426542223565}, "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-2016-50/segments/1480698541886.85/warc/CC-MAIN-20161202170901-00110-ip-10-31-129-80.ec2.internal.warc.gz"} |
https://lavelle.chem.ucla.edu/forum/viewtopic.php?f=142&t=60278&p=228691 | ## nernst equation
$E_{cell} = E_{cell}^{\circ}-\frac{RT}{nF}\ln Q$
Posts: 93
Joined: Thu Jul 11, 2019 12:17 am
### nernst equation
what exactly does the nernst equation show?
Alicia Lin 2F
Posts: 83
Joined: Wed Sep 18, 2019 12:17 am
### Re: nernst equation
It relates the potential of an electrochemical cell to concentrations in the cell reaction. It also is useful because it can be used for nonstandard conditions too.
Pablo 1K
Posts: 118
Joined: Sat Feb 02, 2019 12:15 am
### Re: nernst equation
It ties the max potential of an electrochemical cell to respective cell concentrations that are in the battery and the conditions of it can be assumed regular as @25 degrees C or also for non-regular conditions.
Esha Chawla 2E
Posts: 108
Joined: Thu Jul 25, 2019 12:17 am
### Re: nernst equation
AMahadi wrote:what exactly does the nernst equation show?
The Nernst equation allows us to calculate the Ecell value at non-standard conditions given the reaction quotient. At the same time, the Nernst equation can also be used to calculate the reaction quotient or the equilibrium constant given Ecell and Ecell under standard conditions.
Return to “Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)”
### Who is online
Users browsing this forum: No registered users and 2 guests | 2020-05-26 13:56:01 | {"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": 0, "img_math": 0, "codecogs_latex": 1, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6385476589202881, "perplexity": 6752.151181700805}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-24/segments/1590347390758.21/warc/CC-MAIN-20200526112939-20200526142939-00152.warc.gz"} |
https://uclouvain-cbio.github.io/scp/reference/readSCP.html | Convert tabular quantitative MS data and metadata from a spreadsheet or a data.frame into a QFeatures object containing SingleCellExperiment objects.
readSCP(
featureData,
colData,
batchCol,
channelCol,
suffix = NULL,
removeEmptyCols = FALSE,
verbose = TRUE,
...
)
## Arguments
featureData File or object holding the quantitative data. Can be either a character(1) with the path to a text-based spreadsheet (comma-separated values by default, but see ...) or an object that can be coerced to a data.frame. It is advised not to encode characters as factors. A data.frame or any object that can be coerced to a data.frame. colData is expected to contains all the sample meta information. Required fields are the acquisition batch (given by batchCol) and the acquisition channel within the batch (e.g. TMT channel, given by channelCol). Additional fields (e.g. sample type, acquisition date,...) are allowed and will be stored as sample meta data. A numeric(1) or character(1) pointing to the column of featureData and colData that contain the batch names. Make sure that the column name in both table are either identical and syntactically valid (if you supply a character) or have the same index (if you supply a numeric). Note that characters can be converted to syntactically valid names using make.names A numeric(1) or character(1) pointing to the column of colData that contains the column names of the quantitative data in featureData (see Example). A character() giving the suffix of the column names in each assay. The length of the vector should equal the number of quantification channels and should contain unique character elements. If NULL, the names of the quantification columns in featureData are taken as suffix. A logical(1). If true, the function will remove in each batch the columns that contain only missing values. A logical(1) indicating whether the progress of the data reading and formatting should be printed to the console. Default is TRUE. Further arguments that can be passed on to read.csv except stringsAsFactors, which is always FALSE.
## Value
An instance of class QFeatures. The expression data of each batch is stored in a separate assay as a SingleCellExperiment object.
