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https://www.shaalaa.com/question-bank-solutions/representation-images-formed-spherical-mirrors-using-ray-diagrams-concave-mirror-we-wish-obtain-erect-image-object-using-concave-mirror-focal-length-15-cm-what-should-be-range-distance-object-mirror_6179	[ "Share\nNotifications\n\nView all notifications\nBooks Shortlist\nYour shortlist is empty\n\n# We Wish to Obtain an Erect Image of an Object, Using a Concave Mirror of Focal Length 15 Cm. What Should Be the Range of Distance of the Object from the Mirror? - CBSE Class 10 - Science\n\nLogin\nCreate free account\n\nForgot password?\nConceptRepresentation of Images Formed by Spherical Mirrors Using Ray Diagrams - Concave Mirror\n\n#### Question\n\nWe wish to obtain an erect image of an object, using a concave mirror of focal length 15 cm. What should be the range of distance of the object from the mirror? What is the nature of the image? Is the image larger or smaller than the object? Draw a ray diagram to show the image formation in this case.\n\n#### Solution 1\n\nRange of the distance of the object = 0 to 15 cm from the pole of the mirror.\nNature of the image = virtual, erect and larger than the object.", null, "#### Solution 2\n\nRange of object distance = 0 cm to15 cm\nA concave mirror gives an erect image when an object is placed between its pole (P) and the principal focus (F).\nHence, to obtain an erect image of an object from a concave mirror of focal length 15 cm, the object must be placed anywhere between the pole and the focus. The image formed will be virtual, erect, and magnified in nature, as shown in the given figure.", null, "Is there an error in this question or solution?\n\n#### APPEARS IN\n\nNCERT Solution for Science Textbook for Class 10 (2019 to Current)\nChapter 10: Light – Reflection and Refraction\nQ: 7 \| Page no. 186\nSolution We Wish to Obtain an Erect Image of an Object, Using a Concave Mirror of Focal Length 15 Cm. What Should Be the Range of Distance of the Object from the Mirror? Concept: Representation of Images Formed by Spherical Mirrors Using Ray Diagrams - Concave Mirror.\nS" ]	[ null, "https://www.shaalaa.com/images/_4:ac5a12585d0349fc91f62f6334fd8924.png", null, "https://www.shaalaa.com/images/_4:f1db2eb3a5264e2696270ea4958760a5.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.90484565,"math_prob":0.6441245,"size":882,"snap":"2019-35-2019-39","text_gpt3_token_len":209,"char_repetition_ratio":0.17995444,"word_repetition_ratio":0.071428575,"special_character_ratio":0.23469388,"punctuation_ratio":0.09139785,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9504005,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,3,null,3,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-08-20T03:49:25Z\",\"WARC-Record-ID\":\"<urn:uuid:bd049ace-186f-4798-a5ea-9916524b0fbb>\",\"Content-Length\":\"47546\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4136fc12-b318-4b7d-998f-f966d4b7936e>\",\"WARC-Concurrent-To\":\"<urn:uuid:dbbfc90b-de21-4ef8-8c57-86df20e0f866>\",\"WARC-IP-Address\":\"139.162.10.45\",\"WARC-Target-URI\":\"https://www.shaalaa.com/question-bank-solutions/representation-images-formed-spherical-mirrors-using-ray-diagrams-concave-mirror-we-wish-obtain-erect-image-object-using-concave-mirror-focal-length-15-cm-what-should-be-range-distance-object-mirror_6179\",\"WARC-Payload-Digest\":\"sha1:WQDZ2NMLSVUVAZ2P6MOSFEE7VQ27RRDI\",\"WARC-Block-Digest\":\"sha1:3JGZTRHXSJQWI2VKKRZVDYHLEHN5WFV2\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-35/CC-MAIN-2019-35_segments_1566027315222.14_warc_CC-MAIN-20190820024110-20190820050110-00393.warc.gz\"}"}
https://eprint.iacr.org/2001/094	[ "### Slope packings and coverings, and generic algorithms for the discrete logarithm problem\n\nM. Chateauneuf, A. C. H. Ling, and D. R. Stinson\n\n##### Abstract\n\nWe consider the set of slopes of lines formed by joining all pairs of points in some subset S of a Desarguesian affine plane of prime order, p. If all the slopes are distinct and non-infinite, we have a slope packing; if every possible non-infinite slope occurs, then we have a slope covering. We review and unify some results on these problems that can be derived from the study of Sidon sets and sum covers. Then we report some computational results we have obtained for small values of p. Finally, we point out some connections between slope packings and coverings and generic algorithms for the discrete logarithm problem in prime order (sub)groups. Our results provide a combinatorial characterization of such algorithms, in the sense that any generic algorithm implies the existence of a certain slope packing or covering, and conversely.\n\nAvailable format(s)\nCategory\nFoundations\nPublication info\nPublished elsewhere. preprint\nKeywords\ndiscrete logarithm problemcombinatorial cryptography\nContact author(s)\ndstinson @ uwaterloo ca\nHistory\nShort URL\nhttps://ia.cr/2001/094", null, "CC BY\n\nBibTeX\n\n@misc{cryptoeprint:2001/094,\nauthor = {M. Chateauneuf and A. C. H. Ling and D. R. Stinson},\ntitle = {Slope packings and coverings, and generic algorithms for the discrete logarithm problem},\nhowpublished = {Cryptology ePrint Archive, Paper 2001/094},\nyear = {2001},\nnote = {\\url{https://eprint.iacr.org/2001/094}},\nurl = {https://eprint.iacr.org/2001/094}\n}", null, "Note: In order to protect the privacy of readers, eprint.iacr.org does not use cookies or embedded third party content." ]	[ null, "https://eprint.iacr.org/img/license/CC_BY.svg", null, "https://eprint.iacr.org/img/iacrlogo_small.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.83990467,"math_prob":0.64632624,"size":1682,"snap":"2023-14-2023-23","text_gpt3_token_len":432,"char_repetition_ratio":0.116805725,"word_repetition_ratio":0.06692913,"special_character_ratio":0.25208086,"punctuation_ratio":0.1525974,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95302504,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-06-03T15:31:36Z\",\"WARC-Record-ID\":\"<urn:uuid:ff97cfe7-9f22-4641-a5bb-5ff8c61bf889>\",\"Content-Length\":\"12216\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:950d1b08-6871-4831-94a4-a65258080d47>\",\"WARC-Concurrent-To\":\"<urn:uuid:295c2890-df48-4b81-8cf4-965708e8219f>\",\"WARC-IP-Address\":\"64.227.104.178\",\"WARC-Target-URI\":\"https://eprint.iacr.org/2001/094\",\"WARC-Payload-Digest\":\"sha1:LOQA3D475LFY6W6XYK7BY66RZICOC6FS\",\"WARC-Block-Digest\":\"sha1:I522NT5C5WKRWTQK7HM4Z3CFDVT4Z5JY\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224649293.44_warc_CC-MAIN-20230603133129-20230603163129-00567.warc.gz\"}"}
https://math.libretexts.org/Bookshelves/Applied_Mathematics/Book%3A_Introduction_to_the_Modeling_and_Analysis_of_Complex_Systems_(Sayama)/12%3A_Cellular_Automata_II_-_Analysis	[ "$$\\newcommand{\\id}{\\mathrm{id}}$$ $$\\newcommand{\\Span}{\\mathrm{span}}$$ $$\\newcommand{\\kernel}{\\mathrm{null}\\,}$$ $$\\newcommand{\\range}{\\mathrm{range}\\,}$$ $$\\newcommand{\\RealPart}{\\mathrm{Re}}$$ $$\\newcommand{\\ImaginaryPart}{\\mathrm{Im}}$$ $$\\newcommand{\\Argument}{\\mathrm{Arg}}$$ $$\\newcommand{\\norm}{\\\| #1 \\\|}$$ $$\\newcommand{\\inner}{\\langle #1, #2 \\rangle}$$ $$\\newcommand{\\Span}{\\mathrm{span}}$$\n\n# 12: Cellular Automata II - Analysis\n\n$$\\newcommand{\\vecs}{\\overset { \\rightharpoonup} {\\mathbf{#1}} }$$ $$\\newcommand{\\vecd}{\\overset{-\\!-\\!\\rightharpoonup}{\\vphantom{a}\\smash {#1}}}$$$$\\newcommand{\\id}{\\mathrm{id}}$$ $$\\newcommand{\\Span}{\\mathrm{span}}$$ $$\\newcommand{\\kernel}{\\mathrm{null}\\,}$$ $$\\newcommand{\\range}{\\mathrm{range}\\,}$$ $$\\newcommand{\\RealPart}{\\mathrm{Re}}$$ $$\\newcommand{\\ImaginaryPart}{\\mathrm{Im}}$$ $$\\newcommand{\\Argument}{\\mathrm{Arg}}$$ $$\\newcommand{\\norm}{\\\| #1 \\\|}$$ $$\\newcommand{\\inner}{\\langle #1, #2 \\rangle}$$ $$\\newcommand{\\Span}{\\mathrm{span}}$$ $$\\newcommand{\\id}{\\mathrm{id}}$$ $$\\newcommand{\\Span}{\\mathrm{span}}$$ $$\\newcommand{\\kernel}{\\mathrm{null}\\,}$$ $$\\newcommand{\\range}{\\mathrm{range}\\,}$$ $$\\newcommand{\\RealPart}{\\mathrm{Re}}$$ $$\\newcommand{\\ImaginaryPart}{\\mathrm{Im}}$$ $$\\newcommand{\\Argument}{\\mathrm{Arg}}$$ $$\\newcommand{\\norm}{\\\| #1 \\\|}$$ $$\\newcommand{\\inner}{\\langle #1, #2 \\rangle}$$ $$\\newcommand{\\Span}{\\mathrm{span}}$$\n\n• 12.1: Sizes of Rule Space and Phase Space\nOne of the unique features of typical CA models is that time, space, and states of cells are all discrete. Because of such discreteness, the number of all possible state-transition functions is ﬁnite, i.e., there are only a ﬁnite number of “universes” possible in a given CA setting. Moreover, if the space is ﬁnite, all possible conﬁgurations of the entire system are also enumerable. This means that, for reasonably small CA settings, one can conduct an exhaustive search of the entire rule space o\n• 12.2: Phase Space Visualization\nIf the phase space of a CA model is not too large, you can visualize it using the technique we discussed in Section 5.4. Such visualizations are helpful for understanding the overall dynamics of the system, especially by measuring the number of separate basins of attraction, their sizes, and the properties of the attractors.\n• 12.3: Mean-Field Approximation\nBehaviors of CA models are complex and highly nonlinear, so it isn’t easy to analyze their dynamics in a mathematically elegant way. But still, there are some analytical methods available. Mean-ﬁeld approximation is one such analytical method. It is a powerful analytical method to make a rough prediction of the macroscopic behavior of a complex system.\n• 12.4: Renormalization Group Analysis to Predict Percolation Thresholds\nThe next analytical method is for studying critical thresholds for percolation to occur in spatial contact processes, like those in the epidemic/forest ﬁre CA model discussed in Section 11.5. The percolation threshold may be estimated analytically by a method called renormalization group analysis." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.69766074,"math_prob":0.99886215,"size":2728,"snap":"2020-10-2020-16","text_gpt3_token_len":750,"char_repetition_ratio":0.20778267,"word_repetition_ratio":0.11377245,"special_character_ratio":0.28225806,"punctuation_ratio":0.11136891,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97099847,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-04-07T01:33:58Z\",\"WARC-Record-ID\":\"<urn:uuid:3328dd44-0d16-4df3-a429-b285484445d2>\",\"Content-Length\":\"96730\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:1357dd01-0f83-420d-b755-36dcb704beee>\",\"WARC-Concurrent-To\":\"<urn:uuid:2ad9c541-8db5-4b55-bb88-42744e57e764>\",\"WARC-IP-Address\":\"13.249.44.60\",\"WARC-Target-URI\":\"https://math.libretexts.org/Bookshelves/Applied_Mathematics/Book%3A_Introduction_to_the_Modeling_and_Analysis_of_Complex_Systems_(Sayama)/12%3A_Cellular_Automata_II_-_Analysis\",\"WARC-Payload-Digest\":\"sha1:SWU7SE3EZXAVOOISX465GOWKXCVYQL76\",\"WARC-Block-Digest\":\"sha1:VUGYJMTN4GXSSC6JHZUHTGGHICCHME4I\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-16/CC-MAIN-2020-16_segments_1585371662966.69_warc_CC-MAIN-20200406231617-20200407022117-00217.warc.gz\"}"}
https://www.math.net/octahedron	[ "# Octahedron\n\nAn octahedron is a space figure with 8 faces that are polygons. In the figure below are 3 different types of octahedrons. The prefix \"octa\" means eight.\n\nAn octahedron can be formed by two pyramids with bases in the shape of a quadrilateral, as shown in the figure below.\n\n## Properties of a regular octahedron\n\nA regular octahedron is an octahedron whose faces are all congruent, regular polygons. Otherwise, it is irregular. Regular octahedrons are studied more often.\n\nA regular octahedron, such as the one shown above, is one of the 5 Platonic solids, which are a type of regular polyhedron. A regular octahedron has 8 congruent faces that are congruent equilateral triangles, 12 congruent edges, and 6 vertices; an edge is a line segment formed by the intersection of two adjacent faces; a vertex for a regular octahedron is a point where 4 edges meet.\n\n## Surface area of a regular octahedron\n\nWe can find the area of one of the faces and multiply it by eight to find the total surface area of a regular octahedron. An equilateral triangle with side length, e (also the length of the edges of a regular octahedron), has an area, A, of\n\nThe total surface area, S, of a regular octahedron in terms of its edges, e, is,\n\n## Volume of a regular octahedron\n\nThe volume, V, of a regular octahedron is\n\nwhere e is the length of the edge.\n\nExample:\n\nIf the total surface area of a regular octahedron is", null, ", what is its volume?\n\nWe can find e by substituting the given value in for the total surface area to get\n\n36 = e2\n\ne = 6\n\nSubstituting the length of the edge into the volume formula:" ]	[ null, "https://dfrrh0itwp1ti.cloudfront.net/mj/NzIgXHNxcnR7M30=_100.svg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9387575,"math_prob":0.979551,"size":1570,"snap":"2022-40-2023-06","text_gpt3_token_len":410,"char_repetition_ratio":0.23116219,"word_repetition_ratio":0.035587188,"special_character_ratio":0.21528663,"punctuation_ratio":0.11320755,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99878365,"pos_list":[0,1,2],"im_url_duplicate_count":[null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-02-06T19:41:25Z\",\"WARC-Record-ID\":\"<urn:uuid:92093594-3ed9-41d4-815b-39a5f3f96378>\",\"Content-Length\":\"9661\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b8dbe895-d018-4f87-ab74-d11b92125fac>\",\"WARC-Concurrent-To\":\"<urn:uuid:b89ba3d2-46b8-4c28-b1b4-611e602db9e6>\",\"WARC-IP-Address\":\"69.10.42.198\",\"WARC-Target-URI\":\"https://www.math.net/octahedron\",\"WARC-Payload-Digest\":\"sha1:3RR2WJIS4S2LMFG444DVW2XJH6ZDHEET\",\"WARC-Block-Digest\":\"sha1:G4B64SIRBVZXWA7J5MWMQF6AP7Y3DDNB\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764500357.3_warc_CC-MAIN-20230206181343-20230206211343-00828.warc.gz\"}"}
http://algebra2014.wikidot.com/theorem-8-16	[ "Theorem 8.16\n\nLet $G$ be a group. We show that $G$ is isomorphic to a subgroup of $S_G$. By Lemma 8.15, we need only to define a one-to-one function $\\phi : G \\rightarrow S_G$ such that $\\phi (xy) = \\phi (x) \\phi (y)$ for all $x, y \\in G$. For $x \\in G$, let $\\lambda _x : G \\rightarrow G$ be defined by $\\lambda _x (g) = xg$ for all $g \\in G$. (We think of $\\lambda _x$ as performing left multiplication by $x$.) The equation $\\lambda _x (x^{-1} c) = x(x^{-1} c) =c$ for all $c \\in G$ shows that $\\lambda _x$ maps $G$ onto $G$. If $\\lambda _x (a) = \\lambda _x (b)$, then $xa = xb$ so $a = b$ by cancellation. Thus $\\lambda _x$ is also one-to-one, and is a permutation of $G$. We now define $\\phi : G \\rightarrow S_G$ by defining $\\phi (x) = \\lambda _x$ for all $x \\in G$.\nTo show that $\\phi$ is one-to-one, suppose that $\\phi (x) = \\phi (y)$. Then $\\lambda _x = \\lambda _y$ as functions mapping $G$ into $G$. In particular $\\lambda _x (e) = \\lambda _y (e)$, so $xe = ye$ and $x=y$. Thus $\\phi$ is one-to-one. It only remains to show that $\\phi (xy) = \\phi (x) \\phi (y)$, that is, that $\\lambda _{xy} = \\lambda _x \\lambda _y$. Now for any $g \\in G$, we have $\\lambda _{xy} (g) = (xy)g$. Permutation multiplication is function composition, so $(\\lambda _x \\lambda _y )(g) = \\lambda _x (\\lambda _y(g)) = \\lambda _x (yg) =x(yg)$. Thus by associativity, $\\lambda _{xy} = \\lambda _x \\lambda _y$. $\\blacksquare$" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7719419,"math_prob":1.0000097,"size":1510,"snap":"2019-43-2019-47","text_gpt3_token_len":521,"char_repetition_ratio":0.21779549,"word_repetition_ratio":0.05,"special_character_ratio":0.3807947,"punctuation_ratio":0.115625,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.00001,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-10-21T12:27:27Z\",\"WARC-Record-ID\":\"<urn:uuid:aee0751e-3c99-4902-b224-c530d8808b6c>\",\"Content-Length\":\"30478\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:637f7054-a3fd-442b-9974-845c2c5913bd>\",\"WARC-Concurrent-To\":\"<urn:uuid:485423fb-1923-4e81-aff3-2c2453630d93>\",\"WARC-IP-Address\":\"107.20.139.176\",\"WARC-Target-URI\":\"http://algebra2014.wikidot.com/theorem-8-16\",\"WARC-Payload-Digest\":\"sha1:WLZD2OEIK4RHVZ5BXDIUB5K6H3FKJJDN\",\"WARC-Block-Digest\":\"sha1:FYMMF254J2I6JEUQ2MNBLM32HN2IZEFS\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-43/CC-MAIN-2019-43_segments_1570987773711.75_warc_CC-MAIN-20191021120639-20191021144139-00500.warc.gz\"}"}
https://math.libretexts.org/Bookshelves/Analysis/Book%3A_Real_Analysis_(Boman_and_Rogers)/05%3A_Convergence_of_the_Taylor_Series-_A_%E2%80%9CTayl%E2%80%9D_of_Three_Remainders/5.01%3A_The_Integral_Form_of_the_Remainder	[ "$$\\newcommand{\\id}{\\mathrm{id}}$$ $$\\newcommand{\\Span}{\\mathrm{span}}$$ $$\\newcommand{\\kernel}{\\mathrm{null}\\,}$$ $$\\newcommand{\\range}{\\mathrm{range}\\,}$$ $$\\newcommand{\\RealPart}{\\mathrm{Re}}$$ $$\\newcommand{\\ImaginaryPart}{\\mathrm{Im}}$$ $$\\newcommand{\\Argument}{\\mathrm{Arg}}$$ $$\\newcommand{\\norm}{\\\| #1 \\\|}$$ $$\\newcommand{\\inner}{\\langle #1, #2 \\rangle}$$ $$\\newcommand{\\Span}{\\mathrm{span}}$$\n\n# 5.1: The Integral Form of the Remainder\n\n$$\\newcommand{\\vecs}{\\overset { \\rightharpoonup} {\\mathbf{#1}} }$$ $$\\newcommand{\\vecd}{\\overset{-\\!-\\!\\rightharpoonup}{\\vphantom{a}\\smash {#1}}}$$$$\\newcommand{\\id}{\\mathrm{id}}$$ $$\\newcommand{\\Span}{\\mathrm{span}}$$ $$\\newcommand{\\kernel}{\\mathrm{null}\\,}$$ $$\\newcommand{\\range}{\\mathrm{range}\\,}$$ $$\\newcommand{\\RealPart}{\\mathrm{Re}}$$ $$\\newcommand{\\ImaginaryPart}{\\mathrm{Im}}$$ $$\\newcommand{\\Argument}{\\mathrm{Arg}}$$ $$\\newcommand{\\norm}{\\\| #1 \\\|}$$ $$\\newcommand{\\inner}{\\langle #1, #2 \\rangle}$$ $$\\newcommand{\\Span}{\\mathrm{span}}$$ $$\\newcommand{\\id}{\\mathrm{id}}$$ $$\\newcommand{\\Span}{\\mathrm{span}}$$ $$\\newcommand{\\kernel}{\\mathrm{null}\\,}$$ $$\\newcommand{\\range}{\\mathrm{range}\\,}$$ $$\\newcommand{\\RealPart}{\\mathrm{Re}}$$ $$\\newcommand{\\ImaginaryPart}{\\mathrm{Im}}$$ $$\\newcommand{\\Argument}{\\mathrm{Arg}}$$ $$\\newcommand{\\norm}{\\\| #1 \\\|}$$ $$\\newcommand{\\inner}{\\langle #1, #2 \\rangle}$$ $$\\newcommand{\\Span}{\\mathrm{span}}$$\n\n##### Learning Objectives\n• Explain the integral form of the remainder\n\nNow that we have a rigorous definition of the convergence of a sequence, let’s apply this to Taylor series. Recall that the Taylor series of a function $$f(x)$$ expanded about the point $$a$$ is given by\n\n$\\sum_{n=0}^{\\infty }\\frac{f^{(n)}(a)}{n!}(x-a)^n = f(a) + \\frac{f'(a)}{1!}(x-a) + \\frac{f''(a)}{2!}(x-a)^2 + \\cdots$\n\nWhen we say that $$\\sum_{n=0}^{\\infty }\\frac{f^{(n)}(a)}{n!}(x-a)^n$$ for a particular value of $$x$$,what we mean is that the sequence of partial sums\n\n$\\left (\\sum_{j=0}^{n}\\frac{f^{(j)}(a)}{j!}(x-a)^j \\right )_{n=0}^{\\infty } = \\left ( f(a), f(a) + \\frac{f'(a)}{1!}(x-a), f(a) + \\frac{f'(a)}{1!}(x-a) + \\frac{f''(a)}{2!}(x-a)^2 + \\cdots \\right )$\n\nconverges to the number $$f(x)$$. Note that the index in the summation was changed to $$j$$ to allow $$n$$ to represent the index of the sequence of partial sums. As intimidating as this may look, bear in mind that for a fixed real number $$x$$, this is still a sequence of real numbers so, that saying $$f(x) = \\sum_{n=0}^{\\infty }\\frac{f^{(n)}(a)}{n!}(x-a)^n$$ means that $$\\lim_{n \\to \\infty }\\left (\\sum_{j=0}^{n}\\frac{f^{(j)}(a)}{j!}(x-a)^j \\right ) = f(x)$$ and in the previous chapter we developed some tools to examine this phenomenon. In particular, we know that $$\\lim_{n \\to \\infty }\\left (\\sum_{j=0}^{n}\\frac{f^{(j)}(a)}{j!}(x-a)^j \\right ) = f(x)$$ is equivalent to\n\n$\\lim_{n \\to \\infty }\\left [f(x) - \\left (\\sum_{j=0}^{n}\\frac{f^{(j)}(a)}{j!}(x-a)^j \\right ) \\right ] = 0$\n\nWe saw an example of this in the last chapter with the geometric series $$1 + x + x^2 + x^3+\\cdots$$. Problem Q4 of the last chapter basically had you show that this series converges to $$\\frac{1}{1-x}$$, for $$\|x\| < 1$$ by showing that $$\\lim_{n \\to \\infty }\\left [\\frac{1}{1-x} - \\left (\\sum_{j=0}^{n}x^j \\right ) \\right ] = 0$$.\n\nThere is generally not a readily recognizable closed form for the partial sum for a Taylor series. The geometric series is a special case. Fortunately, for the issue at hand (convergence of a Taylor series), we don’t need to analyze the series itself. What we need to show is that the difference between the function and the $$n^{th}$$ partial sum converges to zero. This difference is called the remainder (of the Taylor series). (Why?)\n\nWhile it is true that the remainder is simply\n\n$f(x) - \\left (\\sum_{j=0}^{n}\\frac{f^{(j)}(a)}{j!}(x-a)^j \\right )$\n\nthis form is not easy to work with. Fortunately, a number of alternate versions of this remainder are available. We will explore these in this chapter. Recall the result from Theorem 3.1.2 from Chapter 3,\n\n$f(x) = f(a) + \\frac{f'(a)}{1!}(x-a) + \\frac{f''(a)}{2!}(x-a)^2 + \\cdots + \\frac{f^{(n)}(a)}{n!}(x-a)^n + \\frac{1}{n!}\\int_{t=a}^{x}f^{(n+1)}(t)(x-t)^n dt$\n\nWe can use this by rewriting it as\n\n$f(x) - \\left (\\sum_{j=0}^{n}\\frac{f^{(j)}(a)}{j!}(x-a)^j \\right ) = \\frac{1}{n!}\\int_{t=a}^{x}f^{(n+1)}(t)(x-t)^n dt \\label{50}$\n\nThe left hand side of Equation \\ref{50} is called the integral form of the remainder for the Taylor series of $$f(x)$$, and the Taylor series will converge to $$f(x)$$ exactly when the sequence $$\\lim_{n \\to \\infty }\\left (\\frac{1}{n!}\\int_{t=a}^{x}f^{(n+1)}(t)(x-t)^n dt \\right )$$ converges to zero. It turns out that this form of the remainder is often easier to handle than the original $$f(x) - \\left (\\sum_{j=0}^{n}\\frac{f^{(j)}(a)}{j!}(x-a)^j \\right )$$ and we can use it to obtain some general results.\n\n##### Theorem $$\\PageIndex{1}$$: Taylor’s Series\n\nIf there exists a real number $$B$$ such that $$\|f^{(n+1)}(t)\|≤ B$$ for all nonnegative integers $$n$$ and for all $$t$$ on an interval containing $$a$$ and $$x$$, then\n\n$\\lim_{n \\to \\infty }\\left (\\frac{1}{n!}\\int_{t=a}^{x}f^{(n+1)}(t)(x-t)^n dt \\right ) = 0$\n\nand so\n\n$f(x) = \\sum_{n=0}^{\\infty }\\frac{f^{(n)}(a)}{n!}(x-a)^n$\n\nIn order to prove this, it might help to first prove the following Lemma.\n\n##### Lemma $$\\PageIndex{1}$$: Triangle Inequality for Integrals\n\nIf $$f$$ and $$\|f\|$$ are integrable functions and $$a ≤ b$$, then\n\n$\\left \| \\int_{t=a}^{b} f(t)dt\\right \| \\leq \\int_{t=a}^{b} \\left \|f(t) \\right \|dt$\n\n##### Exercise $$\\PageIndex{1}$$\n\nProve Lemma $$\\PageIndex{1}$$.\n\nHint\n\n$$-\|f(t)\|≤ f(t) ≤\|f(t)\|$$.\n\n##### Exercise $$\\PageIndex{2}$$\n\nProve Theorem $$\\PageIndex{1}$$.\n\nHint\n\nYou might want to use Problem Q8 of Chapter 4. Also there are two cases to consider: $$a < x$$ and $$x < a$$ (the case $$x = a$$ is trivial). You will find that this is true in general. This is why we will often indicate that $$t$$ is between $$a$$ and $$x$$ as in the theorem. In the case $$x < a$$, notice that \\begin{align} \\left \| \\int_{t=a}^{x}f^{(n+1)}(t)(x-t)^n dt \\right \| &= \\left \| (-1)^{n+1}\\int_{t=a}^{x}f^{(n+1)}(t)(t-x)^n dt \\right \|\\\\ &= \\left \| \\int_{t=a}^{x}f^{(n+1)}(t)(t-x)^n dt \\right \| \\end{align}\n\n##### Exercise $$\\PageIndex{3}$$\n\nUse Theorem $$\\PageIndex{1}$$ to prove that for any real number $$x$$\n\n1. $$\\displaystyle \\sin x = \\sum_{n=0}^{\\infty }\\frac{(-1)^n x^{2n+1}}{(2n+1)!}$$\n2. $$\\displaystyle \\cos x = \\sum_{n=0}^{\\infty }\\frac{(-1)^n x^{2n}}{(2n)!}$$\n3. $$\\displaystyle e^x = \\sum_{n=0}^{\\infty }\\frac{x^n}{n!}$$\n\nPart c of exercise $$\\PageIndex{3}$$ shows that the Taylor series of $$e^x$$ expanded at zero converges to $$e^x$$ for any real number $$x$$. Theorem $$\\PageIndex{1}$$ can be used in a similar fashion to show that\n\n$e^x = \\sum_{n=0}^{\\infty }\\frac{e^a(x-a)^n}{n!}$\n\nfor any real numbers $$a$$ and $$x$$.\n\nRecall that in section 2.1 we showed that if we define the function $$E(x)$$ by the power series $$\\sum_{n=0}^{\\infty }\\frac{x^n}{n!}$$ then $$E(x + y) = E(x)E(y)$$. This, of course, is just the familiar addition property of integer coefficients extended to any real number. In Chapter 2 we had to assume that defining $$E(x)$$ as a series was meaningful because we did not address the convergence of the series in that chapter. Now that we know the series converges for any real number we see that the definition\n\n$f(x) = e^x = \\sum_{n=0}^{\\infty }\\frac{x^n}{n!}$\n\nis in fact valid.\n\nAssuming that we can differentiate this series term-by-term it is straightforward to show that $$f'(x) = f(x)$$. Along with Taylor’s formula this can then be used to show that $$e^{a+b} = e^ae^b$$ more elegantly than the rather cumbersome proof in section 2.1, as the following problem shows.\n\n##### Exercise $$\\PageIndex{4}$$\n\nRecall that if $$f(x) = e^x$$ then $$f'(x) = e^x$$. Use this along with the Taylor series expansion of $$e^x$$ about $$a$$ to show that $e^{a+b} = e^ae^b \\nonumber$\n\nTheorem $$\\PageIndex{1}$$ is a nice “first step” toward a rigorous theory of the convergence of Taylor series, but it is not applicable in all cases. For example, consider the function $$f(x) = \\sqrt{1+x}$$. As we saw in Chapter 2, Exercise 2.2.9, this function’s Maclaurin series (the binomial series for $$(1 + x)^{1/2}$$)appears to be converging to the function for $$x ∈ (-1,1)$$. While this is, in fact, true, the above proposition does not apply. If we consider the derivatives of $$f(t) = (1 + t)^{1/2}$$, we obtain:\n\n$f'(t) = \\frac{1}{2}(1+t)^{\\frac{1}{2}-1}$\n\n$f''(t) = \\frac{1}{2}\\left ( \\frac{1}{2}-1 \\right )(1+t)^{\\frac{1}{2}-2}$\n\n$f'''(t) = \\frac{1}{2}\\left ( \\frac{1}{2}-1 \\right )\\left ( \\frac{1}{2}-2 \\right )(1+t)^{\\frac{1}{2}-2}$\n\n$\\vdots$\n\n$f^{n+1}(t) = \\frac{1}{2}\\left ( \\frac{1}{2}-1 \\right )\\left ( \\frac{1}{2}-2 \\right )\\cdots \\left ( \\frac{1}{2}-n \\right )(1+t)^{\\frac{1}{2}-(n+1)}$\n\nNotice that\n\n$\\left \|f^{n+1}(0) \\right \| = \\frac{1}{2}\\left ( 1 - \\frac{1}{2} \\right )\\left ( 2 - \\frac{1}{2} \\right )\\cdots \\left ( n - \\frac{1}{2} \\right )$\n\nSince this sequence grows without bound as $$n →∞$$, then there is no chance for us to find a number $$B$$ to act as a bound for all of the derviatives of $$f$$ on any interval containing $$0$$ and $$x$$, and so the hypothesis of Theorem $$\\PageIndex{1}$$ will never be satisfied. We need a more delicate argument to prove that\n\n$\\sqrt{1+x} = 1 + \\frac{1}{2}x + \\frac{\\frac{1}{2}\\left ( \\frac{1}{2}-1 \\right )}{2!}x^2 + \\frac{\\frac{1}{2}\\left ( \\frac{1}{2}-1 \\right )\\left ( \\frac{1}{2}-2 \\right )}{3!}x^3 + \\cdots$\n\nis valid for $$x ∈ (-1,1)$$. To accomplish this task, we will need to express the remainder of the Taylor series differently. Fortunately, there are at least two such alternate forms.\n\nThis page titled 5.1: The Integral Form of the Remainder is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Eugene Boman and Robert Rogers (OpenSUNY) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.768199,"math_prob":1.0000058,"size":8092,"snap":"2022-27-2022-33","text_gpt3_token_len":2943,"char_repetition_ratio":0.15838279,"word_repetition_ratio":0.045379538,"special_character_ratio":0.39977756,"punctuation_ratio":0.06852497,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.0000051,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-08-15T22:22:26Z\",\"WARC-Record-ID\":\"<urn:uuid:f67f99c3-b009-408d-acae-fba292e6b312>\",\"Content-Length\":\"120399\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:6a2dc27b-4181-4a3a-9ff1-769bb91a1adb>\",\"WARC-Concurrent-To\":\"<urn:uuid:b992017c-b814-47e6-a488-0da044c85816>\",\"WARC-IP-Address\":\"13.249.39.24\",\"WARC-Target-URI\":\"https://math.libretexts.org/Bookshelves/Analysis/Book%3A_Real_Analysis_(Boman_and_Rogers)/05%3A_Convergence_of_the_Taylor_Series-_A_%E2%80%9CTayl%E2%80%9D_of_Three_Remainders/5.01%3A_The_Integral_Form_of_the_Remainder\",\"WARC-Payload-Digest\":\"sha1:E5HSXDYSI3OJ3CW6VZWZMHFAFTKJJG6M\",\"WARC-Block-Digest\":\"sha1:JEAWWZZR2FDK75XSNAB7MCC2F5ONGQF5\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-33/CC-MAIN-2022-33_segments_1659882572212.96_warc_CC-MAIN-20220815205848-20220815235848-00246.warc.gz\"}"}
https://arizona.pure.elsevier.com/en/publications/remarks-on-the-fourier-coefficients-of-modular-forms	[ "# Remarks on the Fourier coefficients of modular forms\n\nResearch output: Contribution to journalArticle\n\n### Abstract\n\nWe consider a variant of a question of N. Koblitz. For an elliptic curve E/Q which is not Q-isogenous to an elliptic curve with torsion, Koblitz has conjectured that there exists infinitely many primes p such that Np(E)=#E(Fp)=p+1-ap(E) is also a prime. We consider a variant of this question. For a newform f, without CM, of weight k≥4, on Γ 0(M) with trivial Nebentypus χ 0 and with integer Fourier coefficients, let N p(f)=χ 0(p)p k-1+1-a p(f) (here a p(f) is the p-th-Fourier coefficient of f). We show under GRH and Artin's Holomorphy Conjecture that there are infinitely many p such that N p(f) has at most [5k+1+log(k)] distinct prime factors. We give examples of about hundred forms to which our theorem applies. We also show, on GRH, that the number of distinct prime factors of N p(f) is of normal order log(log(p)) and that the distribution of these values is asymptotically a Gaussian distribution (\"Erdo{double acute}s-Kac type theorem\").\n\nOriginal language English (US) 1314-1336 23 Journal of Number Theory 132 6 https://doi.org/10.1016/j.jnt.2011.10.004 Published - Jun 1 2012\n\n### Keywords\n\n• Hecke eigenvalues\n• Koblitz conjecture\n• Modular forms\n• Normal orders\n\n### ASJC Scopus subject areas\n\n• Algebra and Number Theory" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.88579714,"math_prob":0.8301621,"size":1280,"snap":"2020-34-2020-40","text_gpt3_token_len":360,"char_repetition_ratio":0.08934169,"word_repetition_ratio":0.009708738,"special_character_ratio":0.26015624,"punctuation_ratio":0.05928854,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9973166,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-08-05T16:55:05Z\",\"WARC-Record-ID\":\"<urn:uuid:1eff1bdd-d86d-415b-932d-ac7b9766c727>\",\"Content-Length\":\"41403\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:684ccc0e-e503-4e6b-a005-174ac283c394>\",\"WARC-Concurrent-To\":\"<urn:uuid:e37a46b2-3bdd-4903-9ef9-4c713c2085b6>\",\"WARC-IP-Address\":\"52.72.50.98\",\"WARC-Target-URI\":\"https://arizona.pure.elsevier.com/en/publications/remarks-on-the-fourier-coefficients-of-modular-forms\",\"WARC-Payload-Digest\":\"sha1:7CJGQ3CQRAXJWQC6QPWVWWHGXHF4JKND\",\"WARC-Block-Digest\":\"sha1:IFRYXRDWW745NIKDO7H74IZVAIRP4LHP\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-34/CC-MAIN-2020-34_segments_1596439735963.64_warc_CC-MAIN-20200805153603-20200805183603-00349.warc.gz\"}"}
