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https://artofproblemsolving.com/wiki/index.php?title=1985_AJHSME_Problems/Problem_21&diff=next&oldid=72835	[ "# Difference between revisions of \"1985 AJHSME Problems/Problem 21\"\n\n## Problem\n\nMr. Green receives a", null, "$10\\%$ raise every year. His salary after four such raises has gone up by what percent?", null, "$\\text{(A)}\\ \\text{less than }40\\% \\qquad \\text{(B)}\\ 40\\% \\qquad \\text{(C)}\\ 44\\% \\qquad \\text{(D)}\\ 45\\% \\qquad \\text{(E)}\\ \\text{more than }45\\%$\n\n## Solution\n\nAssume his salary is 100 dollars then the next year he would have 110 dollars then the next year he would have 121 dollars then the next year he would have 133.1 dollars so therefore it is A\n\nso the percent is smaller than", null, "$40\\%$ and the only choice left is", null, "$\\boxed{\\text{A}}$. It is 4 raises, not 3. Thus, the correct answer is a 46.41 percent increase, or", null, "$\\boxed{\\text{E}}$.\n\nThe problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.", null, "" ]	[ null, "https://latex.artofproblemsolving.com/2/b/5/2b56bd637cb2c936d539be1b5000a64a898af389.png ", null, "https://latex.artofproblemsolving.com/7/6/9/769d7b59602692648661862a5e8b5611c522cea7.png ", null, "https://latex.artofproblemsolving.com/f/4/e/f4e4e5c6d549ce47918b94180cad5c1644d9ea88.png ", null, "https://latex.artofproblemsolving.com/d/7/9/d79e0e8ce41b5c08729443bcc258d13ca061ff35.png ", null, "https://latex.artofproblemsolving.com/d/1/3/d1302e73ece29d10218238af739577904f6f00a3.png ", null, "https://wiki-images.artofproblemsolving.com//8/8b/AMC_logo.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.835995,"math_prob":0.98879075,"size":1369,"snap":"2021-43-2021-49","text_gpt3_token_len":423,"char_repetition_ratio":0.11941392,"word_repetition_ratio":0.21960784,"special_character_ratio":0.35719502,"punctuation_ratio":0.09558824,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9938856,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-19T20:26:00Z\",\"WARC-Record-ID\":\"<urn:uuid:f8a201bd-9dc0-4ddc-a2b1-e794d120099b>\",\"Content-Length\":\"42739\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:efe69e70-d1b8-4b62-b7b8-018c5f69f98b>\",\"WARC-Concurrent-To\":\"<urn:uuid:68c5a025-48e6-4ef0-baf9-19e385d3923e>\",\"WARC-IP-Address\":\"172.67.69.208\",\"WARC-Target-URI\":\"https://artofproblemsolving.com/wiki/index.php?title=1985_AJHSME_Problems/Problem_21&diff=next&oldid=72835\",\"WARC-Payload-Digest\":\"sha1:GMFJXHFVNRUWMWNHT6P3EA5ZDISUEWZH\",\"WARC-Block-Digest\":\"sha1:77LXTH74W4SEUWWYKIH2XJD73AWBWC4L\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323585281.35_warc_CC-MAIN-20211019202148-20211019232148-00562.warc.gz\"}"}
https://www.mdpi.com/journal/symmetry/special_issues/orthogonal_polynomials	[ "", null, "Journal Browser\n\n# Special Issue \"Symmetry in Orthogonal Polynomials\"\n\nA special issue of Symmetry (ISSN 2073-8994).\n\nDeadline for manuscript submissions: closed (16 August 2016).\n\n## Special Issue Editor\n\nProf. Charles F. Dunkl\nE-Mail Website\nGuest Editor\nDepartment of Mathematics, University of Virginia, Charlottesville, Virginia, USA\nInterests: harmonic analysis; representation theory; special functions of several variables; applications to mathematical physics, especially exactly solvable systems of quantum mechanics\nSpecial Issues and Collections in MDPI journals\n\n## Special Issue Information\n\nDear Colleagues,\n\nThe concept of symmetry has been fundamental and studied for millennia. The ancient geometers already knew the five regular solids. For a long time, symmetry was a part of the discipline of geometry, but in more recent times it has become very important in analysis, mathematical physics, and of course, group theory. Symmetry is a key tool in analyzing functions of several variables. For example, the harmonic homogeneous polynomials, which are invariant under the group of rotations fixing the North Pole on the unit sphere in are essentially the same as Gegenbauer polynomials of index N/2−1. By now, this idea has been vastly generalized, for example, to interpreting Jacobi polynomials of several variables (defined on a simplex and orthogonal with respect to a Dirichlet measure) as harmonic polynomials with certain subgroup invariance properties.\n\nIn mathematical physics there are the quantum-mechanical models of Calogero–Moser–Sutherland type: N identical particles with 1/r2 interaction and possibly an external potential, of which wavefunctions involve Jack polynomials. The symmetric group occurs naturally in any system of identical particles, where the properties are invariant under the interchange of two particles. Recent developments have extended this to “supermodels” and introduced super polynomials with bosonic (commuting) and fermionic (anticommuting) variables. These are involved in the open question of whether SUSY (supersymmetry) manifests in the real world. Another application of orthogonal polynomials is as wavefunctions of isotropic quantum harmonic oscillators. Lie and quadratic algebras are being used to provide more insight into the Askey tableau, a scheme for organizing the classical polynomials of hypergeometric type.\n\nSymmetry appears in algebraic combinatorics, for example in association schemes and distance-regular graphs. These structures are analyzed with the help of orthogonal polynomials, which arise as eigenfunctions of an associated Laplacian operator.\n\nSymmetry in orthogonal polynomials also appears when the domain is a symmetric shape and the weight function is invariant under a group generated by reflections: for example orthogonal polynomials on a regular hexagon find an application in wave-front analysis for hexagonal mirror segments in large astronomical telescopes. Another example is the analysis of trigonometric polynomials, which are periodic on a lattice (or tesselation of space by regular polytopes).\n\nIn this Special Issue we aim to present the newest developments in the interaction of symmetry and orthogonal polynomials, in areas such as quantum physics, combinatorics, and classical analysis problems dealing with convergence of polynomial expansions. In the situations discussed above, much has been discovered, nevertheless, more needs to be done, in more precise formulas, approximation theorems about expansions in orthogonal polynomials of several variables, dependence on the parameters of a weight function, vanishing properties of specific polynomials (such as Jack and Macdonald), and so on.\n\nProf. Charles F. Dunkl\nGuest Editor\n\nManuscript Submission Information\n\nManuscripts should be submitted online at www.mdpi.com by registering and logging in to this website. Once you are registered, click here to go to the submission form. Manuscripts can be submitted until the deadline. All papers will be peer-reviewed. Accepted papers will be published continuously in the journal (as soon as accepted) and will be listed together on the special issue website. Research articles, review articles as well as short communications are invited. For planned papers, a title and short abstract (about 100 words) can be sent to the Editorial Office for announcement on this website.\n\nSubmitted manuscripts should not have been published previously, nor be under consideration for publication elsewhere (except conference proceedings papers). All manuscripts are thoroughly refereed through a single-blind peer-review process. A guide for authors and other relevant information for submission of manuscripts is available on the Instructions for Authors page. Symmetry is an international peer-reviewed open access monthly journal published by MDPI.\n\nPlease visit the Instructions for Authors page before submitting a manuscript. The Article Processing Charge (APC) for publication in this open access journal is 1400 CHF (Swiss Francs). Submitted papers should be well formatted and use good English. Authors may use MDPI's English editing service prior to publication or during author revisions.\n\n## Keywords\n\n• weight functions invariant under reflection groups\n• Calogero-Moser-Sutherland models\n• Jack polynomials\n• orthogonal polynomials in several variables of classical type\n• polynomials periodic on a lattice\n\n## Published Papers (6 papers)\n\nOrder results\nResult details\nSelect all\nExport citation of selected articles as:\n\n# Research\n\nOpen AccessFeature PaperArticle\nPlanar Harmonic and Monogenic Polynomials of Type A\nby", null, "Charles F. Dunkl\nSymmetry 2016, 8(10), 108; https://doi.org/10.3390/sym8100108 - 21 Oct 2016\nAbstract\nHarmonic polynomials of type A are polynomials annihilated by the Dunkl Laplacian associated to the symmetric group acting as a reflection group on $R N$ . The Dunkl operators are denoted by $T j$ for $1 ≤ j ≤ N$ , and the [...] Read more.\nHarmonic polynomials of type A are polynomials annihilated by the Dunkl Laplacian associated to the symmetric group acting as a reflection group on $R N$ . The Dunkl operators are denoted by $T j$ for $1 ≤ j ≤ N$ , and the Laplacian $Δ κ = ∑ j = 1 N T j 2$ . This paper finds the homogeneous harmonic polynomials annihilated by all $T j$ for $j > 2$ . The structure constants with respect to the Gaussian and sphere inner products are computed. These harmonic polynomials are used to produce monogenic polynomials, those annihilated by a Dirac-type operator. Full article\nOpen AccessArticle\nThe Role of Orthogonal Polynomials in Tailoring Spherical Distributions to Kurtosis Requirements\nby", null, "Luca Bagnato ,", null, "Mario Faliva and", null, "Maria Grazia Zoia\nSymmetry 2016, 8(8), 77; https://doi.org/10.3390/sym8080077 - 05 Aug 2016\nAbstract\nThis paper carries out an investigation of the orthogonal-polynomial approach to reshaping symmetric distributions to fit in with data requirements so as to cover the multivariate case. With this objective in mind, reference is made to the class of spherical distributions, given that [...] Read more.\nThis paper carries out an investigation of the orthogonal-polynomial approach to reshaping symmetric distributions to fit in with data requirements so as to cover the multivariate case. With this objective in mind, reference is made to the class of spherical distributions, given that they provide a natural multivariate generalization of univariate even densities. After showing how to tailor a spherical distribution via orthogonal polynomials to better comply with kurtosis requirements, we provide operational conditions for the positiveness of the resulting multivariate Gram–Charlier-like expansion, together with its kurtosis range. Finally, the approach proposed here is applied to some selected spherical distributions. Full article\nShow Figures", null, "Figure 1\n\nOpen AccessArticle\nCubature Formulas of Multivariate Polynomials Arising from Symmetric Orbit Functions\nby", null, "Jiří Hrivnák ,", null, "Lenka Motlochová and", null, "Jiří Patera\nSymmetry 2016, 8(7), 63; https://doi.org/10.3390/sym8070063 - 14 Jul 2016\nCited by 9\nAbstract\nThe paper develops applications of symmetric orbit functions, known from irreducible representations of simple Lie groups, in numerical analysis. It is shown that these functions have remarkable properties which yield to cubature formulas, approximating a weighted integral of any function by a weighted [...] Read more.\nThe paper develops applications of symmetric orbit functions, known from irreducible representations of simple Lie groups, in numerical analysis. It is shown that these functions have remarkable properties which yield to cubature formulas, approximating a weighted integral of any function by a weighted finite sum of function values, in connection with any simple Lie group. The cubature formulas are specialized for simple Lie groups of rank two. An optimal approximation of any function by multivariate polynomials arising from symmetric orbit functions is discussed. Full article\nShow Figures", null, "Figure 1\n\nOpen AccessArticle\nThree New Classes of Solvable N-Body Problems of Goldfish Type with Many Arbitrary Coupling Constants\nby", null, "Francesco Calogero\nSymmetry 2016, 8(7), 53; https://doi.org/10.3390/sym8070053 - 24 Jun 2016\nCited by 7\nAbstract\nThree new classes of N-body problems of goldfish type are identified, with N an arbitrary positive integer ( $N ≥ 2$ ). These models are characterized by nonlinear Newtonian (“accelerations equal forces”) equations of motion describing N equal point-particles moving in the [...] Read more.\nThree new classes of N-body problems of goldfish type are identified, with N an arbitrary positive integer ( $N ≥ 2$ ). These models are characterized by nonlinear Newtonian (“accelerations equal forces”) equations of motion describing N equal point-particles moving in the complex z-plane. These highly nonlinear equations feature many arbitrary coupling constants, yet they can be solved by algebraic operations. Some of these N-body problems are isochronous, their generic solutions being all completely periodic with an overall period T independent of the initial data (but quite a few of these solutions are actually periodic with smaller periods $T / p$ with p a positive integer); other models are isochronous for an open region of initial data, while the motions for other initial data are not periodic, featuring instead scattering phenomena with some of the particles incoming from, or escaping to, infinity in the remote past or future. Full article\nOpen AccessArticle\nOn a Reduction Formula for a Kind of Double q-Integrals\nby", null, "Zhi-Guo Liu\nSymmetry 2016, 8(6), 44; https://doi.org/10.3390/sym8060044 - 08 Jun 2016\nCited by 3\nAbstract\nUsing the q-integral representation of Sears’ nonterminating extension of the q-Saalschütz summation, we derive a reduction formula for a kind of double q-integrals. This reduction formula is used to derive a curious double q-integral formula, and also allows us [...] Read more.\nUsing the q-integral representation of Sears’ nonterminating extension of the q-Saalschütz summation, we derive a reduction formula for a kind of double q-integrals. This reduction formula is used to derive a curious double q-integral formula, and also allows us to prove a general q-beta integral formula including the Askey–Wilson integral formula as a special case. Using this double q-integral formula and the theory of q-partial differential equations, we derive a general q-beta integral formula, which includes the Nassrallah–Rahman integral as a special case. Our evaluation does not require the orthogonality relation for the q-Hermite polynomials and the Askey–Wilson integral formula. Full article\nby", null, "Robert Griffiths\nThis paper defines the multivariate Krawtchouk polynomials, orthogonal on the multinomial distribution, and summarizes their properties as a review. The multivariate Krawtchouk polynomials are symmetric functions of orthogonal sets of functions defined on each of N multinomial trials. The dual multivariate Krawtchouk polynomials, which also have a polynomial structure, are seen to occur naturally as spectral orthogonal polynomials in a Karlin and McGregor spectral representation of transition functions in a composition birth and death process. In this Markov composition process in continuous time, there are N independent and identically distributed birth and death processes each with support $0 , 1 , …$ . The state space in the composition process is the number of processes in the different states $0 , 1 , …$ . Dealing with the spectral representation requires new extensions of the multivariate Krawtchouk polynomials to orthogonal polynomials on a multinomial distribution with a countable infinity of states. 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https://pykeen.readthedocs.io/en/stable/reference/models.html	[ "# Models\n\nA knowledge graph embedding model is capable of computing real-valued scores representing the plausibility of a triple $$(h,r,t) \\in \\mathbb{K}$$, where a larger score indicates a higher plausibility. The interpretation of the score value is model-dependent, and usually it cannot be directly interpreted as a probability.\n\nIn PyKEEN, the API of a model is defined in Model, where the scoring function is exposed as Model.score_hrt(), which can be used to compute plausability scores for (a batch of) triples. In addition, the Model class also offers additional scoring methods, which can be used to (efficiently) compute scores for a large number of triples sharing some parts, e.g., to compute scores for triples (h, r, e) for a given (h, r) pair and all available entities $$e \\in \\mathcal{E}$$.\n\nNote\n\nThe implementations of the knowledge graph embedding models provided here all operate on entity / relation indices rather than string representations, cf. here.\n\nOn top of these scoring methods, there are also corresponding prediction methods, e.g., Model.predict_hrt(). These methods extend the scoring ones, by ensuring the model is in evaluation mode, cf. torch.nn.Module.eval(), and optionally applying a sigmoid activation on the scores to ensure a value range of $$[0, 1]$$.\n\nWarning\n\nDepending on the model at hand, directly applying sigmoid might not always be sensible. For instance, distance-based interaction functions, such as pykeen.nn.modules.TransEInteraction, result in non-positive scores (since they use the negative distance as scoring function), and thus the output of the sigmoid only covers the interval $$[0.5, 1]$$.\n\nMost models derive from ERModel, which is a generic implementation of a knowledge graph embedding model. It combines a variable number of representations for entities and relations, cf. pykeen.nn.representation.Representation, and an interaction function, cf. pykeen.nn.modules.Interaction. The representation modules convert integer entity or relation indices to numeric representations, e.g., vectors. The interaction function takes the representations of the head entities, relations and tail entities as input and computes a scalar plausability score for triples.\n\nNote\n\nAn in-depth discussion of representation modules can be found in the corresponding tutorial.\n\nNote\n\nThe specific models from this module, e.g., RESCAL, package given specific entity and relation representations with an interaction function. For more flexible combinations, consider using ERModel directly.\n\n## Functions\n\n make_model(dimensions, interaction[, ...]) Build a model from an interaction class hint (name or class). make_model_cls(dimensions, interaction[, ...]) Build a model class from an interaction class hint (name or class).\n\n## Classes\n\n Model(, triples_factory[, loss, ...]) A base module for KGE models. ERModel(, triples_factory, interaction[, ...]) A commonly useful base for KGEMs using embeddings and interaction modules. InductiveERModel(, triples_factory[, ...]) A base class for inductive models. LiteralModel(triples_factory, interaction[, ...]) Base class for models with entity literals that uses combinations from pykeen.nn.combinations. EvaluationOnlyModel(triples_factory) A model which only implements the methods used for evaluation. AutoSF([embedding_dim, num_components, ...]) An implementation of AutoSF from [zhang2020]. BoxE([, embedding_dim, tanh_map, p, ...]) An implementation of BoxE from [abboud2020]. CompGCN(, triples_factory[, embedding_dim, ...]) An implementation of CompGCN from [vashishth2020]. ComplEx([, embedding_dim, ...]) An implementation of ComplEx [trouillon2016]. ComplExLiteral(triples_factory[, ...]) An implementation of the LiteralE model with the ComplEx interaction from [kristiadi2018]. ConvE(triples_factory[, input_channels, ...]) An implementation of ConvE from [dettmers2018]. ConvKB([, embedding_dim, ...]) An implementation of ConvKB from [nguyen2018]. CP([embedding_dim, rank, ...]) An implementation of CP as described in [lacroix2018] based on [hitchcock1927]. CrossE([, embedding_dim, ...]) An implementation of CrossE from [zhang2019b]. DistMA([embedding_dim, entity_initializer, ...]) An implementation of DistMA from [shi2019]. DistMult([, embedding_dim, ...]) An implementation of DistMult from [yang2014]. DistMultLiteral(triples_factory[, ...]) An implementation of the LiteralE model with the DistMult interaction from [kristiadi2018]. DistMultLiteralGated(triples_factory[, ...]) An implementation of the LiteralE model with thhe Gated DistMult interaction from [kristiadi2018]. ERMLP([, embedding_dim, hidden_dim, ...]) An implementation of ERMLP from [dong2014]. ERMLPE([, embedding_dim, hidden_dim, ...]) An extension of pykeen.models.ERMLP proposed by [sharifzadeh2019]. HolE([, embedding_dim, entity_initializer, ...]) An implementation of HolE [nickel2016]. KG2E([, embedding_dim, dist_similarity, ...]) An implementation of KG2E from [he2015]. FixedModel(, triples_factory, *_kwargs) A mock model returning fixed scores. MuRE([, embedding_dim, p, power_norm, ...]) An implementation of MuRE from [balazevic2019b]. NodePiece(, triples_factory[, num_tokens, ...]) A wrapper which combines an interaction function with NodePiece entity representations from [galkin2021]. NTN([, embedding_dim, num_slices, ...]) An implementation of NTN from [socher2013]. PairRE([embedding_dim, p, power_norm, ...]) An implementation of PairRE from [chao2020]. ProjE([, embedding_dim, ...]) An implementation of ProjE from [shi2017]. QuatE([, embedding_dim, ...]) An implementation of QuatE from [zhang2019]. RESCAL([, embedding_dim, ...]) An implementation of RESCAL from [nickel2011]. RGCN(, triples_factory[, embedding_dim, ...]) An implementation of R-GCN from [schlichtkrull2018]. RotatE([, embedding_dim, ...]) An implementation of RotatE from [sun2019]. SimplE([, embedding_dim, clamp_score, ...]) An implementation of SimplE [kazemi2018]. SE([, embedding_dim, scoring_fct_norm, ...]) An implementation of the Structured Embedding (SE) published by [bordes2011]. TorusE([embedding_dim, p, power_norm, ...]) An implementation of TorusE from [ebisu2018]. TransD([, embedding_dim, relation_dim, ...]) An implementation of TransD from [ji2015]. TransE([, embedding_dim, scoring_fct_norm, ...]) An implementation of TransE [bordes2013]. TransF([embedding_dim, entity_initializer, ...]) An implementation of TransF from [feng2016]. TransH([, embedding_dim, scoring_fct_norm, ...]) An implementation of TransH [wang2014]. TransR([, embedding_dim, relation_dim, ...]) An implementation of TransR from [lin2015]. TuckER([, embedding_dim, relation_dim, ...]) An implementation of TuckEr from [balazevic2019]. UM([, embedding_dim, scoring_fct_norm, ...]) An implementation of the Unstructured Model (UM) published by [bordes2014]. InductiveNodePiece(, triples_factory, ...) A wrapper which combines an interaction function with NodePiece entity representations from [galkin2021]. InductiveNodePieceGNN([, gnn_encoder]) Inductive NodePiece with a GNN encoder on top. SoftInverseTripleBaseline(triples_factory[, ...]) Score based on relation similarity. MarginalDistributionBaseline(triples_factory) Score based on marginal distributions. CooccurrenceFilteredModel(, triples_factory) A model which filters predictions by co-occurence.\n\n## Class Inheritance Diagram", null, "" ]	[ null, "https://pykeen.readthedocs.io/en/stable/_images/inheritance-502a223ba8fbcf008ea1580bb066dd34fc69a1ec.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7438499,"math_prob":0.98337585,"size":5363,"snap":"2023-14-2023-23","text_gpt3_token_len":1270,"char_repetition_ratio":0.264415,"word_repetition_ratio":0.12607099,"special_character_ratio":0.2496737,"punctuation_ratio":0.14642452,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.984049,"pos_list":[0,1,2],"im_url_duplicate_count":[null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-06-09T11:02:44Z\",\"WARC-Record-ID\":\"<urn:uuid:76485a25-0b49-40ab-a329-fc5fbe6130bb>\",\"Content-Length\":\"61754\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ef254f36-37d6-42fa-8d0e-7f16bbcdc953>\",\"WARC-Concurrent-To\":\"<urn:uuid:b286186b-4796-41d7-9bbd-5b6de5c92898>\",\"WARC-IP-Address\":\"104.17.32.82\",\"WARC-Target-URI\":\"https://pykeen.readthedocs.io/en/stable/reference/models.html\",\"WARC-Payload-Digest\":\"sha1:OM7BAADAEQZMEGDXHL2XR3JPYL5ONGCJ\",\"WARC-Block-Digest\":\"sha1:PUBTEBBVACP2W3XT7O46S2G26S2FY6H3\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224656675.90_warc_CC-MAIN-20230609100535-20230609130535-00734.warc.gz\"}"}
https://www.colorhexa.com/00db79	[ "# #00db79 Color Information\n\nIn a RGB color space, hex #00db79 is composed of 0% red, 85.9% green and 47.5% blue. Whereas in a CMYK color space, it is composed of 100% cyan, 0% magenta, 44.7% yellow and 14.1% black. It has a hue angle of 153.2 degrees, a saturation of 100% and a lightness of 42.9%. #00db79 color hex could be obtained by blending #00fff2 with #00b700. Closest websafe color is: #00cc66.\n\n• R 0\n• G 86\n• B 47\nRGB color chart\n• C 100\n• M 0\n• Y 45\n• K 14\nCMYK color chart\n\n#00db79 color description : Pure (or mostly pure) cyan - lime green.\n\n# #00db79 Color Conversion\n\nThe hexadecimal color #00db79 has RGB values of R:0, G:219, B:121 and CMYK values of C:1, M:0, Y:0.45, K:0.14. Its decimal value is 56185.\n\nHex triplet RGB Decimal 00db79 `#00db79` 0, 219, 121 `rgb(0,219,121)` 0, 85.9, 47.5 `rgb(0%,85.9%,47.5%)` 100, 0, 45, 14 153.2°, 100, 42.9 `hsl(153.2,100%,42.9%)` 153.2°, 100, 85.9 00cc66 `#00cc66`\nCIE-LAB 77.305, -66.421, 35.818 28.781, 52.04, 26.616 0.268, 0.484, 52.04 77.305, 75.463, 151.663 77.305, -68.717, 58.658 72.139, -55.029, 28.622 00000000, 11011011, 01111001\n\n# Color Schemes with #00db79\n\n• #00db79\n``#00db79` `rgb(0,219,121)``\n• #db0062\n``#db0062` `rgb(219,0,98)``\nComplementary Color\n• #00db0b\n``#00db0b` `rgb(0,219,11)``\n• #00db79\n``#00db79` `rgb(0,219,121)``\n• #00d0db\n``#00d0db` `rgb(0,208,219)``\nAnalogous Color\n• #db0b00\n``#db0b00` `rgb(219,11,0)``\n• #00db79\n``#00db79` `rgb(0,219,121)``\n• #db00d0\n``#db00d0` `rgb(219,0,208)``\nSplit Complementary Color\n• #db7900\n``#db7900` `rgb(219,121,0)``\n• #00db79\n``#00db79` `rgb(0,219,121)``\n• #7900db\n``#7900db` `rgb(121,0,219)``\n• #62db00\n``#62db00` `rgb(98,219,0)``\n• #00db79\n``#00db79` `rgb(0,219,121)``\n• #7900db\n``#7900db` `rgb(121,0,219)``\n• #db0062\n``#db0062` `rgb(219,0,98)``\n• #008f4f\n``#008f4f` `rgb(0,143,79)``\n• #00a85d\n``#00a85d` `rgb(0,168,93)``\n• #00c26b\n``#00c26b` `rgb(0,194,107)``\n• #00db79\n``#00db79` `rgb(0,219,121)``\n• #00f587\n``#00f587` `rgb(0,245,135)``\n• #0fff94\n``#0fff94` `rgb(15,255,148)``\n• #29ff9f\n``#29ff9f` `rgb(41,255,159)``\nMonochromatic Color\n\n# Alternatives to #00db79\n\nBelow, you can see some colors close to #00db79. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #00db42\n``#00db42` `rgb(0,219,66)``\n• #00db54\n``#00db54` `rgb(0,219,84)``\n• #00db67\n``#00db67` `rgb(0,219,103)``\n• #00db79\n``#00db79` `rgb(0,219,121)``\n• #00db8b\n``#00db8b` `rgb(0,219,139)``\n• #00db9e\n``#00db9e` `rgb(0,219,158)``\n• #00dbb0\n``#00dbb0` `rgb(0,219,176)``\nSimilar Colors\n\n# #00db79 Preview\n\nThis text has a font color of #00db79.\n\n``<span style=\"color:#00db79;\">Text here</span>``\n#00db79 background color\n\nThis paragraph has a background color of #00db79.\n\n``<p style=\"background-color:#00db79;\">Content here</p>``\n#00db79 border color\n\nThis element has a border color of #00db79.\n\n``<div style=\"border:1px solid #00db79;\">Content here</div>``\nCSS codes\n``.text {color:#00db79;}``\n``.background {background-color:#00db79;}``\n``.border {border:1px solid #00db79;}``\n\n# Shades and Tints of #00db79\n\nA shade is achieved by adding black to any pure hue, while a tint is created by mixing white to any pure color. In this example, #000302 is the darkest color, while #effff8 is the lightest one.\n\n• #000302\n``#000302` `rgb(0,3,2)``\n• #00170d\n``#00170d` `rgb(0,23,13)``\n• #002a17\n``#002a17` `rgb(0,42,23)``\n• #003e22\n``#003e22` `rgb(0,62,34)``\n• #00522d\n``#00522d` `rgb(0,82,45)``\n• #006538\n``#006538` `rgb(0,101,56)``\n• #007943\n``#007943` `rgb(0,121,67)``\n• #008d4e\n``#008d4e` `rgb(0,141,78)``\n• #00a058\n``#00a058` `rgb(0,160,88)``\n• #00b463\n``#00b463` `rgb(0,180,99)``\n• #00c76e\n``#00c76e` `rgb(0,199,110)``\n• #00db79\n``#00db79` `rgb(0,219,121)``\n• #00ef84\n``#00ef84` `rgb(0,239,132)``\n• #03ff8e\n``#03ff8e` `rgb(3,255,142)``\n• #17ff97\n``#17ff97` `rgb(23,255,151)``\n• #2affa0\n``#2affa0` `rgb(42,255,160)``\n• #3effa9\n``#3effa9` `rgb(62,255,169)``\n• #52ffb1\n``#52ffb1` `rgb(82,255,177)``\n• #65ffba\n``#65ffba` `rgb(101,255,186)``\n• #79ffc3\n``#79ffc3` `rgb(121,255,195)``\n• #8dffcc\n``#8dffcc` `rgb(141,255,204)``\n• #a0ffd5\n``#a0ffd5` `rgb(160,255,213)``\n• #b4ffdd\n``#b4ffdd` `rgb(180,255,221)``\n• #c7ffe6\n``#c7ffe6` `rgb(199,255,230)``\n• #dbffef\n``#dbffef` `rgb(219,255,239)``\n• #effff8\n``#effff8` `rgb(239,255,248)``\nTint Color Variation\n\n# Tones of #00db79\n\nA tone is produced by adding gray to any pure hue. In this case, #65766e is the less saturated color, while #00db79 is the most saturated one.\n\n• #65766e\n``#65766e` `rgb(101,118,110)``\n• #5d7e6f\n``#5d7e6f` `rgb(93,126,111)``\n• #548770\n``#548770` `rgb(84,135,112)``\n• #4c8f71\n``#4c8f71` `rgb(76,143,113)``\n• #439872\n``#439872` `rgb(67,152,114)``\n• #3ba073\n``#3ba073` `rgb(59,160,115)``\n• #33a874\n``#33a874` `rgb(51,168,116)``\n• #2ab175\n``#2ab175` `rgb(42,177,117)``\n• #22b975\n``#22b975` `rgb(34,185,117)``\n• #19c276\n``#19c276` `rgb(25,194,118)``\n• #11ca77\n``#11ca77` `rgb(17,202,119)``\n• #08d378\n``#08d378` `rgb(8,211,120)``\n• #00db79\n``#00db79` `rgb(0,219,121)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #00db79 is perceived by people affected by a color vision deficiency. This can be useful if you need to ensure your color combinations are accessible to color-blind users.\n\nMonochromacy\n• Achromatopsia 0.005% of the population\n• Atypical Achromatopsia 0.001% of the population\nDichromacy\n• Protanopia 1% of men\n• Deuteranopia 1% of men\n• Tritanopia 0.001% of the population\nTrichromacy\n• Protanomaly 1% of men, 0.01% of women\n• Deuteranomaly 6% of men, 0.4% of women\n• Tritanomaly 0.01% of the population" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.5327989,"math_prob":0.7306176,"size":3704,"snap":"2019-51-2020-05","text_gpt3_token_len":1596,"char_repetition_ratio":0.13270271,"word_repetition_ratio":0.010989011,"special_character_ratio":0.5537257,"punctuation_ratio":0.23146068,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99106187,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-28T16:48:05Z\",\"WARC-Record-ID\":\"<urn:uuid:0cb2b55a-8b21-4ff5-a611-3f2b50c9ebb1>\",\"Content-Length\":\"36274\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:27cc7384-25b6-4ed6-b15b-a920a73eefdf>\",\"WARC-Concurrent-To\":\"<urn:uuid:e9e440ea-ebdf-44c5-a9b9-04ee243f3102>\",\"WARC-IP-Address\":\"178.32.117.56\",\"WARC-Target-URI\":\"https://www.colorhexa.com/00db79\",\"WARC-Payload-Digest\":\"sha1:VIGHM4KFE5KMBQASXNH7NVJXA5NZLMB7\",\"WARC-Block-Digest\":\"sha1:AFFDCFLMUKZEXIXQXGZX4BFQKKYIGEMM\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579251779833.86_warc_CC-MAIN-20200128153713-20200128183713-00170.warc.gz\"}"}
https://socratic.org/questions/552b4d09581e2a667e2bfe91	[ "# Question #bfe91\n\nApr 13, 2015\n\nSo, you need to prepare a certain volume of a solution that has a certain molarity.\n\nMolarity expresses the number of moles of solute, which you have to determine, dissolved per liter of solution.\n\n$C = \\frac{n}{V}$\n\nIf you have 1 mole of a solute dissolved in 1 L of solution, you'll have a 1 M solution\n\n$C = \\text{1 mole\"/\"1 L\" = \"1 mol/\"L = \"1 M}$\n\nSince you know the volume and the molarity, you can rearrange the equation to solve for the number of moles that would give you that respective molarity\n\n$C = \\frac{n}{V} \\implies n = C \\cdot V$\n\nFor the above example, if you are given 1 L of a 1-M solution, the number of moles of solute will be\n\n$n = 1 \\text{mol\"/cancel(\"L\") * 1cancel(\"L\") = \"1 mole}$", null, "Keep in mind that you must always use liters for the volume; even if you are given the volume in mililiters, you must convert it to liters when applying the molarity equation.\n\nHere's another example. Let's say I give you 100 mL of a 1-M solution and ask you to determine the number of moles of solute present.\n\nYou'd get\n\n$n = C \\cdot V = 1 \\text{mol\"/cancel(\"L\") * 100 * 10^(-3)cancel(\"L\") = \"0.1 moles}$\n\nCompare this with the 1-L, 1-M solution. The volume is now 10 times smaller, but the molarity is the same, which means that you must have 10 times fewer moles present.\n\nIf you add 1 mole to 100 mL, the molarity will be 10 times greater than when you had 1 mole in 1 L because you get the same amount of moles in a smaller volume $\\to$ increased molarity.\n\n$C = \\frac{n}{V} = \\text{1 mole\"/(100 * 10^(-3)\"L\") = \"10 mol/L\" = \"10 M}$" ]	[ null, "https://useruploads.socratic.org/vy0sQt1R8u6Y9mi7uWbg_screen-shot-2013-09-29-at-2-03-31-pm.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.92226213,"math_prob":0.9981027,"size":1259,"snap":"2022-40-2023-06","text_gpt3_token_len":316,"char_repetition_ratio":0.15697211,"word_repetition_ratio":0.021097047,"special_character_ratio":0.23431295,"punctuation_ratio":0.0754717,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9998801,"pos_list":[0,1,2],"im_url_duplicate_count":[null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-10-03T15:15:26Z\",\"WARC-Record-ID\":\"<urn:uuid:6c0a1523-8e8e-49fa-99d6-789ebdda5ce2>\",\"Content-Length\":\"37042\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:685c43d0-082e-4d1f-8ffa-11942b2f71a3>\",\"WARC-Concurrent-To\":\"<urn:uuid:07cb6f02-1720-4903-84f4-d02aa49ab180>\",\"WARC-IP-Address\":\"216.239.32.21\",\"WARC-Target-URI\":\"https://socratic.org/questions/552b4d09581e2a667e2bfe91\",\"WARC-Payload-Digest\":\"sha1:75DFPNCC3UJVGMABYNLXYQOKHEOZTZ53\",\"WARC-Block-Digest\":\"sha1:ZHGEB3TVVIWNASEP5TFQXLQYCEIGHVQR\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030337421.33_warc_CC-MAIN-20221003133425-20221003163425-00663.warc.gz\"}"}
http://slideplayer.com/slide/6412219/	[ "Presentation is loading. Please wait.", null, "# Chapter 1 Exploring Data\n\n## Presentation on theme: \"Chapter 1 Exploring Data\"— Presentation transcript:\n\nChapter 1 Exploring Data\n\nIntroduction Statistics: Individuals: Variable:\nthe science of data. We begin our study of statistics by mastering the art of examining data. Any set of data contains information about some group of individuals. The information is organized in variables. Individuals: The objects described by a set of data. Individuals may be people, but they may also be other things. Variable: Any characteristic of an individual. Can take different values for different individuals.\n\nVariable Types Categorical variable: Quantitative variable:\nplaces an individual into one of several groups of categories. Quantitative variable: takes numerical values for which arithmetic operations such as adding and averaging make sense. Distribution: pattern of variation of a variable tells what values the variable takes and how often it takes these values.\n\nB. The variables given are\nA. The individuals are the BMW 318I, the Buick Century, and the Chevrolet Blazer. B. The variables given are Vehicle type (categorical) Transmission type (categorical) Number of cylinders (quantitative) City MPG (quantitative) Highway MPG (quantitative)\n\n1.1: Displaying Distributions with graphs.\nGraphs used to display data: bar graphs, pie charts, dot plots, stem plots, histograms, and time plots Purpose of a graph: Helps to understand the data. Allows overall patterns and striking deviations from that pattern to be seen. Describing the overall pattern: Three biggest descriptors: shape, center and spread. Next look for outliers and clusters.\n\nShape Concentrate on main features. Types of Shapes:\nMajor peaks, outliers (not just the smallest and largest observations), rough symmetry or clear skewness. Types of Shapes: Symmetric Skewed right Skewed left\n\nHow to make a bar graph.\n\n1.5 How to make a bar graph. Percent of females among people earning doctorates in 1994. 70 60.8% 62.2% 60 Percent 50 40.7% 40 30 21.7% 20 15.4% 11.1% 10 Computer science Physical sciences Education Engineering Life sciences Psychology\n\nNo, a pie chart is used to display one variable with all of its categories totaling 100%\n\nHighway mpg for some 2000 midsize cars\nHow to make a dotplot Highway mpg for some 2000 midsize cars 10 8 Frequency or Count 6 4 2 21 22 23 24 25 26 27 28 29 30 31 32 MPG\n\nHow to make and read a stemplot\nA stemplot is similar to a dotplot but there are some format differences. Instead of dots actual numbers are used. Instead of a horizontal axis, a vertical one is used. Stems Leaves Leaves are single digits only This arrangement would be read as the numbers 523 and 526.\n\nHow to make and read a stemplot\nWith the following data, make a stemplot. Stems Leaves\n\nHow to make and read a stemplot\nLets use the same stemplot but now split the stems Stems Leaves Split stems Leaves, first stem uses number 0-4, second uses numbers 5-9\n\nHow to construct a histogram\nThe most common graph of the distribution of one quantitative variable is a histogram. To make a histogram: Divide the range into equal widths. Then count the number of observations that fall in each group. Label and scale your axes and title your graph. Draw bars that represent each count, no space between bars.\n\nDivide range into equal widths and count\n0 < CEO Salary < 100 100 < CEO Salary < 200 200 < CEO Salary < 300 300 < CEO Salary < 400 400 < CEO Salary < 500 500 < CEO Salary < 600 600 < CEO Salary < 700 700 < CEO Salary < 800 800 < CEO Salary < 900 Scale 1 3 11 10 2 Counts\n\nDraw and label axis, then make bars\nCEO Salary in thousands of dollars Count 1 2 3 4 5 6 7 8 9 10 11 Shape – the graph is skewed right Center – the median is the first value in the \\$300,000 to \\$400,000 range Spread – the range of salaries is from \\$21,000 to \\$862,000. Outliers – there does not look like there are any outliers, I would have to calculate to make sure. 100 200 300 400 500 600 700 800 900 Thousand dollars\n\nSection 1.1 Day 1 Homework: #’s 2, 4, 6, 8, 11a&b, 14, 16\nAny questions on pg. 1-4 in additional notes packet\n\nNew terms used when graphing data.\nRelative frequency: Category count divided by the total count Gives a percentage Cumulative frequency: Sum of category counts up to an including the current category Ogives (pronounced O-Jive) Cumulative frequencies divided by the total count Relative cumulative frequency graph Percentile: The pth percentile of a distribution is the value such that p percent of the observations fall at or below it.\n\nLets look at a table to see what an ogive would refer to.\n\nThe graph of an ogive for this data would look like this.\n\n85th percentile Median 10th percentile Find the age of the 10th percentile, the median, and the 85th percentile? 47 55.5 62.5\n\nLast graph of this section\nTime plots : Graph of each observation against the time at which it was measured. Time is always on the x-axis. Use time plots to analyze what is occurring over time.\n\nDeaths from cancer per 100,000 Deaths Year 45 50 55 60 65 70 75 80 85 90 95 134 144 154 164 174 184 194 204\n\nSection 1.1 Day 2 Homework: #’s 20, 22, 29 (use scale starting at 7 with width of .5), 60, 61, 63, 66a&c Any questions on pg. 5-8 in additional notes packet\n\nSection 1.2: Describing Distributions with Numbers.\nCenter: Mean Median Mode – (only a measure of center for categorical data) Spread: Range Interquartile Range (IQR) Variance Standard Deviation\n\nMeasuring center: Mean: Most common measure of center.\nIs the arithmetic average. Formula: or Not resistant to the influence of extreme observations.\n\nMeasuring center: Median The midpoint of a distribution\nThe number such that half the observations are smaller and the other half are larger. If the number of observations n is odd, the median is the center of the ordered list. If the number of observations n is even, the median M is the mean of the two center observations in the ordered list. Is resistant to the influence of extreme observations.\n\nQuick summary of measures of center.\nDefinition Example using 1,2,3,3,4,5,5,9 Mean Middle value for an odd # of data values For 1,2,3,3,4,5,5,9, the middle values are 3 and 4. The median is: Median Mean of the 2 middle values for an even # of data values The most frequently occurring value (Categorical data only) Two modes: 3 and 5 Mode Set is bimodal.\n\nComparing the Mean and Median.\nThe location of the mean and median for a distribution are effected by the distribution’s shape. Median and Mean Symmetric Median and Mean Skewed right Mean and Median Skewed left\n\nSince zero is an outlier it effects the mean, since the mean is not a resistant measurement of the center of data.\n\nMeasuring spread or variability:\nRange Difference between largest and smallest points. Not resistant to the influence of extreme observations. Interquartile Range (IQR) Measures the spread of the middle half of the data. Is resistant to the influence of extreme observations. Quartile 3 minus Quartile 1.\n\nTo calculate quartiles:\nArrange the observations in increasing order and locate the median M. The first quartile Q1 is the median of the observations whose position in the ordered list is to the left of the overall median. The third quartile Q3 is the median of the observations whose position in the ordered list is to the right of the overall median.\n\nThe five number summary and box plots.\nConsists of the min, Q1, median, Q3, max Offers a reasonably complete description of center and spread. Used to create a boxplot. Boxplot Shows less detail than histograms or stemplots. Best used for side-by-side comparison of more than one distribution. Gives a good indication of symmetry or skewness of a distribution. Regular boxplots conceal outliers. Modified boxplots put outliers as isolated points.\n\nStart by finding the 5 number summary for each of the groups.\nUse your calculator and put the two lists into their own column, then use the 1-var Stats function. Min Q M Q3 Max Women: Men:\n\nHow to construct a side-by-side boxplot\nSSHA Scores for first year college students Women Men Scores 70 80 90 100 110 120 130 140 150 160 170 180 190 200\n\nCalculating outliers Outlier\nAn observation that falls outside the overall pattern of the data. Calculated by using the IQR Anything smaller than or larger than is an outlier Min Q1 Median Q3 Max\n\nConstructing a modified boxplot\nMin Q M Q3 Max Women:\n\nConstructing a modified boxplot\nMin Q M Q3 Max Women: SSHA Scores for first year college students Women Scores 70 80 90 100 110 120 130 140 150 160 170 180 190 200\n\nSection 1.2 Day 1 Homework: #’s 34, 35, 37a-d, 39, 66b, 67, 68, 69\nAny questions on pg in additional notes packet.\n\nMeasuring Spread: Variance (s2) Standard deviation (s)\nThe average of the squares of the deviations of the observations from their mean. In symbols, the variance of n observations x1, x2, …, xn is Standard deviation (s) The square root of variance. or\n\nHow to find the mean and standard deviation from their definitions.\nWith the list of numbers below, calculate the standard deviation. 5, 6, 7, 8, 10, 12\n\nProperties of Variance:\nUses squared deviations from the mean because the sum of all the deviations not squared is always zero. Has square units. Found by taking an average but dividing by n-1. The sum of the deviations is always zero, so the last deviation can be found once the other n-1 deviations are known. Means only n-1 of the squared deviations can vary freely, so the average is found by dividing by n-1. n-1 is called the degrees of freedom.\n\nProperties of Standard Deviation\nMeasures the spread about the mean and should be used only when the mean is chosen as the measure of center. Equals zero when there is no spread, happens when all observations are the same value. Otherwise it is always positive. Not resistant to the influence of extreme observations or strong skewness.\n\nMean & Standard Deviation Vs. Median & the 5-Number Summary\nMost common numerical description of a distribution. Used for reasonably symmetric distributions that are free from outliers. Five-Number Summary Offer a reasonably complete description of center and spread. Used for describing skewed distributions or a distribution with strong outliers.\n\nAlways plot your data. Graphs Numerical measures of center and spread\nGive the best overall picture of a distribution. Numerical measures of center and spread Only give specific facts about a distribution. Do not describe its entire shape. Can give a misleading picture of a distribution or the comparison of two or more distributions.\n\nChanging the unit of measurement.\nLinear Transformations Changes the original variable x into the new variable xnew. xnew = a + bx Do not change the shape of a distribution. Can change one or both the center and spread. The effects of the changes follow a simple pattern. Adding the constant (a) shifts all values of x upward or downward by the same amount. Adds (a) to the measures of center and to the quartiles but does not change measures of spread. Multiplying by the positive constant (b) changes the size of the unit of measurement. Multiplies both the measures of center (mean and median) and the measures of spread (standard deviation and IQR) by (b).\n\nThe table shows an original data set and two different linear transformations for that set.\nOriginal (x) x + 12 3(x) - 7 5 17 8 6 18 11 7 19 14 20 10 22 23 12 24 29 What are the original and transformed mean, median, range, quartiles, IQR, variance and standard deviation?\n\nOriginal Data x + 12 3(x) – 7 Mean: Median: Q1: Q3: IQR: Range:\nVariance: St Dev: x + 12 Mean: Median: Q1: Q3: IQR: Range: Variance: St Dev: 3(x) – 7 Mean: Median: Q1: Q3: IQR: Range: Variance: St Dev:\n\nSection 1.2 Day 2 Homework: #’s (40, 41) find mean and standard deviation, 42 – 46, 54 – 56, 58 Any questions on pg in additional notes packet.\n\nChapter review\n\nChapter 1 Complete Homework: #’s 60, 61, 63, 66 – 69\nAny questions on pg in additional notes packet.\n\nDownload ppt \"Chapter 1 Exploring Data\"\n\nSimilar presentations\n\nAds by Google" ]	[ null, "http://slideplayer.com/static/blue_design/img/slide-loader4.gif", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8532444,"math_prob":0.9558811,"size":12026,"snap":"2021-04-2021-17","text_gpt3_token_len":3019,"char_repetition_ratio":0.12685077,"word_repetition_ratio":0.07717803,"special_character_ratio":0.2652586,"punctuation_ratio":0.122121714,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98755,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-04-23T18:43:30Z\",\"WARC-Record-ID\":\"<urn:uuid:16fdf248-0a88-4616-acb0-c22dc8b0d870>\",\"Content-Length\":\"261154\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e96c9df5-2655-433e-bd5b-19ea61ff1f60>\",\"WARC-Concurrent-To\":\"<urn:uuid:2a929cce-2aca-43fe-9c31-3d646662346b>\",\"WARC-IP-Address\":\"144.76.153.40\",\"WARC-Target-URI\":\"http://slideplayer.com/slide/6412219/\",\"WARC-Payload-Digest\":\"sha1:AKHLVIGLCHCBQBIPV2KDVTECMC5HST3F\",\"WARC-Block-Digest\":\"sha1:JW62STWSOHG7PVHDJ3JXN33MJAHX5D7T\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-17/CC-MAIN-2021-17_segments_1618039596883.98_warc_CC-MAIN-20210423161713-20210423191713-00301.warc.gz\"}"}
https://judaism.stackexchange.com/questions/104741/the-number-8-does-not-appear-in-the-numbering-of-the-tribes-in-bamidbar-except-f	[ "# The number 8 does not appear in the numbering of the tribes in Bamidbar except for the census of Kehos and the total of Leviim. Why not?\n\nIs there any significance in the absence of the number 8 in the individual census of the 12 tribes taken in Chapter 1 of Bamidbar?\n\nReuben: 46,500\n\nSimeon: 59,300\n\nJudah: 74,600\n\nIssachar: 54,400\n\nZebulun: 57,400\n\nEphraim: 40,500\n\nManasseh: 32,200\n\nBenjamin: 35,400\n\nDan: 62,700\n\nAsher: 41,500\n\nNaphtali: 53,400\n\nThe 8 only appears in the separate census of K’hos (8,600 - Chap. 3:28) and the total of Levi’im aged 30-50 years old where the number 8 appears twice (4:48) namely 8,580.\n\nDo the esoteric properties of the number 8 have any significance here, especially relating to the special and privileged role of the Family of K’hos.\n\n• I also only see one 9. Probably this is some application of Bedford's law. Jun 12, 2019 at 20:34\n• @DoubleAA I think that's only for the first digit [/has generalizations for later digits but they show less of a trend - you can just use Benford's law in base 100 and then sum over the possible first digits], but I agree that this question would be much more interesting with a statistical analysis. Jun 12, 2019 at 20:50\n• Because that's how many Jews there were, and how many people in the tribe of Kehas... Jun 12, 2019 at 20:59\n• @DoubleAA Actually no, this isn't Benford's law because none of the numbers start with 1. You'd have to have the population probability distribution, which I don't see how it's possible to get based on odd features like sextuplets and judaism.stackexchange.com/q/13815/11532. What you could do is ignore the first digits (which aren't that relevant to this question anyway because Yehuda is an outlier and the probability of someone being in the 80s is small) and Gad's 50, assume the 2nd and 3rd digits have a uniform distribution, and then plug into Poissons and compare to Monte Carlo. Jun 12, 2019 at 21:00\n• Nobody's perfect? Jun 12, 2019 at 21:26\n\nRemez is much more interesting when you can find a feature that has a low probability of happening by chance. Before asking for the significance, you should look at the probability that this would happen in a random similar text. In other words, we want to define what you find to be \"interesting\", generate random data similar to what we find in the Chumash, and see if we find similar results. If we do, the remez becomes less compelling, because we would have expected something similar to happen in any case.\n\nHow can we generate lists of numbers similar to these? To start with, we don't have to worry about the last digit, which is always 0. The second to last digit is also 0 for everybody except Gad, so let's exclude that as well. Generating lists of numbers for the first digit would be very hard because it would have to involve data about the population growth and size, which was far from natural due to sextuplets, What happened to the bechorim (first-borns)?, and other things. In any case they're not so relevant to this question, because Yehuda, who had unusually large population, still fell short of 80,000 people.\n\nWe can approximate that in populations of people of around this size, the second and third digits would be randomly distributed. There are 24 numbers in these positions - two from each shevet. So, we can generate lists of 24 random digits and see how likely it is that there's a digit that doesn't appear.\n\nRunning this code in python:\n\n``````import numpy as np\nfrom collections import Counter\nfrom scipy.stats import poisson\n\nsimple = True\n\ndef probability(ints):\nctr = Counter(ints)\ncounts = [ctr[_] for _ in range(10)]\nif simple:\nreturn 0 in counts\nelse:\nreturn np.prod([poisson(2.4).pmf(_) for _ in counts])\n\nfor i in range(20):\nprint(probability(np.random.randint(0, 10, size=24)))\n\nprint()\nprint(probability([6,5,9,3,5,6,4,6,4,4,7,4,0,5,2,2,5,4,2,7,1,5,3,4]))\n``````\n\nI first simulate 20 lists of 24 numbers and print whether or not there's a digit that doesn't appear. I get about half `True` and half `False`. Then I print whether there's a digit that doesn't appear in the real numbers, and of course I get `True`.\n\nFor a more complicated analysis, you can set `simple = False`. Then it will print the probability of getting the counts, using a Poisson distribution with a mean of 2.4 for each digit. The probability for the real numbers is about `4e-09`, which is on the low end but within the range I find for the simulated numbers.\n\nObviously, Hashem gave us these numbers for a reason, but I don't think the absence of 8 is that reason.\n\n• How do I get the code to syntax highlight? Jun 12, 2019 at 21:40\n• I am sorry but I don't understand Jun 13, 2019 at 5:48\n• @Heshy 1) is this an answer? 2) can you paraphrase it in plain language please? Jun 13, 2019 at 11:48\n• I think there might be a flaw here. You can't ignore the first digit because everyone \"fell short of 80,000\" and then check against your simulations to see if \"there's a digit that doesn't appear\". Digits other than 0,1,2,8 and 9 do appear as first digits and therefore it isn't fair to count them if they don't appear at all in the second or third digits. Jun 14, 2019 at 17:03\n• @Silver I see your point, hadn't thought of it that way. It's essentially a factor of 2 effect, so it doesn't change the overall picture. I'll try to edit when I get a chance. Jun 14, 2019 at 17:55" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9469886,"math_prob":0.8475867,"size":2550,"snap":"2022-05-2022-21","text_gpt3_token_len":630,"char_repetition_ratio":0.10722702,"word_repetition_ratio":0.011627907,"special_character_ratio":0.24745098,"punctuation_ratio":0.1454219,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97303987,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-27T20:07:14Z\",\"WARC-Record-ID\":\"<urn:uuid:d5d2178c-174a-4681-b98a-498cd876de82>\",\"Content-Length\":\"221836\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:65b2b163-5f6f-4918-8f0f-6a22e11a7d72>\",\"WARC-Concurrent-To\":\"<urn:uuid:8e042dcf-50cc-4eb1-a44b-5d6c4abb2685>\",\"WARC-IP-Address\":\"151.101.193.69\",\"WARC-Target-URI\":\"https://judaism.stackexchange.com/questions/104741/the-number-8-does-not-appear-in-the-numbering-of-the-tribes-in-bamidbar-except-f\",\"WARC-Payload-Digest\":\"sha1:5WVK5D4RGMDFN5D2DNGUDR75V2URBLWQ\",\"WARC-Block-Digest\":\"sha1:G5V36EETM3JG64TJ6BLDEVXWMNIQ5BRG\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652662675072.99_warc_CC-MAIN-20220527174336-20220527204336-00302.warc.gz\"}"}
https://www.quantumdiaries.org/tag/measurement/	[ "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "## Posts Tagged ‘measurement’\n\n### Solving the Measurement Problem (Guest Post)\n\nWednesday, October 5th, 2016\n\nThe following is a guest posting from Ken Krechmer of the College of Engineering at Applied Science at the University of Colorado, at Boulder.", null, "Ken Krechmer\n\nThe dichotomy between quantum measurement theory and classical measurement results has been termed: the measurement disturbance, measurement collapse and the measurement problem. Experimentally it is observed that the measurement of the position of one particle changes the momentum of the same particle instantaneously. This is described as the measurement disturbance. Quantum measurement theory calculates the probability of a measurement result but does not calculate an actual measurement result. What occurs that causes the quantum measurement probability to collapse into a classical measurement result? Different approaches have been proposed to resolve one or both of these issues including hidden variables, non-local variables and decoherence, but none of these approaches appear to fully resolve both these aspects of the measurement problem.\n\nFurther complicating this measurement problem: 1. The quantum effect called entanglement is another measurement disturbance where the measurement of one particle instantaneously impacts a similar measurement of another, far remote, particle. 2. The quantum effect called uncertainty which defines the minimum variation between two measurement results and changes depending on the order of the two measurements.\n\nRelational measurements and uncertainty,” also available at Measurement, resolves both aspects of the measurement problem by expanding the definition of a classical measurement to include sampling and calibration to a reference. Experimentally, it is well known that a measurement must be sampled and calibrated to a reference to establish a measurement result. This paper proves that the measurement collapse is due to the effect of sampling and calibration which is equal to the universal quantum measurement uncertainty. The universal quantum measurement uncertainty has been verified in independent quantum experiments. Next, one quantum measurement is shown to instantaneously disturb another because one sampling and calibration process is applied to both measurement results.\n\nThe paper resolves the dichotomy between quantum theory and classical measurement results, derives the quantum uncertainty relations using classical physics, unifies the measurement process across all scales and formally models calibration and sampling." ]	[ null, "https://www.quantumdiaries.org/wp-content/plugins/quantum_diaries_user_pics_header/user_pics/RNathvani.png", null, "https://www.quantumdiaries.org/wp-content/plugins/quantum_diaries_user_pics_header//user_popup_arrow.gif", null, "https://www.quantumdiaries.org/wp-content/plugins/quantum_diaries_user_pics_header/user_pics/jfelde.png", null, "https://www.quantumdiaries.org/wp-content/plugins/quantum_diaries_user_pics_header//user_popup_arrow.gif", null, "https://www.quantumdiaries.org/wp-content/plugins/quantum_diaries_user_pics_header/user_pics/USLHC.png", null, 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https://www.sanfoundry.com/unix-questions-answers-using-test-evaluate-expressions-1/	[ "# Unix Questions and Answers – Using test and [ ] to Evaluate Expressions – 1\n\nThis set of Unix Multiple Choice Questions & Answers (MCQs) focuses on “Using test and [ ] to Evaluate Expressions – 1”.\n\n1. test command can be used to check which of the following?\na) Compare two numbers\nb) Compare two strings\nc) Check attributes of a file\nd) All of the mentioned\n\nExplanation: When we use if to evaluate expressions, we need the test statement because the true or false values returned by expressions can’t be directly handled by if. The test uses certain operators to evaluate the condition on its right and returns either true or false exit status. test works in three ways:\n• compares two numbers\n• compare two strings or a single one for a null value\n• check a file’s attributes\n\n2. Which of the following operators is used with test for comparison of numeric values?\na) -eq\nb) -ne\nc) -gg\nd) –eq and -ne\n\nExplanation: There are some comparison operators which are used by test. For example,\n\n```-eq - equal to\n-ne - not equal to```\n\n3. We can use comparison operators without a ‘-‘.\na) True\nb) False\n\nExplanation: Every numerical comparison operator used by the test begin a hyphen, followed by a two letter string and enclosed by a whitespace on either side.\nSanfoundry Certification Contest of the Month is Live. 100+ Subjects. Participate Now!\n\n4. ___ implies greater than and ____ implies less than.\na) gt, le\nb) gt, lt\nc) ge,le\nd) ge,lt\n\nExplanation: There are some comparison operators which are used by test. For example,\ngt implies greater than, lt implies less than, le implies less than or equal to, ge implies greater than or equal to.\n\n5. Which of the following operator is used as a shorthand for test?\na) % %\nb) [ ]\nc) & &\nd) ( )\n\nExplanation: UNIX provides a shorthand for test i.e. [ ]. We can use this pair of rectangular brackets enclosing the expression. Thus the following two are equal,\n\n```test \\$x -eq \\$y\n[ \\$x -eq \\$y ]```\n\n6. It is essential to use whitespaces when we use [].\na) True\nb) False\n\nExplanation: We must provide whitespaces around the operators (like -eq), their operands (like \\$x) and inside the [ and ].\n\n7. test and [ ] can be used for string comparison.\na) True\nb) False\n\nExplanation: test can be used to compare strings with another set of operators (like = for equality of strings and != for inequality). For example, [ ! -z \\$string ] negates [ -z \\$string ].\n\n8. Which one of the following option is used for AND operation in test command?\na) -o\nb) -a\nc) -e\nd) -an\n\nExplanation: test also permits the checking of more than one condition in the same line using -a (AND) operator. For example,\n\n```If [ -n “\\$fname -a -n “\\$lname” ]; then\n. . .```\n\n9. Which one of the following option is used for OR operation in test command?\na) -o\nb) -a\nc) -e\nd) -an\n\nExplanation: test also permits the checking of more than one condition in the same line using -o (OR) operator. For example,\n\n```If [ -n “\\$fname -o -n “\\$lname” ] ; then\n. . .```\n\n10. Which one of the following option is used for checking that the string is not null?\na) -a\nb) -o\nc) -z\nd) -n\n\nExplanation: test can be used to compare strings with another set of operators. -n is used for checking if the string is not null. For example,\n\n```If [ -n “\\$fname ] ;\nthen\necho “\\$fname”```\n\nSanfoundry Global Education & Learning Series – Unix.\n\nTo practice all areas of Unix, here is complete set of 1000+ Multiple Choice Questions and Answers.", null, "" ]	[ null, "data:image/svg+xml,%3Csvg%20xmlns=%22http://www.w3.org/2000/svg%22%20viewBox=%220%200%20150%20150%22%3E%3C/svg%3E", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.74288034,"math_prob":0.9106236,"size":3448,"snap":"2023-40-2023-50","text_gpt3_token_len":897,"char_repetition_ratio":0.13937283,"word_repetition_ratio":0.19384615,"special_character_ratio":0.26450115,"punctuation_ratio":0.116959065,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9877337,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-05T07:58:48Z\",\"WARC-Record-ID\":\"<urn:uuid:cde868d6-45e2-4613-a3a9-6e0236e6ae48>\",\"Content-Length\":\"146727\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:3e42edad-7fa8-408a-a242-f5bf06db8ef7>\",\"WARC-Concurrent-To\":\"<urn:uuid:11c9d7c1-af83-4c0f-b4a3-3aafd0e82bba>\",\"WARC-IP-Address\":\"104.25.132.119\",\"WARC-Target-URI\":\"https://www.sanfoundry.com/unix-questions-answers-using-test-evaluate-expressions-1/\",\"WARC-Payload-Digest\":\"sha1:HUXA6KMHVVUOOOEA2XEE6EE4ZLAVSXRB\",\"WARC-Block-Digest\":\"sha1:KGOIJSAARJPJYR5NN46KXKZB7QYQM4ZH\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100550.40_warc_CC-MAIN-20231205073336-20231205103336-00603.warc.gz\"}"}