## Note
The SingleCellExperiment class is built on top of the RangedSummarizedExperiment class. This means that some column names are forbidden in the rowData. Avoid using the following names: seqnames, ranges, strand, start, end, width, element
## Author
Laurent Gatto, Christophe Vanderaa
## Examples
## Load an example table containing MaxQuant output
data("mqScpData")
## Load the (user-generated) annotation table
data("sampleAnnotation")
## Format the tables into a QFeatures object
colData = sampleAnnotation,
batchCol = "Raw.file",
channelCol = "Channel") | 2021-10-19 18:26:36 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.28215163946151733, "perplexity": 4221.741669170672}, "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/1634323585280.84/warc/CC-MAIN-20211019171139-20211019201139-00048.warc.gz"} |
https://www.mariakzurek.com/publication/aaltonen-2015uva/ | # Measurement of central exclusive $\pi^+ \pi^-$ production in $p\bar{p}$ collisions at $\sqrt{s} = 0.9$ and 1.96 TeV at CDF
### Abstract
We measure exclusive $\pi^+\pi^-$ production in proton-antiproton collisions at center-of-mass energies $\sqrt{s}$ = 0.9 and 1.96 TeV in the Collider Detector at Fermilab. We select events with two oppositely charged particles, assumed to be pions, with pseudorapidity $|\eta| < 1.3$ and with no other particles detected in $|\eta| < 5.9$. We require the \pipi system to have rapidity $|y|<$ 1.0. The production mechanism of these events is expected to be dominated by double pomeron exchange, which constrains the quantum numbers of the central state. The data are potentially valuable for isoscalar meson spectroscopy and for understanding the pomeron in a region of transition between nonperturbative and perturbative quantum chromodynamics. The data extend up to dipion mass $M(\pi^+\pi^-)$ = 5000 MeV/$c^2$ and show resonance structures attributed to $f_0$ and $f_2(1270)$ mesons. From the $\pi^+\pi^-$ and $K^+K^-$ spectra, we place upper limits on exclusive $\chi_{c0}(3415)$ production.
Type
Publication
In: Phys. Rev. D91
Date | 2019-11-22 02:57:56 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7546983361244202, "perplexity": 1832.3197013538781}, "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/1573496671106.83/warc/CC-MAIN-20191122014756-20191122042756-00368.warc.gz"} |
https://lambdageeks.com/phase-modulation-frequency-modulations/ | # Phase Modulation And Frequency Modulation: Carson’s Rule
## Define Angle or Phase Modulation:
“Angle modulation is a non-linear process and transmission bandwidth is usually much greater than twice the message bandwidth. Because of larger bandwidth, this modulation provides increased signal to noise ratio without increased transmitted power.”
Basically, angle modulation is divided into two categories namely Frequency Modulation & Phase Modulation.
One significant characteristic of this type is that it can better classify in contrast to noise and interference signal than amplitude modulation. This adjustment in execution is accomplished in the expense of expanded transmission bandwidth; that is, this modulation gives us a method for improved signal to noise ratio.
Besides, this improvement in execution in angle modulation is achieved in the expense of complex circuitry in both the transmitter and receiver section and not possible in Amplitude one.
## Mathematical Expression of Angle Modulation:
Let θi(t) signify the angle of a modulated sinusoidal carrier at time t; it is assumed to be a function of the information-bearing signal or message signal. The resultant angle-modulated signal is,
s(t) = Ac cos [θi(t)]
Where Ac is the carrier amplitude, a complete oscillation happens each and every time the angle θi (t) will changed by the value of 2π radians if θi (t) increases with time, then the average freq in hertz, over a trivial intervals of t to t+∆t.
The angle-modulated signal s(t) as a rotating phasor of length Ac and angle θi (t) respectivelySuch a phasor’s angular velocity is dθi (t)/dt, measured in radians/sec. The angle θi (t) is represented for an unmodulated carrier signal,
θi (t) = 2πfct + kp m(t)
and the corresponding phasor rotating with a constant angular velocity measured in radians/sec. This constant specify the angle of the unmodulated carrier during that period.
There are various methods in which the angle θi (t) could be changed in a manner w.r.t to the message signal.
## Frequency Modulation:
Frequency Modulation is one form of angle modulation in that instantaneous freq of the carrier is changed proportionally with the instantaneous amplitude variation of the modulating signal”.
FM is one sort of angle modulation in with fi (t) is linearly proportional with the message signal m(t) as expressed below,
fi (t) = fc + kf m(t)
The steady value of fc presented to the frequency of the unmodulated carriers signal; the fixed kf termed as modulator’s ‘frequency-sensitivity factor’, measured in hertz per volt on the other hand m(t) is a voltage signal waveform. Integrating w.r.t time and multiply the result by a factor 2π, we can write
where the 2nd term for the increase or decrease in the instant phase θi(t) due to the message m(t) one. The frequency-modulated signal is consequently,
## Phase Modulation:
Phase Modulation is such type of angle modulation in which the instantaneous angle θi(t) is linearly proportional with the message ‘ m(t)’ signal as presented by means of,
θi(t) = 2πfct + kp m(t)
The term 2πfct expresses to the un-modulated carrier angle Øc set to ‘0’ in the phase modulation. The fixed kp value phase sensitivity factor of the modulator, communicated in radians/volt and m(t) is the voltage signal. In the phase modulation, modulated signal s(t) is correspondingly depicted in the time-space by,
s(t) = Ac cos [2πfct + kp m(t)]
Show that FM and PM are basically same:
Let the carrier signal is = Ac cos (2πfct)
Let the message signal is = m(t)
So, the expression of F.M. signal is =
Now if the modulation method is Phase Modulation. then the expression of Phase Modulation signal is
= Acos [2πfct + mp . m(t)]
Where, mp is a constant for the Phase Modulation
Also the Phase Modulation signal can be treated as a Frequency Modulation signal where message signal is dm(t)/dt.