https://matheducators.stackexchange.com/questions/4165/is-the-reciprocal-function-continuous	[ "# Is the reciprocal function continuous?\n\nI'm curious the views of those who teach calculus.\n\nAs you know the continuity of a function at a point is defined in terms of the limit in the typical course. I'd like to ask a pair of questions:\n\n1. Consider $f(x)=1/x$. Is $f$ continuous ?\n2. Let $\\text{dom}(g) = (-\\infty, 0 ) \\cup (0, \\infty)$ where $g(x)=1/x$ for each $x \\in \\text{dom}(g)$. Is $g$ continuous?\n\nI don't want to say much more initially as I'd rather not influence the answers at the outset. The audience I have in mind is a mixed math majors / science majors calculus course at the university and time is not a large issue.\n\nI think the point of the question is that although there is no mathematical ambiguity, in different circumstances one would give different answers. So:\n\nI. From the perspective of a freshman calculus textbook: well, I know my enemy. The textbook answer to (1) is \"A function is 'continuous' if it is continuous at $c$ for every real number $c$. A function is continuous at $c$ if it is defined at $c$, if $\\lim_{x \\rightarrow c} f(x) = L$ exists and $L = f(c)$. Since $f(x)$ is not defined at $0$, it is not continuous there.\" Yikes. To be fair, most calculus textbooks would classify the discontinuity at $0$ as not being removable.\n\nI feel ever so slightly worried that I've unfairly caricatured this addle-pated view of continuity. I know that freshman calculus textbooks would give the answer \"No, because it is not defined at $0$\" to (1). As written, it sounds like they would give a similar answer to the continuity of, say, $f(x) = \\sqrt{x-1}$, and I think that they may not be as legalistic as I implied above that a function can only be called continuous if it is defined at all real numbers: sometimes being defined on a closed interval is regarded as sufficient. I think question (2) outfoxes the freshman calculus text: even by asking it you show understanding that the continuity of a function depends upon what domain you take. The calculus texts are not clear on this.\n\nII. From the perspective of a mathematician: (1) I would ask, \"Well, what is the domain and codomain of the function? If the function is meant to take real or complex values, then it is not defined at $0$. As a function from $\\mathbb{R} \\setminus \\{0\\}$ to $\\mathbb{R}$ or from $\\mathbb{C} \\setminus \\{0\\}$ to $\\mathbb{C}$, it is continuous (and more...). The function cannot be extended continuously to $0$, e.g. because it is unbounded in every neighborhood of $0$. However, it extends continuously [and more...] to a function on the Riemann sphere: $f(0) = \\infty$.\" (2) Yes, of course $g$ is continuous [and more...].\n\nIII. If a freshman calculus student asked me this in class:\n\n(1) \"There are some tricky issues in continuity for functions which are not defined at isolated points. Our intuitive notion of continuity is that the graph is a nice unbroken curve. [Draws the graph of f] As you can see, in this case the graph is not nice at $0$. Now in fact the function is not even defined at $0$. Whether a function is continuous at a point where it is not defined is a bit legalistic! More pertinently, sometimes functions which are not defined at an isolated point can be defined at the point so as to be continuous there. This happens exactly when the limit as you approach the point exists and is called a 'removable discontinuity'. However, in this case, $\\lim_{x \\rightarrow 0^{-}} f(x) = -\\infty$ and $\\lim_{x \\rightarrow 0^{+}} f(x) = \\infty$ so it's pretty bad: each one-sided limit is infinite, but the signs are different, so we can't even say that the overall limit is $\\pm \\infty$ [even if we could, since $\\pm \\infty$ are not real numbers, the limit would not exist]. The upshot is that there is no \"good\" way to define $f(x) = \\frac{1}{x}$ at $0$.\"\n\n(2) \"Yes, the function is definitely continuous on the given domain, as you can see from the graph. Beware though: the given domain is not an interval but rather a union of two non-overlapping intervals. Be very careful with this: most of the big theorems in calculus concern continuous functions defined on an interval. When the domain is 'more than one interval', some funny things can happen. [Depending on where we are in freshman calculus...] For instance the intermediate value theorem fails here: the function takes negative values and positive values but never takes the value $0$. Also one antiderivative of the function is $\\log \|x\|$, but not every antiderivative is of the form $\\log \|x\| + C$! [Maybe say more about this if someone seems interested; it is a tricky point.] It may be more useful to think of the function $g$ as really being 'two different functions', one defined for negative values and one defined for positive values.\"\n\nAdded: The most general antiderivative of $g$ seems like a tricky point indeed, since more than one person has asked about it in the comments. It may be worth explaining the answer. The uniqueness of antiderivatives up to a constant is equivalent to the Zero Velocity Theorem: if $F$ is differentiable and $F' = 0$ (identically) then $F = C$ is constant. This in turn is a consequence of the Mean Value Theorem, and the hypothesis that the domain be an interval is critical: more generally any locally constant function has derivative zero, and if the domain has $N$ connected components, the vector space of locally constant functions has dimension $N$. In this case the domain of $g$ consists of two disjoint intervals, so the most general antiderivative is $G(x) =$ $\\log(x) + C_1$, $x > 0$\n$\\log(-x) + C_2$, $x <0$\nfor any two $C_1,C_2 \\in \\mathbb{R}$.\n\n• +1 especially for how a mathematician would salvage the first question. Aug 9, 2014 at 22:27\n• @metacompactness By the fundamental theorem of calculus if $F(x)$ is the antiderivative of $f(x)$ then any other antiderivative can only differ by a constant. So no, there's no other antiderivatives. Aug 10, 2014 at 9:02\n• @Darksonn Of course. Aug 10, 2014 at 9:04\n• @Darksonn: The Fundamental Theorem of Calculus applies separately on each interval. Thus the most general antiderivative of $g$ is $F(x) = \\log x + C_1$ for $x > 0$ and $\\log (-x) + C_2$ for $x < 0$, where $C_1,C_2$ are two arbitrary constants. (metacompactness and I had some previous exchanges about this which we both deleted.) Aug 10, 2014 at 16:20\n• Every freshman calculus text that I've seen explicitly restricts the definition of a continuous function to considering only points in the domain, such that the answer is \"yes\", $f(x) = 1/x$ is a continuous function. E.g., in Stein/Barcellos Sec. 2.8 that is the very first example. Jul 5, 2017 at 16:36\n\nIn this situation, I'd claim there's the \"prior error\" of seemingly requiring a boolean answer, yes-or-no. Surely we're not so much interested in pranking the students by giving them a function whose domain is not topologically connected, which is continuous on each connected component, but not \"continuous\" in an obvious, intuitive sense \"across the gap\". The situation is clear, because everyone in Calc I knows the graph of this function. What is not clear is the semantics of a formal definition, whose pitfalls will not have caught the students' attention... and maybe don't merit it, besides.\n\nThat is, \"$f(x)=1/x$ has a problem at $x=0$, but otherwise is fine\".\n\n(\"What about $1/x^2$?\")\n\nThat is, I'd propose not over-burdening a boolean \"continuity-or-not\" with this situation in Calc I. It's not so much a mathematical issue as a semantic one, which oughtn't be a primary point, in my opinion.\n\n• I tend to agree with your viewpoint about not trapping students. So, if this was a major point in the course and it was my goal to \"get\" them on it I could see it. But, the reason I discuss this issue (when I do) is to bring attention to the limitations of the definitions. In the same course, at other points, I am quite careful about the $\\epsilon \\delta$ proofs, so, it seems lopsided to just abandon logical clarity for the sake of intuition at this point. I mean, intuitively who needs the $\\epsilon \\delta$ proofs? Of course, the issue can be avoided by focusing on intervals. Aug 12, 2014 at 4:08\n\nI thought it might be interesting to look at a variety of different Calculus textbooks to see how they handle the terminology. Here is an unscientific survey of the half-dozen different Calc textbooks I have on my bookshelf right now:\n\nStewart, Single Variable Calculus, 4e (both Early Transcendentals and Late Transcendentals versions) and 5e:\n\nStewart defines the phrase continuous at a number a in the usual way; the text also defines continuous from the right at a and continuous from the left at a. Armed with these definitions, the text defines continuous on an interval, with one-sided limits used at the endpoints of the interval if appropriate in context.\n\nWith respect to the specific question of the OP, Stewart's Theorem 5 states that \"Any rational function is continuous wherever it is defined; that is, it is continuous on its domain.\"\n\nThe phrase \"a continuous function\" (without qualifiers like \"at a point\" or \"on an interval\") is used in the exposition, but never in the definitions, theorems, proofs, or exercises. The phrase is used solely as a kind of informal shorthand that makes it possible to refer to several different types of continuity at once, for example in statements like \"a sum or product of continuous functions is continuous\". A question like \"Is the reciprocal function continuous?\" would not be found in this text; however, \"Is the reciprocal function continuous on its domain?\" and \"At what points is the reciprocal function discontinuous?\" would be legitimate.\n\nGillett, Introduction to Calculus and Analytic Geometry, 4e\n\nLike Stewart, Gillett defines \"continuous at $x_0$\". This text also defines \"continuous in S\" (where S is a subset of the domain). (Interestingly the definition does not impose any restrictions on what types of subsets may be considered, but every example seems to be a countable union of intervals.) The phrase \"continuous in its domain\" comes next, and Gillett has as Example 6 that \"The reciprocal function is continuous in its domain.\"\n\nRegarding the terminology \"a continuous function\", Gillett has the following remark immediately following Example 6:\n\nWhen a function is continuous in its domain, many writers simply call it continuous, using the global term (unqualified)… This is not unreasonable… but it can be misleading. To say that f(x) = 1/x is continuous (without any qualification) is to risk overlooking the infinite jump at x = 0. It also sounds paradoxical to say \"f(x) is continuous\" in one breath and \"f(x) is discontinuous at x=0\" in the next. That is why we prefer the language \"continuous in its domain.\" It leaves the door open for further remarks.\n\nSo Gillett explicitly rejects the phrase \"a continuous function\" without specifying a point or a subset of its domain, and justifies this using the reciprocal function as its example of why that usage would be poor.\n\nMoise, Calculus: Part I.\n\nThis text (from 1966), like the more modern examples, starts by defining \"continuous at $x_0$\". Unlike Stewart and Gillett, Moise continues by defining \"If $f$ is continuous at every point $x_0$ of its domain, then we say that $f$ is continuous.\" So it uses precisely the terminology that Gillett critiques.\n\nAs an example of how continuous functions can be combined to produce more continuous functions, Moise considers the example of $f(x) = x^2 + 1$ and $h(x) = x^2 - 1$. The text states that $f+h$ and $fh$ are both \"continuous\" (note the absence of any qualifier) and then says that $f/h$ is continuous except at $1$ and $-1$. The example continues, \"Of course, at $x=1$ and $x = -1$ it is not just continuity that breaks down: the quotient function is not even defined at these points, because the denominator of $f/h$ becomes $0$.\" Oddly enough, Moise stops short of saying that $f/h$ is a \"continuous function\", despite the fact that it would be completely in keeping with his own usage conventions to do so. Presumably that is because, as Gillett noted in his remark quoted above, it would seem weird to say that \"$f/h$ is continuous\" and then also say that \"$f/h$ is discontinuous at $1$ and $-1$\".\n\nZill (1993), Calculus (3e)\n\nLike all the others, Zill begins by defining \"continuity at a point\". Zill states that that rational functions are discontinuous at points where they are not defined. The text defines continuity on an interval (with \"continuous from the right/left\" used at end points as appropriate).\n\nWith respect to the usage \"continuous function\", Zill says:\n\nFunctions that are continuous on $(-\\infty, \\infty)$ are said to be \"continuous everywhere\" or simply \"continuous\".\n\nSo Zill reserves the phrase \"continuous function\" for use only with functions whose domain is all of $\\mathbb{R}$. On the other hand (and somewhat in contradiction to this explicit definition), \"continuous function\" is used throughout the exposition (but never in the theorems!) as a shorthand, much in the same way that Stewart does.\n\nSo, to summarize, I'm going to go out on a limb here and, based on the single example of the 1966 textbook, suggest that the use of the phrase \"continuous function\" to mean \"function that is continuous at each point of its domain\" might be an artifact of an earlier generation of textbook authors. More recent texts either avoid giving the phrase any technical meaning at all, using it instead as an informal shorthand, or (in the case of Zill) define it to mean \"continuous on all of $\\mathbb{R}$\".\n\nThis is my experience as a calculus TA for the last 4 years, and I have had no active role in setting the curriculums I'm talking about. The \"continuity of a function\" may be defined and explained thoroughly in an advanced honors rigorous calculus class, but in the trenches that are the first year calculus at most first year universities it does not play a large role. There are essentially 3 options for defining continuity for these students.\n\n1. A function is continuous if its graph can be drawn without picking up the pen. This is the most popular method (in what I've observed) , and it leads to the fallacy you are hinting at. Still, this definition (whether flawed or not) is understandable to calculus students and is of the kind of thinking that mathematicians use.\n\n2. $\\lim_{x\\rightarrow a} f(x)=f(a)$. This definition is in every calculus text and is of course entirely correct. The issue is that this definiton has the giant black box of a \"limit\", which many students simply will not understand unless they are forced to (through HW or lectures). At least in my university the pressure to skip over these more difficult and harder to test aspects of calculus in favor of a larger more impressive-looking curriculum guarantees that they will not ever get experience in things like limits.\n\n3. $\\forall\\epsilon\\ \\exists\\delta$ s.t $\|f(x)-f(x_0)\|<\\epsilon\\ \\forall\|x-x_0\|<\\delta$. Again if an effort is made to teach this kind of a definition, it can be taught to first years. There simply isn't time for that in the curriculums I have seen.\n\nThe goal of most calculus curriculums I have been party to is to get the students calculating integrals and derivatives of elementary functions. What is done after that is not necessairly constant, but almost all of these classes are in a sprint to get to this material. Generally what I think the very best students in such a class understand that continuity of a function allows us to apply the fundamental theorem of calculus to that function and perhaps the first definition I've given. There is a huge expanse of material to cover in a couple of semesters and foundations generally get short shrift. Saying that, without some power to change the curriculum, I cannot expect students to try and understand difficult concepts when they will be given tests that completely ignore these concepts.\n\nEDIT: If this seems as an indirect answer it somewhat is. I understand what continuity means if it were said in a research seminar, but I choose to reinforce definition 1 to calculus students then the others as I think it is much more likely to be understood.\n\n• The definitions you give are silent on endpoints. Is $\\sqrt{x}$ continuous at $x=0$? But, I understand, it's hard to cover all the cases when we're mostly teaching algebra which has yet to be mastered. A compromise I've found is to write the two-sided limit definition and mutter in class that we replace with appropriated one-sided limits for points in the domain. The graph-drawing technique is hard to violate if you add the critera \"on a connected domain\" aka interval. Aug 9, 2014 at 14:17\n\nI cover three different concepts of continuity in my high-school calculus class (separate from advanced concepts such as uniform continuity). This vocabulary is based on the AP textbook \"Calculus: Graphical, Numerical, Algebraic\" by Ross L. Finney et al., pages 75 and 77.\n\n1. A function $y=f(x)$ is continuous at an interior point $c$ of its domain if $$\\mathop {\\lim }\\limits_{x \\to c} f(x) = f(c)$$ And similarly for an endpoint of an interval, using a one-sided limit. A function is not continuous at any point not in its domain. Hence your reciprocal function is continuous at every value of $x$ other than $x=0$, where it is discontinuous.\n\n2. A function is continuous on an interval if and only if it is continuous at every point of the interval. Your reciprocal function is continuous on every interval not containing $x=0$.\n\n3. A continuous function is one that is continuous at every point in its domain. Hence, the answer to your first question is: for $f(x)=\\frac1x$, $f$ is a continuous function. And so is your function $g$.\n\nThese definitions are not perfect but they are indeed helpful in my teaching. I can use theorems about continuous functions without being concerned about points not in the domain.\n\n• See, this to be is logical, but it makes some of my colleagues uneasy as it disagrees with precalculus discussions where the pen-drawing definition makes $y=1/x$ a discontinuous graph. I was discussing this with brother and we agreed to disagree. For him, the nineteenth century conflation of the function with its formula and the intuitive appeal is not worth abandoning for the sake of matching the technical defn. of continuity we give in later courses. I'm generally a fan of doing things right the first time, so I like your approach. But, I see @Pete L. Clark shares my brothers view. Aug 9, 2014 at 14:21\n• To be clear, I don't think it is wrong to teach $1/x$ is discontinuous in a class where domains are not emphasized. However, if serious discussion is made of domains of functions then it seems out of place to me to say $1/x$ is discontinuous at $x=0$. Unless, you give a definition of discontinuity which explicitly defines discontinuity for points outside the domain. Aug 9, 2014 at 14:23\n• @JamesS.Cook: In my class I emphasize that the graph of a continuous function can be drawn with a pen kept on the paper through any point on the graph but not necessarily for the entire graph. This has worked best for me as a compromise in the various, somewhat conflicting ideas about continuity. I do see the point of what you say. Aug 9, 2014 at 15:13\n• I remember that in primary school (ages 5 - 10 I think) our maths textbooks emphasised the importance of the domain and codomain of a function. The massive, heavy, expensive College Calculus textbooks unfortunately throw all that progress away. This particular issue (about the continuity of 1/x) has caused much confusion and consternation among students and colleagues during my time as a University teacher - because they are smart, and can smell a rat. @Pete L. Clark's characterisation of those books as \"Legalistic\" is spot-on, if erring on the side of charitable ! Sep 16, 2020 at 23:51\n\nIt might bear pointing out that there is a rather wild kaleidoscope of different definitions for continuous functions seen in basic calculus and analysis texts (which might possibly inform different answers to this question). For example, this issue is highlighted in Steven Krantz's How to Teach Mathematics (Sec. 2.12; cites \"some calculus books\" that conclude the reciprocal function is discontinuous at $$x = 0$$). And a short paper on the issue was written by J.F. Harper:\n\n• Harper, J. F. \"What Really is a Continuous Function?.\" (2007).\n\nOne can say $$x\\mapsto1/x$$ is continuous on $$\\mathbb R\\cup\\{\\infty\\}$$ where this is the $$\\text{“}\\infty\\text{”}$$ that is approached if one goes in either direction (so that the domain is topologically a circle) and the value of this function at $$0$$ is $$\\infty$$ (not $$\\text{“}{+\\infty}\\text{”}$$ nor $$\\text{“}{-\\infty}{”}$$), and the limit of the function as the argument approached $$0$$ is also $$\\infty,$$ and the value of the function at $$\\infty$$ is $$0.$$\n\nThis way of looking at things makes the tangent function continuous on $$\\mathbb R$$ (but not on $$\\mathbb R\\cup\\{\\infty\\}.$$ If you want a discontinuity that fewer people will argue about, look at the discontinuity of the tangent function at $$\\infty.$$ And if it is objected that $$\\infty$$ is not in the domain so the function cannot have a discontinuity there, I will claim that it sometimes makes sense to speak of continuity on the closure of the domain." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.945162,"math_prob":0.9864772,"size":5400,"snap":"2023-40-2023-50","text_gpt3_token_len":1227,"char_repetition_ratio":0.18254262,"word_repetition_ratio":0.002244669,"special_character_ratio":0.22574075,"punctuation_ratio":0.09281437,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99752367,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-10-04T23:46:50Z\",\"WARC-Record-ID\":\"<urn:uuid:e19cf73c-2822-4d37-b79c-5a84f4cad05f>\",\"Content-Length\":\"251321\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f6d8aa98-9d65-433f-a052-14bebc1dc00f>\",\"WARC-Concurrent-To\":\"<urn:uuid:b694f48d-5b9f-49cb-9138-f13a5f876209>\",\"WARC-IP-Address\":\"104.18.11.86\",\"WARC-Target-URI\":\"https://matheducators.stackexchange.com/questions/4165/is-the-reciprocal-function-continuous\",\"WARC-Payload-Digest\":\"sha1:OBEBONOT4454ZJXOX33UGZTMFXPXODI2\",\"WARC-Block-Digest\":\"sha1:EBW2Q6WTIUCBNXPCLSJZBJFOSGJTTXBG\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233511424.48_warc_CC-MAIN-20231004220037-20231005010037-00622.warc.gz\"}"}
https://www.colorhexa.com/784580	[ "# #784580 Color Information\n\nIn a RGB color space, hex #784580 is composed of 47.1% red, 27.1% green and 50.2% blue. Whereas in a CMYK color space, it is composed of 6.3% cyan, 46.1% magenta, 0% yellow and 49.8% black. It has a hue angle of 291.9 degrees, a saturation of 29.9% and a lightness of 38.6%. #784580 color hex could be obtained by blending #f08aff with #000001. Closest websafe color is: #663399.\n\n• R 47\n• G 27\n• B 50\nRGB color chart\n• C 6\n• M 46\n• Y 0\n• K 50\nCMYK color chart\n\n#784580 color description : Dark moderate magenta.\n\n# #784580 Color Conversion\n\nThe hexadecimal color #784580 has RGB values of R:120, G:69, B:128 and CMYK values of C:0.06, M:0.46, Y:0, K:0.5. Its decimal value is 7882112.\n\nHex triplet RGB Decimal 784580 `#784580` 120, 69, 128 `rgb(120,69,128)` 47.1, 27.1, 50.2 `rgb(47.1%,27.1%,50.2%)` 6, 46, 0, 50 291.9°, 29.9, 38.6 `hsl(291.9,29.9%,38.6%)` 291.9°, 46.1, 50.2 663399 `#663399`\nCIE-LAB 37.496, 32.015, -24.389 13.77, 9.808, 21.589 0.305, 0.217, 9.808 37.496, 40.247, 322.701 37.496, 22.539, -37.608 31.318, 23.674, -18.948 01111000, 01000101, 10000000\n\n# Color Schemes with #784580\n\n• #784580\n``#784580` `rgb(120,69,128)``\n• #4d8045\n``#4d8045` `rgb(77,128,69)``\nComplementary Color\n• #5b4580\n``#5b4580` `rgb(91,69,128)``\n• #784580\n``#784580` `rgb(120,69,128)``\n• #80456b\n``#80456b` `rgb(128,69,107)``\nAnalogous Color\n• #45805b\n``#45805b` `rgb(69,128,91)``\n• #784580\n``#784580` `rgb(120,69,128)``\n• #6b8045\n``#6b8045` `rgb(107,128,69)``\nSplit Complementary Color\n• #458078\n``#458078` `rgb(69,128,120)``\n• #784580\n``#784580` `rgb(120,69,128)``\n• #807845\n``#807845` `rgb(128,120,69)``\n• #454d80\n``#454d80` `rgb(69,77,128)``\n• #784580\n``#784580` `rgb(120,69,128)``\n• #807845\n``#807845` `rgb(128,120,69)``\n• #4d8045\n``#4d8045` `rgb(77,128,69)``\n• #492a4e\n``#492a4e` `rgb(73,42,78)``\n• #59335f\n``#59335f` `rgb(89,51,95)``\n• #683c6f\n``#683c6f` `rgb(104,60,111)``\n• #784580\n``#784580` `rgb(120,69,128)``\n• #884e91\n``#884e91` `rgb(136,78,145)``\n• #9757a1\n``#9757a1` `rgb(151,87,161)``\n• #a365ac\n``#a365ac` `rgb(163,101,172)``\nMonochromatic Color\n\n# Alternatives to #784580\n\nBelow, you can see some colors close to #784580. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #694580\n``#694580` `rgb(105,69,128)``\n• #6e4580\n``#6e4580` `rgb(110,69,128)``\n• #734580\n``#734580` `rgb(115,69,128)``\n• #784580\n``#784580` `rgb(120,69,128)``\n• #7d4580\n``#7d4580` `rgb(125,69,128)``\n• #80457e\n``#80457e` `rgb(128,69,126)``\n• #804579\n``#804579` `rgb(128,69,121)``\nSimilar Colors\n\n# #784580 Preview\n\nThis text has a font color of #784580.\n\n``<span style=\"color:#784580;\">Text here</span>``\n#784580 background color\n\nThis paragraph has a background color of #784580.\n\n``<p style=\"background-color:#784580;\">Content here</p>``\n#784580 border color\n\nThis element has a border color of #784580.\n\n``<div style=\"border:1px solid #784580;\">Content here</div>``\nCSS codes\n``.text {color:#784580;}``\n``.background {background-color:#784580;}``\n``.border {border:1px solid #784580;}``\n\n# Shades and Tints of #784580\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, #010001 is the darkest color, while #f8f3f8 is the lightest one.\n\n• #010001\n``#010001` `rgb(1,0,1)``\n• #0c070d\n``#0c070d` `rgb(12,7,13)``\n• #180e1a\n``#180e1a` `rgb(24,14,26)``\n• #241527\n``#241527` `rgb(36,21,39)``\n• #301c34\n``#301c34` `rgb(48,28,52)``\n• #3c2340\n``#3c2340` `rgb(60,35,64)``\n• #482a4d\n``#482a4d` `rgb(72,42,77)``\n• #54305a\n``#54305a` `rgb(84,48,90)``\n• #603767\n``#603767` `rgb(96,55,103)``\n• #6c3e73\n``#6c3e73` `rgb(108,62,115)``\n• #784580\n``#784580` `rgb(120,69,128)``\n• #844c8d\n``#844c8d` `rgb(132,76,141)``\n• #905399\n``#905399` `rgb(144,83,153)``\n• #9c5aa6\n``#9c5aa6` `rgb(156,90,166)``\n``#a367ad` `rgb(163,103,173)``\n• #ab73b4\n``#ab73b4` `rgb(171,115,180)``\n• #b380bb\n``#b380bb` `rgb(179,128,187)``\n• #ba8dc1\n``#ba8dc1` `rgb(186,141,193)``\n• #c29ac8\n``#c29ac8` `rgb(194,154,200)``\n• #caa6cf\n``#caa6cf` `rgb(202,166,207)``\n• #d1b3d6\n``#d1b3d6` `rgb(209,179,214)``\n• #d9c0dd\n``#d9c0dd` `rgb(217,192,221)``\n• #e1cde4\n``#e1cde4` `rgb(225,205,228)``\n• #e8d9eb\n``#e8d9eb` `rgb(232,217,235)``\n• #f0e6f2\n``#f0e6f2` `rgb(240,230,242)``\n• #f8f3f8\n``#f8f3f8` `rgb(248,243,248)``\nTint Color Variation\n\n# Tones of #784580\n\nA tone is produced by adding gray to any pure hue. In this case, #675c69 is the less saturated color, while #aa01c4 is the most saturated one.\n\n• #675c69\n``#675c69` `rgb(103,92,105)``\n• #6d5471\n``#6d5471` `rgb(109,84,113)``\n• #724d78\n``#724d78` `rgb(114,77,120)``\n• #784580\n``#784580` `rgb(120,69,128)``\n• #7e3d88\n``#7e3d88` `rgb(126,61,136)``\n• #83368f\n``#83368f` `rgb(131,54,143)``\n• #892e97\n``#892e97` `rgb(137,46,151)``\n• #8e279e\n``#8e279e` `rgb(142,39,158)``\n• #941fa6\n``#941fa6` `rgb(148,31,166)``\n``#9918ad` `rgb(153,24,173)``\n• #9f10b5\n``#9f10b5` `rgb(159,16,181)``\n• #a408bd\n``#a408bd` `rgb(164,8,189)``\n• #aa01c4\n``#aa01c4` `rgb(170,1,196)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #784580 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":0.52767384,"math_prob":0.60604835,"size":3715,"snap":"2020-10-2020-16","text_gpt3_token_len":1619,"char_repetition_ratio":0.12449475,"word_repetition_ratio":0.011090573,"special_character_ratio":0.57523555,"punctuation_ratio":0.23783186,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9913363,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-04-09T08:34:58Z\",\"WARC-Record-ID\":\"<urn:uuid:88aa6e25-aa0a-47f8-bb3e-a8533611e0b0>\",\"Content-Length\":\"36312\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:10d8b287-f4ef-44e1-a993-c71cd2ace52c>\",\"WARC-Concurrent-To\":\"<urn:uuid:15be0df8-b2c9-43d8-94ed-0eeeeb89a98e>\",\"WARC-IP-Address\":\"178.32.117.56\",\"WARC-Target-URI\":\"https://www.colorhexa.com/784580\",\"WARC-Payload-Digest\":\"sha1:FDFROKV3MLI32PU4RLNMI56OU4ABDJMR\",\"WARC-Block-Digest\":\"sha1:DJOGVPALSERL2YVNJUGNJDQ2KCZ73V5T\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-16/CC-MAIN-2020-16_segments_1585371830894.88_warc_CC-MAIN-20200409055849-20200409090349-00084.warc.gz\"}"}