https://pdfkul.com/monopoly-pricing-with-dual-capacity-constraints_59bfe1c31723dd9b4393e57b.html	[ "Monopoly pricing with dual capacity constraints Robert Somogyi\n\nJOB MARKET PAPER September 14, 2015\n\nAbstract This paper studies the price-setting behavior of a monopoly facing two capacity constraints: one on the number of consumers it can serve, the other on the total amount of products it can sell. Facing two consumer groups that differ in their demands and the distribution of their willingness-to-pay, the monopoly’s optimal non-linear pricing strategy consists of offering one or two price-quantity bundles. The characterization of the firm’s optimal pricing in the short run as a function of its two capacities reveals a rich structure that also gives rise to some surprising results. In particular, I show that prices are non-monotonic in capacity levels. Moreover, there always exists a range of parameters in which weakening one of the capacity constraints decreases consumer surplus. In the long run, when the firm can choose how much capacity to build, prices and consumer surplus are monotonic in capacity costs.\n\nJEL Classification: D21, D42, L12 Keywords: Capacity constraint, Monopoly pricing, Multi-dimensional capacity\n\nEcole Polytechnique and CREST. Email address: [email protected] The latest version of this paper is available here.\n\n1\n\nIntroduction\n\nThe economics literature typically considers one-dimensional capacity constraints (Kreps and Scheinkman ). This is idealized because in real-world production processes firms typically face several capacity constraints (size of plants, inventories, workforce, etc.). In general, due to the use of supply chains, firms are constrained on even more aspects of their production. The objective of this paper is to develop a theory of monopoly pricing in the presence of multiple capacity constraints. Examples of industries characterized by dual capacity constraints range from hospitals, through restaurants, to the freight transport industry. Hospitals are constrained by the number of beds available in their intensive care unit on the one hand, and operating room time on the other hand. Each consumer (i.e. patient) needs one bed but consumers differ in their need of operating room time. Hospitals’ patients have price-inelastic individual demands; even if a long surgery (e.g., a kidney transplant) is very cheap, someone in need of a short surgery (e.g., fixing of a broken arm) will never prefer having the longer one. Restaurants constitute another prominent example for the co-existence of the two types of capacity constraints. Restaurants have to take into account in their pricing decisions both the number of tables they have at their disposal and the size of their kitchen, that limits the amount of food they can prepare. Patrons of restaurants arrive in groups of different sizes, hence they are typically heterogeneous in their consumption and also in their willingness-to-pay. Dual capacities are key characteristics of the freight transport industry as well. Both in ocean container shipping and air cargo transport, an important concern of the transporting company is optimizing the mix of items according to both their size and their weight. As physical and regulatory limits are present in both dimensions, profit-maximizing firms cannot avoid taking into account both constraints. Another class of examples includes markets where firms are constrained by some physical capacity constraint on the one hand and their workers’ time on the other. For instance airplanes cannot fly at full capacity if the ratio of passengers to flight attendants exceeds a certain regulatory limit. Business class passengers tend to use up more of the flight attendants’ time than economy class passengers and the willingness-to-pay of the two groups are obviously different. Several questions arise if one wants to understand the consequences of the co-existence of the two types of capacity constraints. How much will the predictions\n\n1\n\nof the model change compared to a model with only one capacity constraint? What are the optimal prices a firm must charge to different consumer groups? How will these optimal prices change as a function of the capacity levels? Under which conditions are both capacity constraints binding in optimum? How do aggregate consumer surplus, profit and total welfare vary with the capacity levels? In order to answer these questions, I consider a price-setting monopolist facing two types of capacity constraints. The firm is unable to serve more than K consumers. In addition, it cannot sell more than a quantity Q of its products. Consumers differ in their price-inelastic demands and their willingness-to-pay (WTP). There are two consumer groups: high-types intend to buy a larger amount of the product than the low-types. The monopoly can observe the demand of each consumer but not their WTP. The WTP of high-types and low-types are distributed along two different intervals. Some high-types have a larger WTP than all the low-types while some low-types have a larger per-unit WTP than all the high-types. The monopoly is allowed to offer different prices for the two quantities in order to discriminate between consumers. The existence of a second type of capacity constraint fundamentally changes the monopoly’s optimal behavior. The optimal pricing strategy in the short run (with exogenously given capacities) is the following. When K, the constraint on the number of consumers, is very tight, the monopoly excludes low-types and serves only high-type consumers. Conversely, when the capacity constraint on total production, Q, is very small then only low-types are served. For larger levels of capacities, it is optimal for the monopoly to serve both types. When both constraints are very large, the firm chooses the unconstrained optimal prices. When K is of intermediate value while Q is very large, optimal prices are chosen in a way that K binds and Q does not. Conversely, if Q is of medium value and K is large, then Q is binding and K is slack. Importantly, there exist capacity levels for which the two prices are chosen so that both constraints bind. Intuitively, when only very few consumers can be served, the monopoly will prefer serving those with the highest overall WTP so it excludes all the low-types. Conversely, when the total production is very limited, the firm is concerned by the per-unit surplus it can extract from the consumers, hence it excludes all the high-types. Even when capacities do not take extreme values and some consumers of both types are served, an increase in K ceteris paribus makes the low-types more attractive to the firm therefore it chooses its prices to attract more low-types. In sharp contrast to the standard case with only one capacity constraint,\n\n2\n\nprices are non-monotonic in the size of capacities. In particular, while prices are decreasing in both capacities in most parameter regions, for capacity levels when both constraints bind, the price charged for high-types increases in K while the price for low-types increases in Q. The intuition for this result is the following. For very small values of K, the capacity on the number of people served, the monopoly only serves high-types. As K increases low-types become relatively more valuable for the firm so after a threshold value it starts serving both consumer groups. Below the threshold value the firm decreases the price charged for the high-types so that their number equals K. Above the threshold, in order to make space for the low-types, the firm is interested in serving less high-types so it raises the price charged to them. A similar argument can be made to understand the price increase for the low-types as Q increases. Furthermore, this price increase has a non-negligible effect on aggregate consumer surplus: I show that there always exists a range of parameters in which weakening one of the capacity constraints decreases aggregate consumer surplus. This phenomenon cannot occur in models that consider a single capacity constraint. The decrease of consumer surplus is a consequence of the monopoly adjusting its optimal mix of consumers in the following sense. When K is very small, only high-types get served. As it increases above a threshold value, the firm starts serving some low-type consumers as well. However, as the transition is smooth, close to the turning point the firm still serves many high-types and only a few low-types. The price increase suffered by the numerous high-types dominates the gain of the few low-types that start being served and the aggregate consumer surplus decreases. I also investigate long-run behavior of the monopoly where in addition to prices, it can also choose endogenously how much capacity to build in each dimension. For any positive cost function of capacity building, the firm chooses capacity levels and prices so that both capacities bind. These are exactly the capacity levels for which the prices are increasing and aggregate consumer surplus is decreasing in the short run. I provide a complete characterization of optimal capacity levels for the case of linear cost functions. The outcome of the model gets close to the unconstrained optimum as both costs tend to zero. The optimal consumer mix depends on the relation of the capacity costs to the marginal benefit of serving an additional consumer of a given type. Therefore, depending on parameter values, both low-types and high-types can be excluded in optimum, or some consumers of both groups can be served. Both optimal prices and consumer surplus are\n\n3\n\nmonotonic in the capacity costs. I also consider an extension which allows low-types to buy a large quantity of the product and freely dispose of the amount they do not consume. Such a model suits better some of the industries cited as examples, for instance the restaurants. I show that this opportunity does not alter any of the results obtained in the more restrictive basic model. Intuitively, consumers’ possibility of free disposal may limit the monopoly’s choice. In case it wants to sell the two quantities at different prices, it must charge a lower price for the smaller quantity, otherwise all consumers would buy the larger bundle for the lower price which is obviously unprofitable for the firm. However, the optimal prices of the basic model satisfy this condition, thus optimal firm behavior is not altered by free disposal. Next, I investigate incentive compatibility for the high-type consumers. In this variant of the model, the high-types are allowed to buy several small bundles to satisfy their demand and I assume that the good is perfectly divisible. A monopoly that wants to sell both types of bundles must choose lower per-unit prices for the large bundle than for the small one, otherwise it can sell exclusively small bundles. I show that the monopoly’s optimal prices are not affected by the perfect divisibility and high-types’ ability to purchase the small bundle. Finally, I provide sufficient conditions on the distribution of consumers such that all the qualitative insights of the model with uniform distribution hold. In particular, I find that if both distributions satisfy the monotone hazard rate condition, then the region where both capacity constraints bind has a well-defined shape; moreover, prices are non-monotonic in capacity levels and I prove the existence of a region where consumer surplus is decreasing in a capacity level. Furthermore, I identify hazard rate dominance relations as sufficient conditions for the exclusion of a consumer group.\n\n1.1\n\nRelated literature\n\nThe monopoly’s problem of third-degree price discrimination with a single capacity constraint is a textbook exercise (see Besanko and Braeutigam, 2010, p. 507, Exercise 12.6). Therefore, the literature of capacity-constrained pricing has mainly focused on the case of competition. The seminal paper of Kreps and Scheinkman (1983) shows that under certain assumptions, the outcome of a two-stage game where firms first choose capacities then engage in price competition coincides with the Cournot outcome. Davidson and Deneckere (1986) were the first to point out that this result is not robust to the choice of rationing rule. In particular, they show that the results are more competitive than the Cournot outcome for virtually 4\n\nany rationing rule other than the efficient one. Cripps and Ireland (1988) and Acemoglu et al. (2009) both consider capacityconstrained competition when firms face consumers with price-inelastic demands. They show the existence of pure-strategy subgame perfect equilibria for any capacity levels, which are supported by mixed-strategy equilibria off-path, similarly to Kreps and Scheinkman (1983). Acemoglu et al. (2009) also analyze efficiency properties of the equilibria, i.e., they compare the total social welfare in different equilibria to the welfare-maximizing first-best and they find that some equilibria can be arbitrarily inefficient. Reynolds and Wilson (2000) introduce demand uncertainty into the two-stage game described by Kreps and Scheinkman (1983). They show that symmetric purestrategy equilibria in the capacity choice game do not exist when the variability of demand is high, and they provide a set of assumptions that guarantee the existence of asymmetric pure-strategy equilibria. De Frutos and Fabra (2011) investigate the interaction of demand uncertainty and capacity constraints assuming price-inelastic demand. They identify submodularity of the demand distribution as a sufficient condition of the existence of pure-strategy equilibria and they show that these equilibria are asymmetric. The welfare analysis of the model with general distribution function is related to another stream of recent literature which investigates the welfare effects of a monopoly’s third-degree price discrimination. Cowan (2007) derives two alternative sufficient conditions for the convexity of the slope of demand under which price discrimination enhances social welfare. Aguirre et al. (2010) analyze more general demand functions and identify sufficient conditions on the elasticities and demand curvatures of the two markets that make price discrimination reduce or increase welfare. Cowan (2012) shows how the the cost pass-through coefficient can determine the way discrimination effects aggregate consumer surplus. Finally, Bergemann et al. (2015) investigate welfare effects of additional information a monopoly can use to price discriminate and they show that any combination of consumer and producer surpluses is achievable that satisfies some mild conditions. Models where several capacity constraints co-exist have so far been relegated to the realms of operations research and revenue management. Patient admission planning for scheduled surgeries and patient mix optimization are both important problems hospitals have to face. The multidimensional nature of capacities in hospitals is crucial for such planning. Many recent papers in the operations research literature focus on solving a variety of problems that arise in a context where the\n\n5\n\ntreatment of different categories of patients require different levels of capacities. For example, Adan and Vissers (2002) take into account operating room time, intensive care unit beds, medium care unit beds and nurses’ time to simulate the optimal schedule of a real-world hospital department. Banditori and al. (2013) also consider a multiple capacity setting for their simulation, in addition, they provide a comparison of the recent articles in this area (Banditori and al., 2013, Table 2). Multidimensional capacities are crucial in freight transport, such as the air cargo industry or the container shipping industry. Since dynamic pricing is widespread in these industries, the literature models freight transport pricing by extending standard revenue management models to accommodate multiple capacities. Xiao and Yang (2010) consider revenue management with two capacity dimensions. Their focus is on ocean container shipping, where container sizes are standardized (two varieties dominate the market) and maximal weights of all containers are set by on-road regulations. They derive an analytical solution and show that under some conditions the optimal policy is qualitatively different when considering the second capacity constraint. Kasilingam (1997) describes how air cargo revenue management is different from air passenger revenue management, and one of the key differences he identifies is the multidimensional aspect of capacities: volume, weight and even cargo position may be constraining. Finally, the hospitality industry’s capacity management literature has also recognized the importance of dual capacities. Kimes and Thompson (2004) optimize the table mix for restaurant revenue management taking into account not only the number and distribution of seats but also the size of the service areas. Bertsimas and Romy (2003) consider both sizes of parties to be seated and expected service duration to compare several optimization-based approaches to restaurant revenue management. The rest of the paper is organized as follows. Section 2 discusses a simple benchmark, then outlines the main model. Section 3 describes the results of the main model, the monopoly’s optimal behavior, and provides comparative statics for the capacity levels. Section 4 discusses the consequences of optimal pricing for consumer surplus. Section 5 generalizes the model by allowing capacity levels to be chosen endogenously. Section 6 investigates a variant of the baseline model with incentivecompatibility. Section 7 analyzes the model with general distribution of consumers and Section 8 concludes. All omitted proofs are relegated to the Appendix.\n\n6\n\n2\n\nThe model\n\n2.1\n\nA simple benchmark\n\nIn this section I present a simple model of monopoly characterized by one capacity constraint facing only one consumer group. Consider a price-setting monopolist that can produce for zero cost up to quantity Q of a good, then his costs become infinite, i.e., the monopoly is characterized by capacity constraint Q. All consumers have a unit demand. Consumers are heterogeneous in their willingness-to-pay (WTP), they are uniformly distributed on the interval [0, 1]. The monopoly can only observe the distribution but not the individual WTP values. The net consumer surplus of a buyer with WTP level w ∈ [0, 1] is given by w − p if he buys the product where p denotes its price, and it is 0 otherwise. A consumer is willing to buy if and only if his net consumer surplus is positive. In this setting, the mass of consumers willing to buy the product at a given price p ∈ [0, 1] is 1−p: this is the fraction of consumers with a higher WTP than the price. Since each consumer has unit demand, the total demand for the product is also 1−p. Hence the monopoly solves the following maximization problem: max π = (1 − p)p\n\ns.t.\n\np\n\n1−p≤Q\n\nThe profit-maximizing price is given by p∗ =\n\n( 1 − Q if Q ≤ 1/2, 1/2\n\notherwise.\n\nFirst notice that the capacity constraint is binding up to a certain level (Q ≤ 1/2) then the monopoly can implement its unconstrained optimal strategy. The price is decreasing in the level of capacity, Q, as long as the capacity is small enough to bind, then the price becomes independent of it. Given these prices, total demand is simply Q if Q ≤ 1/2; while for larger values of the capacity it is equal to 1/2. Intuitively, the monopoly chooses prices such that the capacity binds when it is small, then it implements its unconditional optimum. Profit is given by π∗ =\n\n( (1 − Q)Q if Q ≤ 1/2, 1/4\n\notherwise.\n\n7\n\nAs one should expect, both the demand and the profit are increasing in capacity for binding levels of capacity. Finally, consumer surplus in this setting is given by (1 − p)2 /2. Clearly, consumer surplus is always decreasing in the price (p never exceeds 1). Hence the fact that the optimal price is decreasing in Q means that consumer surplus is increasing in the level of capacity. This is not surprising: An increase in the level of capacity means more of the consumers are served for a lower price. The four main insights of this simple model are that an increase in the capacity level Q can decrease prices and can increase demand, consumer surplus and profits.1 Out of these 4 insights only the one concerning profits will remain true for a model with dual capacity constraints presented in the next section.\n\n2.2\n\nThe dual capacity model\n\nConsider a price-setting monopolist serving a market that consists of two consumers groups. Each consumer is characterized by its demand and its total willingness-topay (WTP) for the product. Low-types want to consume a fix amount of qL > 0 products while high-types want to consume a fix amount of qH > qL , i.e., individual demand is price-inelastic. A consumer of type i ∈ {L, H} with total WTP w has a net consumer surplus of w − pi if he buys a quantity qi of the product for price pi , and 0 otherwise.2 Total WTP of consumers of type i is uniformly distributed on the interval [0, vi ]. Consumers maximize their net surplus and they demand the good if and only if their net surplus is positive. Assume that the total mass of high-type and low-type consumers is αvH and (1 − α)vL , respectively, where 0 ≤ α ≤ 1 scales the relative weight of the two consumer groups.3 The monopoly can observe α, the consumers’ individual demand, and the distribution of their WTP but not their individual values.\n\nAssumption 1. Let the WTP of consumers satisfy the following conditions: 0 < vL < vH\n\nand vL /qL > vH /qH\n\n1\n\nThe same insights remain true in case there are 2 consumer groups, as discussed in Section 3.2, “case of very large Q”. 2 In Section 6 I show that all results remain unchanged if one allows low-types to buy the large bundle; or high-types to buy several small bundles. 3 This normalization of the mass of consumers is made to simplify the exposition of results. In particular, it shortens significantly the formulas obtained for optimal prices and profits without altering the qualitative properties of the model.\n\n8\n\nAssumption 1 guarantees that some high-type consumers’ valuation always exceed all the low-types’ valuation, whereas the per-unit-WTP of some low-type consumers is greater then the per-unit-WTP of all high-type consumers. This assumption restricts the analysis to the most interesting cases, because otherwise the monopoly would always prefer to serve consumers of one group first, irrespective of the size of capacity constraints. Also, this corresponds to decreasing marginal value of consumption in the present setting. Finally, notice that high-types are not necessarily more valuable for the firm, it simply refers to the high level of their demand. I analyze a monopoly facing two types of capacity constraints: • K denotes the maximal number of consumers the firm can serve • Q denotes the maximal total production of the firm Both constraints are exogenously given. For simplicity, production is costless up to capacity then it becomes impossible. The monopoly has an optimal pricing structure that consist of offering at most 2 price-quantity bundles. Given the price-inelastic demands, no consumer would buy any bundle that offers them a quantity different from their desired demand. Moreover, if the firm were to offer several bundles with the same quantity for a different price, consumers would only buy the cheapest one. Let pH and pL denote the price of the bundle with high and low quantities, respectively.\n\n3\n\nResults\n\n3.1\n\nOptimal monopoly pricing\n\nIn this section I describe and solve the monopoly’s profit maximization problem. For any price pH ∈ [0, vH ], the high-type consumers willing to buy are the ones H who have a higher WTP than pH . They represent a fraction vHv−p of the high-types, H which means that the total mass of high-type consumers who demand the good is given by α(vH − pH ). Similarly, for any price pL ∈ [0, vL ], the total mass of low-types willing to buy is (1 − α)(vL − pL ).\n\n9\n\nHence the total demand the monopoly faces at such prices is given by α(vH − pH )qH + (1 − α)(vL − pL )qL The monopoly also has the option of not serving one of the consumer groups. Hence, it must choose between serving both consumer groups, excluding low-type consumers, or excluding high-type consumers. Notice that in some cases the latter possibility can be profitable since some low-type consumers have a higher per-unit WTP than all the high-types. I will break down the general optimization problem into 3 separate maximization problems and compare the locally optimal profits to find the monopoly’s profit-maximizing strategy. The maximization problem of the firm if it decides to exclude low-types writes as\n\n4\n\n(P-EL)\n\nmax π = α(vH − pH )pH\n\ns.t.\n\npH\n\nα(vH − pH ) ≤ K\n\n(1a)\n\nα(vH − pH )qH ≤ Q\n\n(1b)\n\npH ≥ 0\n\n(1c)\n\nThe maximization problem if the monopoly serves only low-types is given by\n\n(P-EH)\n\nmax π = (1 − α)(vL − pL )pL pL\n\ns.t.\n\n(1 − α)(vL − pL ) ≤ K\n\n(2a)\n\n(1 − α)(vL − pL )qL ≤ Q\n\n(2b)\n\npL ≥ 0\n\n(2c)\n\nThe maximization problem of the firm when serving some consumers of both groups writes as 4 Throughout the paper, EL stands for “excluding low-types” and EH stands for “excluding high-types”.\n\n10\n\n(P-LH)\n\nmax\n\npL ,pH\n\nπ = α(vH − pH )pH + (1 − α)(vL − pL )pL\n\ns.t.\n\nα(vH − pH ) + (1 − α)(vL − pL ) ≤ K\n\n(3a)\n\nα(vH − pH )qH + (1 − α)(vL − pL )qL ≤ Q\n\n(3b)\n\npL ≥ 0 , pH ≥ 0\n\n(3c)\n\npL < vL , pH < vH\n\n(3d)\n\nIn each case, the firm maximizes the product of the mass of consumers buying and the price. Constraints 1a, 2a and 3a constitute the upper bound on the maximal number of people that the monopoly can serve, whereas 1b, 2b and 3b are the capacity constraints on total production: The mass of people buying times their demand cannot exceed Q. Constraints 1c, 2c and 3c guarantee non-negativity of the prices. Finally, 3d ensures that some consumers of both types are served in the last case. 3.1.1\n\nExcluding one group of consumers\n\nIn problem (P-EL) when the monopoly excludes low types, first notice that the 2 capacity constraints 1a and 1b can be rewritten as α(vH − pH ) ≤ min(K, Q/qH ) Obviously, the unconstrained maximum is attained at pH = vH /2. Thus, the 2 /4 if min(K, Q/qH ) > αvH /2. Otherwise the firm sells up to the profit equals αvH tighter capacity constraint for a price pH = vH − min(K, Q/qH )/α, consequently its profit equals to (min(K, Q/qH ))2 α The non-negativity constraint is trivially satisfied. The solution of the second maximization problem, (P-EH), when the firm excludes high-types, is analogous. Hence the optimal price is given by vH min(K, Q/qH ) −\n\npL =\n\n( vL − vL /2\n\nmin(K,Q/qL ) 1−α\n\nif min(K, Q/qL ) ≤ (1 − α)vL /2, otherwise.\n\nThe optimal profit is\n\n11\n\nπ=\n\n3.1.2\n\n( vL min(K, Q/qL ) −\n\n(min(K,Q/qL ))2 1−α\n\n(1 − α)vL2 /4\n\nif min(K, Q/qL ) ≤ (1 − α)vL /2, otherwise.\n\nServing both consumer groups\n\nSolving the remaining maximization problem, (P-LH) is more complex and requires writing the Karush-Kuhn-Tucker conditions. I omit the constraints 3c and 3d on prices and show ex-post that the solution of the relaxed problem satisfies them. Let λ1 denote the multiplier of constraint 3a and let λ2 denote the multiplier of 3b. The objective function is\n\nL(pL , pH , λ1 , λ2 ) = α(vH − pH )pH + (1 − α)(vL − pL )pL − λ1 [α(vH − pH ) + (1 − α)(vL − pL ) − K] − λ2 [α(vH − pH )qH + (1 − α)(vL − pL )qL − Q] The first order conditions can be written as 2pH = vH + λ1 + λ2 qH\n\nand\n\n2pL = vL + λ1 + λ2 qL There are 4 cases depending on the sign of the multipliers, i.e., depending on which of the two constraints is binding. The following notation simplifies the exposition of results: let E(x) denote the weighted average of any 2 variables xL and xH , i.e., E(x) = αxH + (1 − α)xL . Case 1 When the capacity constraint on the mass of people served is binding while the other constraint is slack (λ1 > 0 and λ2 = 0), the optimal pricing strategy of the monopoly is given by vH − vL −K 2 and its optimal profit is pL = v L + α\n\nand pH = vH − (1 − α)\n\n\u0012 πK ≡ α(1 − α)\n\nvH − vL 2\n\n\u00132\n\nvH − vL −K 2\n\n+ KE(v) − K 2\n\nPrimal and dual feasibility of this local optimum require that K, the capacity constraint on the mass of people served, be of an intermediate size with respect to 12\n\nthe other parameters of the model: α (vH − vL ) ≤ K ≤ min 2 where\n\n1 g(Q) = E(q)\n\n\u0012\n\n\u0013 E(v) , g(Q) 2\n\n\u0012 \u0013 vH − vL Q − α(1 − α)(qH − qL ) . 2\n\nThe non-negativity constraints are satisfied at this solution. Case 2 When the capacity constraint on the total production is binding while the other constraint is slack (λ1 = 0 and λ2 > 0), the optimal pricing strategy of the monopoly is given by \u0012 \u0013 qL α qH pL = (vL + vH )qH + (1 − α)vL qL − Q E(q 2 ) 2 qL and qH pH = E(q 2 )\n\n\u0012\n\n\u0013 1 α qH qH (vL + vH )qH + (1 − α)vL qL − Q − (vL − vH ) 2 qL 2 qL\n\nIts optimal profit is\n\n\u0012 \u00132 ! qH − vL + qL \u0013\u0012 \u0012 \u0013 α 1 α qH qH + vH )qH + (1 − α)vL qL − Q Q + (vL − vH )qH + (vL E(q 2 ) 2 qL 2 qL\n\nα πQ ≡ 4\n\n2 vH\n\nPrimal and dual feasibility of this local optimum require that Q, the capacity constraint on the total production, be of an intermediate size with respect to the other parameters of the model: \u0012 \u0013 qL E(vq) 1−α qL vL − vH ≤Q≤ and K ≥ f (Q) 2 qH 2 where f (Q) =\n\nE(q) 1 Q + α(1 − α)(vL qH − vH qL )(qH − qL ) E(q 2 ) 2E(q 2 )\n\nCase 3 When both capacity constraints are binding (λ1 > 0 and λ2 > 0), the optimal pricing strategy of the monopoly is given by\n\n13\n\npL = vL −\n\nKqH − Q (1 − α)(qH − qL )\n\nand p H = vH −\n\nQ − KqL α(qH − qL )\n\nIts optimal profit is\n\nπKQ\n\n1 ≡ qH − q L\n\n\u0012 \u0013 Q2 + E(q 2 )K 2 − 2KQE(q) (vH − vL )Q + (vL qH − vH qL )K − α(1 − α)(qH − qL )\n\nPrimal and dual feasibility of this local optimum require that max (Q/qH , g(Q)) ≤ K ≤ min (Q/qL , f (Q)) Case 4 Clearly, the global unconstrained optimum (λ1 = 0 and λ2 = 0) is attainable whenever K ≥ E(v)/2 and Q ≥ E(vq)/2. It consists of the firm choosing prices vL /2 and vH /2, and its value is πU = α\n\n2 vH v2 + (1 − α) L 4 4\n\nI characterize the monopoly’s optimal pricing strategy by comparing the local maxima obtained above for all possible range of parameters. Figure 1 depicts the firm’s optimal division of the K-Q space.5 The monopoly’s optimal behavior is different in each of these regions. The firm chooses optimal prices in such a way that only capacity K binds in region K (Case 1), only capacity Q binds in region Q (Case 2), both K and Q bind in region KQ (Case 3). None of the constraints bind in region U which corresponds to the unconstrained optimum (Case 4). Only capacity K binds and the monopoly excludes low-types in region EL. Conversely, only capacity Q binds and the monopoly excludes high-types in region EH. Notice that each of the 4 lines bordering the core region KQ have a different slope, hence the region KQ is not symmetric. Proposition 1 provides a complete characterization of the firm’s optimal choice for any combination of its capacity levels. \u0010 \u0011 Figure 1 depicts the case of (1 − α)qL vL − vH qqHL < αqH (vH − vL ). When this ordering is reversed the figure changes accordingly. However, the coordinates of all lines and critical points remain the same. 5\n\n14\n\nFigure 1: Division of the capacity-space based on the monopoly’s optimal strategy\n\nK\n\nQ\n\nEH\n\nU\n\nE(v) 2\n\nf (Q)\n\nK 1−α 2 (vL\n\ng(Q)\n\n− vH qqHL ) L α vH −v 2\n\nKQ Q/qH Q/qL\n\n1−α 2 qL (vL\n\nEL − vH qqHL )\n\nL αqH vH −v 2\n\nE(vq) 2\n\nQ\n\nProposition 1. The optimal prices of the monopoly are L L • pL = vL + α vH −v − K and pH = vH − (1 − α) vH −v − K in region K where 2 2 K binds, Q is slack, some consumers of both groups are served; \u0010 \u0011 qL qH α • pL = E(q (v + v )q + (1 − α)v q − Q and pH = pL qqHL in reL qL H H L L 2) 2 gion Q where Q binds, K is slack, some consumers of both groups are served;\n\nKqH −Q Q−KqL and pH = vH − α(q in region KQ where both K • pL = vL − (1−α)(q H −qL ) H −qL ) and Q bind, some consumers of both groups are served;\n\n• pH = vH − • p L = vL − excluded;\n\nK α\n\nin region EL where K binds, Q is slack, low-types are excluded;\n\nQ qL (1−α)\n\nin region EH where Q binds, K is slack, high-types are\n\n• pL = vL /2 and pH = vH /2 in region U where both K and Q are slack, some consumers of both groups are served.\n\n15\n\nAssumption 1 guarantees the existence of all 6 regions enumerated in Proposition 1. The main intuition driving the results is the two consumer groups’ varying relative attractiveness for the firm. For any given Q, an increase in K makes the constraint on total production, Q, tighter. This in turn makes low-types, that consume less of the tighter capacity, relatively more valuable for the firm, so in optimum the firm adjusts its prices to attract more of them (except for extreme values of K). Figure 2: Small Q rH Prices\n\n1 vH pH\n\nvH −\n\nQ αqH\n\nvL pL vL −\n\nQ (1−α)qL\n\n0 Q qH\n\nQ qL\n\nK\n\n(a) Prices\n\n3.2\n\nQ qH\n\nQ qL\n\nK\n\n(b) Share of high-types\n\nComparative statics\n\nIn this section I investigate the comparative statics properties of the monopoly’s optimal behavior. The variables of interest are optimal prices charged for the two consumer groups and the share of high-types among all consumers served. I present all the results as a function of the constraint on the number of people served, K, however, analogous statements can be made for the other constraint, Q, as the model is almost symmetric in K and Q. For a detailed analysis of the firm’s pricing behavior, it will be useful to divide the Q-K plane into 4 regions by 3 vertical lines going through the 3 values that appear on the Q axis of Figure 1. I call the 4 resulting cases “small Q”, “medium Q”, “large Q”, and “very large Q”. 3.2.1\n\nSmall Q\n\nThe first region of interest is delimited by \u0012 \u0013 \u0012 1−α qL 0 < Q ≤ min qL vL − vH , 2 qH 16\n\nvH − vL αqH 2\n\n\u0013\n\nFigure 3: Medium Q, Case A rH Prices\n\n1\n\nvH pH\n\nvH −\n\nQ αqH\n\npL\n\nvL\n\nQ qH\n\nf (Q)\n\nQ qL\n\nK\n\n0\n\n(a) Prices\n\nQ qH\n\nf (Q)\n\nQ qL\n\nK\n\n(b) Share of high-types\n\nFor very low levels of K the firm excludes low-types and sells up to K exclusively to high-type consumers, Q is slack here. As K grows larger, the monopoly starts serving some low-types as well in such a way that both capacity constraints bind. This means that as K grows the firm serves more and more low-types and less and less high-types. Finally, as K becomes even larger high-types get excluded and the firm only serves low-types. The intuition is the following. As K grows, Q, the constraint on the production gets relatively stricter. This means that low-type consumers who use up less of the production constraint become more and more valuable for the firm. This is indeed reflected in the share of high-type consumers served decreasing from 1 all the way to 0 as depicted in the right panel of Figure 2. It is interesting to note that while pL is decreasing in K, pH first decreases then increases (see panel a of Figure 2). The decrease occurs in region EL where only high-types are served: the monopoly decreases its prices so that the demand of high-types equals the capacity level. The increase is again a consequence of low-types becoming more valuable for the firm: In order to accommodate more low-types, the firm must serve less high-types so it raises its price for them.\n\n3.2.2\n\nMedium Q\n\nThe monopoly’s pricing strategy changes somewhat when Q is of intermediate size, i.e.,\n\n17\n\nFigure 4: Medium Q, Case B rH Prices\n\n1 vH vL +vH 2\n\npH\n\nvH /2 vL pL\n\n0 L α vH −v 2\n\ng(Q)\n\nK\n\nQ/qL\n\n(a) Prices\n\n\u0012 min\n\nL α vH −v 2\n\ng(Q) Q/qL\n\nK\n\n(b) Share of high-types\n\n\u0013 \u0013 \u0012 \u0013 \u0013 \u0012 \u0012 1−α vH − vL 1−α vH − vL qL qL qL vL − vH , αqH < Q ≤ max qL vL − vH , αqH 2 qH 2 2 qH 2 \u0010\n\nCase A: (1 − α)qL vL −\n\nvH qqHL\n\n\u0011\n\n< αqH (vH − vL ) :\n\nCase A represents the parameter setting depicted in Figure 1. As before, for very low levels of K the firm excludes low-types. For larger levels of K the firm starts serving low-types and serves less and less high-types. However, when K hits the threshold value of f (Q), the prices become independent of the size of K. For these capacity levels the firm serves some consumers of both groups in such a way that Q binds and K is slack. Intuitively, as K grows serving low-types becomes relatively more profitable so there is a region where the mass of high-types served decreases. The profit is always the sum of the profit the firm makes on serving low-types and high-types. It is easy to see that both of those partial profit curves are concave. The profit the monopoly earns on high-types is decreasing at an increasing rate while the profit made on low-types is increasing at a decreasing rate. This means that there is a point (f (Q)) where the marginal revenue of high-types equals the marginal loss on low-types. At this point the monopoly prefers to switch to the Q regime where the share of high-types does not decrease more in K. As f (K) < Q/qL , the share of high-types does not decrease all the way to 0 (see Figure 3).\n\n18\n\nFigure 5: Large Q rH 1\n\nPrices vH vL +vH 2\n\npH\n\nvL pL\n\n0 L α vH −v 2\n\ng(Q)\n\nf (Q)\n\nK\n\n(a) Prices\n\nL α vH −v 2\n\ng(Q) f (Q)\n\nK\n\n(b) Share of high-types\n\n\u0010 \u0011 Case B: αqH (vH − vL ) ≤ (1 − α)qL vL − vH qqHL : As before, for very low levels of K the firm excludes low-types. However, as K grows the capacity levels enter the K region which means that K still remains binding while both types start getting served and both prices keep decreasing in K. Next, K reaches the KQ region where the monopoly starts raising pH and both constraints bind. The increase in K continues until none of the high-types are willing to buy, only low-types are served and Q binds. The mass of high-types firstly increases with K, then the mass still increases but their share starts decreasing hyperbolically. When both constraints bind even the absolute mass of high-types starts decreasing so their share drops even further, according to a steeper hyperbola until it reaches zero, see Figure 4. 3.2.3\n\nLarge Q\n\nThe next region of interest is \u0012 \u0012 \u0013 1−α qL max qL vL − vH , 2 qH\n\nvH − vL αqH 2\n\n\u0013\nE(v) 2\n\nThe evolution of optimal prices and consumer shares is very similar to the one for medium Q, case B, as one can see from Figure 5. Indeed, as K grows the optimal prices are chosen from regions EL, K, KQ, respectively. However, as Q is larger here, the monopoly does not exclude high-types even for very large values of K, the last\n\n19\n\nFigure 6: Very large Q Prices rH\n\nvH 1 vL +vH 2\n\npH\n\nvH /2 vL pL\n\nvL /2 L α vH −v 2\n\nE(v)/2\n\nK\n\nαvH /E(v) 0\n\n(a) Prices\n\nL α vH −v 2\n\nE(v)/2\n\nK\n\n(b) Share of high-types\n\nregion the prices are chosen from is Q. Accordingly, the share of high-types varies as described in Case B up to K = f (Q) where it becomes independent of K and levels off at a strictly positive value. 3.2.4\n\nVery large Q\n\nWhen Q > E(vq)/2, the capacity constraint on the amount of production will never be binding.6 For very low levels of K the firm excludes low-types and serves only high-type consumers. As K grows larger the monopoly starts to serve some low-types as well, although the constraint on the number of consumers still binds. Finally, as K becomes very large neither of the constraints will bind and the monopoly can achieve the unconstrained optimum. Clearly, both prices pH and pL are weakly decreasing in K in this region (Figure 6a). The share of high-types served decreases from 1 to αvH /E(v) which corresponds exactly to the proportion of high-types in the whole market (Figure 6b).\n\n4\n\nWelfare\n\nIn this section, I investigate welfare properties of the monopoly’s optimal pricing behavior in the presence of dual capacity constraints. I first analyze aggregate consumer surplus as a function of the capacity constraints given that the monopoly 6\n\nOne can see the case of very large Q as an alternative benchmark. As Q never binds, this region describes the monopoly’s optimal strategy when there is only one capacity constraint and two consumer groups.\n\n20\n\nchooses its profit-maximizing prices described in Proposition 1. Next, I calculate total welfare as the sum of consumer surplus and the monopoly’s profit, although the most interesting insights come from the study of consumer surplus. Figure 7: Non-monotonicity of consumer surplus\n\nK\n\nQ\n\nEH\n\nU\n\nE(v) 2\n\nK 1−α 2 (vL\n\n− vH qqHL )\n\nKQ\n\nL α vH −v 2\n\nEL\n\nQ/qH 1−α 2 qL (vL\n\n− vH qqHL )\n\nL αqH vH −v 2\n\nE(vq) 2\n\nQ\n\nWhen both consumer groups are served, the general form of total consumer surplus writes as α 1−α (vH − pH )2 + (vL − pL )2 . 2 2 In the EL region, where the monopoly only serves high-types, consumer surplus equals α2 (vH − pH )2 . In the EH region, where the firm serves exclusively the low-types, consumer surplus equals 1−α (vL − pL )2 . 2 CS =\n\nThus consumer surplus is weakly decreasing in both prices pL and pH . As shown in the previous section, with the exception of the KQ region where both constraints bind, prices are decreasing in the size of capacities. Hence consumer surplus is 21\n\nincreasing in capacity levels for any capacity-pairs outside of the KQ region. However, the problem is more complicated in the KQ region where both capacities bind. In this region pH is increasing while pL is decreasing in K. The following proposition sheds light on the effect of this trade-off. Proposition 2. Aggregate consumer surplus is non-monotonic in the size of the capacity constraints in the parameter region where both capacities are binding. In particular, there always exists a region of capacity pairs inside KQ where consumer surplus is decreasing in K and increasing in Q. Moreover, there exists a second region inside KQ where consumer surplus is decreasing in Q and increasing in K. The two regions are always disjoint. The proof of Proposition 2 shows that consumer surplus is decreasing in K in E(q) the KQ region if and only if K ≤ Q E(q 2 ) (light grey area in Figure 7). The intuition for this result is the clearest when Q is relatively low and one can consider the transition between the regions EL and KQ. When K is increasing from a value lower than Q/qH the monopoly first excludes all low-types and serves only high-types. As K reaches Q/qH it starts to be profitable to serve some low-type consumers as well such that both constraints bind. To accommodate low-types, the firm starts serving less high-types thus it increases pH while it lowers pL . Consumer surplus is affected by 3 factors. Firstly, as K is binding in both regions, the total mass of consumers served goes up which ceteris paribus increases consumer surplus. Secondly, the decrease in pL has the same effect also, however, the increase of pH goes in the opposite direction. To see that this last effect dominates the first two, one should consider the mass of consumers affected. Indeed, when K is relatively small, the firm serves a lot more high-types than lowtypes, so the loss suffered by the high-types dominates the gain of the few low-types. Similarly, consumer surplus is decreasing in Q in the KQ region if and only if K ≥ Q/E(q) (dark grey area in Figure 7) and the arguments are analogous to the ones described above. The next proposition provides comparative statics results for total welfare, which in this context simply equals to the sum of the monopoly’s profit and consumer surplus.\n\nProposition 3. Total welfare, i.e., the sum of consumer surplus and the monopoly’s profit, is increasing in both capacity levels for every parameter value. I show in the Appendix that in both areas where consumer surplus is decreasing, profit increases faster than consumer surplus decreases. This property is obviously 22\n\ntrue for other capacity pairs where both consumer surplus and profit are increasing in capacities.\n\n5\n\nEndogenous capacity choice\n\nIn this section I investigate the monopoly’s optimal choice of capacity levels in the long run. Although in the short run it is reasonable to assume that the monopoly chooses its prices facing fixed capacity levels, in the long run firms can extend or shrink both of their capacities. Let cK (K) denote the cost of building capacity K and let cQ (Q) denote the cost of building Q. I assume that the costs are separable. In the hospital example, although there is a fixed cost of constructing the hospital building, the additional costs of adding beds and equipping the operating rooms are separable. Assume that these costs are strictly positive whenever the capacity levels are strictly positive. The monopoly maximizes its profit which is now a function of its two capacities as well as its prices. The optimal choice of prices for any capacity-pair is the one described in Proposition 1. The following Lemma holds under these very general conditions. Lemma 1. If capacity choice is endogenous and the cost of building capacities is strictly positive then the monopoly chooses its prices and capacity levels in such a way that both constraints bind, i.e., the optimal capacities are always chosen from the KQ region. An alternative interpretation of Lemma 1 is that the monopoly never chooses capacities in such a way that only one of the capacities be binding. Intuitively, in a world of deterministic demand it is never profitable for a monopoly to build unused capacity. Notice that choosing capacities from the KQ region does not necessarily mean that the monopoly serves both types of consumers. The firm may choose its capacities at the limit of the region in a way to exclude all consumers of one or the other type. In order to obtain closed-form results and hence a clear intuition, I will focus on the case of linear costs in the remainder of the section, i.e., let cK (K) = cK\n\nand cQ (Q) = dQ.\n\n23\n\nThe following proposition provides a complete characterization of the monopoly’s optimal capacity choice given the linear cost functions. Proposition 4. If cost functions are linear, i.e., cK (K) = cK then the optimal capacity levels are given by\n\nand cQ (Q) = dQ\n\n1. K = 21 (E(v) − c − dE(q)) and Q = 12 (E(vq) − cE(q) − dE(q 2 )) if vH > c + dqH and vL > c + dqL 2. K = α2 (vH − c − dqH ) and Q = α2 qH (vH − c − dqH ) if vH > c + dqH and vL ≤ c + dqL (vL − c − dqL ) and Q = 1−α qL (vL − c − dqL ) 3. K = 1−α 2 2 if vH ≤ c + dqH and vL > c + dqL 4. K = Q = 0 if vH ≤ c + dqH\n\nand vL ≤ c + dqL\n\nThe resulting optimal prices are pi = (vi + c + dqi )/2, i-types are not excluded.\n\ni ∈ {L, H} whenever\n\nNotice that c + dqH (c + dqL ) corresponds to the marginal cost of building capacity to serve an additional high-type (low-type) consumer. Some high-type consumers are served if and only if the marginal cost of capacity necessary to serve a high-type is lower than the WTP of the most valuable high-type consumer, i.e., vH > c + dqH . An analogous result holds for low-types. Hence the four cases of Proposition 4: depending on the relative sizes of the two marginal costs with respect to the WTP of the most valuable consumers, the firm either serves both types or excludes one or both groups of consumers. The first case in Proposition 4 corresponds to the situation when some consumers of both types are served, i.e., capacities are chosen from the inside of the KQ region. The second case describes a situation where the monopoly prefers excluding low-type consumers. This arises when the capacity constraint on production, Q is relatively cheap to build, which makes K relatively stricter which in turn increases the attractively of high-types. Indeed, the two conditions imply d < (vH − vL )/(qH − qL ). The optimal capacities satisfy K = Q/qH which means that the monopoly chooses its capacities from exactly one side of the KQ quadrilateral. Notice also that the resulting price for high-types, pH = (vH + c + dqH )/2, corresponds exactly to the optimal price of a monopoly facing only high-type consumers and whose production costs are equal to c + dqH .\n\n24\n\nThe third case is analogous to the second one: the firm prefers excluding hightypes and chooses its capacity from the K = Q/qL side of the KQ quadrilateral. The fourth case completes the discussion: if both capacities are very expensive to build with respect to the WTP of all consumers then the monopoly prefers to exit the market. As one should expect, capacity levels are decreasing in costs, moreover, transitions between the different cases are smooth in the sense that K and Q are continuous in both c and d. In the limit as both costs go to 0, the monopoly builds enough capacities to achieve its unconstrained optimum, i.e., K → E(v)/2 and Q → E(vq)/2. Both prices pL and pH are an increasing function of both cost parameters c and d. Therefore, the consumer surplus is always decreasing in both c and d when capacities are chosen endogenously. Hence, in the long run both prices and consumer surplus are monotonic in capacity costs.\n\n6\n\nIncentive compatibility\n\nIn this section I relax the assumption that the monopoly is able to observe the quantity demanded by each consumer. Given the self-selection of consumers, the firm faces new incentive compatibility constraints. I investigate two different extensions of the baseline model. In the first scenario low-types are allowed to buy the large bundle and freely dispose of the unused part. In the second scenario, I check the robustness of the baseline model by assuming that qH is a multiple of qL , and that high-types are allowed to buy several small bundles .\n\n6.1\n\nIncentive compatibility for low-types\n\nIn this section I consider an extension of the model where low-types are allowed to buy quantity qH for pH and throw away the quantity qH − qL they do not consume. One can think of this scenario as the case with free disposal. This assumption is realistic if the monopoly cannot tell consumers apart according to their demand. How does consumers’ possibility of free disposal alter the monopoly’s incentives? Firstly notice that free disposal only alters consumers’ (low-types’) incentives in case pH < pL ≤ vL , i.e., whenever the larger quantity is cheaper. Problem (P-EH) where only low-types are served will thus remain unaffected by free disposal.\n\n25\n\nThe maximization problem of serving both consumer groups becomes more complicated in the presence of free disposal. The monopoly must decide whether it chooses its prices in a way that makes both types buy the quantity intended for them, or alternatively, it may choose a price such that all consumers buy the larger quantity, qH . In the former case, the maximization problem is very similar to (P-LH) described previously, with one additional constraint:\n\n(P-ICL)\n\nmax\n\npL ,pH\n\nπ = α(vH − pH )pH + (1 − α)(vL − pL )pL\n\ns.t.\n\nα(vH − pH ) + (1 − α)(vL − pL ) ≤ K α(vH − pH )qH + (1 − α)(vL − pL )qL ≤ Q 0 ≤ pL < vL\n\n, pL ≤ pH < vH\n\nThe monopoly must choose a lower price for the smaller quantity if it wants the two consumer groups separated, hence the new, stricter lower bound on for the high-price: pL ≤ pH . However, we know from the previous section that the solution of (P-LH) for any parameter region satisfies this stricter condition. This means that the solution of (P-ICL) will exactly coincide with the solution of (P-LH) described by Proposition 1. In addition, the monopoly might now choose a price pH < pL ≤ vL that makes all consumers buy qH for pH . In this case the mass of buyers is α(vH − pH ) + (1 − α)(vL − pH ) = E(v) − pH and total demand is equal to (E(v) − pH )qH . Hence the maximization problem writes as\n\n(P-ICL2)\n\nmax π = (E(v) − pH )pH pH\n\ns.t.\n\nE(v) − pH ≤ min(K, Q/qH ) pH < vL\n\nNotice that the two constraints imply that the following two inequalities must be satisfied for (P-ICL2) to have a feasible solution: K > α(vH − vL ) and Q > αqH (vH − vL ) 26\n\nIntuitively, the capacity constraints must be relatively large if the monopoly is able to serve some consumers of both types for a relatively low price. Finally, if the monopoly wants to exclude low-types from buying its products, it must satisfy the stricter condition of vL < pH :\n\n(P-ICLEL)\n\nmax π = α(vH − pH )pH pH\n\ns.t.\n\nα(vH − pH ) ≤ min(K, Q/qH ) vL < pH However, the optimal solution of (P-ICLEL) is obviously dominated by the solution of (P-EL) as it is a constrained version of it. Proposition 5. The monopoly’s optimal pricing strategy is not affected by the possibility of free disposal. Proposition 5 states that the model of dual capacities is robust to the introduction of the free disposal assumption. The new strategies of the monopoly consist of pooling both types at pH and selling them the large quantity qH . The proof in the Appendix shows that this pooling is always dominated by some strategy that was already available in the baseline model.\n\n6.2\n\nIncentive compatibility for high-types\n\nIn this section I consider a model in which the individual demanded of high-types is a multiple of the individual demand of low-types, i.e., qH = kqL , where k is an integer. The monopoly cannot observe consumers’ individual demand. High-types therefore must make a choice between buying one bundle of qH or k bundles of qL . The monopoly must decide whether to choose prices such that high-types prefer buying the large bundle (qH ) or to choose them in a way that everyone buys small bundles (qL ). Given that the monopoly is unable to tell apart consumers, it must choose a low enough price for qH if it wants high-types to buy it instead of several small bundles. In particular, the unit price of the large bundle cannot exceed the unit price of the small bundle: pH ≤ kpL ⇐⇒ pH /qH ≤ pL /qL . Notice that the optimal solutions of (P-LH), as described in Proposition 1 satisfy this condition for any capacity-pair. Hence, similarly to the free disposal case, the additional incentive compatibility constraint does not alter the firm’s optimal 27\n\npricing behavior if it wants the two consumer groups to buy different bundles. The other option of the firm is to choose a relatively low price for the small quantity, resulting in all consumers buying that variety. The total price a high-type buyer must pay to satisfy its demand qH is kpL . The maximization problem when serving some consumers of both types then writes as\n\n(P-ICH)\n\npL pL )qH + (1 − α)(vL − pL )pL s.t. qL qL pL α(vH − qH ) + (1 − α)(vL − pL ) ≤ K qL pL α(vH − qH )qH + (1 − α)(vL − pL )qL ≤ Q qL 0 ≤ kpL < vH\n\nmax π = α(vH − qH pL\n\nThe last inequality ensures a low enough price so that some high-types consume the product. Notice that this maximization problem coincides with (P-LH) with the additional constraint of pH = kpL = qH pqLL . As shown above, the monopoly’s optimal choice with perfect divisibility coincides with the solution of (P-LH). (P-ICH) being a more restricted problem, it can never be more profitable for the firm to make all consumers buy the small bundle than to separate the two consumer groups. Finally, the monopoly can choose to exclude one group of consumers. The firm faces exactly the same problem as (P-EL) if it serves only qH bundles to hightypes. Serving only qL bundles does not necessarily exclude the high-types, if the monopoly wants to serve only low-types, it must choose a relatively high unit price: vH < pqLL < vqLL . However, the solution of this sub-problem either coincides with the qH solution of (P-EH), or it is dominated by serving both consumer groups, hence the following proposition: Proposition 6. The monopoly’s optimal pricing strategy is not affected by high-type consumers’ ability to buy several small bundles despite the monopoly’s inability to observe consumers’ demand. Proposition 6 concludes the analysis of incentive compatibility. Jointly with Proposition 5, it suggests that the results of the model are robust to the relaxation of the observable individual demand assumption.\n\n28\n\n7\n\nGeneral distribution of consumers\n\nIn the baseline model presented in Section 2, consumers’ WTP is distributed uniformly within each group, which implies linear demand that in turn leads to the clear-cut results presented in Proposition 1. In this section I generalize the baseline model by assuming a general distribution function of consumers’ WTP. I identify sufficient conditions on the distribution functions of the two consumer groups that guarantee the existence of the regions discovered in the baseline model, illustrated by Figure 1. Moreover, I show that under these conditions the main results carry over to the general distribution case. Let the WTP of consumers of type i ∈ {L, H} be distributed according to the twice continuously differentiable distribution Fi with support [0, θi ]. Let fi denote the first derivative of Fi . For any p ≥ 0 let Di (p) = 1−Fi (p) be the demand function that measures the proportion of i-type consumers willing to buy at price p. Let α and 1−α denote the total mass of high-types and low-types, respectively. Individual demand of consumers is the same as in the baseline model: qH > qL . As before, the monopoly is constrained by K, the total mass of consumers it can serve on the one hand, and by Q, its maximal production on the other hand. The maximization problem of the monopoly writes as\n\n(P-GEN)\n\nmax\n\npL ,pH\n\nπ = αpH DH (pH ) + (1 − α)pL DL (pL )\n\ns.t.\n\nαDH (pH ) + (1 − α)DL (pL ) ≤ K\n\n(λ1 )\n\nαqH DH (pH ) + (1 − α)qL DL (pL ) ≤ Q\n\n(λ2 )\n\npL ≥ 0 , pH ≥ 0\n\nwhere λ1 is the multiplier of the constraint on the mass of consumers served and λ2 is the multiplier of the constraint on total production. The only restrictions on the demand functions so far are that they be decreasing, Di (0) = 1 and Di (θi ) = 0. Notice that Di (pi ) = 0 for any pi ≥ θi , so choosing a large enough pi excludes consumer group i. Assume that the profits derived from serving the two groups, i.e., pi Di (pi ), is concave. The monotone hazard rate condition, which in this context translates to the function φi (p) ≡\n\nfi (p) D0 (p) =− i 1 − Fi (p) Di (p)\n\nbeing non-decreasing, is a sufficient condition for the concavity of pi Di (pi ). The Karush-Kuhn-Tucker conditions of the maximization problem write as 29\n\n0 (pL −λ1 −λ2 qL )DL0 (pL )+DL (pL ) = 0 and (pH −λ1 −λ2 qH )DH (pH )+DH (pH ) = 0.\n\nThese conditions are trivially satisfied for a consumer group which is excluded from the market. Otherwise, when prices are chosen such that some consumers of both types are served (i.e., pL < θL and pH < θH ) the first order conditions can be rewritten using the distribution functions:\n\nλ1 + λ2 q L = pL −\n\n1 − FL (pL ) fL (pL )\n\nand λ1 + λ2 qH = pH −\n\n1 − FH (pH ) . fH (pH )\n\ni (p) Notice that the term p − 1−F which appears in both equations is the virtual fi (p) valuation function that is widely used in the mechanism design literature.7\n\nThe first order conditions imply that in optimum the virtual valuation of the two consumer groups must coincide whenever the capacity on the number of people served (K) binds and the other constraint is slack (λ1 > 0, λ2 = 0). Conversely, if Q binds and K is slack (λ1 = 0, λ2 > 0) then the per-unit virtual valuation of the two groups must be equal. The monopoly can charge its unconstrained optimal prices (where both virtual valuations equal zero) if none of the constraints bind. Replacing the optimal prices into the two capacity constraints, one gets a unique threshold level for both capacity levels, K and Q that delimit region U’. Next, I identify sufficient conditions for the existence of a core region KQ’ where both constraints bind. Both constraints binding immediately imply that the optimal prices must satisfy DL (pL ) =\n\nKqH − Q (1 − α)(qH − qL )\n\nand DH (pH ) =\n\nQ − KqL . α(qH − qL )\n\nThis means that 2 of the 4 curves delimiting the core region KQ’ remain the same as in the baseline model: in order to serve both consumer groups, K ≥ Q/qH and K ≤ Q/qL must be satisfied. The two other frontiers of the KQ’ region are given by capacity-pairs within the region that also satisfy the equations λ1 = 0 and λ2 = 0. Let K = g(Q) denote the curve above which capacity K is slack (λ1 = 0), and let K = h(Q) be the curve that specifies whether Q is binding or slack (λ2 = 0). Finding an explicit formula for these curves would necessitate knowing the value of Di0 (pi ) for the optimal prices. Although demand levels at the optimal prices are 7\n\nThe connection between auction theory and the monopoly’s problem of third-degree price discrimination was first revealed by Bulow and Roberts (1989).\n\n30\n\nsimple, there is no direct formula to express the derivatives of the demands at the optimal prices. However, using the implicit function theorem, one can prove the following Lemma. Lemma 2. (i) Assume that both distribution functions satisfy Myerson’s regularity condition, i (p) i.e., the virtual valuation functions p − 1−F , i ∈ {L, H} are strictly infi (p) 0 0 creasing in p. Then g (Q) > 0 and h (Q) > 0. (ii) g(Q) < h(Q) for any 0 ≤ Q < Q and g(Q) = h(Q) = K. e > 0 such that g(Q) ≤ Q/qH for ∀ 0 ≤ Q < Q. e (iii) ∃ Q e > 0 such that h(Q) ≥ Q/qL for ∀ 0 ≤ h(Q) < K. e (iv) ∃ K (i) identifies regularity of the distribution functions as sufficient conditions for curves g and h to be increasing. Notice that the monotone hazard rate condition implies Myerson’s regularity condition. (ii) states that g is always below h when at least one of the capacities are binding, and they cross exactly at point (Q, K). (iii) and (iv) state that g(Q) < Q/qH for small values of Q and conversely, h(Q) > Q/qL for small values of K. The combination of these results imply the existence of a KQ’ region delimited by two increasing curves in addition to the two straight lines. Next, I show that the existence of region EL’ where the monopoly excludes low-types is guaranteed if the distribution of high-types hazard rate dominates the distribution of low-types, i.e., fH (p) fL (p) < 1 − FH (p) 1 − FL (p)\n\nfor all p.\n\nNotice that the hazard rate dominance relation rewrites in terms of demand as εH (p) < εL (p) for all p, where εi (p) is the elasticity of demand: pDi0 (p) . εi (p) = − Di (p) Therefore, hazard rate dominance is equivalent to the low-types having a larger demand elasticity than high-types at each price. As hazard rate dominance implies first order stochastic dominance, it also implies θL < θH . The proof consists of showing that firstly, prices are decreasing and demands are increasing in K in the K’ region, and secondly, there is a positive threshold level of K below which optimal prices induce zero demand for the low-types and strictly positive demand for the high-types. Intuitively, the curve separating regions K and EL is a horizontal line because in both regions only constraint K binds and the prices are profits are 31\n\nindependent of Q. Conversely, I show that a sufficient condition for the existence of an EH’ region where the monopoly serves exclusively the low-type consumers is\n\nqL\n\nfL (p) fH (p) < qH 1 − FL (p) 1 − FH (p)\n\n⇐⇒\n\nqL εL (p) < qH εH (p) for all p.\n\nThe sufficient condition requires that for every price, the demand elasticity weighted by the individual demand be higher for the high-types than for the lowtypes. This ensures the existence of a threshold level of Q at which the optimal prices in the Q’ region induce a zero demand for the high-types while demand of low-types remain positive. The curve separating regions Q’ and EH’ is a vertical line because prices and profits in both regions only depend on Q and are independent of K. Consumer surplus in the context of general distributions can be written as Z\n\nθH\n\nZ\n\nθL\n\n(w − pH )fH (w)dw + (1 − α)\n\nCS = α pH\n\n(w − pL )fL (w)dw. pL\n\nObviously, the consumer surplus is a decreasing function of both prices. In most regions both prices are decreasing in both capacity levels, thus aggregate consumer surplus is increases in both K and Q. However, in the KQ’ region pH is increasing in K and decreasing in Q and conversely, pL is increasing in Q and decreasing in K. Therefore, consumer surplus of the high-types is decreasing while consumer surplus of low-types is increasing in K. The following proposition states that there always exists a region of capacity-pairs inside of KQ’ where the first effect is dominant. Proposition 7. Assume that both distribution functions satisfy the monotone hazard rate condition. Then there exist two disjoint regions inside of KQ’ such that the consumer surplus decreases in K in one of the regions, and it decreases in Q in the other region. The monotone hazard rate condition implies both the concavity of profits and increasing virtual valuation functions. In the Appendix, I show that ∂CS DL (pL ) DH (pH ) < 0 ⇐⇒ 0 < 0 ∂Q DL (pL ) DH (pH )\n\nand\n\n∂CS D0 (pL ) D0 (pH ) < 0 ⇐⇒ qL L < qH H . ∂K DL (pL ) DH (pH )\n\nIt follows from Lemma 2 that part of the Q/qL and Q/qH lines border the KQ’ region. From the demand functions, it is obvious that the first condition is satisfied for K = Q/qL and the second for K = Q/qH . Existence of the two regions can be 32\n\nproved by continuity arguments. In order to gain intuition for the above formulas, one can rewrite them in terms of elasticities: ∂CS εL (pL ) pL < 0 ⇐⇒ < ∂Q εH (pH ) pH\n\nand\n\n∂CS εL (pL ) pL /qL < 0 ⇐⇒ > . ∂K εH (pH ) pH /qH\n\nConsumer surplus in decreasing in Q whenever at the optimal prices the demand elasticity to price ratio is larger for high-types than for low-types. Furthermore, consumer surplus is decreasing in K when at the optimal prices the demand elasticity to unit price ratio is higher for the low-types than for the high-types. This formulation also shows clearly that there is no capacity-pair that satisfies both conditions, i.e., there is no situation in which an increase in both capacities decreases consumer surplus.\n\n8\n\nConclusion\n\nSeveral capacity constraints co-exist in various real-world industries. The present paper provides a formal economic analysis of the effects of dual capacity constraints on optimal firm behavior. It reveals a rich structure of optimal monopoly pricing which in the short run is qualitatively different from the predictions of models of firms bound by a single capacity. In particular, prices charged for some consumers increase for some capacity pairs as one of the capacities is enlarged. Moreover, aggregate consumer surplus is decreased by an increase of one capacity level for some capacity pairs. These results are robust to observability of individual demand and also for a fairly general class of distribution of consumers. In the long run, when capacity building is endogenous, prices and consumer surplus are monotonic in capacity costs. Future research can extend the results in several important aspects. One could verify the model’s robustness by approximating the capacity constraints with convex and continuous cost functions. Moreover, the model could accommodate more than two consumer groups. Finally, by extending the model to the case of a duopoly, it would become directly comparable with the various models of Bertrand-Edgeworth competition with one capacity constraint.\n\n33\n\nAppendix Proof of Proposition 1 Consider the following maximization problem that encompasses (P-LH), (P-EL) and (P-EH):\n\n(P-GEN)\n\nmax\n\npL ,pH\n\nπ = α(vH − pH )pH + (1 − α)(vL − pL )pL\n\ns.t.\n\nα(vH − pH ) + (1 − α)(vL − pL ) ≤ K\n\n(λ1 )\n\nα(vH − pH )qH + (1 − α)(vL − pL )qL ≤ Q\n\n(λ2 )\n\npL ≤ v L\n\n(λ3 )\n\npH ≤ v H\n\n(λ4 )\n\npL ≥ 0 , pH ≥ 0 The non-negativity constraints are omitted and verified ex-post. Multiplying the third and fourth constraint by 1 − α and α, respectively, the objective function for deriving the Karush-Kuhn-Tucker conditions writes as\n\nL(pL , pH , λ1 , λ2 , λ3 , λ4 ) = α(vH − pH )pH + (1 − α)(vL − pL )pL − λ1 [α(vH − pH ) + (1 − α)(vL − pL ) − K] − λ2 [α(vH − pH )qH + (1 − α)(vL − pL )qL − Q] − λ3 (1 − α)(vL − pL ) − λ4 α(vH − pH ) Hence the two first order conditions are 2pL = vL + λ1 + λ3 + λ2 qL\n\nand 2pH = vH + λ1 + λ4 + λ2 qH .\n\nAs λ3 > 0 implies pL = vL , this case corresponds to excluding the low-types. Indeed, (P-GEN) is reduced to (P-EL) whose solution is described in the main text. Similarly, λ4 > 0 implies pH = vH i.e., the exclusion of high-types which corresponds to the maximization problem (P-EH), also solved in the main text. Obviously, λ3 > 0 and λ4 > 0 leads to zero profit hence it is never optimal. Finally, λ3 = λ4 = 0 corresponds to the case of serving both consumer groups, described in (P-LH). In the following, I prove the formulas for optimal prices and the borders of the optimal regions described in Case 1 - Case 3. Case 4 is described in the main body of the paper.\n\n34\n\nCase 1 : λ1 > 0 and λ2 = 0 From the FOCs: λ1 = 2pL − vL = 2pH = vH and λ1 > 0 implies that K binds: α(vH − pH ) + (1 − α)(vL − pL ) = K. The optimal prices can be calculated from these 2 equations: vH − vL vH − vL − K and pH = vH − (1 − α) −K 2 2 The capacity constraint on production rewrites as pL = v L + α\n\nvH − vL vH − vL )qH + (1 − α)(K − α )qL ≤ Q, 2 2 which is equivalent to K ≤ g(Q) by definition. λ1 > 0 implies K < E(v)/2 and finally, pL ≤ vL implies α2 (vH − vL ) ≤ K. The non-negativity constraints and pH ≤ vH are always satisfied in this optimum. α(K + (1 − α)\n\nCase 2 : λ1 = 0 and λ2 > 0 L From the FOCs: λ2 = 2pLq−v = 2pHqH−vH and λ2 > 0 implies that Q binds: L α(vH − pH )qH + (1 − α)(vL − pL )qL = Q. The optimal prices can be calculated from these 2 equations. The capacity constraint K must be slack, replacing the optimal \u0010 prices into\u0011 the constraint leads to f (Q) ≤ K. Moreover, pH ≤ vH implies 1−α qL vL − vH qqHL ≤ Q and finally, λ2 > 0 implies Q < E(vq)/2. 2\n\nCase 3 : λ1 > 0 and λ2 > 0 Both K and Q bind, hence the optimal values of the prices follow directly from the two equations. The four borders of the KQ regions can be calculated as follows. The values of the two multipliers can be calculated from the FOCs by replacing the optimal prices: λ2 = 2\n\nvH − vL Q − KqH KqL − Q + − 2 qH − qL α(qH − qL ) (1 − α)(qH − qL )2\n\nand\n\n\u0013 \u0012 Q − KqL vH − vL Q − KqH KqL − Q λ1 = 2vH − 2 − qH 2 + − . α(qH − qL ) qH − qL α(qH − qL )2 (1 − α)(qH − qL )2 It follows that λ1 > 0 is equivalent to K ≥ g(Q) and λ2 > 0 is equivalent to K ≤ f (Q). Furthermore, pL ≤ vL implies K ≤ Q/qL and pH ≤ vH implies K ≥ Q/qH .