So, basically Frequency Modulation and Phase Modulation are basically same.
## Pre-Emphasis and De-Emphasis in FM:
A random undesired signal or noise constantly comes with a triangular spectral distribution in a Frequency Modulation technique, together with the impact that noise happens at the maximum frequency of baseband.
This may be offset, to some restricted selection, by raising the frequencies prior to transmitting and decreasing them with a corresponding receiver number. If we decrease the high frequencies from the receiver, then, in addition, it reduces the high-frequency noise.
These practice of increasing and decreasing of these frequencies are called pre-emphasis and de-emphasis, respectively. Most frequently 50 µs time constant is employed.
The total quantity of pre-emphasis which may be implemented is restricted by the simple fact that lots of kinds of modern sound signal comprise higher frequency energy compared to the musical styles that have predominated at the beginning of FM broadcasting.
They cannot be pre-emphasized since it might cause excess deviation. (systems more contemporary compared to FM broadcasting often utilize either programmed-dependent variable pre-emphasis.)
## What is Narrow Band FM (NBFM) and Wide Band FM (WBFM ?
The expression for the FM signals is given by
and hence the instantaneous frequency ωi is given by,
where, kf = constant of proportionality and kr . em (t) represents the deviation of carrier frequency from the quiescent value ωc. Constant Kf hence controls the frequency deviation. If the Kf is small the frequency deviation is also small and the spectrum of the FM signal is having a narrow band. On the other hand, for higher value of kf, we get wide frequency spectrum corresponding to wideband FM case.
## NARROW BAND FM:
The modulation index for narrow band FM is generally near unity and hence for this case, the maximum deviation δ<<fm and the bandwidth is
B = 2fm.
This bandwidth is same as that occupied by AM signal. The narrowband FM is used wherein intelligible signals for communications are to be transmitted such as in mobile communication used by police, ambulance etc.
## WIDE BAND FM:
The modulation index for wideband FM is greater than unity. The bandwidth of a wideband FM system is given by,
B = 2(δ+fm)
For wideband FM δ<<fm and hence B =
Thus, the bandwidth of wideband FM is twice the maximum frequency deviation. The wideband FM is used where the purpose is to transmit high fidelity signals such as in FM broadcasting and TV sound. | 2022-10-04 20:08:18 | {"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": 0.8488696217536926, "perplexity": 2740.738298931926}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030337524.47/warc/CC-MAIN-20221004184523-20221004214523-00484.warc.gz"} |
https://www.gradesaver.com/textbooks/math/precalculus/precalculus-6th-edition-blitzer/chapter-6-review-exercises-page-798/41 | # Chapter 6 - Review Exercises - Page 798: 41
$y=-x$ See graph.
#### Work Step by Step
Step 1. Using $\frac{y}{x}=tan\theta=tan\frac{3\pi}{4}=-1$, we have $y=-x$ Step 2. We can identify the above equation as a line. Step 3. See graph.
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback. | 2020-06-05 22:25:26 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.7072353363037109, "perplexity": 1393.7965055240177}, "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-2020-24/segments/1590348504341.78/warc/CC-MAIN-20200605205507-20200605235507-00083.warc.gz"} |
https://www.jobilize.com/precalculus/course/8-6-parametric-equations-further-applications-of-trigonometry-by-opens?qcr=www.quizover.com&page=4 | # 8.6 Parametric equations (Page 5/6)
Page 5 / 6
## Verbal
What is a system of parametric equations?
A pair of functions that is dependent on an external factor. The two functions are written in terms of the same parameter. For example, $\text{\hspace{0.17em}}x=f\left(t\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}y=f\left(t\right).$
Some examples of a third parameter are time, length, speed, and scale. Explain when time is used as a parameter.
Explain how to eliminate a parameter given a set of parametric equations.
Choose one equation to solve for $\text{\hspace{0.17em}}t,\text{\hspace{0.17em}}$ substitute into the other equation and simplify.