http://mvphip.org/9-2-practice-measuring-angles-and-arcs-answers/	[ "# 9-2 Practice Measuring Angles And Arcs Answers\n\nPosted on\n\n9-2 Practice Measuring Angles And Arcs Answers. › 9 2 skills practice answers. This online notice lesson 10 2 angles and arcs answers can be one of the options to accompany you considering having additional time. Aanndsawrcesrsskill practice answers can be. › practice and homework lesson 9.2. Lesson 9.2 measuring angles and arcsdone.notebook february 23, 2016 9.2 arc measure and arc length a central angle of a circle is an angle with a vertex in the center of the. An angle that intersects a circle in two points and has a vertex at the center of the circle. › lesson 9.2 prime factorization answers. The sum of the measures of the central angles of a circle with no interior points in common is answer: This is just one of the. This video is intended for sophomores and juniors in a geometry college prep course. Identify each arc as a major arc, minor arc, or semicircle. 10.2 find arc measures homework 409k: 150 ccss precision and are diameters of. Angles and arcs skill practice. Yeah, reviewing a book measuring angles and arcs skill practice answers could build up your close contacts listings.\n\n9-2 Practice Measuring Angles And Arcs Answers Indeed lately is being hunted by consumers around us, maybe one of you. People now are accustomed to using the net in gadgets to see video and image information for inspiration, and according to the name of the post I will discuss about 9-2 Practice Measuring Angles And Arcs Answers.\n\n• Notes And Practice – Student Copy – Lesson 12 – Geometry … : Two Pages Of Notes Do Now:\n• 7.2 Use The Converse Of The Pythagorean Theorem , Feb 1, 2013, 7:41 Am:\n• Ch 9.Pdf – Lesson 9.1 Skills Practice Name Date Three … – Tangents, Secants, Arcs And Angles.\n• 9-2 Angle Relationships Practice Wkst – Youtube . Lintel Angle Iron Span Tables.\n• Geometry Archives – Ms. Hall – Kenowa Hills High School … . Feb 1, 2013, 7:41 Am:\n• Trig Exam.pdf – Math 1316 Test 2Review Sheet Short Answer … : Try Your Best To Answer The Questions Above.\n• Warm-Up_-_Student_Copy_-_Lesson_6_-_Geometryhonors_-_Crm_1 … : Then Tell Whether It Is A Minor Arc, Major Arc, Or A Semicircle.\n• Kami Export – Angles Homework (5).Pdf – Section 2 Angles … , The Not So Safe Category 27.\n• Chapter 5 Practice Test – Geometry Connections Chapter 5 … , Two Pages Of Notes Do Now:\n• Solved: The Numerical Aperture (Na) Of An Optical Fiber Os … , Students Find Positive And Negative Measure Coterminals With Given Angles.\n\nFind, Read, And Discover 9-2 Practice Measuring Angles And Arcs Answers, Such Us:", null, "• 9 Practice Using Visual Cues Use The Triangle Below To … . Shopping The Graph Shows The Results Of A.", null, "• 7.2 Use The Converse Of The Pythagorean Theorem . 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Identify Each Arc As A Major Arc, Minor Arc, Or Semicircle.", null, "• Sb_Activity_14.Pdf – Angles And Angle Measure Activity 14 … – Therefore, The Red Arc In The Picture Below Is Not Used In This Formula.\n\n## 9-2 Practice Measuring Angles And Arcs Answers – Triangles On Act Math: Geometry Guide And Practice Problems\n\nNotes and Practice – Student Copy – Lesson 12 – Geometry …. › practice and homework lesson 9.2. Aanndsawrcesrsskill practice answers can be. 10.2 find arc measures homework 409k: This is just one of the. Angles and arcs skill practice. The sum of the measures of the central angles of a circle with no interior points in common is answer: Yeah, reviewing a book measuring angles and arcs skill practice answers could build up your close contacts listings. › lesson 9.2 prime factorization answers. Lesson 9.2 measuring angles and arcsdone.notebook february 23, 2016 9.2 arc measure and arc length a central angle of a circle is an angle with a vertex in the center of the. Identify each arc as a major arc, minor arc, or semicircle. This video is intended for sophomores and juniors in a geometry college prep course. This online notice lesson 10 2 angles and arcs answers can be one of the options to accompany you considering having additional time. An angle that intersects a circle in two points and has a vertex at the center of the circle. › 9 2 skills practice answers. 150 ccss precision and are diameters of.", null, "HSG-SRT.8 Right Triangles and the Pythagorean Theorem … from s3.amazonaws.com\n\nThe two angles formed at b are 50° and 30° because they are corresponding angles for parallel lines. The sides opposite those angles are congruent. Angles are measured with something called a protractor. Then tell whether it is a minor arc, major arc, or a semicircle. Use arc addition to find measures of arcs arc addition postulate = mlh + mhi = 58 + 32 = 90 answer: Lily crossing chapter 13 questions. What is the measure of \\$\\$ \\angle x \\$\\$?\n\n## An angle that intersects a circle in two points and has a vertex at the center of the circle.\n\nThere are two commonly used units of measurement for angles. In the exam, write your answers in capital letters on tho separate answer sheet. I used it last year during distance learning for my 4th students to practice measuring angles. The not so safe category 27. Angles and arcs skill practice. This is the currently selected item. Remember how one side of the angle traces out a circular arc? Lintel angle iron span tables. Students find positive and negative measure coterminals with given angles. Name the shaded arc in each picture practice: Arc a minor arc is the shortest arc connecting two endpoints on a circle. Lily crossing chapter 13 questions. It is given that zw bisects angle. 10.2 find arc measures homework 409k: The measure of the arc is equal to the measure of the central angle. This video is intended for sophomores and juniors in a geometry college prep course. To identify and use parts of circles and to solve problems involving the circumference of a circle and to recognize major arcs, minor arcs, semicircles, and central angles to find measures of practice problems (cont) determine the circumference of a circle with a radius of 2.5 inches. Lesson 9.2 measuring angles and arcsdone.notebook february 23, 2016 9.2 arc measure and arc length a central angle of a circle is an angle with a vertex in the center of the. For this angles and angle measure worksheet, learners draw given angles and they rewrite degrees in radians. The two angles formed at b are 50° and 30° because they are corresponding angles for parallel lines. Type your answers into the boxes provided leaving no spaces. New vocabulary central angle arc minor arc major arc semicircle congruent arcs adjacent arcs arcs and arc measure. › practice and homework lesson 9.2. Tangents, secants, arcs and their angles. › lesson 9.2 prime factorization answers. Measure angles using a protractor. This online notice lesson 10 2 angles and arcs answers can be one of the options to accompany you considering having additional time. There is an example at the beginning (0). This is just one of the. We welcome your feedback, comments and questions. › 9 2 skills practice answers.\n\n## 9-2 Practice Measuring Angles And Arcs Answers , New Vocabulary Central Angle Arc Minor Arc Major Arc Semicircle Congruent Arcs Adjacent Arcs Arcs And Arc Measure.", null, "## 9-2 Practice Measuring Angles And Arcs Answers , Lintel Angle Iron Span Tables.", null, "" ]	[ null, "https://files.spazioweb.it/e2/f6/e2f64f65-23e1-4fd7-b13d-35ac01b094ed.png", null, "https://files.spazioweb.it/e2/f6/e2f64f65-23e1-4fd7-b13d-35ac01b094ed.png", null, "https://files.spazioweb.it/e2/f6/e2f64f65-23e1-4fd7-b13d-35ac01b094ed.png", null, "https://files.spazioweb.it/e2/f6/e2f64f65-23e1-4fd7-b13d-35ac01b094ed.png", null, "https://files.spazioweb.it/e2/f6/e2f64f65-23e1-4fd7-b13d-35ac01b094ed.png", null, "https://files.spazioweb.it/e2/f6/e2f64f65-23e1-4fd7-b13d-35ac01b094ed.png", null, "https://files.spazioweb.it/e2/f6/e2f64f65-23e1-4fd7-b13d-35ac01b094ed.png", null, "https://files.spazioweb.it/e2/f6/e2f64f65-23e1-4fd7-b13d-35ac01b094ed.png", null, "https://files.spazioweb.it/e2/f6/e2f64f65-23e1-4fd7-b13d-35ac01b094ed.png", null, "https://files.spazioweb.it/e2/f6/e2f64f65-23e1-4fd7-b13d-35ac01b094ed.png", null, "https://s3.amazonaws.com/atlas-production.goalbookapp.com/resource-607b985b-50e1-4703-6bd9-2c40d4a4bd47/G.SRT.C.8_QA_FullPDF+%28dragged%291.png", null, "https://files.spazioweb.it/e2/f6/e2f64f65-23e1-4fd7-b13d-35ac01b094ed.png", null, "https://files.spazioweb.it/e2/f6/e2f64f65-23e1-4fd7-b13d-35ac01b094ed.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.82408756,"math_prob":0.7052921,"size":7518,"snap":"2021-04-2021-17","text_gpt3_token_len":1806,"char_repetition_ratio":0.16156508,"word_repetition_ratio":0.35795027,"special_character_ratio":0.24208567,"punctuation_ratio":0.12722299,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98663765,"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],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,1,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-04-15T16:16:50Z\",\"WARC-Record-ID\":\"<urn:uuid:36257410-cba7-425c-ba2d-fd5292df4ee8>\",\"Content-Length\":\"63169\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:2d10daac-5c63-4401-9113-8b9dd6dea482>\",\"WARC-Concurrent-To\":\"<urn:uuid:83f3e43c-2dfe-4664-ba4e-16e94f40a1ed>\",\"WARC-IP-Address\":\"172.67.147.130\",\"WARC-Target-URI\":\"http://mvphip.org/9-2-practice-measuring-angles-and-arcs-answers/\",\"WARC-Payload-Digest\":\"sha1:WPT2PJOL75UXX3Z5I4U4S6ZMNU2RT2MO\",\"WARC-Block-Digest\":\"sha1:TOVYQCHNDMCL6ZI3YGWWGKH7TKKHMTZY\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-17/CC-MAIN-2021-17_segments_1618038087714.38_warc_CC-MAIN-20210415160727-20210415190727-00184.warc.gz\"}"}
https://www.fxsolver.com/browse/?like=1587&p=3	[ "# Search results\n\nFound 1213 matches\nArea of rhombus (circumscribed)\n\nRhombus is a simple (non-self-intersecting) quadrilateral whose four sides all have the same length. The can be calculated by the semi perimeter and the ... more\n\nEuler's quadrilateral theorem\n\nIn any convex quadrilateral the sum of the squares of the four sides is equal to the sum of the squares of the two diagonals plus four times the square of ... more\n\nArea of a Square\n\nSquare is a regular quadrilateral, which means that it has four equal sides and four equal angles (90-degree angles, or right angles). It can also be ... more\n\nRhombus' side\n\nThe length of the side can computed by the length of the diagonals\n\n... more\n\nPythagorean theorem (arbitrary triangle - acute angle)\n\nGeneralization of the Pythagorean theorem for the side opposite of the acute angle of an arbitrary triangle\n\n... more\n\nSemiperimeter of a triangle\n\nThe semi sum of the length of a triangle’s sides\n\n... more\n\nPerimeter of a rhombus\n\nA rhombus is a simple (non-self-intersecting) quadrilateral all of whose four sides have the same length. A perimeter of a rhombus is a path that surrounds ... more\n\nRelation between the consecutive sides and the diagonals of a trapezoid\n\nTrapezoid is a convex quadrilateral with only one pair of parallel sides. The parallel sides are called the bases of the trapezoid and the other two sides ... more\n\nRelation between inradius,exradii and sides of a right triangle\n\nRight triangle or right-angled triangle is a triangle in which one angle is a right angle (that is, a 90-degree angle). The incircle or inscribed circle of ... more\n\nVarignon's theorem (Varignon parallelogram)\n\nThe Varigons theorem states that :\nThe midpoints of the sides of an arbitrary quadrangle form a parallelogram. If the quadrangle is convex or ... more\n\n...can't find what you're looking for?\n\nCreate a new formula" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8975175,"math_prob":0.9967972,"size":1455,"snap":"2019-35-2019-39","text_gpt3_token_len":336,"char_repetition_ratio":0.17298415,"word_repetition_ratio":0.031620555,"special_character_ratio":0.22130585,"punctuation_ratio":0.147651,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9980228,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-08-19T05:23:51Z\",\"WARC-Record-ID\":\"<urn:uuid:84188aa1-ac98-4d8b-a6b8-0a3790866308>\",\"Content-Length\":\"119524\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:570cad01-f3c3-4f69-9e87-07f7a5372c46>\",\"WARC-Concurrent-To\":\"<urn:uuid:227b1f23-3fdd-44f4-b410-12b7b792297f>\",\"WARC-IP-Address\":\"178.254.54.75\",\"WARC-Target-URI\":\"https://www.fxsolver.com/browse/?like=1587&p=3\",\"WARC-Payload-Digest\":\"sha1:FOIFDGLM3FZBMPI5QGWD63L3FCMWBUD4\",\"WARC-Block-Digest\":\"sha1:6FEOAU5PIW4VASLZLEBHXO6DZ5T3POBF\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-35/CC-MAIN-2019-35_segments_1566027314667.60_warc_CC-MAIN-20190819052133-20190819074133-00447.warc.gz\"}"}
https://developpaper.com/understand-pythons-iterators-iteratable-objects-and-generators/	[ "# Understand Python’s iterators, iteratable objects, and generators\n\nTime：2020-1-10\n\nMany partners are a little confused about the concepts of Python iterator, iteratable object and generator. Let me talk about my understanding and hope to help the friends in need.\n\n# 1 iterator protocol\n\nIterator protocol is the core. If you understand this, the above concepts will be well understood.\n\nThe so-called iterator protocol requires an iterator to implement the following two methods\n\n`iterator.__iter__()`\nReturn the iterator object itself.\n\n`iterator.__next__()`\nReturn the next item from the container.\n\nIn other words, as long as an object supports the above two methods, it is an iterator.`__iter__()`You need to return the iterator itself, and`__next__()`The next element needs to be returned.\n\n# 2 iteratable objects\n\nKnowing the concept of iterator, what is an iterative object?\n\nThis is simpler, as long as the object is implemented`__iter__()`Method and returns an iterator, which is an iteratable object.\n\nFor example, our common list is an iterative object\n\n``````>>> l = [1, 3, 5]\n>>> iter(l)``````\n\nUsing iter() will call the corresponding`__iter__()`Method, which returns a list iterator, so a list is an iteratable object.\n\n# 3 handwriting an iterator\n\nThere are different ways to implement iterators. I believe the first thing you can think of is custom classes. Let’s start from this.\n\nFor illustration, we write an iterator to generate an odd sequence.\n\nAccording to the iterator protocol, we implement the above two methods.\n\n``````class Odd:\ndef __init__(self, start=1):\nself.cur = start\n\ndef __iter__(self):\nreturn self\n\ndef __next__(self):\nret_val = self.cur\nself.cur += 2\nreturn ret_val``````\n\nIn the terminal, we instantiate an odd class to get an object odd\n\n``````>>> odd = Odd()\n>>> odd``````\n\nUsing the ITER () method will call the`__iter__`Method, get it\n\n``>>> iter(odd)``\n\nUsing the next () method will call the corresponding`__next__()`Method to get the next element\n\n``````>>> next(odd)\n1\n>>> next(odd)\n3\n>>> next(odd)\n5``````\n\nIn fact, the odd object is an iterator.\n\nWe can traverse it with for\n\n``````odd = Odd()\nfor v in odd:\nprint(v)``````\n\nCareful partners may find that this will be printed infinitely, so how to solve it?\n\nLet’s experiment with a list and get its iterator object first\n\n``````>>> l = [1, 3, 5]\n>>> li = iter(l)\n>>> li``````\n\nThen get the next element manually until there is no next element, and see what happens\n\n``````>>> next(li)\n1\n>>> next(li)\n3\n>>> next(li)\n5\n>>> next(li)\nTraceback (most recent call last):\nFile \"\", line 1, in\nStopIteration``````\n\nThe original list iterator will throw a stopiteration exception when there is no next element. It is estimated that the for statement is based on this exception to determine whether to end.\n\nLet’s modify the original code to generate odd numbers within the specified range\n\n``````class Odd:\ndef __init__(self, start=1, end=10):\nself.cur = start\nself.end = end\n\ndef __iter__(self):\nreturn self\n\ndef __next__(self):\nif self.cur > self.end:\nraise StopIteration\nret_val = self.cur\nself.cur += 2\nreturn ret_val``````\n\nLet’s try it with for\n\n``````>>> odd = Odd(1, 10)\n>>> for v in odd:\n... print(v)\n...\n1\n3\n5\n7\n9``````\n\nAs expected, it is consistent with the expectation.\n\nWe use the while loop to simulate the execution of for\n\nTarget code\n\n``````for v in iterable:\nprint(v)``````\n\nTranslated code\n\n``````iterator = iter(iterable)\nwhile True:\ntry:\nv = next(iterator)\nprint(v)\nexcept StopIteration:\nbreak``````\n\nIn fact, Python’s for statement principle is just like this. You can understand for as a syntax sugar.\n\n# 4 other ways to create iterators\n\nGenerators are also iterators, so you can use the way generators are created to create iterators.\n\n## 4.1 generator functions\n\nUnlike the return of a normal function, the generator function uses yield.\n\n``````>>> def odd_func(start=1, end=10):\n... for val in range(start, end + 1):\n... if val % 2 == 1:\n... yield val\n...\n>>> of = odd_func(1, 5)\n>>> of\n\n>>> iter(of)\n\n>>> next(of)\n1\n>>> next(of)\n3\n>>> next(of)\n5\n>>> next(of)\nTraceback (most recent call last):\nFile \"\", line 1, in\nStopIteration``````\n\n## 4.2 generator expression\n\n``````>>> g = (v for v in range(1, 5 + 1) if v % 2 == 1)\n>>> g\nat 0x101a142b0>\n>>> iter(g)\nat 0x101a142b0>\n>>> next(g)\n1\n>>> next(g)\n3\n>>> next(g)\n5\n>>> next(g)\nTraceback (most recent call last):\nFile \"\", line 1, in\nStopIteration``````\n\n## 4.3 how to choose\n\nSo far, we know three ways to create iterators, so how to choose?\n\nNeedless to say, the simplest is the generator expression. If the expression can meet the needs, it is it; if you need to add more complex logic, you can choose the generator function; if the first two cannot meet the needs, then you can customize the class implementation. In short, choose the simplest way.\n\n# Characteristics of 5 iterators\n\n## 5.1 inertia\n\nIterators don’t compute all elements in advance, but return when they need to.\n\n## 5.2 support for unlimited elements\n\nFor example, the first odd class we created above, whose instance odd is greater than all the odd numbers of start, and the list and other containers can’t hold infinite elements.\n\n## 5.3 provincial space\n\nLike 10000 elements\n\n``````>>> from sys import getsizeof\n>>> a = * 10000\n>>> getsizeof(a)\n80064``````\n\n``````>>> from itertools import repeat\n>>> b = repeat(1, times=10000)\n>>> getsizeof(b)\n56``````\n\nIt only takes 56 bytes.\n\nBecause of the inertia of the iterator, it has this advantage.\n\n# 6.1 iterators are also iterative objects\n\nBecause of the`__iter__()`Method returns itself, which is an iterator, so iterators are also iteratable objects.\n\n# 6.2 iterator can’t start from scratch after traversing once\n\nLook at a strange example\n\n``````>>> l = [1, 3, 5]\n>>> li = iter(l)\n>>> li\n\n>>> 3 in li\nTrue\n>>> 3 in li\nFalse``````\n\nBecause Li is a list iterator, it was found the first time it looked for 3, so it returned true. However, since the first iteration has skipped the 3 element, it cannot be found the second time, so it will appear false.\n\nTherefore, remember that iterators are “disposable.”.\n\nOf course, lists are iterative objects, and it’s normal to look up them several times. (if it’s hard to understand, think about the execution principle of the for statement above. Every time, a new iterator will be obtained from the iteratable object through the ITER () method.)\n\n``````>>> 3 in l\nTrue\n>>> 3 in l\nTrue``````\n\n# 7 stanzas\n\n• All objects that implement the iterator protocol are iterators\n• Realized`__iter__()`Method and return iterators are iteratable objects\n• Generator is also an iterator\n• There are three ways to create an iterator: generator expression, generator function, and custom class. You can choose the simplest one according to the situation\n• Iterators are also iterative objects\n• Iterators are “disposable”\n\nThe first three small items are the key points. These three points have been understood and others can be understood. It is no problem to understand the concepts of the nouns in the title.\n\n# 8 reference\n\n• https://docs.python.org/3/library/stdtypes.html#iterator-types\n• https://opensource.com/article/18/3/loop-better-deeper-look-iteration-python\n• http://treyhunner.com/2018/06/how-to-make-an-iterator-in-python" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7766273,"math_prob":0.89679015,"size":7067,"snap":"2020-24-2020-29","text_gpt3_token_len":1759,"char_repetition_ratio":0.15220161,"word_repetition_ratio":0.04599659,"special_character_ratio":0.27564737,"punctuation_ratio":0.13874911,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9748783,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-07-05T04:12:47Z\",\"WARC-Record-ID\":\"<urn:uuid:8a08d29e-0f2c-4578-96e3-43928e6c61d1>\",\"Content-Length\":\"50034\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4e805144-b4f5-4aed-991e-35a585ec862d>\",\"WARC-Concurrent-To\":\"<urn:uuid:31bf1837-cdb7-4d10-9fd5-b5d84dec5db6>\",\"WARC-IP-Address\":\"104.28.3.139\",\"WARC-Target-URI\":\"https://developpaper.com/understand-pythons-iterators-iteratable-objects-and-generators/\",\"WARC-Payload-Digest\":\"sha1:C7HMD62LISD2SXFX7OBGVGVBFWK4KM7I\",\"WARC-Block-Digest\":\"sha1:EQTR4BQIX5LQI66U3ONTXWYQ3YDWESL5\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-29/CC-MAIN-2020-29_segments_1593655886865.30_warc_CC-MAIN-20200705023910-20200705053910-00584.warc.gz\"}"}
https://jp.mathworks.com/matlabcentral/cody/problems/12-fibonacci-sequence/solutions/140439	[ "Cody\n\nProblem 12. Fibonacci sequence\n\nSolution 140439\n\nSubmitted on 19 Sep 2012 by Robert\nThis solution is locked. To view this solution, you need to provide a solution of the same size or smaller.\n\nTest Suite\n\nTest Status Code Input and Output\n1 Pass\n%% n = 1; f = 1; assert(isequal(fib(n),f))\n\n2 Pass\n%% n = 6; f = 8; assert(isequal(fib(n),f))\n\n3 Pass\n%% n = 10; f = 55; assert(isequal(fib(n),f))\n\n4 Pass\n%% n = 20; f = 6765; assert(isequal(fib(n),f))" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.51680964,"math_prob":0.9990038,"size":442,"snap":"2019-43-2019-47","text_gpt3_token_len":173,"char_repetition_ratio":0.16438356,"word_repetition_ratio":0.0,"special_character_ratio":0.4638009,"punctuation_ratio":0.16666667,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99567574,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-10-24T05:12:45Z\",\"WARC-Record-ID\":\"<urn:uuid:51e3f853-1e51-4971-a13d-528081933376>\",\"Content-Length\":\"73199\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:0d6598c7-449a-4abe-8af2-61d9852861a1>\",\"WARC-Concurrent-To\":\"<urn:uuid:adb4c02b-1337-4d16-85e5-b525cfadc2f9>\",\"WARC-IP-Address\":\"104.110.193.39\",\"WARC-Target-URI\":\"https://jp.mathworks.com/matlabcentral/cody/problems/12-fibonacci-sequence/solutions/140439\",\"WARC-Payload-Digest\":\"sha1:M5EMS5VC43LOD5JANHG24GDSQ52FXFZI\",\"WARC-Block-Digest\":\"sha1:NI6WMZO4ZA6NT2ZIYLLFQJC3EP33UCQA\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-43/CC-MAIN-2019-43_segments_1570987841291.79_warc_CC-MAIN-20191024040131-20191024063631-00459.warc.gz\"}"}
https://artofproblemsolving.com/wiki/index.php/2003_AIME_I_Problems/Problem_13	[ "# 2003 AIME I Problems/Problem 13\n\n## Problem\n\nLet", null, "$N$ be the number of positive integers that are less than or equal to", null, "$2003$ and whose base-", null, "$2$ representation has more", null, "$1$'s than", null, "$0$'s. Find the remainder when", null, "$N$ is divided by", null, "$1000$.\n\n## Solution 1\n\nIn base-", null, "$2$ representation, all positive numbers have a leftmost digit of", null, "$1$. Thus there are", null, "${n \\choose k}$ numbers that have", null, "$n+1$ digits in base", null, "$2$ notation, with", null, "$k+1$ of the digits being", null, "$1$'s.\n\nIn order for there to be more", null, "$1$'s than", null, "$0$'s, we must have", null, "$k+1 > \\frac{d+1}{2} \\Longrightarrow k > \\frac{d-1}{2} \\Longrightarrow k \\ge \\frac{d}{2}$. Therefore, the number of such numbers corresponds to the sum of all numbers on or to the right of the vertical line of symmetry in Pascal's Triangle, from rows", null, "$0$ to", null, "$10$ (as", null, "$2003 < 2^{11}-1$). Since the sum of the elements of the", null, "$r$th row is", null, "$2^r$, it follows that the sum of all elements in rows", null, "$0$ through", null, "$10$ is", null, "$2^0 + 2^1 + \\cdots + 2^{10} = 2^{11}-1 = 2047$. The center elements are in the form", null, "${2i \\choose i}$, so the sum of these elements is", null, "$\\sum_{i=0}^{5} {2i \\choose i} = 1 + 2 +6 + 20 + 70 + 252 = 351$.\n\nThe sum of the elements on or to the right of the line of symmetry is thus", null, "$\\frac{2047 + 351}{2} = 1199$. However, we also counted the", null, "$44$ numbers from", null, "$2004$ to", null, "$2^{11}-1 = 2047$. Indeed, all of these numbers have at least", null, "$6$", null, "$1$'s in their base-", null, "$2$ representation, as all of them are greater than", null, "$1984 = 11111000000_2$, which has", null, "$5$", null, "$1$'s. Therefore, our answer is", null, "$1199 - 44 = 1155$, and the remainder is", null, "$\\boxed{155}$.\n\n## Solution 2\n\nWe seek the number of allowed numbers which have", null, "$k$ 1's, not including the leading 1, for", null, "$k=0, 1, 2, \\ldots , 10$.\n\nFor", null, "$k=0,\\ldots , 4$, this number is", null, "$\\binom{k}{k}+\\binom{k+1}{k}+\\cdots+\\binom{2k}{k}$.\n\nBy the Hockey Stick Identity, this is equal to", null, "$\\binom{2k+1}{k+1}$. So we get", null, "$\\binom{1}{1}+\\binom{3}{2}+\\binom{5}{3}+\\binom{7}{4}+\\binom{9}{5}=175$.\n\nFor", null, "$k=5,\\ldots , 10$, we end on", null, "$\\binom{10}{k}$ - we don't want to consider numbers with more than 11 digits. So for each", null, "$k$ we get", null, "$\\binom{k}{k}+\\binom{k+1}{k}+\\ldots+\\binom{10}{k}=\\binom{11}{k+1}$\n\nagain by the Hockey Stick Identity. So we get", null, "$\\binom{11}{6}+\\binom{11}{7}+\\binom{11}{8}+\\binom{11}{9}+\\binom{11}{10}+\\binom{11}{11}=\\frac{2^{11}}{2}=2^{10}=1024$.\n\nThe total is", null, "$1024+175=1199$. Subtracting out the", null, "$44$ numbers between", null, "$2003$ and", null, "$2048$ gives", null, "$1155$. Thus the answer is", null, "$155$.\n\n## Solution 3\n\nWe will count the number of it", null, "$< 2^{11}=2048$ instead of", null, "$2003$ (In other words, the length of the base-2 representation is at most", null, "$11$. If there are even digits,", null, "$2n$, then the leftmost digit is", null, "$1$, the rest,", null, "$2n-1$, has odd number of digits. In order for the base-2 representation to have more", null, "$1$'s, we will need more", null, "$1$ in the remaining", null, "$2n-1$ than", null, "$0$'s. Using symmetry, this is equal to", null, "$\\frac{2^9+2^7+..+2^1}{2}$ Using similar argument where there are odd amount of digits. The remaining even amount of digit must contains the number of", null, "$1$'s at least as the number of", null, "$0$'s. So it's equal to", null, "$\\frac{\\binom{10}{5}+2^{10}+\\binom{8}{4}+2^8+\\binom{6}{3}+2^6+...+\\binom{0}{0}+2^0}{2}$ Summing both cases, we have", null, "$\\frac{2^0+2^1+..+2^{10}+\\binom{0}{0}+..+\\binom{10}{5}}{2} = 1199$. There are", null, "$44$ numbers between", null, "$2004$ and", null, "$2047$ inclusive that satisfy it. 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https://quuxplusone.github.io/blog/2018/08/05/mathjax-in-jekyll/	[ "# MathJax in Jekyll\n\nI spent most of today figuring out how to write the mathematics in my previous post on quantum circuits. It turned out to be way easier than I was making it.\n\n(This post assumes that you are already familiar enough with TeX, or should I say, $\\rm\\TeX$. I have found the TeX StackExchange to be super useful.)\n\nDason Kurkiewicz’s blog post from October 2012 is still surprisingly accurate. Step one is to follow the most basic possible instructions from MathJax’s own Getting Started guide: place the snippet\n\n<script type=\"text/javascript\" async\nsrc=\"https://cdnjs.cloudflare.com/ajax/libs/mathjax/2.7.5/MathJax.js?config=TeX-MML-AM_CHTML\">\n</script>\n\n\nsomewhere that Jekyll will pick it up. (You don’t even need the suggested config=TeX-MML-AM_CHTML parameter; we’re going to specify our own config.)\n\nStep two is to specify the config, right above (not below!) the script tag that fetches MathJax.js.\n\n<script type=\"text/x-mathjax-config\">\nMathJax.Hub.Config({\nextensions: [\n\"MathZoom.js\",\n\"AssistiveMML.js\",\n],\njax: [\"input/TeX\", \"output/CommonHTML\"],\nTeX: {\nextensions: [\n\"AMSmath.js\",\n\"AMSsymbols.js\",\n\"noErrors.js\",\n\"noUndefined.js\",\n]\n}\n});\n</script>\n\n\nStep three is to realize that the Markdown processor used by Jekyll — its name is “Kramdown” — has a built-in feature called “math blocks” which can be hooked up to MathJax! This means that the rest of the integration has already been done for you. Having followed steps 1 and 2 above, you can now just write TeX code with double-dollar-sign escapes:\n\n... a given wire happens to be carrying \"$$\\lvert 0\\rangle$$.\"\nBy that we mean that it's carrying the linear combination\n$$\\begin{psmallmatrix} 1 \\\\ 0 \\end{psmallmatrix}$$ ...\n\n\nThis renders as:\n\n… a given wire happens to be carrying “$\\lvert 0\\rangle$.” By that we mean that it’s carrying the linear combination $\\begin{psmallmatrix} 1 \\\\ 0 \\end{psmallmatrix}$\n\nAnd if you make a paragraph that contains nothing but a $$-escaped chunk of math, it will be rendered using MathJax’s mode=display, i.e., TeX display mode. To see how to enable MathJax for your own Jekyll blog, click through to the relevant commit in this blog’s GitHub repository. The rabbit-hole that I went down by accident was that I didn’t realize until very late that Kramdown already supported “math blocks.” So I spent some time trying to use MathJax’s “tex2jax.js” preprocessor — which is very easy to add to the config block above, by the way. //... extensions: [ \"tex2jax.js\", // HERE \"MathMenu.js\", \"MathZoom.js\", \"AssistiveMML.js\", \"a11y/accessibility-menu.js\" ], tex2jax: { // AND HERE inlineMath: [['', '']], displayMath: [['$$', '$$']] }, jax: [\"input/TeX\", \"output/CommonHTML\"], //... But “tex2jax.js” runs on the text of the page after Kramdown gets done with it; which is to say, after Kramdown has already processed out all of the $$s and replaced them with <script type=\"math/tex\"> tags. What’s more, Kramdown overloads $$ to mean both “inline” and “display,” depending on the surrounding linebreaks. So the upshot of that interaction was that I kept writing $$ (with no surrounding linebreaks) expecting display math, and what I got on the rendered page was inline math. It took me forever to figure out that this was due to Kramdown, and not a bug either in my config or in “tex2jax.js”!\n\nThe fix for this issue, of course, was to find out that Kramdown math blocks were a better solution than “tex2jax.js”.\n\nPosted 2018-08-05" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.89041454,"math_prob":0.648711,"size":3385,"snap":"2019-13-2019-22","text_gpt3_token_len":883,"char_repetition_ratio":0.09730849,"word_repetition_ratio":0.019193858,"special_character_ratio":0.2579025,"punctuation_ratio":0.16739447,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9690793,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-03-23T00:48:53Z\",\"WARC-Record-ID\":\"<urn:uuid:eae956c1-c997-4adc-bd68-50c0d5b99401>\",\"Content-Length\":\"17643\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:03c807e3-42cf-43da-867b-57795dd29af9>\",\"WARC-Concurrent-To\":\"<urn:uuid:99e4d8f6-fc9f-4534-a02f-31152a0894c1>\",\"WARC-IP-Address\":\"185.199.108.153\",\"WARC-Target-URI\":\"https://quuxplusone.github.io/blog/2018/08/05/mathjax-in-jekyll/\",\"WARC-Payload-Digest\":\"sha1:NG5LQIRPDMOCERZMKIFAT6VZTP5MLJCZ\",\"WARC-Block-Digest\":\"sha1:UZLJO5H24RQ4SFU3BIQGMOMVXWZY5GCE\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-13/CC-MAIN-2019-13_segments_1552912202704.58_warc_CC-MAIN-20190323000443-20190323022443-00361.warc.gz\"}"}
https://affairscloud.com/basic-concepts-of-vedic-maths/?amp	[ "", null, "# Basic Concepts of Vedic Maths\n\nIt is very important to learn and understand the two basic concepts to make any calculation easy. These two concepts are the pillars of Vedic Maths. They are\n\n1) Base\n\n2) Complements\n\nWhat is a Base?\n\nBases are the numbers which start with 1 and ends with 0’s. For example 10, 100, 1000, 10000, 100000, and so on are called bases. The first 2-digit number is 10 and the first 3-digit number is 100 and so on. Only these kind of numbers are considered as bases. Numbers such as 200, 300, 450, 1500 are not considered as bases.\n\nWhat is a Complement?\n\nWhen two numbers are added with each other and it results in the next nearest base, then they are called Complements of each other.\n\nFor Example,\n\n1) Consider the number 73. The next nearest base of 73 is 100. By adding 27 we will get 100. So 73 and 27 are complements to each other.\n\n2) Consider the number 156. The next nearest base of 156 is 1000. By adding 844 we get 1000. So they are Complements.\n\n3) Similarly,\n\nComplement of 6 is 4 (Base is 10).\n\nComplement of 25 is 75 (Base is 100).\n\nComplement of 6545 is 3455 (Base is 10000).\n\nHow to find the Complement of a number in a easier way?\n\nComplement of a number can be calculated by subtracting all numbers from 9 and the last number by 10.\n\neg: Complement of 358732?\n\nSolution: 9 9 9 9 9 10 3 5 8 7 3 2 = 6 4 1 3 6 8\n\nThe Formula is “All from 9 and the last from 10“.\n\nComplement of 5183?\n\nSolution :\n\n9 9 9 10 – 5 1 8 3 = 4 8 1 7\n\nWhat if the last digit is “Zero”??\n\nIn that case treat the last non-zero number as the last digit and use the same above formula.\n\nEg: Complement of 69560670??\n\nSolution: 9 9 9 9 9 9 10\n\n– 6 9 5 6 0 6 7 0\n\n3 0 4 3 9 3 3 0\n\nExercises:\n\nFind the Complement of\n\n1) 234 2) 3457 3) 12967 4) 2463751 5) 10920" ]	[ null, "data:image/svg+xml;base64,PHN2ZyBoZWlnaHQ9IjI4MiIgd2lkdGg9IjMzNiIgeG1sbnM9Imh0dHA6Ly93d3cudzMub3JnLzIwMDAvc3ZnIiB2ZXJzaW9uPSIxLjEiLz4=", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.877625,"math_prob":0.99638,"size":1767,"snap":"2021-43-2021-49","text_gpt3_token_len":609,"char_repetition_ratio":0.17073171,"word_repetition_ratio":0.038043477,"special_character_ratio":0.39388794,"punctuation_ratio":0.11694511,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9981025,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-11-30T16:01:07Z\",\"WARC-Record-ID\":\"<urn:uuid:6656ca8d-1dd8-4aad-9b41-7d5948ba4f28>\",\"Content-Length\":\"102651\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c7cd2cc2-52a2-412e-8e52-44fbbf949c7f>\",\"WARC-Concurrent-To\":\"<urn:uuid:9c4a079d-337a-489c-bb7e-16582ef7e126>\",\"WARC-IP-Address\":\"104.26.5.172\",\"WARC-Target-URI\":\"https://affairscloud.com/basic-concepts-of-vedic-maths/?amp\",\"WARC-Payload-Digest\":\"sha1:WDEYF754ATZK3IWBLIHRYRHNZIBLQRGA\",\"WARC-Block-Digest\":\"sha1:VP2LSDHM4A52FDEEFAXJFHVSF3M37LAZ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964359037.96_warc_CC-MAIN-20211130141247-20211130171247-00187.warc.gz\"}"}