\n\n35\n\nExcluding one consumer group In the parameter regions delimited by one of the Cases 1 - 4, the optimal prices are such that some consumers of both groups are served. There are two remaining regions where one group of consumers will be excluded. If \u0001 K ≤ min Q/qH , α2 (vH − vL ) then K must bind, so one must compare the profit 2 K2 of vH K − Kα when excluding low-types with the profit of vL K − 1−α obtained by α excluding the high-types. K ≤ 2 (vH −vL ) implies that the first profit, i.e., excluding the low-types is always more profitable for small values of K. An analogous argument shows why is\u0011\u0011 more profitable than excluding low-types \u0010 excluding high-types \u0010 qL 1−α when Q ≤ min KqL , 2 qL vL − vH qH . \u0004 Proof of Proposition 2 Replacing the optimal prices pL and pH of region KQ into the general formula, the consumer surplus equals \u00132 \u0012 \u00132 Q − KqL KqH − Q 1−α + = α(qH − qL ) 2 (1 − α)(qH − qL ) \u0001 1 2 2 2 = Q − 2E(q)KQ + E(q )K . 2α(1 − α)(qH − qL )2\n\nα CS = 2\n\n\u0012\n\nTherefore the first derivative of the consumer surplus with respect to K and Q are \u0001 1 ∂CS 2 = −E(q)Q + E(q )K ∂K α(1 − α)(qH − qL )2 ∂CS 1 = (−E(q)K + Q) . ∂Q α(1 − α)(qH − qL )2\n\nand\n\nThus consumer surplus is decreasing in K whenever K ≤ E(q)/E(q 2 )Q and it is decreasing in Q if K ≥ Q/E(q). The two regions delimited by these lines are disjoint since they both cross the origin and the slope of K = E(q)/E(q 2 )Q is smaller than the slope of K = Q/E(q) as (E(q))2 < E(q 2 ). \u0004 Proof of Proposition 3 Total welfare is given by the sum of consumer surplus and profit:\n\n36\n\n1 ((vH − vL )Q + (vL qH − vH qL )K) − (qH − qL ) \u0001 1 − Q2 − 2E(q)KQ + E(q 2 )K 2 . 2 2α(1 − α)(qH − qL )\n\nT W = CS + πKQ =\n\nNotice that wherever consumer surplus is decreasing in either K or Q, the third term of the above equation is increasing, so do the first two terms which means that total welfare is always an increasing function of both capacity levels. \u0004 Proof of Proposition 4 Given Lemma 1, optimal capacities are chosen from the KQ region which the following maximization problem: (vH − vL )Q (vL qH − vH qL )K Q2 + E(q 2 )K 2 − 2KQE(q) max π = + − − cK − dQ K,Q q H − qL q H − qL α(1 − α)(qH − qL )2 Q/qH ≤ K\n\n(λ1 )\n\nK ≤ Q/qL\n\n(λ2 )\n\ng(Q) ≤K ≤ h(Q) K≥0, Q≥0 ∂π\n\nKQ Firstly, consider the interior solution where λ1 = λ2 = 0.From ∂K = c and ∂πKQ = d one immediately gets the optimal capacity levels described in part 1 ∂Q of Proposition 4. Replacing these optimal values into conditions Q/qH ≤ K and K ≤ Q/qL imply vH > c + dqH and vL > c + dqL , respectively. The remaining four primal feasibility conditions are satisfied at this solution.\n\nSecondly, consider λ1 > 0 and λ2 = 0 that corresponds to the exclusion of low-types. K = Q/qH and the two first order conditions provide the optimal capacity levels described in part 2 of Proposition 4. Positivity of λ1 requires vL ≤ c + dqL , and the positivity of Q implies vH > c + dqH .The remaining four primal feasibility conditions are satisfied. Thirdly, consider λ2 > 0 and λ1 = 0 that corresponds to the exclusion of high-types. K = Q/qL and the two first order conditions provide the optimal capacity levels described in part 3 of Proposition 4. Positivity of λ2 requires vH ≤ c + dqH , in addition, the positivity of K implies vL > c + dqL . The remaining four primal feasibility conditions are satisfied. 37\n\ns.t.\n\nBoth capacity levels are zero if the marginal cost of serving each consumer group is prohibitively high, which corresponds to part 4 of Proposition 4. Finally, replacing the optimal capacity levels into the optimal prices in region KQ, one immediately gets that pi = (vi + c + dqi )/2, i ∈ {L, H}. \u0004 Proof of Proposition 5 Showing that the outcome of (P-ICL) coincides with the outcome of (P-LH) consists of showing that in each local optimum, pL ≤ pH is satisfied. Optimal prices in regions K, KQ and U trivially satisfy this condition. In region Q, it is equivalent to\n\nvL −\n\nKqH − Q Q − KqL 1 ≤ vH − ⇔ K≥ (Q − α(1 − α)(qH − qL )(vH − vL )) , (1 − α)(qH − qL ) α(qH − qL ) E(q)\n\nwhich by the definition of g(Q) is equivalent to K ≥ g(Q) −\n\n1 vH − vL α(1 − α)(qH − qL ) . E(q) 2\n\nThis condition is always satisfied in region Q, since in this region the stronger condition of K ≥ g(Q) is also satisfied. Next, I show that the solutions of (P-ICL2) are always dominated by some solution of (P-ICL). Notice that pH = E(v)/2 is only attainable if K > E(v)/2 and Q > E(v)qH /2 meaning that the capacities are from the U region, where the solution of (P-ICL2) is clearly dominated by π U . The two conditions can be rewritten as E(v) − min(K, Q/qH ) ≤ pH < vL , which immediately implies that a necessary condition for existence of a solution is min(K, Q/qH ) > α(vh − vL ). This in turn implies that the solution cannot be in the EL region. By concavity of the objective function, whenever solution exists, it is given by pH = E(v) − min(K, Q/qH ) thus the profit equals E(v)K − K 2 . For K < Q/qH the necessary condition implies that capacity levels fall in the K region, where πK is attainable in (P-ICL). By definition, \u0012 πK = α(1 − α)\n\nvH − vL 2\n\n\u00132\n\n+ KE(v) − K 2 > KE(v) − K 2 ,\n\nso if K < Q/qH , the optimal solution is dominated. Next, I show that the same 38\n\nis true for K ≥ Q/qH . In this case, for all K we have KE(v) − K 2 > E(v)Q/qH − (Q/qH )2 , where the right-hand side is the optimal profit of (P-ICL2). The left-hand side is smaller than πK , as shown above, therefore it is also dominated by the optimal profits attainable in regions KQ and Q in the (P-ICL) problem.\u0004 Proof of Lemma 2 i (p) (i) Myerson’s regularity condition, i.e., p − 1−F , i ∈ {L, H} being strictly fi (p) increasing in p can be rewritten in terms of the demand function as\n\nDi00 (p) <\n\n2(Di0 (p))2 . Di (p)\n\nFirstly, consider the slope of curve h(Q), defined as the part of the KQ’ region where λ2 = 0. Substituting λ2 = 0 in the first order conditions, one gets\n\npH +\n\nDH (pH ) DL (pL ) = pL + 0 0 DH (pH ) DL (pL )\n\n⇐⇒\n\n0 0 ξ(K, Q) ≡ DL DH −DH DL0 +(pL −pH )DL0 DH =0\n\n∂ξ/∂Q The implicit function theorem ensures that h0 (Q) = ∂K = − ∂ξ/∂K . In the ∂Q following I show that this derivative is positive if the regularity condition is satisfied. ∂x , i.e., for any function x, x denotes its partial derivative with respect to Let x = ∂Q KqH −Q Q−KqL Q. We know that DL (pL ) = (1−α)(q and DH (pH ) = α(q , which implies H −qL ) H −qL )\n\nDL =\n\n−1 (1 − α)(qH − qL )\n\nand DH =\n\n1 α(qH − qL )\n\nand pi =\n\nDi Di0\n\nand Di0 = Di00\n\nDi Di0\n\nfor i ∈ {L, H}. We have ∂ξ ∂DL ξ= = ∂Q ∂Q\n\n\u0012\n\n0 2DH\n\n+ ((pL −\n\n0 pH )DH\n\n\u0013 \u0012 \u0013 00 DL00 ∂DH DH 0 0 − DH ) 0 + 2DL − ((pL − pH )DL + DL ) 0 DL ∂Q DH\n\nSimilarly, ∂ξ ∂DL = ∂K ∂K\n\n\u0012 \u0013 \u0012 \u0013 00 DL00 DH ∂DH 0 0 0 0 2DH + ((pL − pH )DH − DH ) 0 + 2DL − ((pL − pH )DL + DL ) 0 . DL ∂K DH\n\nFrom the implicit function theorem:\n\n39\n\n−A − ∂K (1−α)(qH −qL ) =− AqH ∂Q + (1−α)(qH −qL )\n\nB α(qH −qL ) BqL α(qH −qL )\n\n=\n\nαA + (1 − α)B αAqH + (1 − α)BqL\n\nwhere\n\n0 0 A = 2DH + ((pL − pH )DH − DH )\n\nDL00 DL0\n\nand B = 2DL0 − ((pL − pH )DL0 + DL )\n\n00 DH 0 DH\n\n> 0 is that both A and B be negative. Next, I show A sufficient condition for ∂K ∂Q that the regularity condition for FL and FH are equivalent to A < 0 and B < 0, respectively.\n\nA<0\n\n⇐⇒\n\n0 0 DL00 (DH −(pL −pH )DH ) < 2DL0 DH\n\n⇐⇒\n\nDL00 <\n\n0 2DL0 DH 0 0 DH + pH DH − pL D H\n\nThe first inequality comes from DL0 < 0. The second inequality’s direction follows 0 > 0 which from the positive slope of the profit curve that guarantees DH + pH DH 0 0 0 in turn implies that DH + pH DH − pL DH > 0. Simplifying the ratio by DH leads to\n\nDL00\n\n2DL0 < 0 DH /DH + pH − pL\n\n⇐⇒\n\nDL00\n\n2DL0 < DL /DL0 + pL − pL\n\n⇐⇒\n\nDL00\n\n2(DL0 )2 < DL\n\nwhere the first inequality comes from the first order condition of DH L pH + D = pL + D The last inequality corresponds exactly to the regular0 0 . DL H ity of FL . Similar arguments prove that B < 0 is equivalent to the regularity of FH , which concludes the proof of h0 (Q) > 0. Analogous arguments can be made with obvious modifications that show that g is also increasing under the regularity assumption. \u0004 (ii) By definition K is slack for K > h(Q) and Q is slack for K < g(Q). By b < Q. Then by continuity there contradiction, assume that h < g at some point Q b where h < g. For any Q in this interval and for any exists a neighborhood of Q K < h(Q) we have none of the constraints binding. However, this is impossible, as the unconstrained optimum can only be achieved for (Q, K) ≥ (Q, K). Moreover, g(Q) = h(Q) = K follows directly from the definition of Q and K. \u0004 (iii) By contradiction, assume that ∀ Q > 0 : g(Q) > Q/qH . This means that ∀ Q > 0 : ∃ \u000f(Q) > 0 such that g(Q) > Q/qH + \u000f(Q). By definition of g, the\n\n40\n\nconstraint on Q is slack for all capacity-pairs (Q, Q/qH + \u000f(Q). The constraint on Q being slack translates to qH DH (pH ) + (1 − α)qL DL (pL ) < Q and the constraint on K at point (Q, Q/qH + \u000f(Q) writes as αDH (pH ) + (1 − α)DL (pL ) ≤ K = Q/qH + \u000f(Q) Taking the limit of Q → 0, the first inequality implies DL and DH must also tend to 0. Thus the second constraint must also be slack as the left hand side tends to zero while the right hand side equals \u000f(Q) which is always strictly positive. However, this is a contradiction as both constraints cannot be slack unless they are larger then (Q, K). (iv) can be proven with arguments analogous to (iii). \u0004 Proof of Proposition 6 It is sufficient to show that the optimal solutions of (P-LH) satisfy the additional constraint of pH /qH ≤ pL /qL . In the K region, this condition writes as \u0013 \u0013 \u0012 \u0012 1 1 vH − vL vH − vL −K ≤ −K , vH − (1 − α) vL + α qH 2 qL 2 which can be rewritten as K<\n\nvL qH − vH qL E(q)(vH − vL ) . + qH − qL 2(qH − qL )\n\nElementary algebra shows that the expression on the right-hand side is greater than E(v)/2, so the condition is always satisfied in the K region. In the KQ region, we have\n\npH /qH ≤ pL /qL ⇔ K < f (Q) +\n\n1 α(1 − α)(vL qH − vH qL )(qH − qL ), 2E(q 2 )\n\nwhich is always satisfied as the KQ region is delimited by K ≤ f (Q) and the right-hand side is greater than f (Q). It is straightforward to see that the constraint is also satisfied in regions Q and U. \u0004 Proof of Proposition 7 As the consumer surplus is additive in the consumer surplus of the consumer groups, we have\n\n41\n\n∂CS ∂pH ∂ = ∂K ∂K ∂pH\n\nZ\n\nθH\n\npH\n\n∂pL ∂ α(w − pH )fH (w)dw + ∂K ∂pL\n\nZ\n\nθL\n\n(1 − α)(w − pL )fL (w)dw. pL\n\nUsing the Leibniz-rule it follows that qH −qL ∂CS = (−(1 − α)DL ) + (−αDH ) 0 0 ∂K (1 − α)(qH − qL )DL α(qH − qL )DH which implies that ∂CS <0 ∂K\n\n⇐⇒\n\nqL\n\nDL0 (pL ) D0 (pH ) < qH H . DL (pL ) DH (pH )\n\nAnalogous steps prove the second statement. \u0004\n\nReferences Acemoglu, D., K. Bimpikis, and A. Ozdaglar (2009): “Price and capacity competition,” Games and Economic Behavior, 66(1), 1 – 26. Adan, I., and J. Vissers (2002): “Patient mix optimisation in hospital admission planning: a case study,” International Journal of Operations & Production Management, 22(4), 445–461. Aguirre, I., S. Cowan, and J. Vickers (2010): “Monopoly price discrimination and demand curvature,” The American Economic Review, 100(4), 1601–1615. Banditori, C., P. Cappanera, and F. Visintin (2013): “A combined optimization-simulation approach to the master surgical scheduling problem,” IMA Journal of Management Mathematics. Bergemann, D., B. A. Brooks, and S. Morris (forthcoming): “The limits of price discrimination,” The American Economic Review. Bertsimas, D., and R. Shioda (2003): “Restaurant revenue management,” Operations Research, 51(3), 472–486. Besanko, D., and R. Braeutigam (2010): Microeconomics. John Wiley & Sons. Bulow, J., and J. Roberts (1989): “The Simple Economics of Optimal Auctions,” Journal of Political Economy, 97(5), 1060–1090.\n\n42\n\nCowan, S. (2007): “The welfare effects of third-degree price discrimination with nonlinear demand functions,” RAND Journal of Economics, 38(2), 419–428. (2012): “Third-Degree Price Discrimination and Consumer Surplus,” The Journal of Industrial Economics, 60(2), 333–345. Cripps, M., and N. Ireland (1988): “Equilibrium and capacities in a market of fixed size,” Manuscript, University of Warwick. Davidson, C., and R. Deneckere (1986): “Long-Run Competition in Capacity, Short-Run Competition in Price, and the Cournot Model,” The RAND Journal of Economics, 17(3), 404–415. de Frutos, M.-Á., and N. Fabra (2011): “Endogenous capacities and price competition: The role of demand uncertainty,” International Journal of Industrial Organization, 29(4), 399–411. Kasilingam, R. G. (1997): “Air cargo revenue management: Characteristics and complexities,” European Journal of Operational Research, 96(1), 36–44. Kimes, S. E., and G. M. Thompson (2004): “Restaurant revenue management at Chevys: Determining the best table mix,” Decision Sciences, 35(3), 371–392. Kreps, D. M., and J. A. Scheinkman (1983): “Quantity Precommitment and Bertrand Competition Yield Cournot Outcomes,” The Bell Journal of Economics, 14(2), 326–337. Reynolds, S. S., and B. J. Wilson (2000): “Bertrand-Edgeworth Competition, Demand Uncertainty, and Asymmetric Outcomes,” Journal of Economic Theory, 92(1), 122–141. Xiao, B., and W. Yang (2010): “A revenue management model for products with two capacity dimensions,” European Journal of Operational Research, 205(2), 412– 421.\n\n43\n\n## Monopoly pricing with dual capacity constraints\n\nSep 14, 2015 - Email address: [email protected] The latest version ...... 6One can see the case of very large Q as an alternative benchmark.\n\n#### Recommend Documents\n\nContractual Pricing with Incentive Constraints\nintegral part of a team's organization when individual behavior is subject to incen- tive compatibility. (Without incentive compatibility, there is no need for secrets.).\n\nUpstream capacity constraint and the preservation of monopoly power ...\nMar 12, 2010 - serve its market power and induce the monopoly outcome on the downstream .... For a firm selling music through the internet, the capacity .... beliefs and compare it with the PBE under wary beliefs in two particular cases.\n\nCompetitive Pricing: Capacity View - Clary Business Machines\nCompetitor A. Competitor B. Competitor C. Ports at 1080p30. 16. 7. 12. 7. Ports at 720p60. 16. 7. 0. 0. Ports at 720p30. 16. 15. 12. 15. MSRP. \\$64,999. \\$108,000. \\$159,000. \\$85,000. *Based on estimated MSRP and documented port configuration; cost in U\n\nCapacity Choice and List Pricing in a Duopoly\nwe allow the firms to choose their capacity levels, the equilibrium coincides again with .... discount at the discounting stage while firm 2 sets a high enough list.\n\nLearning with convex constraints\nUnfortunately, the curse of dimensionality, especially in presence of many tasks, makes many complex real-world problems still hard to face. A possi- ble direction to attach those ..... S.: Calculus of Variations. Dover publications, Inc (1963). 5. G\n\nSale or Lease? Durable-Goods Monopoly with Network ...\nNov 1, 2007 - Therefore at a later stage the business strategy would resemble .... 10For instance, internet marketplaces such as Half.com or ...... Bell Journal.\n\nErgodic Capacity and Outage Capacity\nJul 8, 2008 - Radio spectrum is a precious and limited resource for wireless communication ...... Cambridge, UK: Cambridge University Press, 2004.\n\nDual Booting With Virtual Box.pdf\nDownload. Connect more apps... Try one of the apps below to open or edit this item. Dual Booting With Virtual Box.pdf. Dual Booting With Virtual Box.pdf. Open.\n\nofficial monopoly rules pdf\nConnect more apps... Try one of the apps below to open or edit this item. official monopoly rules pdf. official monopoly rules pdf. Open. Extract. Open with. Sign In.\n\nFirm pricing with consumer search\nDec 21, 2016 - products, and cost differences across firms, but search costs are not one of them ...... Unfortunately, accounting for such learning complicates.\n\nFOUNDATION - CHRISTMAS PRICING CALENDAR 2018 with 3C.pdf ...\nDEPENDING ON WHAT DATE YOU ARE LOOKING FOR. CHECK OUT OUR CALENDAR BELOW FOR A FULL LIST OF THESE DATES. AND OUR PRICES! MON. 19 NOV. TUE. 20 NOV. WED. 21 NOV. THU. 22 NOV. FRI. 23 NOV. SAT. 24 NOV. SUN. 25 NOV. 26 NOV 27 NOV 28 NOV 29 NOV 30 NOV 01" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.89976954,"math_prob":0.9774454,"size":82242,"snap":"2022-27-2022-33","text_gpt3_token_len":21423,"char_repetition_ratio":0.18829495,"word_repetition_ratio":0.0962143,"special_character_ratio":0.25555068,"punctuation_ratio":0.08960595,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9912454,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-06-27T15:00:09Z\",\"WARC-Record-ID\":\"<urn:uuid:17b49448-f683-43cc-a099-551045648430>\",\"Content-Length\":\"138558\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:5d1557bc-4a03-472d-86ee-eb017387fc51>\",\"WARC-Concurrent-To\":\"<urn:uuid:4f620031-3f01-4d14-b994-29f1636b06ef>\",\"WARC-IP-Address\":\"104.21.59.76\",\"WARC-Target-URI\":\"https://pdfkul.com/monopoly-pricing-with-dual-capacity-constraints_59bfe1c31723dd9b4393e57b.html\",\"WARC-Payload-Digest\":\"sha1:GOEXLLBB7WFEROTANITLGN5GHUFHXUAL\",\"WARC-Block-Digest\":\"sha1:LHBEEGRMGVFUUDZQNIPJM3NAV2BOPCA6\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656103334753.21_warc_CC-MAIN-20220627134424-20220627164424-00360.warc.gz\"}"}
https://www.clutchprep.com/chemistry/practice-problems/73478/the-equilibrium-constant-kc-for-the-reactionh2-g-i2-g-2h1-g-is-54-3-at-430-c-at-	[ "# Problem: The equilibrium constant Kc for the reactionH2(g) + I2(g) ⇌ 2HI(g)is 54.3 at 430 C. At the start of the reaction, there are 0.714 mole of H_2, 0.984 mole of I2, and 0.886 mole of HI in a 2.40 L reaction chamber. Calculate the concentrations of the gases at equilibrium.\n\n###### FREE Expert Solution\n89% (102 ratings)", null, "###### Problem Details\n\nThe equilibrium constant Kc for the reaction\n\nH2(g) + I2(g) ⇌ 2HI(g)\n\nis 54.3 at 430 C. At the start of the reaction, there are 0.714 mole of H_2, 0.984 mole of I2, and 0.886 mole of HI in a 2.40 L reaction chamber. Calculate the concentrations of the gases at equilibrium.", null, "" ]	[ null, "https://cdn.clutchprep.com/assets/button-view-text-solution.png", null, "https://lightcat-files.s3.amazonaws.com/problem_images/1526903641306.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.90715396,"math_prob":0.98980004,"size":746,"snap":"2020-45-2020-50","text_gpt3_token_len":187,"char_repetition_ratio":0.12803234,"word_repetition_ratio":0.0,"special_character_ratio":0.25201073,"punctuation_ratio":0.11875,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99075687,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-12-03T23:38:06Z\",\"WARC-Record-ID\":\"<urn:uuid:0e7d3858-14d1-4078-8f6d-7c7abe491c8b>\",\"Content-Length\":\"166828\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a11f0eb9-c8a4-4a59-af33-d70558f33125>\",\"WARC-Concurrent-To\":\"<urn:uuid:3a7aad51-49c9-4600-b24c-7e278d730f9d>\",\"WARC-IP-Address\":\"3.90.94.177\",\"WARC-Target-URI\":\"https://www.clutchprep.com/chemistry/practice-problems/73478/the-equilibrium-constant-kc-for-the-reactionh2-g-i2-g-2h1-g-is-54-3-at-430-c-at-\",\"WARC-Payload-Digest\":\"sha1:AWBSCCQSB4FGSMFIFB5AUGFCLMJAJ6BB\",\"WARC-Block-Digest\":\"sha1:W3L3BJE4JVSLB3XBZBFPK5WERSVM5CV2\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141732835.81_warc_CC-MAIN-20201203220448-20201204010448-00167.warc.gz\"}"}
https://edurev.in/course/quiz/attempt/21426_Test-35-Year-JEE-Previous-Year-Questions-Units-Mea/f718af1c-6bc1-488d-84b8-bee9d52191c0	[ "Courses\n\n# Test: 35 Year JEE Previous Year Questions: Units & Measurements\n\n## 20 Questions MCQ Test Physics For JEE \| Test: 35 Year JEE Previous Year Questions: Units & Measurements\n\nDescription\nThis mock test of Test: 35 Year JEE Previous Year Questions: Units & Measurements for JEE helps you for every JEE entrance exam. This contains 20 Multiple Choice Questions for JEE Test: 35 Year JEE Previous Year Questions: Units & Measurements (mcq) to study with solutions a complete question bank. The solved questions answers in this Test: 35 Year JEE Previous Year Questions: Units & Measurements quiz give you a good mix of easy questions and tough questions. JEE students definitely take this Test: 35 Year JEE Previous Year Questions: Units & Measurements exercise for a better result in the exam. You can find other Test: 35 Year JEE Previous Year Questions: Units & Measurements extra questions, long questions & short questions for JEE on EduRev as well by searching above.\nQUESTION: 1\n\n### A dense collection of equal number of electrons and positive ions is called neutral plasma. Certain solids containing fixed positive ions surrounded by free electrons can be treated as neutral plasma. Let ‘N’ be the number density of free electrons, each of mass ‘m’. When the electrons are subjected to an electric field, they are displaced relatively away from the heavy positive ions. If the electric field becomes zero, the electrons begin to oscillate about the positive ions with a natural angular frequency ‘ωp’ which is called the plasma frequency. To sustain the oscillations, a time varying electric field needs to be applied that has an angular frequency ω, where a part of the energy is absorbed and a part of it is reflected. As ω approaches ωp all the free electrons are set to resonance together and all the energy is reflected. This is the explanation of high reflectivity of metals. Q. Taking the electronic charge as ‘e’ and the permittivity as ‘ε0’. Use dimensional analysis to determine the correct expression for ωp.\n\nSolution:", null, "We do not want Ampere [A] in the expression. This is only possible when ∈0 occurs as square. Therefore options a and b are incorrect.", null, "QUESTION: 2\n\n### A dense collection of equal number of electrons and positive ions is called neutral plasma. Certain solids containing fixed positive ions surrounded by free electrons can be treated as neutral plasma. Let ‘N’ be the number density of free electrons, each of mass ‘m’. When the electrons are subjected to an electric field, they are displaced relatively away from the heavy positive ions. If the electric field becomes zero, the electrons begin to oscillate about the positive ions with a natural angular frequency ‘ωp’ which is called the plasma frequency. To sustain the oscillations, a time varying electric field needs to be applied that has an angular frequency ω, where a part of the energy is absorbed and a part of it is reflected. As ω approaches ωp all the free electrons are set to resonance together and all the energy is reflected. This is the explanation of high reflectivity of metals. Q. Estimate the wavelength at which plasma reflection will occur for a metal having the density of electrons N ≈ 4 x 1027 m–3. Taking ε0 = 10–11 and mass m ≈ 10–30, ≈ where these quantities are in proper SI units.\n\nSolution:", null, "QUESTION: 3\n\n### Identify the pair whose dimensions are equal\n\nSolution:", null, "QUESTION: 4\n\nDimen sion of", null, "where symbols have their usual meaning, are\n\nSolution:\n\nWe know that the velocity of light in vacuum is given by", null, "QUESTION: 5\n\nThe physical quantities not having same dimensions are\n\nSolution:\n\nMomentum = mv = [MLT–1]\n\nPlanck’s constant,", null, "QUESTION: 6\n\nWhich one of the following represents the correct dimensions of the coefficient of viscosity?\n\nSolution:\n\nFrom stokes law", null, "", null, "QUESTION: 7\n\nOut of the following pair , which one does NOT have identical dimensions is\n\nSolution:\n\nMoment of Inertia, I = Mr2\n\n[I] = [ML2]", null, "QUESTION: 8\n\nThe dimension of magnetic field in M, L, T and C (coulomb) is given as\n\nSolution:\n\nWe know that F = q v B", null, "QUESTION: 9\n\nA body of mass m = 3.513 kg is moving along the x-axis with a speed of 5.00 ms–1. The magnitude of its momentum is recorded as\n\nSolution:\n\nMomentum, p = m × v = (3.513) × (5.00) = 17.565 kg m/s\n= 17.6 (Rounding off to get three significant figures)\n\nQUESTION: 10\n\nTwo full turns of the circular scale of a screw gauge cover a distance of 1mm on its main scale. The total number of divisions on the circular scale is 50. Further, it is found that the screw gauge has a zero error of – 0.03 mm. While measuring the diameter of a thin wire, a student notes the main scale reading of 3 mm and the number of circular scale divisions in line with the main scale as 35. The diameter of the wire is\n\nSolution:\n\nLeast count of screw gauge = 0.5/50 mm 0.01mm\n\nreading × L.C] – (zero error)\n= [3 + 35 × 0.01] – (–0.03) = 3.38 mm\n\nQUESTION: 11\n\nIn an experiment the angles are required to be measured using an instrument, 29 divisions of the main scale exactly coincide with the 30 divisions of the vernier scale. If the smallest division of the main scale is half- a degree (= 0.5°), then the least count of the instrument is :\n\nSolution:\n\n30 Divisions of vernier scale coincide with 29 divisions of main scales", null, "Least count = 1 MSD – 1VSD", null, "", null, "QUESTION: 12\n\nThe respective number of significant figures for the numbers 23.023, 0.0003 and 2.1 × 10–3 are\n\nSolution:\n\nNumber of significant figures in 23.023 = 5\nNumber of significant figures in 0.0003 = 1\nNumber of significant figures in 2.1 × 10–3 = 2\n\nQUESTION: 13\n\nA screw gauge gives the following reading when used to measure the diameter of a wire.\nMain scale reading : 0 mm\nCircular scale reading : 52 divisions\nGiven that 1mm on main scale corresponds to 100 divisions of the circular scale. The diameter of wire from the above data is:\n\nSolution:", null, "Diameter of wire", null, "= 0.52 mm = 0.052 cm\n\nQUESTION: 14\n\nResistance of a given wire is obtained by measuring the current flowing in it and the voltage difference applied across it. If the percentage errors in the measurement of the current and the voltage difference are 3% each, then error in the value of resistance of the wire is :\n\nSolution:", null, "", null, "QUESTION: 15\n\nA spectrometer gives the following reading when used to measure the angle of a prism.\nMain scale reading : 58.5 degree\nVernier scale reading : 09 divisions\nGiven that 1 division on main scale corresponds to 0.5 degree.\nTotal divisions on the Vernier scale is 30 and match with 29 divisions of the main scale. The angle of the prism from the above data :\n\nSolution:\n\nVernier scale reading = 09 division\nleast count of Vernier = 0.5°/30", null, "R = 58.65\n\nQUESTION: 16\n\nLet [ε0] denote the dimensional formula of the permittivity of vacuum. If M = mass, L = length, T = time and A = electric current, then\n\nSolution:\n\nUsing Coulomb law,", null, "Hence,", null, "Substituting the units in the equation:", null, "=", null, "QUESTION: 17\n\nA student measured the length of a rod and wrote it as 3.50 cm. Which instrument did he use to measure it?\n\nSolution:\n\nMeasured length of rod = 3.50 cm\nFor vernier scale with 1 Main Scale Division = 1 mm\n9 Main Scale Division = 10 Vernier Scale Division,\nLeast count = 1 MSD –1 VSD = 0.1 mm\n\nQUESTION: 18\n\nThe period of oscillation of a simple pendulum is", null, "Measured value of L is 20.0 cm known to 1 mm accuracy and time for 100 oscillations of the pendulum is found to be 90 s using a wrist watch of 1s resolution. The accuracy in the determination of g is :\n\nSolution:", null, "QUESTION: 19\n\nA student measures the time period of 100 oscillations of a simple pendulum four times. The data set is 90s, 91s, 92s and 95s. If the minimum division in the measuring clock is ls, then the reported mean time should be\n\nSolution:\n\nSum of all observation= 90+91+95+92=368\n\nAverage =368/4=92\n\nSum of modulus errors=2+1+3=6\n\nAverage error =6/4=1.5,rounded of 2.\n\nFinal answer =(92 ± 2) s\n\nQUESTION: 20\n\nA screw gauge with a pitch of 0.5 mm and a circular scale with 50 divisions is used to measure the thickness of a thin sheet of Aluminium. Before starting the measurement, it is found that wen the two jaws of the screw gauge are brought in contact, the 45th division coincides with the main scale line and the zero of the main scale is barely visible. What is the thickness of the sheet if the main scale reading is 0.5 mm and the 25th division coincides with the main scale line?\n\nSolution:", null, "Zero error = 5 × 0.01 = 0.05 mm (Negative)\nReading = (0.5 + 25 × 0.01) + 0.05 = 0.80 mm" ]	[ null, "https://cdn3.edurev.in/ApplicationImages/Temp/3066490_106f7856-f410-4d4c-9171-e912e868d900_lg.png", null, "https://cdn3.edurev.in/ApplicationImages/Temp/3066490_f40c2c8d-7568-4e24-82bb-90d309082fda_lg.png", null, "https://cdn3.edurev.in/ApplicationImages/Temp/3066490_5272aae3-d957-40b7-824e-d6be320ba481_lg.png", null, "https://cdn3.edurev.in/ApplicationImages/Temp/3066490_cad669f5-9d14-4ec4-9e5d-67a6eec11da7_lg.png", null, "https://cdn3.edurev.in/ApplicationImages/Temp/3066490_a291b90e-e78c-4c48-90c2-f5044fbcb92e_lg.png", null, "https://cdn3.edurev.in/ApplicationImages/Temp/3066490_64ab521d-2e71-4dce-b212-6e785826219b_lg.png", null, "https://cdn3.edurev.in/ApplicationImages/Temp/3066490_412e1b6b-b362-4d76-9c15-f17a872d9805_lg.png", null, "https://cdn3.edurev.in/ApplicationImages/Temp/3066490_4206a2e5-4692-4952-8a57-89df2784e586_lg.png", null, "https://cdn3.edurev.in/ApplicationImages/Temp/3066490_9dedd0bf-4746-4220-934b-b8d06fab5af5_lg.png", null, "https://cdn3.edurev.in/ApplicationImages/Temp/3066490_e78d51d1-84ff-4c61-9e27-90566b5e08ea_lg.png", null, "https://cdn3.edurev.in/ApplicationImages/Temp/3066490_ed4f095b-a14c-4578-9bba-9182eb3e97a2_lg.png", null, "https://cdn3.edurev.in/ApplicationImages/Temp/3066490_ff7bb688-9c8f-4bf7-8f37-4161e89b0c99_lg.png", null, "https://cdn3.edurev.in/ApplicationImages/Temp/3066490_f1e0df52-7718-4ca6-a7de-90b8e9ac2050_lg.png", null, "https://cdn3.edurev.in/ApplicationImages/Temp/3066490_626bb58d-bebb-42e4-b768-2b12e743e166_lg.png", null, "https://cdn3.edurev.in/ApplicationImages/Temp/3066490_23d4c231-7fa7-4af1-a7a9-e15434181f14_lg.png", null, "https://cdn3.edurev.in/ApplicationImages/Temp/3066490_378547da-eebc-4e99-933e-0290e3a72655_lg.png", null, "https://cdn3.edurev.in/ApplicationImages/Temp/3066490_4c918cd5-801f-42d7-a0f4-a3f4b4f3053e_lg.png", null, "https://cdn3.edurev.in/ApplicationImages/Temp/3066490_e9ec85c4-6999-46a0-bfff-6d379fa9a134_lg.png", null, "https://cdn3.edurev.in/ApplicationImages/Temp/3066490_e077240b-0212-4b23-af29-31e5c1d74531_lg.png", null, "https://cdn3.edurev.in/ApplicationImages/Temp/10129244_f0de2a86-f8e2-4cbc-9bb9-042cb1bf7961_lg.png", null, "https://cdn3.edurev.in/ApplicationImages/Temp/10129244_c5247a44-0764-45a6-be6a-aaf2b7940ee4_lg.png", null, "https://cdn3.edurev.in/ApplicationImages/Temp/10129244_b45a6889-194b-4ed4-920f-b026d37463bd_lg.png", null, "https://cdn3.edurev.in/ApplicationImages/Temp/10129244_2255f3aa-89e2-42f7-a98b-ceef9426e78b_lg.png", null, "https://cdn3.edurev.in/ApplicationImages/Temp/3066490_6a00c099-7940-49c8-b0ce-e74aa43fbd7b_lg.png", null, "https://cdn3.edurev.in/ApplicationImages/Temp/3066490_b9a80a7a-68dd-4b27-8473-bdf273ae9d54_lg.png", null, "https://cdn3.edurev.in/ApplicationImages/Temp/3066490_9143610f-3d32-4a21-89f5-f37634a172ff_lg.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8991244,"math_prob":0.98345137,"size":7288,"snap":"2021-31-2021-39","text_gpt3_token_len":1862,"char_repetition_ratio":0.14017023,"word_repetition_ratio":0.23802505,"special_character_ratio":0.26564217,"punctuation_ratio":0.09256757,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9973292,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50,51,52],"im_url_duplicate_count":[null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-28T06:47:36Z\",\"WARC-Record-ID\":\"<urn:uuid:2f27eefe-ae27-4445-87b7-abe19eb80ea9>\",\"Content-Length\":\"398254\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:46762b49-a740-44bc-b8c5-6842789d122f>\",\"WARC-Concurrent-To\":\"<urn:uuid:103ec828-cb4a-440e-b0d3-dcea74713b28>\",\"WARC-IP-Address\":\"34.87.165.150\",\"WARC-Target-URI\":\"https://edurev.in/course/quiz/attempt/21426_Test-35-Year-JEE-Previous-Year-Questions-Units-Mea/f718af1c-6bc1-488d-84b8-bee9d52191c0\",\"WARC-Payload-Digest\":\"sha1:SMNDACUCIFT77OUUEJKUXN2KBZM4QUNV\",\"WARC-Block-Digest\":\"sha1:5TGYQZADR4RRNXN3WYBT5RWFCD3DF4CF\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780060538.11_warc_CC-MAIN-20210928062408-20210928092408-00489.warc.gz\"}"}
http://www-f1.ijs.si/~pavsic/IntroductionHtml	[ "Matej Pavsic, The Landscape of Theoretical Physics: A Global View\n\nIntroduction\nThe unification of various branches of theoretical physics is a joint project of many researchers, and everyone contributes as much as he can. So far we have accumulated a great deal of knowledge and insight encoded in such marvelous theories as general relativity, quantum mechanics, quantum field theory, the standard model of electroweak interaction, and chromodynamics. In order to obtain a more unified view, various promising theories have emerged, such as those of strings and branes\", induced gravity, the embedding models of gravity, and the brane world\" models, to mention just a few. The very powerful Clifford algebra as a useful tool for geometry and physics is becoming more and more popular. Fascinating are the ever increasing successes in understanding the foundations of quantum mechanics and their experimental verification, together with actual and potential practical applications in cryptography, teleportation, and quantum computing.\n\nIn this book I intend to discuss the conceptual and technical foundations of those approaches which, in my opinion, are most relevant for unification of general relativity and quantum mechanics on the one hand, and fundamental interactions on the other hand. After many years of active research I have arrived at a certain level of insight into the possible interrelationship between those theories. Emphases will be on the exposition and understanding of concepts and basic techniques, at the expense of detailed and rigorous mathematical development. Theoretical physics is considered here as a beautiful landscape. A global view of the landscape will be taken. This will enable us to see forests and mountain ranges as a whole, at the cost of seeing trees and rocks.\n\nPhysicists interested in the foundations of physics, conceptual issues, and the unification program, as well as those working in a special field and desiring to broaden their knowledge and see their speciality from a wider perspective, are expected to profit the most from the present book. They are assumed to possess a solid knowledge at least of quantum mechanics, and special and general relativity.\n\nAs indicated in the subtitle, I will start from point particles. They move along geodesics which are the lines of minimal, or, more generally, extremal length, in spacetime. The corresponding action from which the equations of motion are derived is invariant with respect to reparametrizations of an arbitrary parameter denoting position on the worldline swept by the particle. There are several different, but equivalent, reparametrization invariant point particle actions. A common feature of such an approach is that actually there is no dynamics in spacetime, but only in space. A particle's worldline is frozen in spacetime, but from the 3-dimensional point of view we have a point particle moving in 3-space. This fact is at the roots of all the difficulties we face when trying to quantize the theory: either we have a covariant quantum theory but no evolution in spacetime, or we have evolution in 3-space at the expense of losing manifest covariance in spacetime. In the case of a point particle this problem is not considered to be fatal, since it is quite satisfactorily resolved in relativistic quantum field theory. But when we attempt to quantize extended objects such as branes of arbitrary dimension, or spacetime itself, the above problem emerges in its full power: after so many decades of intensive research we have still not yet arrived at a generally accepted consistent theory of quantum gravity.\n\nThere is an alternative to the usual relativistic point particle action proposed by Fock \\ci{1} and subsequently investigated by Stueckelberg \\ci{2}, Feynman \\ci{3}, Schwinger \\ci{4}, Davidon \\ci{4a}, Horwitz \\ci{5,5a} and many others \\ci{6}--\\ci{12b}. In such a theory a particle or event\" in spacetime obeys a law of motion analogous to that of a nonrelativistic particle in 3-space. The difference is in the dimensionality and signature of the space in which the particle moves. None of the coordinates $x^0, x^1, x^2, x^3$ which parametrize spacetime has the role of evolution parameter. The latter is separately postulated and is Lorentz invariant. Usually it is denoted as $\\tau$ and evolution goes along $\\tau$. There are no constraints in the theory, which can therefore be called {\\it the unconstrained theory}. First and second quantizations of the unconstrained theory are straightforward, very elegant, and manifestly Lorentz covariant. Since $\\tau$ can be made to be related to proper time such a theory is often called a Fock--Schwinger proper time formalism. The value and elegance of the latter formalism is widely recognized, and it is often used, especially when considering quantum fields in curved spaces \\ci{12}. There are two main interpretations of the formalism:\n\n(i) According to the first interpretation, it is considered merely as a useful calculational tool, without any physical significance. Evolution in $\\tau$ and the absence of any constraint is assumed to be fictitious and unphysical. In order to make contact with physics one has to get rid of $\\tau$ in all the expressions considered by integrating them over $\\tau$. By doing so one projects unphysical expressions onto the physical ones, and in particular one projects unphysical states onto physical states.\n\n(ii) According to the second interpretation, evolution in $\\tau$ is genuine and physical. There is, indeed, dynamics in spacetime. Mass is a constant of motion and not a fixed constant in the Lagrangian.\n\nPersonally, I am inclined to the interpretation (ii). In the history of physics it has often happened that a good new formalism also contained good new physics waiting to be discovered and identified in suitable experiments. It is one of the purposes of this book to show a series of arguments in favor of the interpretation (ii). The first has roots in geometric calculus based on Clifford algebra \\ci{13}\n\nClifford numbers can be used to represent vectors, multivectors, and, in general, polyvectors (which are Clifford aggregates). They form a very useful tool for geometry. The well known equations of physics can be cast into elegant compact forms by using the geometric calculus based on Clifford algebra.\n\nThese compact forms suggest the generalization that every physical quantity is a polyvector \\ci{14,14a}. For instance, the momentum polyvector in 4-dimensional spacetime has not only a vector part, but also a scalar, bivector, pseudovector and pseudoscalar part. Similarly for the velocity polyvector. Now we can straightforwardly generalize the conventional constrained action by rewriting it in terms of polyvectors. By doing so we obtain in the action also a term which corresponds to the pseudoscalar part of the velocity polyvector. A consequence of this extra term is that, when confining ourselves, for simplicity, to polyvectors with pseudoscalar and vector part only, the variables corresponding to 4-vector components can all be taken as independent. After a straightforward procedure in which we omit the extra term in the action (since it turns out to be just the total derivative), we obtain Stueckelberg's unconstrained action! This is certainly a remarkable result. The original, constrained action is equivalent to the unconstrained action. Later in the book (Sec. 4.2) I show that the analogous procedure can also be applied to extended objects such as strings, membranes, or branes in general.\n\nWhen studying the problem of how to identify points in a generic curved spacetime, several authors \\ci{15}, and, especially recently Rovelli \\ci{16}, have recognized that one must fill spacetime with a reference fluid. Rovelli considers such a fluid as being composed of a bunch of particles, each particle carrying a clock on it. Besides the variables denoting positions of particles there is also a variable denoting the clock. This extra, clock, variable must enter the action, and the expression Rovelli obtains is formally the same as the expression we obtain from the polyvector action (in which we neglect the bivector, pseudovector, and scalar parts). We may therefore identify the pseudoscalar part of the velocity polyvector with the speed of the clock variable. Thus have a relation between the polyvector generalization of the usual constrained relativistic point particle, the Stueckleberg particle, and the DeWitt-Rovelli particle with clock.\n\nA relativistic particle is known to posses spin, in general. We show how spin arises from the polyvector generalization of the point particle and how the quantized theory contains the Dirac spinors together with the Dirac equation as a particular case. Namely, in the quantized theory a state is naturally assumed to be represented as a polyvector wave function $\\Phi$, which, in particular, can be a spinor. That spinors are just a special kind of polyvectors (Clifford aggregates), namely the elements of the minimal left or right ideals of the Clifford algebra, is an old observation \\ci{17}. Now, scalars, vectors, spinors, etc., can be reshuffled by the elements of the Clifford algebra. This means that scalars, vectors, etc., can be transformed into spinors, and {\\it vice versa}. Within Clifford algebra thus we have transformations which change bosons into fermions. In Secs. 2.5 and 2.7 I discuss the possible relation between the Clifford algebra formulation of the spinning particle and a more widely used formulation in terms of Grassmann variables.\n\nA very interesting feature of Clifford algebra concerns the signature of the space defined by basis vectors which are generators of the Clifford algebra. In principle we are not confined to choosing just a particular set of elements as basis vectors; we may choose some other set. For instance, if $e^0$, $e^1$, $e^2$, $e^3$ are the basis vectors of a space $M_e$ with signature (+ + + +), then we may declare the set ($e^0$, $e^0 e^1$, $e^0 e^2$, $e^0 e^3$) as basis vectors $\\gamma^0$, $\\gamma^1$, $\\gamma^2$, $\\gamma^3$ of some other space $M_{\\gamma}$ with signature $(+ - - -)$. That is, by suitable choice of basis vectors we can obtain within the same Clifford algebra a space of arbitrary signature. This has far reaching implications. For instance, in the case of even-dimensional space we can always take a signature with an equal number of pluses and minuses. A harmonic oscillator in such a space has vanishing zero point energy, provided that we define the vacuum state in a very natural way as proposed in refs. \\ci{18}. An immediate consequence is that there are no central terms and anomalies in string theory living in spacetime with signature $(+ + + ... - - -)$, even if the dimension of such a space is not critical. In other words, spacetime with such a symmetric' signature need not have 26 dimensions \\ci{18}.\n\nThe principle of such a harmonic oscillator in a pseudo-Euclidean space is applied in Chapter 3 to a system of scalar fields. The metric in the space of fields is assumed to have signature $(+ + + ... - - -)$ and it is shown that the vacuum energy, and consequently the cosmological constant, are then exactly zero. However, the theory contains some negative energy fields (exotic matter\") which couple to the gravitational field in a different way than the usual, positive energy, fields: the sign of coupling is reversed, which implies a repulsive gravitational field around such a source. This is the price to be paid if one wants to obtain a small cosmological constant in a straightforward way. One can consider this as a prediction of the theory to be tested by suitably designed experiments.\n\nThe problem of the cosmological constant is one of the toughest problems in theoretical physics. Its resolution would open the door to further understanding of the relation between quantum theory and general relativity. Since all more conventional approaches seem to have been more or less exploited without unambiguous success, the time is right for a more drastic novel approach. Such is the one which relies on the properties of the harmonic oscillator in a pseudo-Euclidean space.\n\nIn Part II \\hs{1mm} I discuss the theory of extended objects, now known as branes\" which are membranes of any dimension and are generalizations of point particles and strings. As in the case of point particles I pay much attention to the unconstrained theory of membranes. The latter theory is a generalization of the Stueckelberg point particle theory. It turns out to be very convenient to introduce the concept of the infinite-dimensional {\\it membrane space} ${\\cal M}$. Every point in ${\\cal M}$ represents an unconstrained membrane. In ${\\cal M}$ we can define distance, metric, covariant derivative, etc., in an analogous way as in a finite-dimensional curved space. A membrane action, the corresponding equations of motion, and other relevant expressions acquire very simple forms, quite similar to those in the point particle theory. We may say that a membrane is a point particle\nin an infinite dimensional space!\n\nAgain we may proceed in two different interpretations of the theory:\n\n(i) We may consider the formalism of the membrane space as a useful calculational tool (a generalization of the Fock--Schwinger proper time formalism) without any genuine physical significance. Physical quantities are obtained after performing a suitable projection.\n\n(ii) The points in ${\\cal M}$-space are physically distinguishable, that is, a membrane can be physically deformed in various ways and such a deformation may change with evolution in $\\tau$.\n\nIf we take the interpretation (ii) then we have a marvelous connection (discussed in Sec. 2.8) with the Clifford algebra generalization of the conventional constrained membrane on the one hand, and the concept of DeWitt--Rovelli reference fluid with clocks on the other hand.\n\nClifford algebra in the infinite-dimensional membrane space ${\\cal M}$ is described in Sec. 6.1. When quantizing the theory of the unconstrained membrane one may represent states by wave functionals which are polyvectors in ${\\cal M}$-space. A remarkable connection with quantum field theory is shown in Sec. 7.2\n\nWhen studying the ${\\cal M}$-space formulation of the membrane theory we find that in such an approach one cannot postulate the existence of a background embedding space independent from a membrane configuration. By `membrane configuration\" I understand a system of (many) membranes, and the membrane configuration is identified with the embedding space. There is no embedding space without the membranes. This suggests that our spacetime is nothing but a membrane configuration. In particular, our spacetime could be just one 4-dimensional membrane (4-brane) amongst many other membranes within the configuration. Such a model is discussed in Part III. The 4-dimensional gravity is due to the induced metric on our 4-brane $V_4$, whilst matter comes from the self-intersections of $V_4$, or the intersections of $V_4$ with other branes. As the intersections there can occur manifolds of various dimensionalities. In particular, the intersection or the self-intersection can be a 1-dimensional worldline. It is shown that such a worldline is a geodesic on $V_4$. So we obtain in a natural way four-dimensional gravity with sources. The quantized version of such a model is also discussed, and it is argued that the kinetic term for the 4-dimensional metric $g_{\\mu \\nu}$ is induced by quantum fluctuations of the 4-brane embedding functions.\n\nIn the last part I discuss mainly the problems related to the foundations and interpretation of quantum mechanics. I show how the brane world view sheds new light on our understanding of quantum mechanics and the role of the observer." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.91873723,"math_prob":0.9799904,"size":15751,"snap":"2021-31-2021-39","text_gpt3_token_len":3326,"char_repetition_ratio":0.13196164,"word_repetition_ratio":0.0072492952,"special_character_ratio":0.2015745,"punctuation_ratio":0.100530975,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.982765,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-21T08:17:37Z\",\"WARC-Record-ID\":\"<urn:uuid:a7f5d417-a82b-4028-ad84-1a55accc27b2>\",\"Content-Length\":\"16767\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ce873539-96fa-4e63-a992-1793cd4d2f53>\",\"WARC-Concurrent-To\":\"<urn:uuid:9b884140-095c-4519-82e5-5bc5030b72d2>\",\"WARC-IP-Address\":\"212.235.211.68\",\"WARC-Target-URI\":\"http://www-f1.ijs.si/~pavsic/IntroductionHtml\",\"WARC-Payload-Digest\":\"sha1:OYGGHRSKY53AQI7E4QCVAIDZDD63JSKD\",\"WARC-Block-Digest\":\"sha1:CRZZQAHSEEHAWS6ZR4CVE6SXOARM2WPQ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780057199.49_warc_CC-MAIN-20210921070944-20210921100944-00063.warc.gz\"}"}
https://book.huihoo.com/data-mining-desktop-survival-guide/Random_Forests0.html	[ "", null, "DATA MINING\nDesktop Survival Guide\nby Graham Williams", null, "# Random Forests: Classification\n\nThe approach taken by http://en.wikipedia.org/wiki/Random_forestrandom forests is to build multiple decision trees (often many hundreds) from different subsets of entities from the dataset and from different subsets of the variables from the dataset, to obtain substantial performance gains over single tree classifiers. Each decision tree is built from a sample of the full dataset, and a random sample of the available variables is used for each node of each tree. Having built an ensemble of models the final decision is the majority vote of the models, or the average value for a regression problem. The generalisation error rate from random forests tends to compare favourably to boosting approaches, yet the approach is more robust to noise in the data.\n\nNote and illustrate with the audit data how building a random forest and evaluate on training dataset gives \"prefect\" results (that might make you wonder about it overfitting) but on test data you get a realistic (and generally good still) performance.\n\nIn RF if there are many noise variables, increase the number of variables considered at each node.\n\nRandom Forrests are implemented:\n\n ```> library(randomForest) > ```\n\nThe classwt option in the current randomForrest package does not fully work and should be avoided. The sampsize and strata options can be used together. Note that if strata is not specified, the class labels will be used.\n\nHere's an example using the iris data:\n\n ```> iris.rf <- randomForest(Species ~ ., iris, sampsize=c(10, 20, 10)) ```\n\nThis will randomly sample 10, 20 and 10 entities from the three classes of species (with replacement) to grow each tree.\n\nYou can also name the classes in the sampsize specification:\n\n ```> samples <- c(setosa=10, versicolor=20, virginica=10) > iris.rf <- randomForest(Species ~ ., iris, sampsize=samples) ```\n\nYou can do a stratified sampling using a different variable than the class labels so that you even up the distribution of the class. Andy Liaw gives an example of the multi-centered clinical trial data where you want to draw the same number of patients per center to grow each tree where you can do something like:\n\n ```> randomForest(..., strata=center, sampsize=rep(min(table(center))), nlevels(center))) ```\n\nThis samples the same number of patients (minimum at any center) from each center to grow each tree.\n\nThe importance option allows us to review the importance of each variable in determining the outcome. The first importance is the scaled average of the prediction accuracy of each variable, and the second is the total decrease in node impurities splitting on the variable over all trees, using the Gini index.\n\nSubsections" ]	[ null, "https://book.huihoo.com/data-mining-desktop-survival-guide/togaware.png", null, "https://www.google.com/logos/Logo_40wht.gif", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.771444,"math_prob":0.93629426,"size":1695,"snap":"2020-10-2020-16","text_gpt3_token_len":374,"char_repetition_ratio":0.11413365,"word_repetition_ratio":0.02255639,"special_character_ratio":0.21828909,"punctuation_ratio":0.11147541,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9842153,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-04-07T14:10:40Z\",\"WARC-Record-ID\":\"<urn:uuid:5002e1a0-1a9e-4d85-ab64-b16ad397cfd4>\",\"Content-Length\":\"14506\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7e0661ba-73d9-446e-b37d-29f2394f8f6d>\",\"WARC-Concurrent-To\":\"<urn:uuid:fa3ea5ad-0d91-4123-a715-8702d68c1164>\",\"WARC-IP-Address\":\"104.18.57.214\",\"WARC-Target-URI\":\"https://book.huihoo.com/data-mining-desktop-survival-guide/Random_Forests0.html\",\"WARC-Payload-Digest\":\"sha1:VCRM7VYA6DX6EOQGA4UG65WJS22XOHCF\",\"WARC-Block-Digest\":\"sha1:PRN67I4FSSEVU32Q7V7KSDLQWCRCIFOY\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-16/CC-MAIN-2020-16_segments_1585371799447.70_warc_CC-MAIN-20200407121105-20200407151605-00543.warc.gz\"}"}
https://tutorialworld.in/how-to-use-equals-method-in-java-with-example/	[ "In Java, there is a equals() method which is used to compare strings. This comparison of strings are based on the content of the string.\n\n## How the equals() method works in Java?\n\nequals() methods compare two strings and if all characters of a string is not matched then it will return false. And If all the characters of the string matched it will return true.\n\nequals() method of String class overrides the equals() method of Object class.\n\n## Below is the internal implementation of the equals() method.\n\npublic boolean equals(Object anObject) {\nif (this == anObject) {\nreturn true;\n}\nif (anObject instanceof String) {\nString anotherString = (String) anObject;\nint n = value.length;\nif (n == anotherString.value.length) {\nchar v1[] = value;\nchar v2[] = anotherString.value;\nint i = 0;\nwhile (n– != 0) {\nif (v1[i] != v2[i])\nreturn false;\ni++;\n}\nreturn true;\n}\n}\nreturn false;\n}\n\n## Signature and Parameter of equals() method.\n\npublic boolean equals(Object anotherObject)\n\n## Return Type\n\nboolean value true or false. if string is equal then return true otherwise return false.\n\n## Another Program to demonstrate equals method in Java\n\npublic class EqualsExample{\npublic static void main(String args[]){\nString s1=”tutorialworld”;\nString s2=”tutorialworld”;\nString s3=”java”;\nSystem.out.println(s1.equals(s2));\nSystem.out.println(s1.equals(s3));\n}}\n\nOutput:\n\ntrue\nfalse\n\nAs you can see in the above example string s1 and s2 are the same as both have the same value “tutorialworld” but string s3 have different content “java”.\n\nSo if we compare s1 and s2 then it returned true and if we have compared s1 and s3 then it has returned false as it is not equal.\n\n## See Below one more example of List element comparision.\n\nimport java.util.ArrayList;\npublic class Main {\npublic static void main(String[] args) {\nString str1 = “Prakash”;\nArrayList name = new ArrayList<>();" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.6101132,"math_prob":0.97040886,"size":2099,"snap":"2020-45-2020-50","text_gpt3_token_len":510,"char_repetition_ratio":0.15083532,"word_repetition_ratio":0.0060975607,"special_character_ratio":0.25964746,"punctuation_ratio":0.1489899,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99055356,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-12-05T21:46:02Z\",\"WARC-Record-ID\":\"<urn:uuid:6f727038-ffa1-497a-90f4-0dc61c3f5e5b>\",\"Content-Length\":\"83189\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:cb6a67aa-5e7b-4ed7-b1b3-b2cc0f4012e0>\",\"WARC-Concurrent-To\":\"<urn:uuid:712d5c8b-bb88-43c0-a7e1-742186974329>\",\"WARC-IP-Address\":\"68.65.122.146\",\"WARC-Target-URI\":\"https://tutorialworld.in/how-to-use-equals-method-in-java-with-example/\",\"WARC-Payload-Digest\":\"sha1:SZXXICDHWFN6MCXGD2TDTRXTUJBHCLYX\",\"WARC-Block-Digest\":\"sha1:LDYM52DEBICEW3JJ7RZ4WJTBGB5BI5T5\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141750841.83_warc_CC-MAIN-20201205211729-20201206001729-00656.warc.gz\"}"}
https://convertoctopus.com/2268-feet-per-second-to-kilometers-per-hour	[ "## Conversion formula\n\nThe conversion factor from feet per second to kilometers per hour is 1.0972799999991, which means that 1 foot per second is equal to 1.0972799999991 kilometers per hour:\n\n1 ft/s = 1.0972799999991 km/h\n\nTo convert 2268 feet per second into kilometers per hour we have to multiply 2268 by the conversion factor in order to get the velocity amount from feet per second to kilometers per hour. We can also form a simple proportion to calculate the result:\n\n1 ft/s → 1.0972799999991 km/h\n\n2268 ft/s → V(km/h)\n\nSolve the above proportion to obtain the velocity V in kilometers per hour:\n\nV(km/h) = 2268 ft/s × 1.0972799999991 km/h\n\nV(km/h) = 2488.631039998 km/h\n\nThe final result is:\n\n2268 ft/s → 2488.631039998 km/h\n\nWe conclude that 2268 feet per second is equivalent to 2488.631039998 kilometers per hour:\n\n2268 feet per second = 2488.631039998 kilometers per hour\n\n## Alternative conversion\n\nWe can also convert by utilizing the inverse value of the conversion factor. In this case 1 kilometer per hour is equal to 0.00040182734359883 × 2268 feet per second.\n\nAnother way is saying that 2268 feet per second is equal to 1 ÷ 0.00040182734359883 kilometers per hour.\n\n## Approximate result\n\nFor practical purposes we can round our final result to an approximate numerical value. We can say that two thousand two hundred sixty-eight feet per second is approximately two thousand four hundred eighty-eight point six three one kilometers per hour:\n\n2268 ft/s ≅ 2488.631 km/h\n\nAn alternative is also that one kilometer per hour is approximately zero times two thousand two hundred sixty-eight feet per second.\n\n## Conversion table\n\n### feet per second to kilometers per hour chart\n\nFor quick reference purposes, below is the conversion table you can use to convert from feet per second to kilometers per hour\n\nfeet per second (ft/s) kilometers per hour (km/h)\n2269 feet per second 2489.728 kilometers per hour\n2270 feet per second 2490.826 kilometers per hour\n2271 feet per second 2491.923 kilometers per hour\n2272 feet per second 2493.02 kilometers per hour\n2273 feet per second 2494.117 kilometers per hour\n2274 feet per second 2495.215 kilometers per hour\n2275 feet per second 2496.312 kilometers per hour\n2276 feet per second 2497.409 kilometers per hour\n2277 feet per second 2498.507 kilometers per hour\n2278 feet per second 2499.604 kilometers per hour" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.70554954,"math_prob":0.985251,"size":2371,"snap":"2023-14-2023-23","text_gpt3_token_len":632,"char_repetition_ratio":0.29615548,"word_repetition_ratio":0.068601586,"special_character_ratio":0.3369886,"punctuation_ratio":0.0794702,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99504834,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-03-28T17:49:45Z\",\"WARC-Record-ID\":\"<urn:uuid:f5845263-a407-490c-aae1-fa64d0f44799>\",\"Content-Length\":\"28257\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:efda8962-46f2-4876-bc59-b0fe4fd647be>\",\"WARC-Concurrent-To\":\"<urn:uuid:240aa4f0-d7ca-45f4-a73c-e57860369a7b>\",\"WARC-IP-Address\":\"104.21.29.10\",\"WARC-Target-URI\":\"https://convertoctopus.com/2268-feet-per-second-to-kilometers-per-hour\",\"WARC-Payload-Digest\":\"sha1:DLN3QCGLC4TKZ7STQVMA2RQP5ZMZJABV\",\"WARC-Block-Digest\":\"sha1:7QKC3EZMCWMBJ6ZUBLL5S2SK3SPTO5UG\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-14/CC-MAIN-2023-14_segments_1679296948868.90_warc_CC-MAIN-20230328170730-20230328200730-00761.warc.gz\"}"}
https://dodgerocksgasmonkey.com/solve-this-math-question-33	[ "# Solve this math problem\n\nMath problem solver will save your results in History tab so you can go back to them anytime. Math problem solver covers levels of math including Arithmetic, Algebras\n\n• Figure out math question\n\nFor those who struggle with math, equations can seem like an impossible task. However, with a little bit of practice, anyone can learn to solve them.\n\n• More than just an app\n\nThe best way to spend your free time is with your family and friends.\n\n• Solve step-by-step\n\nWe are more than just an application, we are a community.\n\n## Microsoft Math Solver\n\nSolve an equation, inequality or a system. Example: 2x-1=y,2y+3=x What can QuickMath do? QuickMath will automatically answer the most common problems in algebra, equations and\n\n• Figure out mathematic\n• Better than just an application\n• Determine mathematic problems\n• Guaranteed Originality\n• Decide math questions\n\n## Online Math Problem Solver\n\nOnline math solver with free step by step solutions to algebra, calculus, and other math problems. Get help on the web or with our math app.\n\nFree time to spend with your family and friends\n\nMath can be tough, but with a little practice, anyone can master it!\n\nDeal with mathematic equations\n\nThis app is more than just a simple task manager.\n\nGet mathematics help online\n\nMath can be tough, but with a little practice, anyone can master it.\n\n## Algebra Calculator\n\nFree math problem solver answers your algebra homework questions with step-by-step explanations.", null, "" ]	[ null, "https://dodgerocksgasmonkey.com/images/ec8eecec12e1360a/hpjdilabqcmkfgone-phone.webp", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9302389,"math_prob":0.9296472,"size":1122,"snap":"2022-40-2023-06","text_gpt3_token_len":247,"char_repetition_ratio":0.11270125,"word_repetition_ratio":0.08421053,"special_character_ratio":0.20499109,"punctuation_ratio":0.118421055,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9935453,"pos_list":[0,1,2],"im_url_duplicate_count":[null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-02-01T03:52:16Z\",\"WARC-Record-ID\":\"<urn:uuid:cb9a832e-0a02-4dc1-b9af-d30bc6f57fb4>\",\"Content-Length\":\"37955\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:846dc60a-da43-45b0-baaf-e4a4ce124261>\",\"WARC-Concurrent-To\":\"<urn:uuid:9e7b0266-93b6-481d-ba62-19383b8e7303>\",\"WARC-IP-Address\":\"170.178.164.167\",\"WARC-Target-URI\":\"https://dodgerocksgasmonkey.com/solve-this-math-question-33\",\"WARC-Payload-Digest\":\"sha1:DJYSHBFIKUVXK5NFPDI3UBURIOZF5SUY\",\"WARC-Block-Digest\":\"sha1:T6VQVTTS2SELVJXHNPGGMDIHIN6J7TUO\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764499899.9_warc_CC-MAIN-20230201013650-20230201043650-00860.warc.gz\"}"}
https://theteachersinstitute.org/curriculum_unit/pascals-triangle-and-the-binomial-theorem/	[ "", null, "", null, "## Pascal’s Triangle and the Binomial Theorem\n\nAuthor: Shuang-Ching Su\n\nSchool/Organization:\n\nYear: 2006\n\nSchool Subject(s): Math\n\nWhich of the following equations are valid forms of the Binomial Theorem?", null, "If you choose D as the answer, congratulations! Please prove your answer then. As high school mathematics teachers, we certainly hope our students are able to handle a challenging question like this.\n\nThis article presents a curriculum unit pertaining to Pascal’s triangle and the Binomial Theorem. It is intended to serve as a reference framework for high school seniors who seek to conduct a senior project in mathematics.\n\nThe curriculum unit is based on lesson plans and class activities in my Discrete Mathematics classes in a period of approximately two weeks. In each lesson during this period, we proved one or two properties of Pascal’s triangle or the Binomial Theorem. At the end of the curriculum unit, students put together their worksheets on a construction paper as a project.", null, "", null, "", null, "", null, "", null, "(No Ratings Yet)", null, "Loading..." ]	[ null, "https://theteachersinstitute.org//wp-content/uploads/2017/08/penn-logo-wht-clean.png", null, "https://theteachersinstitute.org//wp-content/uploads/2017/08/penn-logo-gry-clean.png", null, "https://theteachersinstitute.org/wp-content/uploads/2018/11/Triangle-formula.png", null, "https://theteachersinstitute.org/wp-content/plugins/wp-postratings/images/stars/rating_off.gif", null, "https://theteachersinstitute.org/wp-content/plugins/wp-postratings/images/stars/rating_off.gif", null, "https://theteachersinstitute.org/wp-content/plugins/wp-postratings/images/stars/rating_off.gif", null, "https://theteachersinstitute.org/wp-content/plugins/wp-postratings/images/stars/rating_off.gif", null, "https://theteachersinstitute.org/wp-content/plugins/wp-postratings/images/stars/rating_off.gif", null, "https://theteachersinstitute.org/wp-content/plugins/wp-postratings/images/loading.gif", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9224277,"math_prob":0.67364556,"size":1060,"snap":"2021-31-2021-39","text_gpt3_token_len":213,"char_repetition_ratio":0.08901515,"word_repetition_ratio":0.0,"special_character_ratio":0.19056603,"punctuation_ratio":0.11282051,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9639392,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18],"im_url_duplicate_count":[null,null,null,null,null,4,null,null,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-23T14:13:43Z\",\"WARC-Record-ID\":\"<urn:uuid:8fa92d4f-15f3-42e9-bf6a-f9f0cf67fcef>\",\"Content-Length\":\"180877\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:50772e4d-9ab2-4baa-9edc-7fabb3667327>\",\"WARC-Concurrent-To\":\"<urn:uuid:fedafdc4-6099-4419-b31d-e13fd0493059>\",\"WARC-IP-Address\":\"104.196.39.146\",\"WARC-Target-URI\":\"https://theteachersinstitute.org/curriculum_unit/pascals-triangle-and-the-binomial-theorem/\",\"WARC-Payload-Digest\":\"sha1:GFDKOJB5U53F2A4GPJ3JVYGSY37Q2ASP\",\"WARC-Block-Digest\":\"sha1:7TX7UEUR6MBDOCZCRKYSGBKJRZQBDLGG\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780057424.99_warc_CC-MAIN-20210923135058-20210923165058-00045.warc.gz\"}"}