What is a benefit of writing a system of parametric equations as a Cartesian equation?
What is a benefit of using parametric equations?
Some equations cannot be written as functions, like a circle. However, when written as two parametric equations, separately the equations are functions.
Why are there many sets of parametric equations to represent on Cartesian function?
## Algebraic
For the following exercises, eliminate the parameter $\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ to rewrite the parametric equation as a Cartesian equation.
$\left\{\begin{array}{l}x\left(t\right)=5-t\hfill \\ y\left(t\right)=8-2t\hfill \end{array}$
$y=-2+2x$
$\left\{\begin{array}{l}x\left(t\right)=6-3t\hfill \\ y\left(t\right)=10-t\hfill \end{array}$
$\left\{\begin{array}{l}x\left(t\right)=2t+1\hfill \\ y\left(t\right)=3\sqrt{t}\hfill \end{array}$
$y=3\sqrt{\frac{x-1}{2}}$
$\left\{\begin{array}{l}x\left(t\right)=3t-1\hfill \\ y\left(t\right)=2{t}^{2}\hfill \end{array}$
$\left\{\begin{array}{l}x\left(t\right)=2{e}^{t}\hfill \\ y\left(t\right)=1-5t\hfill \end{array}$
$x=2{e}^{\frac{1-y}{5}}\text{\hspace{0.17em}}$ or $\text{\hspace{0.17em}}y=1-5ln\left(\frac{x}{2}\right)$
$\left\{\begin{array}{l}x\left(t\right)={e}^{-2t}\hfill \\ y\left(t\right)=2{e}^{-t}\hfill \end{array}$
$\left\{\begin{array}{l}x\left(t\right)=4\text{log}\left(t\right)\hfill \\ y\left(t\right)=3+2t\hfill \end{array}$
$x=4\mathrm{log}\left(\frac{y-3}{2}\right)$
$\left\{\begin{array}{l}x\left(t\right)=\text{log}\left(2t\right)\hfill \\ y\left(t\right)=\sqrt{t-1}\hfill \end{array}$
$\left\{\begin{array}{l}x\left(t\right)={t}^{3}-t\hfill \\ y\left(t\right)=2t\hfill \end{array}$
$x={\left(\frac{y}{2}\right)}^{3}-\frac{y}{2}$
$\left\{\begin{array}{l}x\left(t\right)=t-{t}^{4}\hfill \\ y\left(t\right)=t+2\hfill \end{array}$
$\left\{\begin{array}{l}x\left(t\right)={e}^{2t}\hfill \\ y\left(t\right)={e}^{6t}\hfill \end{array}$
$y={x}^{3}$
$\left\{\begin{array}{l}x\left(t\right)={t}^{5}\hfill \\ y\left(t\right)={t}^{10}\hfill \end{array}$
${\left(\frac{x}{4}\right)}^{2}+{\left(\frac{y}{5}\right)}^{2}=1$
$\left\{\begin{array}{l}x\left(t\right)=3\mathrm{sin}\text{\hspace{0.17em}}t\hfill \\ y\left(t\right)=6\mathrm{cos}\text{\hspace{0.17em}}t\hfill \end{array}$
${y}^{2}=1-\frac{1}{2}x$
$\left\{\begin{array}{l}x\left(t\right)=\mathrm{cos}\text{\hspace{0.17em}}t+4\\ y\left(t\right)=2{\mathrm{sin}}^{2}t\end{array}$
$\left\{\begin{array}{l}x\left(t\right)=t-1\\ y\left(t\right)={t}^{2}\end{array}$
$y={x}^{2}+2x+1$
$\left\{\begin{array}{l}x\left(t\right)=-t\\ y\left(t\right)={t}^{3}+1\end{array}$
$\left\{\begin{array}{l}x\left(t\right)=2t-1\\ y\left(t\right)={t}^{3}-2\end{array}$
$y={\left(\frac{x+1}{2}\right)}^{3}-2$
For the following exercises, rewrite the parametric equation as a Cartesian equation by building an $x\text{-}y$ table.