https://www.shaalaa.com/question-bank-solutions/convert-0010-0100-0010-1101-base-2-decimal-convert-134-base-10-hexadecimal-write-steps-conversion-turing-model_58121	[ "# Convert 0010 0100 0010 1101 from Base 2 to Decimal. Convert 134 from Base 10 to Hexadecimal. Write Steps for Conversion. - Structured Programming Approach\n\nConvert 0010 0100 0010 1101 from base 2 to decimal. Convert 134 from base 10 to hexadecimal. Write steps for conversion.\n\n#### Solution\n\nBinary to decimal: The steps to be followed are:\n\n1. Multiply the binary digits (bits) with powers of 2 according to their positionalweight.\n2. The position of the first bit (going from right to left) is 0 and it keeps on incrementing as you go towards left for every bit.\n3.Given: (0010 0100 0010 1101)2 = (?)10\n4.0215+0214+1213+0212+0211+1210+029+028+027+026+125+024+123+122+021+120\n=0+0+8192+0+0+1024+0+0+0+0+32+0+8+4+1\n=9261\n(0010 0100 0010 1101)2 = (9261)10\n\nDecimal to hexadecimal: The steps to be followed are:\n1.Divide the decimal number by 16. Treat the division as an integer division.\n2.Write down the remainder (in hexadecimal).\n3.Divide the result again by 16. Treat the division as an integer division.\n4.Repeat step 2 and 3 until result is 0.\n5.The hexadecimal value is digit sequence of the reminders from last to first. 6.Given: (134)10 = (?)16", null, "(134)10 = (86)16\n\nConcept: Turing Model\nIs there an error in this question or solution?\n2016-2017 (June) CBCGS\n\nShare" ]	[ null, "https://www.shaalaa.com/images/_4:be792f4a939645d9af57e02bce80180d.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.685843,"math_prob":0.9785089,"size":1022,"snap":"2023-40-2023-50","text_gpt3_token_len":355,"char_repetition_ratio":0.11689588,"word_repetition_ratio":0.103896104,"special_character_ratio":0.4295499,"punctuation_ratio":0.12345679,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.998466,"pos_list":[0,1,2],"im_url_duplicate_count":[null,3,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-09T08:55:41Z\",\"WARC-Record-ID\":\"<urn:uuid:14f9c961-edf4-4fcc-8e80-4dabce41477b>\",\"Content-Length\":\"69996\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:2e33af4c-2df7-481b-b28c-d36ab5c841e6>\",\"WARC-Concurrent-To\":\"<urn:uuid:e97462bd-2e29-4ba7-ada0-4e3255f95ba9>\",\"WARC-IP-Address\":\"104.26.12.31\",\"WARC-Target-URI\":\"https://www.shaalaa.com/question-bank-solutions/convert-0010-0100-0010-1101-base-2-decimal-convert-134-base-10-hexadecimal-write-steps-conversion-turing-model_58121\",\"WARC-Payload-Digest\":\"sha1:BRXQAC54SFKIV4IPDN2AZTVPP42PJN7Y\",\"WARC-Block-Digest\":\"sha1:QP65I47HPH5FKYGFQUUYAE42RTEZO3DI\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100873.6_warc_CC-MAIN-20231209071722-20231209101722-00447.warc.gz\"}"}
https://quantumcomputing.stackexchange.com/questions/9287/can-you-use-rz-to-flip-from-rangle-to-rangle	[ "# Can you use Rz to flip from $\|+\\rangle$ to $\|-\\rangle$?\n\nHere's the Rz matrix:\n\n$$Rz(\\theta) = \\begin{bmatrix} e^{-i\\theta/2} & 0 \\\\ 0 & e^{i\\theta/2} \\end{bmatrix}$$\n\nAs I understand it, Rz rotates around the Z axis on the Bloch sphere. Since $$\|+\\rangle$$ and $$\|-\\rangle$$ are both on the Bloch sphere X-Y plane, it seems you should be able to rotate between them; however, I can't figure out a value of $$\\theta$$ to do that.\n\nAm I misunderstanding how the rotation gates work, or is there just a solution I'm not seeing?\n\nThe more elegant approach is to view $$R_z(\\phi)$$ as a linear transformation acting on the basis $$\\{\|0\\rangle, \|1\\rangle\\}$$. $$\|0\\rangle$$ is mapped to $$e^{i\\phi/2}\|0\\rangle$$ and $$\|1\\rangle$$ is mapped to $$e^{-i\\phi/2}\|1\\rangle$$ by the transformation. As, $$\|+\\rangle = \\frac{1}{\\sqrt 2}(\|0\\rangle +\|1\\rangle)$$ it'd be mapped to $$R_z(\\phi)\|+\\rangle = \\frac{1}{\\sqrt 2}(e^{i\\phi/2}\|0\\rangle + e^{-i\\phi/2}\|1\\rangle)$$. Now the question is, for what value of $$\\phi$$ (if at all), $$R_z\|+\\rangle$$ coincides with $$\|-\\rangle = \\frac{1}{\\sqrt 2}(\|0\\rangle - \|1\\rangle)$$. As global phase factors are irrelevant in quantum mechanics, you just need to check for what value of $$\\phi$$ (if at all) the coefficients of the basis vectors are in proportion i.e.,\n\n$$\\frac{e^{i\\phi/2}}{e^{-i\\phi/2}} = \\frac{1}{-1} \\implies e^{i\\phi} = -1 \\implies e^{i\\phi} = e^{i\\pi} \\implies \\phi = \\pi$$\n\nassuming $$\\phi \\in [0, 2\\pi)$$.\n\nObviously, the Bloch sphere visualization (as in @kludg's answer) makes it even more evident. You can clearly see that a anti-clockwise rotation of $$\\phi = \\pi$$ (180 degrees) about the $$z$$-axis ($$\|0\\rangle$$ - $$\|1\\rangle$$ axis) takes the $$\|+\\rangle$$ state to the $$\|-\\rangle$$ state.\n\nP.S: I used $$R_z(\\phi)$$ instead of $$R_z(\\theta)$$ in order to match the convention in the above diagram.\n\nIf you use $$\\theta = \\pi$$, you get the following:\n\n$$Rz(\\pi)\\begin{bmatrix} \\frac{1}{\\sqrt{2}} \\\\ \\frac{1}{\\sqrt{2}} \\end{bmatrix} = \\begin{bmatrix} e^{-i \\pi/2} & 0 \\\\ 0 & e^{i \\pi/2} \\end{bmatrix} \\begin{bmatrix} \\frac{1}{\\sqrt{2}} \\\\ \\frac{1}{\\sqrt{2}} \\end{bmatrix} = \\begin{bmatrix} -i & 0 \\\\ 0 & i \\end{bmatrix} \\begin{bmatrix} \\frac{1}{\\sqrt{2}} \\\\ \\frac{1}{\\sqrt{2}} \\end{bmatrix} = -i\\begin{bmatrix} \\frac{1}{\\sqrt{2}} \\\\ \\frac{-1}{\\sqrt{2}} \\end{bmatrix}$$\n\nWhich is equal to $$\|-\\rangle$$ under the global phase $$-i = e^{-i\\pi/2}$$.\n\n• to visualise the rotations in the Bloch sphere it might be more natural to work directly with density matrices. In this case, you can see quite nicely how $R_z(\\theta)$ acts as a rotation when you consider its action (via conjugation) on states: $\\rho\\mapsto R_z(\\theta)\\rho R_z(\\theta)^\\dagger$.\n– glS\nDec 22, 2019 at 16:41\n\nJust think where $$\|+\\rangle$$ and $$\|-\\rangle$$ states on the Bloch sphere ($$x$$ axis):", null, "On IBM Q the Rz gate is defined as\n\n$$\\begin{equation} Rz(\\theta) = \\begin{pmatrix} 1 & 0 \\\\ 0 & \\mathrm{e}^{i\\theta} \\end{pmatrix} \\end{equation}$$\n\nIf you put $$\\theta = \\pi$$ then the matrix turns to\n\n$$\\begin{equation} Rz(\\pi) = \\begin{pmatrix} 1 & 0 \\\\ 0 & \\mathrm{e}^{i\\pi} \\end{pmatrix} = \\begin{pmatrix} 1 & 0 \\\\ 0 & -1 \\end{pmatrix} =Z \\end{equation}$$ Hence $$Rz(\\pi)\|+\\rangle = \|-\\rangle$$.\n\nYou can rewrite your matrix as\n\n$$\\begin{equation} Rz(\\theta) = \\begin{pmatrix} \\mathrm{e}^{-i\\frac{\\theta}{2}} & 0 \\\\ 0 & \\mathrm{e}^{i\\frac{\\theta}{2}} \\end{pmatrix} = \\mathrm{e}^{-i\\frac{\\theta}{2}} \\begin{pmatrix} 1 & 0 \\\\ 0 & \\mathrm{e}^{i\\theta} \\end{pmatrix} \\end{equation}$$\n\nHence the only difference is a global phase.\n\nOverall, Rz gate matrix as defined on IBM Q seems to more convinient as you do not have to deal with a global phase coeficient.\n\n• I disagree with IBM Q conventions; in QM, spin 1/2 rotation matrices are related to Pauli matrices as $R_i(\\theta)=\\exp(i \\theta/2\\cdot \\sigma_i)$ and this defines $R_z(\\theta)$; and IBM for the sake of petty simplification is making a mess of QM conventions. Dec 22, 2019 at 7:50\n• @kludg: I understand your objection but in the end the result is the same regardless which matrix (either QM approach or IBM approach) is used as the only difference is omission of global phase. Quantum states which differ only in global phase are considered to be same, so both approaches work. Or do I understand anything in wrong way? Dec 22, 2019 at 8:49\n• This is the question of using common conventions. For example, it is possible to define Pauli matrices differently; fortunately physicists are wise to avoid a mess and everybody uses the form proposed by Pauli; IMO IBM Q designers were not wise when defined $R_z(\\theta)$ matrix. Dec 22, 2019 at 9:43\n• @kludg: OK, thanks for explanation. Dec 22, 2019 at 9:53" ]	[ null, "https://i.stack.imgur.com/mX4eD.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.90182316,"math_prob":0.9999709,"size":461,"snap":"2023-40-2023-50","text_gpt3_token_len":140,"char_repetition_ratio":0.10940919,"word_repetition_ratio":0.0,"special_character_ratio":0.30368763,"punctuation_ratio":0.094736844,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.0000038,"pos_list":[0,1,2],"im_url_duplicate_count":[null,4,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-24T06:06:21Z\",\"WARC-Record-ID\":\"<urn:uuid:35e7be82-33f8-4a71-868c-6272eb48d6e2>\",\"Content-Length\":\"195692\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7f3cff37-0111-4015-b542-d1bce105f535>\",\"WARC-Concurrent-To\":\"<urn:uuid:636add0b-384c-4786-9cb5-2ed26984480f>\",\"WARC-IP-Address\":\"104.18.10.86\",\"WARC-Target-URI\":\"https://quantumcomputing.stackexchange.com/questions/9287/can-you-use-rz-to-flip-from-rangle-to-rangle\",\"WARC-Payload-Digest\":\"sha1:CB6FPPKPP3JKWQUJKA7PYI2ZWNUSRSCA\",\"WARC-Block-Digest\":\"sha1:RFPYBIK4CWKRQZJAZ7TGQ3ZZIZG4MLNQ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233506623.27_warc_CC-MAIN-20230924055210-20230924085210-00051.warc.gz\"}"}
https://offbeat.cc/blog/our-trip-to-integer-partitions.html	[ "# Our Next Trip to Integer Partitions\n\nPublished on 18 Sep 2021 by Susam Pal\n\n## Introduction\n\nAfter 114 meetings and 75 hours of studying together, our analytic number theory book club has finally reached the final chapter of the book Introduction to Analytic Number Theory (Apostol, 1976). We have less than 18 pages to read in order to complete reading this book. Considering that we read about 2-3 pages in every meeting, it appears that we will complete reading this book in another 2 weeks.\n\nReading this book has been quite a journey! The previous three blog posts on this blog provide an account of how this journey has been. It has been fun, of course. The best part of hosting a book club like this has been the number of extremely smart people I got an opportunity to meet and interact with. The insights and comments on the study material that some of our book club participants shared during the meetings were very helpful. Special thanks to Andrey who goes by the Libera IRC nick halyavin for joining most of our meetings and sharing his explanations on various steps of the proofs we came across in the book.\n\nThe meeting log shows that our book club started really small with only 4 participants in the first meeting in March 2021 and then it gradually grew to about 10-12 regular members within a month. Then a few months later, the number of participants began dwindling a little. This happened because some members of the book club had to drop out as they got busy with other personal or professional engagements. However, six months later, we still have about 4-5 regular participants meeting consistently. I think it is pretty good that we have made it this far.\n\n## Unrestricted Partitions\n\nThe final chapter on integer partitions is very unlike all the previous 12 chapters. While the previous chapters dealt with multiplicative number theory, this final chapter deals with additive number theory. For example, the first theorem talks about an interesting property of unrestricted partitions. We study the number of ways a positive integer can be expressed as a sum of positive integers. The number of summands is unrestricted, repetition of summands is allowed, and the order of the summands is not taken into account. For example, the number 3 has 3 partitions: 3, 2 + 1, and 1 + 1 + 1. Similarly, the number 4 has 5 partitions: 4, 3 + 1, 2 + 2, 2 + 1 + 1, and 1 + 1 + 1 + 1.\n\nI have always wanted to learn about partitions more deeply, so I am quite happy that this book ends with a chapter on partitions. The subject of partitions is rich with very interesting results obtained by various accomplished mathematicians. In the book, the first theorem about partitions is a very simple one that follows from the geometric representation of partitions. Let us see an illustration first.\n\nHow many partitions of 6 are there? There are 11 partitions of 6. They are 6, 5 + 1, 4 + 2, 4 + 1 + 1, 3 + 3, 3 + 2 + 1, 3 + 1 + 1 + 1, 2 + 2 + 2, 2 + 2 + 1 + 1, 2 + 1 + 1 + 1 + 1, and 1 + 1 + 1 + 1 + 1 + 1. Now how many of these partitions are made up of 5 parts? Each summand is called a part. The answer is 2. There are 2 partitions of 6 that are made up of 5 parts. They are 3 + 1 + 1 + 1 and 2 + 2 + 1 + 1. Let us represent both these partitions as arrangements of lattice points. Here is the representation of the partition 3 + 1 + 1 + 1:\n\n• • •\n•\n•\n•\n\n\nNow if we read this arrangement from left-to-right, column-by-column, we get another partition of 6, i.e., 4 + 1 + 1. Note that the number of parts in 3 + 1 + 1 + 1 (i.e., 4) appears as the largest part in 4 + 1 + 1. Similarly, the number of parts in 4 + 1 + 1 (i.e., 3) appears as the largest part in 3 + 1 + 1 + 1. Let us see one more example of this relationship. Here is the geometric representation of 2 + 2 + 1 + 1:\n\n• •\n• •\n•\n•\n\n\nOnce again, reading this representation from left-to-right, we get 4 + 2, another partition of 6. Once again, we can see that the number of partitions in 2 + 2 + 1 + 1 (i.e., 4) appears as the largest part in 4 + 2, and vice versa. These observations lead to the first theorem in the chapter on partitions:\n\nTheorem 14.1 The number ofpartitions of $$n$$ into $$m$$ parts is equal to the number of partitions of $$n$$ into parts, the largest of which is $$m$$.\n\nThat was a brief introduction to the chapter on partitions. In the next two or so weeks, we will dive deeper into the theory of partitions.\n\n## Next Meeting\n\nIf this blog post was fun for you, consider joining our next meeting. Our next meeting is on Tue, 21 Sep 2021 at 17:00 UTC. Since we are at the beginning of a new chapter, it is a good time for new participants to join us. It is also a good time for members of the club who have been away for a while to join us back. Since this chapter does not depend much on the previous chapters, new participants should be able to join our reading sessions for this chapter and follow along easily without too much effort.\n\nTo join our book club, see our channel details in the home page here. To get the meeting link for the next meeting, visit the analytic number theory book club page.\n\nIt is worth mentioning here that lurking is absolutely fine in our meetings. In fact, most members of the club join in and stay silent throughout the meeting. Only a few members talk via audio/video or chat. This is considered absolutely normal in our meetings, so please do not hesitate to join our meetings!" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9577121,"math_prob":0.95631325,"size":5266,"snap":"2021-43-2021-49","text_gpt3_token_len":1342,"char_repetition_ratio":0.1622957,"word_repetition_ratio":0.114754096,"special_character_ratio":0.27041396,"punctuation_ratio":0.11212397,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9530624,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-21T09:37:02Z\",\"WARC-Record-ID\":\"<urn:uuid:3b9263d9-e53d-48af-b1b2-f482fc36420f>\",\"Content-Length\":\"8027\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:52a40224-d61c-46d6-8a72-b4b79d2b0102>\",\"WARC-Concurrent-To\":\"<urn:uuid:c51c6c13-8e41-40e1-b9a3-961606987982>\",\"WARC-IP-Address\":\"185.199.108.153\",\"WARC-Target-URI\":\"https://offbeat.cc/blog/our-trip-to-integer-partitions.html\",\"WARC-Payload-Digest\":\"sha1:PC3Z47IS2MY4U3RRINYZE4W3MPJDP6AN\",\"WARC-Block-Digest\":\"sha1:TDNHTPERHWDPVCIPNURL62HVDIPNEPMJ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323585382.32_warc_CC-MAIN-20211021071407-20211021101407-00015.warc.gz\"}"}
https://softwareengineering.stackexchange.com/questions/164221/understanding-binary-numbers-in-terms-of-real-world-objects	[ "# Understanding binary numbers in terms of real world objects [closed]\n\nWhen I represent a number in the decimal system, I have an intuitive knowledge of what it amounts to. For example take the number '10': I understand that it means 10 apples or 10 people... i.e I can count in the real world.\n\nBut as soon as the number is converted to any other system, this understanding no longer applies. For example 10 when converted to binary will be 1010...now what does this represent? Is there a way to understand this number 1010 in terms of counting objects in the real world?\n\n• For me, 10 apples means \"one zero\" apples. What do you mean by one zero apples? – Francisco Presencia Oct 22 '13 at 11:03\n\n\"Io sono italiano\" is the italian sentence for \"I am italian\". Barring issues with translation, the two sequence of letters mean to (a reader who is able to speak both) the same concept.\n\nSo a number expressed in the decimal system may, under some condition, represent a counting device (you can count with integer, but then think that zero is a strange number, negative numbers are even stranger, and rational and irrational and imaginary numbers also are interesting).\n\nBut the decimal system is just a notation, which happens to have gained some predominance (well, in the computing world the case could be made for binary, octal and hexadecimal). It is by no means special: a given number can be written in whatever notation you decide to use.\n\nSo 1010 (in binary notation) is precisely the same as 02^0 + 12^1 + 02^2 + 12^3 = 10 in decimal notation. You can say that humans usually have 10 fingers on their two hands (decimal notation) or 1010 finger on their 10 hands (binary notation).\n\n• in any base (in any notation) the number written as 10 is special\n• there are a lot (to use an understatement) of numbers which are properly expressed implicitly: think about e and pi which are surely very famous numbers whose representation we prefer to keep \"in letters\" rather than using a sequence of digits (which should be infinite to be precise).\n• It all comes down to practice. You can read and think fluently in any language or radix as long as you practice it. People who know morse code don't hear .-- --- .-. -.. ..., they hear \"words\". – Hand-E-Food Sep 9 '12 at 5:55\n• Morse code is a nice example of a specific notation (albeit not positional :-): you go with the flow of it rather than consciously thinking \"ok, now the sequence for S was...?\". For this reason it could happen that one is better able to \"write it\" than to \"read it\" or vice versa. – Francesco Sep 9 '12 at 5:59\n• Fun fact: You can count to 1023 with your fingers. – phant0m Sep 9 '12 at 11:28\n• Well, octal doesn't fit so well since we just about standardized on the 8-bit-byte, and it's on the decline. It's most notorious for gotchas by now. – Deduplicator Oct 16 '15 at 21:33\n\nStudents of The Hitch Hiker's Guide to the Galaxy will recall that the Ultimate Question (Answer: 42) was \"What do you get when you multiply 6 by 9\"\n\nSo, whilst the rebuilt Earth uses Base 10 (decimal) the original Earth used base 13.\n\nThis seemingly off-topic prologue is there to illustrate that our concept of numbers is because we use Base 10 (decimal) as a matter of routine. And whilst decimalisation reinforces the 10-ness, this doesn't have to be the case.\n\nAnd since this is programmers.SE, then binary (true/false) and hexadecimal (byte packing) should be equally widely understood... by that I mean, to me, A(hex) is (nearly) as understandable a concept as 10(dec)\n\nOr more esoterically, in days gone by, before grams and kilograms, we used oz/lb/st and people conceptually \"got it\" - in fact some of these other measures were more intuitive.\n\nA binary number denotes a sum of powers of two. 10102 means (from right to left) no one, a two, no four, an eight.\n\nI'll try to give a more expansive explanation of the concept of bases here.\n\nI would argue that a number is an abstract concept. When you read `10`, it is not the real thing, it is merely one possible representation of an idea/concept that you cannot truly grasp.\n\n`10` (that is, `1` followed by a `0`), is an encoding for this idea of \"ten\". Although we have may have developed some sort of intuition for this way of encoding numbers, there is a more fundamental system at work here.\n\n`143` for instance is `100 + 40 + 3`, or: 3 ones, 4 tens and 1 hundred (ten tens). We can decompose any such number by their digits. The position of a digit determines its significance, i.e. what power of ten it counts:", null, "As you can see, the digits denote the coefficients for a sum of powers of ten. We call this number (ten) the base.\n\nTen (as a concept) is not a special number. It just seems natural because we are so used to it.\n\nWe can freely choose virtually any basis we like. The meaning of each digit always depends on the base and is obtained as follows:", null, "Thus, `143` (the encoding) can have a different meaning, depending on which base is implied (usually ten of course) but the base could just as well be 11:", null, "With binary numbers, it works the exact same way, but the base is two of course.", null, "For any encoding in a certain base, you need base-many distinct symbols to encode numbers. That is, for base ten, you need ten distinct symbols, for a base-two system, you need two distinct symbols, or for hexadecimal, you need sixteen distinct symbols. Also, you need to have the idea of a \"successor\", i.e. in the decimal system, the symbol \"2\" is the successor of the symbol \"1\".\n\nIn a system that uses the the special symbols `0` and `1` (with their common meaning), `10` is always the encoding for its base.\n\"common meaning\": 0a = 0, 1a = a\n\nThis system can be used to denote any objects that are enumerable.\n\n## In real world terms - visually\n\nTry to imagine a series of objects starting at some point, continuing indefinitely.\n\nFor the purposes of this document, I'll use dots:\n\n`Zero . . . . . . . . .`\n\nEvery dot represents a number. naturally, each dot is the successor of the dot to its left.\n\nOne way to refer to a dot would be to assign a unique symbol to each dot. As you probably can imagine, without any system, this is going to be extremely cumbersome. Just to be able to count to one thousand, you would have to remember one thousand arbitrary symbols (the Chinese for example prove it's possible, but that's beside the point).\n\nLet's just go with the assumption that it's cumbersome, and provide some shortcut: I'll just replace every fourth dot with a pipe:\n\n`Zero . . . \| . . . \| . . . \| .`\n\nNow, we have simplified the problem a bit, assuming we can reliably name which pipe we mean, we can name dots relatively to a pipe. For instance, I can say: The second point after pipe X. Or the second point after the third pipe to denote the number fourteen.\n\nHowever, as numbers get bigger (or speaking visually: we zoom out, we are left with a series of pipes), it's extremely cumbersome to simply count the pipes and then refer to a dot relatively of one such pipe. We apply the same trick. We can replace every fourth pipe with an ampersand:\n\n`Zero . . . \| . . . \| . . . \| . . . & . . . \| . . . \| . . . \| . . . & . . .`\n\nNow, it becomes easier to specify a certain pipe, we simply say which ampersand we mean. Going from that, we relatively specify the pipe, and going from that, we relatively specify the dot.\n\nThis is exactly what the base-notation does. It places these markers on every n-th object, different markers on every n-th marker, etc Thus, there are never n or more of the same markers in sequence.\n\nIn this case, we have encode numbers with the base four: 1234 refers to the first ampersand, the second pipe after that, and the third dot after that: The number twenty-seven.\n\nApplying the aforementioned rule, we never need more than four distinct symbols to denote an arbitrary number in base four.\n\nProof: Let's assume we need more than four distinct symbols. This would imply, that I need to count four or more markers of the same kind from a given position. However, that is never necessary, since I replaced every fourth of these very markers with a different kind, I must be able to skip ahead. Thus, the premise is incorrect: Four distinct symbols are always enough.\n\n• Good explaination of the concept. Too often we (as in the set of programmers) make big assumptions, because we (as individual programmers) know (and understand) what is meant to be said, and therefore leave it unsaid – Andrew Sep 9 '12 at 10:46\n• @Andrew Thanks. I have added a second part trying to illustrate the concept of bases visually. – phant0m Sep 9 '12 at 10:59\n• Even better... because this just about illustrates how the Roman's (What have they done for us?) came up with their numbering... One is a I... five Is became an V, two Vs an X Five Xs became an L, two Ls a C etc So in effect, they had a binary, pentary (?) system – Andrew Sep 9 '12 at 11:07\n• @Andrew A good observation, they even put these \"markers\" as I called them (I have no idea whether there is any \"official\" terminology on this matter) explicitly into their numbers, whereas we have the implicitly by position. – phant0m Sep 9 '12 at 11:26\n\nBasically, the decimal system is based on the powers of 10 (10⁰, 10¹, 10²...), whilst the binary number system is based off of the powers of two (Place values count up by two, 2⁰, 2¹, 2²...). So here's 101 converted from binary to decimal. [(1x2³)+(0x2²)+(1x2¹)]÷2 which equals to 5. Do you notice the pattern (No, not illuminati). The exponents count down from how many numbers the binary term is, while 0 represents.. well.. a zero in the binary term, when a 1 represents a 1. 1x2 is a one in the term, and 0x2 is a zero." ]	[ null, "https://i.stack.imgur.com/v67iK.png", null, "https://i.stack.imgur.com/KB0DF.png", null, "https://i.stack.imgur.com/2GdVJ.png", null, "https://i.stack.imgur.com/oNhrt.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.93926936,"math_prob":0.94578904,"size":4465,"snap":"2020-34-2020-40","text_gpt3_token_len":1057,"char_repetition_ratio":0.11723829,"word_repetition_ratio":0.06549708,"special_character_ratio":0.25352743,"punctuation_ratio":0.1714876,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9751421,"pos_list":[0,1,2,3,4,5,6,7,8],"im_url_duplicate_count":[null,3,null,3,null,3,null,3,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-08-15T03:06:44Z\",\"WARC-Record-ID\":\"<urn:uuid:2fb0b44f-1921-4435-be3b-4651bdb067df>\",\"Content-Length\":\"168563\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:fed8ba5b-57c4-4cd0-838a-707793a22dbd>\",\"WARC-Concurrent-To\":\"<urn:uuid:1b904307-c542-433d-9abc-4893360d5c90>\",\"WARC-IP-Address\":\"151.101.65.69\",\"WARC-Target-URI\":\"https://softwareengineering.stackexchange.com/questions/164221/understanding-binary-numbers-in-terms-of-real-world-objects\",\"WARC-Payload-Digest\":\"sha1:CU3N63HZM3XIGDTA7CUHABGBAN4W2M27\",\"WARC-Block-Digest\":\"sha1:ZZD25ZP6DJ7TPD5WEV3DEZ6DI46HBXG3\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-34/CC-MAIN-2020-34_segments_1596439740423.36_warc_CC-MAIN-20200815005453-20200815035453-00112.warc.gz\"}"}
https://www.fxsolver.com/browse/?like=2679&p=150	[ "'\n\n# Search results\n\nFound 1529 matches\nStandard Equation of an Ellipse\n\nEllipse is a curve on a plane surrounding two focal points such that a straight line drawn from one of the focal points to any point on the curve and then ... more\n\nLift-to-Drag Ratio - with wetted aspect ratio\n\nIn aerodynamics, the lift-to-drag ratio, or L/D ratio, is the amount of lift generated by a wing or vehicle, divided by the drag it creates by moving ... more\n\nHorizontal curve - Sight obstraction distance (S<L)\n\nHorizontal curve – Sight Distance Properties (S<L)\n\nHorizontal Curves are one of the two important transition elements in geometric ... more\n\nDiffusion Coefficient for two different gases (related to Fick's laws)\n\nDiffusion is the net movement of a substance (e.g., an atom, ion or molecule) from a region of high concentration to a region of low concentration. For two ... more\n\nMenelaus' theorem ( transversal line passes inside triangle )\n\nMenelaus’ theorem, named for Menelaus of Alexandria, is a theorem about triangles in plane geometry. Given a triangle ABC, ... more\n\nCeva's theorem (lines from vertices to the opposite sides of a triangle)\n\nCeva’s theorem is a theorem about triangles in Euclidean plane geometry. Given a triangle ABC, let the lines AO, BO and CO ... more\n\nBall Screw - Preload Drag Torque\n\nA ball screw is a mechanical linear actuator that translates rotational motion to linear motion with little friction. A threaded shaft provides a helical ... more\n\nWorksheet 980\n\nPPI can be calculated from knowing the diagonal size of the screen in inches and the resolution in pixels (width and height). This can be done in two steps\n\nUsing the Pythagorean theorem, for 3 different screen resolutions:\n\nDiagonal Resolution - Pixels\n\nUsing the Diagonal Resolution from the previous formula we calculate the PPI for 3 corresponding screen sizes :\n\nPixels Per Inch (PPI)\n\nResults:\n\n10.1 inch tablet screen of resolution 1024×600 : 117.5PPI\n21.5 inch PC monitor of 1080p resolution : 102.46PPI\n27 inch PC monitor of 1440p resolution : 108.78PPI\n\nHorizontal curve - Sight obstraction distance (S>L)\n\nHorizontal Curves are one of the two important transition elements in geometric design for highways (along with Vertical Curves). A horizontal curve ... more\n\nKnoop hardness test\n\nThe Knoop hardness test /kəˈnuːp/ is a microhardness test – a test for mechanical hardness used particularly for very brittle materials or thin sheets, ... more\n\n...can't find what you're looking for?\n\nCreate a new formula\n\n### Search criteria:\n\nSimilar to formula\nCategory" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8732817,"math_prob":0.9599776,"size":1953,"snap":"2023-40-2023-50","text_gpt3_token_len":444,"char_repetition_ratio":0.11185223,"word_repetition_ratio":0.073846154,"special_character_ratio":0.23195085,"punctuation_ratio":0.16153847,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9746103,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-09T12:41:26Z\",\"WARC-Record-ID\":\"<urn:uuid:40190a40-ec40-4aba-9402-08e3c9e1bf1f>\",\"Content-Length\":\"206483\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:1892e958-2d2c-4939-ad0f-670d6becb4c4>\",\"WARC-Concurrent-To\":\"<urn:uuid:b6ac0c90-1aa6-41e7-84cf-7c557ecc25c9>\",\"WARC-IP-Address\":\"178.254.54.75\",\"WARC-Target-URI\":\"https://www.fxsolver.com/browse/?like=2679&p=150\",\"WARC-Payload-Digest\":\"sha1:MV6YPV22WYML7ETKAZO3FCNS2GPGOHAA\",\"WARC-Block-Digest\":\"sha1:Z46WHHBD5AAPZ74QZPGCJQKO7LJIIITI\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100909.82_warc_CC-MAIN-20231209103523-20231209133523-00303.warc.gz\"}"}