https://www.askiitians.com/rd-sharma-solutions/class-10/chapter-7-statistics/exercise-7-3/	[ "", null, "×", null, "", null, "", null, "", null, "", null, "#### Thank you for registering.\n\nOne of our academic counsellors will contact you within 1 working day.\n\nClick to Chat\n\n1800-1023-196\n\n+91-120-4616500\n\nCART 0\n\n• 0\n\nMY CART (5)\n\nUse Coupon: CART20 and get 20% off on all online Study Material\n\nITEM\nDETAILS\nMRP\nDISCOUNT\nFINAL PRICE\nTotal Price: Rs.\n\nThere are no items in this cart.\nContinue Shopping", null, "• Complete JEE Main/Advanced Course and Test Series\n• OFFERED PRICE: Rs. 15,900\n• View Details\n\n```Chapter 7: Statistics Exercise – 7.3\n\nQuestion: 1\n\nThe following table gives the distribution of total household expenditure (in rupees) of manual workers in a city.\n\nExpenditure (in rupees) (x)\nFrequency (fi)\nExpenditure (in rupees) (xi)\nFrequency (fi)\n\n100 – 150\n24\n300 – 350\n30\n\n150 – 200\n40\n350 – 400\n22\n\n200 – 250\n33\n400 – 450\n16\n\n250 – 300\n28\n450 – 500\n7\n\nFind the average expenditure (in rupees) per household\n\nSolution:\n\nLet the assumed mean (A) = 275\n\nClass interval\nMid value (xi)\ndi = xi – 275\nui = (xi - 275)/50\nFrequency fi\nfiui\n\n100 – 150\n125\n-150\n-3\n24\n-12\n\n150 – 200\n175\n-100\n-2\n40\n-80\n\n200 – 250\n225\n-50\n-1\n33\n-33\n\n250 – 300\n275\n0\n0\n28\n0\n\n300 – 350\n325\n50\n1\n30\n30\n\n350 – 400\n375\n100\n2\n22\n44\n\n400 – 450\n425\n150\n3\n16\n48\n\n450 – 500\n475\n200\n4\n7\n28\n\nN = 200\nSum = -35\n\nWe have A = 275, h = 50\nMean = A + h * sum/N\n= 275 + 50 * - 35/200\n= 275 – 8.75\n= 266.25\n\nQuestion: 2\n\nA survey was conducted by a group of students as a part of their environmental awareness program, in which they collected the following data regarding the number of plants in 200 houses in a locality. Find the mean number of plants per house.\n\nNumber of Number of plants:\n0 - 2\n2 – 4\n4 – 6\n6 – 8\n8 – 10\n10 – 12\n12 – 14\n\nNumber of houses:\n\n1\n2\n5\n6\n2\n3\n\nWhich method did you use for finding the mean, and why?\n\nSolution:\n\nLet us find class marks (xi) = (upper class limit + lower class limit)/2. Now we may compute xi and fixi as following.\n\nNumber of plants\nNumber of house (fi)\nxi\nFixi\n\n0 - 2\n1\n1\n1\n\n2 – 4\n2\n3\n6\n\n4 – 6\n1\n5\n5\n\n6 – 8\n5\n7\n35\n\n8 – 10\n6\n9\n54\n\n10 – 12\n2\n11\n22\n\n12 – 14\n3\n13\n39\n\nTotal\nN = 20\n\nSum = 162\n\nFrom the table we may observe that N = 20, Sum = 162", null, "So mean number of plants per house is 8.1. We have used for the direct method values Xi and fi are very small\n\nQuestion: 3\n\nConsider the following distribution of daily wages of workers of a factory\n\nDaily wages (in Rs)\n100 - 120\n120 - 140\n140 - 160\n160 - 180\n180 - 200\n\nNumber of workers:\n12\n16\n8\n6\n10\n\nFind the mean daily wages of the workers of the factory by using an appropriate method.\n\nSolution:\n\nLet the assume mean (A) = 150\n\nClass interval\nMid value xi\ndi = xi - 150\nui = (xi - 150)/20\nFrequency fi\nFiui\n\n100 – 120\n110\n-40\n-2\n12\n-24\n\n120 – 140\n130\n-20\n-1\n14\n-14\n\n140 – 160\n150\n0\n0\n8\n0\n\n160 – 180\n170\n20\n1\n6\n6\n\n180 – 200\n190\n40\n2\n10\n20\n\nN = 50\nSum = -12\n\nWe have N = 50, h = 20", null, "= 150 - 4.8\n= 145.2\n\nQuestion: 4\n\nThirty women were examined in a hospital by a doctor and the number of heart beats per minute recorded and summarized as follows. Find the mean heart beats per minute for these women, choosing a suitable method. Number of heart\n\nBeats Per minute:\n65 - 68\n68 - 71\n71 - 74\n74 - 77\n77 - 80\n80 - 83\n83 - 86\n\nNumber of women:\n2\n4\n3\n8\n7\n4\n2\n\nSolution:\n\nWe may find marks of each interval (xi) by using the relation (xi¬) = (upper class limit + lower class limit)/2 Class size of this data = 3 Now talking 75.5 as assumed mean (a) We may calculate di, ui, fiui as following\n\nNumber of heart beats per minute\nNumber of women (xi)\nxi\ndi = xi – 75.5\nui = (xi - 755)/h\nfiui\n\n65-68\n2\n66.5\n-9\n-3\n-6\n\n68-71\n9\n69.5\n-6\n-2\n-8\n\n71-74\n3\n72.5\n-3\n-1\n-3\n\n74-77\n8\n75.5\n0\n0\n0\n\n77-80\n7\n78.5\n3\n1\n7\n\n80-83\n4\n81.5\n6\n2\n8\n\n83-86\n2\n84.5\n9\n3\n6\n\nN = 30\n\nSum = 4\n\nNow we may observe from table that N = 30, sum = 4", null, "= 75.5 + 0.4\n= 75.9\nSo mean heart beats per minute for those women are 75.9 beats per minute\n\nQuestion: 5\n\nFind the mean of each of the following frequency distributions: (5 - 14)\n\nClass interval:\n0 - 6\n6 – 12\n12 – 18\n18 – 24\n24 – 30\n\nFrequency:\n6\n8\n10\n9\n7\n\nSolution:\n\nLet us assume mean be 15\n\nClass interval\nMid - value\ndi = xi – 15\nui = (xi – 15)/6\nfi\nfiui\n\n0 - 6\n3\n-12\n-2\n6\n-12\n\n12-Jun\n9\n-6\n-1\n8\n-8\n\n18-Dec\n15\n0\n0\n10\n0\n\n18 - 24\n21\n6\n1\n9\n9\n\n24 - 30\n27\n18\n2\n7\n14\n\nN = 40\nSum = 3\n\nA = 15, h = 6, N = 40", null, "= 15 + 0.45\n\n= 15.47\n\nQuestion: 6\n\nClass interval:\n50 - 70\n70 - 90\n90 - 110\n110 - 130\n130 - 150\n150 - 170\n\nFrequency:\n18\n12\n13\n27\n8\n22\n\nSolution:\n\nLet us assumed mean be 100\n\nClass interval\nMid-value xi\ndi = xi – 100\nui = (xi – 100)/20\nfi\nfiui\n\n50 - 70\n60\n-40\n-2\n18\n-36\n\n70 – 90\n80\n-20\n-1\n12\n-12\n\n90 – 110\n100\n0\n0\n13\n0\n\n110 – 130\n120\n20\n1\n27\n27\n\n130 – 150\n140\n40\n2\n8\n16\n\n150 – 170\n160\n60\n3\n22\n66\n\nN = 100\nSum = 61\n\nA = 100, h = 20", null, "= 100 + 12.2\n= 122.2\n\nQuestion: 7\n\nClass interval:\n0- 8\n8 – 16\n16 - 24\n24 - 32\n32 - 40\n\nFrequency:\n6\n7\n10\n8\n9\n\nSolution:\n\nLet the assumed mean (A) = 20\n\nClass interval\nMid- value\ndi= xi - 20\nui = (xi – 20)/8\nfi\nfiui\n\n0-8\n4\n-16\n-2\n6\n-12\n\n16-Aug\n12\n-8\n-1\n7\n-7\n\n16-24\n20\n0\n0\n10\n0\n\n24-32\n28\n8\n1\n8\n8\n\n32-40\n36\n16\n2\n9\n18\n\nN = 40\nSum = 7\n\nWe have A = 20, h = 8\nMean = A + h (sum/N)\n= 20 + 8 (7/40)\n= 20 + 1.4\n= 21.4\n\nQuestion: 8\n\nClass interval:\n0 - 6\n6 – 12\n12 – 18\n18 - 24\n24 - 30\n\nFrequency:\n7\n5\n10\n12\n6\n\nSolution:\n\nLet the assumed mean be (A) = 15\n\nClass interval\nMid – value\ndi = xi -15\nui = (xi -15)/6\nFrequency fi\nfiui\n\n0 – 6\n3\n-12\n-2\n-1\n-14\n\n6 – 12\n9\n-6\n-1\n5\n-5\n\n12 – 18\n15\n0\n0\n10\n0\n\n18 – 24\n21\n6\n1\n12\n12\n\n24 – 30\n27\n12\n2\n6\n12\n\nN = 40\nSum = 5\n\nWe have A = 15, h = 6\nMean = A + h(sum/N)\n= 15 + 6 (5/40)\n= 15 + 0.75\n= 15.75\n\nQuestion: 9\n\nClass interval:\n0 - 10\n20 – 30\n20 - 30\n30 - 40\n40 - 50\n\nFrequency:\n6\n7\n10\n8\n9\n\nSolution:\n\nLet the assumed mean (A) = 25\n\nClass interval\nMid – value\ndi = xi -25\nui = (xi - 25)/10\nFrequency fi\nfiui\n\n0 - 10\n5\n-20\n-2\n9\n-18\n\n20-Oct\n15\n-10\n-1\n10\n-12\n\n20 - 30\n25\n0\n0\n15\n0\n\n30 - 40\n35\n10\n1\n10\n10\n\n40 - 50\n45\n20\n2\n14\n28\n\nN = 60\nSum = 8\n\nWe have A = 25, h = 10\nMean = A + h (sum/N)\n= 25 + 19 (8/60)\n= 25 + (4/3)\n= 26.333\n\nQuestion: 10\n\nClass interval:\n0- 8\n8 – 16\n16 - 24\n24 - 32\n32 - 40\n\nFrequency:\n5\n9\n10\n8\n8\n\nSolution:\n\nLet the assumed mean (A) = 20\n\nClass interval\nMid value xi\ndi= xi - 20\nui = (xi -20)/8\nFrequency fi\nfiui\n\n0 – 8\n4\n-16\n-2\n5\n-10\n\n8 – 16\n12\n-8\n-1\n9\n-9\n\n16 – 24\n20\n0\n0\n10\n0\n\n24 – 32\n28\n8\n1\n8\n8\n\n32 – 40\n36\n16\n2\n8\n16\n\nN = 40\nSum = 5\n\nWe have, A = 20, h = 8\nMean = A + h (sum/N)\n= 20 + 8 (5/ 40)\n= 20 + 1\n= 21\n\nQuestion: 11\n\nClass interval:\n0- 8\n8 – 16\n16 - 24\n24 - 32\n32 - 40\n\nFrequency:\n5\n6\n4\n3\n2\n\nSolution:\n\nLet the assumed mean (A) = 20\n\nClass interval\nMid value xi\ndi= xi - 20\nui = (xi -20)/8\nFrequency fi\nfiui\n\n0 – 8\n4\n-16\n-2\n-2\n-10\n\n8 – 16\n12\n-8\n-1\n-1\n-6\n\n16 – 24\n20\n0\n0\n0\n0\n\n24 – 32\n28\n8\n1\n1\n3\n\n32 – 40\n36\n16\n2\n2\n4\n\nN = 20\nSum = -9\n\nWe have, A = 20, h = 8\nMean = A + h (sum/N)\n= 20 + 8 (-9/ 20)\n= 20 – (72/20)\n= 20 – 3.6\n= 16.4\n\nQuestion: 12\n\nClass interval:\n20 – 30\n30 - 50\n50 - 70\n70 - 90\n90- 110\n110 - 130\n\nFrequency:\n5\n8\n12\n20\n3\n2\n\nSolution:\n\nLet the assumed mean (A) = 60\n\nClass interval\nMid value xi\ndi= xi - 60\nui = (xi -60)/20\nFrequency fi\nfiui\n\n10 – 30\n20\n-40\n-2\n5\n-10\n\n30 – 50\n40\n-20\n-1\n8\n-8\n\n50 – 70\n60\n0\n0\n12\n0\n\n70 – 90\n80\n20\n1\n20\n20\n\n90 – 110\n100\n40\n2\n3\n6\n\n110 – 130\n120\n60\n3\n2\n6\n\nN = 50\nSum = 14\n\nWe have A = 60, h = 20\nMean = A + h (sum/N)\n= 60 + 20 (14/ 5)\n= 60 + 5.6\n= 65.6\n\nQuestion: 13\n\nClass interval:\n25 - 35\n35 – 45\n45 - 55\n55 - 65\n65 - 75\n\nFrequency:\n6\n10\n8\n10\n4\n\nSolution:\n\nLet the assumed mean (A) = 50\n\nClass interval\nMid value xi\ndi= xi - 50\nui = (xi - 50)/ 10\nFrequency fi\nfiui\n\n25 – 35\n30\n-20\n-2\n6\n-12\n\n35 – 45\n40\n-10\n-1\n10\n-10\n\n45 – 55\n50\n0\n0\n8\n0\n\n55 – 65\n60\n10\n1\n12\n12\n\n65 – 75\n70\n20\n2\n4\n8\n\nN = 40\nSum = -2\n\nWe have A = 50, h = 10\nMean = A + h (sum/N)\n= 50 + 10 (-2/ 40)\n= 50 - 0.5\n= 49.5\n\nQuestion: 14\n\nClass interval:\n25 - 29\n30 - 34\n35 -39\n40 - 44\n45 - 49\n50 - 54\n55 - 59\n\nFrequency:\n14\n22\n16\n6\n5\n3\n4\n\nSolution:\n\nLet the assumed mean (A) = 42\n\nClass interval\nMid value xi\ndi = xi - 42\nui = (xi - 42)/ 5\nFrequency fi\nfiui\n\n25 – 29\n27\n-15\n-3\n14\n-42\n\n30 – 34\n32\n-10\n-2\n22\n-44\n\n35 – 39\n37\n-5\n-1\n16\n-16\n\n40 – 44\n42\n0\n0\n6\n0\n\n45 – 49\n47\n5\n1\n5\n5\n\n50 – 54\n52\n10\n2\n3\n6\n\n55 – 59\n57\n15\n3\n4\n12\n\nN = 70\nSum = -79\n\nWe have A = 42, h = 5\nMean = A + h (sum/N)\n= 42 + 5 (-79/70)\n= 42 – 79/14\n= 36.357\n\nQuestion: 15\n\nFor the following distribution, calculate mean using all suitable methods:\n\nSize of item:\n1 – 4\n4 – 9\n9 – 16\n16 -20\n\nFrequency:\n6\n12\n26\n20\n\nSolution:\n\nBy direct method\n\nMean = (sum/N)\n= 848/ 64= 13.25\n\nBy assuming mean method Let the assumed mean (A) = 65\n\nClass interval\nMid value xi\nui = (xi - A) = xi - 65\nFrequency fi\nfiui\n\n1 – 4\n2.5\n-4\n6\n-25\n\n4 – 9\n6.5\n0\n12\n0\n\n9 – 16\n12.5\n6\n26\n196\n\n16 – 27\n21.5\n15\n20\n300\n\nN = 64\nSum = 432\n\nMean = A + sum/N\n= 6.5 + 6.75\n= 13.25\n\nQuestion: 16\n\nThe weekly observation on cost of living index in a certain city for the year 2004 – 2005 are given below. Compute the weekly cost of living index.\n\nCost of living index\nNumber of students\nCost of living index\nNumber of students\n\n1400 – 1500\n5\n1700 – 1800\n9\n\n1500 – 1600\n10\n1800 – 1900\n6\n\n1600 – 1700\n20\n1900 – 2000\n2\n\nSolution:\n\nLet the assumed mean (A) = 1650\n\nClass interval\nMid value xi\ndi = xi – A = xi– 1650\nui = (xi – 1650)100\nFrequency fi\nfiui\n\n1400 – 1500\n1450\n-200\n-2\n5\n-10\n\n1500 – 1600\n1550\n-100\n-1\n10\n-10\n\n1600 – 1700\n1650\n0\n0\n20\n0\n\n1700 – 1800\n1750\n100\n1\n9\n9\n\n1800 – 1900\n1850\n200\n2\n6\n12\n\n1900 – 2000\n1950\n300\n3\n2\n6\n\nN = 52\nSum = 7\n\nWe have A = 16, h = 100\nMean = A + h (sum/N)\n= 1650 + 100 (7/52)\n= 1650 + (175/13)\n= 21625/13\n= 1663.46\n\nQuestion: 17\n\nThe following table shows the marks scored by 140 students in an examination of a certain paper:\n\nMarks:\n0 - 10\n10 – 20\n20 - 30\n30 - 40\n40 - 50\n\nNumber of students:\n20\n24\n40\n36\n20\n\nSolution:\n\n(i) Direct method:\n\nClass interval\nMid value xi\nFrequency fi\nfixi\n\n0 – 10\n5\n20\n100\n\n10 – 20\n15\n24\n360\n\n20 – 30\n25\n40\n1000\n\n30 – 40\n35\n36\n1260\n\n40 – 50\n45\n20\n900\n\nN = 140\nSum = 3620\n\nMean = sum/ N\n= 3620/ 140\n= 25.857\n\n(ii) Assumed mean method: Let the assumed mean = 25 Mean = A + (sum/ N)\n\nClass interval\nMid value xi\nui = (xi - A)\nFrequency fi\nfiui\n\n0 – 10\n5\n-20\n20\n-400\n\n10 – 20\n15\n-10\n24\n-240\n\n20 – 30\n25\n0\n40\n0\n\n30 – 40\n35\n10\n36\n360\n\n40 – 50\n45\n20\n20\n400\n\nN = 140\nSum = 120\n\nMean = A + (sum/ N)\n= 25 + (120/ 140)\n= 25 + 0.857\n= 25.857\n\n(iii) Step deviation method: Let the assumed mean (A) = 25\n\nClass interval\nMid value xi\ndi= xi – A = xi– 25\nui = (xi – 25)10\nFrequency fi\nfiui\n\n0 – 10\n5\n-20\n-2\n20\n-40\n\n10 – 20\n15\n-10\n-1\n24\n-24\n\n20 – 30\n25\n0\n0\n40\n0\n\n30 – 40\n35\n10\n1\n36\n36\n\n40 – 50\n45\n20\n2\n20\n40\n\nN = 140\nSum = 12\n\nMean = A + h (sum/ N)\n= 25 + 10(12/140)\n= 25 + 0.857\n= 25.857\n\nQuestion: 18\n\nThe mean of the following frequency distribution is 62.8 and the sum of all the frequencies is 50. Compute the miss frequency f1 and f2.\n\nClass:\n0 - 20\n20 – 40\n40 - 60\n60 - 80\n80 - 100\n100 - 120\n\nFrequency:\n5\nf1\n10\nf2\n7\n8\n\nSolution:\n\nClass interval\nMid value xi\nFrequency fi\nfixi\n\n0 – 20\n10\n5\n50\n\n20 – 40\n30\nf1\n30f1\n\n40 – 60\n50\n10\n500\n\n60 – 80\n70\nf2\n70f2\n\n80 – 100\n90\n7\n630\n\n100 – 120\n110\n8\n880\n\nN = 30 +fi + f2\nSum = 30f1+70 f2+ 2060\n\nGiven, sum of frequency\n50 = f1 + f2 + 30\nf1 + f2 = 50 – 30\nf1 +f2 = 20\n3f1 + 3f2 = 60 ---- (1) [multiply both side by 3]\nAnd mean = 62.8 Sum/ N\n= 62.8 = (30f1+ 70f2+ 2060)/50\n= 3140 = 30f1 + 70f2 + 2060\n30f1+ 70f2 = 3140 – 2060\n30f1 + 70f2 = 1080\n3f1 + 7f2 = 108 ---- (2) [divide it by 10]\nSubtract equation (1) from equation (2)\n3f1 + 7f2 - 3f1 - 3f2 = 108 – 60\n4f2 = 48 = f2 = 12\n\nPut value of f2 in equation (1)\n3f1 + 3(12) = 60\nf1 = 24/3 = 8\nf1 = 8, f2 = 12\n\nQuestion: 19\n\nThe following distribution shows the daily pocket allowance given to the children of a multistory building. The average pocket allowance is Rs 18.00. Find out the missing frequency.\n\nClass interval:\n11 – 13\n13- 15\n15 - 17\n17 - 19\n19 - 21\n21 - 23\n23 - 25\n\nFrequency:\n7\n6\n9\n13\n-\n5\n4\n\nSolution:\n\nGiven mean = 18, Let the missing frequency be v\n\nClass interval\nMid value xi\nFrequency fi\nfixi\n\n11 – 13\n12\n7\n84\n\n13 – 15\n14\n6\n88\n\n15 – 17\n16\n9\n144\n\n17 – 19\n18\n13\n234\n\n19 – 21\n20\nx\n20x\n\n21 – 23\n22\n5\n110\n\n23 – 25\n14\n4\n56\n\nN = 44 + x\nSum = 752 + 20x", null, "792 + 18x = 752 + 20x\n20x – 18x = 792 – 752\nx = 40/2\nx = 20\n\nQuestion: 20\n\nIf the mean of the following distribution is 27. Find the value of p.\n\nClass:\n0 -10\n10 – 20\n20 - 30\n30 - 40\n40 - 50\n\nFrequency:\n8\np\n12\n13\n10\n\nSolution:\n\nClass interval\nMid value xi\nFrequency fi\nfixi\n\n0 – 10\n5\n8\n40\n\n10 – 20\n15\nP\n152\n\n20 – 30\n25\n12\n300\n\n30 – 40\n35\n13\n455\n\n40 – 50\n45\n16\n450\n\nN = 43 + P\nSum = 1245 + 15p\n\nGiven Mean = 27\nMean = sum/ N", null, "1161 + 27p = 1245 + 15p\n27p – 15p = 1245 – 1161\n12p = 84\np = 84/12\np = 7\n\nQuestion: 21\n\nIn a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contain varying number of mangoes. The following was the distribution of mangoes according to the number of boxes.\n\nNumber of mangoes:\n50 - 52\n53 - 55\n56 - 58\n59 - 61\n62 - 64\n\nNumber of boxes:\n15\n110\n135\n115\n25\n\nFind the mean number of mangoes kept in packing box. Which method of finding the mean did you choose?\n\nSolution:\n\nNumber of mangoes\nNumber of boxes (fi)\n\n50 – 52\n15\n\n53 – 55\n110\n\n56 – 58\n135\n\n59 – 61\n115\n\n62 – 64\n25\n\nWe may observe that class internals are not continuous There is a gap between two class intervals. So we have to add 1/2 from lower class limit of each interval and class mark (xi) may be obtained by using the relation xi = (upper limit + lower class limit)/2. Class size (h) of this data = 3 Now taking 57 as assumed mean (a) we may calculated di , ui, fiui as follows\n\nClass interval\nFrequency fi\nMid value xi\ndi = xi - A\nui = (xi – A)/3\nfiui\n\n49.5 – 52.5\n15\n51\n-6\n-2\n-30\n\n52.5 – 55.5\n110\n54\n-3\n-1\n-110\n\n55.5 – 58.5\n135\n57\n0\n0\n0\n\n58.5 – 61.5\n115\n60\n3\n1\n115\n\n61.5 – 64.5\n25\n63\n6\n2\n50\n\nTotal\nN = 400\n\nSum = 25\n\nNow we have N = 400, Sum = 25,\nMean = A + h (sum/ N)\n= 57 + 3 (25/400)\n= 57 + 3/16\n= 57+ 0.1875\n= 57.19\nClearly mean number of mangoes kept in packing box is 57.19\n\nQuestion: 22\n\nThe table below shows the daily expenditure on food of 25 households in a locality\n\nDaily expenditure (in Rs):\n100 - 150\n150 - 200\n200 - 250\n250 -300\n300 - 350\n\nNumber of households:\n4\n5\n12\n2\n2\n\nFind the mean daily expenditure on food by a suitable method.\n\nSolution:\n\nWe may calculate class mark (xi) for each interval by using the relation xi = (upper limit + lower class limit)/2. Class size = 50 Now, talking 225 as assumed mean (xi) we may calculate di ,ui, fiui as follows:\n\nDaily expenditure\nFrequency fi\nMid value xi\ndi = xi – 225\nui = (xi – 225)50\nfiui\n\n100 – 150\n4\n125\n-100\n-2\n-8\n\n150 – 200\n5\n175\n-50\n-1\n-5\n\n200 – 250\n12\n225\n0\n0\n0\n\n250 – 300\n2\n275\n50\n1\n2\n\n300 – 350\n2\n325\n100\n2\n4\n\nN = 25\n\nSum = – 7\n\nNow we may observe that N = 25 Sum = -7", null, "225 = 50 + (-7/ 25) × 225\n225 – 14\n= 211\nSo, mean daily expenditure on food is Rs 211\n\nQuestion: 23\n\nTo find out the concentration of SO2 in the air (in parts per million i. e ppm) the data was collected for localities for 30 localities in a certain city and is presented below:\n\nConcentration of SO2 (in ppm)\nFrequency\n\n0.00 – 0.04\n4\n\n0.04 – 0.08\n9\n\n0.08 – 0.12\n9\n\n0.12 – 0.16\n2\n\n0.16 – 0.20\n4\n\n0.20 – 0.24\n2\n\nFind the mean concentration of SO2 in the air\n\nSolution:\n\nWe may find class marks for each interval by using the relation xi = (upper limit + lower class limit)/2. Class size of this data = 0.04 Now taking 0.04 assumed mean (xi) we may calculate di, ui, fiui as follows:\n\nConcentration of SO2\nFrequency fi\nClass interval xi\ndi = xi – 0.14\nui\nfiui\n\n0.00 – 0.04\n4\n0.02\n-0.12\n-3\n-12\n\n0.04 – 0.08\n9\n0.06\n-0.08\n-2\n-18\n\n0.08 – 0.12\n9\n0.1\n-0.04\n-1\n-9\n\n0.12 – 0.16\n2\n0.14\n0\n0\n0\n\n0.16 – 0.20\n4\n0.18\n0.04\n1\n4\n\n0.20 – 0.24\n2\n0.22\n0.08\n2\n4\n\nTotal\nN = 30\n\nSum = -31\n\nFrom the table we may observe that N = 30, Sum = - 31", null, "= 0.04 + (-31/30) × (0.04)\n= 0.099 ppm\nSo mean concentration of SO2 ¬in the air is 0.099 ppm\n\nQuestion: 24\n\nA class teacher has the following absentee record of 40 students of a class for the whole term. Find the mean number of days a student was absent.\n\nNumber of days:\n0 - 6\n6 – 10\n10 – 14\n14 - 20\n20 - 28\n28 - 38\n38 - 40\n\nNumber of students:\n11\n10\n7\n4\n4\n3\n1\n\nSolution:\n\nWe may find class mark of each interval by using the relation xi = (upper limit + lower class limit)/2 Now, taking 16 as assumed mean (a) we may Calculate di and fi di as follows\n\nNumber of days\nNumber of students fi\nXi\nd = xi - 16\nfidi\n\n0 – 6\n11\n3\n-13\n-143\n\n6 – 10\n10\n8\n-8\n-80\n\n10 – 14\n7\n12\n-4\n-28\n\n14 – 20\n7\n16\n0\n0\n\n20 – 28\n8\n24\n8\n64\n\n28 – 36\n3\n33\n17\n51\n\n30 – 40\n1\n39\n23\n23\n\nTotal\nN = 40\n\nSum = -113\n\nNow we may observe that N = 40, Sum = -113", null, "16 + (-113/ 40)\n= 16 – 2.825\n= 13.75\nSo mean number of days is 13.75 days, for which student was absent\n\nQuestion: 25\n\nThe following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate.\n\nLiteracy rate (in %):\n45 -55\n55 - 65\n65 - 75\n75 - 85\n85 - 95\n\nNumber of cites:\n3\n10\n11\n8\n3\n\nSolution:\n\nWe may find class marks by using the relation xi = (upper limit + lower class limit)/2 Class size (h) for this data = 10 Now taking 70 as assumed mean (a) wrong Calculate di ,ui, fiui as follows\n\nLiteracy rate (in %)\nNumber of cities (fi)\nMid value xi\ndi= xi – 70\nui = (di –A)10\nfiui\n\n45 – 55\n3\n50\n-20\n-2\n- 6\n\n55 – 65\n10\n60\n-10\n-1\n-10\n\n65 – 75\n11\n70\n0\n0\n0\n\n75 – 85\n8\n80\n10\n1\n8\n\n85 – 95\n3\n90\n20\n2\n6\n\nTotal\nN = 35\n\nSum = -2\n\nNow we may observe that N = 35, Sum = -2", null, "= 70 + (-2/35) × 10\n= 70 – 4/7\n= 70 – 0.57\n= 69.43\nSo, mean literacy rate is 69.43 %\n```", null, "", null, "### Course Features\n\n• 728 Video Lectures\n• Revision Notes\n• Previous Year Papers\n• Mind Map\n• Study Planner\n• NCERT Solutions\n• Discussion Forum\n• Test paper with Video Solution", null, "" ]	[ null, "https://www.askiitians.com/resources/images/badge.png", null, "https://www.askiitians.com/Resources/images/neet-iit/user.png", null, "https://www.askiitians.com/Resources/images/neet-iit/envelope.png", null, "https://www.askiitians.com/Resources/images/neet-iit/phone-receiver.png", null, "https://www.askiitians.com/Resources/images/neet-iit/reading.png", null, "https://www.askiitians.com/Resources/images/neet-iit/reading.png", null, "https://files.askiitians.com/static/ecom/cms/right/engineering-full-course.jpg", null, "https://files.askiitians.com/cdn/images/2018820-162246975-8258-equation-1.png", null, "https://files.askiitians.com/cdn/images/2018820-162741252-1455-equation-2.png", null, "https://files.askiitians.com/cdn/images/2018820-163319531-2290-equation-3.png", null, "https://files.askiitians.com/cdn/images/2018820-163730417-4195-equation-4.png", null, "https://files.askiitians.com/cdn/images/2018820-164219698-7022-equation-5.png", null, "https://files.askiitians.com/cdn/images/2018820-174724690-2407-equation-6.png", null, "https://files.askiitians.com/cdn/images/2018820-17514263-8302-equation-7.png", null, "https://files.askiitians.com/cdn/images/2018820-18047610-4310-equation-8.png", null, "https://files.askiitians.com/cdn/images/2018820-18047610-4310-equation-8.png", null, "https://files.askiitians.com/cdn/images/2018820-18047610-4310-equation-8.png", null, "https://files.askiitians.com/cdn/images/2018820-181327116-461-equation-9.png", null, "https://www.askiitians.com/Resources/images/youtube_placeholder.jpg", null, "https://files.askiitians.com/static/ecom/cms/bottom/medical-full-course.png", null, "https://www.askiitians.com/Resources/images/wishCross-red.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.86647356,"math_prob":0.9966147,"size":8553,"snap":"2020-45-2020-50","text_gpt3_token_len":2859,"char_repetition_ratio":0.13440168,"word_repetition_ratio":0.15211423,"special_character_ratio":0.3947153,"punctuation_ratio":0.10027248,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9966232,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,4,null,4,null,4,null,4,null,4,null,4,null,4,null,null,null,null,null,null,null,4,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-11-26T07:20:55Z\",\"WARC-Record-ID\":\"<urn:uuid:5189d674-64f5-478c-ac31-cba07d362ac6>\",\"Content-Length\":\"471320\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9dafeffc-9e3f-4426-8508-cf1a2f650d17>\",\"WARC-Concurrent-To\":\"<urn:uuid:6ba7b135-31eb-4910-a234-f65f2d137d81>\",\"WARC-IP-Address\":\"3.7.36.8\",\"WARC-Target-URI\":\"https://www.askiitians.com/rd-sharma-solutions/class-10/chapter-7-statistics/exercise-7-3/\",\"WARC-Payload-Digest\":\"sha1:EPDEIW33EKW4LULFODS5X4JBZ4UDYGX6\",\"WARC-Block-Digest\":\"sha1:UG3FFZXUGR4OYY7BEREKNNKN24IC3AAW\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141186761.30_warc_CC-MAIN-20201126055652-20201126085652-00626.warc.gz\"}"}
https://www.colorhexa.com/7f5300	[ "# #7f5300 Color Information\n\nIn a RGB color space, hex #7f5300 is composed of 49.8% red, 32.5% green and 0% blue. Whereas in a CMYK color space, it is composed of 0% cyan, 34.6% magenta, 100% yellow and 50.2% black. It has a hue angle of 39.2 degrees, a saturation of 100% and a lightness of 24.9%. #7f5300 color hex could be obtained by blending #fea600 with #000000. Closest websafe color is: #666600.\n\n• R 50\n• G 33\n• B 0\nRGB color chart\n• C 0\n• M 35\n• Y 100\n• K 50\nCMYK color chart\n\n#7f5300 color description : Dark orange [Brown tone].\n\n# #7f5300 Color Conversion\n\nThe hexadecimal color #7f5300 has RGB values of R:127, G:83, B:0 and CMYK values of C:0, M:0.35, Y:1, K:0.5. Its decimal value is 8344320.\n\nHex triplet RGB Decimal 7f5300 `#7f5300` 127, 83, 0 `rgb(127,83,0)` 49.8, 32.5, 0 `rgb(49.8%,32.5%,0%)` 0, 35, 100, 50 39.2°, 100, 24.9 `hsl(39.2,100%,24.9%)` 39.2°, 100, 49.8 666600 `#666600`\nCIE-LAB 39.069, 12.387, 47.636 11.846, 10.699, 1.441 0.494, 0.446, 10.699 39.069, 49.22, 75.424 39.069, 35.746, 38.979 32.71, 7.402, 20.284 01111111, 01010011, 00000000\n\n# Color Schemes with #7f5300\n\n• #7f5300\n``#7f5300` `rgb(127,83,0)``\n• #002c7f\n``#002c7f` `rgb(0,44,127)``\nComplementary Color\n• #7f1400\n``#7f1400` `rgb(127,20,0)``\n• #7f5300\n``#7f5300` `rgb(127,83,0)``\n• #6c7f00\n``#6c7f00` `rgb(108,127,0)``\nAnalogous Color\n• #14007f\n``#14007f` `rgb(20,0,127)``\n• #7f5300\n``#7f5300` `rgb(127,83,0)``\n• #006c7f\n``#006c7f` `rgb(0,108,127)``\nSplit Complementary Color\n• #53007f\n``#53007f` `rgb(83,0,127)``\n• #7f5300\n``#7f5300` `rgb(127,83,0)``\n• #007f53\n``#007f53` `rgb(0,127,83)``\n• #7f002c\n``#7f002c` `rgb(127,0,44)``\n• #7f5300\n``#7f5300` `rgb(127,83,0)``\n• #007f53\n``#007f53` `rgb(0,127,83)``\n• #002c7f\n``#002c7f` `rgb(0,44,127)``\n• #332100\n``#332100` `rgb(51,33,0)``\n• #4c3200\n``#4c3200` `rgb(76,50,0)``\n• #664200\n``#664200` `rgb(102,66,0)``\n• #7f5300\n``#7f5300` `rgb(127,83,0)``\n• #996400\n``#996400` `rgb(153,100,0)``\n• #b27400\n``#b27400` `rgb(178,116,0)``\n• #cc8500\n``#cc8500` `rgb(204,133,0)``\nMonochromatic Color\n\n# Alternatives to #7f5300\n\nBelow, you can see some colors close to #7f5300. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #7f3300\n``#7f3300` `rgb(127,51,0)``\n• #7f3e00\n``#7f3e00` `rgb(127,62,0)``\n• #7f4800\n``#7f4800` `rgb(127,72,0)``\n• #7f5300\n``#7f5300` `rgb(127,83,0)``\n• #7f5e00\n``#7f5e00` `rgb(127,94,0)``\n• #7f6800\n``#7f6800` `rgb(127,104,0)``\n• #7f7300\n``#7f7300` `rgb(127,115,0)``\nSimilar Colors\n\n# #7f5300 Preview\n\nThis text has a font color of #7f5300.\n\n``<span style=\"color:#7f5300;\">Text here</span>``\n#7f5300 background color\n\nThis paragraph has a background color of #7f5300.\n\n``<p style=\"background-color:#7f5300;\">Content here</p>``\n#7f5300 border color\n\nThis element has a border color of #7f5300.\n\n``<div style=\"border:1px solid #7f5300;\">Content here</div>``\nCSS codes\n``.text {color:#7f5300;}``\n``.background {background-color:#7f5300;}``\n``.border {border:1px solid #7f5300;}``\n\n# Shades and Tints of #7f5300\n\nA shade is achieved by adding black to any pure hue, while a tint is created by mixing white to any pure color. In this example, #090600 is the darkest color, while #fffbf5 is the lightest one.\n\n• #090600\n``#090600` `rgb(9,6,0)``\n• #1d1300\n``#1d1300` `rgb(29,19,0)``\n• #312000\n``#312000` `rgb(49,32,0)``\n• #442d00\n``#442d00` `rgb(68,45,0)``\n• #583900\n``#583900` `rgb(88,57,0)``\n• #6b4600\n``#6b4600` `rgb(107,70,0)``\n• #7f5300\n``#7f5300` `rgb(127,83,0)``\n• #936000\n``#936000` `rgb(147,96,0)``\n• #a66d00\n``#a66d00` `rgb(166,109,0)``\n• #ba7900\n``#ba7900` `rgb(186,121,0)``\n• #cd8600\n``#cd8600` `rgb(205,134,0)``\n• #e19300\n``#e19300` `rgb(225,147,0)``\n• #f5a000\n``#f5a000` `rgb(245,160,0)``\n• #ffaa09\n``#ffaa09` `rgb(255,170,9)``\n• #ffb11d\n``#ffb11d` `rgb(255,177,29)``\n• #ffb731\n``#ffb731` `rgb(255,183,49)``\n• #ffbe44\n``#ffbe44` `rgb(255,190,68)``\n• #ffc558\n``#ffc558` `rgb(255,197,88)``\n• #ffcc6b\n``#ffcc6b` `rgb(255,204,107)``\n• #ffd37f\n``#ffd37f` `rgb(255,211,127)``\n• #ffd993\n``#ffd993` `rgb(255,217,147)``\n• #ffe0a6\n``#ffe0a6` `rgb(255,224,166)``\n• #ffe7ba\n``#ffe7ba` `rgb(255,231,186)``\n• #ffeecd\n``#ffeecd` `rgb(255,238,205)``\n• #fff5e1\n``#fff5e1` `rgb(255,245,225)``\n• #fffbf5\n``#fffbf5` `rgb(255,251,245)``\nTint Color Variation\n\n# Tones of #7f5300\n\nA tone is produced by adding gray to any pure hue. In this case, #44413b is the less saturated color, while #7f5300 is the most saturated one.\n\n• #44413b\n``#44413b` `rgb(68,65,59)``\n• #494336\n``#494336` `rgb(73,67,54)``\n• #4e4431\n``#4e4431` `rgb(78,68,49)``\n• #53462c\n``#53462c` `rgb(83,70,44)``\n• #584727\n``#584727` `rgb(88,71,39)``\n• #5d4922\n``#5d4922` `rgb(93,73,34)``\n• #624a1d\n``#624a1d` `rgb(98,74,29)``\n• #674c18\n``#674c18` `rgb(103,76,24)``\n• #6b4d14\n``#6b4d14` `rgb(107,77,20)``\n• #704f0f\n``#704f0f` `rgb(112,79,15)``\n• #75500a\n``#75500a` `rgb(117,80,10)``\n• #7a5205\n``#7a5205` `rgb(122,82,5)``\n• #7f5300\n``#7f5300` `rgb(127,83,0)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #7f5300 is perceived by people affected by a color vision deficiency. This can be useful if you need to ensure your color combinations are accessible to color-blind users.\n\nMonochromacy\n• Achromatopsia 0.005% of the population\n• Atypical Achromatopsia 0.001% of the population\nDichromacy\n• Protanopia 1% of men\n• Deuteranopia 1% of men\n• Tritanopia 0.001% of the population\nTrichromacy\n• Protanomaly 1% of men, 0.01% of women\n• Deuteranomaly 6% of men, 0.4% of women\n• Tritanomaly 0.01% of the population" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.51262593,"math_prob":0.8368754,"size":3658,"snap":"2023-14-2023-23","text_gpt3_token_len":1640,"char_repetition_ratio":0.14915161,"word_repetition_ratio":0.007380074,"special_character_ratio":0.56123567,"punctuation_ratio":0.23224352,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98783994,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-06-06T06:31:09Z\",\"WARC-Record-ID\":\"<urn:uuid:5021c2de-3206-43e0-a83e-2f98a395c462>\",\"Content-Length\":\"36096\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:91c59c6b-e148-4024-8128-05ff9dea0a24>\",\"WARC-Concurrent-To\":\"<urn:uuid:60b36597-1837-4058-99c5-bd879087c392>\",\"WARC-IP-Address\":\"178.32.117.56\",\"WARC-Target-URI\":\"https://www.colorhexa.com/7f5300\",\"WARC-Payload-Digest\":\"sha1:5TNSGHVUPWESWRZXZXAYESZAKLVKS2FF\",\"WARC-Block-Digest\":\"sha1:DKFY77LI5GHIUSFDXTNNNXFL2PLVU23G\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224652235.2_warc_CC-MAIN-20230606045924-20230606075924-00551.warc.gz\"}"}
https://www.gcflcm.com/gcf-of-85-and-98	[ "# What is the Greatest Common Factor of 85 and 98?\n\nGreatest common factor (GCF) of 85 and 98 is 1.\n\nGCF(85,98) = 1\n\nWe will now calculate the prime factors of 85 and 98, than find the greatest common factor (greatest common divisor (gcd)) of the numbers by matching the biggest common factor of 85 and 98.\n\nGCF Calculator and\nand\n\n## How to find the GCF of 85 and 98?\n\nWe will first find the prime factorization of 85 and 98. After we will calculate the factors of 85 and 98 and find the biggest common factor number .\n\n### Step-1: Prime Factorization of 85\n\nPrime factors of 85 are 5, 17. Prime factorization of 85 in exponential form is:\n\n85 = 51 × 171\n\n### Step-2: Prime Factorization of 98\n\nPrime factors of 98 are 2, 7. Prime factorization of 98 in exponential form is:\n\n98 = 21 × 72\n\n### Step-3: Factors of 85\n\nList of positive integer factors of 85 that divides 85 without a remainder.\n\n1, 5, 17\n\n### Step-4: Factors of 98\n\nList of positive integer factors of 98 that divides 85 without a remainder.\n\n1, 2, 7, 14, 49\n\n#### Final Step: Biggest Common Factor Number\n\nWe found the factors and prime factorization of 85 and 98. The biggest common factor number is the GCF number.\nSo the greatest common factor 85 and 98 is 1.\n\nAlso check out the Least Common Multiple of 85 and 98" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.88223445,"math_prob":0.92459285,"size":1630,"snap":"2021-31-2021-39","text_gpt3_token_len":538,"char_repetition_ratio":0.2712177,"word_repetition_ratio":0.09883721,"special_character_ratio":0.37055215,"punctuation_ratio":0.083102494,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9998404,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-08-05T23:37:10Z\",\"WARC-Record-ID\":\"<urn:uuid:c0c08347-e253-4901-9a9b-53e84a9a1c97>\",\"Content-Length\":\"21922\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:643e718e-4725-47c6-940f-af897dbef647>\",\"WARC-Concurrent-To\":\"<urn:uuid:ab3b8dde-a67c-4ee6-b8e1-04cc3bc3e750>\",\"WARC-IP-Address\":\"104.154.21.107\",\"WARC-Target-URI\":\"https://www.gcflcm.com/gcf-of-85-and-98\",\"WARC-Payload-Digest\":\"sha1:QQJX76KRCJTUREA3HZUVA3NNGCWAEDH5\",\"WARC-Block-Digest\":\"sha1:PUGRHQZATECUJ4CIMMCVQMZMV4OLQ4YA\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-31/CC-MAIN-2021-31_segments_1627046152085.13_warc_CC-MAIN-20210805224801-20210806014801-00187.warc.gz\"}"}
http://fahry.info/go-math-4th-grade-printable-worksheets/vampire-maze-grade-math-worksheet-for-division-3rd-and-4th-printable-worksheets/	[ "", null, "# Vampire Maze Grade Math Worksheet For Division 3rd And 4th Printable Worksheets", null, "vampire maze grade math worksheet for division 3rd and 4th printable worksheets.\n\nfree printable math worksheets for 3rd and 4th grade go ch 5 lesson 3 word problems common core,free printable math worksheets for 3rd and 4th grade go unit 1 lesson 4 6 activities common core with answers,3rd and 4th grade math printable worksheets free word problems pdf go 4 download them try to solve,math word problems 4th grade printable worksheets vampire maze worksheet for division common core envision,4th grade math worksheets printable pdf go chapter angles free for 3rd and common core,free printable math worksheets for 3rd and 4th grade envision go version worksheet vocabulary by the,free printable math worksheets for 3rd and 4th grade division go kids graders second word problems,go math printable worksheets the best image collection 4th grade division 3rd and,kids go math worksheets adding fractions like denominators 4th grade printable free word problems common core,4th grade math worksheets printable pdf go the best image collection with answers 3rd and." ]	[ null, "http://fahry.info/go-math-4th-grade-printable-worksheets/vampire-maze-grade-math-worksheet-for-division-3rd-and-4th-printable-worksheets/", null, "http://fahry.info/wp-content/uploads/2017/11/vampire-maze-grade-math-worksheet-for-division-3rd-and-4th-printable-worksheets.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.87849295,"math_prob":0.76140445,"size":1062,"snap":"2019-13-2019-22","text_gpt3_token_len":231,"char_repetition_ratio":0.22495274,"word_repetition_ratio":0.16149068,"special_character_ratio":0.18926553,"punctuation_ratio":0.05945946,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97946775,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-05-19T15:22:55Z\",\"WARC-Record-ID\":\"<urn:uuid:6334c1b5-a714-4e8e-8e2f-b108b0172264>\",\"Content-Length\":\"39210\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:2acfe8b4-e210-4ef9-800a-01cc13d712f5>\",\"WARC-Concurrent-To\":\"<urn:uuid:c9d9c4f6-66b7-455d-b61d-1935d83de5f2>\",\"WARC-IP-Address\":\"104.18.33.243\",\"WARC-Target-URI\":\"http://fahry.info/go-math-4th-grade-printable-worksheets/vampire-maze-grade-math-worksheet-for-division-3rd-and-4th-printable-worksheets/\",\"WARC-Payload-Digest\":\"sha1:SGR2UHNGQBAYUSIEJ6HGTJFKBAINHL5N\",\"WARC-Block-Digest\":\"sha1:JACMXZLQU7BZRI4MYUVQ6HJ2HNJJF7OA\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-22/CC-MAIN-2019-22_segments_1558232254889.43_warc_CC-MAIN-20190519141556-20190519163556-00327.warc.gz\"}"}
https://shrew.app/show/tim/speaky-boy	[ "```face = Circle(color='tan')\npupil1 = Circle(width=20, height=20, x=10, color='white')\npupil2 = Circle(width=20, height=20, x=50, color='white')\neye1 = Circle(width=5, height=5, x=10, color='black')\neye2 = Circle(width=5, height=5, x=50, color='black')\nmouth = Circle(width=10, height=5, x=50, y=80, color='black')\n\nwith animation(duration=2):\neye1.x = pupil1.x = 50\neye2.x = pupil2.x = 90\nmouth.height = 10\n\nwith animation(duration=0.2):\nmouth.height = 15\n\nwith animation(duration=0.2):\nmouth.height = 12\n\nwith animation(duration=0.2):\nmouth.height = 16\n\nwith animation(duration=0.2):\nmouth.height = 10\n\nwith animation(duration=0.2):\nmouth.height = 14\n\nwith animation(duration=1):\neye1.x = pupil1.x = 10\neye2.x = pupil2.x = 50\nmouth.height = 5```\n\n# Speaky Boy\n\nby tim\n\nCreated 4 years, 6 months ago.\nBased on Bouncing eyes by ludwik." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.82698905,"math_prob":0.9998485,"size":911,"snap":"2022-40-2023-06","text_gpt3_token_len":286,"char_repetition_ratio":0.17640573,"word_repetition_ratio":0.05042017,"special_character_ratio":0.3677278,"punctuation_ratio":0.23004694,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9882675,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-02-09T08:58:11Z\",\"WARC-Record-ID\":\"<urn:uuid:3b4daebd-0e14-42ae-90c5-9fc3ae0ee4c0>\",\"Content-Length\":\"8646\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:261e8d66-43e9-41b6-893b-2a35ce55f12d>\",\"WARC-Concurrent-To\":\"<urn:uuid:2bc803e8-fa12-4ad3-8cbf-429fb01f798a>\",\"WARC-IP-Address\":\"172.67.194.122\",\"WARC-Target-URI\":\"https://shrew.app/show/tim/speaky-boy\",\"WARC-Payload-Digest\":\"sha1:MS5WMRVYUXEFZWJOE2OVTOLCCDLMYJUA\",\"WARC-Block-Digest\":\"sha1:TJJ37NBTTDPO53GYVOTG4T35YYJLKMAM\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764501555.34_warc_CC-MAIN-20230209081052-20230209111052-00866.warc.gz\"}"}
http://www.csgnetwork.com/surfareacalc.html	[ "", null, "# CSG Surface Area Calculator\n\nThis calculator requires the use of Javascript enabled and capable browsers. It is to determine the surface area of an object. Select the object from the options then answer the questions about length and width. All entries should be numeric. The answer is returned as a number, based on your entries. In the listed formulae for this set of calculations, PI here is used as 3.1415929203539825. Enjoy! If you desire calculations on volume of these same objects, try our volume calculator.\n\nOther area calculators for your use are our Ellipse Area Calculator, Parabola Area Calculator, Parallelogram Area Calculator, Pyramid Area Calculator, Trapezoid Area Calculator, our Area Calculator, Irregular Polygon Area Calculator and our Land Parcel Area Calculator. At least you are in the right area...\n\n## Cone\n\nFormula - SA = PI x r x SH\nEnter the radius of the base\nEnter the slope height of the cone\nsquare units\n\n## Cube\n\nFormula - SA = 6 x L x L\nEnter the length of one edge\nsquare units\n\n## Cylinder\n\nFormula - SA = (2 x PI x r x H) + (2 x PI x (r x r))\nEnter the radius of the base\nEnter the height of the cylinder\nsquare units\n\n## Rectangular Prism\n\nFormula - SA = (2 x L x H) + (2 x W x L) + (2 x W x H)\nEnter the length\nEnter the width\nEnter the height\nsquare units\n\n## Sphere\n\nFormula - A = (4 x PI) x (r x r)\nEnter the radius of the sphere\nsquare units\nVersion 1.6.1\n\n Leave us a question or comment on Facebook", null, "Search or Browse Our Site", null, "", null, "" ]	[ null, "http://storage.googleapis.com/csgnetworkstatic/allbar_header.png", null, "http://storage.googleapis.com/csgnetworkstatic/facebook-icon.png", null, "http://storage.googleapis.com/csgnetworkstatic/allbottom.png", null, "http://storage.googleapis.com/csgnetworkstatic/cslogo.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8949595,"math_prob":0.99434763,"size":487,"snap":"2023-14-2023-23","text_gpt3_token_len":101,"char_repetition_ratio":0.124223605,"word_repetition_ratio":0.0,"special_character_ratio":0.2238193,"punctuation_ratio":0.12765957,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99852103,"pos_list":[0,1,2,3,4,5,6,7,8],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-03-28T18:05:03Z\",\"WARC-Record-ID\":\"<urn:uuid:ffbe8ef6-8c22-4ed6-877d-85d1977cd67c>\",\"Content-Length\":\"20371\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7ebfa785-10c8-4372-b20e-53b28df12d0b>\",\"WARC-Concurrent-To\":\"<urn:uuid:da1f80df-a5ba-4a30-b024-61d9ac7d064e>\",\"WARC-IP-Address\":\"35.201.91.113\",\"WARC-Target-URI\":\"http://www.csgnetwork.com/surfareacalc.html\",\"WARC-Payload-Digest\":\"sha1:SDQXGP3KP7Z3QO37EFN7BVLJKYMDGOOJ\",\"WARC-Block-Digest\":\"sha1:BNXO5ZX2JRIWSYUFDVVQZBJ62TTSUP2X\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-14/CC-MAIN-2023-14_segments_1679296948868.90_warc_CC-MAIN-20230328170730-20230328200730-00644.warc.gz\"}"}
https://www.wisegeek.com/what-is-an-implied-rate.htm	[ "# What is an Implied Rate?\n\nMalcolm Tatum\n\nAn implied rate is the interest rate that represents the difference between the forward rate and the spot rate associated with a specific investment. Typically, this particular interest rate is calculated by subtracting the present or spot rate from the forward or futures rate. The amount of the implied rate can provide the investor with valuable clues regarding the advantages or disadvantages of entering into a futures contract associated with that particular security or commodity.\n\nIn order to understand how the implied rate functions, it is necessary to know what is meant by a spot rate and a forward rate. Essentially, the spot rate is the interest rate that will remain in place over the next couple of trading days. The forward or futures rate is the interest rate that will apply to the security at some specific date over the next several months. Depending on the nature of the investment, the amount of difference between these two rates may indicate that there is a significant benefit to completing the transaction now versus arranging a date to settle in the future.\n\nOne example is to consider the London Interbank Offered Rate (LIBOR) that is offered with a specific investment. If the present LIBOR is set at seven percent and the forward LIBOR rate is at nine percent, that means the implied rate is two percent. This calculation indicates to the investor that borrowing will be somewhat more expensive in the future, and may prompt the investor to go with a transaction that is settled while the present rate is still in effect rather than going with some type of futures or forward contract. By contrast, if the present rate is more than the forward rate, this indicates that going with a futures contract may allow the investor to secure the commodity at a better rate later on, while locking in that better rate here and now.", null, "Get started" ]	[ null, "https://www.wisegeek.com/images/wikibuy/wikibuy_logo.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9500896,"math_prob":0.97141165,"size":2758,"snap":"2020-24-2020-29","text_gpt3_token_len":514,"char_repetition_ratio":0.14887436,"word_repetition_ratio":0.017391304,"special_character_ratio":0.17947789,"punctuation_ratio":0.060606062,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9527417,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-07-04T09:22:23Z\",\"WARC-Record-ID\":\"<urn:uuid:b8a7153c-07fb-40fd-ba72-471ced032032>\",\"Content-Length\":\"63005\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:90c693b9-8669-4f5c-9e8d-ba4d80b6d7be>\",\"WARC-Concurrent-To\":\"<urn:uuid:e91e38c3-c2a7-42ef-b335-c04c426727af>\",\"WARC-IP-Address\":\"162.210.232.130\",\"WARC-Target-URI\":\"https://www.wisegeek.com/what-is-an-implied-rate.htm\",\"WARC-Payload-Digest\":\"sha1:KPXPFYUOO3POW2FGLRHCGNTHEESRPOQM\",\"WARC-Block-Digest\":\"sha1:FNQPMQXY4VZ56XPVXGMEJDT62RCS5HRO\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-29/CC-MAIN-2020-29_segments_1593655886095.7_warc_CC-MAIN-20200704073244-20200704103244-00193.warc.gz\"}"}
https://electronics.stackexchange.com/questions/357910/formula-for-bjt-common-emitter-output-characteristics-curve/357987#357987	[ "# Formula for BJT common emitter output characteristics curve\n\nI collected experimental data of input and output characteristics of a BJT NPN in Common Emitter configuration. I need to find the right function to fit the Ic vs Vce, but I'm not finding it. I've used the Shockley Equation with no luck. I have tried the Early model and it worked fine on the linear region, but I need to fit the whole curve. Could you help me finding the right formula? Thanks, Alberto.\n\nNote: this is the function I've used with no success.\n\nIc = a+bx+c(1-exp((x/(de))) Where x is Vce\n\nUpdate: My bibliography is Jacob Millman's Analog and Digital Circuits.\n\n• I suggest that you look at how transistor models used in circuit simulators like Pspice look like. Google for \"BSIM\" or \"MEXTRAM\" and have a look at how insanely complex these models are. The Shockley equation is more like a \"first order\" approach to the real behavior. Regarding fitting a model to the measurements, there are engineers that have this as their day job. It is complex. I don't believe you can expect the Shockley equation to fit the complete behavior of a BJT. Feb 23 '18 at 12:30\n• How far out was reality from theory? Show the discepancies if you want a complete answer. Feb 23 '18 at 13:10\n• You can find the expression for Vce as a function of Ic/Ib on the following editions of Millman's books: Electronic Devices and Circuits,\"Voltages as functions of currents\", p. 250 and Integrated Electronics, p. 148. Perhaps you can fit the logarithmic function to the inverted data. (Edit: now that I think of it, that model does not account for the Early effect). Feb 25 '18 at 21:32\n\nThere are books on the subject of extracting parameters for BJTs from experimental test setups. What you do depends entirely on the parameter of interest at the time. And that depends upon the model. (There are more than a few.)\n\nIn the CE configuration, the BJT is in active mode. So on the surface it might seem that you can just use:\n\n$$I_\\text{C}=I_\\text{SAT}\\cdot\\left(e^{V_\\text{BE}\\over n\\cdot V_T}-1\\right)$$\n\nBut, unfortunately, there are three horribly glaring holes in this active mode model. (At least three. Someone else will certainly point out my errors and add another few.) One of them is that $$\\I_\\text{SAT}\\$$ is a function and not a constant. And the other is the Early Effect, which is unaccounted there. And a third is $$\\\\beta\\$$, which of course is also a function and not a constant (and if you made a more sophisticated model based on 3D integrals and design details would itself no longer be needed.)\n\nHere's an example of the dependence of $$\\I_\\text{SAT}\\$$ on temperature:\n\n$$I_\\text{SAT}\\left(T\\right)=I_{\\text{SAT}_{T_\\text{nom}}}\\cdot\\left[\\left(\\frac{T}{T_\\text{nom}}\\right)^{3}e^{\\frac{E_\\text{g}}{k}\\cdot\\frac{T-T_\\text{nom}}{T\\cdot T_\\text{nom}}}\\right]$$\n\nNote that $$\\E_\\text{g}\\$$ is the effective energy gap (in eV) and $$\\k\\$$ is Boltzmann's constant (in appropriate units.) $$\\T_\\text{nom}\\$$ is the temperature at which the equation was calibrated, of course, and $$\\I_{\\text{SAT}_{T_\\text{nom}}}\\$$ is the extrapolated saturation current at that calibration temperature.\n\nThe power of 3 used in the equation above is actually a problem, because of the temperature dependence of diffusivity, $$\\\\frac{k T}{q} \\mu_T\\$$. And even that, itself, ignores the bandgap narrowing caused by heavy doping. In practice, the power of 3 is itself turned into a model parameter.\n\nHere is a curve that you would need to be able to approximate accurately for any given BJT (which still is just the BJT outside of a circuit and where you still need to develop something for the Early Effect.) Taken from Figure 2.15 from Ian Getreu's \"Modeling The Bipolar Transistor\":", null, "In Region I, the decline in $$\\\\beta\\$$ is due to three components which can be ignored in the other regions but cannot be ignored with low currents involved. These are:\n\n1. The formation of emitter-base surface channels (which can be reduced by the careful application of processing/manufacturing); and,\n2. The recombination of surface carriers (which also can be reduced by the careful application of processing/manufacturing, but still remains a dominant part of the problem); and,\n3. The recombination of carriers in the emitter-base space-charge layer.\n\nAll three of these have similar variations with the base-emitter voltage so that you wind up with something akin to the following typical component equations. (These are showing $$\\I_\\text{B}\\$$ and not $$\\I_\\text{C}\\$$, but this just means that the value of $$\\I_\\text{SAT}\\$$ in the following equations isn't taken to be the same one discussed for $$\\I_\\text{C}\\$$ -- keep in mind that $$\\I_\\text{SAT}\\$$ is just a projected y-axis intercept and that concept can be applied equally well in a variety of contexts related to currents):\n\n\\begin{align} I_{\\text{B}_\\text{channel}}&=I_{\\text{SAT}_\\text{channel}}\\cdot\\left(e^\\frac{V_\\text{BE}}{4\\cdot V_T}-1\\right)\\\\\\\\ I_{\\text{B}_\\text{surface}}&=I_{\\text{SAT}_\\text{surface}}\\cdot\\left(e^\\frac{V_\\text{BE}}{2\\cdot V_T}-1\\right)\\\\\\\\ I_{\\text{B}_\\text{space-charge}}&=I_{\\text{SAT}_\\text{space-charge}}\\cdot\\left(e^\\frac{V_\\text{BE}}{2\\cdot V_T}-1\\right) \\end{align}\n\nAlthough summed exponentials are not exactly equivalent to any single resulting equivalent exponential, it is practical (and done) to combine the above into a single modeled exponential that uses $$\\\\eta_{_\\text{EL}}\\$$ values often close to 2:\n\n\\begin{align} I_{\\text{B}_\\text{summed}}&=I_{\\text{SAT}_\\text{summed}}\\cdot\\left(e^\\frac{V_\\text{BE}}{\\eta_{_\\text{EL}}\\cdot V_T}-1\\right) \\end{align}\n\nFor most BJTs, the above equation can be made to approximate the reality well enough for practical purposes (and it sums into the usual current equations.)\n\nIn Region III, the injection of minority carriers into the base region starts becoming increasingly important in comparison against the majority carrier concentrations. Because the space-charge neutrality is maintained in the base, the majority concentration has to increase by the same amount.\n\nThe finding is:\n\n\\begin{align} I_{\\text{C}_{\\text{high level}}}&\\propto e^\\frac{V_\\text{BE}}{2\\cdot V_T} \\end{align*}\n\nThe other factor in Region III is, of course, an 'Ohmic resistance' and is already modeled as $$\\r_c\\$$ so it isn't included above.\n\nA model constant is usually applied to the above equation and the resulting term then appears in the divisor used for the usual model saturation current equation.\n\nThe upshot for Region III is:\n\n1. The increasing importance of the injected minority carriers into the base; and,\n2. Ohmic resistance.\n\nAnd so far, we're just talking about one part of one quadrant of operation -- the active mode for your CE amplifier.\n\nAs I said, there are books on the topic. Whole product lines developed, with lots of software, to do this. (I own two working instruments from Tektronix STS now-defunct product line and several others from other manufacturers that are designed for the purpose of extracting parameters and providing charts.)\n\nI don't want to discourage you from creating useful models of actual devices for the purposes of a real circuit that depends upon them. Tektronix made products that were based upon knowing more about BJTs than others and pushing the absolute limits of those devices in circuits that did actually depend upon device-selection for behavior. I've developed LED standard lamps by throwing away more than 99% of them and just keeping a few that \"worked well\" for the purpose.\n\nMost people just design their circuits to \"not overly depend\" upon a specific model behavior, though. So you really need a reason to push like this. I think that's part of why you don't see a lot on the web about the subject.\n\n• Thanks for the detailed description. My experiment wasn't a necessity of mine but it is an experiment we have to work on following the laboratory course in a degree in Physics. Talking with the teacher, he has said we have to watch only the Early Effect and don't mind the first part of the curve. He also suggested to fit the curve with the Shockley Equation and then the active region with the Early Equation. Feb 24 '18 at 10:40\n• @AlbertoPerro Are you still in need of a way of estimating the Early Effect? (I'm pretty certain you already know how to estimate the saturation current by just drawing a line back to the y-axis on a log scale graph.)\n– jonk\nFeb 24 '18 at 20:44\n• Yes, with the Early Effect is a simple linear fit. Thanks Feb 25 '18 at 22:42\n• @AlbertoPerro Okay. So everything is good. Just wanted to be sure you had what you felt you needed for now. Thanks for the update.\n– jonk\nFeb 25 '18 at 22:50" ]	[ null, "https://i.stack.imgur.com/rsFhJ.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.899496,"math_prob":0.9584326,"size":6053,"snap":"2021-31-2021-39","text_gpt3_token_len":1534,"char_repetition_ratio":0.11836667,"word_repetition_ratio":0.013969732,"special_character_ratio":0.24698497,"punctuation_ratio":0.06766257,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9943547,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-22T20:30:32Z\",\"WARC-Record-ID\":\"<urn:uuid:93fc9531-9a27-49f7-beff-10609304cce4>\",\"Content-Length\":\"183306\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:68df96bd-e813-4c3c-9f79-e2cee3d0fc65>\",\"WARC-Concurrent-To\":\"<urn:uuid:a2159d34-a689-43ba-9150-40d5408599aa>\",\"WARC-IP-Address\":\"151.101.129.69\",\"WARC-Target-URI\":\"https://electronics.stackexchange.com/questions/357910/formula-for-bjt-common-emitter-output-characteristics-curve/357987#357987\",\"WARC-Payload-Digest\":\"sha1:TWQLZ5HYAXWS3VOTKWFPAELDPACPCF4T\",\"WARC-Block-Digest\":\"sha1:EF6GFPRXL7AH5XDR4P3I4JLBEPG7WJ45\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780057388.12_warc_CC-MAIN-20210922193630-20210922223630-00581.warc.gz\"}"}
https://edurev.in/studytube/Integer-Type-ques-of-Carbohydrates--Amino-Acids--P/327b314b-f473-428a-bd0d-52a85c3e2b7f_t	[ "Courses\n\n# Integer Type ques of Carbohydrates, Amino Acids, Polymers & Miscellaneous Match the Following, Past JEE Notes \| EduRev\n\n## Class 12 Chemistry 35 Years JEE Mains &Advance Past yr Paper\n\nCreated by: Abhishek Kapoor\n\n## JEE : Integer Type ques of Carbohydrates, Amino Acids, Polymers & Miscellaneous Match the Following, Past JEE Notes \| EduRev\n\nThe document Integer Type ques of Carbohydrates, Amino Acids, Polymers & Miscellaneous Match the Following, Past JEE Notes \| EduRev is a part of the JEE Course Class 12 Chemistry 35 Years JEE Mains &Advance Past yr Paper.\nAll you need of JEE at this link: JEE\n\nQ. 1. The total number of basic groups in the following form of lysine is (2010)", null, "Ans. 2\n\nSol. The basic groups in the given form of lysine is NH2", null, "Q.2. A decapeptide (Mol. wt. 796) on complete hydrolysis gives glycine (Mol. wt. 75), alanine and phenylalanine. Glycine contributes 47.0% to the total weight of the hydrolysed products. The number of glycine units present in the decapeptide is (2011)\n\nAns. 6\n\nSol. Molecular weight of decapeptide = 796 g/mol\nTotal bonds to be hydrolysed = (10 – 1) = 9 per molecule\nTotal weight of H2O added = 9 × 18 = 162 g/mol\nTotal weight of hydrolysis products = 796 + 162 = 958 g\nTotal weight % of glycine (given) = 47%\nTotal weight of glycine in product", null, "Molecular weight of glycine = 75 g/mol\n\nNumber of glycine molecules", null, "Q.3. When the following aldohexose exists in its D-configuration, the total number of stereoisomers in its pyranose form is : (2012)\nCHO — CH2 — CHOH — CHOH — CHOH — CH2OH\n\nAns. 8\n\nSol.", null, "Thus, total number of stereoisomers in pyranose form of D-configuration = 23 = 8\n\nQ.4. The substituents R1 and R2 for nine peptides are listed in the table given below. How many of these peptides are positively charged at pH = 7.0 ? (2012)", null, "", null, "Ans. 4\n\nSol. (4) Peptides with isoelectric point (pI) more than 7, would exist as cation in neutral solution (pH = 7) which means the given polypeptide is of basic nature, so it must contain two or more amino groups. Hence IV, VI, VIII and IX are correct options.\n\nQ.5. A tetrapeptide has —COOH group on alanine. This produces glycine (Gly), valine (Val), phenyl alanine (Phe) and alanine (Ala), on complete hydrolysis. For this tetrapeptide, the number of possible sequences (primary structures) with — NH2 group attached to a chiral center is (JEE Adv. 2013)\n\nAns.\n\nSol. (4) According to question C – Terminal must be alanine and N – Terminal do have chiral carbon means it should not be glycine. So possible sequence is :\nVal Phe Gly Ala\nVal Gly Phe Ala\nPhe Val Gly Ala\nPhe Gly Val Ala\n\nQ.6. The total number of distinct naturally occurring amino acids obtained by complete acidic hydrolysis of the peptide shown below is (JEE Adv. 2014)", null, "Ans. 1\n\nSol. (1) On hydrolysis, the given peptide gives only one naturally occurring amino acid (glycine).\n\nOffer running on EduRev: Apply code STAYHOME200 to get INR 200 off on our premium plan EduRev Infinity!\n\n69 docs\|36 tests\n\n,\n\n,\n\n,\n\n,\n\n,\n\n,\n\n,\n\n,\n\n,\n\n,\n\n,\n\n,\n\n,\n\n,\n\n,\n\n,\n\n,\n\n,\n\n,\n\n,\n\n,\n\n,\n\n,\n\n,\n\n,\n\n,\n\n,\n\n,\n\n,\n\n,\n\n;" ]	[ null, "https://cdn3.edurev.in/ApplicationImages/Temp/3066490_381af9e5-534f-4e40-ab94-49170f20d955_lg.png", null, "https://cdn3.edurev.in/ApplicationImages/Temp/3066490_907bd16a-8661-446d-8440-4e61640b1b65_lg.png", null, "https://cdn3.edurev.in/ApplicationImages/Temp/3066490_dd469ca1-df39-40fe-b67d-43708f24ee8c_lg.png", null, "https://cdn3.edurev.in/ApplicationImages/Temp/3066490_719c9732-58e0-46f1-82a1-fa3fe03efb9c_lg.png", null, "https://cdn3.edurev.in/ApplicationImages/Temp/3066490_1d38bb16-cf8c-4144-be1c-b2ba06d79277_lg.png", null, "https://cdn3.edurev.in/ApplicationImages/Temp/3066490_fa6eed1f-5392-4a0b-84b2-1abd442ce423_lg.png", null, "https://cdn3.edurev.in/ApplicationImages/Temp/3066490_0276f4df-5e95-4886-9cc8-140892789eb7_lg.png", null, "https://cdn3.edurev.in/ApplicationImages/Temp/3066490_97faca52-5bd9-4dc7-ad58-91c52fb83d01_lg.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8630189,"math_prob":0.9131298,"size":2219,"snap":"2020-34-2020-40","text_gpt3_token_len":658,"char_repetition_ratio":0.1268623,"word_repetition_ratio":0.010025063,"special_character_ratio":0.27895448,"punctuation_ratio":0.12641084,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9701704,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16],"im_url_duplicate_count":[null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-08-10T00:26:01Z\",\"WARC-Record-ID\":\"<urn:uuid:62069538-899c-49b0-a47c-6696f53fb1d8>\",\"Content-Length\":\"273257\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e7cc08e2-cf65-48be-a659-0e3be454f01d>\",\"WARC-Concurrent-To\":\"<urn:uuid:1a843c86-ab87-436f-9e70-c1a476f86419>\",\"WARC-IP-Address\":\"35.198.207.72\",\"WARC-Target-URI\":\"https://edurev.in/studytube/Integer-Type-ques-of-Carbohydrates--Amino-Acids--P/327b314b-f473-428a-bd0d-52a85c3e2b7f_t\",\"WARC-Payload-Digest\":\"sha1:6XSFJKNY4QEB4AGJWPMQQDAFLRH3VMCS\",\"WARC-Block-Digest\":\"sha1:VI3ZYTVTP4ETFEIEH6EG6YPIUGJY77EV\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-34/CC-MAIN-2020-34_segments_1596439738595.30_warc_CC-MAIN-20200809222112-20200810012112-00374.warc.gz\"}"}
https://www.tensorflow.org/lattice/api_docs/python/tfl/lattice_lib/evaluate_with_hypercube_interpolation	[ "# tfl.lattice_lib.evaluate_with_hypercube_interpolation\n\nEvaluates a lattice using hypercube interpolation.\n\nLattice function is multi-linearly interpolated between the 2^d vertices of a hypercube. This interpolation method is typically slower than simplex interpolation, since each value is interpolated from 2^d hypercube corners, rather than d+1 simplex corners. For details, see e.g. \"Dissection of the hypercube into simplices\", D.G. Mead, Proceedings of the AMS, 76:2, Sep. 1979.\n\n`inputs` Tensor of shape: `(batch_size, ..., len(lattice_sizes))` or list of `len(lattice_sizes)` tensors of same shape `(batch_size, ..., 1)` which represents points to apply lattice interpolation to. A typical shape is `(batch_size, len(lattice_sizes))`.\n`kernel` Lattice kernel of shape (num_params_per_lattice, units).\n`units` Output dimension of the lattice.\n`lattice_sizes` List or tuple of integers which represents lattice sizes of layer for which interpolation is being computed.\n`clip_inputs` Whether inputs should be clipped to the input range of the lattice." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7298011,"math_prob":0.96790594,"size":1132,"snap":"2020-34-2020-40","text_gpt3_token_len":268,"char_repetition_ratio":0.1746454,"word_repetition_ratio":0.0,"special_character_ratio":0.2270318,"punctuation_ratio":0.21105528,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9950481,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-08-06T23:00:59Z\",\"WARC-Record-ID\":\"<urn:uuid:fd69d350-2043-4325-8e28-78c1dbe7a15f>\",\"Content-Length\":\"105492\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7548e7f9-ecd4-4432-831a-8fe6cb92f022>\",\"WARC-Concurrent-To\":\"<urn:uuid:97387033-27d4-406b-925f-dcb7b947a9f5>\",\"WARC-IP-Address\":\"172.217.15.78\",\"WARC-Target-URI\":\"https://www.tensorflow.org/lattice/api_docs/python/tfl/lattice_lib/evaluate_with_hypercube_interpolation\",\"WARC-Payload-Digest\":\"sha1:FEQVQYPJBC7BQV4GEF4Y7MJTLJBEIYTB\",\"WARC-Block-Digest\":\"sha1:NEVUEFU6IVFL265EKQJWO4VPVGNVUZ7Z\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-34/CC-MAIN-2020-34_segments_1596439737039.58_warc_CC-MAIN-20200806210649-20200807000649-00147.warc.gz\"}"}
https://www.zebra.com/us/en/support-downloads/knowledge-articles/evm/how-do-i-transmit-the-code-39-check-digit.html	[ "# How Do I Transmit the Code 39 Check Digit?\n\nArticle ID: 34049753\n\n### Issue / Question\n\nHow do I make the scanner transmit the Code 39 check digit?\n\n### Applicable To\n\nBarcode scanners\n\nTo enable the transmission of the Code 39 Check Digit, the Enable Code 39 Check Digit programming bar code must be scanned followed by the Transmit Code 39 Check Digit.", null, "", null, "The code with check digit is referred to as Code 39 mod 43.", null, "To compute this, each character is assigned a value. The assignments are listed in the table above, and almost, but not quite, systematic.\n\nHere is how to do the checksum calculation:\n\n1. Take the value (0 through 42) of each character in the barcode excluding start and stop codes.\n2. Sum the values.\n3. Divide the result by 43.\n4. The remainder is the value of the checksum character to be appended.\n\nExample 1:\nThe\ncheck digit of Code 39 string data converts the numbers from Code 39’s character values table.\n\n• C = 12\n• O = 24\n• D = 13\n• E = 14\n• 3 = 3\n• 9 = 9\n\n12(C) + 24(O) + 13(D) + 14(E) + 3 + 9 = 75 - 43 = 32\n\nResults: 32\n= W based on the character values table. The check digit for Code 39's data is W.\n\nExample 2:\nIn case of scan data 04465AZ120:\n\n0 + 4 + 4 + 6 + 5 + 10(A) + 35(Z) + 1 + 2 + 0 + 38(space) + 38(space) = 143 - 43 = 100 - 43 = 57 - 43 = 14\n\nResults: 14 = E based on the character values table. The check digit of the above data is E.\n\nTest Scenario 1:\n\n1. Scan the Factory default barcode", null, "2. Connect to Host PC with HID mode > Scan CODE39W barcode > Check the output data from the Notepad app.", null, "", null, "3. If you want to ignore the check digit value, enable the Code 39 Check digit verification and disable the Transmit Code 39 Check Digit.\n\nTest Scenario 2:\n1. Scan the Code 39 Check Digit Verification barcode.", null, "2. Scan the Do Not Transmit Code 39 Check Digit barcode.", null, "", null, "NOTE Code 39 Check Digit Verification must be enabled for this parameter to function.\n\nThen scan test with CODE39W, the result’s CODE39 without W.\n1. If you want to have the Check digit(W) again, scan the barcode below.", null, "Then scan test with CODE39W, the result’s CODE39W.", null, "This same setting can be configured via 123Scan.", null, "" ]	[ null, "https://www.zebra.com/content/dam/zebra_new_ia/en-us/knowledge-articles/11476-1.jpg", null, "https://www.zebra.com/content/dam/zebra_new_ia/en-us/knowledge-articles/11476-2.jpg", null, "https://www.zebra.com/content/dam/zebra_new_ia/en-us/knowledge-articles/11476-3.jpg", null, "https://www.zebra.com/content/dam/zebra_new_ia/en-us/knowledge-articles/11476-4.jpg", null, "https://www.zebra.com/content/dam/zebra_new_ia/en-us/knowledge-articles/11476-5.jpg", null, "https://www.zebra.com/content/dam/zebra_new_ia/en-us/knowledge-articles/11476-6.jpg", null, "https://www.zebra.com/content/dam/zebra_new_ia/en-us/knowledge-articles/11476-7.jpg", null, "https://www.zebra.com/content/dam/zebra_new_ia/en-us/knowledge-articles/11476-8.jpg", null, "https://www.zebra.com/content/dam/zebra_new_ia/en-us/knowledge-articles/kaicon-1.jpg", null, "https://www.zebra.com/content/dam/zebra_new_ia/en-us/knowledge-articles/11476-9.jpg", null, "https://www.zebra.com/content/dam/zebra_new_ia/en-us/knowledge-articles/11476-10.jpg", null, "https://www.zebra.com/content/dam/zebra_new_ia/en-us/knowledge-articles/11476-11.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.6572756,"math_prob":0.9626044,"size":2003,"snap":"2021-43-2021-49","text_gpt3_token_len":548,"char_repetition_ratio":0.16558279,"word_repetition_ratio":0.04511278,"special_character_ratio":0.31802297,"punctuation_ratio":0.09975062,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9872105,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24],"im_url_duplicate_count":[null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,null,null,2,null,2,null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-12-02T01:16:20Z\",\"WARC-Record-ID\":\"<urn:uuid:12ed76fe-7f2f-4d8c-bcee-8dac9827cb7a>\",\"Content-Length\":\"278945\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:0d5cc716-7033-42be-b0cb-59942ed0134b>\",\"WARC-Concurrent-To\":\"<urn:uuid:8c42f98c-9cd8-4938-a0fc-864012b9e98b>\",\"WARC-IP-Address\":\"23.208.55.83\",\"WARC-Target-URI\":\"https://www.zebra.com/us/en/support-downloads/knowledge-articles/evm/how-do-i-transmit-the-code-39-check-digit.html\",\"WARC-Payload-Digest\":\"sha1:VA6CX3DAIGCPUBRQZ5CQJSPA253ICIGS\",\"WARC-Block-Digest\":\"sha1:H2MTAAH44BBEJTJNY5IWJUAKLLGC2O7M\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964361064.58_warc_CC-MAIN-20211201234046-20211202024046-00490.warc.gz\"}"}