$\left\{\begin{array}{l}x\left(t\right)=2t-1\\ y\left(t\right)=t+4\end{array}$
$\left\{\begin{array}{l}x\left(t\right)=4-t\\ y\left(t\right)=3t+2\end{array}$
$y=-3x+14$
$\left\{\begin{array}{l}x\left(t\right)=2t-1\\ y\left(t\right)=5t\end{array}$
$\left\{\begin{array}{l}x\left(t\right)=4t-1\\ y\left(t\right)=4t+2\end{array}$
$y=x+3$
For the following exercises, parameterize (write parametric equations for) each Cartesian equation by setting $x\left(t\right)=t$ or by setting $\text{\hspace{0.17em}}y\left(t\right)=t.$
$y\left(x\right)=3{x}^{2}+3$
$y\left(x\right)=2\mathrm{sin}\text{\hspace{0.17em}}x+1$
$\left\{\begin{array}{l}x\left(t\right)=t\hfill \\ y\left(t\right)=2\mathrm{sin}t+1\hfill \end{array}$
$x\left(y\right)=3\mathrm{log}\left(y\right)+y$
$x\left(y\right)=\sqrt{y}+2y$
$\left\{\begin{array}{l}x\left(t\right)=\sqrt{t}+2t\hfill \\ y\left(t\right)=t\hfill \end{array}$
For the following exercises, parameterize (write parametric equations for) each Cartesian equation by using $x\left(t\right)=a\mathrm{cos}\text{\hspace{0.17em}}t$ and $\text{\hspace{0.17em}}y\left(t\right)=b\mathrm{sin}\text{\hspace{0.17em}}t.\text{\hspace{0.17em}}$ Identify the curve.
$\frac{{x}^{2}}{4}+\frac{{y}^{2}}{9}=1$
$\frac{{x}^{2}}{16}+\frac{{y}^{2}}{36}=1$
$\left\{\begin{array}{l}x\left(t\right)=4\mathrm{cos}\text{\hspace{0.17em}}t\hfill \\ y\left(t\right)=6\mathrm{sin}\text{\hspace{0.17em}}t\hfill \end{array};\text{\hspace{0.17em}}$ Ellipse
${x}^{2}+{y}^{2}=16$
${x}^{2}+{y}^{2}=10$
$\left\{\begin{array}{l}x\left(t\right)=\sqrt{10}\mathrm{cos}t\hfill \\ y\left(t\right)=\sqrt{10}\mathrm{sin}t\hfill \end{array};\text{\hspace{0.17em}}$ Circle
Parameterize the line from $\text{\hspace{0.17em}}\left(3,0\right)\text{\hspace{0.17em}}$ to $\text{\hspace{0.17em}}\left(-2,-5\right)\text{\hspace{0.17em}}$ so that the line is at $\text{\hspace{0.17em}}\left(3,0\right)\text{\hspace{0.17em}}$ at $\text{\hspace{0.17em}}t=0,\text{\hspace{0.17em}}$ and at $\text{\hspace{0.17em}}\left(-2,-5\right)\text{\hspace{0.17em}}$ at $\text{\hspace{0.17em}}t=1.$
Parameterize the line from $\text{\hspace{0.17em}}\left(-1,0\right)\text{\hspace{0.17em}}$ to $\text{\hspace{0.17em}}\left(3,-2\right)\text{\hspace{0.17em}}$ so that the line is at $\text{\hspace{0.17em}}\left(-1,0\right)\text{\hspace{0.17em}}$ at $\text{\hspace{0.17em}}t=0,\text{\hspace{0.17em}}$ and at $\text{\hspace{0.17em}}\left(3,-2\right)\text{\hspace{0.17em}}$ at $\text{\hspace{0.17em}}t=1.$
$\left\{\begin{array}{l}x\left(t\right)=-1+4t\hfill \\ y\left(t\right)=-2t\hfill \end{array}$
Parameterize the line from $\text{\hspace{0.17em}}\left(-1,5\right)\text{\hspace{0.17em}}$ to $\text{\hspace{0.17em}}\left(2,3\right)$ so that the line is at $\text{\hspace{0.17em}}\left(-1,5\right)\text{\hspace{0.17em}}$ at $\text{\hspace{0.17em}}t=0,\text{\hspace{0.17em}}$ and at $\text{\hspace{0.17em}}\left(2,3\right)\text{\hspace{0.17em}}$ at $\text{\hspace{0.17em}}t=1.$
Parameterize the line from $\text{\hspace{0.17em}}\left(4,1\right)\text{\hspace{0.17em}}$ to $\text{\hspace{0.17em}}\left(6,-2\right)\text{\hspace{0.17em}}$ so that the line is at $\text{\hspace{0.17em}}\left(4,1\right)\text{\hspace{0.17em}}$ at $\text{\hspace{0.17em}}t=0,\text{\hspace{0.17em}}$ and at $\text{\hspace{0.17em}}\left(6,-2\right)\text{\hspace{0.17em}}$ at $\text{\hspace{0.17em}}t=1.$
$\left\{\begin{array}{l}x\left(t\right)=4+2t\hfill \\ y\left(t\right)=1-3t\hfill \end{array}$
## Technology
For the following exercises, use the table feature in the graphing calculator to determine whether the graphs intersect.
yes, at $t=2$
For the following exercises, use a graphing calculator to complete the table of values for each set of parametric equations.