https://www.cbseguess.com/papers/question_papers/xii/2004/2004_economics_outside_delhi_set1.php	[ "Wednesday 26th February 2020\n\nCBSE Guess > Papers > Question Papers > Class XII > 2004 > Economics > Outside Delhi Set-I\n\nECONOMICS (Set I—Outside Delhi)\n\nSECTION – A\n\nQ. 1. Answer the following questions: 1x4\n(i) What is meant by price elasticity of demand?\n(ii) In which market form are the products homogeneous?\n(iii) Define marginal revenue.\n(iv) State the law of supply.\n\nQ. 2. Mention any three factors that affect the price elasticity of demand of a commodity. 3\n\nQ. 3. Distinguish between ‘change in demand’ and ‘change in quantity demanded’ of a commodity. 3\n\nQ. 4. List any three determinants of supply of a commodity. 3\n\nQ. 5. State any three main features of monopolistic competition. Describe any one. 3\n\nQ. 6. The quantity supplied of a commodity at a price of Rs. 8 per unit is 400 units. Its price elasticity of supply is 2. Calculate the price at which its quantity supplied will be 600 units. 4\n\nQ. 7. Complete the following table: 4\n\n Output (unit) Price (Rs.) Total Revenue (Rs.) Margial Revenue (Rs.) 1 2 3 4 7 6 4 2 — — — — — — — —\n\nQ. 8. Explain the relationship between marginal cost and average cost with the help of a cost schedule. 4\nOr\nDistinguish between fixed costs and variable costs. Give two examples of each.\n\nQ. 9. What are the three central problems of an economy? Why do they arise? 4\n\nQ. 10. Explain with the help of diagrams the effect of the following changes on the demand of a commodity: 6\n(i) A fall in the price of complementary good\n(ii) A rise in the income of its buyer\n\nQ. 11. Explain the law of variable proportions with the help of total product and marginal product curves. 6\n\nQ. 12. If at a given price of a commodity there is, excess supply, how will the equilibrium price be reached? Explain with the help of a diagram. 6\nOr\nExplain the effect of a leftward shift of demand curve of a commodity on its equilibrium price and quantity, with the help of a diagram.\n\nSECTION - B\n\nQ. 13. Answer the following questions: 1x4\n(i) What is macro-economics?\n(ii) Give an example of a micro-economic study.\n(iii) What is meant by fiscal deficit?\n(iv) When is there a deficit in the balance of trade?\n\nQ. 14. What is meant by revenue deficit? What are the implications of this deficit? 3\n\nQ. 15. Calculate Net National Disposable Income from the following data: 3\n\n Rs. (Crores) (i) Gross national product at factor cost (ii) Net current transfers from rest of the world (iii) Net indirect tax (iv) Consumption of fixed capital (v) Net factor income from abroad 800 50 70 60 (-) 10\n\nQ. 16. Give the meaning of marginal propensity to save and aver age propensity to save. Can the value of average propensity to save benegative? If yes, when?3\n\nQ. 17. In an economy, the marginal propensity to consume is 0.75. Investment is increased by Rs. 200 crores. Calculate the total increase in income and consumption expenditures. 3\n\nQ. 18. How does a central bank control the availability of credit by open market operations? Explain. 4\n\nQ. 19. Explain briefly any two objectives of a government budget. 4\n\nQ. 20. State the four functions of money. Describe any one. 4\n\nQ. 21. Distinguish between current account and capital account of balance of payments account. Mention any two transactions of capital account. 4\nOr\nHow is the foreign exchange market rate determined? Explain with the help of a diagram.\n\nQ. 22. From the following data calculate National Income by (i) income method and (ii) expenditure method: 3, 3\n\n Rs. (Crores) (i) Compensation of employees (ii) Net factor income from abroad (iii) Net indirect tax (iv) Profits (v) Private final consumption expenditure (vi) Net domestic capital formation (vii) Consumption of fixed capital (viii) Rent (ix) Interest (x) Mixed income of self-employed (xi) Net exports (xii) Government final consumption expenditure 1,200 (-) 20 120 800 2,000 770 130 400 620 700 (-) 30 1, 100\n\nQ. 23. Will the following be included in domestic factor Income of India? Give reasons for your answer. 6\n(i) Profits earned by a foreign bank from its branches Ii India.\n(ii) Scholarships given by Government of India.\n(iii) Profits earned by a resident of India from his company In Singapore.\n(iv) Salaries received by Indians working In American Embassy in India.\n\nQ. 24. Explain the concept of under-employment equilibrium with the help of a diagram. Show on the same diagram the additional in vestment expenditure required to reach full employment equilibrium. 6\nOr\nExplain the equilibrium level of income with the help of Consumption + Investment (C + I) curve If planned saving is greater than planned Investment, what adjustments will bring about equality between the two?\n\n Economics 2004 Question Papers Class XII Delhi Outside Delhi Compartment Delhi Compartment Outside Delhi", null, "Set I", null, "Set I", null, "Set I", null, "Set I", null, "Set II", null, "Set II", null, "Set II", null, "Set II", null, "Set III", null, "Set III\n\n CBSE 2004 Question Papers Class XII", null, "English", null, "Sociology", null, "Functional English", null, "Psychology", null, "Mathematics", null, "Philosopy", null, "Physics", null, "Computer Science", null, "Chemistry", null, "Entrepreneurship", null, "Biology", null, "Informatics Practices", null, "Geography", null, "Multimedia & Web Technology", null, "Economics", null, "Biotechnology", null, "Business Studies", null, "Physical Education", null, "Accountancy", null, "Fine Arts", null, "Political Science", null, "History", null, "Agriculture" ]	[ null, "https://www.cbseguess.com/images/arrow1.gif", null, "https://www.cbseguess.com/images/arrow1.gif", null, "https://www.cbseguess.com/images/arrow1.gif", null, "https://www.cbseguess.com/images/arrow1.gif", null, "https://www.cbseguess.com/images/arrow1.gif", null, "https://www.cbseguess.com/images/arrow1.gif", null, "https://www.cbseguess.com/images/arrow1.gif", null, "https://www.cbseguess.com/images/arrow1.gif", null, "https://www.cbseguess.com/images/arrow1.gif", null, "https://www.cbseguess.com/images/arrow1.gif", null, "https://www.cbseguess.com/images/bullet1.gif", null, "https://www.cbseguess.com/images/bullet1.gif", null, "https://www.cbseguess.com/images/bullet1.gif", null, "https://www.cbseguess.com/images/bullet1.gif", null, "https://www.cbseguess.com/images/bullet1.gif", null, "https://www.cbseguess.com/images/bullet1.gif", null, "https://www.cbseguess.com/images/bullet1.gif", null, "https://www.cbseguess.com/images/bullet1.gif", null, "https://www.cbseguess.com/images/bullet1.gif", null, "https://www.cbseguess.com/images/bullet1.gif", null, "https://www.cbseguess.com/images/bullet1.gif", null, "https://www.cbseguess.com/images/bullet1.gif", null, "https://www.cbseguess.com/images/bullet1.gif", null, "https://www.cbseguess.com/images/bullet1.gif", null, "https://www.cbseguess.com/images/bullet1.gif", null, "https://www.cbseguess.com/images/bullet1.gif", null, "https://www.cbseguess.com/images/bullet1.gif", null, "https://www.cbseguess.com/images/bullet1.gif", null, "https://www.cbseguess.com/images/bullet1.gif", null, "https://www.cbseguess.com/images/bullet1.gif", null, "https://www.cbseguess.com/images/bullet1.gif", null, "https://www.cbseguess.com/images/bullet1.gif", null, "https://www.cbseguess.com/images/bullet1.gif", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.88583916,"math_prob":0.7540444,"size":2653,"snap":"2020-10-2020-16","text_gpt3_token_len":681,"char_repetition_ratio":0.1208003,"word_repetition_ratio":0.029978586,"special_character_ratio":0.25329816,"punctuation_ratio":0.16934046,"nsfw_num_words":1,"has_unicode_error":false,"math_prob_llama3":0.9826251,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50,51,52,53,54,55,56,57,58,59,60,61,62,63,64,65,66],"im_url_duplicate_count":[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,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-02-26T08:01:00Z\",\"WARC-Record-ID\":\"<urn:uuid:26d1e845-5368-4526-bec8-e920154ea5e2>\",\"Content-Length\":\"29750\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:5056a738-fe96-4f10-851f-3bcaf3b2a3fc>\",\"WARC-Concurrent-To\":\"<urn:uuid:530489a7-0a3f-4d8c-a9fa-f36a04f69ea9>\",\"WARC-IP-Address\":\"104.28.8.45\",\"WARC-Target-URI\":\"https://www.cbseguess.com/papers/question_papers/xii/2004/2004_economics_outside_delhi_set1.php\",\"WARC-Payload-Digest\":\"sha1:HLJ44VYIPS2RGROFELOAVCT2U6EFOLG6\",\"WARC-Block-Digest\":\"sha1:6KSRLBVTHG553XKAL4Y5Y57NY5TC7W3O\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-10/CC-MAIN-2020-10_segments_1581875146187.93_warc_CC-MAIN-20200226054316-20200226084316-00232.warc.gz\"}"}
https://www.colorhexa.com/03839f	[ "# #03839f Color Information\n\nIn a RGB color space, hex #03839f is composed of 1.2% red, 51.4% green and 62.4% blue. Whereas in a CMYK color space, it is composed of 98.1% cyan, 17.6% magenta, 0% yellow and 37.6% black. It has a hue angle of 190.8 degrees, a saturation of 96.3% and a lightness of 31.8%. #03839f color hex could be obtained by blending #06ffff with #00073f. Closest websafe color is: #009999.\n\n• R 1\n• G 51\n• B 62\nRGB color chart\n• C 98\n• M 18\n• Y 0\n• K 38\nCMYK color chart\n\n#03839f color description : Dark cyan.\n\n# #03839f Color Conversion\n\nThe hexadecimal color #03839f has RGB values of R:3, G:131, B:159 and CMYK values of C:0.98, M:0.18, Y:0, K:0.38. Its decimal value is 230303.\n\nHex triplet RGB Decimal 03839f `#03839f` 3, 131, 159 `rgb(3,131,159)` 1.2, 51.4, 62.4 `rgb(1.2%,51.4%,62.4%)` 98, 18, 0, 38 190.8°, 96.3, 31.8 `hsl(190.8,96.3%,31.8%)` 190.8°, 98.1, 62.4 009999 `#009999`\nCIE-LAB 50.398, -19.585, -23.38 14.41, 18.754, 35.659 0.209, 0.272, 18.754 50.398, 30.499, 230.047 50.398, -35.839, -32.235 43.306, -16.388, -18.508 00000011, 10000011, 10011111\n\n# Color Schemes with #03839f\n\n• #03839f\n``#03839f` `rgb(3,131,159)``\n• #9f1f03\n``#9f1f03` `rgb(159,31,3)``\nComplementary Color\n• #039f6d\n``#039f6d` `rgb(3,159,109)``\n• #03839f\n``#03839f` `rgb(3,131,159)``\n• #03359f\n``#03359f` `rgb(3,53,159)``\nAnalogous Color\n• #9f6d03\n``#9f6d03` `rgb(159,109,3)``\n• #03839f\n``#03839f` `rgb(3,131,159)``\n• #9f0335\n``#9f0335` `rgb(159,3,53)``\nSplit Complementary Color\n• #839f03\n``#839f03` `rgb(131,159,3)``\n• #03839f\n``#03839f` `rgb(3,131,159)``\n• #9f0383\n``#9f0383` `rgb(159,3,131)``\n• #039f1f\n``#039f1f` `rgb(3,159,31)``\n• #03839f\n``#03839f` `rgb(3,131,159)``\n• #9f0383\n``#9f0383` `rgb(159,3,131)``\n• #9f1f03\n``#9f1f03` `rgb(159,31,3)``\n• #024554\n``#024554` `rgb(2,69,84)``\n• #025a6d\n``#025a6d` `rgb(2,90,109)``\n• #036e86\n``#036e86` `rgb(3,110,134)``\n• #03839f\n``#03839f` `rgb(3,131,159)``\n• #0398b8\n``#0398b8` `rgb(3,152,184)``\n• #04acd1\n``#04acd1` `rgb(4,172,209)``\n• #04c1ea\n``#04c1ea` `rgb(4,193,234)``\nMonochromatic Color\n\n# Alternatives to #03839f\n\nBelow, you can see some colors close to #03839f. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #039f94\n``#039f94` `rgb(3,159,148)``\n• #039d9f\n``#039d9f` `rgb(3,157,159)``\n• #03909f\n``#03909f` `rgb(3,144,159)``\n• #03839f\n``#03839f` `rgb(3,131,159)``\n• #03769f\n``#03769f` `rgb(3,118,159)``\n• #03699f\n``#03699f` `rgb(3,105,159)``\n• #035c9f\n``#035c9f` `rgb(3,92,159)``\nSimilar Colors\n\n# #03839f Preview\n\nThis text has a font color of #03839f.\n\n``<span style=\"color:#03839f;\">Text here</span>``\n#03839f background color\n\nThis paragraph has a background color of #03839f.\n\n``<p style=\"background-color:#03839f;\">Content here</p>``\n#03839f border color\n\nThis element has a border color of #03839f.\n\n``<div style=\"border:1px solid #03839f;\">Content here</div>``\nCSS codes\n``.text {color:#03839f;}``\n``.background {background-color:#03839f;}``\n``.border {border:1px solid #03839f;}``\n\n# Shades and Tints of #03839f\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, #000405 is the darkest color, while #f1fcff is the lightest one.\n\n• #000405\n``#000405` `rgb(0,4,5)``\n• #001418\n``#001418` `rgb(0,20,24)``\n• #01242b\n``#01242b` `rgb(1,36,43)``\n• #01343f\n``#01343f` `rgb(1,52,63)``\n• #024452\n``#024452` `rgb(2,68,82)``\n• #025365\n``#025365` `rgb(2,83,101)``\n• #026378\n``#026378` `rgb(2,99,120)``\n• #03738c\n``#03738c` `rgb(3,115,140)``\n• #03839f\n``#03839f` `rgb(3,131,159)``\n• #0393b2\n``#0393b2` `rgb(3,147,178)``\n• #04a3c6\n``#04a3c6` `rgb(4,163,198)``\n• #04b3d9\n``#04b3d9` `rgb(4,179,217)``\n• #04c2ec\n``#04c2ec` `rgb(4,194,236)``\n• #0acffa\n``#0acffa` `rgb(10,207,250)``\n• #1dd3fb\n``#1dd3fb` `rgb(29,211,251)``\n• #30d7fb\n``#30d7fb` `rgb(48,215,251)``\n• #43dafb\n``#43dafb` `rgb(67,218,251)``\n• #57defc\n``#57defc` `rgb(87,222,252)``\n• #6ae2fc\n``#6ae2fc` `rgb(106,226,252)``\n• #7de6fd\n``#7de6fd` `rgb(125,230,253)``\n• #90e9fd\n``#90e9fd` `rgb(144,233,253)``\n• #a4edfd\n``#a4edfd` `rgb(164,237,253)``\n• #b7f1fe\n``#b7f1fe` `rgb(183,241,254)``\n• #caf5fe\n``#caf5fe` `rgb(202,245,254)``\n• #ddf8fe\n``#ddf8fe` `rgb(221,248,254)``\n• #f1fcff\n``#f1fcff` `rgb(241,252,255)``\nTint Color Variation\n\n# Tones of #03839f\n\nA tone is produced by adding gray to any pure hue. In this case, #4e5354 is the less saturated color, while #03839f is the most saturated one.\n\n• #4e5354\n``#4e5354` `rgb(78,83,84)``\n• #48575a\n``#48575a` `rgb(72,87,90)``\n• #415b61\n``#415b61` `rgb(65,91,97)``\n• #3b5f67\n``#3b5f67` `rgb(59,95,103)``\n• #35636d\n``#35636d` `rgb(53,99,109)``\n• #2f6773\n``#2f6773` `rgb(47,103,115)``\n• #286b7a\n``#286b7a` `rgb(40,107,122)``\n• #226f80\n``#226f80` `rgb(34,111,128)``\n• #1c7386\n``#1c7386` `rgb(28,115,134)``\n• #16778c\n``#16778c` `rgb(22,119,140)``\n• #0f7b93\n``#0f7b93` `rgb(15,123,147)``\n• #097f99\n``#097f99` `rgb(9,127,153)``\n• #03839f\n``#03839f` `rgb(3,131,159)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #03839f 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":0.5506238,"math_prob":0.81108797,"size":3685,"snap":"2019-43-2019-47","text_gpt3_token_len":1626,"char_repetition_ratio":0.1252377,"word_repetition_ratio":0.011111111,"special_character_ratio":0.56580734,"punctuation_ratio":0.23809524,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9920629,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-11-12T01:58:42Z\",\"WARC-Record-ID\":\"<urn:uuid:1b927c1e-f7be-45c7-9914-15319a03a164>\",\"Content-Length\":\"36252\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b49fefee-2e29-43b1-891e-e0065bfd2d32>\",\"WARC-Concurrent-To\":\"<urn:uuid:b985d6f3-a584-45fa-80ff-9440178d39fc>\",\"WARC-IP-Address\":\"178.32.117.56\",\"WARC-Target-URI\":\"https://www.colorhexa.com/03839f\",\"WARC-Payload-Digest\":\"sha1:2BTZXFUPJMDMZT64MOJIDZGNZ4ZE2WRK\",\"WARC-Block-Digest\":\"sha1:PTCACY7WO3QHN5B5SEFVMSXOSLZBOHCB\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-47/CC-MAIN-2019-47_segments_1573496664469.42_warc_CC-MAIN-20191112001515-20191112025515-00309.warc.gz\"}"}
https://hackaday.io/project/11732-code	[ "0%\n0%\n\n# µδ code\n\nA binary maths hack, for use in DSP code and lossless compression\n\nSimilar projects worth following\n172 views\nFor your DSP/DCT/FFT needs, µδ codes are the next enhancement, beyond the lifting scheme and dyadic multiplies.\nAs far as I know, it has not yet been documented openly or used because most of the research is focused on lossy compression (think: MPEG).\nµδ codes make DSP algos more bit-efficient and will be used in my lossless video and audio CODECs, along with the 3R compaction algorithm ( https://hackaday.io/project/7385-recursive-range-reduction-3r-hwsw-codec )\nMore details will appear here soon :-)\n\nA µδ CODEC adds only one XOR gate to the critical datapath of a circuit that computes the simultaneous sum and difference of two integer numbers. The transform is perfectly bit-reversible and the reverse transform is almost identical to the classic version.\n\nThe two results (µ and δ) are as large as the original numbers, whereas a classical sum&difference expands the data by 2 bits (yet the result can only use 1/4 of the coding space).\n\nThe results are distorted : the MSB of µ gets disturbed by the sign of δ, which is simply truncated. The bet is that this distorsion is not critical for certain types of lossless compression CODECs, while reducing the size and consumption of hardware circuits.\n\nYann Guidon / YGDES05/24/2017 at 18:12 0 comments\n\nLook at these new logs:\n\nThe have some similarities with µδ but they differ in critical ways:\n\n• µδ works with median and difference while the new transform creates a sum (σ) and one of the operands (let's call it α). The new transform shall then called σα.\n• µδ works inside a ring (a closed set with wrap-around) but σα \"escapes\" this (the output is not constrained inside the original boundaries), which allows actual compression for some combinations of values.\n• actually σα has an initial value range that is a ring, however each operand can have its own original ring, they both are combined and the boundaries remain as auxiliary data in the background.\n• σα is used with VLC (variable length codes) while µδ uses fixed-length words.\n\nσα already has a fantastic application in the color decomposition of RGB pictures and will make 3R a truly entropy coder... Follow the developments there !\n\n• ### X+X = 2X\n\nYann Guidon / YGDES07/09/2016 at 16:01 0 comments\n\nBefore we start, let's just remember and examine one of the fundamental rules of arithmetics. If you take 2 non-negative numbers and add them, the sum is equal or greater than any of the addends. This is the rule that makes the #Recursive Range Reduction (3R) HW&SW CODEC work.\n\nIn a more general way, if you have two numbers of N bits, you need N+1 bits to uniquely represent the sum. Since X+X=2X, 2^N + 2^N = 2^(N+1). This +1 is the carry bit and is required to restore the original addends if you ever want to subtract it back from the other.\n\nSame goes for the subtraction : the difference requires a \"borrow\" bit. You can't avoid this, even though there are some cases where you can work modulo 2^N.\n\nWe are used to dealing with the carry and borrow bits but things can quickly get out of hand! Imagine you want to compute a 1024-tap integer FFT: each result will be the sum of 2^10 numbers, adding 10 bits to the original sample size. If you're dealing with CD quality, 16+10=26 bits so it fits in the 32-bits registers of common CPUs or DSPs.\n\nNow if you want to use 24-bits samples, you're screwed. 34 bits don't easily fit in 32-bits registers. Will you resort to slower 32-bits floating points ? 40-bits integers ? 64-bits integers ?\n\nTake the classical 8×8 DCT square now. The original 8-bits samples of a picture get fatter during each of the 3+3=6 passes, resulting in 14 bits for the results. The integer units have almost doubled the precision of the source data and this considerably increases gate count, latency, power consumption...\n\nNow you start to see where I'm getting to : the classical filters and time-frequency transforms have a fundamental problem of size.\n\n• ### First publication\n\nYann Guidon / YGDES07/08/2016 at 08:11 8 comments\n\nOpenSilicium#19 is out !\n\nI'll publish the code soon.\n\n• 1\nStep 1\n\n# Step 1: encoding\n\n• Take A and B, two signed numbers made of N bits\n• Compute the sum µ = A + B (with N+1 resolution)\n• Compute the difference δ = A - B (with N+1 bits of resolution)\n• Adjust the sign bit : µ = µ xor MSB(δ)\n• Remove the LSB of µ : µ = µ shr 1 (µ now uses N bits)\n• Remove the MSB of δ : δ = δ mod 2^N (δ now uses N bits)\n• 2\nStep 2\n\n# Step 2: decoding\n\n• Take the coded values µ and δ, with N bits each\n• Restore the LSB of µ by copying it from δ: µ = (µ shl 1) + (δ mod 2)\n• Restore A but drop the MSB: A = (µ + δ) mod 2^N\n• Restore B but drop the MSB: B = (µ - δ) mod 2^N\n\nShare\n\n## Discussions\n\nYann Guidon / YGDES wrote 06/30/2016 at 16:31 point\n\nThe founding article will be published in french kiosks in a few days now :-) I can't wait !\n\nAre you sure? yes \| no\n\n# Does this project spark your interest?\n\nBecome a member to follow this project and never miss any updates", null, "" ]	[ null, "https://analytics.supplyframe.com/trackingservlet/impression", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.91293204,"math_prob":0.90107185,"size":867,"snap":"2020-34-2020-40","text_gpt3_token_len":194,"char_repetition_ratio":0.09733488,"word_repetition_ratio":0.0,"special_character_ratio":0.21453287,"punctuation_ratio":0.08928572,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98446065,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-09-24T05:25:31Z\",\"WARC-Record-ID\":\"<urn:uuid:173756c6-0275-4145-9a45-53d9924da217>\",\"Content-Length\":\"93836\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:1c31cb8e-2dae-4650-bee5-2222dbef97c1>\",\"WARC-Concurrent-To\":\"<urn:uuid:83d1c239-8e2e-48a4-814d-35e5d4eb97de>\",\"WARC-IP-Address\":\"198.54.96.98\",\"WARC-Target-URI\":\"https://hackaday.io/project/11732-code\",\"WARC-Payload-Digest\":\"sha1:OR3AQLCZ5CPT7BMZEM6LWZ3MTBXZCSLT\",\"WARC-Block-Digest\":\"sha1:WTRWO3H6LAEXFWGQYHTBWOHBRRYY6AKA\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-40/CC-MAIN-2020-40_segments_1600400213454.52_warc_CC-MAIN-20200924034208-20200924064208-00609.warc.gz\"}"}
https://learn.microsoft.com/en-us/cpp/cpp/tile-static-keyword?view=msvc-170	[ "# tile_static Keyword\n\nThe tile_static keyword is used to declare a variable that can be accessed by all threads in a tile of threads. The lifetime of the variable starts when execution reaches the point of declaration and ends when the kernel function returns. For more information on using tiles, see Using Tiles.\n\nThe tile_static keyword has the following limitations:\n\n• It can be used only on variables that are in a function that has the `restrict(amp)` modifier.\n\n• It cannot be used on variables that are pointer or reference types.\n\n• A tile_static variable cannot have an initializer. Default constructors and destructors are not invoked automatically.\n\n• The value of an uninitialized tile_static variable is undefined.\n\n• If a tile_static variable is declared in a call graph that is rooted by a non-tiled call to `parallel_for_each`, a warning is generated and the behavior of the variable is undefined.\n\n## Example\n\nThe following example shows how a tile_static variable can be used to accumulate data across several threads in a tile.\n\n``````// Sample data:\nint sampledata[] = {\n2, 2, 9, 7, 1, 4,\n4, 4, 8, 8, 3, 4,\n1, 5, 1, 2, 5, 2,\n6, 8, 3, 2, 7, 2};\n\n// The tiles:\n// 2 2 9 7 1 4\n// 4 4 8 8 3 4\n//\n// 1 5 1 2 5 2\n// 6 8 3 2 7 2\n\n// Averages:\nint averagedata[] = {\n0, 0, 0, 0, 0, 0,\n0, 0, 0, 0, 0, 0,\n0, 0, 0, 0, 0, 0,\n0, 0, 0, 0, 0, 0,\n};\n\narray_view<int, 2> sample(4, 6, sampledata);\narray_view<int, 2> average(4, 6, averagedata);\n\nparallel_for_each(\n// Create threads for sample.extent and divide the extent into 2 x 2 tiles.\nsample.extent.tile<2,2>(),\n[=](tiled_index<2,2> idx) restrict(amp)\n{\n// Create a 2 x 2 array to hold the values in this tile.\ntile_static int nums;\n// Copy the values for the tile into the 2 x 2 array.\nnums[idx.local][idx.local] = sample[idx.global];\n// When all the threads have executed and the 2 x 2 array is complete, find the average.\nidx.barrier.wait();\nint sum = nums + nums + nums + nums;\n// Copy the average into the array_view.\naverage[idx.global] = sum / 4;\n}\n);\n\nfor (int i = 0; i < 4; i++) {\nfor (int j = 0; j < 6; j++) {\nstd::cout << average(i,j) << \" \";\n}\nstd::cout << \"\\n\";\n}\n\n// Output:\n// 3 3 8 8 3 3\n// 3 3 8 8 3 3\n// 5 5 2 2 4 4\n// 5 5 2 2 4 4\n// Sample data.\nint sampledata[] = {\n2, 2, 9, 7, 1, 4,\n4, 4, 8, 8, 3, 4,\n1, 5, 1, 2, 5, 2,\n6, 8, 3, 2, 7, 2};\n\n// The tiles are:\n// 2 2 9 7 1 4\n// 4 4 8 8 3 4\n//\n// 1 5 1 2 5 2\n// 6 8 3 2 7 2\n\n// Averages.\nint averagedata[] = {\n0, 0, 0, 0, 0, 0,\n0, 0, 0, 0, 0, 0,\n0, 0, 0, 0, 0, 0,\n0, 0, 0, 0, 0, 0,\n};\n\narray_view<int, 2> sample(4, 6, sampledata);\narray_view<int, 2> average(4, 6, averagedata);\n\nparallel_for_each(\n// Create threads for sample.grid and divide the grid into 2 x 2 tiles.\nsample.extent.tile<2,2>(),\n[=](tiled_index<2,2> idx) restrict(amp)\n{\n// Create a 2 x 2 array to hold the values in this tile.\ntile_static int nums;\n// Copy the values for the tile into the 2 x 2 array.\nnums[idx.local][idx.local] = sample[idx.global];\n// When all the threads have executed and the 2 x 2 array is complete, find the average.\nidx.barrier.wait();\nint sum = nums + nums + nums + nums;\n// Copy the average into the array_view.\naverage[idx.global] = sum / 4;\n}\n);\n\nfor (int i = 0; i < 4; i++) {\nfor (int j = 0; j < 6; j++) {\nstd::cout << average(i,j) << \" \";\n}\nstd::cout << \"\\n\";\n}\n\n// Output.\n// 3 3 8 8 3 3\n// 3 3 8 8 3 3\n// 5 5 2 2 4 4\n// 5 5 2 2 4 4\n``````" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.555115,"math_prob":0.9971537,"size":3549,"snap":"2023-40-2023-50","text_gpt3_token_len":1328,"char_repetition_ratio":0.13822284,"word_repetition_ratio":0.6566951,"special_character_ratio":0.41983658,"punctuation_ratio":0.22888888,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9610016,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-22T09:46:32Z\",\"WARC-Record-ID\":\"<urn:uuid:9a7f8a4a-067f-43e8-aae3-e4f2cc28e886>\",\"Content-Length\":\"45744\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:27164cf6-bf8e-473f-a7b5-418c1f482b05>\",\"WARC-Concurrent-To\":\"<urn:uuid:748cdf89-9827-498d-9384-231a49d73ce4>\",\"WARC-IP-Address\":\"23.50.126.168\",\"WARC-Target-URI\":\"https://learn.microsoft.com/en-us/cpp/cpp/tile-static-keyword?view=msvc-170\",\"WARC-Payload-Digest\":\"sha1:KHHBZT2RB2B6FL54CVYGPTQB3NPCOBP5\",\"WARC-Block-Digest\":\"sha1:UVBBVSAHI3FRX3SX4REEJJYSORA5F4HY\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233506339.10_warc_CC-MAIN-20230922070214-20230922100214-00749.warc.gz\"}"}
https://www.brightstorm.com/science/chemistry/kinetic-molecular-theory/ideal-gas-law/	[ "", null, "###### Kendal Orenstein\n\nRutger's University\nM.Ed., Columbia Teachers College\n\nKendal founded an academic coaching company in Washington D.C. and teaches in local area schools. In her spare time she loves to explore new places.\n\n##### Thank you for watching the video.\n\nTo unlock all 5,300 videos, start your free trial.\n\n# Ideal Gas Law - Concept\n\nKendal Orenstein", null, "###### Kendal Orenstein\n\nRutger's University\nM.Ed., Columbia Teachers College\n\nKendal founded an academic coaching company in Washington D.C. and teaches in local area schools. In her spare time she loves to explore new places.\n\nShare\n\nThe Ideal Gas Law mathematically relates the pressure, volume, amount and temperature of a gas with the equation pressure x volume = moles x ideal gas constant x temperature; PV=nRT. The Ideal Gas Law is ideal because it ignores interactions between the gas particles in order to simplify the equation. There is also a Real Gas Law which is much more complicated and produces a result which, under most circumstances, is almost identical to that predicted by the Ideal Gas Law.\n\nAlright. So we're going to talk about the ideal gas law and one word that might stick out to you is the word ideal. So we're going to talk about ideal gases versus real gases and gases in real life.\n\nSo ideally, when thinking about this and the kinetic molcular theory, there are two things postulates within that kinetic molcular theory that were kind of iffy or untrue. One of them was that gas particles have virtually no volume and we're going to basically have calculations which they do volume of the gas particles does not count. They're negligible. We know that's not true. Gas particles do have some sort of volume. We know it's really small and we know that a ga- real gas particles actually do contain some sort of volume. And also ideally in an ideal world, we're saying that gas particles have no intermolcular forces meaning that they don't attract or repel from each other and we know that's not true. All gas particles and all molcules actually have some sort of way that they are attracted or repelling from each other called IMFs. So really that's not really true. But in most cases, real gases actually behave extremely similarly to ideal gases and so we can actually use this ideal gas conditions in when we're making our calculations, they're pretty accurate. However, the only time that they are not accurate is when we're dealing with high pressure situations or low temperature situations. The reason high pressure situations are different is because most when you have high pressure, those gas particles they're being pushed together. And these intermolcular forces are going to start playing a major role. Also when you're dealing with low temperatures, those gas particles will start slowing down and they will start containing some sort of volume that is negligible and these IMFs will start playing a part too. So in these two scenarios, we can't use the ideal conditions. Otherwise, we can use them all the time which we are going to start doing.\n\nAlright. So let's compare, let's say you have a gas. Using our combined gas law, let's say you have a gas and you don't have anything to compare it to. We can compare, we can always compare the gas law, any gas to its conditions at STP or standard temperature and pressure. And don't forget these conditions are one atmosphere or 101.3 kilo pascals or 760 milimetres of mercury for your pressure. Our molar volume will always be no matter what gas we're talking about 22.4 litres and our temperature will always be 273 kelvin or 0 degrees celsius. But we like things in kelvin because it's always positive that way.\n\nSo if we were to replace one of these guys, one of these pressure times volume over temperature with our conditions at STP, we get a certain number. And depending on our pressure, whatever unit of pressure we're talking about, we get different numbers. So we get, if we're using atmospheres we get 0.0821, if we're dealing with kilo pascals we get 8.314, if we're dealing with milimetres of mercury we get 62.4. And we're going to, this is always, will always be the case. We're just going ot make it a constant. And we're going to use the letter r to denote that constant. So, when we're dealing with, I'm going to grab a pen. When we're dealing with the combined gas law we want to actually compare something to a situation in a at STP, I can just replace this pv over t with the letter r. So in this case I'm going to say p1 times v1 over t1 equals r. Okay. Great. But let's say we're talking about one mol of gas. We're talking about one mol of gas, this all works fine because at one mol of gas, our volume is 22.4 litres.\n\nGreat. But what if we're talking about two mols of gas? We're going to have to multiply this number by two because we're multiplying that 22.4 litres by 2. Let's say we're talking about 1000 litres. 1000 mols. We have to multiply this whole thing by 1000. If we're talking about 0.55 mols, multiply this whole thing by 0.55. So multiply by the number of mols that we have in our sample.\n\nSo we're going to say the letter n denotes the number of mols we have in our sample. Okay, so if I rearrange this to make it much more easy to, easy to write down or remember, I'm going to rearrange this and bring the t over. So I'm going to say pv=nrt. Some people call it pivnert to remember the ideal gas law. So this combination of things is the ideal gas law. It's basically just the combined gas law we've rearranged using conditions at STP for r. Okay. So this actually uses, the ideal gas law uses all four variables. We have volume, pressure, number of mols and temperature. This number of mols, it's the first time it's been introduced. The other gas laws don't have the number of mols within them. So this actually is very very useful. Let's go over here and do a problem with them.\n\nAlright, so our problem is what is the pressure in atmospheres of a 0.108 mol sample of helium gas at a temperature of 20 degrees celsius if its volume is 0.505 litres. Okay. I know right away this is an ideal gas law problem. how do I know that? Well, my problem had a number of mols in it. Now, remember I told you ideal gas laws, the only gas laws that actually can contain the number of mols within it, within it. So, I know when I'm given a mol sample or I'm asked about the number of mols, I know I'm going to always be using the ideal gas law.\n\nOkay. So let's pull everything out. We have ideal gas law just to rewrite is pv=nrt. So our pressure in this case, we're looking for. We're looking for the pressure. So I'm going to say pressure is our variable. Our volume in this case is 0.505 litres. The number of mols we found is 0.108 mols and the r that we're going to use three r's to choose from. We're going to use r dealing with atmospheres. So pressure is wanted in the unit of atmospheres. So where r unit of atmosphere is this point 0821. And then our temperature in this case is 20 degrees celsius. Don't forget we always want things in kelvin. So we're going to add 273 to make it 293 kelvin. So if we multiplied all these together and divide by 5, sorry, 0.505 to isolate our pressure value, we're actually going to get 5.14 atmospheres. So we've just found out using ideal gas law that our new pressure or that our pressure in this scenario is 5.14 atmospheres. Awesome. Great. So this actually can be used, all you have to do is that make, you have to make sure that you get all your variables out and just plug this within this gas, within this ideal gas law.\n\nBut there's also lots of fun things you can do with the ideal gas law and one of them is to find the density of the gas. So we know density is mass over volume or grams over litres. And actually, if you rearranged the ideal gas law you're going to get this to find density. Let's actually do that together. Let's derive this. Alright. So we know what density is. So we know our mol grams per mol is equal to our molar mss. And I'm going to just substitute m m as molar mass because it's not milimetres. I'm just going to say, shorten it instead of having to write molar mass. Okay.\n\nI'm going to isolate mols because in my ideal gas law, I have mols. So I'm going to isolate mols to just figure out what it solves for. So I'm going to multiply both sides by mol. So we know our grams equals molar mass times mol. But I want to isolate that mol so I'm going to divide by molar mass on both sides to cancel. So when I, our mol equals grams per molar mass. So in our ideal gas law, I'm going to substitute this n for grams per molar mass. And I have pressure times volume equals grams per molar mass times rt. Okay.\n\nI want to get it so it's grams per litre, because that's our density value. So I'm going to mul- I'm going to divide both sides by rt to isolate the grams. So those cross out and I'll have pv over rt equals grams per molar mass. Okay. So now I want to get grams by itself so I'm going to multiply both sides by molar mass. So now my new formula is molar mass times pressure times volume over rt is equal to grams. And then I want to bring out my volume. Please don't forget my volume is measured in litres. So I want to divide both sides by volume, divide both sides by volume and I get molar mass. I can cross out my volume. Molar mass times pressure over rt equals grams over volume which is what I have written here which in other words is grams over litres. The new volume's in litres, which is our density. Yay. We found density using pv=nrt or ideal gas law.\n\nSo, there's lots of fun things you can do to find the ideal gas with the ideal gas law. Lots of fun different things you can do density being one of them and the ideal gas law is used in a lot of different ways." ]	[ null, "https://d3a0jx1tkrpybf.cloudfront.net/img/teachers/teacher-6.png", null, "https://d3a0jx1tkrpybf.cloudfront.net/img/teachers/teacher-6.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9531713,"math_prob":0.96058345,"size":9602,"snap":"2021-43-2021-49","text_gpt3_token_len":2262,"char_repetition_ratio":0.14336321,"word_repetition_ratio":0.051454138,"special_character_ratio":0.2299521,"punctuation_ratio":0.10893033,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98398244,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-21T06:40:47Z\",\"WARC-Record-ID\":\"<urn:uuid:dae435da-b963-4a45-a5c0-774dcf1bee9b>\",\"Content-Length\":\"91755\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a60078a6-5fe6-47c5-8f65-6b3ab64251fb>\",\"WARC-Concurrent-To\":\"<urn:uuid:b7c4f2a0-0772-49a5-aed8-45e8171ebd88>\",\"WARC-IP-Address\":\"23.23.139.119\",\"WARC-Target-URI\":\"https://www.brightstorm.com/science/chemistry/kinetic-molecular-theory/ideal-gas-law/\",\"WARC-Payload-Digest\":\"sha1:VCJHPXP7BBTSDPCHP3MVO6SCNNAZ6PFF\",\"WARC-Block-Digest\":\"sha1:AJFLFHLZH5KDDP5UMMA3U75N35B3IAGN\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323585381.88_warc_CC-MAIN-20211021040342-20211021070342-00312.warc.gz\"}"}