http://models.rhaskell.org/firms/facebook	[ "## Analyst Listing\n\nThe following analysts provide coverage for the subject firm as of May 2016:\n\n Broker Analyst Analyst Email Raymond James Aaron Kessler [email protected] Nomura Research Anthony DiClemente [email protected] CRT Capital Group Arvind Bhatia [email protected] Sterne Agee & Leach Arvind Bhatia [email protected] Pivotal Research Group Brian Wieser [email protected] Bernstein Research Carlos Kirjner [email protected] Atlantic Equities James Cordwell [email protected] Oppenheimer Jason Helfstein [email protected] Cowen & Company John Blackledge [email protected] Needham Laura Martin [email protected] Rosenblatt Securities Martin Pyykkonen [email protected] Wedbush Securities Michael Pachter [email protected] Wells Fargo Securities Peter Stabler [email protected] William Blair Ralph Schackart [email protected] SunTrust Robinson Humphrey Robert S. Peck [email protected] JMP Securities Ronald V. Josey [email protected] Deutsche Bank Research Ross Sandler [email protected] Stifel Nicolaus Scott W. Devitt [email protected] Susquehanna Financial Group Shyam Patil [email protected] Credit Suisse Stephen Ju [email protected] Hilliard Lyons Stephen Turner [email protected] Cantor Fitzgerald Youssef H. Squali [email protected]\n\n## Primary Input Data", null, "## Derived Input Data\n\n### Equational Form\n\nNet Operating Profit Less Adjusted Taxes NOPLAT 4,046 8,120", null, "$NOPLAT\\, =\\, EBIT\\, x\\, (1 \\,-\\, Avg \\,\\,Tax\\,\\, Rate\\,\\, on\\,\\, EBIT)$\nFree Cash Flow FCF 6,076 11,617", null, "$FCF\\,=NOPLAT\\,+\\,Non-Cash\\,Expenses-\\Delta NWC\\,-\\,NCS$\nTax Shield TS 9 2", null, "$TS\\,=\\,Interest\\,\\,Paid\\,\\,x\\,\\, Avg \\,\\,Tax\\,\\,Rate\\,\\, on\\,\\, Pre-Tax\\,\\, Income$\nInvested Capital IC 47,482 62,086", null, "$IC\\,=\\,Fixed\\,\\,Operating\\,\\,Assets\\,\\,+\\,\\,Net\\,\\, Working\\,\\, Capital$\nReturn on Invested Capital ROIC 8.52% 13.08%", null, "$ROIC\\,=\\,\\frac { NOPLAT }{ IC }$\nNet Investment NetInv 10,667 16,946", null, "$NetInv\\,=\\,{ {IC}_{1}}-{{IC}_{0}}+Depreciation$\nInvestment Rate IR 263.63% 208.68%", null, "$IR\\,=\\,\\frac {NetInv}{NOPLAT}$\nWeighted Average Cost of Capital WACCMarket 16.81% 19.99%", null, "$WACC\\,=\\,\\frac { E }{ V } { R }_{ E }\\,+\\,\\frac { P }{ V } { R }_{ P }\\,+\\,\\frac { D }{ V } { R }_{ D }\\left( 1- Avg\\,\\, Tax\\,\\,Rate\\,\\,on\\,\\,Pre-Tax\\,\\,Income \\right)$\nWACCBook 8.58% 9.24%\nEnterprise value EVMarket 279,324 303,276", null, "$EV\\,=\\,Market\\,\\,Cap\\,\\,Equity\\,+\\,\\,Long\\,\\,Term\\,\\,Debt\\,-\\,Cash$\nEVBook 279,324 303,276\nEV/EBIT Multiple", null, "$\\frac{EV_{Market}}{EBIT}$ 44.87 24.28", null, "$EV/EBIT\\,=\\,\\frac { EV}{ EBIT}$\nLong-Run Growth g = IR x ROIC\n22.47% 27.29% Long-run growth rates of the income variable are used in the Continuing Value portion of the valuation models.\ng = %", null, "$\\Delta$ GDP 2.50% 2.50%\n\n## Valuation Model Outcomes\n\nThe outcomes presented in this study are the result of original input data, derived data, and synthesized inputs and, depending on the equational form of any particular valuation model, may result in irrelevant or implausible results. For example, in the event WACC < g, the value of this term, often found in the denominator of an equation’s continuation value term, will be expressly negative and may result in a negative overall valuation for the firm. In the event of a WACC < g relation, the model form as applied to the subject firm offers an irrelevant outcome.\n\n### Equational form\n\nKey Value Driver (NOPLAT) KVD (NOPLAT)", null, "${ Value }_{ DCF/KVD }=\\sum { \\frac { NOPLAT_{ t } }{ { \\left( 1+WACC \\right) }^{ t } } +\\frac { \\frac { { NOPLAT }_{ 1 }\\left( 1-\\frac { g }{ ROIC } \\right) }{ WACC-g } }{ { \\left( 1+WACC \\right) }^{ t } } }$", null, "Key Value Driver (FCF) KVD (FCF)", null, "${ Value }_{ DCF/KVD }=\\sum { \\frac { FCF_{ t } }{ { \\left( 1+WACC \\right) }^{ t } } +\\frac { \\frac { { NOPLAT }_{ 1 }\\left( 1-\\frac { g }{ ROIC } \\right) }{ WACC-g } }{ { \\left( 1+WACC \\right) }^{ t } } }$", null, "Free Cash Flow FCF", null, "${ Value }_{ DCF/FCF }=\\sum { \\frac { FCF_{ t } }{ { \\left( 1+WACC \\right) }^{ t } } +\\frac { \\frac { { FCF }_{ 1 }}{ WACC-g } }{ { \\left( 1+WACC \\right) }^{ t } } }$", null, "Economic Profit ECON π", null, "${ Value }_{ { ECON\\pi } }= I{ C }_{ 0 }+\\sum { \\frac { { IC }_{ t-1 }(ROI{ C }_{t}-WAC{C}_{t}) }{ { \\left( 1+WACC \\right) }^{ t } }+ \\frac {\\frac { I{C}_{0}\\ x\\ (ROI{C}_{1}\\ -\\ WAC{C}_{1}) }{ WACC-g } }{ { \\left( 1+WACC \\right) }^{ t } } }$", null, "Adjusted Present Value APV", null, "${ Value }_{ APV }=\\sum { \\frac { FCF_{ t } }{ { \\left( 1+{ k }_{ u } \\right) }^{ t } } +\\frac { \\frac { { FCF }_{ 1 }}{ { k }_{ u }-g } }{ { \\left( 1+{ k }_{ u } \\right) }^{ t } } } +\\sum { \\frac { { TS }_{ t } }{ { \\left( 1+{ k }_{ tax } \\right) }^{ t } } +\\frac { \\frac { { TS }_{ 1 }}{ { k }_{ tax }-g } }{ { \\left( 1+{ k }_{ tax } \\right) }^{ t } } }$", null, "Forward Market Multiple FMM", null, "${ Value }_{ DCF/FMM}=\\sum { \\frac { FCF_{ t } }{ { \\left( 1+WACC \\right) }^{ t } } +\\frac { { EBIT }_{ 1 }\\,{x}\\,{FMM}}{ { \\left( 1+WACC \\right) }^{ t } } }{\\,\\,\\,; \\,\\,FMM\\,=\\,\\frac{{EV}_{t=0}}{{EBIT}_{t=0}}}$", null, "" ]	[ null, "http://models.rhaskell.org/wp-content/uploads/2016/07/FB.bmp", null, "http://s0.wp.com/latex.php", null, "http://s0.wp.com/latex.php", null, "http://s0.wp.com/latex.php", null, "http://s0.wp.com/latex.php", null, "http://s0.wp.com/latex.php", null, "http://s0.wp.com/latex.php", null, "http://s0.wp.com/latex.php", null, "http://s0.wp.com/latex.php", null, "http://s0.wp.com/latex.php", null, "http://s0.wp.com/latex.php", null, "http://s0.wp.com/latex.php", null, "http://s0.wp.com/latex.php", null, "http://s0.wp.com/latex.php", null, "http://models.rhaskell.org/wp-content/uploads/2016/06/FB-KVD.bmp", null, "http://s0.wp.com/latex.php", null, "http://models.rhaskell.org/wp-content/uploads/2016/06/FB-KVD-FCF.bmp", null, "http://s0.wp.com/latex.php", null, "http://models.rhaskell.org/wp-content/uploads/2016/06/FB-FCF.bmp", null, "http://s0.wp.com/latex.php", null, "http://models.rhaskell.org/wp-content/uploads/2016/06/FB-ECON.bmp", null, "http://s0.wp.com/latex.php", null, "http://models.rhaskell.org/wp-content/uploads/2016/06/FB-APV.bmp", null, "http://s0.wp.com/latex.php", null, "http://models.rhaskell.org/wp-content/uploads/2016/06/FB-FMM.bmp", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.56719303,"math_prob":0.99382603,"size":3189,"snap":"2019-51-2020-05","text_gpt3_token_len":954,"char_repetition_ratio":0.09010989,"word_repetition_ratio":0.0,"special_character_ratio":0.25807464,"punctuation_ratio":0.139823,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99807996,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50],"im_url_duplicate_count":[null,5,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,5,null,null,null,5,null,null,null,5,null,null,null,5,null,null,null,5,null,null,null,5,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-28T16:07:51Z\",\"WARC-Record-ID\":\"<urn:uuid:366f08af-1e5f-419b-853e-6d89e8efd277>\",\"Content-Length\":\"63967\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:2d7a1868-719a-4119-bcb1-bbc35f4224b6>\",\"WARC-Concurrent-To\":\"<urn:uuid:871603a8-9b79-4518-8284-885f37ce58bb>\",\"WARC-IP-Address\":\"74.208.236.206\",\"WARC-Target-URI\":\"http://models.rhaskell.org/firms/facebook\",\"WARC-Payload-Digest\":\"sha1:U23ORA363SQSL37JFSWI6OHRL5IAUTAS\",\"WARC-Block-Digest\":\"sha1:YWK5Y6PVI6JTCUF4QIMJQ55EQDMZLGB3\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579251779833.86_warc_CC-MAIN-20200128153713-20200128183713-00532.warc.gz\"}"}
http://e.biohackers.net/Polymorphic_information_content	[ "# Polymorphic information content#Find similar titles\n\nPIC has become the most widely applied formula for genetic studies to measure the information content of molecular markers.\n\n마커의 heterozygosity는 개체별로 해당 좌위가 hetero한지를 설명함\n\n$$H = 1 - \\sum\\limits_{i = 1}^{l} {P_{i}^{2} }$$\n\n마커의 PIC는 집단내에서 얼마나 다형인지를 설명함 - 마커의 검정력\n\n$${\\mathbf{PIC}} = 1 - \\sum\\limits_{i = 1}^{l} {\\mathop P\\nolimits_{i}^{2} } - \\sum\\limits_{i = 1}^{l - 1} {\\sum\\limits_{j = i + 1}^{l} 2 } \\mathop P\\nolimits_{i}^{2} \\mathop P\\nolimits_{j}^{2} = 1 - \\sum\\limits_{i = 1}^{l} {\\mathop P\\nolimits_{i}^{2} } - \\left( \\sum\\limits_{i = 1}^{l} {\\mathop P\\nolimits_{i}^{2} } \\right)^{2} + \\sum\\limits_{i = 1}^{l} {\\mathop P\\nolimits_{i}^{4} }$$\n\n관련논문" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.5412517,"math_prob":0.9998882,"size":1209,"snap":"2020-45-2020-50","text_gpt3_token_len":479,"char_repetition_ratio":0.17925312,"word_repetition_ratio":0.08648649,"special_character_ratio":0.37386268,"punctuation_ratio":0.07009346,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9999392,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-10-22T12:54:15Z\",\"WARC-Record-ID\":\"<urn:uuid:29bb24dd-3f75-4186-9264-ff7789e11aa1>\",\"Content-Length\":\"14908\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:49d02dd6-8608-4978-bcaf-52ebd6a8cccd>\",\"WARC-Concurrent-To\":\"<urn:uuid:1170aec3-0b88-4ab8-b71b-abdf89521d9c>\",\"WARC-IP-Address\":\"142.250.73.243\",\"WARC-Target-URI\":\"http://e.biohackers.net/Polymorphic_information_content\",\"WARC-Payload-Digest\":\"sha1:47774M5OSN7I6GALT2Z464EDN3JUYQ7I\",\"WARC-Block-Digest\":\"sha1:PAWO6HEROREWZJLCUDM5JBCUAZSUZ5AD\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-45/CC-MAIN-2020-45_segments_1603107879537.28_warc_CC-MAIN-20201022111909-20201022141909-00708.warc.gz\"}"}
https://www.vedantu.com/question-answer/ways-are-there-to-arrange-the-letters-in-class-11-maths-cbse-5ee34d8c63ff30423e34350e	[ "Courses\nCourses for Kids\nFree study material\nFree LIVE classes\nMore", null, "LIVE\nJoin Vedantu’s FREE Mastercalss\n\n# How many ways are there to arrange the letters in the word GARDEN with the vowels in alphabetical order?(a) 120(b) 240(c) 360(d) 480", null, "Verified\n361.2k+ views\nHint: Divide the total ways of arrangement of all the letters by ways of arrangement of vowels.\n\nWe are given a word GARDEN. We have to find the number of ways to arrange the letters in this word with the vowels in alphabetical order.\nThe total number of letters in given word = 6 {G, A, R, D, E, N}\nTherefore, the total ways in which all letters can be arranged is\n$\\underline{6}\\times \\underline{5}\\times \\underline{4}\\times \\underline{3}\\times \\underline{2}\\times \\underline{1}$\n1. There are a total 6 ways to fill the first place.\n2. Now, the second place can be filled by remaining 5 letters\n3. Third place can be filled by remaining 4 letters\n4. Similarly, fourth, fifth and sixth places can be filled by remaining 3, 2 and 1 letters respectively.\nTherefore, we get total ways in which all letters can be arranged\n\\begin{align} & =6! \\\\ & =6\\times 5\\times 4\\times 3\\times 2\\times 1 \\\\ & =720 \\\\ \\end{align}\nNow, total number of vowels in the given word = 2 (A, E)\nThe total ways in which these vowels can be arranged $=\\underline{2}\\times \\underline{1}$\n1. There are 2 ways to fill the first place.\n2. Now second place can be filled in just one way.\nTherefore, we get the total ways in which these vowels can be arranged = 2! = 2\nNow, the total number of ways in which the letters can be arranged such that vowels are in alphabetical order are\n\\begin{align} & =\\dfrac{\\text{Total number of ways of arrangements of all letters}}{\\text{Total number of ways of arrangement of vowels}} \\\\ & =\\dfrac{6!}{2!}=\\dfrac{720}{2}=360 \\\\ \\end{align}\nHence, option (c) is correct.\n\nNote: Students can also solve this question in this way.\n_ _ _ _ _ _\nFirst we select 2 places out of 6 places in which vowels will be arranged in alphabetical order that is only 1 way (A, E)\nTherefore, the total number of ways of selecting 2 out of 6 places for vowels in alphabetical order are $6{{C}_{2}}$.\n\\begin{align} & =\\dfrac{6!}{2!4!}=\\dfrac{6\\times 5\\times 4!}{2\\times 4!} \\\\ & =3\\times 5=15\\text{ ways} \\\\ \\end{align}\nNow, remaining 4 words (G, R, D, N) will be arranged in 4 remaining places such that number of ways will be\n$=4!=4\\times 3\\times 2\\times 1=24$\nTherefore, the total number of ways of arranging the letters of word such that the vowels are in alphabetical order\n\\begin{align} & =6{{C}_{2}}.4! \\\\ & =15\\times 24 \\\\ & =360\\text{ ways} \\\\ \\end{align}\nHence, option (c) is correct.\n\nLast updated date: 22nd Sep 2023\nTotal views: 361.2k\nViews today: 10.61k" ]	[ null, "https://www.vedantu.com/cdn/images/seo-templates/seo-qna.svg", null, "https://www.vedantu.com/cdn/images/seo-templates/green-check.svg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7882727,"math_prob":0.9933804,"size":5284,"snap":"2023-40-2023-50","text_gpt3_token_len":1564,"char_repetition_ratio":0.16193181,"word_repetition_ratio":0.8789238,"special_character_ratio":0.3154807,"punctuation_ratio":0.11040146,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9955691,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-25T10:45:13Z\",\"WARC-Record-ID\":\"<urn:uuid:2b2a6d3e-d57a-4d49-969e-57d97c9e9f6c>\",\"Content-Length\":\"154313\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:bb2f8f17-d72c-483a-8f9d-fa53f5be2de3>\",\"WARC-Concurrent-To\":\"<urn:uuid:969df173-14ec-43c3-8fde-5ee0dc915116>\",\"WARC-IP-Address\":\"108.138.64.12\",\"WARC-Target-URI\":\"https://www.vedantu.com/question-answer/ways-are-there-to-arrange-the-letters-in-class-11-maths-cbse-5ee34d8c63ff30423e34350e\",\"WARC-Payload-Digest\":\"sha1:IZKQ5YSNTAOYB7SLZOCVM2RVB7E42EKV\",\"WARC-Block-Digest\":\"sha1:SGJ55BYMUHGBRMW2ERQ2T3ZSJGHLY7IM\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233508959.20_warc_CC-MAIN-20230925083430-20230925113430-00547.warc.gz\"}"}
https://catholiccharitiesdal.org/inspiration-array-worksheets-for-2nd-graders/	[ "Awesome Array Worksheets For 2nd Graders\n\nArrays Worksheet 3rd Grade. 2 Students explore arrays as a means of solving multiplication problems.", null, "Multiplication arrays worksheets for 2nd and 3rd grade free pdf.", null, "Array worksheets for 2nd graders. This second grade math resource includes printable and digital math worksheets that give your second graders practice with arrays and repeated addition 2OA4. Free printable repeated addition and multiplication arrays worksheets are great for learning and understanding how to write the multiplication equations. Free grade 2 math worksheets.\n\nAccess the most comprehensive library of K-8 resources for learning at school and at home. Your desktop choose images below and share printable 3rd grade math worksheets wallpapers if you love it help you third grader master new skills in reading writing grammar math science and social stu s with. Students will interact with a digital model or manipulative to record their responses.\n\nImage result for teaching arrays 3rd grade Third grade math 3rd grade math Second grade math. Start with some of the easier 1-digit number worksheets. Kids will practice writing number sentences for arrays before applying their knowledge to array word problems.\n\nThis second grade math resource includes printable and digital math worksheets that give your second graders practice with arrays and repeated addition 2OA4. Be sure to introduce multiplication by explaining its relationship to addition. Thousands of parents and educators are turning to the kids learning app that makes real learning truly fun.\n\nUse this worksheet to practice describing an array with repeated addition. This page contains all our printable worksheets in section Multiplication and Division of Second Grade MathAs you scroll down you will see many worksheets for multiplication as repeated addition multiplication and addition sentences skip – count equal groups model with arrays multiply in any order multiply with 1 and 0 times table to 12 multiplication sentences model. All Multiplication Worksheets for 2nd Graders.\n\nThese Common Core aligned math quick checks are perfect for morning work assessment homework review fast-finisher activities exit ticke. 2nd grade Multiplication with Arrays Printable Worksheets. Use these free multiplication worksheets to help your second grader practice and build fluency in multiplication.\n\nAll worksheets are printable pdf documents. Cover the standards with problem solving repeated addition and picture representations. Students will create an array with the given number of rows and columns.\n\nThen start with multiplying by one or even zero. After your second graders connect all the dots and finish coloring they will each have a beautiful. Understand Rows in an Array.\n\nWith row and columns students from 3rd grade and 2nd grade able to identify the concepts of arrays. Multiplication arrays worksheets for 2nd and 3rd grade free pdf. Featured here is a compilation of printable division array worksheets designed to familiarize kids of grade 3 and grade 4 with the concept of division.\n\nClick the checkbox for the options to print and add to Assignments and Collections. The worksheet invites learners to work with a set of problems on equal groups and find the missing numbers in math sentences. Access the most comprehensive library of K-8 resources for learning at school and at home.\n\nOur grade 2 math worksheets emphasize numeracy as well as a conceptual understanding of math concepts. Second Grade Math Worksheets. Our 3rd grade one digit multiplication worksheets are designed by teachers to help children build a strong foundation in math.\n\nHelp you second grader master new skills in reading writing grammar math science and social studies with our collection of second grade worksheets. These Common Core aligned math quick checks are perfect for morning work assessment homework review fast-finisher activities exit ticke. Arrays can be used to better understand multiplication.\n\nTry Kids Academy with 3-day FREE TRIAL. These worksheets cover an array of topics including historical figures mathematical operations. Place Value Rounding.\n\nExplore Building Arrays Gr. Common Core State Standards. Multiplication Arrays Multiplication Arrays Worksheets Multiplication Arrays Worksheets Extension of CCSS 2OA4 CCSS 3OA3 Common Core State Standards.\n\nUse addition to find the total number of objects arranged in rectangular arrays with up to 5 rows and up to 5 columns. Students are asked to view each array and fill in the appropriate value for each of three questions related to the example. The number of rows of an array can represent the number of groups while the number of columns represents the number of objects per group.\n\nBuilding a strong foundation in equal groups is an important step in helping your child become proficient and confident. The multiplication sentence is the number of rows times the number of columns. The game has a set of problems logically crafted for your child so that they practice the previously learned concepts of arrays.\n\n2nd grade 3rd grade math coloring worksheets. Perfect for 2nd-graders or review for 3rd. CCSS 2OA4 and extension CCSS 3OA3 Use multiplication and division within 100 to solve.\n\nNov 24 2016 – Easily teach beginning arrays with plenty of practice pages task cards and an assessment. Once your students understands this introduce. 20 Envision Math 2nd Grade Worksheets.\n\nWrite an equation to express the total as a sum of equal addends. Arrays In Math 2Nd Grade Worksheets. They are great for the classroom homeschool or after school activity and help students build the.\n\nThe arrays covered in this worksheet include a tomato garden a bookshelf a model car case and a photo album.", null, "", null, "", null, "", null, "" ]	[ null, "https://i.pinimg.com/originals/c0/ab/2f/c0ab2fa07faee974504e0dde47d0b635.jpg", null, "https://i.pinimg.com/originals/ab/a9/f3/aba9f31939990dde3b3b6c733f54b61e.jpg", null, "https://i.pinimg.com/originals/3a/7a/b1/3a7ab1cc8b0fbf5b9d231f1b106419c3.jpg", null, "https://i.pinimg.com/originals/0c/94/9f/0c949f8c3b27924d42460c3b1ada922f.jpg", null, "https://i.pinimg.com/originals/ab/a9/f3/aba9f31939990dde3b3b6c733f54b61e.jpg", null, "https://i.pinimg.com/originals/93/8a/7c/938a7c997a2e3e4c36b1221d2e2f3a5f.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.897183,"math_prob":0.73911816,"size":6695,"snap":"2022-05-2022-21","text_gpt3_token_len":1273,"char_repetition_ratio":0.19414139,"word_repetition_ratio":0.1178744,"special_character_ratio":0.17804332,"punctuation_ratio":0.06115108,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9760448,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-01-20T00:01:17Z\",\"WARC-Record-ID\":\"<urn:uuid:8421e4d7-c70b-4c4c-ae2d-7e8c240637b2>\",\"Content-Length\":\"41697\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:286fff3d-dd6f-4adf-9d2d-864bff80605b>\",\"WARC-Concurrent-To\":\"<urn:uuid:5e2ebc39-fede-406b-8ada-aff961d34db4>\",\"WARC-IP-Address\":\"172.67.133.200\",\"WARC-Target-URI\":\"https://catholiccharitiesdal.org/inspiration-array-worksheets-for-2nd-graders/\",\"WARC-Payload-Digest\":\"sha1:RTB7UHGXTTMY63AGMCLEACPMQAMKI4HK\",\"WARC-Block-Digest\":\"sha1:ZEAFZ43FNPYYJWBYHN6GCSZINBEUY6H7\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-05/CC-MAIN-2022-05_segments_1642320301592.29_warc_CC-MAIN-20220119215632-20220120005632-00570.warc.gz\"}"}
https://www.aqion.de/site/71	[ "# Examples\n\nValidation of Water Samples\n\n• Problem. How to validate a water sample (charge balance and other parameters)?\n• Problem. What is the equilibrium composition and alkalinity of a water sample with incomplete analytical data?\n• Problem. What is the optimal parameter (cation or anion) for charge balance adjustment?\n\nThe Carbonate System\n\n• Problem. What are the equivalence points of the following three solutions: (i) 0.001 molar H2CO3, (ii) 0.001 molar NaHCO3, and (iii) 0.001 molar Na2CO3?\n• Problem. Carbonate Species vs. pH – Closed System\n• Problem. Carbonate Species vs. pH – Open System\n• Problem. Calcite Carbonate System\n\npH Calculations\n\n• Problem. What is the pH of a given phosphate buffer before/after addition of NaOH and HCl?\n• Problem. Given is a 0.001 molar NH4Cl solution. What is the pH after addition of 0.2 mmol NH3?\n• Problem. What is the pH of an acid rain sample?\n• Problem. What is the pH of an extremely dilute acid (10-8 M HCl)?\n• Problem. Given is a NaOH solution. How does the pH change when the solution is in contact with air for a long time? How much CO2 from air is captured in the solution?\n\nAcids and Bases\n\n• Problem. What is the pH and the concentration of hydrogen sulfate of a 1.2 mM sulfuric acid?\n• Problem. Addition of 0.2 mM HCl to a given water sample.\n• Problem. Strong acids (HCl, HNO3, H2SO4) without and with ionic activity corrections.\n• Problem. Phosphoric acid as a polyprotic acid.\n• Problem. Phosphoric acid and sodium phosphates.\n• Problem. Calculate the pH value for all acids given here for three different concentrations (1, 10, and 100 mM).\n\nDilution and Mixing\n\n• Problem. 40 mL of 2.5 molar H2SO4 solution will be diluted to 2 L. What is the pH of the final solution?\n• Problem. 100 mL of 0.01 molar HCl and 50 mL of 0.05 molar NaCl are mixed. What is the pH of the final solution?\n\nTitration\n\n• Problem. Titration of HCl with NaOH.\n• Problem. Carbonate Species vs. pH – Closed System\n• Problem. Carbonate Species vs. pH – Open System\n\nDosage / Addition of Chemicals\n\n• Problem. Given is a groundwater with pH 6.9. How much HNO3 and how much H2SO4 are needed to obtain pH 3?\n• Problem. Given is a water sample with pH 7.9. What is the pH after addition of 0.3 mM KOH? What is the impact of calcite precipitation?\n\nMineral Phases\n\n• Problem. Dissolution of Calcite in a closed and in an open CO2 system\n• Problem. Dissolution of Gypsum\n• Problem. Temperature dependence of calcite precipitation\n\nEnvironmental Chemistry\n\n• Problem. Given is pH and total ammonia. What is the NH3 concentration (which is toxic to fish)?\n• Problem. What is the chemical composition of pristine rainwater?\n• Problem. What is the pH of a given acid rain sample?" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.82490695,"math_prob":0.9563089,"size":2718,"snap":"2022-40-2023-06","text_gpt3_token_len":731,"char_repetition_ratio":0.22660281,"word_repetition_ratio":0.14435147,"special_character_ratio":0.26085356,"punctuation_ratio":0.15970962,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9962691,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-10-02T19:42:13Z\",\"WARC-Record-ID\":\"<urn:uuid:3c4015a7-a695-4b41-ab6c-e6d2e71a3120>\",\"Content-Length\":\"9641\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9787291b-1693-43ed-a423-4fe27d067c2b>\",\"WARC-Concurrent-To\":\"<urn:uuid:1f4b2770-63ab-4428-80d9-c7fd2383b0ba>\",\"WARC-IP-Address\":\"164.92.237.53\",\"WARC-Target-URI\":\"https://www.aqion.de/site/71\",\"WARC-Payload-Digest\":\"sha1:BAN2PKCXNAPPZCZ5RMSZWD2A2GWM6ZOM\",\"WARC-Block-Digest\":\"sha1:QRVB6GALBAOITTBTRQJYXVI6NHGNVW4W\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030337339.70_warc_CC-MAIN-20221002181356-20221002211356-00529.warc.gz\"}"}
https://ir.cwi.nl/pub/29225	[ "Following the breakthrough work of Tardos in the bit-complexity model, Vavasis and Ye gave the first exact algorithm for linear programming in the real model of computation with running time depending only on the constraint matrix. For solving a linear program (LP) $\\max\\, c^\\top x,\\: Ax = b,\\: x \\geq 0,\\: A \\in \\mathbb{R}^{m \\times n}$, Vavasis and Ye developed a primal-dual interior point method using a 'layered least squares' (LLS) step, and showed that $O(n^{3.5} \\log (\\bar{\\chi}_A+n))$ iterations suffice to solve (LP) exactly, where $\\bar{\\chi}_A$ is a condition measure controlling the size of solutions to linear systems related to $A$. Monteiro and Tsuchiya, noting that the central path is invariant under rescalings of the columns of $A$ and $c$, asked whether there exists an LP algorithm depending instead on the measure $\\bar{\\chi}^_A$, defined as the minimum $\\bar{\\chi}_{AD}$ value achievable by a column rescaling $AD$ of $A$, and gave strong evidence that this should be the case. We resolve this open question affirmatively. Our first main contribution is an $O(m^2 n^2 + n^3)$ time algorithm which works on the linear matroid of $A$ to compute a nearly optimal diagonal rescaling $D$ satisfying $\\bar{\\chi}_{AD} \\leq n(\\bar{\\chi}^)^3$. This algorithm also allows us to approximate the value of $\\bar{\\chi}_A$ up to a factor $n (\\bar{\\chi}^)^2$. As our second main contribution, we develop a scaling invariant LLS algorithm, together with a refined potential function based analysis for LLS algorithms in general. With this analysis, we derive an improved $O(n^{2.5} \\log n\\log (\\bar{\\chi}^_A+n))$ iteration bound for optimally solving (LP) using our algorithm. The same argument also yields a factor $n/\\log n$ improvement on the iteration complexity bound of the original Vavasis-Ye algorithm.\n\nNetworks and Optimization\n\nDadush, D.N, Huiberts, S, Natura, B, & Végh, L.A. (2019). A scaling-invariant algorithm for linear programming whose running time depends only on the constraint matrix." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8059383,"math_prob":0.9988385,"size":2201,"snap":"2022-40-2023-06","text_gpt3_token_len":563,"char_repetition_ratio":0.12608102,"word_repetition_ratio":0.06811146,"special_character_ratio":0.2521581,"punctuation_ratio":0.09975669,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99981624,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-10-04T10:36:40Z\",\"WARC-Record-ID\":\"<urn:uuid:bc2081d8-d536-4f30-8add-ab61285d7b2f>\",\"Content-Length\":\"21448\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9d22b8d6-31f4-4c5e-9475-8985d864afef>\",\"WARC-Concurrent-To\":\"<urn:uuid:b5ace8ab-76f9-4065-b14f-4c92b44c8177>\",\"WARC-IP-Address\":\"5.79.73.6\",\"WARC-Target-URI\":\"https://ir.cwi.nl/pub/29225\",\"WARC-Payload-Digest\":\"sha1:P6O3VJX3RJHGJGZPL2V57AWWKAG4PJC5\",\"WARC-Block-Digest\":\"sha1:GBBMSLL7OXMGKU4VEDJVMFCGAPSBPYDN\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030337490.6_warc_CC-MAIN-20221004085909-20221004115909-00377.warc.gz\"}"}
https://rupress.org/jgp/article/112/3/263/11020/Effects-of-Partial-Sarcoplasmic-Reticulum-Calcium	[ "Resting sarcoplasmic reticulum (SR) Ca content ([CaSR]R) was varied in cut fibers equilibrated with an internal solution that contained 20 mM EGTA and 0–1.76 mM Ca. SR Ca release and [CaSR]R were measured with the EGTA–phenol red method (Pape et al. 1995. J. Gen. Physiol. 106:259–336). After an action potential, the fractional amount of Ca released from the SR increased from 0.17 to 0.50 when [CaSR]R was reduced from 1,200 to 140 μM. This increase was associated with a prolongation of release (final time constant, from 1–2 to 10–15 ms) and of the action potential (by 1–2 ms). Similar changes in release were observed with brief stimulations to −20 mV in voltage-clamped fibers, in which charge movement (Qcm) could be measured. The peak values of Qcm and the fractional rate of SR Ca release, as well as their ON time courses, were little affected by reducing [CaSR]R from 1,200 to 140 μM. After repolarization, however, the OFF time courses of Qcm and the rate of SR Ca release were slowed by factors of 1.5–1.7 and 6.5, respectively. These and other results suggest that, after action potential stimulation of fibers in normal physiological condition, the increase in myoplasmic free [Ca] that accompanies SR Ca release exerts three negative feedback effects that tend to reduce additional release: (a) the action potential is shortened by current through Ca-activated potassium channels in the surface and/or tubular membranes; (b) the OFF kinetics of Qcm is accelerated; and (c) Ca inactivation of Ca release is increased. Some of these effects of Ca on an SR Ca channel or its voltage sensor appear to be regulated by the value of [Ca] within 22 nm of the mouth of the channel.\n\n## Introduction\n\nPape et al. (1995) described a method for the direct measurement of sarcoplasmic reticulum (SR)1 Ca release and [CaSR]R in frog cut muscle fibers equilibrated with an internal solution that contains 20 mM EGTA and phenol red. With 1.76 mM Ca in the internal solution, the value of [CaSR]R varied threefold, between 1,391 and 4,367 μM, whereas that of the fractional amount of [CaSR]R released from the SR by a single action potential (f1) remained relatively constant, between 0.13 and 0.17. During the course of those experiments, some of the fibers were partially depleted of Ca by decreasing the concentration of Ca in the internal solution and by repetitive stimulation. After the value of [CaSR]R fell below 1,000–1,200 μM, the value of f1 began to increase and, with [CaSR]R = 150 μM, it was 0.5. This increase in f1 helped stabilize the amount of Ca released from the SR. Since such stabilization may be important in helping to maintain normal contractile activation, experiments were undertaken to determine the factors involved in the regulation of f1 by [CaSR]R.\n\nThis article describes our results. They show that three factors participate in this regulation: (a) Ca-activated potassium channels in the surface and/or tubular membranes that, when opened, can shorten the duration of the action potential; (b) the OFF kinetics of Qcm that determines the rate of removal of voltage-dependent activation during the repolarization phase of an action potential; and (c) inhibition of SR Ca release caused by an elevation of myoplasmic free [Ca] (called Ca inactivation of Ca release).\n\nThe discussion presents a comparison of the effects on SR Ca release produced by the addition of high affinity Ca buffers to the myoplasm and by a reduction of SR Ca content (which is expected to reduce the single-channel Ca flux). appendixes b and c give derivations of steady state and transient solutions of Δ[Ca] near a point source in the presence of different kinds of Ca buffers. With these derivations, a shows that some of the effects of Ca on release appear to be controlled by the concentration of Ca ions within a region ≤22 nm from the mouth of the channel. Such effects of Ca could be mediated by a receptor or receptors located on the SR Ca channel (including the foot structure) and/or its voltage sensor. The appendixes also show that, in the absence of extrinsic Ca buffers, the concentration of Ca ions near the mouth of a channel can contain a large contribution from Ca ions that are released from neighboring channels and that this contribution is markedly decreased by the addition of 0.5– 1.0 mM fura-2 or 20 mM EGTA to the myoplasm. The general features of this theoretical analysis may apply to other nonmuscle cells.\n\n## Materials And Methods\n\nAction potential and voltage-clamp experiments were carried out on frog cut muscle fibers mounted in a double Vaseline-gap chamber and equilibrated with an internal solution that contained 20 mM EGTA and phenol red. The theoretical and experimental basis for the measurement of SR Ca release with EGTA– phenol red is described in Pape et al. (1995). After stimulation by an action potential, EGTA is expected to capture almost all (∼96%) of the Ca released from the SR, to capture it rapidly (<0.1 ms), and to exchange it for protons with a 1:2 stoichiometry. Δ[CaT] is calculated from the associated decrease in myoplasmic pH (which is monitored with phenol red) and the value of myoplasmic buffering power (which was determined in a separate set of experiments with EGTA, phenol red, and fura-2). This method for the measurement of SR Ca release has the advantage that it is direct and little influenced by the presence of intrinsic myoplasmic Ca buffers. Indeed, for reasons that are not completely clear, the amount of Ca that appears to be complexed by the Ca-regulatory sites on troponin after an action potential is negligibly small and not statistically significant, 2.9 μM (SEM, 3.8 μM) out of a total site concentration of 240 μM (column 8 in Table II in Pape et al., 1995). The expectation that the EGTA–phenol red method gives an accurate measurement of the amplitude and time course of SR Ca release is validated by the finding that values of the peak rate of release and time to half-peak measured with this method are not significantly different from those estimated from Ca transients measured with low affinity Ca indicators (antipyrylazo III, tetramethylmurexide, and purpurate-3,3′diacetic acid [PDAA]) in cut fibers with only 0.1 mM EGTA (Pape et al., 1995). On the other hand, the half-width of the rate of SR Ca release measured with EGTA–phenol red is 0.7–0.9 ms longer than that estimated with 0.1 mM EGTA, an increase likely caused by the ability of 20 mM EGTA to reduce Ca inactivation of Ca release (Pape et al., 1995). Another feature of the EGTA–phenol red method is that [CaSR]R can be monitored routinely by the measurement of Δ[CaT] after a train of action potentials or after a long lasting step depolarization that releases essentially all of the readily releasable Ca from the SR. Additional information about the EGTA–phenol red method is given in Pape et al. (1995).\n\nThe methods used for the determination of Qcm are described in Chandler and Hui (1990), Hui and Chandler (1990, 1991), and Jong et al. (1995b). For the estimation of Icm associated with a brief depolarization (see Fig. 8 A), the final level of the OFF ITESTICONTROL was used for the template of the OFF ionic current, and the template of the ON ionic current was adjusted to make the amplitudes of ON and OFF charge equal.\n\nIn the action potential experiments, a K-glutamate internal solution was used in the end pools. It contained (mM): 45 K-glutamate, 20 EGTA as a combination of K and Ca salts, 6.8 MgSO4, 5.5 Na2-ATP, 20 K2-creatine phosphate, 5 K3-phospho(enol)pyruvate, and 5 MOPS, with pH adjusted to 7.0 by the addition of KOH. The concentration of Ca-complexed EGTA was 0, 0.44, or 1.76 mM; at pH 7.0, the calculated concentration of free Ca was 0, 0.008, or 0.036 μM, respectively, and the calculated concentration of free Mg was 1 mM. A NaCl Ringer's solution was used in the central pool in the action potential experiments (see Table I of Irving et al., 1987).\n\nIn the voltage-clamp experiments, a Cs-glutamate internal solution was used in the end pools. It was similar to the K-glutamate internal solution except that Cs replaced K and Na. The composition of the external solution used in the central pool was 110 mM tetraethylammonium-gluconate, 10 mM MgSO4, 10 mM MOPS, and 1 μM tetrodotoxin; it was nominally Ca-free and had a pH of 7.1.\n\nIn the experiments, the sarcomere length of the fibers was 3.3– 3.7 μm, the holding potential was −90 mV, and the temperature was 14–15°C. The difference between the mean values of two sets of results was assessed with Student's two-tailed t test and considered to be significant if P < 0.05.\n\n## Results\n\n### Effect of a Reduction of [CaSR]R on SR Ca Release During a Train of Action Potentials\n\nFig. 1 A shows traces from a fiber that was initially equilibrated with an end-pool solution that contained 20 mM EGTA and 1.76 mM Ca. The top traces show superimposed voltage records taken during four trials in which the fiber was stimulated to give a train of action potentials with different amounts of Ca inside the SR. The other four traces show the associated Δ[CaT] signals.\n\nThe first Δ[CaT] signal (Fig. 1 A, a) was obtained 125 min after saponin treatment. The first action potential elicited an abrupt increase in Δ[CaT] of 445 μM. This amount of release is similar to that obtained from other fibers equilibrated with the same internal solution and having a similar value of [CaSR]R (Pape et al., 1995). On the other hand, it may be somewhat larger than that obtained in fibers not equilibrated with EGTA because of a reduction of Ca inactivation of Ca release caused by the ability of EGTA to reduce myoplasmic free [Ca]. Ca inactivation of Ca release represents an inhibition of Ca flux through SR Ca channels that is produced by an increase in myoplasmic free [Ca] (Baylor et al., 1983; Simon et al., 1985, 1991; Schneider and Simon, 1988). According to Schneider and Simon (1988) (see also Jong et al., 1995a), its properties can be described by a model in which Ca equilibrates rapidly with a receptor on the release channel, after which the Ca receptor–channel complex is able to undergo a slower transition to an inactivated state.\n\nThe amplitude of the Δ[CaT] signal elicited by the second action potential in Fig. 1 A, a, was only half that elicited by the first action potential. Part of the reduction can be attributed to a decrease in [CaSR] after the first response. Most of the reduction, however, is probably caused by Ca inactivation of Ca release produced by the first response, even though the internal solution contained 20 mM EGTA (Jong et al., 1995a). The responses elicited by the third and subsequent action potentials progressively decreased, and, after 20–30 action potentials, Δ[CaT] had reached a maximal value of 2,452 μM, which is taken for the value of [CaSR]R.\n\n4 min after Fig. 1 A, a, was taken, Ca was removed from the end-pool solution. b–d were obtained after the SR Ca content had been reduced by repeated stimulation with a train of action potentials, delivered first every 5 min and then every 2 min. During this period, the value of [CaSR]R decreased from 2,452 μM in a to 1,760 μM in b, 1,224 μM in c, and 708 μM in d.\n\nFig. 1 B shows voltage and Δ[CaT] during the first 24 action potentials in A; the Δ[CaT] signals have been normalized by [CaSR]R to give a maximal value of unity and have been plotted on expanded horizontal and vertical gains. f1 was 0.181 in a and 0.194 in b. These values are slightly larger than the mean value of 0.144 (SEM, 0.004) found by Pape et al. (1995) in 12 fibers with [CaSR]R = 1,391–4,367 μM. The normalized increase in Δ[CaT] produced by the first few action potentials became progressively larger from a to d, indicating a more rapid fractional depletion of Ca from the SR. In addition, the stepwise appearance of the Δ[CaT] signal after each action potential became progressively more rounded from a to d. The experiments described below were undertaken to study possible causes of these changes, such as a reduction of Ca inactivation of Ca release associated with partial SR Ca depletion.\n\n### Changes in the Action Potential when [CaSR]R Is Varied between 150 and 1,200 μM\n\nThe purpose of the experiment described in Figs. 24 was to reduce [CaSR]R to values smaller than those used in Fig. 1 and to study the effect on the action potential, as described in this section, and on SR Ca release, as described in the following section. The value of [CaSR]R was varied reversibly between 140 and 1,154 μM by the use of 0.44 mM Ca in the end-pool solution and by variation of the recovery period between successive trials between 0.5 and 10 min.\n\nFig. 2 A, top, shows superimposed voltage records from four trials. A single action potential was followed by a 150-ms recovery period. Then, a train of action potentials was used to deplete the SR of its readily releasable Ca so that the value of [CaSR]R could be determined. During the first part of each trace, the interval of time between data points was 0.12 ms, which is sufficiently brief to resolve the time courses of the first action potential and of the associated rate of SR Ca release. After the recovery period, the interval of time between data points was increased to 0.6 ms, which allowed resolution of the relatively slow changes in Δ[CaT], but not the time courses of the individual action potentials in the train.\n\nFig. 2,A, a–d, show the Δ[CaT] signals associated with the four trials. The value of [CaSR]R was decreased from 512 μM in a to 296 μM in b to 146 μM in c by the use of progressively shorter recovery periods between trials (see legend). The recovery period was increased after c, and, when d was taken, the value of [CaSR]R was 876 μM. More information about the Δ[CaT] signals is given in the legend of Fig. 2 and in the following section.\n\nIn Fig. 2 B, top, superimposed traces show the first action potential of each trial, plotted on expanded vertical and horizontal gains. Action potentials a and d are similar to each other and have a shorter duration than b, which has a shorter duration than c.\n\nThe difference between action potentials is seen more clearly in Fig. 2 C, c and d, where the time scale has been expanded. The top pair of traces shows that, during the first 2 ms after stimulation, the two action potentials were virtually identical. During the next millisecond, the traces started to diverge as the slope of d became more negative than that of c.\n\nFor a membrane action potential in an axon with constant surface capacitance (C), a more negative slope would be expected to have been caused by an increase in outward ionic current equal to −CdV/dt (Hodgkin and Huxley, 1952). In a muscle fiber, however, the situation is complicated by the presence of the transverse tubular system. As a consequence, if C represents the capacitance of the surface membrane of the fiber, −CdV/dt would equal the sum of the ionic current through the surface membrane and the current from the mouths of the transverse tubules where they invaginate from the surface membrane. Another complication of the muscle fiber experiment is that the ideal of a membrane action potential is only approximately realized in a fiber mounted in a double Vaseline-gap chamber. In spite of these complications, however, the idea that the more negative slope of d is caused by an increase in outward ionic current across the surface and/or tubular membranes is still expected to apply. During the period when c and d diverged, the maximal value of the difference between the derivatives of the traces (not shown) occurred 3.2 ms after stimulation and was 17 mV/ms. This indicates that the outward ionic current at this time was larger in d than in c by ∼17 mV/ms = 17 μA/μF. With the internal and external solutions used in this experiment, such an outward ionic current could have been carried by potassium ions leaving the fiber or chloride ions entering the fiber.\n\nAt about the time that action potentials c and d in Fig. 2 started to diverge, the associated dΔ[CaT]/dt signals became noticeably different; dΔ[CaT]/dt represents the estimated rate of SR Ca release. A possible explanation for the extra outward ionic current in action potential d is that SR Ca release produced an increase in myoplasmic free [Ca], which is expected to be approximately proportional to the rate of SR Ca release (Pape et al., 1995), and that this, in turn, activated ionic channels permeable to potassium or chloride. A comparison (not shown) of the differences between the two dV/dt signals and the two dΔ[CaT]/dt signals in Fig. 2 C indicates that such channel activation by Ca must have been rapid, with a delay no greater than 1–2 ms.\n\nFig. 3 shows the amplitude of the first action potential of a stimulation (A) and its half-width (B) plotted as a function of [CaSR]R, from the experiment in Fig. 2. The half-width is taken to be the interval between the times to half-peak on the rising and falling phases of the signal. In this figure, filled circles (which include a–c) denote values obtained during the first part of the experiment, when the value of [CaSR]R was decreased from 1,154 to 146 μM by a progressive decrease in the recovery period between successive trials from 5 to 0.5– 0.6 min. Open circles (including d) denote values obtained thereafter, when the recovery period was progressively increased to 10 min, decreased to 0.5 min, and finally increased again to 10 min.\n\nIn Fig. 3 A, the first stimulation occurred when [CaSR]R = 1,154 μM and the amplitude of the first action potential was 135.0 mV (• at extreme right). The amplitude showed little change during the experiment, with a small progressive decrease, ≤3 mV, that can be reasonably attributed to fiber run down.\n\nIn contrast, Fig. 3,B shows that the half-width of the action potential was increased when the value of [CaSR]R was decreased from its initial value of 1,154 to 146 μM (Fig. 3 B, •). The increase in half-width was most pronounced for [CaSR]R ≤ 500 μM. Most of this increase in half-width was reversed when the value of [CaSR]R was allowed to increase (○).\n\nThe tentative conclusions of this section are that (a) an increase in the value of [CaSR]R from 150 to 800 μM produces a reversible 1–2-ms decrease in the half-width of the action potential, (b) the outward ionic current responsible for this decrease flows through Ca-activated potassium or chloride channels, and (c) the activation of these channels by Ca is normally rapid, developing within 1–2 ms after SR Ca release begins.\n\n### Changes in SR Ca Release Elicited by a Single Action Potential when [CaSR]R Is Varied between 150 and 1,200 μM\n\nFig. 2,A, a–d, shows four Δ[CaT] signals obtained with [CaSR]R = 146–876 μM. The value of [CaSR]R and the amplitude of the Δ[CaT] signal after the first action potential progressively decreased from a to b to c, and then increased in d to values larger than those in a. The initial segments of the Δ[CaT] traces are shown in Fig. 2 B, plotted on expanded horizontal and vertical gains. These signals show that, as the value of [CaSR]R was decreased, the Δ[CaT] signal became more rounded, indicating a longer period of SR Ca release.\n\nFig. 4 A shows the value of Δ[CaT] after the first action potential in each trial, plotted as a function of [CaSR]R. The concave curvature of the relation between Δ[CaT] and [CaSR]R indicates that the reduction in Δ[CaT] was less marked than that in [CaSR]R. For example, a threefold decrease in [CaSR]R from 1,200 to 400 μM resulted in a reduction of Δ[CaT] of only 30%. Thus, under these conditions, the amount of Ca released by an action potential is relatively insensitive to SR Ca content. On the other hand, a steeper dependence was observed for values of [CaSR]R < 400 μM. The open and filled circles track the same relation between Δ[CaT] and [CaSR]R, showing that the effect of [CaSR]R on Δ[CaT] was reversible.\n\nFig. 4 B shows the corresponding dependence of f1 on [CaSR]R. The value of f1 increased threefold from 0.165 at [CaSR]R = 1,154 μM to 0.51–0.52 at [CaSR]R = 140–150 μM. Remarkably, at the smallest values of [CaSR]R obtained in this experiment, 140–150 μM, slightly more than half of the readily releasable Ca inside the SR was released by the first action potential.\n\nFig. 4,C shows the peak value of dΔ[CaT]/dt plotted as a function of [CaSR]R. The relation is slightly convex, indicating that the increase in f1 associated with decreasing [CaSR]R (Fig. 4 B) is not caused by an increase in the fractional rate of SR Ca release; rather, it occurs in spite of a small decrease in the peak fractional rate of release.\n\nSince the dΔ[CaT]/dt signals are noisier than the Δ[CaT] signals (compare, for example, Fig. 2, B and C), the data in Fig. 4,C have more fractional scatter than those in Fig. 4,A, especially at small values of [CaSR]R. Noise reduces the reliability of estimates of the half-width of dΔ[CaT]/dt and its final time constant when the value of [CaSR]R is small (Fig. 2,C; but see Fig. 7, C and D). In this situation, the time constant associated with the final half of the Δ[CaT] signal (τΔ[CaT]) can be used as an estimate of the duration of SR Ca release. Fig. 4 D shows τΔ[CaT] plotted as a function of [CaSR]R. From [CaSR]R = 1,154 to 140–150 μM, the value of τΔ[CaT] increased by an order of magnitude, from 0.9 to 11–13 ms.\n\nThe value 11–13 ms is similar to the time constant expected for Ca dissociation from the Ca-regulatory sites on troponin, estimated to be 8.7 ms (column 5 in Table I, model 2, in Baylor et al., 1983). This similarity raises the possibility that, with [CaSR]R = 140–150 μM, the Ca complexed by EGTA (which determines Δ[CaT]) came from Ca that had just dissociated from troponin rather than from Ca that had just left the SR. This seems unlikely for the following reason. In fibers with [CaSR]R ≥ 1,391 μM, the increase in [CaEGTA] that accompanies Ca dissociation from troponin after an action potential appears to be negligibly small and not statistically significant, 2.9 μM (SEM, 3.8 μM) (Table II in Pape et al., 1995). Since a decrease in [CaSR]R would be expected to produce a decrease, not an increase, in the amount of Ca complexed by troponin, it seems unlikely that Ca that dissociated from troponin made a significant contribution to the Ca that was complexed by EGTA during the 140–150 μM Δ[CaT] signals that had the 11–13-ms time constant.\n\nThese results show that, when the SR is partially depleted of Ca, the amount of Ca released by a single action potential does not decrease in proportion to the value of [CaSR]R (Fig. 4,A). Rather, its value is partially stabilized by an increase in f1 (Fig. 4,B) that is caused by a prolongation of Ca release (increase in τΔ[CaT], Fig. 4,D). This prolongation of Ca release may be due, at least in part, to the accompanying prolongation of the action potential (Figs. 2,C and 3 B). The order-of-magnitude increase in τΔ[CaT], however, suggests that some other effect may also be involved (next section).\n\n### The Effect of Partial SR Ca Depletion on Action Potential–stimulated SR Ca Release Can Be Mimicked by a Brief Voltage-Clamp Depolarization\n\nTo study further the effect of [CaSR] on the rate of turn-off of SR Ca release, experiments were carried out on voltage-clamped fibers. One advantage of this method is that, unlike experiments with action potential stimulation, the voltage waveform is constant and does not depend on the value of [CaSR]R; thus, any changes in release that are observed cannot be attributed to changes in voltage. Another advantage is that measurements can be made of intramembranous charge movement (Qcm), which is thought to arise from movement of the voltage sensors for SR Ca release. Such information might help determine whether changes in the turn-off of SR Ca release are caused by changes in the voltage sensor. Before describing the effects of [CaSR] on Qcm, however, it is important to establish that the effect of SR Ca depletion on Ca release is similar with voltage-clamp and action potential stimulation.\n\nFig. 5 shows the results of a voltage-clamp experiment that was designed to mimic the action potential experiment illustrated in Fig. 2. For this purpose, the same concentration of Ca, 0.44 mM, was used in the end-pool solution. A 10-ms pulse to −20 mV was used to release a small fraction of the readily releasable Ca from the SR, similar to that released by the first action potential in a trial in Fig. 2. This pulse was followed by a 200-ms repolarization to −90 mV, and then a 420-ms depolarization to −40 mV to deplete the SR of its remaining Ca so that the value of [CaSR]R could be determined.\n\nFig. 5,A, a, shows the Δ[CaT] signal that was obtained 78 min after saponin treatment. After the first depolarization, Δ[CaT] reached a value of 229 μM. By the end of the second depolarization, Δ[CaT] had increased to a quasi-steady value of 1,152 μM, which is taken for the value of [CaSR]R. Thus, the first depolarization released 229/1,152 = 0.199 of the readily releasable Ca from the SR, similar to the fraction f1 released by the first action potential in Fig. 2,d (0.221 with [CaSR]R = 876 μM). Fig. 5,A, b and c, were obtained later in the experiment after the value of [CaSR]R had decreased to 430 and 213 μM, respectively. The Δ[CaT] traces in Fig. 5, A and B, are similar to those in Fig. 2, A and B.\n\nFig. 6 shows the effect of [CaSR]R on Δ[CaT] (A), f1 (B), peak dΔ[CaT]/dt (C), and τΔ[CaT] (D), from the experiment illustrated in Fig. 5. It is plotted with the same format used in Fig. 4. Filled circles represent measurements made when the recovery period between successive trials was progressively decreased from 5 to 1 min. After the value of [CaSR]R had decreased to 141 μM by three successive stimulations with recovery periods of 1 min (three • at extreme left in each panel), the recovery period was increased to 10 min for two trials (○), and then, 35 min later in the experiment, to 5 min for two trials (□). The relations between release parameters and [CaSR]R in Fig. 6 are similar to those in Fig. 4.\n\nIn the experiment in Fig. 6, after the value of [CaSR]R had been decreased to 141 μM, its value was increased to only 205–212 μM by increasing the recovery period to 10 min (○). A similar small increase in [CaSR]R, from 148 μM after two 1-min recovery periods to 174 μM after a 10-min recovery period, was observed in the other voltage-clamp experiment in which 0.44 mM Ca was used in the end-pool solution (fiber O08912). This small recovery of [CaSR]R is unlike the large recovery observed in the action potential experiment in Fig. 2, from 146 μM in c to 876 μM in d. Although the reason for this difference is unknown, an important factor may be the presence of 1.8 mM Ca in the external solution in the action potential experiments and the absence of external Ca in the voltage-clamp experiments. Whatever the reason, the poor recovery of [CaSR]R in the voltage-clamp experiments makes it difficult to study the reversibility of the effect of [CaSR]R on SR Ca release in these experiments, as was done in the action potential experiments (Fig. 2).\n\nExperiments similar to the one in Figs. 5 and 6 were carried out on three other voltage-clamped fibers in which a 10–12-ms pulse to −20 mV was used for the first stimulation. In one of the experiments (O08912), 0.44 mM Ca was used in the end-pool solution, and the value of [CaSR]R was reduced by progressively decreasing the recovery period between successive trials from 5 to 1 min, as was done in the experiment in Figs. 5 and 6. In the other two experiments (N14911 and N15911), the fibers were first equilibrated with an end-pool solution that contained 1.76 mM Ca. Then, Ca was removed from the end-pool solution and the value of [CaSR]R was reduced by successive stimulations every 1.5 min. The results of all four experiments were similar; the mean value of f1 was 0.147 (SEM, 0.022) with [CaSR]R = 1,000–1,200 μM and 0.267 (SEM, 0.082) with [CaSR]R = 140–300 μM; the mean value of the ratio f1([CaSR]R = 140–300 μM)/f1([CaSR]R = 1,000–1,200 μM) was 1.71 (SEM, 0.33), which is not significantly different from unity. The mean value of τΔ[CaT] was 3.63 ms (SEM, 0.54 ms) with [CaSR]R = 1,000–1,200 μM and 16.9 ms (SEM, 4.0 ms) with [CaSR]R = 140–300 μM; the mean value of the ratio τΔ[CaT]([CaSR]R = 140–300 μM)/τΔ[CaT]([CaSR]R = 1,000–1,200 μM) was 4.51 (SEM, 0.59), which is significantly different from unity.\n\nThe results in Fig. 6 and those described in the preceding paragraph are qualitatively similar to those in Fig. 4. This suggests that the effect of [CaSR]R on Δ[CaT] signals elicited by action potential stimulation cannot be explained primarily by the effect of [CaSR]R on the duration of the action potential.\n\n### The Effect of Partial SR Ca Depletion on dΔ[CaT]/dt\n\nFurther analysis of the Δ[CaT] signals in Fig. 5 is shown in Fig. 7. Fig. 7 A, middle, shows dΔ[CaT]/dt expressed in units of micromolar per millisecond. As the value of [CaSR]R decreased from 1,152 (a) to 430 (b) to 213 (c) μM (only a and c are labeled), the amplitude of dΔ[CaT]/dt was decreased and its duration was increased.\n\nFig. 7 A, bottom, shows dΔ[CaT]/dt corrected for SR Ca depletion. At each moment in time, the value of dΔ[CaT]/dt in units of micromolar per millisecond was divided by [CaSR]R − Δ[CaT] and multiplied by 100 to give units in percent per millisecond (Jacquemond et al., 1991). Because of the division by [CaSR]R − Δ[CaT], each trace becomes progressively noisier with time, and the noise increases from a to b to c. Within the noise of the signals, the initial time courses are similar. After the peak value was reached, however, the duration became progressively longer from a to b to c.\n\nFig. 7 B shows the peak amplitude of dΔ[CaT]/dt in units of percent per millisecond, plotted as a function of [CaSR]R. The values increased slightly from [CaSR]R = 1,152 to 300–400 μM, and then decreased as the value of [CaSR]R became smaller than 300 μM. In 25 measurements of dΔ[CaT]/dt in four fibers, the mean peak value of dΔ[CaT]/dt with [CaSR]R = 140–300 μM was 0.861 (SEM, 0.016) times that measured in the same fiber with [CaSR]R = 600–1,200 μM; the value 0.861 is significantly different from unity.\n\nFig. 7,C shows the half-width of dΔ[CaT]/dt (•, ○, and □) and its time to half-peak (▪), plotted as a function of [CaSR]R. Fig. 7,D shows the final time constant of dΔ[CaT]/dtdΔ[CaT]/dt). As the value of [CaSR]R decreased from 1,152 to 141 μM, the half-width of dΔ[CaT]/dt (C) and the value of τdΔ[CaT]/dt(D) progressively increased; in both panels, the open symbols and filled circles superimpose within the scatter of the points. On the other hand, the relative constancy of Fig. 7 C, ▪, shows that the rising phase of dΔ[CaT]/dt was little affected by [CaSR]R.\n\nIt is clear from Fig. 7,A, bottom, that the dΔ[CaT]/dt signal became noisier as the value of [CaSR]R became smaller, making the determinations of the half-width of dΔ[CaT]/dt (Fig. 7,C) and of the value of τdΔ[CaT]/dt(Fig. 7,D) more difficult. For this reason, the filled circles at the left-hand side of Fig. 7, C and D, with small values of [CaSR]R, show considerable scatter. In spite of this, the relation between τdΔ[CaT]/dtand [CaSR]R in Fig. 7,D is similar to that between τΔ[CaT] and [CaSR]R in Fig. 6,D, which suggests that the relation in Fig. 7,D is reliable within the noise of the dΔ[CaT]/dt signals. This being the case, it seems likely that the relation between the half-width of dΔ[CaT]/dt and [CaSR]R in Fig. 7 C is also reliable.\n\nIn four fibers, the mean value of τdΔ[CaT]/dtwas 2.48 ms (SEM, 0.53 ms) with [CaSR]R = 1,000–1,200 μM and 16.2 ms (SEM, 4.0 ms) with [CaSR]R = 140–300 μM; the mean value of the ratio τdΔ[CaT]/dt([CaSR]R = 140–300 μM)/ τdΔ[CaT]/dt([CaSR]R = 1,000–1,200 μM) was 6.47 (SEM, 0.76), which is significantly different from unity.\n\nThe results in this section, obtained from voltage-clamped fibers, confirm those obtained with action potential stimulation (Fig. 4). In particular, when the value of [CaSR]R is decreased from 1,000–1,200 to 140–300 μM, the observed increase in f1 (Fig. 6,B) is caused by a prolongation of SR Ca release (Fig. 7, C and D), and not by an increase in the fractional rate of release (Fig. 7 B).\n\n### The Effect of Partial SR Ca Depletion on Qcm\n\nThe next question to consider is whether partial SR Ca depletion affects the voltage sensor for SR Ca release. Fig. 8,A, middle, shows the currents from intramembranous charge movement (Icm) associated with Fig. 5, a–c (only a and c are labeled). The ON waveforms of a and b superimpose, with an amplitude that is slightly larger than that of c, owing to a small decrease in the amount of charge in c (see below). On the other hand, the OFF waveforms of Icm are clearly different. From a to b to c, as the value of [CaSR]R decreased from 1,152 to 430 to 213 μM, the amplitude of the OFF response became smaller and its duration became longer.\n\nFig. 8 A, bottom, shows Qcm, the running integral of Icm, which is expected to be more closely related than Icm to the state of activation of the SR Ca channels. Except for a small reduction in the amplitude of c, the initial time courses of the three traces are similar until the time to peak; thereafter, the return to baseline became progressively slower from a to b to c.\n\nFig. 8,B, •, shows the peak value of Qcm plotted as a function of [CaSR]R. As expected from Fig. 8 A, bottom, the peak value of Qcm was similar in a and b and was slightly smaller in c. The 10–15% decrease in Qcm that was associated with the decrease in [CaSR]R from 600 to 141 μM may have been caused by a direct effect of the reduction in [CaSR]R or by the decrease in the recovery period between successive trials that was used to reduce [CaSR]R from 4 to 1 min. Since the decrease in Qcm was reversed by increasing the recovery period to 5 or 10 min (□ and ○, respectively), which increased [CaSR]R only slightly, the decrease in recovery period is the more likely explanation. Such a decrease in Qcm could have been caused by slow inactivation of charge movement (Chandler et al., 1976). According to this idea, a small amount of slow inactivation would develop during the 10-ms pulse to −20 mV and the subsequent 420-ms pulse to −40 mV. This inactivation would then be removed during the recovery period at −90 mV, and removal would be more complete with a 3–4-min recovery period than with a 1-min period.\n\nFig. 8 C shows the half-width of Qcm (•, ○, and □) and the time to half-peak of Qcm (▪) plotted as a function of [CaSR]R. The filled circles show that the half-width progressively increased when the value of [CaSR]R was decreased from 1,152 to 141 μM (○ and □ are discussed in the following section). The increase in half-width is caused by a prolongation of the falling phase of the signal since the time to half-peak of the rising phase of the signal was constant (▪).\n\nThe falling phase of Qcm was analyzed by fitting an exponential function to the final half of the signal, starting at the time from half-peak and ending 164 ms after repolarization (not shown). Fig. 8 D, •, shows the values of the time constant of the exponential function that were obtained in this manner (τQcm). The value of τQcm increased from 12 to 27 ms as the value of [CaSR]R was decreased from 1,152 to 141 μM.\n\nResults similar to those shown in Fig. 8, •, were obtained in four fibers in which the value of [CaSR]R was progressively decreased from 1,000–1,200 to 140 μM. In the two fibers with 0.44 mM Ca in the end-pool solution (O08911 and O08912), this depletion was accomplished by decreasing the recovery period between trials from 5 to 1 min. In the two fibers with 0 mM Ca in the end-pool solution (N14911 and N15911), the recovery period was 1.5 min throughout. The mean value of τQcm was 10.0 ms (SEM, 1.1 ms) with [CaSR]R = 1,000–1,200 μM and 17.7 ms (SEM, 3.7 ms) with [CaSR]R = 140–300 μM; the mean value of the ratio τQcm([CaSR]R = 140–300 μM)/τQcm([CaSR]R = 1,000– 1,200 μM) was 1.71 (SEM, 0.20), which is significantly different from unity.\n\n### The Effect of the Duration of the Recovery Period on OFF Qcm\n\nFig. 8, B–D, •, denotes measurements made when the value of [CaSR]R was decreased by reducing the duration of the recovery period between successive trials. With the shortest recovery period used, 1 min, the value of [CaSR]R was 141–159 μM (three left-most • in each panel). When the recovery period was then increased to 5 or 10 min, the value of [CaSR]R was increased to 180–185 μM (□) or 205–212 (○), respectively. The associated values of the half-width of Qcm (Fig. 8 C) and of the value of τQcm (D) were reduced by the longer recovery periods so that the open symbols lie below the relation defined by the filled circles; the reduction was larger with a 10-min recovery period (○) than with a 5-min period (□). Recovery periods >10 min were not studied.\n\nIn two fibers, a 10-min recovery period was used after the value of [CaSR]R had decreased to 141 (O08911) or 148 (O08912) μM. The mean value of τQcm was 11.9 ms (SEM, 0.1 ms) with [CaSR]R = 1,000–1,200 μM and 17.4 ms (SEM, 0.2 ms) with a 10-min recovery period and [CaSR]R = 140–300 μM; the mean value of the ratio τQcm([CaSR]R = 140–300 μM)/τQcm([CaSR]R = 1,000– 1,200 μM) was 1.47 (SEM, 0.03), which is significantly different from unity.\n\nThis dependence of the half-width of Qcm and of the value of τQcm on recovery period was not observed in fibers with [CaSR]R > 1,200 μM. (In two experiments, not shown, fibers N14911 and N15911 were first equilibrated with an end-pool solution that contained 1.76 mM Ca and had initial values of [CaSR]R that were >2,000 μM. Later in the experiment, Ca was removed from the end-pool solution and intramembranous charge movement and SR Ca release were monitored with 10-ms pulses to −20 mV, as was used in Fig. 5. In fiber N14911, the values of Qcm half-width and τQcm were 9.66 and 7.37 ms, respectively, after a 6.9-min recovery period ([CaSR]R = 2,442 μM) and were 9.94 and 7.12 ms after a 1.57-min recovery period ([CaSR]R = 1,729 μM). In fiber N15911, the values of Qcm half-width and τQcm were 10.02 and 6.64 ms, respectively, after a 13-min recovery period ([CaSR]R = 2,268 μM) and were 10.88 and 8.01 ms after a 1.5-min recovery period ([CaSR]R = 1,289 μM). These small changes in Qcm half-width and τQcm are probably within the error of measurement.)\n\nThe results described above suggest that, as the value of [CaSR]R is reduced from 1,200 to 140–210 μM, the OFF kinetics of Qcm is slowed in a use-dependent manner. With [CaSR]R >1,200 μM, either the slowing effect does not occur or, if it does, it is removed by a 1.5-min recovery period. Unfortunately, recovery periods <1 min could not be studied with our voltage-clamp method owing to the time required to take and process data during the CONTROL and TEST pulses of each trial.