$\left\{\begin{array}{l}{x}_{1}\left(t\right)=3{t}^{2}-3t+7\hfill \\ {y}_{1}\left(t\right)=2t+3\hfill \end{array}$
$t$ $x$ $y$
–1
0
1
$\left\{\begin{array}{l}{x}_{1}\left(t\right)={t}^{2}-4\hfill \\ {y}_{1}\left(t\right)=2{t}^{2}-1\hfill \end{array}$
$t$ $x$ $y$
1
2
3
$t$ $x$ $y$
1 -3 1
2 0 7
3 5 17
$\left\{\begin{array}{l}{x}_{1}\left(t\right)={t}^{4}\hfill \\ {y}_{1}\left(t\right)={t}^{3}+4\hfill \end{array}$
$t$ $x$ $y$
-1
0
1
2
## Extensions
Find two different sets of parametric equations for $\text{\hspace{0.17em}}y={\left(x+1\right)}^{2}.$
Find two different sets of parametric equations for $\text{\hspace{0.17em}}y=3x-2.$
Find two different sets of parametric equations for $\text{\hspace{0.17em}}y={x}^{2}-4x+4.$
answers may vary: ,
#### Questions & Answers
is precalculus needed to take caculus
It depends on what you already know. Just test yourself with some precalculus questions. If you find them easy, you're good to go.
Spiro
the solution doesn't seem right for this problem
what is the domain of f(x)=x-4/x^2-2x-15 then
x is different from -5&3
Seid
how to prroved cos⁴x-sin⁴x= cos²x-sin²x are equal
Don't think that you can.
Elliott
how do you provided cos⁴x-sin⁴x = cos²x-sin²x are equal
What are the question marks for?
Elliott
Someone should please solve it for me Add 2over ×+3 +y-4 over 5 simplify (×+a)with square root of two -×root 2 all over a multiply 1over ×-y{(×-y)(×+y)} over ×y
For the first question, I got (3y-2)/15 Second one, I got Root 2 Third one, I got 1/(y to the fourth power) I dont if it's right cause I can barely understand the question.
Is under distribute property, inverse function, algebra and addition and multiplication function; so is a combined question
Abena
find the equation of the line if m=3, and b=-2
graph the following linear equation using intercepts method. 2x+y=4
Ashley
how
Wargod
what?
John
ok, one moment
UriEl
how do I post your graph for you?
UriEl
it won't let me send an image?
UriEl
also for the first one... y=mx+b so.... y=3x-2
UriEl
y=mx+b you were already given the 'm' and 'b'. so.. y=3x-2
Tommy
Please were did you get y=mx+b from
Abena
y=mx+b is the formula of a straight line. where m = the slope & b = where the line crosses the y-axis. In this case, being that the "m" and "b", are given, all you have to do is plug them into the formula to complete the equation.
Tommy
thanks Tommy
Nimo
0=3x-2 2=3x x=3/2 then . y=3/2X-2 I think
Given
co ordinates for x x=0,(-2,0) x=1,(1,1) x=2,(2,4)
neil
"7"has an open circle and "10"has a filled in circle who can I have a set builder notation
x=-b+_Гb2-(4ac) ______________ 2a
I've run into this: x = r*cos(angle1 + angle2) Which expands to: x = r(cos(angle1)*cos(angle2) - sin(angle1)*sin(angle2)) The r value confuses me here, because distributing it makes: (r*cos(angle2))(cos(angle1) - (r*sin(angle2))(sin(angle1)) How does this make sense? Why does the r distribute once
so good
abdikarin
this is an identity when 2 adding two angles within a cosine. it's called the cosine sum formula. there is also a different formula when cosine has an angle minus another angle it's called the sum and difference formulas and they are under any list of trig identities
strategies to form the general term
carlmark
consider r(a+b) = ra + rb. The a and b are the trig identity.