https://mathgoodies.com/lessons/vol4/fractions_to_percents	[ "# Writing Fractions as Percents\n\n## Learn About Writing Fractions as Percents With The Following Examples And Interactive Exercises", null, "Problem: Last marking period, Ms. Jones gave an A grade to 15 out of every 100 students and Mr. McNeil gave an A grade to 3 out of every 20 students. What percent of each teacher's students received an A?\n\n Solution Teacher Ratio Fraction Percent Ms. Jones 15 to 100", null, "15% Mr. McNeil 3 to 20", null, "15%", null, "Solution: Both teachers gave 15% of their students an A last marking period.\n\nIn the problem above, the fraction for Ms. Jones was easily converted to a percent. This is because It is easy to convert a fraction to a percent when the denominator is 100. If a fraction does not have a denominator of 100, you can convert it to an equivalent fraction with a denominator of 100, and then write the equivalent fraction as a percent. This is what was done in the problem above for Mr. McNeil. Let's look at some problems in which we use equivalent fractions to help us convert a fraction to a percent.\n\nExample 1: Write each fraction as a percent:", null, "Solution Fraction Equivalent Fraction Percent", null, "", null, "50%", null, "", null, "90%", null, "", null, "80%\n\nExample 2: One team won 19 out of every 20 games played, and a second team won 7 out of every 8 games played. Which team has a higher percentage of wins?\n\n Solution Team Fraction Equivalent Fraction Percent 1", null, "", null, "95% 2", null, "", null, "87.5%\n\nSolution: The first team has a higher percentage of wins.\n\nIn Examples 1 and 2, we used equivalent fractions to help us convert each fraction to a percent. Another way to do this is to convert each fraction to a decimal, and then convert each decimal to a percent. To convert a fraction to a decimal, divide its numerator by its denominatorLook at Example 3 below to see how this is done.\n\nExample 3: Write each fraction as a percent:", null, "Solution Fraction Decimal Percent", null, "", null, "87.5%", null, "", null, "95%", null, "", null, "1.5%\n\nNow that you are familiar with writing fractions as percents, do you see a pattern in the problem below?\n\n Problem: If 165% equals", null, ", and 16.5% equals", null, ", then what fraction is equal to 1.65%?\n Solution Percent Fraction 165%", null, "16.5%", null, "1.65%", null, "Summary: To write a fraction as a percent, we can convert it to an equivalent fraction with a denominator of 100. Another way to write a fraction as a percent is to divide its numerator by its denominator, then convert the resulting decimal to a percent.\n\n### Exercises\n\nDirections: Read each question below. Select your answer by clicking on its button. Feedback to your answer is provided in the RESULTS BOX. If you make a mistake, choose a different button.\n\n 1. Which of the following is equal to 36%", null, "", null, "", null, "None of the above. RESULTS BOX:\n 2. Which of the following is equal to 62.5%?", null, "", null, "", null, "None of the above. RESULTS BOX:\n 3. Which of the following is equal to", null, "? .583% 5.83% 58.3% None of the above. RESULTS BOX:\n 4. Which of the following is equal to", null, "? 11% 5.5% 200% None of the above. RESULTS BOX:\n 5. What fraction is equal to .42%?", null, "", null, "", null, "All of the above. RESULTS BOX:" ]	[ null, "https://mathgoodies.com/sites/default/files/lesson_images/per_centum.gif", null, "https://mathgoodies.com/sites/all/modules/custom/lessons/images/percent/fifteen_100ths.gif", null, "https://mathgoodies.com/sites/all/modules/custom/lessons/images/percent/equiv_3_20ths.gif", null, "https://mathgoodies.com/sites/default/files/lesson_images/gradebook.gif", null, "https://mathgoodies.com/sites/all/modules/custom/lessons/images/percent/fractions_example1.gif", null, "https://mathgoodies.com/sites/all/modules/custom/lessons/images/percent/one_half.gif", null, "https://mathgoodies.com/sites/all/modules/custom/lessons/images/percent/equiv_one_half.gif", null, "https://mathgoodies.com/sites/all/modules/custom/lessons/images/percent/eighteen_20ths.gif", null, "https://mathgoodies.com/sites/all/modules/custom/lessons/images/percent/equiv_18_20ths.gif", null, "https://mathgoodies.com/sites/all/modules/custom/lessons/images/percent/four_fifths.gif", null, "https://mathgoodies.com/sites/all/modules/custom/lessons/images/percent/equiv_4_5ths.gif", null, "https://mathgoodies.com/sites/all/modules/custom/lessons/images/percent/nineteen_20ths.gif", null, "https://mathgoodies.com/sites/all/modules/custom/lessons/images/percent/equiv_19_20ths.gif", null, "https://mathgoodies.com/sites/all/modules/custom/lessons/images/percent/seven_8ths.gif", null, "https://mathgoodies.com/sites/all/modules/custom/lessons/images/percent/equiv_7_8ths.gif", null, "https://mathgoodies.com/sites/all/modules/custom/lessons/images/percent/fractions_example2.gif", null, "https://mathgoodies.com/sites/all/modules/custom/lessons/images/percent/seven_8ths.gif", null, "https://mathgoodies.com/sites/all/modules/custom/lessons/images/percent/seven_div8.gif", null, "https://mathgoodies.com/sites/all/modules/custom/lessons/images/percent/nineteen_20ths.gif", null, "https://mathgoodies.com/sites/all/modules/custom/lessons/images/percent/nineteen_div20.gif", null, "https://mathgoodies.com/sites/all/modules/custom/lessons/images/percent/three_200ths.gif", null, "https://mathgoodies.com/sites/all/modules/custom/lessons/images/percent/three_div200.gif", null, "https://mathgoodies.com/sites/all/modules/custom/lessons/images/percent/thirty3_20ths.gif", null, "https://mathgoodies.com/sites/all/modules/custom/lessons/images/percent/thirty3_200ths.gif", null, "https://mathgoodies.com/sites/all/modules/custom/lessons/images/percent/thirty3_20ths.gif", null, "https://mathgoodies.com/sites/all/modules/custom/lessons/images/percent/thirty3_200ths.gif", null, "https://mathgoodies.com/sites/all/modules/custom/lessons/images/percent/thirty3_2000ths.gif", null, "https://mathgoodies.com/sites/all/modules/custom/lessons/images/percent/nine_25ths.gif", null, "https://mathgoodies.com/sites/all/modules/custom/lessons/images/percent/five_8ths.gif", null, "https://mathgoodies.com/sites/all/modules/custom/lessons/images/percent/six_25ths.gif", null, "https://mathgoodies.com/sites/all/modules/custom/lessons/images/percent/four_fifths.gif", null, "https://mathgoodies.com/sites/all/modules/custom/lessons/images/percent/five_8ths.gif", null, "https://mathgoodies.com/sites/all/modules/custom/lessons/images/percent/eighteen_20ths.gif", null, "https://mathgoodies.com/sites/all/modules/custom/lessons/images/percent/fifty8_point3_100ths.gif", null, "https://mathgoodies.com/sites/all/modules/custom/lessons/images/percent/eleven_200ths.gif", null, "https://mathgoodies.com/sites/all/modules/custom/lessons/images/percent/four_point2_1000ths.gif", null, "https://mathgoodies.com/sites/all/modules/custom/lessons/images/percent/forty_two_10000ths.gif", null, "https://mathgoodies.com/sites/all/modules/custom/lessons/images/percent/twenty1_5000ths.gif", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9199037,"math_prob":0.98213714,"size":2723,"snap":"2022-27-2022-33","text_gpt3_token_len":696,"char_repetition_ratio":0.1636631,"word_repetition_ratio":0.11340206,"special_character_ratio":0.26625046,"punctuation_ratio":0.12743363,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9986824,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50,51,52,53,54,55,56,57,58,59,60,61,62,63,64,65,66,67,68,69,70,71,72,73,74,75,76],"im_url_duplicate_count":[null,9,null,3,null,3,null,3,null,3,null,3,null,3,null,6,null,3,null,6,null,3,null,6,null,3,null,6,null,3,null,3,null,6,null,3,null,6,null,3,null,3,null,3,null,6,null,6,null,6,null,6,null,3,null,3,null,6,null,3,null,6,null,6,null,6,null,3,null,3,null,3,null,3,null,3,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-08-17T19:12:08Z\",\"WARC-Record-ID\":\"<urn:uuid:0021c461-7a77-494e-b10e-e0dbb7b4b7e4>\",\"Content-Length\":\"82756\",\"Content-Type\":\"application/http; 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https://www.mbeckler.org/blog/?tag=cpp	[ "# cpp\n\n## Pretty printing a number of bytes in C/C++\n\nI spend a lot of time write and debugging command line scripts and programs. As much as I like looking at large numbers (millions, billions, trillions, etc) it can be difficult to read a big number and quickly parse how large it is, i.e. “is that 12 megabytes or 1.2 gigabytes?”.\n\nA long time ago I wrote a small function that does pretty printing of a number of bytes. It can handle from bytes to exabytes, and properly handles integer numbers of a unit by printing it as an integer instead of float. Should be easy enough to adjust to your specific needs or style desires.\n\n```// Prints to the provided buffer a nice number of bytes (KB, MB, GB, etc)\nvoid pretty_bytes(char* buf, uint bytes)\n{\nconst char* suffixes;\nsuffixes = \"B\";\nsuffixes = \"KB\";\nsuffixes = \"MB\";\nsuffixes = \"GB\";\nsuffixes = \"TB\";\nsuffixes = \"PB\";\nsuffixes = \"EB\";\nuint s = 0; // which suffix to use\ndouble count = bytes;\nwhile (count >= 1024 && s < 7)\n{\ns++;\ncount /= 1024;\n}\nif (count - floor(count) == 0.0)\nsprintf(buf, \"%d %s\", (int)count, suffixes[s]);\nelse\nsprintf(buf, \"%.1f %s\", count, suffixes[s]);\n}```" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8087208,"math_prob":0.9739436,"size":1099,"snap":"2020-45-2020-50","text_gpt3_token_len":321,"char_repetition_ratio":0.14977169,"word_repetition_ratio":0.0,"special_character_ratio":0.322111,"punctuation_ratio":0.17167382,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9557192,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-10-26T03:41:39Z\",\"WARC-Record-ID\":\"<urn:uuid:74219561-02de-45fd-b787-0a3aa58b8c1d>\",\"Content-Length\":\"16070\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:3cf35ad2-bdcc-46f5-9f29-8dd5a9fda5d2>\",\"WARC-Concurrent-To\":\"<urn:uuid:f815ac19-acd8-4183-a04f-7e288143b463>\",\"WARC-IP-Address\":\"66.33.210.224\",\"WARC-Target-URI\":\"https://www.mbeckler.org/blog/?tag=cpp\",\"WARC-Payload-Digest\":\"sha1:JFDZVL3QMDJ4KCRUPJK4KMT33YJVDEYE\",\"WARC-Block-Digest\":\"sha1:ICBXTCDO3GO6YVEMHN4KWIMHVOYFQWTB\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-45/CC-MAIN-2020-45_segments_1603107890273.42_warc_CC-MAIN-20201026031408-20201026061408-00557.warc.gz\"}"}
https://simplyans.com/mathematics/what-is-the-positive-square-root-of-12656457	[ "", null, ", 14.05.2019 20:39 cesarcastellan9\n\n# What is the positive square root of 0.81enter your answer in the box.", null, "", null, "", null, "### Another question on Mathematics", null, "Mathematics, 04.02.2019 19:52\nPart a: select all of the ordered pairs that are located on the graph of the equation. part b: does the graph of the equation represent a function? select all correct answers for part a and one answer for part b.", null, "Mathematics, 04.02.2019 06:30\nYou are given a convex polygon. you are asked to draw a new polygon by increasing the sum of the interior angle measures by 540°. how many sides does your new polygon have?", null, "Mathematics, 04.02.2019 05:14\n43lbs of tomatos cost \\$387. how much would 41lbs cost", null, "Mathematics, 03.02.2019 17:10\nAgarden consists of an apple tree, a pear tree, cauliflowers, and heads of cabbage. there are 40 vegetables in the garden. 24 of them are cauliflowers. what is the ratio of the number of cauliflowers to the number of heads of cabbage?\nWhat is the positive square root of 0.81\n...\nQuestions", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "Questions on the website: 6616171" ]	[ null, "https://simplyans.com/tpl/images/cats/mat.png", null, "https://simplyans.com/tpl/images/cats/User.png", null, "https://simplyans.com/tpl/images/ask_question.png", null, "https://simplyans.com/tpl/images/ask_question_mob.png", null, "https://simplyans.com/tpl/images/cats/mat.png", null, "https://simplyans.com/tpl/images/cats/mat.png", null, "https://simplyans.com/tpl/images/cats/mat.png", null, "https://simplyans.com/tpl/images/cats/mat.png", null, "https://simplyans.com/tpl/images/cats/mir.png", null, "https://simplyans.com/tpl/images/cats/health.png", null, "https://simplyans.com/tpl/images/cats/en.png", null, "https://simplyans.com/tpl/images/cats/en.png", null, "https://simplyans.com/tpl/images/cats/ekonomika.png", null, "https://simplyans.com/tpl/images/cats/fizika.png", null, "https://simplyans.com/tpl/images/cats/mat.png", null, "https://simplyans.com/tpl/images/cats/istoriya.png", null, "https://simplyans.com/tpl/images/cats/obshestvoznanie.png", null, "https://simplyans.com/tpl/images/cats/istoriya.png", null, "https://simplyans.com/tpl/images/cats/health.png", null, "https://simplyans.com/tpl/images/cats/istoriya.png", null, "https://simplyans.com/tpl/images/cats/fr.png", null, "https://simplyans.com/tpl/images/cats/biologiya.png", null, "https://simplyans.com/tpl/images/cats/fizika.png", null, "https://simplyans.com/tpl/images/cats/ekonomika.png", null, "https://simplyans.com/tpl/images/cats/en.png", null, "https://simplyans.com/tpl/images/cats/mat.png", null, "https://simplyans.com/tpl/images/cats/obshestvoznanie.png", null, "https://simplyans.com/tpl/images/cats/mat.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.81871957,"math_prob":0.9283398,"size":1291,"snap":"2020-45-2020-50","text_gpt3_token_len":425,"char_repetition_ratio":0.13519813,"word_repetition_ratio":0.115384616,"special_character_ratio":0.36948103,"punctuation_ratio":0.248503,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9674881,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50,51,52,53,54,55,56],"im_url_duplicate_count":[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,null,null,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-10-24T05:52:27Z\",\"WARC-Record-ID\":\"<urn:uuid:3f73f90a-91f1-4ee3-ae3d-bd6a37e8ef57>\",\"Content-Length\":\"129330\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:017eb1d3-e8d1-493f-ace6-565e5554f95f>\",\"WARC-Concurrent-To\":\"<urn:uuid:d4f0f5fa-7c47-4ba8-8175-384ee174da26>\",\"WARC-IP-Address\":\"64.74.160.240\",\"WARC-Target-URI\":\"https://simplyans.com/mathematics/what-is-the-positive-square-root-of-12656457\",\"WARC-Payload-Digest\":\"sha1:2QC2TX3CNQR4QMT743DXVDWUOBJGFGWT\",\"WARC-Block-Digest\":\"sha1:WSPK5AZJ5HNXB62PAUW6P36Z6AHOTKH7\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-45/CC-MAIN-2020-45_segments_1603107882102.31_warc_CC-MAIN-20201024051926-20201024081926-00466.warc.gz\"}"}
https://schoolgraders.com/evaluate-the-function-at-the-indicated-value-of-x-round-your-result-to-three-decimal-places-function-fx-0-5x-value-x-1-7/	[ "# Evaluate the function at the indicated value of x. Round your result to three decimal places. Function: f(x) = 0.5x Value: x = 1.7\n\nQUESTION 1\n1. Evaluate the function at the indicated value of x. Round your result to three decimal places.\nFunction: f(x) = 0.5x Value: x = 1.7\n\n-0.308\n\n1.7\n\n0.308\n\n0.5\n\n-1.7\nQUESTION 2\n1. Match the graph with its exponential function.\n\ny = 2-x – 3\n\ny = -2x + 3\n\ny = 2x + 3\n\ny = 2x – 3\n\ny = -2x – 3\n\nQUESTION 3\n1. Select the graph of the function.\nf(x) = 5x-1\n\nQUESTION 4\n1. Evaluate the function at the indicated value of x. Round your result to three decimal places.\nFunction: f(x) = 500e0.05x Value: x=17\n\n1169.823\n\n1369.823\n\n1569.823\n\n1269.823\n\n1469.823\n\nQUESTION 5\n1. Use the One-to-One property to solve the equation for x.\ne3x+5 = e6\n\nx = -1/3\n\nx2 = 6\n\nx = -3\n\nx = 1/3\n\nx = 3\n\nQUESTION 6\n1. Write the logarithmic equation in exponential form.\nlog8 64 = 2\n\n648 = 2\n\n82 = 16\n\n82 = 88\n\n82 = 64\n\n864 = 2\n\nQUESTION 7\n1. Write the logarithmic equation in exponential form.\nlog7 343 = 3\n\n7343 = 2\n\n73 = 77\n\n73 = 343\n\n73 = 14\n\n3437 = 2\n\nQUESTION 8\n1. Write the exponential equation in logarithmic form.\n43 = 64\n\nlog64 4 = 3\n\nlog4 64 = 3\n\nlog4 64 = -3\n\nlog4 3 = 64\n\nlog4 64 = 1/3\n\nQUESTION 9\n1. Use the properties of logarithms to simplify the expression.\nlog20 209\n\n0\n\n-1/9\n\n1/9\n\n-9\n\n9\n\nQUESTION 10\n1. Use the One-to-One property to solve the equation for x.\nlog2(x+4) = log2 20\n\n19\n\n17\n\n18\n\n16\n\n20\n\nQUESTION 11\n1. Find the exact value of the logarithmic expression.\nlog6 36\n\n2\n\n6\n\n36\n\n-2\n\nnone of these\n\nQUESTION 12\n1. Use the properties of logarithms to expand the expression as a sum, difference, and/or constant multiple of logarithms. (Assume all variables are positive.)\nlog3 9x\n\nlog3 9 x log3 x\n\nlog3 9 + log3 x\n\nlog3 9 log3\n\nnone of these\n\nQUESTION 13\n1. Condense the expression to a logarithm of a single quantity.\nlogx – 2logy + 3logz\n\nQUESTION 14\n1. Evaluate the logarithm using the change-of-base formula. Round your result to three decimal places.\nlog4 9\n\n1.585\n\n5.585\n\n3.585\n\n4.585\n\n2.585\nQUESTION 15\n1. Determine whether the given x-value is a solution (or an approximate solution) of the equation.\n42x-7 = 16\nx = 5\n\nno\n\nyes\nQUESTION 16\n1. Solve for x.\n3x = 81\n\n7\n\n3\n\n4\n\n-4\n\n-3\nQUESTION 17\n1. Solve the exponential equation algebraically. Approximate the resulte to three decimal places.\ne5x = ex2-14\n\n-7, -2\n\n7, -2\n\n5, -14\n\n7, 2\n\n-7, 2\nQUESTION 18\n1. Solve the logarithmic equation algebraically. Approximate the result to three decimal places.\nlog3(6x-8) = log3(5x + 10)\n\n18\n\n20\n\n17\n\n19\n\n-2\nQUESTION 19\n1. Find the magnitude R of each earthquake of intensity I (let I0=1).\nI = 19000\n\n3.28\n\n5.28\n\n4.28\n\n2.38\n\n6.28\nQUESTION 20\n1. \\$2500 is invested in an account at interest rate r, compounded continuously. Find the time required for the amount to double. (Approximate the result to two decimal places.)\nr = 0.0570\n\n13.16 years\n\n10.16 years\n\n11.16 years\n\n12.16 years\n\n14.16 years" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7656984,"math_prob":0.99945027,"size":2696,"snap":"2020-45-2020-50","text_gpt3_token_len":1005,"char_repetition_ratio":0.15156017,"word_repetition_ratio":0.13108614,"special_character_ratio":0.41357568,"punctuation_ratio":0.14037855,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99986434,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-11-29T03:49:17Z\",\"WARC-Record-ID\":\"<urn:uuid:2551aa22-36e8-4019-8f62-7fcdb0f5f4c0>\",\"Content-Length\":\"122503\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7df15222-3b29-46d9-913f-6ebcd8c7bf82>\",\"WARC-Concurrent-To\":\"<urn:uuid:b34288b8-f73e-456f-97d7-1ce145aac5c9>\",\"WARC-IP-Address\":\"172.67.152.131\",\"WARC-Target-URI\":\"https://schoolgraders.com/evaluate-the-function-at-the-indicated-value-of-x-round-your-result-to-three-decimal-places-function-fx-0-5x-value-x-1-7/\",\"WARC-Payload-Digest\":\"sha1:BHA3SNYVVLYIY2KMRNAVTL5ZS4GXUPXV\",\"WARC-Block-Digest\":\"sha1:PNLEIV7MRNAFLB6CKHAGKJ5WZWXDRGNR\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141196324.38_warc_CC-MAIN-20201129034021-20201129064021-00223.warc.gz\"}"}
http://www.neuralnoise.com/2016/gaussian-fields/	[ "# neuralnoise.com\n\nHomepage of Dr. Pasquale Minervini, Ph.D.\nResearcher at University College London\nLondon, United Kingdom\n\nIn several occasions, we find ourselves in need of propagating information among nodes in an undirected graph.\n\nFor instance, consider graph-based Semi-Supervised Learning (SSL): here, labeled and unlabeled examples are represented by an undirected graph, referred to as the similarity graph.\n\nThe task consists in finding a label assignment to all examples, such that:\n\n1. The final labeling is consistent with training data (e.g. positive training examples are still classified as positive at the end of the learning process), and\n2. Similar examples are assigned similar labels: this is referred to as the semi-supervised smoothness assumption.\n\nSimilarly, in networked data such as social networks, we might assume that related entities (such as friends) are associated to similar attributes (such as political and religious views, musical tastes and so on): in social network analysis, this phenomenon is commonly referred to as homophily (love of the same).\n\nIn both cases, propagating information from a limited set of nodes in a graph to all nodes provides a method for predicting the attributes of such nodes, when this information is missing.\n\nIn the following, we introduce a really clever method for efficiently propagating information about nodes in undirected graphs, known as the Gaussian Fields method.\n\n### Propagation as a Cost Minimization Problem\n\nWe now cast the propagation problem as a binary classification task. Let $X = \\{ x_{1}, x_{2}, \\ldots, x_{n} \\}$ be a set of $n$ instances, of which only $l$ are labeled: $X^{+}$ are positive examples, while $X^{-}$ are negative examples\n\nSimilarity relations between instances can be represented by means of an undirected similarity graph having adjacency matrix $\\mathbf{W} \\in \\mathbb{R}^{n \\times n}$: if two instances are connected in the similarity graph, it means that they are considered similar, and should be assigned the same label. Specifically, $\\mathbf{W}_{ij} > 0$ iff the instances $x_{i}, x_{j} \\in X$ are connected by an edge in the similarity graph, and $\\mathbf{W}_{ij} = 0$ otherwise.\n\nLet $y_{i} \\in \\{ \\pm 1 \\}$ be the label assigned to the $i$-th instance $x_{i} \\in X$. We can encode our assumption that similar instances should be assigned similar labels by defining a quadratic cost function over labeling functions in the form $f : X \\mapsto \\{ \\pm 1 \\}$:\n\nGiven an input labeling function $f$, the cost function $E(\\cdot)$ associates, for each pair of instances $x_{i}, x_{j} \\in X$, a non-negative cost $\\mathbf{W}_{ij} \\left[ f(x_{i}) - f(x_{j}) \\right]$: this quantity is $0$ when $\\mathbf{W}_{ij} = 0$ (i.e. $x_{i}$ and $X_{j}$ are not linked in the similarity graph), or when $f(x_{i}) = f(x_{j})$ (i.e. they are assigned the same label).\n\nFor such a reason, the cost function $E(\\cdot)$ favors labeling functions that are more likely to assign the same labels to instances that are linked by an edge in the similarity graph.\n\nNow, the problem of finding a labeling function that is both consistent with training labels, and assigns similar labels to similar instances, can be cast as a cost minimization problem. Let’s represent a labeling function $f$ by a vector $\\mathbf{f} \\in \\mathbb{R}^{n}$, $L \\subset X$ denote labeled instances, and $\\mathbf{y}_{i} \\in \\{ \\pm 1 \\}$ denote the label of the $x_{i}$-th instance. The optimization problem can be defined as follows:\n\nThe constraint $\\forall x \\in L : \\mathbf{f}_{i} = \\mathbf{y}_{i}$ enforces the label of each labeled example $x_{i} \\in L$ to $\\mathbf{f}_{i} = +1$ if the instance has a positive label, and to $\\mathbf{f}_{i} = -1$ if the instance has a negative label, so to achieve consistency with training labels.\n\nHowever, constraining labeling functions $f$ to only take discrete values has two main drawbacks:\n\n• Each function $f$ can only provide hard classifications, without yielding any measure of confidence in the provided classification.\n• The cost term $E(\\cdot)$ can be hard to optimize in a multi-label classification setting.\n\nFor overcoming such limitations, Zhu et al. propose a continuous relaxation of the previous optimization problem:\n\nwhere the term $\\sum_{x_{i} \\in X} \\mathbf{f}_{i}^{2} = \\mathbf{f}^{T} \\mathbf{f}$ is a $L_{2}$ regularizer over $\\mathbf{f}$, weighted by a parameter $\\epsilon > 0$ which ensures that the optimization problem has a unique global solution.\n\nThe parameter $\\epsilon$ can be interpreted as the decay of the propagation process: as the distance from a labeled instance within the similarity graph increases, the confidence in the classification (as measured by the continuous label) gets closer to zero.\n\nThis optimization problem has a unique, global solution that can be calculated in closed-form. Specifically, the optimal (relaxed) discriminant function $f : X \\mapsto \\mathbb{R}$ is given by $\\mathbf{\\hat{f}} = \\left[ \\mathbf{f}_{L}, \\mathbf{f}_{U} \\right]^{T}$, where $\\mathbf{\\hat{f}}_{L} = \\mathbf{y}_{L}$ (i.e. labels for labeled examples in $L$ coincide with training labels), while $\\mathbf{\\hat{f}}_{U}$ is given by:\n\nwhere $\\mathbf{L} = \\mathbf{D} - \\mathbf{W}$ is the graph Laplacian of the similarity graph with adjacency matrix $\\mathbf{W}$, and $\\mathbf{D}$ is a diagonal matrix such that $\\mathbf{D}_{ii} = \\sum_{j} \\mathbf{W}_{ij}$." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8657781,"math_prob":0.9989867,"size":5242,"snap":"2021-21-2021-25","text_gpt3_token_len":1332,"char_repetition_ratio":0.14852998,"word_repetition_ratio":0.010012516,"special_character_ratio":0.25982448,"punctuation_ratio":0.095918365,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9999528,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-05-16T21:12:34Z\",\"WARC-Record-ID\":\"<urn:uuid:90a7ca33-73fd-40d6-a75c-bce08213fcad>\",\"Content-Length\":\"17596\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7322c4f2-cf05-441b-b9f4-08b3c1641535>\",\"WARC-Concurrent-To\":\"<urn:uuid:8b1226cd-0c2a-4817-a152-f4e7a056bd22>\",\"WARC-IP-Address\":\"148.251.85.242\",\"WARC-Target-URI\":\"http://www.neuralnoise.com/2016/gaussian-fields/\",\"WARC-Payload-Digest\":\"sha1:E5FZWZHKVCCMHGS7J4S5WSXXIDLFMG2K\",\"WARC-Block-Digest\":\"sha1:V7EISJDUXPXGMGK25V3LALZKXXJW2SCI\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-21/CC-MAIN-2021-21_segments_1620243989914.60_warc_CC-MAIN-20210516201947-20210516231947-00519.warc.gz\"}"}