\n\nThe main conclusion from this and the preceding section is that a reduction in [CaSR]R from 1,000–1,200 to 140–300 μM is able to prolong the duration of OFF Qcm after a brief depolarization in a use-dependent manner with little, if any, effect on ON Qcm. In four fibers in which a recovery period as brief as 1–1.5 min was used to determine τQcm with [CaSR]R = 140–300 μM, the final time constant of OFF Qcm was increased by the factor 1.71 (SEM, 0.33) (see preceding section). In the two fibers in which a 10-min recovery period was also used, the factor was 1.47 (SEM, 0.03) (this section). As mentioned above, recovery periods >10 min were not studied.\n\n### A Comparison of the Effect of Partial SR Ca Depletion on the Time Courses of Qcm and dΔ[CaT]/dt\n\nFig. 9,A shows three pairs of Qcm and dΔ[CaT]/dt traces, replotted from Figs. 8,A and 7 A, bottom, respectively; the same calibration factors, 4 nC/μF and 1%/ms, apply, respectively, to all the Qcm and dΔ[CaT]/dt traces. Within each pair of traces, the rising phase of the noisier dΔ[CaT]/dt signal lagged that of the Qcm signal by 2–3 ms so that its peak value was reached after that of Qcm.\n\nSimon and Hill (1992) measured the time course of charge movement and SR Ca release during and after 100-ms depolarizations to potentials between −60 and 20 mV. A depolarizing prepulse was used to produce Ca inactivation of Ca release so that the time course of dΔ[CaT]/dt would, according to them, correspond to the time course of voltage activation of the noninactivating component of dΔ[CaT]/dt. They found that the time course of dΔ[CaT]/dt matched that of Qcm4. Based on this result, they proposed that four identical voltage sensors, which move independently, act in concert to gate the SR Ca release channel. In our experiments, with 10-15 ms depolarizations, the time courses of Qcm4 and of dΔ[CaT]/dt were clearly different (not shown). For example, the peak value of dΔ[CaT]/dt occurred after that of Qcm4 or, for that matter, Qcm raised to any positive power (see Fig. 9 A). Moreover, the later time to peak of dΔ[CaT]/dt cannot be attributed to the development of Ca inactivation of Ca release since this would be expected to decrease, not increase, the time to peak of dΔ[CaT]/dt. Thus, if four voltage sensors act in concert to gate the SR Ca channel, as proposed by Simon and Hill (1992), the opening of the channel does not occur immediately with the movement of the voltage sensors, but after a 2–3-ms delay.\n\nAfter the peak values of dΔ[CaT]/dt in Fig. 9 A were reached, both Qcm and dΔ[CaT]/dt returned to zero with rates that progressively decreased from a to b to c. In a, the final return of dΔ[CaT]/dt to zero was more rapid than that of Qcm. The relative difference in the final time courses of Qcm and dΔ[CaT]/dt was less marked in b. In c, the final time courses of the two signals were similar, with the Qcm trace lying within the noise of the dΔ[CaT]/dt trace.\n\nSince, with small values of [CaSR]R, the value of τQcm with a 1.5-min recovery period was larger than that with a 10-min recovery period (Fig. 8,D, • c and ○), whereas the corresponding values of τdΔ[CaT]/dtwere within the scatter of the points (Fig. 7 D), it was of interest to compare Qcm and dΔ[CaT]/dt signals with a 10-min recovery period. Within the noise of the Δ[CaT]/dt signal, the final time courses of the two signals were similar (not shown).\n\nFig. 9 B shows the ratio τQcmdΔ[CaT]/dt plotted as a function of [CaSR]R. Its value decreased from 4.5 with [CaSR]R = 1,152 μM (a) to values that fluctuated around unity with values of [CaSR]R < 400 μM. With [CaSR]R < 400 μM, the mean value of τQcmdΔ[CaT]/dt was 1.07 (SEM, 0.05) for the filled circles and 0.91 (SEM, 0.07) for the open symbols; these two values are not significantly different from each other or from unity. The comparisons described in this and the preceding paragraph show that, with [CaSR]R < 400 μM, the noise in the dΔ[CaT]/dt signals makes it difficult to assess quantitatively the effect of increasing the recovery period from 1–2 to 10 min on the relation between Qcm and dΔ[CaT]/dt.\n\nResults similar to those in Fig. 9,B were obtained in three other fibers, and the combined data are shown in Fig. 9 C. The mean value of τQcmdΔ[CaT]/dtwith [CaSR]R = 1,000–1,500 μM was 4.66 (SEM, 0.35) and, with [CaSR]R = 140–300 μM, was 1.06 (SEM, 0.04). The first, but not the second, value is significantly different from unity.\n\nJong et al. (1995a) also measured Qcm and dΔ[CaT]/dt after brief depolarizations (10–15 ms to −20 mV) in fibers equilibrated with an end-pool solution similar to that used here with 1.76 mM Ca. In five fibers in which [CaSR]R = 2,003–2,759 μM, the mean values of τQcm and τdΔ[CaT]/dtwere 8.45 ms (SEM, 1.76 ms) and 1.64 ms (SEM, 0.17 ms), respectively; the mean value of τQcm/ τdΔ[CaT]/dt, 4.96 (SEM, 0.61), is significantly different from unity. These values with [CaSR]R = 2,003–2,759 μM are generally consistent with the results in Figs. 7,D, 8,D, and 9 C.\n\nThe main conclusion of this section is that, after a brief depolarization, (a) with [CaSR]R = 1,000–1,500 μM, dΔ[CaT]/dt returns to its resting state several times more rapidly than Qcm; and (b) with [CaSR]R = 140–300 μM and the associated increase in noise of the dΔ[CaT]/dt signal, the final time courses of Qcm and dΔ[CaT]/dt, appropriately scaled, are indistinguishable from each other.\n\n## Discussion\n\nThis article describes the effects of partial SR Ca depletion on SR Ca release in cut muscle fibers that have been equilibrated with an internal solution that contains 20 mM EGTA. One of the most interesting of these effects is that the fractional amount of SR Ca released by a single action potential is increased when the value of [CaSR]R is decreased. In the experiment in Fig. 4, for example, when the value of [CaSR]R was reduced from 1,154 to 140–150 μM, the value of f1 was increased threefold, from 0.165 to 0.51–0.52. Three factors appear to contribute to this marked effect: a prolongation of the action potential, a slowing of the OFF kinetics of Qcm, and a decrease in Ca inactivation of Ca release. Because some or all of these factors may be caused by alterations in the increase in myoplasmic free [Ca] that occurs during release, it is instructive to compare the two methods that have been used by others and by us to attenuate myoplasmic free Ca transients: the addition of high affinity Ca buffers to the myoplasm and a reduction of SR Ca content. As shown below, such information can be used to estimate an upper limit for the distance between a putative regulatory Ca receptor and the mouth of the SR Ca channel.\n\n### Comparison of the Use of High Affinity Ca Buffers and Partial SR Ca Depletion to Decrease Myoplasmic Free [Ca] during SR Ca Release\n\nOne way to reduce the increase in myoplasmic free [Ca] that accompanies SR Ca release is with high affinity Ca buffers such as fura-2, the Ca indicator developed by Grynkiewicz et al. (1985) (Baylor and Hollingworth, 1988; Jacquemond et al., 1991; Hollingworth et al., 1992; Csernoch et al., 1993; Jong et al., 1993; Pape et al., 1993). Another way is with partial SR Ca depletion, which is expected to reduce single-channel Ca flux (this article). These two methods are expected to reduce [Ca] in spatially different ways. This can be illustrated by the theoretical example of a single SR Ca channel that behaves as a point source of Ca immersed in an infinite medium of myoplasm that is isotropic and isopotential. In the absence of freely diffusible Ca buffers, the steady-state increase in myoplasmic free [Ca] at a distance r from the mouth of the channel (Δ[Ca]ps) satisfies the well-known relation\n\n\\begin{equation}{\\Delta}[Ca]_{ps}=\\frac{{\\mathit{{\\phi}}}_{point}}{4{\\pi}D_{Ca}r}.\\end{equation}\n1\n\nφpoint represents the flux of Ca ions through the point source or channel, and DCa represents the diffusion coefficient of Ca in myoplasm. The subscript “ps” refers to a single point source.\n\nIn the presence of a large concentration of a freely diffusible Ca buffer such as EGTA or fura-2, Eq. 1 can be replaced by\n\n\\begin{equation}{\\Delta}[Ca]_{ps}=\\frac{{\\mathit{{\\phi}}}_{point}}{4{\\pi}D_{Ca}r}{\\cdot}exp(-r/{\\mathit{{\\lambda}}}_{Ca})\\end{equation}\n2\n\nif the fractional change in concentration of Ca-free buffer is small (Neher, 1986; Stern, 1992; Appendix B in Pape et al., 1995) and DCa/[B]R << DB/([Ca]R + Kd) (Appendix B in Pape et al., 1995). DB represents the diffusion coefficient of the Ca buffer in myoplasm (which is assumed to be independent of the state of Ca complexation), Kd represents the Ca buffer's dissociation constant, and [B]R represents the resting concentration of Ca-free buffer. The value of λCa, which determines the distance that Ca diffuses before capture by the buffer, is given by\n\n\\begin{equation}{\\mathit{{\\lambda}}}_{Ca}=\\sqrt{\\frac{D_{Ca}}{k_{1}[B]_{R}}},\\end{equation}\n3\n\nwhere k1 is the forward rate constant for Ca binding to the Ca buffer. The steady-state value of Δ[Ca]ps given by either Eq. 1 or 2 is unaffected by the presence of immobile Ca buffers such as troponin.\n\nIn fibers equilibrated with 20 mM EGTA and 1.76 mM Ca, the value of λCa is estimated to be 81 nm (Pape et al., 1995). Although the exact value of φpoint for an SR Ca channel inside a muscle fiber is unknown, measurements on canine cardiac SR Ca channels incorporated into lipid bilayers indicate that, under physiological conditions, the single-channel current is ≤0.5 pA (Mejia-Alvarez et al., 1998), which gives φpoint ≤ 1.5 × 106 ions s−1. In this case, for r > 400 nm, the value of Δ[Ca]ps calculated from Eq. 2 with λCa = 81 nm is expected to be negligible, ≤0.01 μM. Thus, in fibers equilibrated with 20 mM EGTA and 1.76 mM Ca, the spread of Δ[Ca]ps from an open SR Ca channel is expected to be confined to a region that is no farther than 400 nm from the mouth of the channel. Since the value of λCa is decreased if the amount of Ca added to the EGTA is decreased, this condition holds for 20 mM EGTA and 0–1.76 mM Ca.\n\nFura-2 complexes Ca much more rapidly than EGTA. According to Jong et al. (1995a), the myoplasmic value of k1 for fura-2 is 28 times that for EGTA. Consequently, 0.65 mM Ca-free fura-2 is expected to be equivalent to 18.24 mM Ca-free EGTA (the concentration of Ca-free EGTA in an internal solution containing 20 mM EGTA and 1.76 mM Ca) in its ability to decrease Δ[Ca]ps. With 0.65, 2, 4, 6, and 8 mM Ca-free fura-2, the estimated values of λCa are 81, 46, 33, 27, and 23 nm, respectively.\n\nComparison of Eqs. 1 and 2 shows that the fractional reduction of Δ[Ca]ps produced by high affinity Ca buffers depends on distance from the mouth of a channel according to the factor exp(−rCa). In contrast, with partial SR Ca depletion, the fractional reduction of Δ[Ca]ps is independent of r and is given by the fractional reduction of φpoint. As a result, high affinity Ca buffers would be expected to be more effective than SR Ca depletion in reducing Δ[Ca]ps at large distances from the mouth of an open SR Ca channel, whereas partial SR Ca depletion would be expected to be more effective near the mouth of the channel.\n\nAlthough the spatial profile of Δ[Ca]ps near an open SR Ca channel is not expected to be as simple as that described by Eq. 1 or 2, these equations may help evaluate, at least qualitatively, the differences between reducing Δ[Ca]ps by high affinity Ca buffers and by SR Ca depletion. To apply these equations to our results, a relation between φpoint and [CaSR] must be specified. In principle, this requires knowledge of (a) the relation between φpoint and free [Ca] inside the SR, which depends on the properties of the SR Ca channel, and (b) the relation between free [Ca] and [CaSR], which depends on the amount and properties of luminal Ca buffers such as calsequestrin. Since these relations inside a muscle fiber are unknown, a direct proportionality between φpoint and [CaSR] has been assumed in the discussion below. This assumption is consistent with the finding that the peak fractional rate of SR Ca release is relatively independent of the value of [CaSR]R between 140 and 1,200 μM (Figs. 4,C, 6,C, and 7 B).\n\nIn addition to the Δ[Ca]ps signal produced by Ca flux through a particular channel of interest, an increase in free [Ca] can also be produced by Ca flux through other open channels. The total increase in free [Ca] (Δ[Ca]) is given by\n\n\\begin{equation}{\\Delta}[Ca]={\\Delta}[Ca]_{ps}+{\\Delta}[Ca]_{os}.\\end{equation}\n4\n\nΔ[Ca]os represents the sum of the contributions from all other open channels, with the contribution from each channel described by an equation similar to Eq. 1 or 2; the subscript “os” refers to other sources. The principle of superposition applies to cases in which Ca buffers are either absent or, as is the case with EGTA in the experiments reported here, present in a sufficiently large concentration to ensure that, after an action potential or brief voltage pulse, the fractional change in concentration of Ca-free buffer is small. Estimates of Δ[Ca]ps and Δ[Ca]os with various Ca buffers are given in a. These show that the contribution of Δ[Ca]os to Δ[Ca] can be substantial, especially in the absence of high affinity Ca buffers such as EGTA or, during the early part of a transient, in the absence of troponin (an immobile Ca buffer that makes no contribution to Δ[Ca] in the steady state).\n\n### Effect of SR Ca Depletion on the Time Course of the Action Potential\n\nOne of the effects of partial SR Ca depletion is a prolongation of the action potential (Fig. 2 C), which is caused by a decrease in net outward ionic current during repolarization. The ion whose current is decreased could be either potassium or chloride since the internal solution contained potassium as the predominant cation and the external solution contained chloride as the predominant anion. Because the affected current depends on SR Ca release and develops soon after release begins, it seems likely that it is carried by ions that move through channels in the surface and/or tubular membranes that are activated rapidly, within 1–2 ms, by the increase in myoplasmic free [Ca] that accompanies release. With the large concentration of EGTA used in these experiments, the increase in free [Ca] is expected to be confined to a region ≤400 nm from the SR Ca channels (preceding section). Thus, any Ca-activated channels that are opened would be expected to be located in the transverse tubular membrane or in the surface membrane within 400 nm of the Z line.\n\nThe surface and/or tubular membranes of skeletal muscle contain both Ca-activated potassium channels (Pallotta et al., 1981) and Ca-activated chloride channels (Hui and Chen, 1994). In cut fibers equilibrated with an internal solution that contained only 0.1 mM EGTA, the current through Ca-activated chloride channels was no more than 1–2 μA/μF at the voltages where action potentials c and d in Fig. 2 C diverged (Hui and Chen, 1994). This is an order of magnitude smaller than the 15–20 μA/μF current that is required to explain the difference in the action potentials. Consequently, the affected current is probably carried by Ca-activated potassium channels, which, at least in myocytes from guinea pig urinary bladder, can be activated by Ca rapidly, within milliseconds (Markwardt and Isenberg, 1992). These results support the suggestion of Blatz and Magleby (1987) that Ca-activated potassium channels help stabilize and repolarize the transverse tubular membrane after an action potential.\n\n### Comparison of the Effect of High Affinity Ca Buffers and SR Ca Depletion on the Value of f1 Elicited by an Action Potential\n\nIn fibers equilibrated with 20 mM EGTA and partially depleted of SR Ca, a single action potential is able to release a large fraction of the SR Ca content, with values of f1 as large as 0.5 (Fig. 4,B). The increase in f1 with decreasing [CaSR]R is associated with a prolongation of the action potential (Fig. 3,B) and a slowing of the turn-off of SR Ca release (Fig. 4,D). Since a substantial increase in f1 was also observed in voltage-clamp experiments (Fig. 6 B), the prolongation of the action potential cannot explain all of the increase in f1 that accompanies a reduction in [CaSR]R.\n\nIn previous studies with action potential stimulation of fibers equilibrated with 20 mM EGTA and 1.76 mM Ca, the value of [CaSR]R was 1,391–4,367 μM; f1 varied between 0.13 and 0.17 and had a mean value of 0.144 (Pape et al., 1995). Although direct measurement of [CaSR]R, and consequently of f1, is only possible in fibers with a large myoplasmic concentration of an extrinsic Ca buffer such as EGTA or fura-2 to complex the Ca released from the SR, there is reason to believe that f1 is larger with 20 mM than with 0–0.1 mM EGTA. In fibers stimulated by an action potential, the estimated amount of SR Ca release is increased by as much as twofold by the introduction of 0.5–1 mM fura-2 into the myoplasm (Baylor and Hollingworth, 1988; Hollingworth et al., 1992; Pape et al., 1993), which is expected to reduce myoplasmic free [Ca] to approximately the same extent as 20 mM EGTA (see above). In voltage-clamped fibers, the estimated peak rate of SR Ca release is increased by as much as twofold by 0.5–1 mM fura-2 (Jong et al., 1993) or by as much as threefold by 20 mM EGTA (Jong et al., 1995a). This ability of 0.5–1 mM fura-2 or 20 mM EGTA to increase SR Ca release is probably caused by a decrease in Ca inactivation of Ca release associated with the reduction in myoplasmic Δ[Ca] produced by the Ca buffer. Although the reduction in Δ[Ca] could be caused by a reduction in either Δ[Ca]ps or Δ[Ca]os (Eq. 4), the analysis in a suggests that 0.5–1 mM fura-2 or 20 mM EGTA reduces Δ[Ca]os more than Δ[Ca]ps.\n\nThe amount of Ca inactivation that is able to develop in the presence of 0.5–1 mM fura-2 or 20 mM EGTA does not appear to be reduced substantially by 6 mM fura-2. In fibers with [CaSR]R = 1,306–2,852 μM, the value of f1 elicited by an action potential was constant at about 0.13 for values of [fura-2] between 2 and 4 mM. Moreover, the value of f1, as well as that of the peak fractional rate of release, decreased about twofold when [fura-2] was increased from 4 to 6 mM (Fig. 9 in Jong et al., 1995a). As the value of [fura-2] was increased from 2 to 4 to 6 mM, the turn-off of SR Ca release remained rapid and showed little or none of the slowing effect that was routinely observed when the value of [CaSR]R was decreased to values <1,000 μM.\n\nAnother measure of Ca inactivation is provided by the ratio f2/f1. f2, the fraction of SR Ca content that is released by the second action potential in a train, is less than f1, probably because Ca inactivation develops with the first stimulation and is only partially removed before the second stimulation. With action potentials 20 ms apart, the value of f2/f1 was 0.57 with 20 mM EGTA and 0.64 with 2–6 mM fura-2 (Fig. 10 A in Jong et al., 1995a); if Ca inactivation of Ca release had been eliminated by fura-2, the value of f2/f1 should have been unity.\n\nThe general conclusion from these results is that in fibers with [CaSR]R ≥ 1,300 μM, high affinity Ca buffers (20 mM EGTA or ≤6 mM fura-2) are able to increase the value of f1 by a modest amount but are unable either to increase the value of f1 above 0.17 or to slow the turn-off of SR Ca release. On the other hand, as shown in this article, the latter effects are routinely observed in fibers equilibrated with 20 mM EGTA when the value of [CaSR]R is reduced below 1,000 μM. If these effects are mediated by a Ca receptor located at a distance rCaR from the mouth of an SR Ca channel, the value of rCaR is expected to be small. For example, the value of 0.21 for f1 with [CaSR]R = 900 μM (Fig. 4 B) suggests that the value of Δ[Ca] at rCaR is smaller with [CaSR]R = 900 μM and 20 mM EGTA in the myoplasm than with [CaSR]R (null) 2,000 μM and 6 mM fura-2 in the myoplasm. With Eq. 2 and the assumption that φpoint is proportional to [CaSR]R, this means that 900exp(−rCaR/ 78) < 2,000exp(−rCaR/27) or, with rearrangement, exp(−rCaR/27)/exp(−rCaR/78) > 900/2,000 = 0.45; 27 and 78 nm are the values of λCa for 6 mM fura-2 and 20 mM EGTA plus 0.44 mM Ca, respectively (Eq. 4). From this it follows that rCaR is < 33 nm. If contributions from Δ[Ca]os (Eq. 4) are taken into account, the upper limit for rCaR is expected to be even smaller. For example, according to appendix a, if φpoint ≤ 1.5 × 106 ions s−1, rCaR is ≤ 22 nm. Such close proximity to the mouth of the channel would be consistent with a regulatory Ca receptor or receptors located on the SR Ca channel (including the foot structure) and/or its voltage sensor.\n\nSome of the assumptions used in this analysis may not hold exactly. In particular, the diffusion of Ca from an SR Ca channel may be more complicated than that assumed in Eq. 1 or 2; EGTA may not be uniformly distributed in the myoplasmic space that surrounds a channel; and more than one Ca receptor, each located at a different distance from the mouth of a channel, may be involved in the regulation of the processes that determine f1. Nonetheless, the inequality rCaR ≤ 22 nm suggests that at least some action of Ca on an SR Ca channel or its voltage sensor, which acts to decrease Ca release, is regulated by a Ca receptor that is close to the mouth of the channel itself. It is not known whether the Ca receptor (or receptors) responsible for this effect also participates in the increase in SR Ca release that is produced by the introduction of 0.5–1 mM fura-2 or 20 mM EGTA into the myoplasm (see above). Such participation is possible since this amount of Ca buffering is able to substantially decrease Δ[Ca]os and, as a consequence, Δ[Ca] near the mouth of an SR Ca channel ( a).\n\n### Effect of SR Ca Depletion on the OFF Kinetics of Qcm and the Decay Kinetics of dΔ[CaT]/dt\n\nAdditional information about the factors that cause the increase in f1 with partial SR Ca depletion can be obtained from voltage-clamp experiments. Partial SR Ca depletion slows the OFF kinetics of Qcm and the decay kinetics of dΔ[CaT]/dt after a brief depolarization, with a more marked effect on dΔ[CaT]/dt than on Qcm (Figs. 79). There is little or no effect on the ON kinetics of either signal.\n\nWith the smallest values of [CaSR]R studied, 140–300 μM, Qcm and dΔ[CaT]/dt returned to baseline with time courses that were similar within the noise of the dΔ[CaT]/dt signal (traces c in Fig. 9,A and associated text); with 1–2-min recovery periods, the mean values of τQcm and τdΔ[CaT]/dtwere 17.7 and 16.2 ms, respectively. Under these conditions, the value of τQcm was use-dependent; it decreased when the recovery period between trials was increased to 5–10 min (compare open symbols and filled circles in Fig. 8 D).\n\nThe general similarity between the values of τQcm and τdΔ[CaT]/dt with [CaSR]R = 140–300 μM is illustrated by a plot of the values ofτQcm/τdΔ[CaT]/dt, which are about 1 with some scatter (Fig. 9, B and C). Under these conditions, at late times after a repolarization, the rate constant that underlies the return of charge to its resting state may be the same as the rate constant for the closing of SR Ca channels. If charge can be described by a sequential model with n + 1 states x0, x1, . . , xn−1, xn, and the SR Ca channel is closed for states 0, 1, . . , n − 1 and open only for state n (see Jong et al., 1995b), closure of the SR Ca channel would be associated with the transition of charge from xn to xn−1. During repolarization to the holding potential (−90 mV), the forward rate constants from xi to xi+1, denoted by ki,i+1 with 0 ≤ i ≤ n − 1, can probably be neglected so that the rate constant for channel closure would be given simply by the backward rate constant from xn to xn−1, denoted by kn,n−1. If the value of kn,n−1 were sufficiently small compared with the values of the other reverse rate constants ki,i−1, with 1 ≤ i ≤ n − 1, kn,n−1 would also be the apparent rate constant for the return of charge to its resting state during the final stages of repolarization.\n\nWith [CaSR]R = 1,000–1,200 μM, the OFF kinetics of Qcm and the decay kinetics of dΔ[CaT]/dt were more rapid than those observed with [CaSR]R = 140–300 μM (Figs. 7,D, 8,D, and 9 A). The mean value of τQcm was 10.0 ms (compared with 17.7 ms) and the mean value of τdΔ[CaT]/dt was 2.5 ms (compared with 16.2 ms); thus, with larger [CaSR]R, a single rate constant no longer appears to describe both the closure of the SR Ca channels and the return of charge to its resting state. In terms of the sequential model for charge movement described in the preceding paragraph, it seems likely that the value of kn,n−1 is increased by an increase in [CaSR]R from 140–300 to 1,000–1,200 μM and that the time course of OFF Qcm may no longer depend only on the value of kn,n−1 but also on the values of other reverse rate constants. Without additional information, it is difficult to decide whether a decrease in kn,n−1 can explain all of the changes in the OFF kinetics of Qcm that occur when the value of [CaSR]R is decreased from 1,000–1,200 to 140–300 μM. It is also difficult to know which rate constants participate in the use dependence of τQcm that is observed with [CaSR]R = 140–300 μM.\n\nThe OFF kinetics of Qcm is expected to underlie the time course of the removal of voltage-dependent activation of SR Ca release after repolarization from a brief depolarization. Since Ca inactivation of Ca release may also contribute to the turn-off of dΔ[CaT]/dt, it is of interest to consider the relative importance of these two processes when the value of [CaSR]R is decreased from 1,000–1,200 to 140–300 μM in fibers equilibrated with 20 mM EGTA. Several observations, taken together, are consistent with the idea that the strength of Ca inactivation decreases with decreasing values of [CaSR]R in this range. First, Ca inactivation of Ca release appears to contribute to the turn-off of SR Ca release after action potential stimulation under normal conditions without fura-2 or EGTA in the myoplasm (preceding section). Second, although Ca inactivation appears to be reduced by 0.5–1 mM fura-2 or 20 mM EGTA, it is by no means removed. In cut fibers equilibrated with 20 mM EGTA and studied with two stimulations separated by a variable period of recovery, either an action potential or a 10–15-ms prepulse to −20 mV appeared to inactivate 90% of the ability of the SR to release Ca immediately after the first period of release ([CaSR]R = 1,813– 2,562 μM in the action potential experiments and 1,996–2,759 μM in the voltage-clamp experiments; Jong et al., 1995a). Third, since, with 20 mM EGTA, the Δ[Ca] signal at any myoplasmic location is expected to be proportional to dΔ[CaT]/dt (Pape et al., 1995), and the value of dΔ[CaT]/dt is approximately proportional to [CaSR] (Figs. 4,C, 6,C, and 7 B), a decrease in [CaSR]R from 1,000–1,200 to 140–300 μM would be expected to reduce the local myoplasmic Δ[Ca] signal severalfold; this, in turn, would likely decrease the development of Ca inactivation. Fourth, a decrease in [CaSR]R from 1,000–1,200 to 140–300 μM increased τdΔ[CaT]/dt by a factor of 6.47 but increased τQcm by only a factor of 1.71 if the recovery period was 1–2 min or 1.47 if recovery was 10 min. Although these observations do not prove rigorously that Ca inactivation is diminished when the value of [CaSR]R is decreased from 1,000–1,200 to 140– 300 μM in fibers equilibrated with 20 mM EGTA, they clearly show that the idea is reasonable.\n\nIn summary, if the value of [CaSR]R is increased from 140–300 to 1,000–1,200 μM, the turn-off of dΔ[CaT]/dt that occurs after an action potential or after repolarization from a brief depolarization is accelerated. This appears to be caused, at least in part, by an acceleration of the OFF kinetics of Qcm, at least if the recovery period between trials is ≤10 min as used in the experiments reported here. Ca inactivation of Ca release may also play a role. Although it is difficult to assess the relative contributions of charge movement and Ca inactivation to the turn-off of dΔ[CaT]/dt, the line of reasoning presented in the last two paragraphs of the preceding section suggests that at least some of the acceleration of the turn-off of SR Ca release through a particular SR Ca channel is regulated by a Ca receptor that is located ≤22 nm from the mouth of the channel. This receptor could act on the channel's voltage sensor to accelerate the OFF kinetics of Qcm and/or it could act on the channel to cause Ca inactivation of Ca release.\n\n### Effect of SR Ca Content on the Peak Value of dΔ[CaT]/dt\n\nIn the experiments described in this article, which used both action potential and voltage-clamp stimulation, the value of peak dΔ[CaT]/dt (in units of %/ms) was relatively constant when the value of [CaSR]R was varied between 140 and 1,200 μM (Figs. 4,C, 6,C, and 7 B). This observation may be relevant to the question of whether some SR Ca channels are gated open by Ca (Ca-induced Ca release; Ford and Podolsky, 1968; Endo et al., 1968; Endo, 1977), an idea that has received new attention with the electron microscope studies of Block et al. (1988) on muscle from toadfish swimbladder. Block et al. (1988) observed tetrads in the junctional tubular membrane, each of which is thought to consist of four dihydropyridine receptors, which function as the voltage sensors for SR Ca release (Ríos and Brum, 1987; Tanabe et al., 1988). Block et al. (1988) compared the spatial arrangement of these tetrads with that of the foot structures that span the region between the tubular and SR membranes (Franzini-Armstrong, 1975) and form the Ca channels in the SR membrane (Inui et al., 1987; Lai et al., 1988). The distance between tetrads along a row is twice that between foot structures, suggesting that only every other foot structure apposes a tetrad. This arrangement raises the possibility that the Ca channels in foot structures apposed to tetrads might be gated by voltage and that the Ca channels in alternate foot structures might be either silent or gated by a different mechanism (Block et al., 1988). Ríos and Pizarró (1988) expanded on this idea and speculated that the Ca channels in the nonapposed foot structures are gated by Ca itself. Under normal physiological conditions, this Ca would come from neighboring open Ca channels, which, immediately after depolarization, would be expected to be those gated by voltage.\n\nIn our experiments on fibers stimulated by brief pulses to −20 mV, dΔ[CaT]/dt (in units of %/ms) was decreased only slightly by reducing the value of [CaSR]R from 600–1,200 to 140–300 μM; the peak value of dΔ[CaT]/dt (in units of %/ms) with [CaSR]R = 140–300 μM was 0.861 (SEM, 0.016) times that with [CaSR]R = 600–1,200 μM (Fig. 7,B and associated text). The time course of ON dΔ[CaT]/dt (Fig. 7,A) and ON Qcm (Fig. 8 A) was little affected by these changes in [CaSR]R. The simplest interpretation of the relative constancy of the time course of ON dΔ[CaT]/dt and its peak value (in units of %/ms) under these conditions is that the Ca flux through an open SR Ca channel is approximately proportional to the value of [CaSR]R, and that the number of open SR Ca channels at the time of peak changes little when [CaSR]R is decreased from 600– 1,200 to 140–300 μM. Thus, if Ca-gated channels participate in SR Ca release when [CaSR]R = 600–1,200 μM, it seems likely that most of them also participate when [CaSR]R = 140–300 μM. If so, any model of SR Ca release (for example see Stern et al., 1997) must be able to explain how, in the presence of 20 mM EGTA, Ca gating is able to occur with [CaSR]R = 140–300 μM and the associated small rates of single-channel Ca flux but, with [EGTA] ≤ 0.1 mM, not spread regeneratively along the Z line when the value of [CaSR]R is almost certainly an order of magnitude larger (for example, in the local activation experiments of Huxley and Taylor and in the Ca spark experiments of Klein et al. ).\n\n## Acknowledgments\n\nWe thank the staff of the Biomedical Instrumentation Laboratory of the Yale Department of Cellular and Molecular Physiology for help with the design and construction of equipment. We also thank Drs. Steve Baylor and Stephen Hollingworth for many helpful discussions and for critically reading the manuscript.\n\nThis work was supported by the U.S. Public Health Service grant AM-37643.\n\n## Abbreviations used in this paper\n\n• EGTA\n\nethyleneglycol-bis-(β-aminoethyl ether)-N,N′-tetraacetic acid\n\n•\n• MOPS\n\n3-[N-Morpholino]-propanesulfonic acid\n\n•\n• pHR\n\nresting value of myoplasmic pH\n\n•\n• SR\n\nsarcoplasmic reticulum\n\n•\n• [Ca]R\n\nresting concentration of myoplasmic free Ca\n\n•\n• [CaSR]\n\nreadily releasable SR Ca content, expressed in terms of myoplasmic concentration\n\n•\n• [CaSR]R\n\nresting value of [CaSR]\n\n•\n• Δ[CaT]\n\nthe amount of Ca released from the SR\n\n•\n• f1\n\nthe fractional amount of [CaSR]R released from the SR by a single action potential\n\n•\n• Qcm\n\nintramembranous charge movement\n\n•\n• Icm\n\ncurrent from intramembranous charge movement\n\n### appendix a\n\nEqs. 1 and 2 predict the steady-state increase in free [Ca] (Δ[Ca]ps) that develops near the mouth of an open SR Ca channel that behaves as a point source in an infinite isotropic medium. In many situations of physiological interest, however, the value of [Ca] near a particular channel is increased substantially by Ca flux from neighboring channels (Δ[Ca]os). In this case, the total increase in free [Ca] (Δ[Ca]) is given by the sum of Δ[Ca]ps and Δ[Ca]os (Eq. 4). The aim of this appendix is to estimate Δ[Ca]ps and Δ[Ca]os, both in the absence of Ca buffers and in the presence of ATP, troponin, EGTA, or fura-2.\n\nAs it turns out, at distances ≤30 nm from the mouth of an SR Ca channel, the transient change in Δ[Ca]ps is nearly complete within 0.1 ms after a change in Ca flux. Consequently, steady-state equations can be used to estimate Δ[Ca]ps. Steady-state equations can also be used to estimate the contributions of individual channels to Δ[Ca]os in the presence of 20 mM EGTA or ≥0.5 mM fura-2. Without such Ca buffers, however, time-dependent solutions must be used for the estimation of Δ[Ca]os.\n\n#### General Description of the Estimation of Δ[Ca]os\n\nIn frog muscle, the transverse tubular system is located at the Z line and is arranged as a lattice surrounding individual myofibrils; the mean lattice spacing is 0.6–0.7 μm (Peachey, 1965). In electron micrographs, the tubules appear flattened and the membrane on each side is separated from the apposing membrane of the SR by two parallel rows of foot structures (Franzini-Armstrong, 1975; Block et al., 1988). The foot structures form the Ca release channels in the SR membrane (Inui et al., 1987; Lai et al., 1988). Half of the foot structures lie on each side of the transverse tubular system. From symmetry, we assume that the Ca that leaves a channel on one side of the plane of the Z line remains on that side and does not, on average, cross the plane to the other side. This restriction, while somewhat arbitrary, does not influence the general conclusions from these calculations. The diffusion of Ca from each individual channel is treated as that of Ca from a point source into a semi-infinite medium, which represents the myoplasm that extends from the appropriate side of the plane of the Z line to the end of the fiber. For this reason, any equations for Δ[Ca] that have been derived for a point source of Ca immersed in an infinite medium must be modified for the calculation of Δ[Ca]os. This modification, which requires replacing the factor φpoint/4π with φpoint/2π, must be done for the equations in the discussion of this paper, in appendixes b and c of this paper, and in appendixes b and c of Pape et al. (1995).\n\nIn the calculations described below,\n\n\\begin{equation}{\\Delta}[Ca]_{os}={ \\,\\substack{ \\\\ {\\sum} \\\\ _{i} }\\, }{\\Delta}[Ca]_{i},\\end{equation}\nA1\n\nwhere Δ[Ca]i represents the contribution of the ith Ca channel to Δ[Ca]os. The summation extends over all open Ca channels that lie on the specified side of the plane of the Z line, except the channel responsible for Δ[Ca]ps. The myoplasmic concentration of SR Ca channels or foot structures was taken to be 0.27 μM (Pape et al., 1995). With a sarcomere spacing of 3.6 μm, this concentration corresponds to a density of 585 SR Ca channels per μm2 in the plane of the Z line, half of which lie on each side of the Z line. The initial choice of the value of φpoint, 0.5 × 106 ions s−1 (which corresponds to 0.16 pA), was selected so that the rate of Ca release with all channels open, 135 μM/ms, would be similar to the mean peak rate of release determined in our action potential experiments on fibers equilibrated with 20 mM EGTA plus 1.76 mM Ca, 143 μM/ms (Pape et al., 1995). The values of the parameters associated with the Ca buffers used in the calculations are given in Table I; DCa = 3 × 10−6 cm2 s−1.\n\nEstimates were made with two different spatial arrangements of Ca channels, each suggested by a particular aspect of a muscle's structure. If the value of Δ[Ca]os is influenced by Ca flux through channels that are located as far away as several myofibrils from the channel responsible for Δ[Ca]ps, the estimate of Δ[Ca]os should be based on a two-dimensional arrangement of Ca channels in the plane of the Z line, such as the one illustrated in Fig. A1,A. On the other hand, if Δ[Ca]os contains contributions from Ca channels that are only a short distance away, Δ[Ca]os would depend mainly on Ca flux through the channels in the two parallel rows of foot structures that extend along one side of a short segment of tubule. In this case, Δ[Ca]os should be estimated with a quasi–one-dimensional arrangement of Ca channels, such as that illustrated in Fig. A1 B.\n\n#### Δ[Ca]ps and Δ[Ca]os Calculated with SR Ca Channels Arranged in Concentric Rings (Fig. A1 A) in the Absence of Ca Buffers and with 20 mM EGTA\n\nThe simplest example of Δ[Ca]ps and Δ[Ca]os is that without intrinsic Ca buffers; a more complete analysis with the main intrinsic Ca buffers in muscle will be considered in the following section. The calculations in this section use the hypothetical arrangement of Ca channels illustrated in Fig. A1,A. The filled circle indicates the location of the channel used to calculate Δ[Ca]ps. The open circles, arranged on concentric rings around the filled circle (only the first two rings are shown in the figure), indicate the locations of the channels used to calculate Δ[Ca]os. The radii of the rings are integer multiples of l, and the number of channels on each ring is eight times that integer; the channels are equally spaced on each ring. This symmetrical arrangement gives a uniform density of channels in the following sense: inside any circle of radius (n + 0.5)l, where n = 0, 1, 2, . . . , with the origin at the filled circle, the density of channels is constant, equal to one channel per πl2/4. In the calculations described in Figs. A2 and A3, the value of l was set to 66 nm to give a density of 585/2 channels per μm2. (See above. The factor 2 is used because only half the channels in a Z line are assumed to contribute Ca to the myoplasm on a particular side of the plane of the Z line.) Fig. A2 A shows the steady-state value of Δ[Ca]ps plotted as a function of r, the distance from the mouth of the channel. The details of the calculations for this and subsequent figures are given in the figure legends. The curve labeled “No Buffer” assumes the absence of any freely diffusible Ca buffer (Eq. 1). With the value of DCa used for these calculations (3 × 10−6 cm2 s−1) and with r ≤ 30 nm, Δ[Ca]ps reaches >90% of its steady level within 0.1 ms after a step change in Ca flux (from calculations with Eq. C20 with [Bi]R = 0 and DCa,app = DCa). Consequently, the steady-state relation between Δ[Ca]ps and r provides a good approximation of Δ[Ca]ps during an SR Ca release transient.\n\nThe curve labeled “EGTA” was calculated for the case of Ca in the presence of 18.24 mM EGTA (Eq. 2), the concentration that was used in our internal solution that contained 20 mM total EGTA plus 1.76 mM Ca. This steady-state relation can also be used to estimate Δ[Ca]ps during an SR Ca release transient since the steady state is reached within 0.1 ms after a change in Ca flux (Appendix C in Pape et al., 1995). Since the right-hand side of Eq. 2 is equal to exp(−rCa) times the right-hand side of Eq. 1, the two curves in Fig. A2 A are similar for small values of r and diverge at larger values.\n\nIn the absence of Ca buffers, the steady-state value of Δ[Ca]os that is obtained from Eq. A1 is not finite for either of the (infinite) spatial arrangements of channels illustrated in Fig. A1. In this situation, the time-dependent solution of the diffusion equation, Eq. C20 (with [Bi]R = 0 and DCa,app = DCa), can be used to estimate each individual channel contribution (Δ[Ca]i) to Δ[Ca]os in Eq. A1. For this purpose, r in Eq. C20 represents the distance from the mouth of the channel to the point of reference for Δ[Ca]os, and φpoint/4π on the right-hand side of Eq. C20 is replaced with φpoint/2π.\n\nThe continuous curve labeled “No Buffer” in Fig. A2,B shows the value of Δ[Ca]os during the first 10 ms after a step change in Ca flux from zero to φpoint through all of the neighboring sites (Fig. A1 A). Although this calculation was performed at r = 0 nm (corresponding to the mouth of the channel responsible for Δ[Ca]ps), closely similar curves of Δ[Ca]os vs. t were obtained at other nearby locations on the line that originates at the mouth of the Δ[Ca]ps channel and extends perpendicularly from the plane of the Z line. In this context, r represents the distance from the mouth of the Δ[Ca]ps channel to the point of reference. For 0.1 ≤ t ≤ 10 ms, the value of Δ[Ca]os at r = 30 nm is only 7–8 μM less than the value at r = 0 nm. Part of the reason for the small variation in Δ[Ca]os with r arises from considerations of symmetry, which require that ∂ Δ[Ca]os/∂ r = 0 at r = 0.\n\nThe dotted curve slightly above the “No Buffer” curve shows the value of Δ[Ca]x=0 calculated from Eq. C36 for a step of uniform Ca flux in the plane of the Z line, with a density φplane = φpoint/(πl2/4); the factor πl2/4 is equal to the area per channel for the arrangement of channels illustrated in Fig. A1 A. From 0.1 to 10 ms, the dotted curve lies 25.8–26.7 μM above the continuous curve. The close similarity of the two curves shows that, in the absence of Ca buffers, Δ[Ca]x=0 calculated with Eq. C36 provides a good approximation of Δ[Ca]os at r = 0.\n\nThe continuous curve labeled “EGTA” in Fig. A2 B shows Δ[Ca]os at r = 0 nm with 18.24 mM EGTA and no other Ca buffers (Eq. C17). As expected from the value of 22 μs for τCa (Pape et al., 1995), Δ[Ca]os had almost reached its final level of 42.6 μM by 0.1 ms, when the first point was plotted.\n\nThe dotted curve labeled “EGTA” shows the steady-state level of Δ[Ca]x=0, calculated for a uniform flux of Ca in the plane of the Z line in the presence of 18.24 mM EGTA (Eq. C31). The value of φplane was the same as that used for the “No Buffer” dotted curve. The steady level of Δ[Ca]os is 65.6 μM. The absolute difference between the final values of Δ[Ca]r=0 and Δ[Ca]os with EGTA, 23 μM, is similar to that without Ca buffers, 26– 27 μM (see above), although the relative difference is more pronounced.\n\nThe curves in Fig. A2,C show the range of distances over which Δ[Ca]i contributes to Δ[Ca]os in Eq. A1. The abscissa is d and the ordinate is fd, the fractional value of Δ[Ca]os that is contributed by channels that are located within a distance d from the channel used to calculate Δ[Ca]ps (indicated by the filled circle in Fig. A1,A); r = 0 nm. The curves marked by numbers were calculated without Ca buffers, at the times indicated (in ms). The curves become progressively broader with time because Ca ions are able to diffuse farther from their sources to contribute to Δ[Ca]os. At t = 3 ms, which is the approximate duration of SR Ca release after an action potential in fibers equilibrated with 20 mM EGTA and 1.76 mM Ca (Pape et al., 1995), fd = 0.50 at 660 nm (denoted by d0.5). This value of d0.5 is roughly equal to the mean lattice spacing of the myofibrils (see above). Thus, the value of Δ[Ca]os is determined by Ca ions that leave sites arranged along the perimeter of several myofibrils. On the other hand, with 18.24 mM EGTA, the value of fd was 0.56 with the first ring of sites at d = 66 nm (Fig. A1,A) and was 0.92 after inclusion of the third ring of sites at d = 198 nm. In this case, the value of Δ[Ca]os is determined primarily by Ca ions that leave sites along a short segment at the perimeter of a single myofibril, as illustrated by the arrangement of sites in Fig. A1 B (see below).\n\nIn the estimates of Δ[Ca]os described thus far, Eq. A1 was used with either Eq. C20 or C17, and the principle of superposition was implicitly assumed to hold. This is clearly the case in the absence of any Ca buffers. It is also expected to apply to fibers equilibrated with 20 mM EGTA because the fractional change in concentration of Ca-free buffer is expected to be small during the period of SR Ca release elicited by an action potential or a brief voltage pulse (Pape et al., 1995).\n\nThis section shows that steady-state equations can be used to estimate Δ[Ca]ps without Ca buffers and with 18.24 mM EGTA. Steady-state equations can also be used to estimate the individual contributions of Δ[Ca]i to Δ[Ca]os with 18.24 mM EGTA. On the other hand, in the absence of Ca buffers, transient equations must be used for Δ[Ca]os. Δ[Ca]os can make substantial contributions to Δ[Ca], especially in the absence of Ca buffers, when the value of d0.5 is large. Without Ca buffers, the value of Δ[Ca]os calculated with the arrangement of Ca channels in Fig. A1 A is very similar to that given by the analytical solution for a uniform Ca flux with φplane = φpoint/(πl2/4) (Eq. C36). These same general features apply with 5.5 mM ATP, as shown in the following section.\n\n#### Effect of ATP and Troponin on Δ[Ca]ps and Δ[Ca]os Calculated with SR Ca Channels Arranged in Concentric Rings (Fig. A1 A)\n\nThe principles that were used to estimate Δ[Ca]ps and Δ[Ca]os in Fig. A2 can also be applied to a situation that resembles muscle in which mobile and immobile intrinsic Ca buffers are present.\n\nFreely diffusible intrinsic Ca buffers will be considered first. According to Baylor and Hollingworth (1998), most of this kind of buffering power in muscle is due to ATP. ATP acts as a rapidly equilibrating, low affinity Ca buffer with an apparent forward rate constant for Ca complexation of 0.136 × 108 M−1s−1, a backward rate constant of 30,000 s−1, and an apparent dissociation constant of 2.2 mM (Table I); the effects of Mg and K complexation by ATP have been included in the estimation of these parameters. The total concentration of ATP used for the calculations here, 5.5 mM, is the same as that used in our internal end-pool solutions.\n\nThe continuous curves in Fig. A3,A show steady-state Δ[Ca]ps plotted as a function of r. Curve a shows Δ[Ca]ps without Ca buffers, replotted from the “No Buffer” curve in Fig. A2,A. Curve b was calculated with 5.5 mM ATP and the values of parameters in Table I, from the relation\n\n\\begin{equation}{\\Delta}[Ca]_{ps}=\\frac{{\\mathit{{\\phi}}}_{point}}{2{\\pi}D_{Ca}r}{\\cdot} \\left\\{ exp(-r/{\\lambda})+\\frac{{\\lambda}^{2}}{{\\lambda}_{ATP}^{2}}{\\cdot}[1-exp(-r/{\\lambda})] \\right\\} ,\\end{equation}\nA2\n\nin which\n\n\\begin{equation}{\\lambda}_{Ca}=\\sqrt{\\frac{D_{Ca}}{k_{1,ATP}[ATP]_{R}}}\\;,\\end{equation}\nA3\n\\begin{equation}{\\lambda}_{ATP}=\\sqrt{\\frac{D_{ATP}}{k_{1,ATP}[Ca]_{R}+k_{-1,ATP}}}\\;,\\end{equation}\nA4\n\nand\n\n\\begin{equation}1/{\\lambda}^{2}=1/{\\lambda}_{Ca}^{2}+1/{\\lambda}_{ATP}^{2}\\end{equation}\nA5\n\n(with the parameters in Table I, λCa = 63.2 nm, λATP = 68.3 nm, and λ = 46.4 nm). The dotted curve shows the negative change in [ATP] associated with curve b,\n\n\\begin{equation}-{\\Delta}[ATP]=\\frac{{\\mathit{{\\phi}}}_{point}}{2{\\pi}D_{ATP}r}{\\cdot}[1-exp(-r/{\\lambda})]{\\cdot} \\left[ 1-\\frac{{\\lambda}^{2}}{{\\lambda}_{ATP}^{2}} \\right] .\\end{equation}\nA6\n\nEqs. A2A6 are similar to equations derived for EGTA in Appendix B in Pape et al. (1995).\n\nCurve c in Fig. A3,A was also calculated with 5.5 mM ATP but with the simplifying assumptions that (a) the value of Δ[CaATP] is directly proportional to Δ[Ca] and (b) the equilibration of Ca with ATP is instantaneous. Assumption a follows from the relatively large value of Kd,ATP, 2.2 mM, and the relative constancy of [ATP]. The maximal change of 10.8 μM for −Δ[ATP] (dotted curve in Fig. A3 A) corresponds to only 0.2% of the value of [ATP]R, 5.5 mM. Assumption b follows approximately from the large value of k-1,ATP, 30,000 s−1, which sets a lower limit for the apparent rate constant of Ca equilibration with ATP. c gives equations for Δ[Ca] that have been derived with these assumptions and with the assumption that the diffusion coefficient of the Ca buffer (in this case, ATP) is the same with and without Ca. Curve c was calculated from Eq. C19 with αATP, the proportionality constant that relates Δ[CaATP] to [ATP]RΔ[Ca], equal to Kd,ATP−1 (Eq. C4).\n\nCurve b, which gives the exact solution for Δ[Ca]ps, is similar to curve a for small values of r. This is expected because, when Ca is near the mouth of the channel, little time has elapsed since it left the channel and had a chance to become complexed by ATP. At larger values of r, more time has elapsed for Ca to diffuse from its source and equilibrate with ATP. Consequently, curve b becomes similar to c, which was calculated with the assumption that the concentrations of Ca and ATP were in equilibrium.\n\nEq. C20 gives transient solutions at different values of r that are associated with the steady-state curves a and c. Within 0.1 ms after a step change in Ca flux, Δ[Ca]ps at r ≤ 30 nm reaches >90% of its steady level in a (with DCa,app = DCa = 3 × 10−6 cm2s−1; see text discussion of the “No Buffer” curve in Fig. A2 A) and >87% of its steady level in c (with DCa,app = 1.86 × 10−6 cm2s−1). Although an exact transient solution is not available for the conditions used to calculate curve b, it seems likely that such a solution would be bracketed by the transient solutions for a and c and that, consequently, at r ≤ 30 nm, it would reach >87% of the steady level within 0.1 ms. Thus, during SR Ca release, the steady-state curve b is expected to provide a good approximation of Δ[Ca]ps at the small values of r that are usually of interest (typically ≤30 nm).\n\nOn the other hand, the channels that contribute to Δ[Ca]os are located sufficiently far from the Δ[Ca]ps channel that the assumptions used for curve c are expected to apply reasonably well. Consequently, Eq. C20 can be used to estimate the transient solutions of Δ[Ca]i that are used to calculate Δ[Ca]os from Eq. A1. The continuous curve labeled “ATP” in Fig. A3,C shows Δ[Ca]os at r = 0 nm calculated in this manner. The neighboring dotted curve shows Δ[Ca]x=0, calculated for a uniform Ca flux in the plane of the Z line; from 0.1 to 10 ms, the dotted curve lies 11.8–12.4 μM above the continuous curve. Similar to the comparison of the continuous and dotted “No Buffer” curves in Fig. A2,B, Δ[Ca]x=0 in Fig. A3,C provides a good approximation of Δ[Ca]os at r = 0. The Δ[Ca]x=0 curves in Figs. A2,B and A3,C were both calculated from Eq. C36 and differ only by the value of the scaling factor c. According to Eq. C37, c = 1 without buffer (Fig. A2,B) and c = 0.36 with 5.5 mM ATP (Fig. A3 C). Thus, to a good approximation, the effect of 5.5 mM ATP on Δ[Ca]os is to reduce its value according to the factor 0.36.\n\nFig. A3,B shows the effect of 18.24 mM EGTA on the steady-state Δ[Ca]ps vs. r relation with 5.5 mM ATP. Curve a was calculated with ATP alone, redrawn from curve b in Fig. A3,A. Curve b was calculated with ATP and EGTA from the exact solution derived in b. Curve c was also calculated with ATP and EGTA but with the same simplifying assumptions that were used for curve c in Fig. A3 A, namely that the value of Δ[CaATP] is directly and instantaneously proportional to Δ[Ca] (Eq. C16). The transient solution with these assumptions, given by Eqs. C17 and C18, shows that, at r ≤ 30 nm, Δ[Ca]ps reaches >99% of its steady-state within 0.1 ms. Thus, for the same reasons used above for ATP alone, Δ[Ca]ps with ATP and EGTA can be estimated from the steady-state solution ( b) and the individual Δ[Ca]i used to estimate Δ[Ca]os can be calculated from the approximate transient solution, Eqs. C17 and C18.\n\nThe continuous curve labeled “EGTA” in Fig. A3,C shows Δ[Ca]os at r = 0 nm for 5.5 mM ATP and 18.24 mM EGTA, calculated from Eq. A1 with Δ[Ca]i calculated from Eqs. C17 and C18. Similar to the EGTA curve in Fig. A2 B, the final level is reached rapidly, within 0.2 ms. The final value, 33.5 μM, is similar to the exact steady-state value, 35.5 μM, that is obtained from Eq. A1 when Δ[Ca]i is calculated from the equations in b. This similarity validates the use of (the approximate) Eqs. C17 and C18. The corresponding Δ[Ca]x=0 curve (not shown), calculated with Eq. C31, lies above the continuous curve by the amount 11.2 μM (similar to the difference between the continuous and dotted ATP curves).\n\nThe continuous curves labeled 0.1, 0.3, and 1.0 in Fig. A3,C show Δ[Ca]os calculated for different concentrations of troponin, under the hypothetical condition that the concentration of sites available for Ca complexation does not change with time. With this restriction, Δ[Ca]os can be evaluated from Eq. A1 with the individual Δ[Ca]i calculated from Eqs. C17 and C18 with k1,Trop = 0.575 × 108M−1s−1 and [Trop]R = 0.1, 0.3, and 1.0 × 0.42 mM, respectively (Table I). The final values of the curves are 182.2, 100.3, and 49.9 μM, respectively. As the value of [Trop]R is increased, the time course becomes more rapid, and the final level becomes smaller. The Δ[Ca]x=0 curves lie 11.4–12.1 μM above the corresponding continuous curves; only the dotted 0.1 curve is shown.\n\nFig. A3,D shows fd plotted as a function of d, for the curves in Fig. A3 C. The curves labeled “ATP” and 0.1–1 were calculated at t = 3 ms. The curve labeled EGTA was calculated for steady-state conditions. The value of d0.5 is 527 nm with ATP and 66 nm with ATP and EGTA; the values for troponin are intermediate.\n\nIn the calculations with 5.5 mM ATP illustrated in Fig. A3, all the Δ[Ca]ps vs. r relations lie on or between curves a and b in Fig. A3,B. The difference between these curves is small, <3.1 μM. Thus, the decreases in Δ[Ca]os produced by 18.24 mM EGTA shown in Fig. A3 C are much larger than the associated decreases in Δ[Ca]ps. The significance of these results is considered in the final section of this appendix.\n\n#### Δ[Ca]os Calculated with SR Ca Channels Arranged in Two Parallel Rows (Fig. A1 B)\n\nWith the arrangement of SR Ca channels illustrated in Fig. A1,A, and with 5.5 mM ATP plus either 18.24 mM EGTA or ≥0.5 mM fura-2 in the myoplasm, fd ≥ 0.46 with the first ring of sites (see Fig. A3,D for EGTA). In this situation, a more reliable estimate of Δ[Ca]os is obtained with the arrangement of SR Ca channels in Fig. A1,B, with two rows of Ca channels on either side of a transverse tubule, similar to the two rows of feet observed by Franzini-Armstrong (1975) and Block et al. (1988). The spacing between channels in each row and the distance between rows is assumed to be the same, denoted by l, with l 30 nm (Franzini-Armstrong, 1975; Block et al., 1988). Although only 14 channels are shown in Fig. A1 B, the rows are assumed to extend indefinitely in both directions. The filled and open circles indicate the locations of the channels used to calculate Δ[Ca]ps and Δ[Ca]os, respectively.\n\nThe two curves labeled EGTA in Fig. A4,A show steady-state Δ[Ca]ps with 18.24 mM EGTA, with (continuous) and without (dotted) 5.5 mM ATP, plotted as a function of r. The curves were redrawn from curve b in Fig. A3,B and the EGTA curve in Fig. A2,A, respectively. The curves labeled “fura-2” are similar except that 6 mM fura-2 was used instead of EGTA. With either EGTA or fura-2, the continuous and dotted curves are similar. This indicates that ATP has little effect on Δ[Ca]ps in the presence of 18.24 mM EGTA or 6 mM fura-2 and, by inference, in the presence of ≥0.65 mM fura-2. The curve labeled “No Buffer” is plotted for comparison; it shows steady-state Δ[Ca]ps in the absence of any Ca buffer and is redrawn from Fig. A2,A. All Δ[Ca]ps curves considered in this article lie on or between the “No Buffer” and the continuous 6 mM fura-2 curve in Fig. A4 A.\n\nFig. A4 B shows Δ[Ca]os at r = 0, with (continuous) and without (dotted) 5.5 mM ATP, plotted as a function of [fura-2]R. These curves were calculated from Eq. A1; the steady-state relation was used for each Δ[Ca]i, an approximation that is reasonable with 18.24 mM EGTA (see above) and, for similar reasons, with ≥0.5 mM fura-2. If the concentration of a high affinity Ca buffer (with a small rate constant for Ca dissociation) is much larger than the amount of Ca released from the SR, the ability of the buffer to reduce free [Ca] near release sites is determined primarily by the product of its concentration and the forward rate constant for Ca complexation (Appendix B in Pape et al., 1995). The forward rate constant for fura-2 is 28 times that for EGTA, so that 18.24 mM EGTA is expected to be equivalent to 0.65 mM fura-2 in its ability to reduce free [Ca] and, consequently, Δ[Ca]os (Jong et al., 1995a). The diamond and square symbols were calculated with 18.24 mM EGTA, without and with ATP, respectively, and plotted with their abscissa values equal to 0.65 mM; Δ[Ca]os = 69.3 and 56.3 μM, respectively. As expected, the symbols lie very near the corresponding fura-2 curves.\n\n#### Application to Experimental Results\n\nThe calculations described above can be applied to two experimental results (see discussion): (a) The addition of 0.5–1.0 mM fura-2 or 20 mM EGTA to myoplasm appears to increase action potential–stimulated SR Ca release by as much as twofold and to increase the peak rate of SR Ca release under voltage-clamp by two to threefold. (b) Action potential–stimulated f1 is increased when the value of [CaSR]R is decreased in a manner that indicates that the value of Δ[Ca] at rCaR (the location of a putative regulatory Ca receptor) appears to be smaller with [CaSR]R = 900 μM and 20 mM EGTA in the myoplasm than with [CaSR]R 2,000 μM and 6 mM fura-2 in the myoplasm. These two experimental results will be considered in terms of a decrease in Δ[Ca] produced by an increase in [EGTA] from 0 to 18.24 mM (a) and by a change in Ca buffers from 18.24 mM EGTA to 6 mM fura-2 (b). The intrinsic Ca buffers used in this analysis are ATP and the Ca-regulatory sites on troponin, the main mobile and immobile myoplasmic Ca buffers, respectively.\n\n##### Result a.\n\nThis result is consistent with the idea that Δ[Ca] and, as a result, Ca inactivation of Ca release are reduced when the value of [EGTA] is increased from 0 to 18.24 mM. Since Δ[Ca]ps is reduced by <3.1 μM by 18.24 mM EGTA in the presence of 5.5 mM ATP (Fig. A3,B), any substantial decrease in Δ[Ca] would require a decrease in Δ[Ca]os. The two-dimensional arrangement of Ca channels in Fig. A1,A was used to analyze the effect of EGTA on Δ[Ca]os because the value of d0.5 is large without EGTA, 527 nm with 5.5 mM ATP and 660 nm without ATP, t = 3 ms (Figs. A3,D and A2 C, respectively).\n\nAccording to Fig. A3,C, at t = 3 ms (the approximate duration of SR Ca release elicited by an action potential in fibers equilibrated with 20 mM EGTA plus 1.76 mM Ca), Δ[Ca]os is 303 μM if 5.5 mM ATP is the only Ca buffer present. With the addition of Ca-regulatory sites on troponin, Δ[Ca]os is expected to be reduced during a transient but to be unchanged in the steady state when Ca complexation by the immobile sites is in equilibrium with free [Ca]. Although the transient reduction in Δ[Ca]os is difficult to evaluate analytically, it can be viewed qualitatively with the help of the curves labeled “0.1”–“1.0” in Fig. A3 C. After an action potential, when SR Ca release just begins, the curve labeled “1.0” is expected to apply. (The number refers to the fraction of sites that are unoccupied by Ca.) As Ca sites become occupied, the value of Δ[Ca]os is expected to progressively increase until a peak fractional occupancy of ∼0.9 is achieved (Hirota et al., 1989). The curves labeled 0.1–1.0 provide a rough idea of how this increase might take place: Δ[Ca]os moves from the 1.0 curve to the 0.1 curve through a series of intermediate states, including the 0.3 curve. At t = 3 ms, Δ[Ca]os = 49.9, 100.2, and 174.0 μM on the curves labeled 1.0, 0.3, and 0.1, respectively. Although this description of events is qualitative, it seems reasonable to expect the value of Δ[Ca]os to eventually lie near the 0.1 curve at t = 3 ms with a value of ∼174 μM.\n\nIf 18.24 mM EGTA is introduced, the development of steady-state conditions after a change in release is rapid, <0.1 ms, so that steady-state equations can be used. This gives Δ[Ca]os = 35.5 μM for the arrangement of Ca channels in Fig. A1,A and 56.3 μM for the arrangement in Fig. A1,B, which is probably more reliable. This large reduction in Δ[Ca]os, from 174 to 56.3 μM, provides a plausible basis for result a. Moreover, for values of r more than a few nanometers, the values of Δ[Ca]os are larger than those of Δ[Ca]ps. At r = 5 nm, for example, Δ[Ca]ps = 80.3–83.2 μM (values from curves a and b, respectively, in Fig. A3 B). Without EGTA, Δ[Ca] = Δ[Ca]ps + Δ[Ca]os = 80–83 + 174 = 254–257 μM and, with EGTA, Δ[Ca] = 80 + 56 = 136 μM. In this case, 18.24 mM EGTA reduces Δ[Ca]os 118 μM, and this reduction is mainly responsible for the almost twofold reduction in Δ[Ca], from 254–257 to 136 μM. As the value of r is increased above 5 nm, the relative contribution of Δ[Ca]os to Δ[Ca] becomes progressively larger. (The binding of Ca by troponin was not included in the estimate of Δ[Ca]os with EGTA because, for reasons that are not entirely clear, troponin appears to bind little if any Ca released by an action potential in fibers equilibrated with 20 mM EGTA plus 1.76 mM Ca; pages 293–294 in Pape et al., 1995.)\n\n##### Result b.\n\nThis result can be used to estimate an upper limit for rCaR, the distance of a putative regulatory Ca receptor from the mouth of an SR Ca channel. Steady-state equations are used to calculate Δ[Ca]ps and Δ[Ca]os with the arrangement of SR Ca channels in Fig. A1,B for Δ[Ca]os (Fig. A4).\n\nThe continuous curves in Fig. A5 show Δ[Ca] (= Δ[Ca]ps + Δ[Ca]os) calculated with 5.5 mM ATP and either 18.24 mM EGTA or 6 mM fura-2, as indicated, plotted as a function of r. The upper limit for rCaR is given by the value of r that satisfies the condition that Δ[Ca] with fura-2 is equal to 0.45 times Δ[Ca] with EGTA. The value 0.45 comes from the condition that Δ[Ca] at rCaR is smaller with [CaSR]R = 900 μM and 20 mM EGTA in the myoplasm than with [CaSR]R = 2,000 μM and 6 mM fura-2 in the myoplasm and from the assumption that φpoint is proportional to [CaSR]R (see discussion). The dotted curve shows the EGTA curve scaled by the factor 0.45. This curve intersects the fura-2 curve at r = 14.7 nm, which gives the inequality rCaR ≤ 14.7 nm. At the intersection, Δ[Ca] = 33.2 μM (Δ[Ca]ps = 16.6 μM, Δ[Ca]os = 16.6 μM with fura-2; Δ[Ca]ps = 10.3 μM, Δ[Ca]os = 22.9 μM with 18.24 mM EGTA and the scaling factor 0.45). At r = 5 nm (the value of r used for calculations above in a), Δ[Ca] = 91.7 μM (Δ[Ca]ps = 72.0 μM, Δ[Ca]os = 19.7 μM) with 6 mM fura-2.\n\nThe value of the upper limit of rCaR depends on the value used for φpoint. In the calculations illustrated in Figs. A2A5, φpoint = 0.5 × 106 ions s−1, which corresponds to a single channel current of 0.16 pA. This value of φpoint represents a lower limit since it was obtained from the peak rate of SR Ca release after an action potential on the assumption that all of the SR Ca channels were open. If some channels were closed, perhaps because they had not been activated by the voltage sensor or because they had been turned off by Ca inactivation, the value of φpoint would be larger.\n\nMejia-Alvarez et al. (1998) carried out experiments on single canine cardiac ryanodine receptor channels incorporated into planar lipid bilayers. From their results, they estimate that, under normal physiological conditions, the single channel current is <0.5 pA, corresponding to φpoint ≤ 1.5 × 106 ions s−1. Although an increase in the value of φpoint would increase the value of Δ[Ca]ps, the value of Δ[Ca]os, on average, should be unchanged because the spatially averaged rate of SR Ca release is unchanged; thus, any increase in φpoint used for the determination of Δ[Ca]os must be offset by a decrease in mean open probability so that the product of φpoint times mean open probability remains constant. Calculations similar to those in Fig. A5 give the inequality rCaR ≤ 21.6 nm for φpoint = 1.5 × 106 ions s−1. Thus, with the simplifying assumptions used for these calculations, it seems reasonable to conclude that rCaR ≤ 22 nm.\n\n### appendix b\n\nAppendix B in Pape et al. (1995) describes the steady-state changes in concentrations of free Ca, Ca-free EGTA, and Ca-complexed EGTA that are expected to develop in the myoplasm near an open SR Ca channel in a fiber equilibrated with EGTA. Although the equations were developed for EGTA, they apply to any Ca buffer that binds Ca with a 1:1 stoichiometry.\n\nThis appendix extends the theory to include any finite number of Ca buffers and, in particular, EGTA and ATP. The SR Ca channel is approximated by a point source of Ca immersed in an infinite isotropic medium, as was done in Appendix B in Pape et al. (1995). An important assumption in the derivations in Appendix B in Pape et al. (1995) and those given below is that the concentration of each Ca-free buffer decreases by only a relatively small amount, even at the Ca source.\n\nConsider N species of buffers. [Bi] and [CaBi] represent, respectively, the concentration of Ca-free and Ca-complexed buffer, i = 1, 2, .., N. Di represents the diffusion coefficient of the buffer, which is assumed to be independent of the state of Ca complexation; ki and k-i represent the forward and backward rate constants for Ca complexation, respectively.