Mike
How can you tell what type of parent function a graph is ?
generally by how the graph looks and understanding what the base parent functions look like and perform on a graph
William
if you have a graphed line, you can have an idea by how the directions of the line turns, i.e. negative, positive, zero
William
y=x will obviously be a straight line with a zero slope
William
y=x^2 will have a parabolic line opening to positive infinity on both sides of the y axis vice versa with y=-x^2 you'll have both ends of the parabolic line pointing downward heading to negative infinity on both sides of the y axis
William
y=x will be a straight line, but it will have a slope of one. Remember, if y=1 then x=1, so for every unit you rise you move over positively one unit. To get a straight line with a slope of 0, set y=1 or any integer.
Aaron
yes, correction on my end, I meant slope of 1 instead of slope of 0
William
what is f(x)=
I don't understand
Joe
Typically a function 'f' will take 'x' as input, and produce 'y' as output. As 'f(x)=y'. According to Google, "The range of a function is the complete set of all possible resulting values of the dependent variable (y, usually), after we have substituted the domain."
Thomas
Sorry, I don't know where the "Â"s came from. They shouldn't be there. Just ignore them. :-)
Thomas
Darius
Thanks.
Thomas
Â
Thomas
It is the  that should not be there. It doesn't seem to show if encloses in quotation marks. "Â" or 'Â' ... Â
Thomas
Now it shows, go figure?
Thomas
what is this?
i do not understand anything
unknown
lol...it gets better
Darius
I've been struggling so much through all of this. my final is in four weeks 😭
Tiffany
this book is an excellent resource! have you guys ever looked at the online tutoring? there's one that is called "That Tutor Guy" and he goes over a lot of the concepts
Darius
thank you I have heard of him. I should check him out.
Tiffany
is there any question in particular?
Joe
I have always struggled with math. I get lost really easy, if you have any advice for that, it would help tremendously.
Tiffany
Sure, are you in high school or college?
Darius
Hi, apologies for the delayed response. I'm in college.
Tiffany
how to solve polynomial using a calculator
So a horizontal compression by factor of 1/2 is the same as a horizontal stretch by a factor of 2, right? | 2020-02-19 00:58:22 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 100, "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": 0.7195873856544495, "perplexity": 1025.472624619352}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-10/segments/1581875143963.79/warc/CC-MAIN-20200219000604-20200219030604-00371.warc.gz"} |
http://questionpaper.org/calculus/ | # Calculus Solutions, Important Examples, Formulas and Videos
Calculus Important Questions - Page 6
Calculus Video Lecture - Page 7
Calculus is that branch of mathematics which mainly deals with the study of change in the value of a function as the points in the domain change. First, we give an intuitive idea of derivative (without act actually defining it). Then we give a naive definition of limit and study some algebra of limits. Then we come back to a definition of derivative and study some algebra of derivatives. We also obtain derivatives of certain standard function.
Limits: The above discussion clearly points towards the fact that we need to understand limiting process in greater clarity. We study a few illustrative examples to gain some familiarity with the concept of limit . Consider the function f(x ) = $x^{2}$ bserve that as x takes values very close to 0, the value of f ( x ) also moves towards 0.
$\lim_{x\rightarrow 0}f(x)$
In genral as $x\rightarrow a,f(x)\rightarrow l$ , then l is called limit of the function f(x) which is symbolically written as $\lim_{x\rightarrow o}f(x)=l$ .
Consider the following function g(x) = |x| , x $\neq$ . observer that g(0) is not defined. Computing the value of g(x) for the value of x very near to 0 we see that the value of g (x) moves towards zero .
So, $\lim_{x\rightarrow 0}g(x)=0$ .
This intuitively clear from the graph of y = |x | for x ≠ 0
Note : we say $\lim_{x\rightarrow a}f(x)$ is the expected value of f at x = a given the values of f near x to the left of a. This value is called the left hand limit of f at a.
We say $\lim_{x\rightarrow a}f(x)$ is the expected value of f at x = a given the values of f near x to the left of a. This value is called the right hand limit of of f(x) at a if the right and left limit coincide , we call common value as the limit of f(x) at x = a and denote it by $\lim_{x\rightarrow a}f(x)$
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10. Type | 2019-10-23 03:06:12 | {"extraction_info": {"found_math": true, "script_math_tex": 9, "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": 9, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.723339855670929, "perplexity": 1906.3754460461416}, "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/1570987828425.99/warc/CC-MAIN-20191023015841-20191023043341-00529.warc.gz"} |
https://www.zbmath.org/?q=an%3A1335.03047 | # zbMATH — the first resource for mathematics
Minimal solutions of generalized fuzzy relational equations: probabilistic algorithm based on greedy approach. (English) Zbl 1335.03047
Summary: The paper deals with generalized fuzzy relational equations that are defined within a recently introduced framework of sup-preserving aggregation structures. Generalized fuzzy relational equations subsume all previously studied types of fuzzy relational equations, namely those that are based on sup-t-norm and inf-residuum compositions. The paper contributes to previous studies of generalized fuzzy relational equations by presenting a method for constructing all minimal solutions and, consequently, for determining the whole solution set for any given generalized fuzzy relational equation that is solvable and for which every solution is bounded from below by a minimal solution. Moreover, in the paper we present a simple probabilistic algorithm for finding all minimal solutions.