https://data.epo.org/publication-server/html-document?PN=EP3550726%20EP%203550726&iDocId=6391744&disclaimer=true&citations=true	[ "(19)", null, "(11)EP 3 550 726 B1\n\n (12) EUROPEAN PATENT SPECIFICATION\n\n (45) Mention of the grant of the patent: 04.11.2020 Bulletin 2020/45\n\n (21) Application number: 19172900.3\n\n (22) Date of filing: 20.05.2011\n(51)International Patent Classification (IPC):\n H03M 7/40(2006.01)\n\n (54) METHODS AND DEVICES FOR REDUCING SOURCES IN BINARY ENTROPY CODING AND DECODINGVERFAHREN UND VORRICHTUNGEN ZUR REDUZIERUNG VON QUELLEN FÜR BINÄRE ENTROPIECODIERUNG UND -DECODIERUNGPROCÉDÉS ET DISPOSITIFS DE RÉDUCTION DE SOURCES DANS LE CODAGE ET LE DÉCODAGE D'ENTROPIE BINAIRE\n\n (84) Designated Contracting States: AL AT BE BG CH CY CZ DE DK EE ES FI FR GB GR HR HU IE IS IT LI LT LU LV MC MK MT NL NO PL PT RO RS SE SI SK SM TR\n\n (30) Priority: 21.05.2010 US 34702710 P\n\n (43) Date of publication of application: 09.10.2019 Bulletin 2019/41\n\n (62) Application number of the earlier application in accordance with Art. 76 EPC: 11782837.6 / 2572455\n\n (73) Proprietor: BlackBerry Limited Waterloo, Ontario N2K 0A7 (CA)\n\n (72) Inventors: HE, DakeWaterloo, Ontario N2K 0A7 (CA)KORODI, Gergely FerencWaterloo, Ontario N2K 0A7 (CA)\n\n (74) Representative: Hanna Moore + Curley Garryard House 25/26 Earlsfort TerraceDublin 2, D02 PX51Dublin 2, D02 PX51 (IE)\n\n(56)References cited: :\n EP-A1- 2 124 343\n\n• KARWOWSKI DAMIAN ET AL: \"Improved context-adaptive arithmetic coding in H.264/AVC\", 2009 17TH EUROPEAN SIGNAL PROCESSING CONFERENCE, IEEE, 24 August 2009 (2009-08-24), pages 2216-2220, XP032758765, ISBN: 978-1-61738-876-7 [retrieved on 2015-04-01]\n• MRAK M ET AL: \"Comparison of context-based adaptive binary arithmetic coders in video compression\", VIDEO/IMAGE PROCESSING AND MULTIMEDIA COMMUNICATIONS, 2003. 4TH EURASI P CONFERENCE FOCUSED ON 2-5 JULY 2003, PISCATAWAY, NJ, USA,IEEE, vol. 1, 2 July 2003 (2003-07-02), pages 277-286, XP010650143, ISBN: 978-953-184-054-5\n• KORODI G ET AL: \"Source selection for V2V entropy coding in HEVC\", 2. JCT-VC MEETING; 21-7-2010 - 28-7-2010; GENEVA; (JOINT COLLABORATIVETEAM ON VIDEO CODING OF ISO/IEC JTC1/SC29/WG11 AND ITU-T SG.16 ); URL:HTTP://WFTP3.ITU.INT/AV-ARCH/JCTVC-SIT E/,, no. JCTVC-B034, 18 July 2010 (2010-07-18), XP030007614, ISSN: 0000-0048\n• KARWOWSKI DAMIAN: \"Improved arithmetic coding in H.264/AVC using Context-Tree Weighting and prediction by Partial Matching\", 2007 15TH EUROPEAN SIGNAL PROCESSING CONFERENCE, IEEE, 3 September 2007 (2007-09-03), pages 1270-1274, XP032772888, ISBN: 978-83-921340-4-6 [retrieved on 2015-04-30]\n\n Note: Within nine months from the publication of the mention of the grant of the European patent, any person may give notice to the European Patent Office of opposition to the European patent granted. Notice of opposition shall be filed in a written reasoned statement. It shall not be deemed to have been filed until the opposition fee has been paid. (Art. 99(1) European Patent Convention).\n\nCROSS-REFERENCE TO RELATED APPLICATIONS\n\nFIELD\n\n The present application generally relates to data compression and, in particular, to an encoder, a decoder and methods for reducing sources in binary entropy coding and decoding.\n\nBACKGROUND\n\n A number of coding schemes have been developed to encode binary data. For example, JPEG images may be encoded using Huffman codes. The H.264 standard allows for two possible entropy coding processes: Context Adaptive Variable Length Coding (CAVLC) or Context Adaptive Binary Arithmetic Coding (CABAC). CABAC results in greater compression than CAVLC, but CABAC is more computationally demanding.\n\n Recent advances in binary entropy coding have made it increasingly attractive as an encoding scheme. For example, it is possible to engage in parallel encoding, where a source input sequence is separated into parallel subsequences based on context modeling, where each subsequence contains the symbols associated with a particular probability from the context model. In another example, variable-to-variable coding with a primary codeset and a secondary codeset can improve coding efficiency. The secondary codeset is used only during \"end\" or \"flush\" events, where a sequence or subsequence of symbols does not complete a primary codeword. The subsequences of symbols associated with particular probabilities defined in the context model may be considered as being produced by different \"sources\".\n\n Nevertheless, there are still certain drawbacks to binary entropy encoding that can arise in certain circumstances. For example, in the context of parallel encoding the output bitstream may include overhead information about the length of each of the encoded subsequences. For a context model having a large number of \"sources\", i.e. probabilities, this can mean a sizable overhead. In the context of two-codeset variable-to-variable encoding, the use of the less-efficient secondary codeset can be costly, particular where there are a large number of \"sources\" and, thus, a large number of partial subsequences that need to be encoded using secondary codewords after a flush event.\n\n Karwowski Damian et al, \"Improved context-adaptive arithmetic coding in H.264/AVC\", 17th European Signal Processing Conference, 24-28 August 2009, page 2216 - 2220, discloses an improved Context-based Adaptive Binary Arithmetic Coding (CABAC), for applications in video compression, having a higher coding efficiency obtained by application of more exact data statistics estimation based on the Context-Tree Weighting (CTW) data modeling algorithm. Finite-state machines in the context modeler are replaced by binary context trees. The improved context modeler generates values of conditional probabilities in a significantly greater set of numbers as compare to the original CABAC. The core of the fast binary arithmetic codec (M-codec) is adopted to work properly with a limited set of only 128 predefined probabilities, therefore each probability calculated with the improved context modeler has to be mapped to a value from the smaller set of 128 predefined probabilities.\n\n Mrak M et al, \"Comparison of context-based adaptive binary arithmetic coders in video compression\", Video/Image Processing and Multimedia Communications, 2-5 July 2003, pages 277-286, discloses a comparison two context-based modeling methods, CABCA and Growth by Reordering and Selection by Pruning (GRASP). GRASP is a two-step approach. In the first step, a full balanced tree of a pre-defined depth is grown and in the second step, the resulting full balanced tree is pruned; branches are pruned that result in a higher cost than that of their corresponding root nodes.\n\nBRIEF SUMMARY\n\n The present application describes architectures, methods and processes for decoding data.\n\n In one aspect, the present application describes a method for decoding an encoded bitstream as specified by claim 1. Further detailed embodiments are specified in the dependent claims.\nIn yet a further aspect, the present application describes a decoder having a memory, processor, and decoding application executable by the processor that, when executed, configures the processor to perform one of the decoding processes described herein.\nIn another aspect, the present application describes a tangible computer-readable medium storing computer-executable instructions\nwhich, when executed by one or more processors, implement the steps of the decoding processes described herein.\nIt will be understood that the reference herein to a processor is not limited to a single processor computer architecture and can include multi-core processors, multiple processors, and distributed\narchitectures, in some embodiments. Only the embodiments for the method for decoding, decoder and associated computer-readable medium storing computer-executable instructions implementing the steps of the method for decoding are to be considered embodiments of the invention. The other so called embodiments are to be considered only examples.\n\n Other aspects and features of the present application will be understood by those of ordinary skill in the art from a review of the following description of examples in conjunction with the accompanying figures.\n\nBRIEF DESCRIPTION OF THE DRAWINGS\n\n Reference will now be made, by way of example, to the accompanying drawings which show example embodiments of the present application, and in which:\n\nFigure 1 shows, in block diagram form, an encoder for encoding video;\n\nFigure 2 shows, in block diagram form, a decoder for decoding video;\n\nFigure 3 shows a block diagram of an encoding process;\n\nFigure 4 shows, in flowchart form, a method of encoding input symbols using a new set of probabilities and a mapping;\n\nFigure 5 shows, in flowchart form, a method of decoding an encoded bitstream based upon a new set of probabilities and a mapping;\n\nFigure 6 shows a simplified block diagram of an example embodiment of an encoder; and\n\nFigure 7 shows a simplified block diagram of an example embodiment of a decoder.\n\n Similar reference numerals may have been used in different figures to denote similar components.\n\nDESCRIPTION OF EXAMPLE EMBODIMENTS\n\n The following description relates to data compression in general and, in particular, to the encoding of finite alphabet sources, such as a binary source. In many of the examples given below, particular applications of such an encoding and decoding scheme are given. For example, many of the illustrations below make reference to video coding. It will be appreciated that the present application is not limited to video coding or image coding.\n\n In the description that follows, example embodiments are described with reference to the H.264 standard. Those ordinarily skilled in the art will understand that the present application is not limited to H.264 but may be applicable to other video coding/decoding standards. It will also be appreciated that the present application is not necessarily limited to video coding/decoding and may be applicable to coding/decoding of any binary sources.\n\n In the description that follows, in the context of video applications the terms frame and slice are used somewhat interchangeably. Those of skill in the art will appreciate that, in the case of the H.264 standard, a frame may contain one or more slices. It will also be appreciated that certain encoding/decoding operations arc performed on a frame-by-frame basis and some arc performed on a slice-by-slice basis, depending on the particular requirements of the applicable video coding standard. In any particular embodiment, the applicable video coding standard may determine whether the operations described below are performed in connection with frames and/or slices, as the case may be. Accordingly, those ordinarily skilled in the art will understand, in light of the present disclosure, whether particular operations or processes described herein and particular references to frames, slices, or both are applicable to frames, slices, or both for a given embodiment.\n\n Reference is now made to Figure 1, which shows, in block diagram form, an encoder 10 for encoding video. Reference is also made to Figure 2, which shows a block diagram of a decoder 50 for decoding video. It will be appreciated that the encoder 10 and decoder 50 described herein may each be implemented on an application-specific or general purpose computing device, containing one or more processing elements and memory. The operations performed by the encoder 10 or decoder 50, as the case may be, may be implemented by way of application-specific integrated circuit, for example, or by way of stored program instructions executable by a general purpose processor. The device may include additional software, including, for example, an operating system for controlling basic device functions. The range of devices and platforms within which the encoder 10 or decoder 50 may be implemented will be appreciated by those ordinarily skilled in the art having regard to the following description.\n\n The encoder 10 receives a video source 12 and produces an encoded bitstream 14. The decoder 50 receives the encoded bitstream 14 and outputs a decoded video frame 16. The encoder 10 and decoder 50 may be configured to operate in conformance with a number of video compression standards. For example, the encoder 10 and decoder 50 may be H.264/AVC compliant. In other embodiments, the encoder 10 and decoder 50 may conform to other video compression standards, including evolutions of the H.264/AVC standard.\n\n The encoder 10 includes a spatial predictor 21, a coding mode selector 20, transform processor 22, quantizer 24, and entropy coder 26. As will be appreciated by those ordinarily skilled in the art, the coding mode selector 20 determines the appropriate coding mode for the video source, for example whether the subject frame/slice is of I, P, or B type, and whether particular macroblocks within the frame/slice are inter or intra coded. The transform processor 22 performs a transform upon the spatial domain data. In particular, the transform processor 22 applies a block-based transform to convert spatial domain data to spectral components. For example, in many embodiments a discrete cosine transform (DCT) is used. Other transforms, such as a discrete sine transform or others may be used in some instances. Applying the block-based transform to a block of pixel data results in a set of transform domain coefficients. The set of transform domain coefficients is quantized by the quantizer 24. The quantized coefficients and associated information, such as motion vectors, quantization parameters, etc., are then encoded by the entropy coder 26.\n\n Intra-coded frames/slices (i.e. type I) are encoded without reference to other frames/slices. In other words, they do not employ temporal prediction. However intra-coded frames do rely upon spatial prediction within the frame/slice, as illustrated in Figure 1 by the spatial predictor 21. That is, when encoding a particular block the data in the block may be compared to the data of nearby pixels within blocks already encoded for that frame/slice. Using a prediction algorithm, the source data of the block may be converted to residual data. The transform processor 22 then encodes the residual data. H.264, for example, prescribes nine spatial prediction modes for 4x4 transform blocks. In some embodiments, each of the nine modes may be used to independently process a block, and then rate-distortion optimization is used to select the best mode.\n\n The H.264 standard also prescribes the use of motion prediction/compensation to take advantage of temporal prediction. Accordingly, the encoder 10 has a feedback loop that includes a de-quantizer 28, inverse transform processor 30, and deblocking processor 32. These elements mirror the decoding process implemented by the decoder 50 to reproduce the frame/slice. A frame store 34 is used to store the reproduced frames. In this manner, the motion prediction is based on what will be the reconstructed frames at the decoder 50 and not on the original frames, which may differ from the reconstructed frames due to the lossy compression involved in encoding/decoding. A motion predictor 36 uses the frames/slices stored in the frame store 34 as source frames/slices for comparison to a current frame for the purpose of identifying similar blocks. Accordingly, for macroblocks to which motion prediction is applied, the \"source data\" which the transform processor 22 encodes is the residual data that comes out of the motion prediction process. The residual data is pixel data that represents the differences (if any) between the reference block and the current block. Information regarding the reference frame and/or motion vector may not be processed by the transform processor 22 and/or quantizer 24, but instead may be supplied to the entropy coder 26 for encoding as part of the bitstream along with the quantized coefficients.\n\n Those ordinarily skilled in the art will appreciate the details and possible variations for implementing H.264 encoders.\n\n The decoder 50 includes an entropy decoder 52, dequantizer 54, inverse transform processor 56, spatial compensator 57, and deblocking processor 60. A frame buffer 58 supplies reconstructed frames for use by a motion compensator 62 in applying motion compensation. The spatial compensator 57 represents the operation of recovering the video data for a particular intra-codcd block from a previously decoded block.\n\n The bitstream 14 is received and decoded by the entropy decoder 52 to recover the quantized coefficients. Side information may also be recovered during the entropy decoding process, some of which may be supplied to the motion compensation loop for use in motion compensation, if applicable. For example, the entropy decoder 52 may recover motion vectors and/or reference frame information for inter-coded macroblocks.\n\n The quantized coefficients are then dequantized by the dequantizer 54 to produce the transform domain coefficients, which arc then subjected to an inverse transform by the inverse transform processor 56 to recreate the \"video data\". It will be appreciated that, in some cases, such as with an intra-coded macroblock, the recreated \"video data\" is the residual data for use in spatial compensation relative to a previously decoded block within the frame. The spatial compensator 57 generates the video data from the residual data and pixel data from a previously decoded block. In other cases, such as inter-coded macroblocks, the recreated \"video data\" from the inverse transform processor 56 is the residual data for use in motion compensation relative to a reference block from a different frame. Both spatial and motion compensation may be referred to herein as \"prediction operations\".\n\n The motion compensator 62 locates a reference block within the frame buffer 58 specified for a particular inter-coded macroblock. It does so based on the reference frame information and motion vector specified for the inter-coded macroblock. It then supplies the reference block pixel data for combination with the residual data to arrive at the recreated video data for that macroblock.\n\n A deblocking process may then be applied to a reconstructed frame/slice, as indicated by the deblocking processor 60. After deblocking, the frame/slice is output as the decoded video frame 16, for example for display on a display device. It will be understood that the video playback machine, such as a computer, set-top box, DVD or Blu-Ray player, and/or mobile handheld device, may buffer decoded frames in a memory prior to display on an output device.\n\n Entropy coding is a fundamental part of all lossless and lossy compression schemes, including the video compression described above. The purpose of entropy coding is to represent a presumably decorrelated signal, often modeled by an independent, but not identically distributed process, as a sequence of bits. The technique used to achieve this must not depend on how the decorrelated signal was generated, but may rely upon relevant probability estimations for each upcoming symbol.\n\n There arc two common approaches for entropy coding used in practice: the first one is variable-length coding, which identifies input symbols or input sequences by codewords, and the second one is range (or arithmetic) coding, which encapsulates a sequence of subintervals of the [0, 1) interval, to arrive at a single interval, from which the original sequence can be reconstructed using the probability distributions that defined those intervals. Typically, range coding methods tend to offer better compression, while VLC methods have the potential to be faster. In either case, the symbols of the input sequence are from a finite alphabet.\n\n A special case of entropy coding is when the input alphabet is restricted to binary symbols. Here VLC schemes must group input symbols together to have any potential for compression, but since the probability distribution can change after each bit, efficient code construction is difficult. Accordingly, range encoding is considered to have greater compression due to its greater flexibility, but practical applications are hindered by the higher computational requirements of arithmetic codes.\n\n One of the techniques used in some entropy coding schemes, such as CAVLC and CABAC, both of which are used in H.264/AVC, is context modeling. With context modeling, each bit of the input sequence has a context, where the context is given by the bits that preceded it. In a first-order context model, the context may depend entirely upon the previous bit (symbol). In many cases, the context models may be adaptive, such that the probabilities associated with symbols for a given context may change as further bits of the sequence are processed. In yet other cases, such as those described herein, the context model is based on i.i.d. (independent and identically distributed) binary sources, wherein the context model defines a set of probabilities of producing a least probable symbol (LPS), and each bit of the input sequence is assigned one of the probabilities from the set. The bits associated with a given one of the probabilities are considered to be a sequence of symbols produced by an i.i.d. source.\n\n Reference is made to Figure 3, which shows a block diagram of an encoding process 100. The encoding process 100 includes a context modeling component 104 and an entropy coder 108. The context modeling component 104 receives the input sequence x 102, which in this example is a bit sequence (b0, b1, ..., bn). The context modeling component 104 determines a context for each bit bi based on one or more previous bits in the sequence, and determines, based on the adaptive context model, a probability pi associated with that bit bi, where the probability is the probability that the bit will be the Least Probable Symbol (LPS). The LPS may be \"0\" or \"1\" in a binary embodiment, depending on the convention or application. The context modeling component outputs the input sequence, i.e. the bits (b0, b1, ..., bn) along with their respective probabilities (p0, p1, ..., pn). The probabilities arc an estimated probability determined by the context model. This data is then input to the entropy coder 106, which encodes the input sequence using the probability information. For example, the entropy coder 106 may be a binary arithmetic coder. The entropy coder 106 outputs a bitstream 108 of encoded data.\n\n It will be appreciated each bit of the input sequence is processed serially to update the context model, and the serial bits and probability information are supplied to the entropy coder 106, which then entropy codes the bits to create the bitstream 108. Those ordinarily skilled in the art will appreciate that, in some embodiments, explicit probability information may not be passed from the context modeling component 104 to the entropy coder 106; rather, in some instances, for each bit the context modeling component 104 may send the entropy coder 106 an index or other indicator that reflects the probability estimation made be the context modeling component 104 based on the context model and the current context of the input sequence 102. The index or other indicator is indicative of the probability estimate associated with its corresponding bit.\n\n In US patent application 12/713,613, filed February 26, 2010, and owned in common herewith, a process and devices were described for encoding an input sequence using dual codesets. The primary codeset efficiently encoded the input sequence in a buffer-based encoding process. When a flush event occurred, for example due to end-of-frame, end of group-of-pictures, etc., secondary codewords from a secondary codeset were used to encode the partial subsequences associated with each probability defined by the context model.\n\n In US patent application 12/707,797, filed February 18, 2010, and owned in common herewith, methods and devices for parallel entropy coding are described. In one example, an input sequence is de-multiplexed into subsequences based on the associated probability of each input symbol (e.g. bit). The subsequences are each encoded by a respective one of the parallel entropy encoders. In similar manner at the decoder, the encoded bitstream is separated into encoded subsequences, which are each decoded by a respective one of the parallel decoders. The encoded bitstream may include overhead information regarding the length of each of the encoded subsequences to allow the decoder to separate them for parallel decoding.\n\n In both of these applications, the input sequence is first evaluated using the applicable context model to assign a probability to each input symbol, where the assigned probability is one of a predefined set of probabilities specified by the context model. In some instances, the number of probabilities can result in inefficiencies, either in the use of a large number of secondary codewords due to a flush event, or in the need to send a large amount of overhead specifying the length of each encoded subsequence for each probability value (or index). Each subsequence associated with each of the probabilities may be considered as originating from a \"source\" as that term is used herein.\n\n In accordance with one aspect, the present application proposes an encoder and decoder in which a new probability set is selected for the purposes of entropy encoding and decoding. The context modeling occurs as before using the predefined probability set associated with the context model: however, prior to entropy encoding and decoding, a new probability set is selected. A mapping defines the relationship between the predefined probability set and the new probability set. Each symbol that has been assigned a probability from the predefined probability set by the context model is then re-assigned or mapped to a new probability selected from the new probability set based on the mapping.\n\n The symbols are then binary entropy encoded using the new probabilities. In many embodiments, there are fewer new probabilities than predefined probabilities. In other words, the new probability set is smaller than the predefined probability set. In some instances, the new probability set is a subset of the probabilities in the predefined probability set, but not necessarily. With fewer probabilities, there are fewer subsequences, and thus less overhead and/or fewer secondary codewords. Accordingly, conceptually by using fewer probabilities we are allowing for fewer \"sources\". Each of the \"sources\" is still considered to be producing a random sequence of symbols as an i.i.d. binary source.\n\n Provided the loss of efficiency in merging symbols from more than one of the predefined probabilities into a single new probability is offset by the gain in reduced overhead or fewer secondary codewords, then the mapping improves the efficiency of the encoder and/or decoder.\n\n To illustrate one example embodiment, consider a finite number of probability sets P0, P1, ..., PK-1, where P0=P. For each Pk, we define a mapping Tk: P→Pk by the following rule:", null, "We denote R(p,q) = plog(p/q) + (1-p)log[(1-p)/(1-q)]. It will be understood that the quantity R(p,q) is a measure of relative entropy. It will be noted that the mapping expression for Tk(p) results in mapping each probability p of the probability set P to a probability q of the new probability set Pk, on the basis that the selection of probability q minimizes the relative entropy expression.\n\n In one embodiment, the Tk mappings are computed by the following algorithm, which for any probability set Pk runs in O(N) time, provided that the sets P and Pk are sorted according to decreasing probability:\n1. 1. Let i=0, j=0, p=P.\n2. 2. If i==N, the algorithm terminates.\n3. 3. While j<\|Pk\|-1 and Pk[j] >= p, set j=j+1\n4. 4. If j==0 or R(p, Pk[j]) < R(p, Pk[j-1]), set Tk(p) = Pk[j]. Otherwise, set Tk(p) = Pk[j-1].\n5. 5. Let i=i+1, p=P[i] and go to 2.\n\n In the off-line setting, the probability sets and mappings are known to both the encoder and the decoder. In the on-line setting, the sets and the mappings between them are updated on the run, optionally adaptively based on the input data.\n\n In the description below various example embodiments are described. The examples vary in how the probability set is selected. As a general rule, each variation introduces some overhead to the encoded sequences. This overhead may be taken into account when finding the best probability set for encoding. It will be appreciated that the underlying entropy coding method may incur some overhead on its own. For the purposes of this description the encoded sequences may be regarded as inseparable, their overhead included, and we discuss only the overhead for this framework. As a consequence for this discussion we may not use the entropy to compute the sequence lengths, but use the actual encoded length, or at least a reliable upper bound.\n\nOffline Solution\n\n In one exemplary application, the probability sets and mappings may be determined off-line (prior to both the encoding and decoding process, thereby not affecting encoding and decoding times). In one embodiment, the encoding process can be described as follows:\n\n For each k = 0, ..., K-1, the entropy coder encodes the N sources using the probability mappings Tk(pj) for pj ∈ P. That is, instead of using the probability pj for source j, the output of source j will be merged into the output of the source with probability Tk(pj), and the interleaved output of all such sources are encoded with probability Tk(pj). For index k this results in \|Pk\| encoded sequences of total length Lk. The encoder then picks index m that minimizes Lk (k=0, ..., K-1), writes C(m) into the output, followed by the encoded sequences for mapping m (the definition of the prefix code C() is described in US patent application 12/707,797). The overhead in this example is \|C(m)\|.\n\n If the number K of candidate sets is low, choosing the best one can be done by an exhaustive search, that is, trying out each of P0, ..., PK-1. This increases encoding computational complexity, but it does not affect the decoder. If the number of candidate sets is too high for an exhaustive linear search, we may use a heuristic search, such as a search based on gradients evaluated from neighbouring encoded length values.\n\n The decoder reads out the value of m from the decoded file, after which the probability value Tm(pj) is used instead of pj ∈ P for the sources Pm instead of P. That is, whenever the decoder needs to determine a symbol from source j, it looks up the source j' to which j is mapped based on Pm and Tm, then decodes the next bit from the encoded data for j' using probability Tm(pj).\n\n One embodiment of the present application may be illustrated by way of the following example. Let P be the probability values defined for CABAC: pk = 0.5αk for k=0, ..., 63, α = (20.01875)1/63. Let P0=P={pk\|k=0, ..., 63}, and for k = 0, ..., 9:", null, "This creates ten probability sets, each of them the subset of P, and with decreasing size, evenly distributed probability values, but retaining the original range (p0 to p63). Since the size of Pk is smaller than the size of P0, the overhead associated to encoding with Pk is also smaller, but the encoded length may be higher. The proposed encoder encodes the input symbols with each of P0, ..., P9, and chooses the one that gives the shortest (overhead + payload) length.\n\nOnline Solution with Fixed Source Probabilities and Dynamic Probability Sets\n\n In another exemplary application, the probability sets and/or the mappings may be determined online, i.e. dynamically. Thus for video encoding the sets and mappings may be different for different slices or frames or groups of frames (GOPs). As opposed to the offline solution outlined above, here we have to specify how to determine the sets P1, ..., PK-1 and how to inform the decoder about the mappings Tm(pj).\n\n Choosing the candidate sets P1, ..., PK-1 may be done based on probabilities, so that the most frequent values are collected in each set (see the example below). Once these sets are known, the mappings Tk are determined with the algorithm above, and the best candidate m is selected using the methods described above in the offline example. After this, the encoder may write C(\|Pm\|), C(Pm(0)), C(Pm(1)), ..., C(Pm(\|Pm\|-1)) into the encoded file, followed by the encoded sequences. We note that a probability is encoded by its index in P. For example, if Pm(0)=0.4, and P={0.5, 0.4, 0.3, ...}, then C(Pm(0)) is analogous to C(1). The overhead of this variation is \|C(\|Pm\|)\| + \|C(Pm(0))\| + ... + \|C(Pm(\|Pm\|-1))\|. This also means that for this exemplary application the set Pk is a subset of P.\n\n The decoder reads the number of merged sources from the file, followed by their probability indices, and from this it reconstructs Pm. Then it computes the mapping Tm.\n\n One example embodiment may be illustrated by way of the following example. Let P be the probability values defined for CABAC: pk = 0.5αk for k=0, ..., 63, α = (20.01875)1/63. For each slice or each frame or each GOP, let Pk denote the subset of P consisting of the (64-k) most frequently used probabilities in P for the current slice or frame, where k=0, 1, ..., 63. The mappings Tk: P→Pk are determined by the algorithm above.\n\n In this example, we have 64 probability sets. Assuming that a fixed length code is used to communicate Pk to the decoder, we find the index number m at which the overhead of using Pm is the smallest among the 64 candidate choices k=0, 1, ..., 63.\n\nOnline Solution with Adaptive Source Probabilities and Static Probability Sets\n\n In another exemplary application, the probability values of P are not known in advance, but rather calculated empirically from the observed data. The context model separates the input symbols into different source buffers, and the source probability is estimated as (number of ones in buffer) / (number of symbols in buffer) for each source. We denote the empirical probabilities by Q = {q0, ..., qN-1}. Then, for the fixed P0, ..., PK-1 probability sets, the encoder first determines the mappings between Pk and Q, then finds the one Pm that gives the smallest encoded length Lm, based on cither an exhaustive or heuristical search, as explained above.\n\n Since Q is in general different from P. the actual mappings Tk(qj) may differ from the static mappings Tk(pj), which are known to both the encoder and decoder. Since the decoder needs Tk(qj), these must be transmitted. In one application, the encoder may just include C(Tm(q0)), C(Tm(q1)), ..., C(Tm(qN-1)) in the overhead, as was described above. However, we point out that the correlation between the two mappings is usually high. Therefore, in another embodiment, the encoder transmits Tm(qj) with a run-length scheme based on the known mappings Tm(pj). In one example, this may work as follows: first, set M to 0. Let j be the first index from M for which Tm(qM+j) ≠ Tm(pM+j), or j=N-M if all of them are equal. Then encode C(j), and if j < N-M, encode C(Tm(qM+j)) as well. Then set M to j+1, and if M<N, repeat the procedure, otherwise stop. The overhead of this variation is \|C(m)\| + \|C(jl)\| + \|C(Tm(qj1))\| + \|C(j2)\| + \|C(Tm(qj2))\| + ... + \|C(jr)\|.\n\n To illustrate by way of an example, let P={0.5, 0.4, 0.3, 0.25, 0.2, 0.15, 0.12, 0.10}, and Pm = {0.5, 0.32, 0.18, 0.12}. Then the mapping algorithm gives Tm(0.5)=0.5, Tm(0.4)=0.32, Tm(0.3)=0.32, Tm(0.25)=0.32, Tm(0.2)=0.18, Tm(0.15)=0.18, Tm(0.12)=0.12 and Tm(0.10)=0.12. Suppose that the empirical probabilities are Q={0.48, 0.45, 0.31, 0.20, 0.18, 0.26, 0.08, 0.12}. Then the actual mapping, by the same algorithm, is Tm(0.48)=0.5, Tm(0.45)=0.5, Tm(0.31)=0.32, Tm(0.20)=0.18, Tm(0.18)=0.18, Tm(0.26)=0.32, Tm(0.08)=0.12 and Tm(0.12)=0.12. Then the run-length sequence is (1, 0.5), (1, 0.18), (1, 0.32), (2). For prefix encoding the probability values are represented by their indices in Pm: 0.5 is 0, 0.32 is 1, 0.18 is 2, 0.12 is 3. The encoded header is therefore C(m), C(1), C(0), C(1), C(2), C(1), C(1), C(2).