\n\nThe steady-state differential equation for the diffusion and complexation of Ca by the buffers, in spherical coordinates without angular dependence, is given by\n\n\\begin{equation}D_{Ca}\\{d^{2}[Ca]/{\\mathit{dr}}^{2}+(2/r)d[Ca]/{\\mathit{dr}}\\}-{ \\,\\substack{ ^{N} \\\\ {\\sum} \\\\ _{i=1} }\\, }\\{k_{i}[Ca][B_{i}]-k_{-i}[CaB_{i}]\\}=0.\\end{equation}\nB1\n\nFor each species of Ca buffer, there are two equations,\n\n\\begin{equation}D_{i}\\{d^{2}[B_{i}]/{\\mathit{dr}}^{2}+(2/r)d[B_{i}]/{\\mathit{dr}}\\}-\\{k_{i}[Ca][B_{i}]-k_{-i}[CaB_{i}]\\}=0\\end{equation}\nB2\n\nand\n\n\\begin{equation}D_{i}\\{d^{2}[CaB_{i}]/{\\mathit{dr}}^{2}+(2/r)d[CaB_{i}]/{\\mathit{dr}}\\}+\\{k_{i}[Ca][B_{i}]-k_{-i}[CaB_{i}]\\}=0,\\;\\;i=1,\\;2,\\;..,\\;N.\\end{equation}\nB3\n\nThe concentrations of free Ca and the Ca-free buffers can each be expressed as a change with respect to the resting concentration, denoted by the subscript R,\n\n\\begin{equation}{\\Delta}[Ca]=[Ca]-[Ca]_{R}\\end{equation}\nB4\n\nand\n\n\\begin{equation}{\\Delta}[B_{i}]=[B_{i}]-[B_{i}]_{R},\\;\\;\\;\\;i=1,\\;2,\\;..,\\;N.\\end{equation}\nB5\n\nAddition of Eqs. B2 and B3 shows that, in the absence of any sources or sinks for buffer, [Bi] + [CaBi] is constant and, consequently, Δ[CaBi] = −Δ[Bi].\n\nIf \|Δ[Bi]\| << [Bi]R, Eqs. B4 and B5 can be combined with B1 and B2 to give\n\n\\begin{equation}D_{Ca}\\{d^{2}{\\Delta}[Ca]/{\\mathit{dr}}^{2}+(2/r)d{\\Delta}[Ca]/{\\mathit{dr}}\\}={ \\,\\substack{ ^{N} \\\\ {\\sum} \\\\ _{i=1} }\\, }k_{i}[B_{i}]_{R}{\\Delta}[Ca]+{ \\,\\substack{ ^{N} \\\\ {\\sum} \\\\ _{i=1} }\\, }\\{k_{i}[Ca]_{R}+k_{-i}\\}{\\cdot}{\\Delta}[B_{i}]\\end{equation}\nB6\n\nand\n\n\\begin{equation}D_{i}\\{d^{2}{\\Delta}[B_{{\\mathit{i}}}]/{\\mathit{dr}}^{2}+(2/r)d{\\Delta}[B_{i}]/{\\mathit{dr}}\\}=k_{i}[B_{i}]_{R}{\\Delta}[Ca]+\\{k_{i}[Ca]_{R}+k_{-i}\\}{\\cdot}{\\Delta}[B_{i}],\\;\\;\\;\\;i=1,\\;2,\\;\\;..,\\;N.\\end{equation}\nB7\n\nAt any value of r, the combined flux of Ca and CaBi must equal that of the source, φpoint. Consequently, remembering that Δ[CaBi] = −Δ[Bi],\n\n\\begin{equation}{\\mathit{{\\phi}}}_{point}=-4{\\pi}r^{2}\\{D_{Ca}d{\\Delta}[Ca]/{\\mathit{dr}}\\;-{ \\,\\substack{ ^{N} \\\\ {\\sum} \\\\ _{i=1} }\\, }D_{i}d{\\Delta}[B_{i}]/dr\\}.\\end{equation}\nB8\n\nMultiplication by dr/(4πr2) and integration from r to ∞ gives\n\n\\begin{equation}\\frac{{\\mathit{{\\phi}}}_{point}}{4{\\pi}r}=D_{Ca}{\\Delta}[Ca]-{ \\,\\substack{ ^{N} \\\\ {\\sum} \\\\ _{i=1} }\\, }D_{i}{\\Delta}[B_{i}].\\end{equation}\nB9\n\nAt this stage, it is convenient to introduce the variable yi defined as follows,\n\n\\begin{equation}y_{0}={\\Delta}[Ca]r\\end{equation}\nB10\n\nand\n\n\\begin{equation}y_{i}={\\Delta}[B_{i}]r,\\;\\;\\;\\;i=1,\\;2,\\;..,\\;N.\\end{equation}\nB11\n\nSubstitution into Eqs. B6 and B7 gives\n\n\\begin{equation}D_{Ca}y_{0}^{\\prime\\prime}={ \\,\\substack{ ^{N} \\\\ {\\sum} \\\\ _{i=1} }\\, }k_{i}[B_{i}]_{R}y_{0}+{ \\,\\substack{ ^{N} \\\\ {\\sum} \\\\ _{i=1} }\\, }\\{k_{i}[Ca]_{R}+k_{-i}\\}{\\cdot}y_{i}\\end{equation}\nB12\n\nand\n\n\\begin{equation}D_{i}y_{i}^{\\prime\\prime}=k_{i}[B_{i}]_{R}y_{0}+\\{k_{i}[Ca]_{R}+k_{-i}\\}y_{i},\\;\\;\\;\\;i=1,\\;2,\\;..,\\;N,\\end{equation}\nB13\n\nwhere yi ″ denotes d2y/dr 2. Eqs. B12 and B13 can be rearranged to give\n\n\\begin{equation}y_{0}^{\\prime\\prime}=y_{0}{ \\,\\substack{ ^{N} \\\\ {\\sum} \\\\ _{i=1} }\\, }a_{i}+{ \\,\\substack{ ^{N} \\\\ {\\sum} \\\\ _{i=1} }\\, }b_{i}c_{i}y_{i}\\end{equation}\nB14\n\nand\n\n\\begin{equation}y_{i}^{\\prime\\prime}=y_{0}(a_{i}/c_{i})+b_{i}y_{i},\\;\\;\\;\\;i=1,\\;2,\\;..,\\;N,\\end{equation}\nB15\n\nwhere\n\n\\begin{equation}a_{i}=k_{i}[B_{i}]_{R}/D_{Ca},\\end{equation}\nB16\n\\begin{equation}b_{i}=(k_{i}[Ca]_{R}+k_{-i})/D_{i},\\end{equation}\nB17\n\nand\n\n\\begin{equation}c_{i}=D_{i}/D_{Ca}.\\end{equation}\nB18\n\nEq. B9, with Eqs. B10 and B11, can be rewritten,\n\n\\begin{equation}y_{0}=\\frac{{\\mathit{{\\phi}}}_{point}}{4{\\pi}D_{Ca}}+{ \\,\\substack{ ^{N} \\\\ {\\sum} \\\\ _{i=1} }\\, }c_{i}y_{i}.\\end{equation}\nB19\n\nEq. B14 and the N equations represented by Eq. B15 are not independent. N independent equations can be obtained by combining Eq. B19 with Eq. B15 to give\n\n\\begin{equation}y_{i}^{\\prime\\prime}=(a_{i}/c_{i})(\\frac{{\\mathit{{\\phi}}}_{point}}{4{\\pi}D_{Ca}}+{ \\,\\substack{ ^{N} \\\\ {\\sum} \\\\ _{i=1} }\\, }c_{i}y_{i})+b_{i}y_{i},\\;\\;\\;\\;i=1,\\;2,\\;..,\\;N.\\end{equation}\nB20\n\nEq. B20 can be written in matrix notation:\n\n\\begin{equation}Y^{\\prime\\prime}=MY+\\frac{{\\mathit{{\\phi}}}_{point}}{4{\\pi}D_{Ca}}F.\\end{equation}\nB21\n\nY is a column vector with N elements yi, i = 1, 2, .., N; M is an N × N matrix with the diagonal elements given by mii = ai + bi and the off diagonal elements given by mij = aicj/ci; F is a column vector with N elements given by fi = ai/ci. With\n\n\\begin{equation}{\\mathit{Z}}=Y+\\frac{{\\mathit{{\\phi}}}_{point}}{4{\\pi}D_{Ca}}M^{-1}F,\\end{equation}\nB22\n\nwhere M1 is the inverse of M, Eq. B21 becomes\n\n\\begin{equation}{\\mathit{Z}}^{\\prime\\prime}=M{\\mathit{Z}}.\\end{equation}\nB23\n\nM, in general, is nonsingular and can be diagonalized by a similarity transformation to give\n\n\\begin{equation}D=G^{-1}MG.\\end{equation}\nB24\n\nD is a diagonal matrix of the eigenvalues (dii) of M, and G is a matrix whose columns are the components of the unit eigenvectors of M. Since\n\n\\begin{equation}M=GDG^{-1},\\end{equation}\nB25\n\nEqs. B23 and B25 can be combined to give\n\n\\begin{equation}G^{-1}{\\mathit{Z}}^{\\prime\\prime}=G^{-1}GDG^{-1}{\\mathit{Z}}\\;\\;=DG^{-1}{\\mathit{Z}}.\\end{equation}\nB26\n\nSince G1Z″ = (G1Z)″, and the solutions of Eq. B26 must be finite at large values of r, G1Z has the general solution\n\n\\begin{equation}G^{-1}{\\mathit{Z}}=HE,\\end{equation}\nB27\n\nwhere H is an N × N diagonal matrix with elements hii and E is a column vector with elements ei = exp(-rdii). Premultiplying both sides by G gives\n\n\\begin{equation}{\\mathit{Z}}=GHE,\\end{equation}\nB28\n\nwhich, when combined with Eq. B22, gives\n\n\\begin{equation}{\\mathit{Y}}=GHE-\\frac{{\\mathit{{\\phi}}}_{point}}{4{\\pi}D_{Ca}}M^{-1}F.\\end{equation}\nB29\n\nSince the flux of Bi is zero at r = 0 (i = 1, 2, .., N), the value of Δ[Bi] must be finite everywhere (see, for example, the dotted curve in Fig. A3 A). Consequently, according to Eq. B11 and the definition of Y,\n\n\\begin{equation}y_{i}(r=0)=0,\\;\\;\\;\\;i=1,\\;2,\\;..,\\;N,\\end{equation}\nB30\n\\begin{equation}Y(r=0)=0,\\end{equation}\nB31\n\nand Eq. B29 (with Eq. B25) can be rearranged to give\n\n\\begin{equation}HE(r=0)=\\frac{{\\mathit{{\\phi}}}_{point}}{4{\\pi}D_{Ca}}D^{-1}G^{-1}F,\\end{equation}\nB32\n\nfrom which the values of hii can be determined. The yi's can be evaluated from Eq. B29 and y0 can be obtained from Eq. B19. Division of the yi's by r gives Δ[Ca] and Δ[Bi] (Eqs. B10 and B11).\n\nThe functional form of Δ[Ca] and of each Δ[Bi] consists of the sum of N terms of the form [constant × exp(-rdii)/r] plus a term of the form [constant/r]. The [constant × exp(-rdii)/r] terms can be summed over infinite distributions of Ca sources such as those illustrated in Fig. A1, A and B, to give finite contributions to Δ[Ca]os. On the other hand, a similar summation of the [constant/r] term is not finite.\n\n### appendix c\n\nJunge and McLaughlin (1987) and Irving et al. (1990), each with different assumptions, derived a partial differential equation for the apparent diffusion of a substance such as protons or Ca ions in the presence of mobile and immobile rapidly equilibrating buffers. For Ca buffers, an important limitation of these derivations is the assumption that the concentration of Ca-complexed buffer is directly proportional to free [Ca] at all times, an assumption that requires instantaneous equilibration of Ca and buffer. This requirement holds reasonably well for many low affinity Ca buffers, such as ATP (for r ≥ 60 nm, Fig. A3 A), but clearly doesn't hold for high affinity buffers such as EGTA that have a small rate constant for Ca dissociation.\n\nThis appendix extends the analyses of Junge and McLaughlin (1987) and Irving et al. (1990) in two ways: first, to include mobile high affinity Ca buffers such as EGTA from which Ca dissociates slowly (at least compared with the duration of SR Ca release) and second, to give the time course of free [Ca] that is expected to develop after a step change in Ca flux (a) near an open SR Ca channel, which is approximated by a point source of Ca immersed in an infinite isotropic medium and (b) near a planar source of Ca that can diffuse into a semi-infinite medium. As was the case with the analyses of Junge and McLaughlin (1987) and Irving et al. (1990), the equations given below can be applied to the diffusion of other substances such as protons that are buffered inside cells.\n\nAccording to Junge and McLaughlin (1987) and Irving et al. (1990), the partial differential equation that describes the diffusion of Ca in an isotropic medium in the presence of mobile and immobile rapidly equilibrating buffers is\n\n\\begin{equation}{\\partial}[Ca]/{\\partial}t+{ \\,\\substack{ ^{N} \\\\ {\\sum} \\\\ _{i=0} }\\, }{\\partial}[CaB_{i}]/{\\partial}t=D_{Ca}{\\nabla}^{2}[Ca]+{ \\,\\substack{ ^{N} \\\\ {\\sum} \\\\ _{i=1} }\\, }D_{i}{\\nabla}^{2}[CaB_{i}].\\end{equation}\nC1\n\nN denotes the number of species of mobile Ca buffers, each with a resting concentration [Bi]R and diffusion coefficient Di, which is assumed to be independent of the state of Ca complexation, i = 1, 2, .., N; [B0]R denotes the concentration of fixed or immobile Ca buffers. If [Ca] and [CaBi] are expressed as changes with respect to their resting values, as was done in b, Eq. C1 becomes\n\n\\begin{equation}{\\partial}{\\Delta}[Ca]/{\\partial}t+{ \\,\\substack{ ^{N} \\\\ {\\sum} \\\\ _{i=0} }\\, }{\\partial}{\\Delta}[CaB_{i}]/{\\partial}t=D_{Ca}{\\nabla}^{2}{\\Delta}[Ca]+{ \\,\\substack{ ^{N} \\\\ {\\sum} \\\\ _{i=1} }\\, }D_{i}{\\nabla}^{2}{\\Delta}[CaB_{i}].\\end{equation}\nC2\n\nAccording to Appendixes b and c in Pape et al. (1995), if a large concentration of a high affinity, slowly dissociating Ca buffer (such as EGTA) is used in the solution, any changes in buffer concentration are expected to be negligibly small, so that the buffer behaves as a sink that removes Ca at the rate kon[Buf]RΔ[Ca]. [Buf]R denotes the resting concentration of Ca-free buffer and kon denotes the forward rate constant for Ca complexation. Thus, if a high affinity, slowly dissociating Ca buffer is added to the solution of mobile and immobile rapidly equilibrating Ca buffers, Eq. C2 becomes\n\n\\begin{equation}{\\partial}{\\Delta}[Ca]/{\\partial}t+{ \\,\\substack{ ^{N} \\\\ {\\sum} \\\\ _{i=0} }\\, }{\\partial}{\\Delta}[CaB_{i}]/{\\partial}t=D_{Ca}{\\nabla}^{2}{\\Delta}[Ca]+{ \\,\\substack{ ^{N} \\\\ {\\sum} \\\\ _{i=1} }\\, }D_{i}{\\nabla}^{2}{\\Delta}[CaB_{i}]-k_{on}[Buf]_{R}{\\Delta}[Ca].\\end{equation}\nC3\n\nIf Δ[CaBi] is sufficiently small, it is expected to be proportional to Δ[Ca]. The proportionality constant (αi) can be defined by\n\n\\begin{equation}{\\Delta}[CaB_{i}]={\\mathit{{\\alpha}}}_{i}[B_{i}]_{R}{\\Delta}[Ca],\\;\\;\\;\\;i=0,\\;1,\\;..,\\;N.\\end{equation}\nC4\n\nIf the stoichiometry of Ca binding is 1:1 and [Ca]R is much smaller than the dissociation constant of the buffer (Kd,i), αiKd,i−1.\n\nReplacement of Δ[CaBi] in Eq. C3 by the right-hand side of Eq. C4 gives\n\n\\begin{equation}{\\partial}{\\Delta}[Ca]/{\\partial}t=D_{Ca,\\;app}{\\nabla}^{2}{\\Delta}[Ca]-{\\tau}_{Ca,app}^{-1}{\\Delta}[Ca].\\end{equation}\nC5\n\nThe apparent diffusion coefficient of Ca (DCa,app) is given by\n\n\\begin{equation}D_{Ca,app}=D_{Ca}\\frac{1+{ \\,\\substack{ ^{N} \\\\ {\\sum} \\\\ _{i=1} }\\, }{\\mathit{{\\alpha}}}_{i}[B_{i}]_{R}D_{i}/D_{Ca}}{1+{ \\,\\substack{ ^{N} \\\\ {\\sum} \\\\ _{i=0} }\\, }{\\alpha}_{i}[B_{i}]_{R}},\\end{equation}\nC6\n\nand the apparent time constant for Ca complexation by the high affinity Ca buffer (τCa,app) is given by\n\n\\begin{equation}{\\tau}_{Ca,app}={\\tau}_{Ca}(1+{ \\,\\substack{ ^{N} \\\\ {\\sum} \\\\ _{i=0} }\\, }{\\mathit{{\\alpha}}}_{i}[B_{i}]_{R}),\\end{equation}\nC7\n\nwhere\n\n\\begin{equation}{\\tau}_{Ca}=(k_{on}[Buf]_{R})^{-1}.\\end{equation}\nC8\n\nThe characteristic distance associated with Ca diffusion (λCa,app) is given by\n\n\\begin{equation}{\\lambda}_{Ca,app}=\\sqrt{{\\tau}_{Ca,app}D_{Ca,app}}\\end{equation}\nC9\n\\begin{equation}={\\lambda}_{Ca}\\sqrt{1+{ \\,\\substack{ ^{N} \\\\ {\\sum} \\\\ _{i=1} }\\, }{\\mathit{{\\alpha}}}_{{\\mathit{i}}}[B_{i}]_{R}D_{i}/D_{Ca}},\\end{equation}\nC10\n\nwhere\n\n\\begin{equation}{\\lambda}_{Ca}=\\sqrt{\\frac{D_{Ca}}{k_{on}[Buf]_{R}}}.\\end{equation}\nC11\n\nIn the absence of rapidly equilibrating Ca buffers, DCa,app = DCa, τCa,app = τCa, and λCa,app = λCa. Since the diffusion coefficient of Ca buffers is smaller than that of free Ca, Eq. C6 shows that rapidly equilibrating Ca buffers, both mobile and immobile, decrease the value of DCa,app. In addition, rapidly equilibrating Ca buffers, both mobile and immobile, increase the time required for the high affinity Ca buffer to complex Ca (Eq. C7). Rapidly equilibrating mobile Ca buffers increase the distance that Ca is able to diffuse before capture by the high affinity Ca buffer (Eq. C10).\n\n#### Δ[Ca] Near a Point Source of Ca Flux\n\nIn spherical coordinates without angular dependence, Eq. C5 can be written\n\n\\begin{equation}{\\partial}{\\Delta}[Ca]/{\\partial}t=D_{Ca,app}\\{{\\partial}^{2}{\\Delta}[Ca]/{\\partial}r^{2}+(2/r){\\partial}{\\Delta}[Ca]/{\\partial}r\\}-{\\tau}_{Ca,app}^{-1}{\\Delta}[Ca],\\end{equation}\nC12\n\nwhere r is the distance from the point source of Ca flux. Eq. C12 is similar to Eq. C3 in Pape et al. (1995).\n\nThe steady-state solution of Eq. C12 for an infinite medium is given by\n\n\\begin{equation}{\\Delta}[Ca]=\\frac{A}{r}exp(-r/{\\lambda}_{Ca,app})\\end{equation}\nC13\n\n(Neher, 1986; Stern, 1992; Pape et al., 1995). The value of the coefficient A is determined by the boundary condition at the Ca source,\n\n\\begin{equation}{\\mathit{{\\phi}}}_{point}=-Lim_{r{\\rightarrow}0}4{\\pi}r^{2}\\{D_{Ca}{\\partial}{\\Delta}[Ca]/{\\partial}r+{ \\,\\substack{ ^{N} \\\\ {\\sum} \\\\ _{i=1} }\\, }D_{i}{\\partial}{\\Delta}[CaB_{i}]/{\\partial}r\\}\\end{equation}\nC14\n\\begin{equation}\\;=4{\\pi}A\\{D_{Ca}+{ \\,\\substack{ ^{N} \\\\ {\\sum} \\\\ _{i=1} }\\, }{\\mathit{{\\alpha}}}_{i}[B_{i}]_{R}D_{i}\\}.\\end{equation}\nC15\n\nEqs. C13 and C15 give the desired steady-state solution,\n\n\\begin{equation}{\\Delta}[Ca]=\\frac{{\\mathit{{\\phi}}}_{point}}{4{\\pi}D_{Ca}r\\{1+{ \\,\\substack{ ^{N} \\\\ {\\sum} \\\\ _{i=1} }\\, }{\\alpha}_{i}[B_{i}]_{R}D_{i}/D_{Ca}\\}}exp(-r/{\\lambda}_{Ca,app}).\\end{equation}\nC16\n\nAfter a step change in φpoint, the transient solution of Eq. C12 is given by the product of the steady-state solution (right-hand side of Eq. C16) and F(r,t) (which equals unity as t → ∞),\n\n\\begin{equation}{\\Delta}[Ca]=\\frac{{\\mathit{{\\phi}}}_{point}}{4{\\pi}D_{Ca}r\\{1+{ \\,\\substack{ ^{N} \\\\ {\\sum} \\\\ _{i=1} }\\, }{\\alpha}_{i}[B_{i}]_{R}D_{i}/D_{Ca}\\}}{\\cdot}exp(-r/{\\lambda}_{Ca,app})F(r,t),\\end{equation}\nC17\n\nwhere\n\n\\begin{equation}F(r,t)=0.5[erfc \\left( \\frac{r}{\\sqrt{4D_{Ca,app}t}}-\\sqrt{t/{\\tau}_{Ca,app}} \\right) +exp(2r/{\\lambda}_{Ca,app}){\\cdot}erfc \\left( \\frac{r}{\\sqrt{4D_{Ca,app}t}}+\\sqrt{t/{\\tau}_{Ca,app}} \\right) ]\\end{equation}\nC18\n\n(Appendix C in Pape et al., 1995).\n\nIn the special case in which [Buf]R = 0, the values of λCa,app and τCa,app → ∞ and Eqs. C16 and C17 are replaced by\n\n\\begin{equation}{\\Delta}[Ca]=\\frac{{\\mathit{{\\phi}}}_{point}}{4{\\pi}D_{Ca}r\\{1+{ \\,\\substack{ ^{N} \\\\ {\\sum} \\\\ _{i=1} }\\, }{\\alpha}_{i}[B_{i}]_{R}D_{i}/D_{Ca}\\}}\\end{equation}\nC19\n\nand\n\n\\begin{equation}{\\Delta}[Ca]=\\frac{{\\mathit{{\\phi}}}_{point}}{4{\\pi}D_{Ca}r\\{1+{ \\,\\substack{ ^{N} \\\\ {\\sum} \\\\ _{i=1} }\\, }{\\alpha}_{i}[B_{i}]_{R}D_{i}/D_{Ca}\\}}{\\cdot}erfc \\left( \\frac{r}{\\sqrt{4D_{Ca,app}t}} \\right) ,\\end{equation}\nC20\n\nrespectively (see section 10.4 in Carslaw and Jaeger, 1959).\n\n#### Δ[Ca] Near a Planar Source of Ca Flux\n\nThe one-dimensional form of Eq. C5,\n\n\\begin{equation}{\\partial}{\\Delta}[Ca]/{\\partial}t=D_{Ca,\\;app}{\\partial}^{2}{\\Delta}[Ca]/{\\partial}x^{2}-{\\tau}_{Ca,\\;app}^{-1}{\\Delta}[Ca],\\;\\end{equation}\nC21\n\nis solved with the boundary condition that Ca enters the semi-infinite medium at x = 0.\n\nThe Laplace transform of Eq. C21 with the initial condition Δ[Ca] = 0 at t = 0 is given by\n\n\\begin{equation}p\\bar {{\\Delta}[Ca]}=D_{Ca,app}d^{2}\\bar {{\\Delta}[Ca]}/{\\mathit{dx}}^{2}-{\\tau}_{Ca,app}^{-1}\\bar {{\\Delta}[Ca]},\\end{equation}\nC22\n\nwhere Δ[Ca] is the Laplace transform of Δ[Ca] and p represents the transform variable. The general solution that is finite at large values of x is given by\n\n\\begin{equation}\\bar {{\\Delta}[Ca]}=A\\;exp(-x\\sqrt{(p+{\\tau}_{Ca,app}^{-1})/D_{Ca,app}}),\\end{equation}\nC23\n\nwhere the value of A is determined from the boundary condition that the Ca flux at x = 0 undergoes a step change from 0 to φplane at t = 0. This gives\n\n\\begin{equation}{\\mathit{{\\phi}}}_{plane}/p=-Lim_{x{\\rightarrow}0}\\{D_{Ca}d\\bar {{\\Delta}[Ca]}/{\\mathit{dx}}+{ \\,\\substack{ ^{N} \\\\ {\\sum} \\\\ _{i=1} }\\, }D_{i}d\\bar {{\\Delta}[CaB_{i}]}{\\mathit{dx}}\\}\\end{equation}\nC24\n\\begin{equation}=A\\{D_{Ca}+{ \\,\\substack{ ^{N} \\\\ {\\sum} \\\\ _{i=1} }\\, }{\\mathit{{\\alpha}}}_{i}[B_{i}]_{R}D_{i}\\}{\\cdot}\\sqrt{(p+{\\tau}_{Ca,app}^{-1})/D_{Ca,app}}\\end{equation}\nC25\n\\begin{equation}=A\\{1+{ \\,\\substack{ ^{N} \\\\ {\\sum} \\\\ _{i=0} }\\, }{\\mathit{{\\alpha}}}_{i}[B_{i}]_{R}\\}D_{Ca,app}{\\cdot}\\sqrt{(p+{\\tau}_{Ca,app}^{-1})/D_{Ca,app}}.\\end{equation}\nC26\n\nConsequently,\n\n\\begin{equation}\\bar {{\\Delta}[Ca]}=\\frac{{\\mathit{{\\phi}}}_{plane}exp(-x\\sqrt{(p+{\\tau}_{Ca,app}^{-1})/D_{Ca,app}})}{p\\{1+{ \\,\\substack{ ^{N} \\\\ {\\sum} \\\\ _{i=0} }\\, }{\\mathit{{\\alpha}}}_{i}[B_{i}]_{R}\\}\\sqrt{(p+{\\tau}_{Ca,app}^{-1})D_{Ca,app}}}.\\end{equation}\nC27\n\nFor the purposes of this paper, it is sufficient to consider the value of Δ[Ca] and its inverse Laplace transform at x = 0. If [Buf]R > 0 and, consequently, τCa,app is finite,\n\n\\begin{equation}\\bar {{\\Delta}[Ca]}_{x=0}=\\frac{{\\mathit{{\\phi}}}_{plane}}{p\\{1+{ \\,\\substack{ ^{N} \\\\ {\\sum} \\\\ _{i=0} }\\, }{\\mathit{{\\alpha}}}_{i}[B_{i}]_{R}\\}\\sqrt{(p+{\\tau}_{Ca,app}^{-1})D_{Ca,app}}}.\\end{equation}\nC28\n\nThe inverse transform is given by\n\n\\begin{equation}{\\Delta}[Ca]_{x=0}=\\frac{{\\mathit{{\\phi}}}_{plane}erf(\\sqrt{t/{\\tau}_{Ca,app}})}{\\{1+{ \\,\\substack{ ^{N} \\\\ {\\sum} \\\\ _{i=0} }\\, }{\\mathit{{\\alpha}}}_{i}[B_{i}]_{R}\\}\\sqrt{{\\tau}_{Ca,app}^{-1}D_{Ca,app}}}\\end{equation}\nC29\n\\begin{equation}=\\frac{{\\mathit{{\\phi}}}_{plane}{\\tau}_{Ca,app}}{\\{1+{ \\,\\substack{ ^{N} \\\\ {\\sum} \\\\ _{i=0} }\\, }{\\mathit{{\\alpha}}}_{i}[B_{i}]_{R}\\}{\\lambda}_{Ca,app}}{\\cdot}erf(\\sqrt{t/{\\tau}_{Ca,app}})\\end{equation}\nC30\n\\begin{equation}\\;=c\\frac{{\\mathit{{\\phi}}}_{plane}{\\tau}_{Ca}}{{\\lambda}_{Ca}}{\\cdot}erf(\\sqrt{t/{\\tau}_{Ca,app}}),\\end{equation}\nC31\n\nwhere\n\n\\begin{equation}c=\\frac{1}{\\sqrt{1+{ \\,\\substack{ ^{N} \\\\ {\\sum} \\\\ _{i=1} }\\, }{\\mathit{{\\alpha}}}_{i}[B_{i}]_{R}D_{i}/D_{Ca}}}.\\end{equation}\nC32\n\nThe steady-state solution is given by\n\n\\begin{equation}{\\Delta}[Ca]_{x=0}=c{\\cdot}\\frac{{\\mathit{{\\phi}}}_{plane}{\\tau}_{Ca}}{{\\lambda}_{Ca}}.\\end{equation}\nC33\n\nOn the other hand, if [Buf]R = 0 and τCa,app−1 = 0,\n\n\\begin{equation}\\bar {{\\Delta}[Ca]}_{x=0}=\\frac{{\\mathit{{\\phi}}}_{plane}}{p\\{1+{ \\,\\substack{ ^{N} \\\\ {\\sum} \\\\ _{i=0} }\\, }{\\mathit{{\\alpha}}}_{i}[B_{i}]_{R}\\}\\sqrt{pD_{Ca,app}}},\\end{equation}\nC34\n\nand the inverse transform is\n\n\\begin{equation}{\\Delta}[Ca]_{x=0}=\\frac{{\\mathit{{\\phi}}}_{plane}}{1+{ \\,\\substack{ ^{N} \\\\ {\\sum} \\\\ _{i=0} }\\, }{\\mathit{{\\alpha}}}_{i}[B_{i}]_{R}}\\sqrt{\\frac{4t}{{\\pi}D_{Ca,app}}}\\end{equation}\nC35\n\\begin{equation}\\;=c{\\cdot}{\\mathit{{\\phi}}}_{plane}\\sqrt{\\frac{4t}{{\\pi}D_{Ca}}},\\end{equation}\nC36\n\nwhere\n\n\\begin{equation}c=\\frac{1}{\\sqrt{1+{ \\,\\substack{ ^{N} \\\\ {\\sum} \\\\ _{i=1} }\\, }{\\mathit{{\\alpha}}}_{i}[B_{i}]_{R}D_{i}/D_{Ca}}\\sqrt{1+{ \\,\\substack{ ^{N} \\\\ {\\sum} \\\\ _{i=0} }\\, }{\\mathit{{\\alpha}}}_{i}[B_{{\\mathit{i}}}]_{R}}}.\\end{equation}\nC37\n\nIn this case, Δ[Ca]x=0 → ∞ as t → ∞, indicating that there is no steady-state solution. The value of c is unity in the absence of rapidly equilibrating Ca buffers and is 0.36 with 5.5 mM [ATP].\n\n## references\n\nBaylor\nSM\n,\nChandler\nWK\n,\nMarshall\nMW\nSarcoplasmic reticulum calcium release in frog skeletal muscle fibres estimated from arsenazo III calcium transients\nJ Physiol (Camb)\n1983\n344\n625\n666\n[PubMed]\nBaylor\nSM\n,\nHollingworth\nS\nFura-2 calcium transients in frog skeletal muscle fibres\nJ Physiol (Camb)\n1988\n403\n151\n192\n[PubMed]\nBaylor\nSM\n,\nHollingworth\nS\nModel of sarcomeric Ca2+ movements, including ATP Ca2+binding and diffusion, during activation of frog skeletal muscle\nJ Gen Physiol\n1998\n112\n297\n316\n[PubMed]\nBlatz\nAL\n,\nMagleby\nKL\nCalcium-activated potassium channels\nTrends Neurosci\n1987\n10\n463\n467\nBlock\nBA\n,\nImagawa\nT\n,\nCampbell\nKP\n,\nFranzini-Armstrong\nC\nStructural evidence for direct interaction between the molecular components of the transverse tubule/sarcoplasmic reticulum junction in skeletal muscle\nJ Cell Biol\n1988\n107\n2587\n2600\n[PubMed]\nCarslaw, H.S., and J.C. Jaeger. 1959. Conduction of Heat in Solids. 2nd edition. Oxford University Press, London, UK.\nChandler\nWK\n,\nHui\nCS\nMembrane capacitance in frog cut twitch fibers mounted in a double Vaseline-gap chamber\nJ Gen Physiol\n1990\n96\n225\n256\n[PubMed]\nChandler\nWK\n,\nRakowski\nRF\n,\nSchneider\nMF\nEffects of glycerol treatment and maintained depolarization on charge movement in skeletal muscle\nJ Physiol (Camb)\n1976\n254\n285\n316\n[PubMed]\nCsernoch\nL\n,\nJacquemond\nV\n,\nSchneider\nMF\nMicroinjection of strong calcium buffers suppresses the peak of calcium release during depolarization in frog skeletal muscle fibers\nJ Gen Physiol\n1993\n101\n297\n333\n[PubMed]\nEndo\nM\nCalcium release from the sarcoplasmic reticulum\nPhysiol Rev\n1977\n57\n71\n108\n[PubMed]\nEndo\nM\n,\nTanaka\nM\n,\nEbashi\nS\nRelease of calcium from sarcoplasmic reticulum in skinned fibers of the frog\nProc Intl Congr Physiol Sci\n1968\n7\n126\nFord\nLE\n,\nPodolsky\nRJ\nForce development and calcium movements in skinned muscle fibers\nFed Proc\n1968\n27\n375\nFranzini-Armstrong\nC\nMembrane particles and transmission at the triad\nFed Proce\n1975\n34\n1382\n1389\n[PubMed]\nGrynkiewicz\nG\n,\nPoenie\nM\n,\nTsien\nRY\nA new generation of Ca indicators with greatly improved fluorescence properties\nJ Biol Chem\n1985\n260\n3440\n3450\n[PubMed]\nHirota\nA\n,\nChandler\nWK\n,\nSouthwick\nPL\n,\nWaggoner\nAS\nCalcium signals recorded from two new purpurate indicators inside frog cut twitch fibers\nJ Gen Physiol\n1989\n94\n597\n631\n[PubMed]\nHodgkin\nAL\n,\nHuxley\nAF\nA quantitative description of membrane current and its application to conduction and excitation in nerve\nJ Physiol (Camb)\n1952\n117\n500\n544\n[PubMed]\nHollingworth\nS\n,\nHarkins\nAB\n,\nKurebayashi\nN\n,\nKonishi\nM\n,\nBaylor\nSM\nExcitation–contraction coupling in intact frog skeletal muscle fibers injected with mmolar concentrations of fura-2\nBiophys J\n1992\n63\n224\n234\n[PubMed]\nHui\nCS\n,\nChandler\nWK\nIntramembranous charge movement in frog cut twitch fibers mounted in a double Vaseline-gap chamber\nJ Gen Physiol\n1990\n96\n257\n297\n[PubMed]\nHui\nCS\n,\nChandler\nWK\nQβ and Qγcomponents of intramembranous charge movement in frog cut twitch fibers\nJ Gen Physiol\n1991\n98\n429\n464\n[PubMed]\nHui\nCS\n,\nChen\nW\nOrigin of delayed outward ionic current in charge movement traces from skeletal muscle\nJ Physiol (Camb)\n1994\n479\n109\n125\n[PubMed]\nHuxley\nAF\n,\nTaylor\nRE\nLocal activation of striated muscle fibres\nJ Physiol (Camb)\n1958\n144\n426\n441\n[PubMed]\nInui\nM\n,\nSaito\nA\n,\nFleischer\nS\nPurification of the ryanodine receptor and identity with feet structures of junctional terminal cisternae of sarcoplasmic reticulum from fast skeletal muscle\nJ Biol Chem\n1987\n262\n1740\n1747\n[PubMed]\nIrving\nM\n,\nMaylie\nJ\n,\nSizto\nNL\n,\nChandler\nWK\nPassive electrical and intrinsic optical properties of cut frog twitch fibers\nJ Gen Physiol\n1987\n89\n1\n40\n[PubMed]\nIrving\nM\n,\nMaylie\nJ\n,\nSizto\nNL\n,\nChandler\nWK\nIntracellular diffusion in the presence of mobile buffers—application to proton movement in muscle\nBiophys J\n1990\n57\n717\n721\n[PubMed]\nJacquemond\nV\n,\nCsernoch\nL\n,\nKlein\nMG\n,\nSchneider\nMF\nVoltage-gated and calcium-gated calcium release during depolarization of skeletal muscle fibers\nBiophys J\n1991\n60\n867\n873\n[PubMed]\nJong\nD-S\n,\nPape\nPC\n,\nBaylor\nSM\n,\nChandler\nWK\nCalcium inactivation of calcium release in frog cut muscle fibers that contain millimolar EGTA of fura-2\nJ Gen Physiol\n1995a\n106\n337\n388\n[PubMed]\nJong\nD-S\n,\nPape\nPC\n,\nChandler\nWK\nEffect of sarcoplasmic reticulum calcium depletion on intramembranous charge movement in frog cut muscle fibers\nJ Gen Physiol\n1995b\n106\n659\n704\n[PubMed]\nJong\nD-S\n,\nPape\nPC\n,\nChandler\nWK\n,\nBaylor\nSM\nReduction of calcium inactivation of sarcoplasmic reticulum calcium release in voltage-clamped cut twitch fibers with fura-2\nJ Gen Physiol\n1993\n102\n333\n370\n[PubMed]\nJunge\nW\n,\nMcLaughlin\nS\nThe role of fixed and mobile buffers in the kinetics of proton movement\nBiochim Biophys Acta\n1987\n890\n1\n5\n[PubMed]\nKlein\nMG\n,\nCheng\nH\n,\nSantana\nLF\n,\nJiang\nY-H\n,\nLederer\nWJ\n,\nSchneider\nMF\nTwo mechanisms of quantized calcium release in skeletal muscle\nNature\n1996\n379\n455\n458\n[PubMed]\nLai\nFA\n,\nErickson\nHP\n,\nRousseau\nE\n,\nLiu\nQ-Y\n,\nMeissner\nG\nPurification and reconstruction of the calcium release channel from skeletal muscle\nNature\n1988\n331\n315\n319\n[PubMed]\nMarkwardt\nF\n,\nIsenberg\nG\nGating of maxi K+ channels studied by Ca2+concentration jumps in excised inside-out multi-channel patches (myocytes from guinea pig urinary bladder)\nJ Gen Physiol\n1992\n99\n841\n862\n[PubMed]\nMejia-Alvarez\nR\n,\nKettlun\nC\n,\nRios\nE\n,\nStern\nM\n,\nFill\nM\nUnitary calcium currents through cardiac ryanodine receptors under physiological conditions\nBiophys J\n1998\n74\nA58\nNeher, E. 1986. Concentration profiles of intracellular calcium in the presence of a diffusible chelator. In Calcium Electrogenesis and Neuronal Functioning. U. Heinemann, M. Klee, E. Neher, and W. Singer, editors. Springer-Verlag, Berlin, Germany. 80–96.\nPage\nS\n,\nHuxley\nHE\nFilament lengths in striated muscle\nJ Cell Biol\n1963\n19\n369\n390\n[PubMed]\nPallotta\nBS\n,\nMagleby\nKL\n,\nBarrett\nJN\nSingle channel recordings of Ca2+ activated K+currents in rat muscle cell culture\nNature\n1981\n293\n471\n474\n[PubMed]\nPape\nPC\n,\nJong\nD-S\n,\nChandler\nWK\nCalcium release and its voltage dependence in frog cut muscle fibers equilibrated with 20 mM EGTA\nJ Gen Physiol\n1995\n106\n259\n336\n[PubMed]\nPape\nPC\n,\nJong\nD-S\n,\nChandler\nWK\n,\nBaylor\nSM\nEffect of fura-2 on action-potential-stimulated calcium release in cut twitch fibers from frog muscle\nJ Gen Physiol\n1993\n102\n295\n332\n[PubMed]\nPeachey\nLD\nThe sarcoplasmic reticulum and transverse tubules of the frog's sartorius\nJ Cell Biol\n1965\n25\n209\n231\n[PubMed]\nRíos\nE\n,\nBrum\nG\nInvolvement of dihydropyridine receptors in excitation-contraction coupling in skeletal muscle\nNature\n1987\n325\n717\n720\n[PubMed]\nRíos\nE\n,\nPizarró\nG\nVoltage sensors and calcium channels of excitation-contraction coupling\nNews Physiol Sci\n1988\n3\n223\n227\nSchneider\nMF\n,\nSimon\nBJ\nInactivation of calcium release from the sarcoplasmic reticulum in frog skeletal muscle\nJ Physiol (Camb)\n1988\n405\n727\n745\n[PubMed]\nSimon\nBJ\n,\nHill\nDA\nCharge movement and SR calcium release in frog skeletal muscle can be related by a Hodgkin-Huxley model with four gating particles\nBiophys J\n1992\n61\n1109\n1116\n[PubMed]\nSimon\nBJ\n,\nKlein\nMG\n,\nSchneider\nMF\nCalcium dependence of inactivation of calcium release from the sarcoplasmic reticulum in skeletal muscle fibers\nJ Gen Physiol\n1991\n97\n437\n471\n[PubMed]\nSimon\nBJ\n,\nSchneider\nMF\n,\nSzücs\nG\nInactivation of sarcoplasmic reticulum calcium release in frog skeletal muscle is mediated by calcium\nJ Gen Physiol\n1985\n86\n36a\nStern\nMD\nBuffering of calcium in the vicinity of a channel pore\nCell Calcium\n1992\n13\n183\n192\n[PubMed]\nStern\nMD\n,\nPizarro\nG\n,\nRíos\nE\nLocal control model of excitation-contraction coupling in skeletal muscle\nJ Gen Physiol\n1997\n110\n415\n440\n[PubMed]\nTanabe\nT\n,\nBeam\nKG\n,\nPowell\nJA\n,\nNuma\nS\nRestoration of excitation-contraction coupling and slow calcium current in dysgenic muscle by dihydropyridine receptor complementary DNA\nNature\n1988\n336\n134\n139\n[PubMed]\n\nDr. Pape's current address is Département de Physiologie et Biophysique, Faculté de Médecine, Université de Sherbrooke, Sherbrooke, Québec J1H5N4, Canada. Dr. Jong's current address is Department of Animal Science, National Taiwan University, Taipei 106, Taiwan, R.O.C\n\nPortions of this work were previously published in abstract form (Pape, P.C., D.-S. Jong, and W.K. Chandler. 1997. Biophys. J. 72:A120).\n\nNote added in proof. After the submission of the final manuscript for this article, the authors became aware of an article that describes a theoretical analysis of Ca diffusion in the presence of mobile Ca buffers that is similar to some of the theoretical results in our appendixes a–c (Naraghi, M., and E. Neher. 1997. Linearized buffered Ca2+ diffusion in microdomains and its implications for calculation of [Ca2+] at the mouth of a calcium channel. J. Neurosi. 17:6961–6973).\n\n## Author notes\n\nAddress correspondence to Dr. W.K. Chandler, Department of Cellular and Molecular Physiology, Yale University School of Medicine, 333 Cedar Street, New Haven, Connecticut 06510-8026. 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https://forum.espruino.com/conversations/315821/?offset=25	[ "# How to send a large amount of data faster ?\n\nPosted on\nPage\nof 2\n/ 2\n• Since you are using a NodeMCU the flashing map is described here:\n\nFlashNodeMCU\n\nI'm testing on a ESP8266-01 with 1MB of flash, I think the NodeMCU has 4MB.\nThe latest test program follows and allows for you to refresh the page to get multiple timings.\nI've also change the HTML script to display data.length instead of data.\nI've added process.memory() to fix a bug in E.toString() and asked for the bug to be examined in another post.\n\nOnce the page is online you might try from a command prompt ping IP address\nexample ping 192.168.1.13\nThis will give you an idea of the wifi speed and if there is a lot of traffic or errors occurring.\n\n``````//WSDesp8266B 28Jan2018\n\n//http//192.168.1.6:8080\n//ESP8266 with Espruino flahed\n\n// list of Wifi and passwords\nvar SSID=\"ssid\";\nvar key=\"keykey\";\nvar Startagain=0;\nvar myinterval;\n\nvar n =4, chk=21024;\nvar tdata=new Uint8Array(chk);\nvar i;\nfor(i=0;i<1024;i++){\ntdata[i]=0x30;\ntdata[i+1024]=0x31;\n}//next i\n\nvar page =\n\"<html>\\r\\n<body>\\r\\n<textarea id=\\\"demo\\\" rows=\\\"32\\\" cols=\\\"64\\\"></textarea>\\r\\n<script>\\r\\nvar ws;\\r\\nvar data=\\\"\\\";\\r\\n//setTimeout(function(){\\r\\nws = new WebSocket(\\\"ws://\\\" + location.host + \\\"/my_websocket\\\", \\\"protocolOne\\\");\\r\\nws.onmessage = function (event) { \\r\\ndata+=event.data;\\r\\ndocument.getElementById(\\\"demo\\\").innerHTML = data.length;\\r\\nws.send(\\\"Hello to Espruino!\\\"); \\r\\n };\\r\\n//setTimeout(function() { ws.send(\\\"Hello to Espruino!\\\"); }, //1000);}\\r\\n//,1000);\\r\\n\\r\\n</script></body>\\r\\n</html>\\r\\n\"\n;\nfunction onPageRequest(req, res) {\nres.end(page);\n}\n\nfunction test(){\n\nconsole.log(\"Start connection process\");\nvar wifi = require(\"Wifi\");\nconsole.log(\"Try Connecting to WiFi \",SSID);\nconsole.log(err);\nif (err){Startagain=1;return;\n}\nconsole.log(\"connected? err=\", err, \"info=\", wifi.getIP());\nconsole.log(\"Wi-Fi Connected\");\nclearInterval( myinterval);\nvar t=getTime();\nvar server = require('ws').createServer(onPageRequest);\nserver.listen(8080);\nserver.on(\"websocket\", function(ws) {\nws.on('message',function(msg) {\nn--;\nprint(\"n=\",n,\"t:\",(getTime()-t).toFixed(3),\"sec\");\nprint(\"[WS] \"+JSON.stringify(msg));\nif(n){\nws.send(E.toString(tdata));\nprocess.memory();\n}else{\nprocess.memory();\nn=4;\n}//end else\n});\nt=getTime();\nws.send(E.toString(tdata));\nprocess.memory();\nprint(\"n=\",n,\"t:\",(getTime()-t).toFixed(3),\"sec\");\n});\n});//end connect\n}//end test\n\nmyinterval=setInterval(function () {\nconsole.log(\"Test for error\");\nif(Startagain){\nStartagain=0;\ntest();\n}//end of Startagain\n}, 2000);\n\ntest();\n``````\n\nHere is some output:\n\n`````` 1v95 Copyright 2017 G.Williams\nEspruino is Open Source. Our work is supported\nonly by sales of official boards and donations:\nhttp://espruino.com/Donate\nFlash map 1MB:512/512, manuf 0xe0 chip 0x4014\n>Start connection process\nTry Connecting to WiFi faux\n=undefined\nTest for error\nTest for error\nnull\nconnected? err= null info= {\n\"ip\": \"192.168.1.13\",\n\"gw\": \"192.168.1.1\",\n\"mac\": \"18:fe:34:cb:2c:23\"\n}\nWi-Fi Connected\nn= 4 t: 0.016 sec\nn= 3 t: 0.098 sec\n[WS] \"Hello to Espruino!\"\nn= 2 t: 0.169 sec\n[WS] \"Hello to Espruino!\"\nn= 1 t: 0.239 sec\n[WS] \"Hello to Espruino!\"\nn= 0 t: 0.309 sec\n[WS] \"Hello to Espruino!\"\nn= 4 t: 0.012 sec\nn= 3 t: 0.083 sec\n[WS] \"Hello to Espruino!\"\nn= 2 t: 0.151 sec\n[WS] \"Hello to Espruino!\"\nn= 1 t: 0.881 sec\n[WS] \"Hello to Espruino!\"\nn= 0 t: 0.951 sec\n[WS] \"Hello to Espruino!\"\nn= 4 t: 0.012 sec\nn= 3 t: 0.089 sec\n[WS] \"Hello to Espruino!\"\nn= 2 t: 0.164 sec\n[WS] \"Hello to Espruino!\"\nn= 1 t: 0.233 sec\n[WS] \"Hello to Espruino!\"\nn= 0 t: 0.307 sec\n[WS] \"Hello to Espruino!\"\nn= 4 t: 0.012 sec\nn= 3 t: 0.089 sec\n[WS] \"Hello to Espruino!\"\nn= 2 t: 0.846 sec\n[WS] \"Hello to Espruino!\"\nn= 1 t: 0.909 sec\n[WS] \"Hello to Espruino!\"\nn= 0 t: 0.994 sec\n[WS\n``````\n\nSome variation in the speed\nHere are some ping timings:\n\n``````Microsoft Windows [Version 10.0.16299.192]\n\nC:\\Users\\jj>ping 192.168.1.13\n\nPinging 192.168.1.13 with 32 bytes of data:\nReply from 192.168.1.13: bytes=32 time=2ms TTL=128\nReply from 192.168.1.13: bytes=32 time=2ms TTL=128\nReply from 192.168.1.13: bytes=32 time=3ms TTL=128\nReply from 192.168.1.13: bytes=32 time=3ms TTL=128\n\nPing statistics for 192.168.1.13:\nPackets: Sent = 4, Received = 4, Lost = 0 (0% loss),\nApproximate round trip times in milli-seconds:\nMinimum = 2ms, Maximum = 3ms, Average = 2ms\n\nC:\\Users\\jj>ping 192.168.1.13\n\nPinging 192.168.1.13 with 32 bytes of data:\nReply from 192.168.1.13: bytes=32 time=2ms TTL=128\nReply from 192.168.1.13: bytes=32 time=1ms TTL=128\nReply from 192.168.1.13: bytes=32 time=2ms TTL=128\nReply from 192.168.1.13: bytes=32 time=3ms TTL=128\n\nPing statistics for 192.168.1.13:\nPackets: Sent = 4, Received = 4, Lost = 0 (0% loss),\nApproximate round trip times in milli-seconds:\nMinimum = 1ms, Maximum = 3ms, Average = 2ms\n``````\n• What is your physical distance to the router? Is it possible the antenna on the board is damaged?\n\nHave you tried the arduino software recently? Perhaps the hardware is damaged?\n\n• What about powesave setting for Wifi, check with `require('Wifi').getStatus().powersave`, output should be \"none\".\n\n• @Wilberforce, I started this conversation after seeing that on arduino my boards were doing ~100kb/sec, while on espruino it was only 1kb/sec, for server a file. In fact, I am using espruino since the v1.91, and all that time I thought it was a hardware limitation of esp8266. But it seems to me that the problem comes from the dSnd buffer sent slowly, every 200m, in small pieces of 256bytes.\n\nAs it works faster on arduino, the board is not defective. It works the same way on the 2 nodemcu-e12 v2 & v3 that I have. I restarted the router, and tried AP alone (at 20cm from my laptop), station alone and AP+ST, still the same damn result !\n\n@MaBe, I found it at ps-poll even though it was in AP+ST mode. No change when I set it to \"none\" using with Wifi.setConfig({powersave:\"none\"}).\n\n@ClearMemory041063, I tried without improvement also flashing the last expressif firmware with the expressif flashing tool like you did, just in case. I used \"esp_init_data_default_v05.bin\", because there was no \"esp_init_data_default.bin\". I could not get it working without also changing addresses of this file and of \"blank.bin\" to 0x3fc000 and 0x37e000. I got a nice :\n\n``````AT+GMR\nAT version:1.4.0.0(May 5 2017 16:10:59)\nSDK version:2.1.0(116b762)\ncompile time:May 5 2017 16:37:48\nOK\n``````\n\nThen I reflashed espruino, this time with expressif flashing tool. I used the files from this url. I used the addresses from your link and I tried with \"blank .bin\" at 0x3FE000 like I used to do, both works same way. Still as slow as before.\n\nI tried other browsers too, older espruino versions (v1.91, v1.94). The strangest that no one else ever have or noticed this problem before.\n\n`````` _____ _\n\| __\|___ ___ ___ _ _\|_\|___ ___\n\| __\|_ -\| . \| _\| \| \| \| \| . \|\n\|_____\|___\| _\|_\| \|___\|_\|_\|_\|___\|\n\|_\| http://espruino.com\nEspruino is Open Source. Our work is supported\nonly by sales of official boards and donations:\nhttp://espruino.com/Donate\nFlash map 4MB:1024/1024, manuf 0xef chip 0x4016\n>Start connection process\nTry Connecting to WiFi Gardening, cheaper than therapy\n=undefined\nTest for error\nTest for error\nTest for error\nnull\nconnected? err= null info= {\n\"ip\": \"192.168.2.6\",\n\"gw\": \"192.168.2.1\",\n\"mac\": \"2c:3a:e8:0e:9f:1c\"\n}\nWi-Fi Connected\nn= 4 t: 0.018 sec\nn= 3 t: 1.711 sec\n[WS] \"Hello to Espruino!\"\nn= 2 t: 3.186 sec\n[WS] \"Hello to Espruino!\"\nn= 1 t: 4.691 sec\n[WS] \"Hello to Espruino!\"\nn= 0 t: 6.165 sec\n[WS] \"Hello to Espruino!\"\n``````\n• #### Finaly got a 4MB ESP8266-12 (AI-Thinker ESP8266 MOD) assembled and working.\n\nI used 10k pullup resistors on the reset, Enable and GPIO 0 pins and a 10k pulldown resistor on GPIO 15.\nI used three batch files to run the ESPtool.py\nI first copied the batch files to the folder:\n\"C:\\Users\\jj\\Documents\\espruinoEsp8266Flash\\espruino_1v95_esp8266_4mb\"\nRun the batch files by clicking on them in file explorer after grounding GPIO 0 and resetting the chip.\nRun ESPinfo.bat and get the following.\n\n``````C:\\Users\\jj\\Documents\\espruinoEsp8266Flash\\espruino_1v95_esp8266_4mb>set /p pport=Enter a Com port\nEnter a Com port com3\n\nC:\\Users\\jj\\Documents\\espruinoEsp8266Flash\\espruino_1v95_esp8266_4mb>echo com3\ncom3\n\nC:\\Users\\jj\\Documents\\espruinoEsp8266Flash\\espruino_1v95_esp8266_4mb>pause\nPress any key to continue . . .\n\nC:\\Users\\jj\\Documents\\espruinoEsp8266Flash\\espruino_1v95_esp8266_4mb>esptool.py --port com3 --baud 115200 --no-stub chip_id\nesptool.py v2.2.1\nConnecting....\nDetecting chip type... ESP8266\nChip is ESP8266EX\nEnabling default SPI flash mode...\nChip ID: 0x00134e51\nHard resetting...\n\nC:\\Users\\jj\\Documents\\espruinoEsp8266Flash\\espruino_1v95_esp8266_4mb>esptool.py --port com3 --baud 115200 --no-stub read_mac\nesptool.py v2.2.1\nConnecting...\nDetecting chip type... ESP8266\nChip is ESP8266EX\nEnabling default SPI flash mode...\nMAC: 5c:cf:7f:13:4e:51\nHard resetting...\n\nC:\\Users\\jj\\Documents\\espruinoEsp8266Flash\\espruino_1v95_esp8266_4mb>esptool.py --port com3 --baud 115200 --no-stub flash_id\nesptool.py v2.2.1\nConnecting...\nDetecting chip type... ESP8266\nChip is ESP8266EX\nEnabling default SPI flash mode...\nManufacturer: e0\nDevice: 4016\nDetected flash size: 4MB\nHard resetting...\n\nC:\\Users\\jj\\Documents\\espruinoEsp8266Flash\\espruino_1v95_esp8266_4mb>pause\nPress any key to continue . . .\n``````\n\nThen I did a reset of the chip and ran EraseFlash.bat.\nThen I did a reset of the chip and ran\n\"C:\\Users\\jj\\Documents\\espruinoEsp8266Flash\\espruino_1v95_esp8266_4mb\\FlashESP8266_4M_40mhz.bat\"\nI un-grounded GPIO 0 and reset the chip.\nReconnect to WEBIDE and try the test program WSDesp8266C.js\nIt connects to my router and serves the HTML page but fails to transmit data via WebSockets.\n\n#### Try again\n\nPut the batch files into the following folder\n\"C:\\Users\\jj\\Documents\\espruinoEsp8266Flash\\espruino_1v95_esp8266\"\nRerun the ESPinfo.bat, EraseFlash.bat and FlashESP8266_4M_40mhz.bat.\nThe WSDesp8266C.js program produces the following output:\n\n`````` 1v95 Copyright 2017 G.Williams\nEspruino is Open Source. Our work is supported\nonly by sales of official boards and donations:\nhttp://espruino.com/Donate\nFlash map 4MB:512/512, manuf 0xe0 chip 0x4016\n>Start connection process\nTry Connecting to WiFi faux\n=undefined\nTest for error\nnull\nconnected? err= null info= {\n\"ip\": \"192.168.1.15\",\n\"gw\": \"192.168.1.1\",\n\"mac\": \"5c:cf:7f:13:4e:51\"\n}\nWi-Fi Connected\nn= 4 t: 0.018 sec\nn= 3 t: 0.095 sec\n[WS] \"2048\"\nn= 2 t: 0.179 sec\n[WS] \"4096\"\nn= 1 t: 0.256 sec\n[WS] \"6144\"\nn= 0 t: 1.133 sec\n[WS] \"8192\"\nn= 0 t: 1.135 sec\nn= 3 t: 1.241 sec\n[WS] \"10240\"\nn= 2 t: 1.312 sec\n[WS] \"12288\"\nn= 1 t: 1.383 sec\n[WS] \"14336\"\nn= 0 t: 1.453 sec\n[WS] \"16384\"\nn= 0 t: 1.455 sec\nn= 3 t: 1.527 sec\n[WS] \"18432\"\nn= 2 t: 2.325 sec\n[WS] \"20480\"\nn= 1 t: 2.394 sec\n[WS] \"22528\"\nn= 0 t: 2.470 sec\n[WS] \"24576\"\nn= 0 t: 2.473 sec\nn= 3 t: 2.541 sec\n[WS] \"26624\"\nn= 2 t: 2.617 sec\n[WS] \"28672\"\nn= 1 t: 2.685 sec\n[WS] \"30720\"\nn= 0 t: 2.757 sec\n[WS] \"32768\"\nn= 0 t: 2.759 sec\nn= 3 t: 2.839 sec\n[WS] \"34816\"\nn= 2 t: 2.940 sec\n[WS] \"36864\"\nn= 1 t: 3.013 sec\n[WS] \"38912\"\nn= 0 t: 3.089 sec\n[WS] \"40960\"\nn= 0 t: 3.092 sec\nn= 3 t: 3.168 sec\n[WS] \"43008\"\nn= 2 t: 3.243 sec\n[WS] \"45056\"\nn= 1 t: 3.317 sec\n[WS] \"47104\"\nn= 0 t: 3.388 sec\n[WS] \"49152\"\nn= 0 t: 3.391 sec\nn= 3 t: 3.457 sec\n[WS] \"51200\"\nn= 2 t: 3.526 sec\n[WS] \"53248\"\nn= 1 t: 4.493 sec\n[WS] \"55296\"\nn= 0 t: 4.567 sec\n[WS] \"57344\"\nn= 0 t: 4.569 sec\nn= 3 t: 4.646 sec\n[WS] \"59392\"\nn= 2 t: 4.715 sec\n[WS] \"61440\"\nn= 1 t: 4.780 sec\n[WS] \"63488\"\nn= 0 t: 4.853 sec\n[WS] \"65536\"\nn= 0 t: 4.855 sec\nn= 3 t: 4.932 sec\n[WS] \"67584\"\nn= 2 t: 4.998 sec\n[WS] \"69632\"\nn= 1 t: 5.075 sec\n[WS] \"71680\"\nn= 0 t: 5.146 sec\n[WS] \"73728\"\nn= 0 t: 5.148 sec\nn= 3 t: 5.221 sec\n[WS] \"75776\"\nn= 2 t: 5.294 sec\n[WS] \"77824\"\nn= 1 t: 5.366 sec\n[WS] \"79872\"\nn= 0 t: 5.432 sec\n[WS] \"81920\"\nn= 0 t: 5.434 sec\nn= 3 t: 5.499 sec\n[WS] \"83968\"\nn= 2 t: 6.244 sec\n[WS] \"86016\"\nn= 1 t: 6.310 sec\n[WS] \"88064\"\nn= 0 t: 6.389 sec\n[WS] \"90112\"\nn= 0 t: 6.391 sec\nn= 3 t: 6.459 sec\n[WS] \"92160\"\nn= 2 t: 6.528 sec\n[WS] \"94208\"\nn= 1 t: 6.597 sec\n[WS] \"96256\"\nn= 0 t: 6.668 sec\n[WS] \"98304\"\nn= 0 t: 6.669 sec\nn= 3 t: 6.739 sec\n[WS] \"100352\"\nn= 2 t: 6.810 sec\n[WS] \"102400\"\nn= 1 t: 7.732 sec\n[WS] \"104448\"\nn= 0 t: 7.800 sec\n[WS] \"106496\"\nn= 0 t: 7.802 sec\nn= 3 t: 7.872 sec\n[WS] \"108544\"\nn= 2 t: 7.944 sec\n[WS] \"110592\"\nn= 1 t: 8.013 sec\n[WS] \"112640\"\nn= 0 t: 8.083 sec\n[WS] \"114688\"\nn= 0 t: 8.085 sec\nn= 3 t: 8.155 sec\n[WS] \"116736\"\nn= 2 t: 8.233 sec\n[WS] \"118784\"\nn= 1 t: 8.299 sec\n[WS] \"120832\"\nn= 0 t: 8.367 sec\n[WS] \"122880\"\nn= 0 t: 8.369 sec\nn= 3 t: 8.446 sec\n[WS] \"124928\"\nn= 2 t: 8.513 sec\n[WS] \"126976\"\nn= 1 t: 8.587 sec\n[WS] \"129024\"\nn= 0 t: 8.659 sec\n[WS] \"131072\"\nn= 0 t: 8.662 sec\nn= 3 t: 8.730 sec\n[WS] \"133120\"\nn= 2 t: 8.797 sec\n[WS] \"135168\"\nn= 1 t: 8.871 sec\n[WS] \"137216\"\nn= 0 t: 8.939 sec\n[WS] \"139264\"\nn= 0 t: 8.941 sec\nn= 3 t: 9.012 sec\n[WS] \"141312\"\nn= 2 t: 9.079 sec\n[WS] \"143360\"\nn= 1 t: 9.152 sec\n[WS] \"145408\"\nn= 0 t: 9.221 sec\n[WS] \"147456\"\nn= 0 t: 9.224 sec\nn= 3 t: 9.290 sec\n[WS] \"149504\"\nn= 2 t: 9.363 sec\n[WS] \"151552\"\nn= 1 t: 9.434 sec\n[WS] \"153600\"\nn= 0 t: 12.071 sec\n[WS] \"155648\"\nn= 0 t: 12.073 sec\nn= 3 t: 12.153 sec\n[WS] \"157696\"\nn= 2 t: 12.229 sec\n[WS] \"159744\"\nn= 1 t: 12.302 sec\n[WS] \"161792\"\nn= 0 t: 12.372 sec\n[WS] \"163840\"\nn= 0 t: 12.374 sec\nn= 3 t: 12.449 sec\n[WS] \"165888\"\nn= 2 t: 12.513 sec\n[WS] \"167936\"\nn= 1 t: 12.584 sec\n[WS] \"169984\"\nn= 0 t: 12.653 sec\n[WS] \"172032\"\nn= 0 t: 12.655 sec\nn= 3 t: 12.738 sec\n[WS] \"174080\"\nn= 2 t: 12.821 sec\n[WS] \"176128\"\nn= 1 t: 12.889 sec\n[WS] \"178176\"\nn= 0 t: 12.960 sec\n[WS] \"180224\"\nn= 0 t: 12.963 sec\nn= 3 t: 13.033 sec\n[WS] \"182272\"\nn= 2 t: 13.104 sec\n[WS] \"184320\"\nn= 1 t: 13.171 sec\n[WS] \"186368\"\nn= 0 t: 13.248 sec\n[WS] \"188416\"\nn= 0 t: 13.250 sec\nn= 3 t: 13.319 sec\n[WS] \"190464\"\nn= 2 t: 13.385 sec\n[WS] \"192512\"\nn= 1 t: 14.278 sec\n[WS] \"194560\"\nn= 0 t: 14.349 sec\n[WS] \"196608\"\nn= 0 t: 14.352 sec\nn= 3 t: 14.425 sec\n[WS] \"198656\"\nn= 2 t: 14.500 sec\n[WS] \"200704\"\nn= 1 t: 14.568 sec\n[WS] \"202752\"\nn= 0 t: 14.635 sec\n[WS] \"204800\"\nn= 0 t: 14.637 sec\nn= 3 t: 14.716 sec\n[WS] \"206848\"\nn= 2 t: 14.792 sec\n[WS] \"208896\"\nn= 1 t: 14.860 sec\n[WS] \"210944\"\nn= 0 t: 14.935 sec\n[WS] \"212992\"\nn= 0 t: 14.938 sec\nn= 3 t: 15.015 sec\n[WS] \"215040\"\nn= 2 t: 15.079 sec\n[WS] \"217088\"\nn= 1 t: 15.147 sec\n[WS] \"219136\"\nn= 0 t: 15.213 sec\n[WS] \"221184\"\nn= 0 t: 15.215 sec\nn= 3 t: 15.286 sec\n[WS] \"223232\"\nn= 2 t: 15.355 sec\n[WS] \"225280\"\nn= 1 t: 16.028 sec\n[WS] \"227328\"\nn= 0 t: 16.100 sec\n[WS] \"229376\"\nn= 0 t: 16.103 sec\nn= 3 t: 16.175 sec\n[WS] \"231424\"\nn= 2 t: 16.249 sec\n[WS] \"233472\"\nn= 1 t: 16.323 sec\n[WS] \"235520\"\nn= 0 t: 16.391 sec\n[WS] \"237568\"\nn= 0 t: 16.393 sec\nn= 3 t: 16.462 sec\n[WS] \"239616\"\nn= 2 t: 16.535 sec\n[WS] \"241664\"\nn= 1 t: 16.604 sec\n[WS] \"243712\"\nn= 0 t: 16.673 sec\n[WS] \"245760\"\nn= 0 t: 16.675 sec\nn= 3 t: 16.743 sec\n[WS] \"247808\"\nn= 2 t: 16.810 sec\n[WS] \"249856\"\nn= 1 t: 16.878 sec\n[WS] \"251904\"\nn= 0 t: 16.948 sec\n[WS] \"253952\"\nn= 0 t: 16.950 sec\nn= 3 t: 17.023 sec\n[WS] \"256000\"\nn= 2 t: 17.089 sec\n[WS] \"258048\"\nn= 1 t: 17.166 sec\n[WS] \"260096\"\nn= 0 t: 17.244 sec\n[WS] \"262144\"\nn= 0 t: 17.246 sec\nn= 3 t: 17.316 sec\n[WS] \"264192\"\nn= 2 t: 18.124 sec\n[WS] \"266240\"\nn= 1 t: 18.194 sec\n[WS] \"268288\"\nn= 0 t: 18.265 sec\n[WS] \"270336\"\nn= 0 t: 18.268 sec\nn= 3 t: 18.339 sec\n[WS] \"272384\"\nn= 2 t: 18.407 sec\n[WS] \"274432\"\nn= 1 t: 18.477 sec\n[WS] \"276480\"\nn= 0 t: 18.543 sec\n[WS] \"278528\"\nn= 0 t: 18.546 sec\nn= 3 t: 18.612 sec\n[WS] \"280576\"\nn= 2 t: 19.313 sec\n[WS] \"282624\"\nn= 1 t: 19.378 sec\n[WS] \"284672\"\nn= 0 t: 19.447 sec\n[WS] \"286720\"\nn= 0 t: 19.449 sec\nn= 3 t: 19.528 sec\n[WS] \"288768\"\nn= 2 t: 19.599 sec\n[WS] \"290816\"\nn= 1 t: 19.666 sec\n[WS] \"292864\"\nn= 0 t: 19.734 sec\n[WS] \"294912\"\nn= 0 t: 19.736 sec\nn= 3 t: 19.815 sec\n[WS] \"296960\"\nn= 2 t: 19.886 sec\n[WS] \"299008\"\nn= 1 t: 19.955 sec\n[WS] \"301056\"\nn= 0 t: 20.031 sec\n[WS] \"303104\"\nn= 0 t: 20.034 sec\nn= 3 t: 20.109 sec\n[WS] \"305152\"\nn= 2 t: 20.179 sec\n[WS] \"307200\"\nn= 1 t: 20.245 sec\n[WS] \"309248\"\nn= 0 t: 20.316 sec\n[WS] \"311296\"\nn= 0 t: 20.318 sec\nn= 3 t: 20.391 sec\n[WS] \"313344\"\nn= 2 t: 20.464 sec\n[WS] \"315392\"\nn= 1 t: 20.532 sec\n[WS] \"317440\"\nn= 0 t: 20.602 sec\n[WS] \"319488\"\nn= 0 t: 20.604 sec\nn= 3 t: 20.685 sec\n[WS] \"321536\"\nn= 2 t: 20.753 sec\n[WS] \"323584\"\nn= 1 t: 20.830 sec\n[WS] \"325632\"\nn= 0 t: 20.899 sec\n[WS] \"327680\"\nn= 0 t: 20.901 sec\nn= 3 t: 20.967 sec\n[WS] \"329728\"\nn= 2 t: 21.037 sec\n[WS] \"331776\"\nn= 1 t: 21.104 sec\n[WS] \"333824\"\nn= 0 t: 21.172 sec\n[WS] \"335872\"\nn= 0 t: 21.174 sec\nn= 3 t: 21.245 sec\n[WS] \"337920\"\nn= 2 t: 21.345 sec\n[WS] \"339968\"\nn= 1 t: 21.440 sec\n[WS] \"342016\"\nn= 0 t: 21.509 sec\n[WS] \"344064\"\nn= 0 t: 21.512 sec\nn= 3 t: 21.586 sec\n[WS] \"346112\"\nn= 2 t: 21.661 sec\n[WS] \"348160\"\nn= 1 t: 21.731 sec\n[WS] \"350208\"\nn= 0 t: 21.798 sec\n[WS] \"352256\"\nn= 0 t: 21.801 sec\nn= 3 t: 21.876 sec\n[WS] \"354304\"\nn= 2 t: 21.971 sec\n[WS] \"356352\"\nn= 1 t: 22.041 sec\n[WS] \"358400\"\nn= 0 t: 22.109 sec\n[WS] \"360448\"\nn= 0 t: 22.112 sec\nn= 3 t: 22.185 sec\n[WS] \"362496\"\nn= 2 t: 22.253 sec\n[WS] \"364544\"\nn= 1 t: 22.321 sec\n[WS] \"366592\"\nn= 0 t: 22.388 sec\n[WS] \"368640\"\nn= 0 t: 22.389 sec\nn= 3 t: 22.455 sec\n[WS] \"370688\"\nn= 2 t: 22.522 sec\n[WS] \"372736\"\nn= 1 t: 22.589 sec\n[WS] \"374784\"\nn= 0 t: 22.657 sec\n[WS] \"376832\"\nn= 0 t: 22.659 sec\nn= 3 t: 22.730 sec\n[WS] \"378880\"\nn= 2 t: 22.801 sec\n[WS] \"380928\"\nn= 1 t: 22.869 sec\n[WS] \"382976\"\nn= 0 t: 22.939 sec\n[WS] \"385024\"\nn= 0 t: 22.941 sec\nn= 3 t: 23.008 sec\n[WS] \"387072\"\nn= 2 t: 23.078 sec\n[WS] \"389120\"\nn= 1 t: 23.149 sec\n[WS] \"391168\"\nn= 0 t: 23.219 sec\n[WS] \"393216\"\nn= 0 t: 23.221 sec\nn= 3 t: 24.147 sec\n[WS] \"395264\"\nn= 2 t: 24.227 sec\n[WS] \"397312\"\nn= 1 t: 24.296 sec\n[WS] \"399360\"\nn= 0 t: 24.361 sec\n[WS] \"401408\"\nn= 0 t: 24.364 sec\nn= 3 t: 24.439 sec\n[WS] \"403456\"\nn= 2 t: 24.518 sec\n[WS] \"405504\"\nn= 1 t: 25.341 sec\n[WS] \"407552\"\nn= 0 t: 25.408 sec\n[WS] \"409600\"\nn= 0 t: 25.410 sec\nn= 3 t: 25.481 sec\n[WS] \"411648\"\nn= 2 t: 25.556 sec\n[WS] \"413696\"\nn= 1 t: 25.621 sec\n[WS] \"415744\"\nn= 0 t: 25.691 sec\n[WS] \"417792\"\nn= 0 t: 25.693 sec\nn= 3 t: 25.765 sec\n[WS] \"419840\"\nn= 2 t: 25.830 sec\n[WS] \"421888\"\nn= 1 t: 28.289 sec\n[WS] \"423936\"\nn= 0 t: 28.356 sec\n[WS] \"425984\"\nn= 0 t: 28.358 sec\nn= 3 t: 28.439 sec\n[WS] \"428032\"\nn= 2 t: 29.252 sec\n[WS] \"430080\"\nn= 1 t: 29.325 sec\n[WS] \"432128\"\nn= 0 t: 29.396 sec\n[WS] \"434176\"\nn= 0 t: 29.398 sec\nn= 3 t: 29.467 sec\n[WS] \"436224\"\nn= 2 t: 29.541 sec\n[WS] \"438272\"\nn= 1 t: 29.609 sec\n[WS] \"440320\"\nn= 0 t: 29.679 sec\n[WS] \"442368\"\nn= 0 t: 29.682 sec\nn= 3 t: 29.758 sec\n[WS] \"444416\"\nn= 2 t: 31.375 sec\n[WS] \"446464\"\nn= 1 t: 31.443 sec\n[WS] \"448512\"\nn= 0 t: 31.510 sec\n[WS] \"450560\"\nn= 0 t: 31.512 sec\nn= 3 t: 31.589 sec\n[WS] \"452608\"\nn= 2 t: 31.655 sec\n[WS] \"454656\"\nn= 1 t: 31.724 sec\n[WS] \"456704\"\nn= 0 t: 32.612 sec\n[WS] \"458752\"\nn= 0 t: 32.614 sec\nn= 3 t: 32.683 sec\n[WS] \"460800\"\nn= 2 t: 32.757 sec\n[WS] \"462848\"\nn= 1 t: 32.833 sec\n[WS] \"464896\"\nn= 0 t: 32.905 sec\n[WS] \"466944\"\nn= 0 t: 32.907 sec\nn= 3 t: 32.975 sec\n[WS] \"468992\"\nn= 2 t: 33.041 sec\n[WS] \"471040\"\nn= 1 t: 33.803 sec\n[WS] \"473088\"\nn= 0 t: 33.875 sec\n[WS] \"475136\"\nn= 0 t: 33.877 sec\nn= 3 t: 33.953 sec\n[WS] \"477184\"\nn= 2 t: 34.017 sec\n[WS] \"479232\"\nn= 1 t: 34.087 sec\n[WS] \"481280\"\nn= 0 t: 34.162 sec\n[WS] \"483328\"\nn= 0 t: 34.164 sec\nn= 3 t: 34.241 sec\n[WS] \"485376\"\nn= 2 t: 34.307 sec\n[WS] \"487424\"\nn= 1 t: 34.373 sec\n[WS] \"489472\"\nn= 0 t: 35.107 sec\n[WS] \"491520\"\nn= 0 t: 35.109 sec\nn= 3 t: 35.182 sec\n[WS] \"493568\"\nn= 2 t: 35.251 sec\n[WS] \"495616\"\nn= 1 t: 35.318 sec\n[WS] \"497664\"\nn= 0 t: 35.394 sec\n[WS] \"499712\"\nn= 0 t: 35.397 sec\nn= 3 t: 35.478 sec\n[WS] \"501760\"\nn= 2 t: 35.545 sec\n[WS] \"503808\"\nn= 1 t: 35.618 sec\n[WS] \"505856\"\nn= 0 t: 35.689 sec\n[WS] \"507904\"\nn= 0 t: 35.691 sec\nn= 3 t: 36.604 sec\n[WS] \"509952\"\nn= 2 t: 36.672 sec\n[WS] \"512000\"\nn= 1 t: 36.738 sec\n[WS] \"514048\"\nn= 0 t: 36.814 sec\n[WS] \"516096\"\nn= 0 t: 36.816 sec\nn= 3 t: 36.886 sec\n[WS] \"518144\"\nn= 2 t: 36.951 sec\n[WS] \"520192\"\nn= 1 t: 37.026 sec\n[WS] \"522240\"\nn= 0 t: 37.095 sec\n[WS] \"524288\"\nn= 0 t: 37.097 sec\nn= 3 t: 37.168 sec\n[WS] \"526336\"\nn= 2 t: 37.240 sec\n[WS] \"528384\"\nn= 1 t: 37.309 sec\n[WS] \"530432\"\nn= 0 t: 37.377 sec\n[WS] \"532480\"\nn= 0 t: 37.379 sec\nn= 3 t: 37.445 sec\n[WS] \"534528\"\nn= 2 t: 37.518 sec\n[WS] \"536576\"\nn= 1 t: 37.589 sec\n[WS] \"538624\"\nn= 0 t: 37.656 sec\n[WS] \"540672\"\nn= 0 t: 37.658 sec\nn= 3 t: 38.473 sec\n[WS] \"542720\"\nn= 2 t: 38.547 sec\n[WS] \"544768\"\nn= 1 t: 38.622 sec\n[WS] \"546816\"\nn= 0 t: 38.689 sec\n[WS] \"548864\"\nn= 0 t: 38.691 sec\nn= 3 t: 38.767 sec\n[WS] \"550912\"\nn= 2 t: 38.835 sec\n[WS] \"552960\"\nn= 1 t: 38.900 sec\n[WS] \"555008\"\nn= 0 t: 38.974 sec\n[WS] \"557056\"\nn= 0 t: 38.977 sec\nn= 3 t: 40.535 sec\n[WS] \"559104\"\nn= 2 t: 40.609 sec\n[WS] \"561152\"\nn= 1 t: 40.682 sec\n[WS] \"563200\"\nn= 0 t: 40.747 sec\n[WS] \"565248\"\nn= 0 t: 40.749 sec\nn= 3 t: 40.825 sec\n[WS] \"567296\"\nn= 2 t: 40.905 sec\n[WS] \"569344\"\nn= 1 t: 40.969 sec\n[WS] \"571392\"\nn= 0 t: 41.039 sec\n[WS] \"573440\"\nn= 0 t: 41.041 sec\nn= 3 t: 41.117 sec\n[WS] \"575488\"\nn= 2 t: 41.185 sec\n[WS] \"577536\"\nn= 1 t: 41.253 sec\n[WS] \"579584\"\nn= 0 t: 41.331 sec\n[WS] \"581632\"\nn= 0 t: 41.333 sec\nn= 3 t: 41.457 sec\n[WS] \"583680\"\nn= 2 t: 41.524 sec\n[WS] \"585728\"\nn= 1 t: 41.592 sec\n[WS] \"587776\"\nn= 0 t: 43.310 sec\n[WS] \"589824\"\nn= 0 t: 43.312 sec\nn= 3 t: 43.384 sec\n[WS] \"591872\"\nn= 2 t: 43.464 sec\n[WS] \"593920\"\nn= 1 t: 43.535 sec\n[WS] \"595968\"\nn= 0 t: 43.604 sec\n[WS] \"598016\"\nn= 0 t: 43.607 sec\nn= 3 t: 43.685 sec\n[WS] \"600064\"\nn= 2 t: 43.749 sec\n[WS] \"602112\"\nn= 1 t: 43.818 sec\n[WS] \"604160\"\nn= 0 t: 43.889 sec\n[WS] \"606208\"\nn= 0 t: 43.892 sec\nn= 3 t: 43.971 sec\n[WS] \"608256\"\nn= 2 t: 44.038 sec\n[WS] \"610304\"\nn= 1 t: 44.104 sec\n[WS] \"612352\"\nn= 0 t: 44.194 sec\n[WS] \"614400\"\nn= 0 t: 44.196 sec\nn= 3 t: 45.118 sec\n[WS] \"616448\"\nn= 2 t: 45.184 sec\n[WS] \"618496\"\nn= 1 t: 45.252 sec\n[WS] \"620544\"\nn= 0 t: 45.324 sec\n[WS] \"622592\"\nn= 0 t: 45.326 sec\nn= 3 t: 45.396 sec\n[WS] \"624640\"\nn= 2 t: 45.471 sec\n[WS] \"626688\"\nn= 1 t: 45.538 sec\n[WS] \"628736\"\nn= 0 t: 46.349 sec\n[WS] \"630784\"\nn= 0 t: 46.352 sec\nn= 3 t: 46.471 sec\n[WS] \"632832\"\nn= 2 t: 46.549 sec\n[WS] \"634880\"\nn= 1 t: 46.613 sec\n[WS] \"636928\"\nn= 0 t: 46.679 sec\n[WS] \"638976\"\nn= 0 t: 46.681 sec\nn= 3 t: 46.754 sec\n[WS] \"641024\"\nn= 2 t: 46.821 sec\n[WS] \"643072\"\nn= 1 t: 47.575 sec\n[WS] \"645120\"\nn= 0 t: 47.641 sec\n[WS] \"647168\"\nn= 0 t: 47.643 sec\nn= 3 t: 47.714 sec\n[WS] \"649216\"\nn= 2 t: 47.787 sec\n[WS] \"651264\"\nn= 1 t: 47.862 sec\n[WS] \"653312\"\nn= 0 t: 47.926 sec\n[WS] \"655360\"\nn= 0 t: 47.928 sec\nn= 3 t: 48.006 sec\n[WS] \"657408\"\nn= 2 t: 48.079 sec\n[WS] \"659456\"\nn= 1 t: 48.145 sec\n[WS] \"661504\"\n>reset();\n=undefined\n``````\n\n661,504 8 / 48.125 = 109,964 Bits per second\n\n5 Attachments\n\n• Thanks, very handy the .bat files. I followed the procedure with no change in speed. I placed the files in the \"espruino_1v95_esp8266\" directory. I ran info, erase and flash batch files.\n\n``````C:\\Users\\Pok\\myfolder\\esp\\tools\\espruino_1v95_esp8266>esptool.py --port COM20 --\nbaud 115200 --no-stub chip_id\nesptool.py v2.0.1\nConnecting....\nDetecting chip type... ESP8266\nChip is ESP8266\nEnabling default SPI flash mode...\nChip ID: 0x000e9f1c\nHard resetting...\n\nC:\\Users\\Pok\\myfolder\\esp\\tools\\espruino_1v95_esp8266>esptool.py --port COM20 --\nesptool.py v2.0.1\nConnecting....\nDetecting chip type... ESP8266\nChip is ESP8266\nEnabling default SPI flash mode...\nMAC: 2c:3a:e8:0e:9f:1c\nHard resetting...\n\nC:\\Users\\Pok\\myfolder\\esp\\tools\\espruino_1v95_esp8266>esptool.py --port COM20 --\nbaud 115200 --no-stub flash_id\nesptool.py v2.0.1\nConnecting....\nDetecting chip type... ESP8266\nChip is ESP8266\nEnabling default SPI flash mode...\nManufacturer: ef\nDevice: 4016\nDetected flash size: 4MB\nHard resetting...\n``````\n``````C:\\Users\\Pok\\myfolder\\esp\\tools\\espruino_1v95_esp8266>esptool.py --port COM20 er\nase_flash\nesptool.py v2.0.1\nConnecting....\nDetecting chip type... ESP8266\nChip is ESP8266\nRunning stub...\nStub running...\nErasing flash (this may take a while)...\nChip erase completed successfully in 7.4s\nHard resetting...\n``````\n``````C:\\Users\\Pok\\myfolder\\esp\\tools\\espruino_1v95_esp8266>esptool.py --port COM20 --\nbaud 115200 write_flash --flash_freq 40m --flash_mode qio --flash_size 4MB 0x000\n0 \"boot_v1.6.bin\" 0x1000 espruino_esp8266_user1.bin 0x3FC000 esp_init_data_defau\nlt.bin 0x3FE000 blank.bin\nesptool.py v2.0.1\nConnecting....\nDetecting chip type... ESP8266\nChip is ESP8266\nRunning stub...\nStub running...\nConfiguring flash size...\nFlash params set to 0x0040\nCompressed 3856 bytes to 2763...\nWrote 3856 bytes (2763 compressed) at 0x00000000 in 0.2 seconds (effective 124.4\nkbit/s)...\nHash of data verified.\nCompressed 456756 bytes to 320993...\nWrote 456756 bytes (320993 compressed) at 0x00001000 in 28.3 seconds (effective\n129.1 kbit/s)...\nHash of data verified.\nCompressed 128 bytes to 75...\nWrote 128 bytes (75 compressed) at 0x003fc000 in 0.0 seconds (effective 78.8 kbi\nt/s)...\nHash of data verified.\nCompressed 4096 bytes to 26...\nWrote 4096 bytes (26 compressed) at 0x003fe000 in 0.0 seconds (effective 4095.8\nkbit/s)...\nHash of data verified.\n\nLeaving...\nHard resetting...\n``````\n\nI replugged and reflashed the WSdesp8266C. It does not go faster than before.\n\n`````` _____ _\n\| __\|___ ___ ___ _ _\|_\|___ ___\n\| __\|_ -\| . \| _\| \| \| \| \| . \|\n\|_____\|___\| _\|_\| \|___\|_\|_\|_\|___\|\n\|_\| http://espruino.com\nEspruino is Open Source. Our work is supported\nonly by sales of official boards and donations:\nhttp://espruino.com/Donate\nFlash map 4MB:512/512, manuf 0xef chip 0x4016\n>Start connection process\nTry Connecting to WiFi Gardening, cheaper than therapy\n=undefined\nTest for error\nTest for error\nTest for error\nnull\nconnected? err= null info= {\n\"ip\": \"192.168.2.5\",\n\"gw\": \"192.168.2.1\",\n\"mac\": \"2c:3a:e8:0e:9f:1c\"\n}\nWi-Fi Connected\n>\n=undefined\nn= 4 t: 0.018 sec\nn= 3 t: 1.688 sec\n[WS] \"2048\"\nn= 2 t: 3.165 sec\n[WS] \"4096\"\nn= 1 t: 4.655 sec\n[WS] \"6144\"\nn= 0 t: 6.156 sec\n[WS] \"8192\"\nn= 0 t: 6.159 sec\nn= 3 t: 7.659 sec\n[WS] \"10240\"\nn= 2 t: 9.126 sec\n[WS] \"12288\"\nn= 1 t: 10.600 sec\n[WS] \"14336\"\nn= 0 t: 12.089 sec\n[WS] \"16384\"\nn= 0 t: 12.091 sec\nn= 3 t: 13.568 sec\n[WS] \"18432\"\nn= 2 t: 15.080 sec\n[WS] \"20480\"\nn= 1 t: 16.564 sec\n``````\n\n1 Attachment\n\n•", null, "•", null, "Post a reply" ]	[ null, "https://espruino.microco.sm/api/v1/files/58c67aa84b299e97ef7b484383f4ef90c0c40604.JPG", null, "https://forum.espruino.com/static/img/avatar.gif", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.65858173,"math_prob":0.86622727,"size":22061,"snap":"2021-43-2021-49","text_gpt3_token_len":10019,"char_repetition_ratio":0.33381692,"word_repetition_ratio":0.0988075,"special_character_ratio":0.5879153,"punctuation_ratio":0.21415667,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9673134,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,2,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-27T17:16:47Z\",\"WARC-Record-ID\":\"<urn:uuid:7e4ee172-c9cc-4e1e-85a6-148c74645d2f>\",\"Content-Length\":\"64592\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a1589744-0644-4fde-ad1c-b656532c082e>\",\"WARC-Concurrent-To\":\"<urn:uuid:9048c81e-527c-4936-81a4-da434f29eccb>\",\"WARC-IP-Address\":\"176.58.105.63\",\"WARC-Target-URI\":\"https://forum.espruino.com/conversations/315821/?offset=25\",\"WARC-Payload-Digest\":\"sha1:HNIWJZHH4WM4HWE5SSP2EU5TTDNYABRN\",\"WARC-Block-Digest\":\"sha1:UXVQBB2LWMFQNI75W24ZSUK6G2YQOTMM\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323588216.48_warc_CC-MAIN-20211027150823-20211027180823-00420.warc.gz\"}"}
https://ras.football/2019/12/27/demario-minter-ras/	[ "# DeMario Minter RAS\n\n### DeMario Minter RAS\n\nDeMario Minter was drafted by Browns with pick 152 in round 5 in the 2006 NFL Draft out of Georgia.\n\nHe recorded a Relative Athletic Score of 8.72, out of a possible 10.0. RAS is a composite metric on a 0 to 10 scale based on the average of all of the percentile for each of the metrics the player completed either at the Combine or pro day.\n\nHe had a recorded height of 5111 that season, recorded as XYYZ where X is feet, YY is inches, and Z is eighths of an inch. That correlates to 5 feet, 11 and 1/8 of an inch or 71.125 inches, or 180.6575 centimeters. This correlates to a 6.5 score out of 10.0.\n\nHe recorded a weight of 190 in pounds, which is approximately 86 kilograms. This correlates to a 5.8 score out of 10.0.\n\nBased on his weight, he has a projected 40 yard dash time of 4.49. This is calculated by taking 0.00554 multiplied by his weight and then adding 3.433.\n\nAt the Combine, he recorded a 40 yard dash of 4.52 seconds. This was a difference of 0.03 seconds from his projected time. This forty time correlates to a 6.69 score out of 10.0.\n\nUsing Bill Barnwell’s calculation, this Combine 40 time gave him a Speed Score of 91.04.\n\nAt his pro day, he recorded a 40 yard dash of 4.49 seconds. Because he also recorded this metric at the Combine, his pro day did not count towards his RAS.\n\nUsing Bill Barnwell’s calculation, his Combine 40 time gave him a Speed Score of 2.15.\n\nThe time traveled between the 20 and 40 yard lines is known as the Flying Twenty. As the distance is also known, we can calculate the player’s speed over that distance. The time he traveled the last twenty yards at the Combine was 1.92 seconds. Over 20 yards, we can calculate his speed in yards per second to 10.42. Taking into account the distance in feet (60 feet), we can calculate his speed in feet per second to 31.25. Breaking it down further, we can calculate his speed in inches per second to 375.0. Knowing the feet per second of 31.25, we can calculate the approximate miles per hour by multiplying that value by 0.681818 to give us a calculated MPH of 21.3 in the last 20 yards of his run.\n\nAt the Combine, he recorded a 20 yard split of 2.6 seconds. This correlates to a 8.52 score out of 10.0.\n\nWe can calculate the speed traveled over the second ten yards of the 40 yard dash easily, as the distance and time are both known. The time he traveled the second ten yards at the Combine was 1.02 seconds. Over 10 yards, we can calculate his speed in yards per second to 9.8. Taking into account the distance in feet (30 feet), we can calculate his speed in feet per second to 29.41. Breaking it down further, we can calculate his speed in inches per second to 352.94. Knowing the feet per second of 29.41, we can calculate the approximate miles per hour by multiplying that value by 0.681818 to give us a calculated MPH of 20.1 in the second ten yards of his run.\n\nAt the Combine, he recorded a 10 yard split of 1.58 seconds. This correlates to a 7.37 score out of 10.0.\n\nThe time he traveled the first ten yards at the Combine was 1.58 seconds. Over 10 yards, we can calculate his speed in yards per second to 6.0. Taking into account the distance in feet (30 feet), we can calculate his speed in feet per second to 19.0. Breaking it down further, we can calculate his speed in inches per second to 228.0. Knowing the feet per second of 19.0, we can calculate the approximate miles per hour by multiplying that value by 0.681818 to give us a calculated MPH of 13.0 in the first ten yards of his run.\n\nAt the Combine, he recorded a bench press of 10 repetitions of 225 pounds. This correlates to a 3.54 score out of 10.0.\n\nAt the Combine, he recorded a vertical jump of 38.5 inches. This correlates to a 8.69 score out of 10.0.\n\nAt the Combine, he recorded a broad jump of 1011, which is recorded as FII or FFII . where F is feet and I is inches. This correlates to a 9.79 score out of 10.0.\n\nAt the Combine, he recorded a 5-10-5 or 20 yard short shuttle of 4.12 seconds. This correlates to a 6.69 score out of 10.0.\n\nAt his pro day, he recorded a 5-10-5 or 20 yard short shuttle of 4.32 seconds. Because he also recorded this metric at the Combine, his pro day did not count towards his RAS.\n\nAt the Combine, he recorded a 3 cone L drill of 6.88 seconds. This correlates to a 7.66 score out of 10.0.\n\nThis site uses Akismet to reduce spam. Learn how your comment data is processed." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.97635514,"math_prob":0.9866895,"size":4332,"snap":"2023-40-2023-50","text_gpt3_token_len":1175,"char_repetition_ratio":0.16959335,"word_repetition_ratio":0.3446191,"special_character_ratio":0.29570636,"punctuation_ratio":0.14285715,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9801804,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-08T08:59:17Z\",\"WARC-Record-ID\":\"<urn:uuid:6bc2dae6-e198-417c-82e8-9092ad8170a1>\",\"Content-Length\":\"265081\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:700163cd-beb6-474b-aee8-d04210d9f3b9>\",\"WARC-Concurrent-To\":\"<urn:uuid:3a2bf157-d9e0-4e3b-949a-e233f04d661c>\",\"WARC-IP-Address\":\"3.210.81.252\",\"WARC-Target-URI\":\"https://ras.football/2019/12/27/demario-minter-ras/\",\"WARC-Payload-Digest\":\"sha1:QEVDCDU7RLO6E52YWV4DCKBIAFY5NONP\",\"WARC-Block-Digest\":\"sha1:BSCBQBUMYPME4J7VJVPJKTRC7JX4VS6T\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100739.50_warc_CC-MAIN-20231208081124-20231208111124-00230.warc.gz\"}"}