##### MSC:
03E72 Theory of fuzzy sets, etc. 68T37 Reasoning under uncertainty in the context of artificial intelligence
Full Text:
##### References:
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Syst., 40, 599-609, (2011) · Zbl 1259.03065 [5] Bartl, E.; Belohlavek, R.; Vychodil, V., Bivalent and other solutions of fuzzy relational equations via linguistic hedges, Fuzzy Sets Syst., 187, 103-112, (2012) · Zbl 1258.03080 [6] Bartl, E.; Klir, G. J., Fuzzy relational equations in general framework, Int. J. Gen. Syst., 43, 1-18, (2014) · Zbl 1286.93004 [7] Belohlavek, R., Fuzzy relational systems: foundations and principles, (2002), Kluwer/Plenum New York · Zbl 1067.03059 [8] Belohlavek, R., Sup-t-norm and inf-residuum are one type of relational product: unifying framework and consequences, Fuzzy Sets Syst., 197, 45-58, (2012) · Zbl 1266.03056 [9] Belohlavek, R.; Vychodil, V., What is a fuzzy concept lattice?, (3rd Int. Conference on Concept Lattices and Their Applications, CLA 2005, (2005)), 34-45 [10] De Baets, B., Analytical solution methods for fuzzy relation equations, (Fundamentals of Fuzzy Sets, (2000), Kluwer Dordrecht), 291-340 · Zbl 0970.03044 [11] Di Nola, A.; Sanchez, E.; Pedrycz, W.; Sessa, S., Fuzzy relation equations and their applications to knowledge engineering, (1989), Kluwer Dordrecht · Zbl 0694.94025 [12] Galatos, N.; Jipsen, P.; Kowalski, T.; Ono, H., Residuated lattices: an algebraic glimpse at substructural logics, Stud. Logic Found. Math., vol. 151, (2007), Elsevier Amsterdam · Zbl 1171.03001 [13] Goguen, J., L-fuzzy sets, J. Math. Anal. 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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. | 2021-08-02 02:45:18 | {"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": 0.8691115379333496, "perplexity": 14032.424260708862}, "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/1627046154302.46/warc/CC-MAIN-20210802012641-20210802042641-00019.warc.gz"} |
https://cs.stackexchange.com/questions/125748/all-total-functions-form-a-prc-class | # All total functions form a PRC class
A class of total functions is a PRC class if:
1. The class includes all projection functions $$p_i(x_1,\ldots,x_n) = x_i$$ and the initial functions $$n(x) = 0$$ and $$s(x) = x+1$$.
2. The class is closed under composition and primitive recursion.
Prove that the class of all total functions is a PRC class.
Is it true if I say every total function is computable? And then say because the class of computable functions is a PRC class, so is the class of total functions.
• Not every total function is computable. – Yuval Filmus May 15 '20 at 11:21
• The exercise is much simpler. The closure operations in the definition of a PRC class all trivially hold for the class fo all functions. – Yuval Filmus May 15 '20 at 11:22
• Perhaps you can include the definition of PRC class (I had to look it up). – Yuval Filmus May 15 '20 at 11:22
• @YuvalFilmus The PRC class definition : We call a class of total functions a PRC class if: 1. It includes initial functions 2. It is close under composition and recursion – Z.Gh May 15 '20 at 11:28
• @YuvalFilmus the initial functions are S(x) = x+1 n(x) =0. And the projection functions – Z.Gh May 15 '20 at 11:29
You cannot say that every total function is computable, since that is not true. For example, the function $$f(n)$$ which returns $$1$$ if the $$n$$'th Turing machine halts and $$0$$ otherwise is not computable. | 2021-01-20 07:34:00 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 7, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6814794540405273, "perplexity": 781.6990081567869}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-04/segments/1610703519923.26/warc/CC-MAIN-20210120054203-20210120084203-00613.warc.gz"} |