https://www.ademcetinkaya.com/2022/10/boh-options-futures-prediction.html	[ "Predictions on stock market prices are a great challenge due to the fact that it is an immensely complex, chaotic and dynamic environment. There are many studies from various areas aiming to take on that challenge and Machine Learning approaches have been the focus of many of them. There are many examples of Machine Learning algorithms been able to reach satisfactory results when doing that type of prediction. This article studies the usage of LSTM networks on that scenario, to predict future trends of stock prices based on the price history, alongside with technical analysis indicators. We evaluate Bank of Hawaii prediction models with Deductive Inference (ML) and Multiple Regression1,2,3,4 and conclude that the BOH stock is predictable in the short/long term. According to price forecasts for (n+6 month) period: The dominant strategy among neural network is to Hold BOH stock.\n\nKeywords: BOH, Bank of Hawaii, stock forecast, machine learning based prediction, risk rating, buy-sell behaviour, stock analysis, target price analysis, options and futures.\n\n## Key Points\n\n1. How do you pick a stock?\n2. What is the best way to predict stock prices?\n3. Stock Forecast Based On a Predictive Algorithm", null, "## BOH Target Price Prediction Modeling Methodology\n\nAccurate prediction of stock market returns is a very challenging task due to volatile and non-linear nature of the financial stock markets. With the introduction of artificial intelligence and increased computational capabilities, programmed methods of prediction have proved to be more efficient in predicting stock prices. We consider Bank of Hawaii Stock Decision Process with Multiple Regression where A is the set of discrete actions of BOH stock holders, F is the set of discrete states, P : S × F × S → R is the transition probability distribution, R : S × F → R is the reaction function, and γ ∈ [0, 1] is a move factor for expectation.1,2,3,4\n\nF(Multiple Regression)5,6,7= $\\begin{array}{cccc}{p}_{a1}& {p}_{a2}& \\dots & {p}_{1n}\\\\ & ⋮\\\\ {p}_{j1}& {p}_{j2}& \\dots & {p}_{jn}\\\\ & ⋮\\\\ {p}_{k1}& {p}_{k2}& \\dots & {p}_{kn}\\\\ & ⋮\\\\ {p}_{n1}& {p}_{n2}& \\dots & {p}_{nn}\\end{array}$ X R(Deductive Inference (ML)) X S(n):→ (n+6 month) $R=\\left(\\begin{array}{ccc}1& 0& 0\\\\ 0& 1& 0\\\\ 0& 0& 1\\end{array}\\right)$\n\nn:Time series to forecast\n\np:Price signals of BOH stock\n\nj:Nash equilibria\n\nk:Dominated move\n\na:Best response for target price\n\nFor further technical information as per how our model work we invite you to visit the article below:\n\nHow do AC Investment Research machine learning (predictive) algorithms actually work?\n\n## BOH Stock Forecast (Buy or Sell) for (n+6 month)\n\nSample Set: Neural Network\nStock/Index: BOH Bank of Hawaii\nTime series to forecast n: 26 Oct 2022 for (n+6 month)\n\nAccording to price forecasts for (n+6 month) period: The dominant strategy among neural network is to Hold BOH stock.\n\nX axis: Likelihood% (The higher the percentage value, the more likely the event will occur.)\n\nY axis: Potential Impact% (The higher the percentage value, the more likely the price will deviate.)\n\nZ axis (Yellow to Green): Technical Analysis%\n\n## Adjusted IFRS Prediction Methods for Bank of Hawaii\n\n1. In applying the effective interest method, an entity identifies fees that are an integral part of the effective interest rate of a financial instrument. The description of fees for financial services may not be indicative of the nature and substance of the services provided. Fees that are an integral part of the effective interest rate of a financial instrument are treated as an adjustment to the effective interest rate, unless the financial instrument is measured at fair value, with the change in fair value being recognised in profit or loss. In those cases, the fees are recognised as revenue or expense when the instrument is initially recognised.\n2. An entity shall apply the amendments to IFRS 9 made by IFRS 17 as amended in June 2020 retrospectively in accordance with IAS 8, except as specified in paragraphs 7.2.37–7.2.42.\n3. To be eligible for designation as a hedged item, a risk component must be a separately identifiable component of the financial or the non-financial item, and the changes in the cash flows or the fair value of the item attributable to changes in that risk component must be reliably measurable.\n4. Accordingly the date of the modification shall be treated as the date of initial recognition of that financial asset when applying the impairment requirements to the modified financial asset. This typically means measuring the loss allowance at an amount equal to 12-month expected credit losses until the requirements for the recognition of lifetime expected credit losses in paragraph 5.5.3 are met. However, in some unusual circumstances following a modification that results in derecognition of the original financial asset, there may be evidence that the modified financial asset is credit-impaired at initial recognition, and thus, the financial asset should be recognised as an originated credit-impaired financial asset. This might occur, for example, in a situation in which there was a substantial modification of a distressed asset that resulted in the derecognition of the original financial asset. In such a case, it may be possible for the modification to result in a new financial asset which is credit-impaired at initial recognition.\n\nInternational Financial Reporting Standards (IFRS) are a set of accounting rules for the financial statements of public companies that are intended to make them consistent, transparent, and easily comparable around the world.\n\n## Conclusions\n\nBank of Hawaii assigned short-term Ba2 & long-term B2 forecasted stock rating. We evaluate the prediction models Deductive Inference (ML) with Multiple Regression1,2,3,4 and conclude that the BOH stock is predictable in the short/long term. According to price forecasts for (n+6 month) period: The dominant strategy among neural network is to Hold BOH stock.\n\n### Financial State Forecast for BOH Bank of Hawaii Stock Options & Futures\n\nRating Short-Term Long-Term Senior\nOutlookBa2B2\nOperational Risk 6853\nMarket Risk7835\nTechnical Analysis8142\nFundamental Analysis7949\nRisk Unsystematic3878\n\n### Prediction Confidence Score\n\nTrust metric by Neural Network: 88 out of 100 with 798 signals.\n\n## References\n\n1. Künzel S, Sekhon J, Bickel P, Yu B. 2017. Meta-learners for estimating heterogeneous treatment effects using machine learning. arXiv:1706.03461 [math.ST]\n2. Hill JL. 2011. Bayesian nonparametric modeling for causal inference. J. Comput. Graph. Stat. 20:217–40\n3. E. Collins. Using Markov decision processes to optimize a nonlinear functional of the final distribution, with manufacturing applications. In Stochastic Modelling in Innovative Manufacturing, pages 30–45. Springer, 1997\n4. Lai TL, Robbins H. 1985. Asymptotically efficient adaptive allocation rules. Adv. Appl. Math. 6:4–22\n5. Y. Chow and M. Ghavamzadeh. Algorithms for CVaR optimization in MDPs. In Advances in Neural Infor- mation Processing Systems, pages 3509–3517, 2014.\n6. T. Morimura, M. Sugiyama, M. Kashima, H. Hachiya, and T. Tanaka. Nonparametric return distribution ap- proximation for reinforcement learning. In Proceedings of the 27th International Conference on Machine Learning, pages 799–806, 2010\n7. Abadir, K. M., K. Hadri E. Tzavalis (1999), \"The influence of VAR dimensions on estimator biases,\" Econometrica, 67, 163–181.\nFrequently Asked QuestionsQ: What is the prediction methodology for BOH stock?\nA: BOH stock prediction methodology: We evaluate the prediction models Deductive Inference (ML) and Multiple Regression\nQ: Is BOH stock a buy or sell?\nA: The dominant strategy among neural network is to Hold BOH Stock.\nQ: Is Bank of Hawaii stock a good investment?\nA: The consensus rating for Bank of Hawaii is Hold and assigned short-term Ba2 & long-term B2 forecasted stock rating.\nQ: What is the consensus rating of BOH stock?\nA: The consensus rating for BOH is Hold.\nQ: What is the prediction period for BOH stock?\nA: The prediction period for BOH is (n+6 month)" ]	[ null, "https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhGtuWgrT7v7mhQG9t6gmuoswzM5z9nish6bESbyRnNfQvf5zXnBndysoOHfOl3HYAxSTHbFinEIq4k0P23kRlFrHf7sLDoMgUVEgAFQ2aFeNMblTLPzRF-cNKeO4mcUD3n5EhWoTNu0OnAwnjTqWo197k2l8T8EtWmQ_l1EbVlVrRbSvvFSIhV3VtH6w/s16000/20220829_123228_0000.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8884896,"math_prob":0.8593608,"size":7703,"snap":"2023-14-2023-23","text_gpt3_token_len":1743,"char_repetition_ratio":0.12118457,"word_repetition_ratio":0.110648915,"special_character_ratio":0.2178372,"punctuation_ratio":0.12876137,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9674725,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-06-08T04:08:50Z\",\"WARC-Record-ID\":\"<urn:uuid:1b7ebaa9-a402-430c-b5ec-6b1b1643e9a1>\",\"Content-Length\":\"322085\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4d00ed18-0f56-4985-ab4f-74b2ec08f384>\",\"WARC-Concurrent-To\":\"<urn:uuid:c72b17f1-9ac2-441f-932e-6710a11f10aa>\",\"WARC-IP-Address\":\"142.251.167.121\",\"WARC-Target-URI\":\"https://www.ademcetinkaya.com/2022/10/boh-options-futures-prediction.html\",\"WARC-Payload-Digest\":\"sha1:3COAJZCLX3AV27N2EGLOZAO53LXLXGIK\",\"WARC-Block-Digest\":\"sha1:SPJUH7PDGAG3UX7GXQZIMSXXHBC4QVT7\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224654097.42_warc_CC-MAIN-20230608035801-20230608065801-00724.warc.gz\"}"}
https://isabelle.in.tum.de/repos/isabelle/rev/5e6296afe08d?revcount=30	[ "author hoelzl Tue, 05 Mar 2013 15:43:08 +0100 changeset 51340 5e6296afe08d parent 51339 04ebef4ee844 child 51341 8c10293e7ea7\nmove Liminf / Limsup lemmas on complete_lattices to its own file\n```--- a/src/HOL/Library/Extended_Real.thy\tTue Mar 05 15:27:08 2013 +0100\n+++ b/src/HOL/Library/Extended_Real.thy\tTue Mar 05 15:43:08 2013 +0100\n@@ -8,7 +8,7 @@\nheader {* Extended real number line }\n\ntheory Extended_Real\n-imports Complex_Main Extended_Nat\n+imports Complex_Main Extended_Nat Liminf_Limsup\nbegin\n\ntext {\n@@ -18,26 +18,6 @@\n\n}\n\n-lemma SUPR_pair:\n- \"(SUP i : A. SUP j : B. f i j) = (SUP p : A \\<times> B. f (fst p) (snd p))\"\n- by (rule antisym) (auto intro!: SUP_least SUP_upper2)\n-\n-lemma INFI_pair:\n- \"(INF i : A. INF j : B. f i j) = (INF p : A \\<times> B. f (fst p) (snd p))\"\n- by (rule antisym) (auto intro!: INF_greatest INF_lower2)\n-\n-lemma le_Sup_iff_less:\n- fixes x :: \"'a :: {complete_linorder, inner_dense_linorder}\"\n- shows \"x \\<le> (SUP i:A. f i) \\<longleftrightarrow> (\\<forall>y<x. \\<exists>i\\<in>A. y \\<le> f i)\" (is \"?lhs = ?rhs\")\n- unfolding le_SUP_iff\n- by (blast intro: less_imp_le less_trans less_le_trans dest: dense)\n-\n-lemma Inf_le_iff_less:\n- fixes x :: \"'a :: {complete_linorder, inner_dense_linorder}\"\n- shows \"(INF i:A. f i) \\<le> x \\<longleftrightarrow> (\\<forall>y>x. \\<exists>i\\<in>A. f i \\<le> y)\"\n- unfolding INF_le_iff\n- by (blast intro: less_imp_le less_trans le_less_trans dest: dense)\n-\nsubsection { Definition and basic properties }\n\ndatatype ereal = ereal real \| PInfty \| MInfty\n@@ -1718,6 +1698,31 @@\nshow thesis by auto\nqed\n\n+instance ereal :: perfect_space\n+proof (default, rule)\n+ fix a :: ereal assume a: \"open {a}\"\n+ show False\n+ proof (cases a)\n+ case MInf\n+ then obtain y where \"{..<ereal y} \\<le> {a}\" using a open_MInfty2[of \"{a}\"] by auto\n+ then have \"ereal (y - 1) \\<in> {a}\" apply (subst subsetD[of \"{..<ereal y}\"]) by auto\n+ then show False using `a = -\\<infinity>` by auto\n+ next\n+ case PInf\n+ then obtain y where \"{ereal y<..} \\<le> {a}\" using a open_PInfty2[of \"{a}\"] by auto\n+ then have \"ereal (y + 1) \\<in> {a}\" apply (subst subsetD[of \"{ereal y<..}\"]) by auto\n+ then show False using `a = \\<infinity>` by auto\n+ next\n+ case (real r)\n+ then have fin: \"\\<bar>a\\<bar> \\<noteq> \\<infinity>\" by simp\n+ from ereal_open_cont_interval[OF a singletonI this] guess e . note e = this\n+ then obtain b where b_def: \"a<b \\<and> b<a+e\"\n+ using fin ereal_between dense[of a \"a+e\"] by auto\n+ then have \"b: {a-e <..< a+e}\" using fin ereal_between[of a e] e by auto\n+ then show False using b_def e by auto\n+ qed\n+qed\n+\nsubsubsection { Convergent sequences }\n\nlemma lim_ereal[simp]:\n@@ -1981,123 +1986,15 @@\nusing ereal_LimI_finite[of x] assms by auto\nqed\n\n-\n-subsubsection { @{text Liminf} and @{text Limsup} }\n-\n-definition\n- \"Liminf F f = (SUP P:{P. eventually P F}. INF x:{x. P x}. f x)\"\n-\n-definition\n- \"Limsup F f = (INF P:{P. eventually P F}. SUP x:{x. P x}. f x)\"\n-\n-abbreviation \"liminf \\<equiv> Liminf sequentially\"\n-\n-abbreviation \"limsup \\<equiv> Limsup sequentially\"\n-\n-lemma Liminf_eqI:\n- \"(\\<And>P. eventually P F \\<Longrightarrow> INFI (Collect P) f \\<le> x) \\<Longrightarrow>\n- (\\<And>y. (\\<And>P. eventually P F \\<Longrightarrow> INFI (Collect P) f \\<le> y) \\<Longrightarrow> x \\<le> y) \\<Longrightarrow> Liminf F f = x\"\n- unfolding Liminf_def by (auto intro!: SUP_eqI)\n-\n-lemma Limsup_eqI:\n- \"(\\<And>P. eventually P F \\<Longrightarrow> x \\<le> SUPR (Collect P) f) \\<Longrightarrow>\n- (\\<And>y. (\\<And>P. eventually P F \\<Longrightarrow> y \\<le> SUPR (Collect P) f) \\<Longrightarrow> y \\<le> x) \\<Longrightarrow> Limsup F f = x\"\n- unfolding Limsup_def by (auto intro!: INF_eqI)\n-\n-lemma liminf_SUPR_INFI:\n- fixes f :: \"nat \\<Rightarrow> 'a :: complete_lattice\"\n- shows \"liminf f = (SUP n. INF m:{n..}. f m)\"\n- unfolding Liminf_def eventually_sequentially\n- by (rule SUPR_eq) (auto simp: atLeast_def intro!: INF_mono)\n-\n-lemma limsup_INFI_SUPR:\n- fixes f :: \"nat \\<Rightarrow> 'a :: complete_lattice\"\n- shows \"limsup f = (INF n. SUP m:{n..}. f m)\"\n- unfolding Limsup_def eventually_sequentially\n- by (rule INFI_eq) (auto simp: atLeast_def intro!: SUP_mono)\n-\n-lemma Limsup_const:\n- assumes ntriv: \"\\<not> trivial_limit F\"\n- shows \"Limsup F (\\<lambda>x. c) = (c::'a::complete_lattice)\"\n-proof -\n- have : \"\\<And>P. Ex P \\<longleftrightarrow> P \\<noteq> (\\<lambda>x. False)\" by auto\n- have \"\\<And>P. eventually P F \\<Longrightarrow> (SUP x : {x. P x}. c) = c\"\n- using ntriv by (intro SUP_const) (auto simp: eventually_False )\n- then show ?thesis\n- unfolding Limsup_def using eventually_True\n- by (subst INF_cong[where D=\"\\<lambda>x. c\"])\n- (auto intro!: INF_const simp del: eventually_True)\n-qed\n+lemma ereal_Limsup_uminus:\n+ fixes f :: \"'a => ereal\"\n+ shows \"Limsup net (\\<lambda>x. - (f x)) = -(Liminf net f)\"\n+ unfolding Limsup_def Liminf_def ereal_SUPR_uminus ereal_INFI_uminus ..\n\n-lemma Liminf_const:\n- assumes ntriv: \"\\<not> trivial_limit F\"\n- shows \"Liminf F (\\<lambda>x. c) = (c::'a::complete_lattice)\"\n-proof -\n- have : \"\\<And>P. Ex P \\<longleftrightarrow> P \\<noteq> (\\<lambda>x. False)\" by auto\n- have \"\\<And>P. eventually P F \\<Longrightarrow> (INF x : {x. P x}. c) = c\"\n- using ntriv by (intro INF_const) (auto simp: eventually_False )\n- then show ?thesis\n- unfolding Liminf_def using eventually_True\n- by (subst SUP_cong[where D=\"\\<lambda>x. c\"])\n- (auto intro!: SUP_const simp del: eventually_True)\n-qed\n-\n-lemma Liminf_mono:\n- fixes f g :: \"'a => 'b :: complete_lattice\"\n- assumes ev: \"eventually (\\<lambda>x. f x \\<le> g x) F\"\n- shows \"Liminf F f \\<le> Liminf F g\"\n- unfolding Liminf_def\n-proof (safe intro!: SUP_mono)\n- fix P assume \"eventually P F\"\n- with ev have \"eventually (\\<lambda>x. f x \\<le> g x \\<and> P x) F\" (is \"eventually ?Q F\") by (rule eventually_conj)\n- then show \"\\<exists>Q\\<in>{P. eventually P F}. INFI (Collect P) f \\<le> INFI (Collect Q) g\"\n- by (intro bexI[of _ ?Q]) (auto intro!: INF_mono)\n-qed\n-\n-lemma Liminf_eq:\n- fixes f g :: \"'a \\<Rightarrow> 'b :: complete_lattice\"\n- assumes \"eventually (\\<lambda>x. f x = g x) F\"\n- shows \"Liminf F f = Liminf F g\"\n- by (intro antisym Liminf_mono eventually_mono[OF _ assms]) auto\n-\n-lemma Limsup_mono:\n- fixes f g :: \"'a \\<Rightarrow> 'b :: complete_lattice\"\n- assumes ev: \"eventually (\\<lambda>x. f x \\<le> g x) F\"\n- shows \"Limsup F f \\<le> Limsup F g\"\n- unfolding Limsup_def\n-proof (safe intro!: INF_mono)\n- fix P assume \"eventually P F\"\n- with ev have \"eventually (\\<lambda>x. f x \\<le> g x \\<and> P x) F\" (is \"eventually ?Q F\") by (rule eventually_conj)\n- then show \"\\<exists>Q\\<in>{P. eventually P F}. SUPR (Collect Q) f \\<le> SUPR (Collect P) g\"\n- by (intro bexI[of _ ?Q]) (auto intro!: SUP_mono)\n-qed\n-\n-lemma Limsup_eq:\n- fixes f g :: \"'a \\<Rightarrow> 'b :: complete_lattice\"\n- assumes \"eventually (\\<lambda>x. f x = g x) net\"\n- shows \"Limsup net f = Limsup net g\"\n- by (intro antisym Limsup_mono eventually_mono[OF _ assms]) auto\n-\n-lemma Liminf_le_Limsup:\n- fixes f :: \"'a \\<Rightarrow> 'b::complete_lattice\"\n- assumes ntriv: \"\\<not> trivial_limit F\"\n- shows \"Liminf F f \\<le> Limsup F f\"\n- unfolding Limsup_def Liminf_def\n- apply (rule complete_lattice_class.SUP_least)\n- apply (rule complete_lattice_class.INF_greatest)\n-proof safe\n- fix P Q assume \"eventually P F\" \"eventually Q F\"\n- then have \"eventually (\\<lambda>x. P x \\<and> Q x) F\" (is \"eventually ?C F\") by (rule eventually_conj)\n- then have not_False: \"(\\<lambda>x. P x \\<and> Q x) \\<noteq> (\\<lambda>x. False)\"\n- using ntriv by (auto simp add: eventually_False)\n- have \"INFI (Collect P) f \\<le> INFI (Collect ?C) f\"\n- by (rule INF_mono) auto\n- also have \"\\<dots> \\<le> SUPR (Collect ?C) f\"\n- using not_False by (intro INF_le_SUP) auto\n- also have \"\\<dots> \\<le> SUPR (Collect Q) f\"\n- by (rule SUP_mono) auto\n- finally show \"INFI (Collect P) f \\<le> SUPR (Collect Q) f\" .\n-qed\n+lemma liminf_bounded_iff:\n+ fixes x :: \"nat \\<Rightarrow> ereal\"\n+ shows \"C \\<le> liminf x \\<longleftrightarrow> (\\<forall>B<C. \\<exists>N. \\<forall>n\\<ge>N. B < x n)\" (is \"?lhs <-> ?rhs\")\n+ unfolding le_Liminf_iff eventually_sequentially ..\n\nlemma\nfixes X :: \"nat \\<Rightarrow> ereal\"\n@@ -2105,220 +2002,6 @@\nand ereal_decseq_uminus[simp]: \"decseq (\\<lambda>i. - X i) = incseq X\"\nunfolding incseq_def decseq_def by auto\n\n-lemma Liminf_bounded:\n- fixes X Y :: \"'a \\<Rightarrow> 'b::complete_lattice\"\n- assumes ntriv: \"\\<not> trivial_limit F\"\n- assumes le: \"eventually (\\<lambda>n. C \\<le> X n) F\"\n- shows \"C \\<le> Liminf F X\"\n- using Liminf_mono[OF le] Liminf_const[OF ntriv, of C] by simp\n-\n-lemma Limsup_bounded:\n- fixes X Y :: \"'a \\<Rightarrow> 'b::complete_lattice\"\n- assumes ntriv: \"\\<not> trivial_limit F\"\n- assumes le: \"eventually (\\<lambda>n. X n \\<le> C) F\"\n- shows \"Limsup F X \\<le> C\"\n- using Limsup_mono[OF le] Limsup_const[OF ntriv, of C] by simp\n-\n-lemma le_Liminf_iff:\n- fixes X :: \"_ \\<Rightarrow> _ :: complete_linorder\"\n- shows \"C \\<le> Liminf F X \\<longleftrightarrow> (\\<forall>y<C. eventually (\\<lambda>x. y < X x) F)\"\n-proof -\n- { fix y P assume \"eventually P F\" \"y < INFI (Collect P) X\"\n- then have \"eventually (\\<lambda>x. y < X x) F\"\n- by (auto elim!: eventually_elim1 dest: less_INF_D) }\n- moreover\n- { fix y P assume \"y < C\" and y: \"\\<forall>y<C. eventually (\\<lambda>x. y < X x) F\"\n- have \"\\<exists>P. eventually P F \\<and> y < INFI (Collect P) X\"\n- proof cases\n- assume \"\\<exists>z. y < z \\<and> z < C\"\n- then guess z ..\n- moreover then have \"z \\<le> INFI {x. z < X x} X\"\n- by (auto intro!: INF_greatest)\n- ultimately show ?thesis\n- using y by (intro exI[of _ \"\\<lambda>x. z < X x\"]) auto\n- next\n- assume \"\\<not> (\\<exists>z. y < z \\<and> z < C)\"\n- then have \"C \\<le> INFI {x. y < X x} X\"\n- by (intro INF_greatest) auto\n- with `y < C` show ?thesis\n- using y by (intro exI[of _ \"\\<lambda>x. y < X x\"]) auto\n- qed }\n- ultimately show ?thesis\n- unfolding Liminf_def le_SUP_iff by auto\n-qed\n-\n-lemma lim_imp_Liminf:\n- fixes f :: \"'a \\<Rightarrow> _ :: {complete_linorder, linorder_topology}\"\n- assumes ntriv: \"\\<not> trivial_limit F\"\n- assumes lim: \"(f ---> f0) F\"\n- shows \"Liminf F f = f0\"\n-proof (intro Liminf_eqI)\n- fix P assume P: \"eventually P F\"\n- then have \"eventually (\\<lambda>x. INFI (Collect P) f \\<le> f x) F\"\n- by eventually_elim (auto intro!: INF_lower)\n- then show \"INFI (Collect P) f \\<le> f0\"\n- by (rule tendsto_le[OF ntriv lim tendsto_const])\n-next\n- fix y assume upper: \"\\<And>P. eventually P F \\<Longrightarrow> INFI (Collect P) f \\<le> y\"\n- show \"f0 \\<le> y\"\n- proof cases\n- assume \"\\<exists>z. y < z \\<and> z < f0\"\n- then guess z ..\n- moreover have \"z \\<le> INFI {x. z < f x} f\"\n- by (rule INF_greatest) simp\n- ultimately show ?thesis\n- using lim[THEN topological_tendstoD, THEN upper, of \"{z <..}\"] by auto\n- next\n- assume discrete: \"\\<not> (\\<exists>z. y < z \\<and> z < f0)\"\n- show ?thesis\n- proof (rule classical)\n- assume \"\\<not> f0 \\<le> y\"\n- then have \"eventually (\\<lambda>x. y < f x) F\"\n- using lim[THEN topological_tendstoD, of \"{y <..}\"] by auto\n- then have \"eventually (\\<lambda>x. f0 \\<le> f x) F\"\n- using discrete by (auto elim!: eventually_elim1)\n- then have \"INFI {x. f0 \\<le> f x} f \\<le> y\"\n- by (rule upper)\n- moreover have \"f0 \\<le> INFI {x. f0 \\<le> f x} f\"\n- by (intro INF_greatest) simp\n- ultimately show \"f0 \\<le> y\" by simp\n- qed\n- qed\n-qed\n-\n-lemma lim_imp_Limsup:\n- fixes f :: \"'a \\<Rightarrow> _ :: {complete_linorder, linorder_topology}\"\n- assumes ntriv: \"\\<not> trivial_limit F\"\n- assumes lim: \"(f ---> f0) F\"\n- shows \"Limsup F f = f0\"\n-proof (intro Limsup_eqI)\n- fix P assume P: \"eventually P F\"\n- then have \"eventually (\\<lambda>x. f x \\<le> SUPR (Collect P) f) F\"\n- by eventually_elim (auto intro!: SUP_upper)\n- then show \"f0 \\<le> SUPR (Collect P) f\"\n- by (rule tendsto_le[OF ntriv tendsto_const lim])\n-next\n- fix y assume lower: \"\\<And>P. eventually P F \\<Longrightarrow> y \\<le> SUPR (Collect P) f\"\n- show \"y \\<le> f0\"\n- proof cases\n- assume \"\\<exists>z. f0 < z \\<and> z < y\"\n- then guess z ..\n- moreover have \"SUPR {x. f x < z} f \\<le> z\"\n- by (rule SUP_least) simp\n- ultimately show ?thesis\n- using lim[THEN topological_tendstoD, THEN lower, of \"{..< z}\"] by auto\n- next\n- assume discrete: \"\\<not> (\\<exists>z. f0 < z \\<and> z < y)\"\n- show ?thesis\n- proof (rule classical)\n- assume \"\\<not> y \\<le> f0\"\n- then have \"eventually (\\<lambda>x. f x < y) F\"\n- using lim[THEN topological_tendstoD, of \"{..< y}\"] by auto\n- then have \"eventually (\\<lambda>x. f x \\<le> f0) F\"\n- using discrete by (auto elim!: eventually_elim1 simp: not_less)\n- then have \"y \\<le> SUPR {x. f x \\<le> f0} f\"\n- by (rule lower)\n- moreover have \"SUPR {x. f x \\<le> f0} f \\<le> f0\"\n- by (intro SUP_least) simp\n- ultimately show \"y \\<le> f0\" by simp\n- qed\n- qed\n-qed\n-\n-lemma Liminf_eq_Limsup:\n- fixes f0 :: \"'a :: {complete_linorder, linorder_topology}\"\n- assumes ntriv: \"\\<not> trivial_limit F\"\n- and lim: \"Liminf F f = f0\" \"Limsup F f = f0\"\n- shows \"(f ---> f0) F\"\n-proof (rule order_tendstoI)\n- fix a assume \"f0 < a\"\n- with assms have \"Limsup F f < a\" by simp\n- then obtain P where \"eventually P F\" \"SUPR (Collect P) f < a\"\n- unfolding Limsup_def INF_less_iff by auto\n- then show \"eventually (\\<lambda>x. f x < a) F\"\n- by (auto elim!: eventually_elim1 dest: SUP_lessD)\n-next\n- fix a assume \"a < f0\"\n- with assms have \"a < Liminf F f\" by simp\n- then obtain P where \"eventually P F\" \"a < INFI (Collect P) f\"\n- unfolding Liminf_def less_SUP_iff by auto\n- then show \"eventually (\\<lambda>x. a < f x) F\"\n- by (auto elim!: eventually_elim1 dest: less_INF_D)\n-qed\n-\n-lemma tendsto_iff_Liminf_eq_Limsup:\n- fixes f0 :: \"'a :: {complete_linorder, linorder_topology}\"\n- shows \"\\<not> trivial_limit F \\<Longrightarrow> (f ---> f0) F \\<longleftrightarrow> (Liminf F f = f0 \\<and> Limsup F f = f0)\"\n- by (metis Liminf_eq_Limsup lim_imp_Limsup lim_imp_Liminf)\n-\n-lemma liminf_bounded_iff:\n- fixes x :: \"nat \\<Rightarrow> ereal\"\n- shows \"C \\<le> liminf x \\<longleftrightarrow> (\\<forall>B<C. \\<exists>N. \\<forall>n\\<ge>N. B < x n)\" (is \"?lhs <-> ?rhs\")\n- unfolding le_Liminf_iff eventually_sequentially ..\n-\n-lemma liminf_subseq_mono:\n- fixes X :: \"nat \\<Rightarrow> 'a :: complete_linorder\"\n- assumes \"subseq r\"\n- shows \"liminf X \\<le> liminf (X \\<circ> r) \"\n-proof-\n- have \"\\<And>n. (INF m:{n..}. X m) \\<le> (INF m:{n..}. (X \\<circ> r) m)\"\n- proof (safe intro!: INF_mono)\n- fix n m :: nat assume \"n \\<le> m\" then show \"\\<exists>ma\\<in>{n..}. X ma \\<le> (X \\<circ> r) m\"\n- using seq_suble[OF `subseq r`, of m] by (intro bexI[of _ \"r m\"]) auto\n- qed\n- then show ?thesis by (auto intro!: SUP_mono simp: liminf_SUPR_INFI comp_def)\n-qed\n-\n-lemma limsup_subseq_mono:\n- fixes X :: \"nat \\<Rightarrow> 'a :: complete_linorder\"\n- assumes \"subseq r\"\n- shows \"limsup (X \\<circ> r) \\<le> limsup X\"\n-proof-\n- have \"\\<And>n. (SUP m:{n..}. (X \\<circ> r) m) \\<le> (SUP m:{n..}. X m)\"\n- proof (safe intro!: SUP_mono)\n- fix n m :: nat assume \"n \\<le> m\" then show \"\\<exists>ma\\<in>{n..}. (X \\<circ> r) m \\<le> X ma\"\n- using seq_suble[OF `subseq r`, of m] by (intro bexI[of _ \"r m\"]) auto\n- qed\n- then show ?thesis by (auto intro!: INF_mono simp: limsup_INFI_SUPR comp_def)\n-qed\n-\n-definition (in order) mono_set:\n- \"mono_set S \\<longleftrightarrow> (\\<forall>x y. x \\<le> y \\<longrightarrow> x \\<in> S \\<longrightarrow> y \\<in> S)\"\n-\n-lemma (in order) mono_greaterThan [intro, simp]: \"mono_set {B<..}\" unfolding mono_set by auto\n-lemma (in order) mono_atLeast [intro, simp]: \"mono_set {B..}\" unfolding mono_set by auto\n-lemma (in order) mono_UNIV [intro, simp]: \"mono_set UNIV\" unfolding mono_set by auto\n-lemma (in order) mono_empty [intro, simp]: \"mono_set {}\" unfolding mono_set by auto\n-\n-lemma (in complete_linorder) mono_set_iff:\n- fixes S :: \"'a set\"\n- defines \"a \\<equiv> Inf S\"\n- shows \"mono_set S \\<longleftrightarrow> (S = {a <..} \\<or> S = {a..})\" (is \"_ = ?c\")\n-proof\n- assume \"mono_set S\"\n- then have mono: \"\\<And>x y. x \\<le> y \\<Longrightarrow> x \\<in> S \\<Longrightarrow> y \\<in> S\" by (auto simp: mono_set)\n- show ?c\n- proof cases\n- assume \"a \\<in> S\"\n- show ?c\n- using mono[OF _ `a \\<in> S`]\n- by (auto intro: Inf_lower simp: a_def)\n- next\n- assume \"a \\<notin> S\"\n- have \"S = {a <..}\"\n- proof safe\n- fix x assume \"x \\<in> S\"\n- then have \"a \\<le> x\" unfolding a_def by (rule Inf_lower)\n- then show \"a < x\" using `x \\<in> S` `a \\<notin> S` by (cases \"a = x\") auto\n- next\n- fix x assume \"a < x\"\n- then obtain y where \"y < x\" \"y \\<in> S\" unfolding a_def Inf_less_iff ..\n- with mono[of y x] show \"x \\<in> S\" by auto\n- qed\n- then show ?c ..\n- qed\n-qed auto\n-\nsubsubsection { Tests for code generator }\n\n( A small list of simple arithmetic expressions )```\n```--- /dev/null\tThu Jan 01 00:00:00 1970 +0000\n+++ b/src/HOL/Library/Liminf_Limsup.thy\tTue Mar 05 15:43:08 2013 +0100\n@@ -0,0 +1,320 @@\n+( Title: HOL/Library/Liminf_Limsup.thy\n+ Author: Johannes Hölzl, TU München\n+)\n+\n+header { Liminf and Limsup on complete lattices }\n+\n+theory Liminf_Limsup\n+imports \"~~/src/HOL/Complex_Main\"\n+begin\n+\n+lemma le_Sup_iff_less:\n+ fixes x :: \"'a :: {complete_linorder, inner_dense_linorder}\"\n+ shows \"x \\<le> (SUP i:A. f i) \\<longleftrightarrow> (\\<forall>y<x. \\<exists>i\\<in>A. y \\<le> f i)\" (is \"?lhs = ?rhs\")\n+ unfolding le_SUP_iff\n+ by (blast intro: less_imp_le less_trans less_le_trans dest: dense)\n+\n+lemma Inf_le_iff_less:\n+ fixes x :: \"'a :: {complete_linorder, inner_dense_linorder}\"\n+ shows \"(INF i:A. f i) \\<le> x \\<longleftrightarrow> (\\<forall>y>x. \\<exists>i\\<in>A. f i \\<le> y)\"\n+ unfolding INF_le_iff\n+ by (blast intro: less_imp_le less_trans le_less_trans dest: dense)\n+\n+lemma SUPR_pair:\n+ \"(SUP i : A. SUP j : B. f i j) = (SUP p : A \\<times> B. f (fst p) (snd p))\"\n+ by (rule antisym) (auto intro!: SUP_least SUP_upper2)\n+\n+lemma INFI_pair:\n+ \"(INF i : A. INF j : B. f i j) = (INF p : A \\<times> B. f (fst p) (snd p))\"\n+ by (rule antisym) (auto intro!: INF_greatest INF_lower2)\n+\n+subsubsection { @{text Liminf} and @{text Limsup} }\n+\n+definition\n+ \"Liminf F f = (SUP P:{P. eventually P F}. INF x:{x. P x}. f x)\"\n+\n+definition\n+ \"Limsup F f = (INF P:{P. eventually P F}. SUP x:{x. P x}. f x)\"\n+\n+abbreviation \"liminf \\<equiv> Liminf sequentially\"\n+\n+abbreviation \"limsup \\<equiv> Limsup sequentially\"\n+\n+lemma Liminf_eqI:\n+ \"(\\<And>P. eventually P F \\<Longrightarrow> INFI (Collect P) f \\<le> x) \\<Longrightarrow>\n+ (\\<And>y. (\\<And>P. eventually P F \\<Longrightarrow> INFI (Collect P) f \\<le> y) \\<Longrightarrow> x \\<le> y) \\<Longrightarrow> Liminf F f = x\"\n+ unfolding Liminf_def by (auto intro!: SUP_eqI)\n+\n+lemma Limsup_eqI:\n+ \"(\\<And>P. eventually P F \\<Longrightarrow> x \\<le> SUPR (Collect P) f) \\<Longrightarrow>\n+ (\\<And>y. (\\<And>P. eventually P F \\<Longrightarrow> y \\<le> SUPR (Collect P) f) \\<Longrightarrow> y \\<le> x) \\<Longrightarrow> Limsup F f = x\"\n+ unfolding Limsup_def by (auto intro!: INF_eqI)\n+\n+lemma liminf_SUPR_INFI:\n+ fixes f :: \"nat \\<Rightarrow> 'a :: complete_lattice\"\n+ shows \"liminf f = (SUP n. INF m:{n..}. f m)\"\n+ unfolding Liminf_def eventually_sequentially\n+ by (rule SUPR_eq) (auto simp: atLeast_def intro!: INF_mono)\n+\n+lemma limsup_INFI_SUPR:\n+ fixes f :: \"nat \\<Rightarrow> 'a :: complete_lattice\"\n+ shows \"limsup f = (INF n. SUP m:{n..}. f m)\"\n+ unfolding Limsup_def eventually_sequentially\n+ by (rule INFI_eq) (auto simp: atLeast_def intro!: SUP_mono)\n+\n+lemma Limsup_const:\n+ assumes ntriv: \"\\<not> trivial_limit F\"\n+ shows \"Limsup F (\\<lambda>x. c) = (c::'a::complete_lattice)\"\n+proof -\n+ have : \"\\<And>P. Ex P \\<longleftrightarrow> P \\<noteq> (\\<lambda>x. False)\" by auto\n+ have \"\\<And>P. eventually P F \\<Longrightarrow> (SUP x : {x. P x}. c) = c\"\n+ using ntriv by (intro SUP_const) (auto simp: eventually_False )\n+ then show ?thesis\n+ unfolding Limsup_def using eventually_True\n+ by (subst INF_cong[where D=\"\\<lambda>x. c\"])\n+ (auto intro!: INF_const simp del: eventually_True)\n+qed\n+\n+lemma Liminf_const:\n+ assumes ntriv: \"\\<not> trivial_limit F\"\n+ shows \"Liminf F (\\<lambda>x. c) = (c::'a::complete_lattice)\"\n+proof -\n+ have : \"\\<And>P. Ex P \\<longleftrightarrow> P \\<noteq> (\\<lambda>x. False)\" by auto\n+ have \"\\<And>P. eventually P F \\<Longrightarrow> (INF x : {x. P x}. c) = c\"\n+ using ntriv by (intro INF_const) (auto simp: eventually_False )\n+ then show ?thesis\n+ unfolding Liminf_def using eventually_True\n+ by (subst SUP_cong[where D=\"\\<lambda>x. c\"])\n+ (auto intro!: SUP_const simp del: eventually_True)\n+qed\n+\n+lemma Liminf_mono:\n+ fixes f g :: \"'a => 'b :: complete_lattice\"\n+ assumes ev: \"eventually (\\<lambda>x. f x \\<le> g x) F\"\n+ shows \"Liminf F f \\<le> Liminf F g\"\n+ unfolding Liminf_def\n+proof (safe intro!: SUP_mono)\n+ fix P assume \"eventually P F\"\n+ with ev have \"eventually (\\<lambda>x. f x \\<le> g x \\<and> P x) F\" (is \"eventually ?Q F\") by (rule eventually_conj)\n+ then show \"\\<exists>Q\\<in>{P. eventually P F}. INFI (Collect P) f \\<le> INFI (Collect Q) g\"\n+ by (intro bexI[of _ ?Q]) (auto intro!: INF_mono)\n+qed\n+\n+lemma Liminf_eq:\n+ fixes f g :: \"'a \\<Rightarrow> 'b :: complete_lattice\"\n+ assumes \"eventually (\\<lambda>x. f x = g x) F\"\n+ shows \"Liminf F f = Liminf F g\"\n+ by (intro antisym Liminf_mono eventually_mono[OF _ assms]) auto\n+\n+lemma Limsup_mono:\n+ fixes f g :: \"'a \\<Rightarrow> 'b :: complete_lattice\"\n+ assumes ev: \"eventually (\\<lambda>x. f x \\<le> g x) F\"\n+ shows \"Limsup F f \\<le> Limsup F g\"\n+ unfolding Limsup_def\n+proof (safe intro!: INF_mono)\n+ fix P assume \"eventually P F\"\n+ with ev have \"eventually (\\<lambda>x. f x \\<le> g x \\<and> P x) F\" (is \"eventually ?Q F\") by (rule eventually_conj)\n+ then show \"\\<exists>Q\\<in>{P. eventually P F}. SUPR (Collect Q) f \\<le> SUPR (Collect P) g\"\n+ by (intro bexI[of _ ?Q]) (auto intro!: SUP_mono)\n+qed\n+\n+lemma Limsup_eq:\n+ fixes f g :: \"'a \\<Rightarrow> 'b :: complete_lattice\"\n+ assumes \"eventually (\\<lambda>x. f x = g x) net\"\n+ shows \"Limsup net f = Limsup net g\"\n+ by (intro antisym Limsup_mono eventually_mono[OF _ assms]) auto\n+\n+lemma Liminf_le_Limsup:\n+ fixes f :: \"'a \\<Rightarrow> 'b::complete_lattice\"\n+ assumes ntriv: \"\\<not> trivial_limit F\"\n+ shows \"Liminf F f \\<le> Limsup F f\"\n+ unfolding Limsup_def Liminf_def\n+ apply (rule complete_lattice_class.SUP_least)\n+ apply (rule complete_lattice_class.INF_greatest)\n+proof safe\n+ fix P Q assume \"eventually P F\" \"eventually Q F\"\n+ then have \"eventually (\\<lambda>x. P x \\<and> Q x) F\" (is \"eventually ?C F\") by (rule eventually_conj)\n+ then have not_False: \"(\\<lambda>x. P x \\<and> Q x) \\<noteq> (\\<lambda>x. False)\"\n+ using ntriv by (auto simp add: eventually_False)\n+ have \"INFI (Collect P) f \\<le> INFI (Collect ?C) f\"\n+ by (rule INF_mono) auto\n+ also have \"\\<dots> \\<le> SUPR (Collect ?C) f\"\n+ using not_False by (intro INF_le_SUP) auto\n+ also have \"\\<dots> \\<le> SUPR (Collect Q) f\"\n+ by (rule SUP_mono) auto\n+ finally show \"INFI (Collect P) f \\<le> SUPR (Collect Q) f\" .\n+qed\n+\n+lemma Liminf_bounded:\n+ fixes X Y :: \"'a \\<Rightarrow> 'b::complete_lattice\"\n+ assumes ntriv: \"\\<not> trivial_limit F\"\n+ assumes le: \"eventually (\\<lambda>n. C \\<le> X n) F\"\n+ shows \"C \\<le> Liminf F X\"\n+ using Liminf_mono[OF le] Liminf_const[OF ntriv, of C] by simp\n+\n+lemma Limsup_bounded:\n+ fixes X Y :: \"'a \\<Rightarrow> 'b::complete_lattice\"\n+ assumes ntriv: \"\\<not> trivial_limit F\"\n+ assumes le: \"eventually (\\<lambda>n. X n \\<le> C) F\"\n+ shows \"Limsup F X \\<le> C\"\n+ using Limsup_mono[OF le] Limsup_const[OF ntriv, of C] by simp\n+\n+lemma le_Liminf_iff:\n+ fixes X :: \"_ \\<Rightarrow> _ :: complete_linorder\"\n+ shows \"C \\<le> Liminf F X \\<longleftrightarrow> (\\<forall>y<C. eventually (\\<lambda>x. y < X x) F)\"\n+proof -\n+ { fix y P assume \"eventually P F\" \"y < INFI (Collect P) X\"\n+ then have \"eventually (\\<lambda>x. y < X x) F\"\n+ by (auto elim!: eventually_elim1 dest: less_INF_D) }\n+ moreover\n+ { fix y P assume \"y < C\" and y: \"\\<forall>y<C. eventually (\\<lambda>x. y < X x) F\"\n+ have \"\\<exists>P. eventually P F \\<and> y < INFI (Collect P) X\"\n+ proof cases\n+ assume \"\\<exists>z. y < z \\<and> z < C\"\n+ then guess z ..\n+ moreover then have \"z \\<le> INFI {x. z < X x} X\"\n+ by (auto intro!: INF_greatest)\n+ ultimately show ?thesis\n+ using y by (intro exI[of _ \"\\<lambda>x. z < X x\"]) auto\n+ next\n+ assume \"\\<not> (\\<exists>z. y < z \\<and> z < C)\"\n+ then have \"C \\<le> INFI {x. y < X x} X\"\n+ by (intro INF_greatest) auto\n+ with `y < C` show ?thesis\n+ using y by (intro exI[of _ \"\\<lambda>x. y < X x\"]) auto\n+ qed }\n+ ultimately show ?thesis\n+ unfolding Liminf_def le_SUP_iff by auto\n+qed\n+\n+lemma lim_imp_Liminf:\n+ fixes f :: \"'a \\<Rightarrow> _ :: {complete_linorder, linorder_topology}\"\n+ assumes ntriv: \"\\<not> trivial_limit F\"\n+ assumes lim: \"(f ---> f0) F\"\n+ shows \"Liminf F f = f0\"\n+proof (intro Liminf_eqI)\n+ fix P assume P: \"eventually P F\"\n+ then have \"eventually (\\<lambda>x. INFI (Collect P) f \\<le> f x) F\"\n+ by eventually_elim (auto intro!: INF_lower)\n+ then show \"INFI (Collect P) f \\<le> f0\"\n+ by (rule tendsto_le[OF ntriv lim tendsto_const])\n+next\n+ fix y assume upper: \"\\<And>P. eventually P F \\<Longrightarrow> INFI (Collect P) f \\<le> y\"\n+ show \"f0 \\<le> y\"\n+ proof cases\n+ assume \"\\<exists>z. y < z \\<and> z < f0\"\n+ then guess z ..\n+ moreover have \"z \\<le> INFI {x. z < f x} f\"\n+ by (rule INF_greatest) simp\n+ ultimately show ?thesis\n+ using lim[THEN topological_tendstoD, THEN upper, of \"{z <..}\"] by auto\n+ next\n+ assume discrete: \"\\<not> (\\<exists>z. y < z \\<and> z < f0)\"\n+ show ?thesis\n+ proof (rule classical)\n+ assume \"\\<not> f0 \\<le> y\"\n+ then have \"eventually (\\<lambda>x. y < f x) F\"\n+ using lim[THEN topological_tendstoD, of \"{y <..}\"] by auto\n+ then have \"eventually (\\<lambda>x. f0 \\<le> f x) F\"\n+ using discrete by (auto elim!: eventually_elim1)\n+ then have \"INFI {x. f0 \\<le> f x} f \\<le> y\"\n+ by (rule upper)\n+ moreover have \"f0 \\<le> INFI {x. f0 \\<le> f x} f\"\n+ by (intro INF_greatest) simp\n+ ultimately show \"f0 \\<le> y\" by simp\n+ qed\n+ qed\n+qed\n+\n+lemma lim_imp_Limsup:\n+ fixes f :: \"'a \\<Rightarrow> _ :: {complete_linorder, linorder_topology}\"\n+ assumes ntriv: \"\\<not> trivial_limit F\"\n+ assumes lim: \"(f ---> f0) F\"\n+ shows \"Limsup F f = f0\"\n+proof (intro Limsup_eqI)\n+ fix P assume P: \"eventually P F\"\n+ then have \"eventually (\\<lambda>x. f x \\<le> SUPR (Collect P) f) F\"\n+ by eventually_elim (auto intro!: SUP_upper)\n+ then show \"f0 \\<le> SUPR (Collect P) f\"\n+ by (rule tendsto_le[OF ntriv tendsto_const lim])\n+next\n+ fix y assume lower: \"\\<And>P. eventually P F \\<Longrightarrow> y \\<le> SUPR (Collect P) f\"\n+ show \"y \\<le> f0\"\n+ proof cases\n+ assume \"\\<exists>z. f0 < z \\<and> z < y\"\n+ then guess z ..\n+ moreover have \"SUPR {x. f x < z} f \\<le> z\"\n+ by (rule SUP_least) simp\n+ ultimately show ?thesis\n+ using lim[THEN topological_tendstoD, THEN lower, of \"{..< z}\"] by auto\n+ next\n+ assume discrete: \"\\<not> (\\<exists>z. f0 < z \\<and> z < y)\"\n+ show ?thesis\n+ proof (rule classical)\n+ assume \"\\<not> y \\<le> f0\"\n+ then have \"eventually (\\<lambda>x. f x < y) F\"\n+ using lim[THEN topological_tendstoD, of \"{..< y}\"] by auto\n+ then have \"eventually (\\<lambda>x. f x \\<le> f0) F\"\n+ using discrete by (auto elim!: eventually_elim1 simp: not_less)\n+ then have \"y \\<le> SUPR {x. f x \\<le> f0} f\"\n+ by (rule lower)\n+ moreover have \"SUPR {x. f x \\<le> f0} f \\<le> f0\"\n+ by (intro SUP_least) simp\n+ ultimately show \"y \\<le> f0\" by simp\n+ qed\n+ qed\n+qed\n+\n+lemma Liminf_eq_Limsup:\n+ fixes f0 :: \"'a :: {complete_linorder, linorder_topology}\"\n+ assumes ntriv: \"\\<not> trivial_limit F\"\n+ and lim: \"Liminf F f = f0\" \"Limsup F f = f0\"\n+ shows \"(f ---> f0) F\"\n+proof (rule order_tendstoI)\n+ fix a assume \"f0 < a\"\n+ with assms have \"Limsup F f < a\" by simp\n+ then obtain P where \"eventually P F\" \"SUPR (Collect P) f < a\"\n+ unfolding Limsup_def INF_less_iff by auto\n+ then show \"eventually (\\<lambda>x. f x < a) F\"\n+ by (auto elim!: eventually_elim1 dest: SUP_lessD)\n+next\n+ fix a assume \"a < f0\"\n+ with assms have \"a < Liminf F f\" by simp\n+ then obtain P where \"eventually P F\" \"a < INFI (Collect P) f\"\n+ unfolding Liminf_def less_SUP_iff by auto\n+ then show \"eventually (\\<lambda>x. a < f x) F\"\n+ by (auto elim!: eventually_elim1 dest: less_INF_D)\n+qed\n+\n+lemma tendsto_iff_Liminf_eq_Limsup:\n+ fixes f0 :: \"'a :: {complete_linorder, linorder_topology}\"\n+ shows \"\\<not> trivial_limit F \\<Longrightarrow> (f ---> f0) F \\<longleftrightarrow> (Liminf F f = f0 \\<and> Limsup F f = f0)\"\n+ by (metis Liminf_eq_Limsup lim_imp_Limsup lim_imp_Liminf)\n+\n+lemma liminf_subseq_mono:\n+ fixes X :: \"nat \\<Rightarrow> 'a :: complete_linorder\"\n+ assumes \"subseq r\"\n+ shows \"liminf X \\<le> liminf (X \\<circ> r) \"\n+proof-\n+ have \"\\<And>n. (INF m:{n..}. X m) \\<le> (INF m:{n..}. (X \\<circ> r) m)\"\n+ proof (safe intro!: INF_mono)\n+ fix n m :: nat assume \"n \\<le> m\" then show \"\\<exists>ma\\<in>{n..}. X ma \\<le> (X \\<circ> r) m\"\n+ using seq_suble[OF `subseq r`, of m] by (intro bexI[of _ \"r m\"]) auto\n+ qed\n+ then show ?thesis by (auto intro!: SUP_mono simp: liminf_SUPR_INFI comp_def)\n+qed\n+\n+lemma limsup_subseq_mono:\n+ fixes X :: \"nat \\<Rightarrow> 'a :: complete_linorder\"\n+ assumes \"subseq r\"\n+ shows \"limsup (X \\<circ> r) \\<le> limsup X\"\n+proof-\n+ have \"\\<And>n. (SUP m:{n..}. (X \\<circ> r) m) \\<le> (SUP m:{n..}. X m)\"\n+ proof (safe intro!: SUP_mono)\n+ fix n m :: nat assume \"n \\<le> m\" then show \"\\<exists>ma\\<in>{n..}. (X \\<circ> r) m \\<le> X ma\"\n+ using seq_suble[OF `subseq r`, of m] by (intro bexI[of _ \"r m\"]) auto\n+ qed\n+ then show ?thesis by (auto intro!: INF_mono simp: limsup_INFI_SUPR comp_def)\n+qed\n+\n+end```\n```--- a/src/HOL/Multivariate_Analysis/Extended_Real_Limits.thy\tTue Mar 05 15:27:08 2013 +0100\n+++ b/src/HOL/Multivariate_Analysis/Extended_Real_Limits.thy\tTue Mar 05 15:43:08 2013 +0100\n@@ -22,35 +22,13 @@\n\nlemma ereal_open_uminus:\nfixes S :: \"ereal set\"\n- assumes \"open S\"\n- shows \"open (uminus ` S)\"\n- unfolding open_ereal_def\n-proof (intro conjI impI)\n- obtain x y where\n- S: \"open (ereal -` S)\" \"\\<infinity> \\<in> S \\<Longrightarrow> {ereal x<..} \\<subseteq> S\" \"-\\<infinity> \\<in> S \\<Longrightarrow> {..< ereal y} \\<subseteq> S\"\n- using `open S` unfolding open_ereal_def by auto\n- have \"ereal -` uminus ` S = uminus ` (ereal -` S)\"\n- proof safe\n- fix x y\n- assume \"ereal x = - y\" \"y \\<in> S\"\n- then show \"x \\<in> uminus ` ereal -` S\" by (cases y) auto\n- next\n- fix x\n- assume \"ereal x \\<in> S\"\n- then show \"- x \\<in> ereal -` uminus ` S\"\n- by (auto intro: image_eqI[of _ _ \"ereal x\"])\n- qed\n- then show \"open (ereal -` uminus ` S)\"\n- using S by (auto intro: open_negations)\n- { assume \"\\<infinity> \\<in> uminus ` S\"\n- then have \"-\\<infinity> \\<in> S\" by (metis image_iff ereal_uminus_uminus)\n- then have \"uminus ` {..<ereal y} \\<subseteq> uminus ` S\" using S by (intro image_mono) auto\n- then show \"\\<exists>x. {ereal x<..} \\<subseteq> uminus ` S\" using ereal_uminus_lessThan by auto }\n- { assume \"-\\<infinity> \\<in> uminus ` S\"\n- then have \"\\<infinity> : S\" by (metis image_iff ereal_uminus_uminus)\n- then have \"uminus ` {ereal x<..} <= uminus ` S\" using S by (intro image_mono) auto\n- then show \"\\<exists>y. {..<ereal y} <= uminus ` S\" using ereal_uminus_greaterThan by auto }\n-qed\n+ assumes \"open S\" shows \"open (uminus ` S)\"\n+ using `open S`[unfolded open_generated_order]\n+proof induct\n+ have \"range uminus = (UNIV :: ereal set)\"\n+ by (auto simp: image_iff ereal_uminus_eq_reorder)\n+ then show \"open (range uminus :: ereal set)\" by simp\n+qed (auto simp add: image_Union image_Int)\n\nlemma ereal_uminus_complement:\nfixes S :: \"ereal set\"\n@@ -61,32 +39,7 @@\nfixes S :: \"ereal set\"\nassumes \"closed S\"\nshows \"closed (uminus ` S)\"\n- using assms unfolding closed_def\n- using ereal_open_uminus[of \"- S\"] ereal_uminus_complement by auto\n-\n-instance ereal :: perfect_space\n-proof (default, rule)\n- fix a :: ereal assume a: \"open {a}\"\n- show False\n- proof (cases a)\n- case MInf\n- then obtain y where \"{..<ereal y} <= {a}\" using a open_MInfty2[of \"{a}\"] by auto\n- then have \"ereal(y - 1):{a}\" apply (subst subsetD[of \"{..<ereal y}\"]) by auto\n- then show False using `a=(-\\<infinity>)` by auto\n- next\n- case PInf\n- then obtain y where \"{ereal y<..} <= {a}\" using a open_PInfty2[of \"{a}\"] by auto\n- then have \"ereal(y+1):{a}\" apply (subst subsetD[of \"{ereal y<..}\"]) by auto\n- then show False using `a=\\<infinity>` by auto\n- next\n- case (real r) then have fin: \"\\<bar>a\\<bar> \\<noteq> \\<infinity>\" by simp\n- from ereal_open_cont_interval[OF a singletonI this] guess e . note e = this\n- then obtain b where b_def: \"a<b & b<a+e\"\n- using fin ereal_between dense[of a \"a+e\"] by auto\n- then have \"b: {a-e <..< a+e}\" using fin ereal_between[of a e] e by auto\n- then show False using b_def e by auto\n- qed\n-qed\n+ using assms unfolding closed_def ereal_uminus_complement[symmetric] by (rule ereal_open_uminus)\n\nlemma ereal_closed_contains_Inf:\nfixes S :: \"ereal set\"\n@@ -303,113 +256,9 @@\nthen show \"x = -\\<infinity>\" by (simp add: bot_ereal_def atLeast_eq_UNIV_iff)\nqed\n\n-lemma ereal_open_mono_set:\n- fixes S :: \"ereal set\"\n- shows \"(open S \\<and> mono_set S) \\<longleftrightarrow> (S = UNIV \\<or> S = {Inf S <..})\"\n- by (metis Inf_UNIV atLeast_eq_UNIV_iff ereal_open_atLeast\n- ereal_open_closed mono_set_iff open_ereal_greaterThan)\n-\n-lemma ereal_closed_mono_set:\n- fixes S :: \"ereal set\"\n- shows \"(closed S \\<and> mono_set S) \\<longleftrightarrow> (S = {} \\<or> S = {Inf S ..})\"\n- by (metis Inf_UNIV atLeast_eq_UNIV_iff closed_ereal_atLeast\n- ereal_open_closed mono_empty mono_set_iff open_ereal_greaterThan)\n-\n-lemma ereal_Liminf_Sup_monoset:\n- fixes f :: \"'a => ereal\"\n- shows \"Liminf net f =\n- Sup {l. \\<forall>S. open S \\<longrightarrow> mono_set S \\<longrightarrow> l \\<in> S \\<longrightarrow> eventually (\\<lambda>x. f x \\<in> S) net}\"\n- (is \"_ = Sup ?A\")\n-proof (safe intro!: Liminf_eqI complete_lattice_class.Sup_upper complete_lattice_class.Sup_least)\n- fix P assume P: \"eventually P net\"\n- fix S assume S: \"mono_set S\" \"INFI (Collect P) f \\<in> S\"\n- { fix x assume \"P x\"\n- then have \"INFI (Collect P) f \\<le> f x\"\n- by (intro complete_lattice_class.INF_lower) simp\n- with S have \"f x \\<in> S\"\n- by (simp add: mono_set) }\n- with P show \"eventually (\\<lambda>x. f x \\<in> S) net\"\n- by (auto elim: eventually_elim1)\n-next\n- fix y l\n- assume S: \"\\<forall>S. open S \\<longrightarrow> mono_set S \\<longrightarrow> l \\<in> S \\<longrightarrow> eventually (\\<lambda>x. f x \\<in> S) net\"\n- assume P: \"\\<forall>P. eventually P net \\<longrightarrow> INFI (Collect P) f \\<le> y\"\n- show \"l \\<le> y\"\n- proof (rule dense_le)\n- fix B assume \"B < l\"\n- then have \"eventually (\\<lambda>x. f x \\<in> {B <..}) net\"\n- by (intro S[rule_format]) auto\n- then have \"INFI {x. B < f x} f \\<le> y\"\n- using P by auto\n- moreover have \"B \\<le> INFI {x. B < f x} f\"\n- by (intro INF_greatest) auto\n- ultimately show \"B \\<le> y\"\n- by simp\n- qed\n-qed\n-\n-lemma ereal_Limsup_Inf_monoset:\n- fixes f :: \"'a => ereal\"\n- shows \"Limsup net f =\n- Inf {l. \\<forall>S. open S \\<longrightarrow> mono_set (uminus ` S) \\<longrightarrow> l \\<in> S \\<longrightarrow> eventually (\\<lambda>x. f x \\<in> S) net}\"\n- (is \"_ = Inf ?A\")\n-proof (safe intro!: Limsup_eqI complete_lattice_class.Inf_lower complete_lattice_class.Inf_greatest)\n- fix P assume P: \"eventually P net\"\n- fix S assume S: \"mono_set (uminus`S)\" \"SUPR (Collect P) f \\<in> S\"\n- { fix x assume \"P x\"\n- then have \"f x \\<le> SUPR (Collect P) f\"\n- by (intro complete_lattice_class.SUP_upper) simp\n- with S(1)[unfolded mono_set, rule_format, of \"- SUPR (Collect P) f\" \"- f x\"] S(2)\n- have \"f x \\<in> S\"\n- by (simp add: inj_image_mem_iff) }\n- with P show \"eventually (\\<lambda>x. f x \\<in> S) net\"\n- by (auto elim: eventually_elim1)\n-next\n- fix y l\n- assume S: \"\\<forall>S. open S \\<longrightarrow> mono_set (uminus ` S) \\<longrightarrow> l \\<in> S \\<longrightarrow> eventually (\\<lambda>x. f x \\<in> S) net\"\n- assume P: \"\\<forall>P. eventually P net \\<longrightarrow> y \\<le> SUPR (Collect P) f\"\n- show \"y \\<le> l\"\n- proof (rule dense_ge)\n- fix B assume \"l < B\"\n- then have \"eventually (\\<lambda>x. f x \\<in> {..< B}) net\"\n- by (intro S[rule_format]) auto\n- then have \"y \\<le> SUPR {x. f x < B} f\"\n- using P by auto\n- moreover have \"SUPR {x. f x < B} f \\<le> B\"\n- by (intro SUP_least) auto\n- ultimately show \"y \\<le> B\"\n- by simp\n- qed\n-qed\n-\nlemma open_uminus_iff: \"open (uminus ` S) \\<longleftrightarrow> open (S::ereal set)\"\nusing ereal_open_uminus[of S] ereal_open_uminus[of \"uminus`S\"] by auto\n\n-lemma ereal_Limsup_uminus:\n- fixes f :: \"'a => ereal\"\n- shows \"Limsup net (\\<lambda>x. - (f x)) = -(Liminf net f)\"\n-proof -\n- { fix P l\n- have \"(\\<exists>x. (l::ereal) = -x \\<and> P x) \\<longleftrightarrow> P (-l)\"\n- by (auto intro!: exI[of _ \"-l\"]) }\n- note Ex_cancel = this\n- { fix P :: \"ereal set \\<Rightarrow> bool\"\n- have \"(\\<forall>S. P S) \\<longleftrightarrow> (\\<forall>S. P (uminus`S))\"\n- apply auto\n- apply (erule_tac x=\"uminus`S\" in allE)\n- apply (auto simp: image_image)\n- done }\n- { fix x S\n- have \"(x::ereal) \\<in> uminus`S \\<longleftrightarrow> -x\\<in>S\"\n- by (auto intro!: image_eqI[of _ _ \"-x\"]) }\n- note remove_uminus_image = this\n- show ?thesis\n- unfolding ereal_Limsup_Inf_monoset ereal_Liminf_Sup_monoset\n- unfolding ereal_Inf_uminus_image_eq[symmetric] image_Collect Ex_cancel\n-qed\n-\nlemma ereal_Liminf_uminus:\nfixes f :: \"'a => ereal\"\nshows \"Liminf net (\\<lambda>x. - (f x)) = -(Limsup net f)\"\n@@ -423,14 +272,6 @@\nereal_lim_mult[of \"\\<lambda>x. - (f x)\" \"-f0\" net \"- 1\"]\nby (auto simp: ereal_uminus_reorder)\n\n-lemma lim_imp_Limsup:\n- fixes f :: \"'a => ereal\"\n- assumes \"\\<not> trivial_limit net\"\n- and lim: \"(f ---> f0) net\"\n- shows \"Limsup net f = f0\"\n- using ereal_Lim_uminus[of f f0] lim_imp_Liminf[of net \"(%x. -(f x))\" \"-f0\"]\n- ereal_Liminf_uminus[of net f] assms by simp\n-\nlemma convergent_ereal_limsup:\nfixes X :: \"nat \\<Rightarrow> ereal\"\nshows \"convergent X \\<Longrightarrow> limsup X = lim X\"\n@@ -461,43 +302,10 @@\nusing assms ereal_Lim_uminus[of f \"-\\<infinity>\"] Liminf_PInfty[of _ \"\\<lambda>x. - (f x)\"]\nereal_Liminf_uminus[of _ f] by (auto simp: ereal_uminus_eq_reorder)\n\n-lemma ereal_Liminf_eq_Limsup:\n- fixes f :: \"'a \\<Rightarrow> ereal\"\n- assumes ntriv: \"\\<not> trivial_limit net\"\n- and lim: \"Liminf net f = f0\" \"Limsup net f = f0\"\n- shows \"(f ---> f0) net\"\n-proof (cases f0)\n- case PInf\n- then show ?thesis using Liminf_PInfty[OF ntriv] lim by auto\n-next\n- case MInf\n- then show ?thesis using Limsup_MInfty[OF ntriv] lim by auto\n-next\n- case (real r)\n- show \"(f ---> f0) net\"\n- proof (rule topological_tendstoI)\n- fix S\n- assume \"open S\" \"f0 \\<in> S\"\n- then obtain a b where \"a < Liminf net f\" \"Limsup net f < b\" \"{a<..<b} \\<subseteq> S\"\n- using ereal_open_cont_interval2[of S f0] real lim by auto\n- then have \"eventually (\\<lambda>x. f x \\<in> {a<..<b}) net\"\n- unfolding Liminf_def Limsup_def less_SUP_iff INF_less_iff\n- by (auto intro!: eventually_conj elim: eventually_elim1 dest: less_INF_D SUP_lessD)\n- with `{a<..<b} \\<subseteq> S` show \"eventually (%x. f x : S) net\"\n- by (rule_tac eventually_mono) auto\n- qed\n-qed\n-\n-lemma ereal_Liminf_eq_Limsup_iff:\n- fixes f :: \"'a \\<Rightarrow> ereal\"\n- assumes \"\\<not> trivial_limit net\"\n- shows \"(f ---> f0) net \\<longleftrightarrow> Liminf net f = f0 \\<and> Limsup net f = f0\"\n- by (metis assms ereal_Liminf_eq_Limsup lim_imp_Liminf lim_imp_Limsup)\n-\nlemma convergent_ereal:\nfixes X :: \"nat \\<Rightarrow> ereal\"\nshows \"convergent X \\<longleftrightarrow> limsup X = liminf X\"\n- using ereal_Liminf_eq_Limsup_iff[of sequentially]\n+ using tendsto_iff_Liminf_eq_Limsup[of sequentially]\nby (auto simp: convergent_def)\n\nlemma limsup_INFI_SUPR:\n@@ -535,45 +343,6 @@\nshows \"X N >= L\"\nusing dec by (intro ereal_lim_mono[of N, OF _ lim tendsto_const]) (simp add: decseq_def)\n\n-lemma liminf_bounded_open:\n- fixes x :: \"nat \\<Rightarrow> ereal\"\n- shows \"x0 \\<le> liminf x \\<longleftrightarrow> (\\<forall>S. open S \\<longrightarrow> mono_set S \\<longrightarrow> x0 \\<in> S \\<longrightarrow> (\\<exists>N. \\<forall>n\\<ge>N. x n \\<in> S))\"\n- (is \"_ \\<longleftrightarrow> ?P x0\")\n-proof\n- assume \"?P x0\"\n- then show \"x0 \\<le> liminf x\"\n- unfolding ereal_Liminf_Sup_monoset eventually_sequentially\n- by (intro complete_lattice_class.Sup_upper) auto\n-next\n- assume \"x0 \\<le> liminf x\"\n- { fix S :: \"ereal set\"\n- assume om: \"open S & mono_set S & x0:S\"\n- { assume \"S = UNIV\" then have \"EX N. (ALL n>=N. x n : S)\" by auto }\n- moreover\n- { assume \"~(S=UNIV)\"\n- then obtain B where B_def: \"S = {B<..}\" using om ereal_open_mono_set by auto\n- then have \"B<x0\" using om by auto\n- then have \"EX N. ALL n>=N. x n : S\"\n- unfolding B_def using `x0 \\<le> liminf x` liminf_bounded_iff by auto\n- }\n- ultimately have \"EX N. (ALL n>=N. x n : S)\" by auto\n- }\n- then show \"?P x0\" by auto\n-qed\n-\n-lemma limsup_subseq_mono:\n- fixes X :: \"nat \\<Rightarrow> ereal\"\n- assumes \"subseq r\"\n- shows \"limsup (X \\<circ> r) \\<le> limsup X\"\n-proof -\n- have \"(\\<lambda>n. - X n) \\<circ> r = (\\<lambda>n. - (X \\<circ> r) n)\" by (simp add: fun_eq_iff)\n- then have \"- limsup X \\<le> - limsup (X \\<circ> r)\"\n- using liminf_subseq_mono[of r \"(%n. - X n)\"]\n- ereal_Liminf_uminus[of sequentially X]\n- ereal_Liminf_uminus[of sequentially \"X o r\"] assms by auto\n- then show ?thesis by auto\n-qed\n-\nlemma bounded_abs:\nassumes \"(a::real)<=x\" \"x<=b\"\nshows \"abs x <= max (abs a) (abs b)\"\n@@ -652,85 +421,6 @@\nshow \"f ----> x\" by auto }\nqed\n\n-lemma Liminf_within:\n- fixes f :: \"'a::metric_space \\<Rightarrow> 'b::complete_lattice\"\n- shows \"Liminf (at x within S) f = (SUP e:{0<..}. INF y:(S \\<inter> ball x e - {x}). f y)\"\n- unfolding Liminf_def eventually_within\n-proof (rule SUPR_eq, simp_all add: Ball_def Bex_def, safe)\n- fix P d assume \"0 < d\" \"\\<forall>y. y \\<in> S \\<longrightarrow> 0 < dist y x \\<and> dist y x < d \\<longrightarrow> P y\"\n- then have \"S \\<inter> ball x d - {x} \\<subseteq> {x. P x}\"\n- by (auto simp: zero_less_dist_iff dist_commute)\n- then show \"\\<exists>r>0. INFI (Collect P) f \\<le> INFI (S \\<inter> ball x r - {x}) f\"\n- by (intro exI[of _ d] INF_mono conjI `0 < d`) auto\n-next\n- fix d :: real assume \"0 < d\"\n- then show \"\\<exists>P. (\\<exists>d>0. \\<forall>xa. xa \\<in> S \\<longrightarrow> 0 < dist xa x \\<and> dist xa x < d \\<longrightarrow> P xa) \\<and>\n- INFI (S \\<inter> ball x d - {x}) f \\<le> INFI (Collect P) f\"\n- by (intro exI[of _ \"\\<lambda>y. y \\<in> S \\<inter> ball x d - {x}\"])\n- (auto intro!: INF_mono exI[of _ d] simp: dist_commute)\n-qed\n-\n-lemma Limsup_within:\n- fixes f :: \"'a::metric_space => 'b::complete_lattice\"\n- shows \"Limsup (at x within S) f = (INF e:{0<..}. SUP y:(S \\<inter> ball x e - {x}). f y)\"\n- unfolding Limsup_def eventually_within\n-proof (rule INFI_eq, simp_all add: Ball_def Bex_def, safe)\n- fix P d assume \"0 < d\" \"\\<forall>y. y \\<in> S \\<longrightarrow> 0 < dist y x \\<and> dist y x < d \\<longrightarrow> P y\"\n- then have \"S \\<inter> ball x d - {x} \\<subseteq> {x. P x}\"\n- by (auto simp: zero_less_dist_iff dist_commute)\n- then show \"\\<exists>r>0. SUPR (S \\<inter> ball x r - {x}) f \\<le> SUPR (Collect P) f\"\n- by (intro exI[of _ d] SUP_mono conjI `0 < d`) auto\n-next\n- fix d :: real assume \"0 < d\"\n- then show \"\\<exists>P. (\\<exists>d>0. \\<forall>xa. xa \\<in> S \\<longrightarrow> 0 < dist xa x \\<and> dist xa x < d \\<longrightarrow> P xa) \\<and>\n- SUPR (Collect P) f \\<le> SUPR (S \\<inter> ball x d - {x}) f\"\n- by (intro exI[of _ \"\\<lambda>y. y \\<in> S \\<inter> ball x d - {x}\"])\n- (auto intro!: SUP_mono exI[of _ d] simp: dist_commute)\n-qed\n-\n-lemma Liminf_within_UNIV:\n- fixes f :: \"'a::metric_space => _\"\n- shows \"Liminf (at x) f = Liminf (at x within UNIV) f\"\n- by simp ( TODO: delete )\n-\n-\n-lemma Liminf_at:\n- fixes f :: \"'a::metric_space => _\"\n- shows \"Liminf (at x) f = (SUP e:{0<..}. INF y:(ball x e - {x}). f y)\"\n- using Liminf_within[of x UNIV f] by simp\n-\n-\n-lemma Limsup_within_UNIV:\n- fixes f :: \"'a::metric_space => _\"\n- shows \"Limsup (at x) f = Limsup (at x within UNIV) f\"\n- by simp ( TODO: delete )\n-\n-\n-lemma Limsup_at:\n- fixes f :: \"'a::metric_space => _\"\n- shows \"Limsup (at x) f = (INF e:{0<..}. SUP y:(ball x e - {x}). f y)\"\n- using Limsup_within[of x UNIV f] by simp\n-\n-lemma Lim_within_constant:\n- assumes \"ALL y:S. f y = C\"\n- shows \"(f ---> C) (at x within S)\"\n- unfolding tendsto_def Limits.eventually_within eventually_at_topological\n- using assms by simp (metis open_UNIV UNIV_I)\n-\n-lemma Liminf_within_constant:\n- fixes f :: \"'a::topological_space \\<Rightarrow> ereal\"\n- assumes \"ALL y:S. f y = C\"\n- and \"~trivial_limit (at x within S)\"\n- shows \"Liminf (at x within S) f = C\"\n- by (metis Lim_within_constant assms lim_imp_Liminf)\n-\n-lemma Limsup_within_constant:\n- fixes f :: \"'a::topological_space \\<Rightarrow> ereal\"\n- assumes \"ALL y:S. f y = C\"\n- and \"~trivial_limit (at x within S)\"\n- shows \"Limsup (at x within S) f = C\"\n- by (metis Lim_within_constant assms lim_imp_Limsup)\n-\nlemma islimpt_punctured: \"x islimpt S = x islimpt (S-{x})\"\nunfolding islimpt_def by blast\n\n@@ -1317,4 +1007,205 @@\nthen show \"\\<forall>i. f i = 0\" by auto\nqed simp\n\n+lemma Liminf_within:\n+ fixes f :: \"'a::metric_space \\<Rightarrow> 'b::complete_lattice\"\n+ shows \"Liminf (at x within S) f = (SUP e:{0<..}. INF y:(S \\<inter> ball x e - {x}). f y)\"\n+ unfolding Liminf_def eventually_within\n+proof (rule SUPR_eq, simp_all add: Ball_def Bex_def, safe)\n+ fix P d assume \"0 < d\" \"\\<forall>y. y \\<in> S \\<longrightarrow> 0 < dist y x \\<and> dist y x < d \\<longrightarrow> P y\"\n+ then have \"S \\<inter> ball x d - {x} \\<subseteq> {x. P x}\"\n+ by (auto simp: zero_less_dist_iff dist_commute)\n+ then show \"\\<exists>r>0. INFI (Collect P) f \\<le> INFI (S \\<inter> ball x r - {x}) f\"\n+ by (intro exI[of _ d] INF_mono conjI `0 < d`) auto\n+next\n+ fix d :: real assume \"0 < d\"\n+ then show \"\\<exists>P. (\\<exists>d>0. \\<forall>xa. xa \\<in> S \\<longrightarrow> 0 < dist xa x \\<and> dist xa x < d \\<longrightarrow> P xa) \\<and>\n+ INFI (S \\<inter> ball x d - {x}) f \\<le> INFI (Collect P) f\"\n+ by (intro exI[of _ \"\\<lambda>y. y \\<in> S \\<inter> ball x d - {x}\"])\n+ (auto intro!: INF_mono exI[of _ d] simp: dist_commute)\n+qed\n+\n+lemma Limsup_within:\n+ fixes f :: \"'a::metric_space => 'b::complete_lattice\"\n+ shows \"Limsup (at x within S) f = (INF e:{0<..}. SUP y:(S \\<inter> ball x e - {x}). f y)\"\n+ unfolding Limsup_def eventually_within\n+proof (rule INFI_eq, simp_all add: Ball_def Bex_def, safe)\n+ fix P d assume \"0 < d\" \"\\<forall>y. y \\<in> S \\<longrightarrow> 0 < dist y x \\<and> dist y x < d \\<longrightarrow> P y\"\n+ then have \"S \\<inter> ball x d - {x} \\<subseteq> {x. P x}\"\n+ by (auto simp: zero_less_dist_iff dist_commute)\n+ then show \"\\<exists>r>0. SUPR (S \\<inter> ball x r - {x}) f \\<le> SUPR (Collect P) f\"\n+ by (intro exI[of _ d] SUP_mono conjI `0 < d`) auto\n+next\n+ fix d :: real assume \"0 < d\"\n+ then show \"\\<exists>P. (\\<exists>d>0. \\<forall>xa. xa \\<in> S \\<longrightarrow> 0 < dist xa x \\<and> dist xa x < d \\<longrightarrow> P xa) \\<and>\n+ SUPR (Collect P) f \\<le> SUPR (S \\<inter> ball x d - {x}) f\"\n+ by (intro exI[of _ \"\\<lambda>y. y \\<in> S \\<inter> ball x d - {x}\"])\n+ (auto intro!: SUP_mono exI[of _ d] simp: dist_commute)\n+qed\n+\n+lemma Liminf_at:\n+ fixes f :: \"'a::metric_space => _\"\n+ shows \"Liminf (at x) f = (SUP e:{0<..}. INF y:(ball x e - {x}). f y)\"\n+ using Liminf_within[of x UNIV f] by simp\n+\n+lemma Limsup_at:\n+ fixes f :: \"'a::metric_space => _\"\n+ shows \"Limsup (at x) f = (INF e:{0<..}. SUP y:(ball x e - {x}). f y)\"\n+ using Limsup_within[of x UNIV f] by simp\n+\n+lemma min_Liminf_at:\n+ fixes f :: \"'a::metric_space => 'b::complete_linorder\"\n+ shows \"min (f x) (Liminf (at x) f) = (SUP e:{0<..}. INF y:ball x e. f y)\"\n+ unfolding inf_min[symmetric] Liminf_at\n+ apply (subst inf_commute)\n+ apply (subst SUP_inf)\n+ apply (intro SUP_cong[OF refl])\n+ apply (cut_tac A=\"ball x b - {x}\" and B=\"{x}\" and M=f in INF_union)\n+ apply (simp add: INF_def del: inf_ereal_def)\n+ done\n+\n+subsection { monoset *}\n+\n+definition (in order) mono_set:\n+ \"mono_set S \\<longleftrightarrow> (\\<forall>x y. x \\<le> y \\<longrightarrow> x \\<in> S \\<longrightarrow> y \\<in> S)\"\n+\n+lemma (in order) mono_greaterThan [intro, simp]: \"mono_set {B<..}\" unfolding mono_set by auto\n+lemma (in order) mono_atLeast [intro, simp]: \"mono_set {B..}\" unfolding mono_set by auto\n+lemma (in order) mono_UNIV [intro, simp]: \"mono_set UNIV\" unfolding mono_set by auto\n+lemma (in order) mono_empty [intro, simp]: \"mono_set {}\" unfolding mono_set by auto\n+\n+lemma (in complete_linorder) mono_set_iff:\n+ fixes S :: \"'a set\"\n+ defines \"a \\<equiv> Inf S\"\n+ shows \"mono_set S \\<longleftrightarrow> (S = {a <..} \\<or> S = {a..})\" (is \"_ = ?c\")\n+proof\n+ assume \"mono_set S\"\n+ then have mono: \"\\<And>x y. x \\<le> y \\<Longrightarrow> x \\<in> S \\<Longrightarrow> y \\<in> S\" by (auto simp: mono_set)\n+ show ?c\n+ proof cases\n+ assume \"a \\<in> S\"\n+ show ?c\n+ using mono[OF _ `a \\<in> S`]\n+ by (auto intro: Inf_lower simp: a_def)\n+ next\n+ assume \"a \\<notin> S\"\n+ have \"S = {a <..}\"\n+ proof safe\n+ fix x assume \"x \\<in> S\"\n+ then have \"a \\<le> x\" unfolding a_def by (rule Inf_lower)\n+ then show \"a < x\" using `x \\<in> S` `a \\<notin> S` by (cases \"a = x\") auto\n+ next\n+ fix x assume \"a < x\"\n+ then obtain y where \"y < x\" \"y \\<in> S\" unfolding a_def Inf_less_iff ..\n+ with mono[of y x] show \"x \\<in> S\" by auto\n+ qed\n+ then show ?c ..\n+ qed\n+qed auto\n+\n+lemma ereal_open_mono_set:\n+ fixes S :: \"ereal set\"\n+ shows \"(open S \\<and> mono_set S) \\<longleftrightarrow> (S = UNIV \\<or> S = {Inf S <..})\"\n+ by (metis Inf_UNIV atLeast_eq_UNIV_iff ereal_open_atLeast\n+ ereal_open_closed mono_set_iff open_ereal_greaterThan)\n+\n+lemma ereal_closed_mono_set:\n+ fixes S :: \"ereal set\"\n+ shows \"(closed S \\<and> mono_set S) \\<longleftrightarrow> (S = {} \\<or> S = {Inf S ..})\"\n+ by (metis Inf_UNIV atLeast_eq_UNIV_iff closed_ereal_atLeast\n+ ereal_open_closed mono_empty mono_set_iff open_ereal_greaterThan)\n+\n+lemma ereal_Liminf_Sup_monoset:\n+ fixes f :: \"'a => ereal\"\n+ shows \"Liminf net f =\n+ Sup {l. \\<forall>S. open S \\<longrightarrow> mono_set S \\<longrightarrow> l \\<in> S \\<longrightarrow> eventually (\\<lambda>x. f x \\<in> S) net}\"\n+ (is \"_ = Sup ?A\")\n+proof (safe intro!: Liminf_eqI complete_lattice_class.Sup_upper complete_lattice_class.Sup_least)\n+ fix P assume P: \"eventually P net\"\n+ fix S assume S: \"mono_set S\" \"INFI (Collect P) f \\<in> S\"\n+ { fix x assume \"P x\"\n+ then have \"INFI (Collect P) f \\<le> f x\"\n+ by (intro complete_lattice_class.INF_lower) simp\n+ with S have \"f x \\<in> S\"\n+ by (simp add: mono_set) }\n+ with P show \"eventually (\\<lambda>x. f x \\<in> S) net\"\n+ by (auto elim: eventually_elim1)\n+next\n+ fix y l\n+ assume S: \"\\<forall>S. open S \\<longrightarrow> mono_set S \\<longrightarrow> l \\<in> S \\<longrightarrow> eventually (\\<lambda>x. f x \\<in> S) net\"\n+ assume P: \"\\<forall>P. eventually P net \\<longrightarrow> INFI (Collect P) f \\<le> y\"\n+ show \"l \\<le> y\"\n+ proof (rule dense_le)\n+ fix B assume \"B < l\"\n+ then have \"eventually (\\<lambda>x. f x \\<in> {B <..}) net\"\n+ by (intro S[rule_format]) auto\n+ then have \"INFI {x. B < f x} f \\<le> y\"\n+ using P by auto\n+ moreover have \"B \\<le> INFI {x. B < f x} f\"\n+ by (intro INF_greatest) auto\n+ ultimately show \"B \\<le> y\"\n+ by simp\n+ qed\n+qed\n+\n+lemma ereal_Limsup_Inf_monoset:\n+ fixes f :: \"'a => ereal\"\n+ shows \"Limsup net f =\n+ Inf {l. \\<forall>S. open S \\<longrightarrow> mono_set (uminus ` S) \\<longrightarrow> l \\<in> S \\<longrightarrow> eventually (\\<lambda>x. f x \\<in> S) net}\"\n+ (is \"_ = Inf ?A\")\n+proof (safe intro!: Limsup_eqI complete_lattice_class.Inf_lower complete_lattice_class.Inf_greatest)\n+ fix P assume P: \"eventually P net\"\n+ fix S assume S: \"mono_set (uminus`S)\" \"SUPR (Collect P) f \\<in> S\"\n+ { fix x assume \"P x\"\n+ then have \"f x \\<le> SUPR (Collect P) f\"\n+ by (intro complete_lattice_class.SUP_upper) simp\n+ with S(1)[unfolded mono_set, rule_format, of \"- SUPR (Collect P) f\" \"- f x\"] S(2)\n+ have \"f x \\<in> S\"\n+ by (simp add: inj_image_mem_iff) }\n+ with P show \"eventually (\\<lambda>x. f x \\<in> S) net\"\n+ by (auto elim: eventually_elim1)\n+next\n+ fix y l\n+ assume S: \"\\<forall>S. open S \\<longrightarrow> mono_set (uminus ` S) \\<longrightarrow> l \\<in> S \\<longrightarrow> eventually (\\<lambda>x. f x \\<in> S) net\"\n+ assume P: \"\\<forall>P. eventually P net \\<longrightarrow> y \\<le> SUPR (Collect P) f\"\n+ show \"y \\<le> l\"\n+ proof (rule dense_ge)\n+ fix B assume \"l < B\"\n+ then have \"eventually (\\<lambda>x. f x \\<in> {..< B}) net\"\n+ by (intro S[rule_format]) auto\n+ then have \"y \\<le> SUPR {x. f x < B} f\"\n+ using P by auto\n+ moreover have \"SUPR {x. f x < B} f \\<le> B\"\n+ by (intro SUP_least) auto\n+ ultimately show \"y \\<le> B\"\n+ by simp\n+ qed\n+qed\n+\n+lemma liminf_bounded_open:\n+ fixes x :: \"nat \\<Rightarrow> ereal\"\n+ shows \"x0 \\<le> liminf x \\<longleftrightarrow> (\\<forall>S. open S \\<longrightarrow> mono_set S \\<longrightarrow> x0 \\<in> S \\<longrightarrow> (\\<exists>N. \\<forall>n\\<ge>N. x n \\<in> S))\"\n+ (is \"_ \\<longleftrightarrow> ?P x0\")\n+proof\n+ assume \"?P x0\"\n+ then show \"x0 \\<le> liminf x\"\n+ unfolding ereal_Liminf_Sup_monoset eventually_sequentially\n+ by (intro complete_lattice_class.Sup_upper) auto\n+next\n+ assume \"x0 \\<le> liminf x\"\n+ { fix S :: \"ereal set\"\n+ assume om: \"open S & mono_set S & x0:S\"\n+ { assume \"S = UNIV\" then have \"EX N. (ALL n>=N. x n : S)\" by auto }\n+ moreover\n+ { assume \"~(S=UNIV)\"\n+ then obtain B where B_def: \"S = {B<..}\" using om ereal_open_mono_set by auto\n+ then have \"B<x0\" using om by auto\n+ then have \"EX N. ALL n>=N. x n : S\"\n+ unfolding B_def using `x0 \\<le> liminf x` liminf_bounded_iff by auto\n+ }\n+ ultimately have \"EX N. (ALL n>=N. x n : S)\" by auto\n+ }\n+ then show \"?P x0\" by auto\n+qed\n+\nend```\n```--- a/src/HOL/Probability/Lebesgue_Integration.thy\tTue Mar 05 15:27:08 2013 +0100\n+++ b/src/HOL/Probability/Lebesgue_Integration.thy\tTue Mar 05 15:43:08 2013 +0100\n@@ -2190,7 +2190,7 @@\nusing diff positive_integral_positive[of M]\nby (subst integral_eq_positive_integral[of _ M]) (auto simp: ereal_real integrable_def)\nthen show ?lim_diff\n- using ereal_Liminf_eq_Limsup[OF trivial_limit_sequentially liminf_limsup_eq]\n+ using Liminf_eq_Limsup[OF trivial_limit_sequentially liminf_limsup_eq]\nby simp\n\nshow ?lim```" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.56489486,"math_prob":0.96426195,"size":53370,"snap":"2021-31-2021-39","text_gpt3_token_len":18160,"char_repetition_ratio":0.18487427,"word_repetition_ratio":0.5840009,"special_character_ratio":0.37764663,"punctuation_ratio":0.15376645,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9994634,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-07-26T15:49:13Z\",\"WARC-Record-ID\":\"<urn:uuid:56b3030d-21f0-4221-bfa6-80f562e8131b>\",\"Content-Length\":\"164464\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c3e3d81b-a2a3-4f75-9016-175e9fcf1538>\",\"WARC-Concurrent-To\":\"<urn:uuid:c7efd416-44c8-4274-a825-e28e9760a121>\",\"WARC-IP-Address\":\"131.159.46.82\",\"WARC-Target-URI\":\"https://isabelle.in.tum.de/repos/isabelle/rev/5e6296afe08d?revcount=30\",\"WARC-Payload-Digest\":\"sha1:NPEYEHGVW3DI723OE2ZTVWOOE7XZ5KM7\",\"WARC-Block-Digest\":\"sha1:33ZZGFDRGMYBWJDZRK4QG3PTPWTIGVXE\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-31/CC-MAIN-2021-31_segments_1627046152144.81_warc_CC-MAIN-20210726152107-20210726182107-00672.warc.gz\"}"}