\n\nOnline Solution with Adaptive Source Probabilities and Dynamic Probability Sets\n\n In another exemplary application, the probability values Q are calculated from the observed data, as described immediately above, and the probability sets P1, ..., PK-1 are determined dynamically. Hence, first the empirical probabilities Q are determined from the (number of ones in buffer) / (number of symbols in buffer) for each source. From this point, this variation follows the one given above: first the candidate sets P0, ..., PK-1 are determined (where P0 is the predetermined probability set and P1,...PK-1 are the dynamically determined probability sets), the mappings Tk are computed, and the best candidate m is selected using the methods described above. After this, the encoder writes C(\|Pm\|), C(Pm(0)), C(Pm(1)), ..., C(Pm(\|Pm\|-1)) into the encoded file, followed by the actual mapping Tm(qj), followed by the encoded sequences. The overhead of this variation is \|C(\|Pm\|)\| + \|C(Pm(0))\| + ... + \|C(Pm(\|Pm\|-1))\| + \|C(j1)\| + \|C(Tm(qj1))\| + \|C(j2)\| + \|C(Tm(qj2))\| + ... + \|C(jr)\|. In this specific example, it is presumed that Pk is a subset of P. The actual mapping may be written into the overhead in the same manner as described above. For example, run-length coding is an option.\n\n The decoder first reads the number of merged sources from the encoded file, then the merged source probabilities indexed by elements of P, as given by C(Pm(0)), ..., C(Pm(\|Pm\|-1)). After this, the decoder can compute the mapping Tm(pj) between P and Pm. Using this mapping and the run-length coded actual mapping, it then proceeds to compute the actual mapping Tm(qj). Once ready, it can start decoding the sequences, decoding from the buffer of Tm(qj) any time a request for a new symbol is made from source j.\n\n The selection of the probability set to use from amongst a number of possible candidate probability sets may be application dependent. In some instances, the mapping for each of the candidate sets may be predefined and known. In some instances, the mapping may be determined dynamically, particularly if there are a large number of candidate probability sets.\n\n To select the probability set to use for a given video, frame, etc., in one embodiment each of the candidates may be tested by using that set to encode the data and the optimal probability set selected based on the testing. This is the brute force approach.\n\n In another embodiment, the selection may be based on the data type or characteristics. For example, one or more particular candidate probabilities sets may be designated for I or P frames, whereas a different candidate probability set or sets may be designated for B frames. Yet other characteristics of the video data may used as the basis for determining the candidate set or sets.\n\n In yet another embodiment, as described above, the selection may be based on an actual distribution of bits amongst the predefined probabilities, i.e. the statistics for the actual data, for example for the frame, etc.. For example, any probabilities that have no bits assigned may dropped; or any probabilities having few bits may be merged with neighbouring probabilities.\n\n In one embodiment, the dynamic merging or consolidation of predetermined probabilities based on actual data statistics may be based upon obtaining approximately the same number of bits for each probability, i.e. balancing.\n\n In yet another embodiment, empirical probability data may be used to dynamically determine the mapping as opposed to the predefined probabilities, since in some instances (particularly at higher order indices) the actual probability statistics may vary considerably from the probabilities predicted by the context model.\n\n Reference is now made to Figure 4, which shows, in flowchart form, one example process 200 for encoding an input sequence by reducing sources. The process 200 begins in operation 202 with application of the context model. In particular, each symbol of the input sequence is processed and assigned an estimated probability. The estimated probability is selected from a defined probability set using the context of the symbol and the context model. As described above, the defined probability set may be predefined and associated with the context model. An example is the 64 states defined in CABAC.\n\n In operation 204, a new probability set is selected. In a simple embodiment, this may mean selecting the sole candidate new probability set. In other embodiments, this operation may involve testing candidate probability sets to see which of them results in the shortest overall length of encoded sequences. In some instances, this operation may involve dynamically generating a new candidate probability set based on actual statistics from the input sequence, as described above.\n\n Operation 206 involves defining or generating the mapping between the predefined probability set and the new probability set. It will be understood that the mapping may be predetermined and stored if the candidate probability sets and the predetermined probability set are predetermined. It will also be understood that the mapping operation may form part of operation 204, i.e. the selection of the new probability set from amongst candidate sets, since a mapping is defined for each candidate set. It will also be appreciated that the mapping means that cach symbol that was assigned a probability from the predefined probability set in operation 202 may be assigned a probability from the new probability set based upon the mapping.\n\n The symbols are entropy encoded in operation 208 to produce an encoded bitstream. The entropy encoding is based upon the new probabilities assigned in operation 206. In other words, the entropy encoding is based upon the new, and in many embodiments reduced-size, probability set.\n\n Figure 5 shows, in flowchart form, a process 300 for decoding an encoded bitstream in accordance with the present application. The process 300 begins in operation 302 with reading information from the bitstream regarding the new probability set. As described above, in some embodiments, this information may simply indicate the selection of one of a plurality of predefined candidate probability sets already known to the decoder. In some embodiments, this information may identify particular probabilities from the predefined probability set that are included in the new probability set, i.e. the information may define or identify a subset of the predefined probability set. In yet other embodiments, this information may provide explicit probability information. In all embodiments, through operation 302 the decoder obtains the new probability set.\n\n Operation 304 includes determining the mapping between the new probability set and the predetermined probability set. The mapping may be predefined such that operation 304 involves retrieving the predefined mapping from memory or other storage. In some embodiments, the mapping may be specified in the bitstream and the decoder may read the mapping from the bitstream. In yet other embodiments, the mapping may be generated by the decoder, for example using the relative entropy algorithm described herein.\n\n In operation 306, the encoded bitstream is entropy decoded. Entropy decoding includes obtaining decoded subsequences, where each decoded subsequence is associated with a probability from the new set of probabilities. The entropy decoding in operation 306 is based upon the context model, which provides a probability estimate for each symbol, and the mapping, which maps the probability estimate to its corresponding probability from the new set of probabilities and, thus, to one of the decoded subsequenes.\n\n Reference now made to Figure 6, which shows a simplified block diagram of an example embodiment of an encoder 900. The encoder 900 includes a processor 902, memory 904, and an encoding application 906. The encoding application 906 may include a computer program or application stored in memory 904 and containing instructions for configuring the processor 902 to perform steps or operations such as those described herein. For example, the encoding application 906 may encode and output video bitstreams encoded in accordance with the probability mapping encoding process described herein. The encoding application 906 may include an entropy encoder 26 configured to entropy encode input sequences and output a bitstream using one or more of the processes described herein. It will be understood that the encoding application 906 may be stored in on a computer readable medium, such as a compact disc, flash memory device, random access memory, hard drive, etc.\n\n In some embodiments, the processor 902 in the encoder 900 may be a single processing unit configured to implement the instructions of the encoding application 906. In some other embodiments, the processor 902 may include more than one processing unit capable of executing instructions in parallel. The multiple processing units may be logically or physically separate processing units. In some instances, the encoder 900 may include N or more processing units, wherein N of the processing units are configured by the encoding application 906 to operate as parallel entropy coders for implementing the methods described herein. It will further be appreciated that in some instances, some or all operations of the encoding application 906 and one or more processing units may be implemented by way of application-specific integrated circuit (ASIC), etc.\n\n Reference is now also made to Figure 7, which shows a simplified block diagram of an example embodiment of a decoder 1000. The decoder 1000 includes a processor 1002, a memory 1004, and a decoding application 1006. The decoding application 1006 may include a computer program or application stored in memory 1004 and containing instructions for configuring the processor 1002 to perform steps or operations such as those described herein. The decoding application 1006 may include an entropy decoder 1008 configured to receive a bitstream encoded in accordance with the probability mapping entropy encoding process described herein, and to extract encoded subsequences from the bitstream and decode them in using mapped probabilities. The decoding application 1006 may configure the processor to decode the encoded subsequences in parallel to produce parallel decoded sequences and to interleave the symbols of the decode sequences to produce a reconstructed sequences. It will be understood that the decoding application 1006 may be stored in on a computer readable medium, such as a compact disc, flash memory device, random access memory, hard drive, etc.\n\n In some embodiments, the processor 1002 in the decoder 1000 may be a single processing unit configured to implement the instructions of the decoding application 1006. In some other embodiments, the processor 1002 may include more than one processing unit capable of executing instructions in parallel. The multiple processing units may be logically or physically separate processing units. In some instances, the decoder 1000 may include d, N or more or fewer processing units, wherein the processing units are configured by the decoding application 1006 to operate as parallel entropy decoders for implementing the methods described herein. It will further be appreciated that in some instances, some or all operations of the decoding application 1006 and one or more processing units may be implemented by way of application-specific integrated circuit (ASIC), etc.\n\n It will be appreciated that the decoder and/or encoder according to the present application may be implemented in a number of computing devices, including, without limitation, servers, suitably programmed general purpose computers, set-top television boxes, television broadcast equipment, and mobile devices. The decoder or encoder may be implemented by way of software containing instructions for configuring a processor to carry out the functions described herein. The software instructions may be stored on any suitable tangible computer-readable medium or memory, including CDs, RAM, ROM, Flash memory, etc.\n\n It will be understood that the encoder and decoder described herein and the module, routine, process, thread, or other software component implementing the described method/process for configuring the encoder may be realized using standard computer programming techniques and languages. The present application is not limited to particular processors, computer languages, computer programming conventions, data structures, other such implementation details. Those skilled in the art will recognize that the described processes may be implemented as a part of computer-executable code stored in volatile or non-volatile memory, as part of an application-specific integrated chip (ASIC), etc.\n\n Certain adaptations and modifications of the described embodiments can be made. Therefore, the above discussed embodiments are considered to be illustrative and not restrictive.\n1. A method (300) for decoding an encoded bitstream to obtain a sequence of symbols, the symbols belonging to a finite alphabet, wherein a context model specifies a predefined probability set, and wherein each symbol of the sequence of symbols is assigned a probability from the predefined probability set on the basis of the context model (202), the method comprising:\n\nreading, from the bitstream, information identifying a new probability set (302), wherein the new probability set is not identical to the predefined probability set and is a subset of the predefined probability set;\n\nassigning, to each of the symbols of the sequence of symbols, a respective probability from the new probability set based upon a mapping (304), wherein the mapping maps each of the probabilities of the predefined probability set to a respective one of the probabilities from the new probability set, wherein assigning to each of the symbols a respective probability, includes defining the mapping by determining, for each probability of the predefined probability set, to which one of the probabilities from the new probability set it is mapped, and wherein determining is based on selecting the one of the probabilities from the new probability set that minimizes a relative entropy expression; and\n\nentropy decoding the encoded bitstream on the basis of their respective assigned probabilities from the new probability set (306).\n\n2. The method claimed in any one of claim 1, wherein the information identifying the new probability set comprises information identifying one of a predefined set of candidate probability sets.\n\n3. The method claimed in any one of claim 1, wherein the information identifying the new probability set specifies the probabilities from the predefined probability set that are included in the new probability set.\n\n4. The method claimed in claim 3, wherein reading further comprises dynamically generating the mapping by determining, for each probability of the predefined probability set, to which one of the probabilities from the new probability set it is mapped based on minimizing a relative entropy expression.\n\n5. A decoder (1000) for decoding an encoded bitstream to obtain a sequence of symbols, the symbols belonging to a finite alphabet, wherein a context model specifies a predefined probability set, and wherein each symbol of the sequence of symbols assigned with a probability from the predefined probability set on the basis of the context model, the decoder comprising:\n\na memory (1004);\n\na processor (1002);\n\na decoding application (1006) executable by the processor and which, when executed, configures the processor to perform the method claimed in any one of claims 1 to 4.\n\n6. A tangible computer-readable medium storing computer-executable instructions which, when executed by one or more processors, implement the steps of the method claimed in any one of claims 1 to 4.\n\n1. Verfahren (300) zum Decodieren für einen codierten Bitstrom, um eine Folge von Symbolen zu erhalten, wobei die Symbole zu einem endlichen Alphabet gehören, wobei ein Kontextmodell einen vordefinierten Wahrscheinlichkeitssatz spezifiziert, und wobei jedem Symbol der Folge von Symbolen basierend auf dem Kontextmodell (202) eine Wahrscheinlichkeit aus dem vordefinierten Wahrscheinlichkeitssatz zugewiesen wird, wobei das Verfahren umfasst:\n\nLesen, aus dem Bitstrom, von Informationen die einen neuen Wahrscheinlichkeitssatz (302) identifizieren, wobei der neue Wahrscheinlichkeitssatz nicht mit dem vordefinierten Wahrscheinlichkeitssatz identisch ist und eine Teilmenge des vordefinierten Wahrscheinlichkeitssatzes ist;\n\nZuweisen, zu jedem der Symbole der Folge von Symbolen, einer jeweiligen Wahrscheinlichkeit aus dem neuen Wahrscheinlichkeitssatz basierend auf einer Abbildung (304), wobei die Abbildung jede der Wahrscheinlichkeiten des vordefinierten Wahrscheinlichkeitssatzes auf eine jeweilige eine der Wahrscheinlichkeiten aus dem neuen Wahrscheinlichkeitssatz abbildet, wobei das Zuweisen einer jeweiligen Wahrscheinlichkeit zu jedem der Symbole ein Definieren der Abbildung durch Bestimmen, für jede Wahrscheinlichkeit des vordefinierten Wahrscheinlichkeitssatzes, in der eine der Wahrscheinlichkeiten aus dem neuen Wahrscheinlichkeitssatz diese abbildet, beinhaltet, und wobei das Bestimmen auf einem Auswählen der einen der Wahrscheinlichkeiten aus dem neuen Wahrscheinlichkeitssatz, das einen relativen Entropieausdruck minimiert, basiert; und\n\nEntropiedecodieren für den codierten Bitstrom basierend auf seinen jeweiligen zugewiesenen Wahrscheinlichkeiten aus dem neuen Wahrscheinlichkeitssatz (306).\n\n2. Verfahren nach einem der Ansprüche 1, wobei die Informationen, die den neuen Wahrscheinlichkeitssatz identifizieren, Informationen umfassen, die einen von einem vordefinierten Satz von Kandidatenwahrscheinlichkeitssätzen identifizieren.\n\n3. Verfahren nach einem der Ansprüche 1, wobei die Informationen, die den neuen Wahrscheinlichkeitssatz identifizieren, die Wahrscheinlichkeiten aus dem vordefinierten Wahrscheinlichkeitssatz spezifizieren, die in dem neuen Wahrscheinlichkeitssatz beinhaltet sind.\n\n4. Verfahren nach Anspruch 3, wobei das Lesen ferner ein dynamisches Erzeugen der Abbildung durch Bestimmen für jede Wahrscheinlichkeit des vordefinierten Wahrscheinlichkeitssatzes, in der eine der Wahrscheinlichkeiten aus dem neuen Wahrscheinlichkeitssatz diese basierend auf einem Minimieren eines relativen Entropieausdrucks abbildet, umfasst.\n\n5. Decodierer (1000) zum Decodieren für einen codierten Bitstrom, um eine Folge von Symbolen zu erhalten, wobei die Symbole zu einem endlichen Alphabet gehören, wobei ein Kontextmodell einen vordefinierten Wahrscheinlichkeitssatz spezifiziert, und wobei jedem Symbol der Folge von Symbolen basierend auf dem Kontextmodell eine Wahrscheinlichkeit aus dem vordefinierten Wahrscheinlichkeitssatz zugewiesen ist, wobei der Decoder umfasst:\n\neinen Speicher (1004);\n\neinen Prozessor (1002);\n\neine Decodierungsanwendung (1006), die durch den Prozessor ausführbar ist und die, wenn sie ausgeführt wird, den Prozessor konfiguriert, um das beanspruchte Verfahren nach einem der Ansprüche 1 bis 4 durchzuführen.\n\n6. Greifbares computerlesbares Medium, das computerausführbare Anweisungen speichert, die, wenn sie durch einen oder mehrere Prozessoren aufgeführt werden, die Schritte des beanspruchten Verfahrens nach einem der Ansprüche 1 bis 4 implementieren.\n\n1. Procédé (300) de décodage d'un train de bits codé pour obtenir une séquence de symboles, les symboles appartenant à un alphabet fini, un modèle de contexte spécifiant un ensemble prédéfini de probabilités, et chaque symbole de la séquence de symboles se voyant attribuer une probabilité à partir de l'ensemble prédéfini de probabilités sur la base du modèle de contexte (202), le procédé comprenant :\n\nla lecture, à partir du train de bits, d'informations identifiant un nouvel ensemble de probabilités (302), le nouvel ensemble de probabilités n'étant pas identique à l'ensemble prédéfini de probabilités et étant un sous-ensemble de l'ensemble prédéfini de probabilités ;\n\nl'attribution, à chacun des symboles de la séquence de symboles, d'une probabilité respective à partir du nouvel ensemble de probabilités sur la base d'un mappage (304), le mappage mappant chacune des probabilités de l'ensemble prédéfini de probabilités à une probabilité respective des probabilités du nouvel ensemble de probabilités, l'attribution à chacun des symboles d'une probabilité respective comprenant la définition du mappage en déterminant, pour chaque probabilité de l'ensemble prédéfini de probabilités, à laquelle des probabilités du nouvel ensemble de probabilités il est mappé, et la détermination étant basée sur la sélection de la probabilité, parmi les probabilités du nouvel ensemble de probabilités, qui réduit au minimum l'expression de l'entropie relative ; et\n\nle décodage entropique du train de bits codé sur la base de leurs probabilités attribuées respectives à partir du nouvel ensemble de probabilités (306).\n\n2. Procédé selon l'une quelconque des revendications 1, dans lequel les informations identifiant le nouvel ensemble de probabilités comprennent des informations identifiant un ensemble candidat de probabilités dans un ensemble prédéfini d'ensembles candidats de probabilités.\n\n3. Procédé selon l'une quelconque des revendications 1, dans lequel les informations identifiant le nouvel ensemble de probabilités spécifient les probabilités, à partir de l'ensemble prédéfini de probabilités, qui font partie du nouvel ensemble de probabilités.\n\n4. Procédé selon la revendication 3, dans lequel la lecture comprend en outre la génération dynamique du mappage en déterminant, pour chaque probabilité de l'ensemble prédéfini de probabilités, à laquelle des probabilités du nouvel ensemble de probabilités il est mappé sur la base de la réduction au minimum d'une expression de l'entropie relative.\n\n5. Décodeur (1000) servant à décoder un train de bits codé pour obtenir une séquence de symboles, les symboles appartenant à un alphabet fini, un modèle de contexte spécifiant un ensemble prédéfini de probabilités, et chaque symbole de la séquence de symboles se voyant attribuer une probabilité à partir de l'ensemble prédéfini de probabilités sur la base du modèle de contexte, le décodeur comprenant :\n\nune mémoire (1004) ;\n\nun processeur (1002) ;\n\nune application de décodage (1006) exécutable par le processeur et qui, lorsqu'elle est exécutée, configure le processeur de façon à ce qu'il réalise le procédé selon l'une quelconque des revendications 1 à 4.\n\n6. Support tangible lisible par ordinateur stockant des instructions exécutables par ordinateur qui, lorsqu'elles sont exécutées par un ou plusieurs processeurs, mettent en oeuvre les étapes du procédé selon l'une quelconque des revendications 1 à 4.", null, "", null, "", null, "", null, "", null, "", null, "", null, "REFERENCES CITED IN THE DESCRIPTION\n\nThis list of references cited by the applicant is for the reader's convenience only. It does not form part of the European patent document. Even though great care has been taken in compiling the references, errors or omissions cannot be excluded and the EPO disclaims all liability in this regard.\n\nPatent documents cited in the description\n\nNon-patent literature cited in the description\n\n• KARWOWSKI DAMIAN et al.Improved context-adaptive arithmetic coding in H.264/AVC17th European Signal Processing Conference, 2009, 2216-2220 \n• MRAK M et al.Comparison of context-based adaptive binary arithmetic coders in video compressionVideo/Image Processing and Multimedia Communications, 2003, 277-286" ]	[ null, "https://data.epo.org/publication-server/img/EPO_BL_WORD.jpg", null, "https://data.epo.org/publication-server/image", null, "https://data.epo.org/publication-server/image", null, "https://data.epo.org/publication-server/image", null, "https://data.epo.org/publication-server/image", null, "https://data.epo.org/publication-server/image", null, "https://data.epo.org/publication-server/image", null, "https://data.epo.org/publication-server/image", null, "https://data.epo.org/publication-server/image", null, "https://data.epo.org/publication-server/image", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8948221,"math_prob":0.8830393,"size":46175,"snap":"2022-05-2022-21","text_gpt3_token_len":9968,"char_repetition_ratio":0.18227892,"word_repetition_ratio":0.104211666,"special_character_ratio":0.22551164,"punctuation_ratio":0.12274327,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97343856,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-01-28T07:43:16Z\",\"WARC-Record-ID\":\"<urn:uuid:496757dc-37b2-46cc-aae8-1bf4c6829f69>\",\"Content-Length\":\"87609\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c9cde661-a913-4d0b-aff9-33d8a98ba4d9>\",\"WARC-Concurrent-To\":\"<urn:uuid:5c8ed537-fdae-4519-87d8-f4158608bc1a>\",\"WARC-IP-Address\":\"195.6.57.68\",\"WARC-Target-URI\":\"https://data.epo.org/publication-server/html-document?PN=EP3550726%20EP%203550726&iDocId=6391744&disclaimer=true&citations=true\",\"WARC-Payload-Digest\":\"sha1:URPC6MZBO73EEDX6DXA244GZRE3LSXD7\",\"WARC-Block-Digest\":\"sha1:ZJYEPF6T5SUDCJSZ67DZYRYB2MDVW55P\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-05/CC-MAIN-2022-05_segments_1642320305423.58_warc_CC-MAIN-20220128074016-20220128104016-00375.warc.gz\"}"}
https://arc2.nesa.nsw.edu.au/view/byarea/course/15240/question/1414	[ "# Question 8\n\nSummary: Calculate values of constants in equation modelling exponential population growth. Find the probability of a single-stage event and of a multi-stage event. Determine the maximum and minimum values of the rate of change of a quantity. Sketch the graph of the quantity as a function of time and identify any points on the graph where the concavity changes." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8270815,"math_prob":0.9947029,"size":439,"snap":"2022-40-2023-06","text_gpt3_token_len":87,"char_repetition_ratio":0.121839084,"word_repetition_ratio":0.0,"special_character_ratio":0.19134396,"punctuation_ratio":0.06329114,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99140143,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-02-08T00:05:46Z\",\"WARC-Record-ID\":\"<urn:uuid:30c0d948-b173-4831-a86d-a52414e3b35f>\",\"Content-Length\":\"29740\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c67b6dc5-8608-4b68-892a-ebdd043a0a93>\",\"WARC-Concurrent-To\":\"<urn:uuid:9f0d65aa-e736-44b1-ae45-c05236fd46fd>\",\"WARC-IP-Address\":\"18.154.227.124\",\"WARC-Target-URI\":\"https://arc2.nesa.nsw.edu.au/view/byarea/course/15240/question/1414\",\"WARC-Payload-Digest\":\"sha1:AJRKK7TWCJ2CV67DHOQYMMHVWEQ6PCVV\",\"WARC-Block-Digest\":\"sha1:75GB2RBIXV7W6BZMSQ3NXWXZRKFUKU6C\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764500664.85_warc_CC-MAIN-20230207233330-20230208023330-00595.warc.gz\"}"}
https://www.colorhexa.com/3a7474	[ "# #3a7474 Color Information\n\nIn a RGB color space, hex #3a7474 is composed of 22.7% red, 45.5% green and 45.5% blue. Whereas in a CMYK color space, it is composed of 50% cyan, 0% magenta, 0% yellow and 54.5% black. It has a hue angle of 180 degrees, a saturation of 33.3% and a lightness of 34.1%. #3a7474 color hex could be obtained by blending #74e8e8 with #000000. Closest websafe color is: #336666.\n\n• R 23\n• G 45\n• B 45\nRGB color chart\n• C 50\n• M 0\n• Y 0\n• K 55\nCMYK color chart\n\n#3a7474 color description : Dark moderate cyan.\n\n# #3a7474 Color Conversion\n\nThe hexadecimal color #3a7474 has RGB values of R:58, G:116, B:116 and CMYK values of C:0.5, M:0, Y:0, K:0.55. Its decimal value is 3830900.\n\nHex triplet RGB Decimal 3a7474 `#3a7474` 58, 116, 116 `rgb(58,116,116)` 22.7, 45.5, 45.5 `rgb(22.7%,45.5%,45.5%)` 50, 0, 0, 55 180°, 33.3, 34.1 `hsl(180,33.3%,34.1%)` 180°, 50, 45.5 336666 `#336666`\nCIE-LAB 45.152, -18.881, -5.861 11.142, 14.651, 18.763 0.25, 0.329, 14.651 45.152, 19.769, 197.245 45.152, -25.037, -5.408 38.276, -15.023, -2.27 00111010, 01110100, 01110100\n\n# Color Schemes with #3a7474\n\n• #3a7474\n``#3a7474` `rgb(58,116,116)``\n• #743a3a\n``#743a3a` `rgb(116,58,58)``\nComplementary Color\n• #3a7457\n``#3a7457` `rgb(58,116,87)``\n• #3a7474\n``#3a7474` `rgb(58,116,116)``\n• #3a5774\n``#3a5774` `rgb(58,87,116)``\nAnalogous Color\n• #74573a\n``#74573a` `rgb(116,87,58)``\n• #3a7474\n``#3a7474` `rgb(58,116,116)``\n• #743a57\n``#743a57` `rgb(116,58,87)``\nSplit Complementary Color\n• #74743a\n``#74743a` `rgb(116,116,58)``\n• #3a7474\n``#3a7474` `rgb(58,116,116)``\n• #743a74\n``#743a74` `rgb(116,58,116)``\n• #3a743a\n``#3a743a` `rgb(58,116,58)``\n• #3a7474\n``#3a7474` `rgb(58,116,116)``\n• #743a74\n``#743a74` `rgb(116,58,116)``\n• #743a3a\n``#743a3a` `rgb(116,58,58)``\n• #214141\n``#214141` `rgb(33,65,65)``\n• #295252\n``#295252` `rgb(41,82,82)``\n• #326363\n``#326363` `rgb(50,99,99)``\n• #3a7474\n``#3a7474` `rgb(58,116,116)``\n• #438585\n``#438585` `rgb(67,133,133)``\n• #4b9696\n``#4b9696` `rgb(75,150,150)``\n• #54a7a7\n``#54a7a7` `rgb(84,167,167)``\nMonochromatic Color\n\n# Alternatives to #3a7474\n\nBelow, you can see some colors close to #3a7474. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #3a7466\n``#3a7466` `rgb(58,116,102)``\n• #3a746a\n``#3a746a` `rgb(58,116,106)``\n• #3a746f\n``#3a746f` `rgb(58,116,111)``\n• #3a7474\n``#3a7474` `rgb(58,116,116)``\n• #3a6f74\n``#3a6f74` `rgb(58,111,116)``\n• #3a6a74\n``#3a6a74` `rgb(58,106,116)``\n• #3a6674\n``#3a6674` `rgb(58,102,116)``\nSimilar Colors\n\n# #3a7474 Preview\n\nThis text has a font color of #3a7474.\n\n``<span style=\"color:#3a7474;\">Text here</span>``\n#3a7474 background color\n\nThis paragraph has a background color of #3a7474.\n\n``<p style=\"background-color:#3a7474;\">Content here</p>``\n#3a7474 border color\n\nThis element has a border color of #3a7474.\n\n``<div style=\"border:1px solid #3a7474;\">Content here</div>``\nCSS codes\n``.text {color:#3a7474;}``\n``.background {background-color:#3a7474;}``\n``.border {border:1px solid #3a7474;}``\n\n# Shades and Tints of #3a7474\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, #060b0b is the darkest color, while #fdfefe is the lightest one.\n\n• #060b0b\n``#060b0b` `rgb(6,11,11)``\n• #0c1818\n``#0c1818` `rgb(12,24,24)``\n• #132626\n``#132626` `rgb(19,38,38)``\n• #193333\n``#193333` `rgb(25,51,51)``\n• #204040\n``#204040` `rgb(32,64,64)``\n• #264d4d\n``#264d4d` `rgb(38,77,77)``\n• #2d5a5a\n``#2d5a5a` `rgb(45,90,90)``\n• #336767\n``#336767` `rgb(51,103,103)``\n• #3a7474\n``#3a7474` `rgb(58,116,116)``\n• #418181\n``#418181` `rgb(65,129,129)``\n• #478e8e\n``#478e8e` `rgb(71,142,142)``\n• #4e9b9b\n``#4e9b9b` `rgb(78,155,155)``\n• #54a8a8\n``#54a8a8` `rgb(84,168,168)``\n• #60b0b0\n``#60b0b0` `rgb(96,176,176)``\n• #6db6b6\n``#6db6b6` `rgb(109,182,182)``\n• #7bbdbd\n``#7bbdbd` `rgb(123,189,189)``\n• #88c3c3\n``#88c3c3` `rgb(136,195,195)``\n• #95caca\n``#95caca` `rgb(149,202,202)``\n• #a2d0d0\n``#a2d0d0` `rgb(162,208,208)``\n• #afd7d7\n``#afd7d7` `rgb(175,215,215)``\n• #bcdddd\n``#bcdddd` `rgb(188,221,221)``\n• #c9e4e4\n``#c9e4e4` `rgb(201,228,228)``\n• #d6ebeb\n``#d6ebeb` `rgb(214,235,235)``\n• #e3f1f1\n``#e3f1f1` `rgb(227,241,241)``\n• #f0f8f8\n``#f0f8f8` `rgb(240,248,248)``\n• #fdfefe\n``#fdfefe` `rgb(253,254,254)``\nTint Color Variation\n\n# Tones of #3a7474\n\nA tone is produced by adding gray to any pure hue. In this case, #555959 is the less saturated color, while #04aaaa is the most saturated one.\n\n• #555959\n``#555959` `rgb(85,89,89)``\n• #4e6060\n``#4e6060` `rgb(78,96,96)``\n• #476767\n``#476767` `rgb(71,103,103)``\n• #416d6d\n``#416d6d` `rgb(65,109,109)``\n• #3a7474\n``#3a7474` `rgb(58,116,116)``\n• #337b7b\n``#337b7b` `rgb(51,123,123)``\n• #2d8181\n``#2d8181` `rgb(45,129,129)``\n• #268888\n``#268888` `rgb(38,136,136)``\n• #1f8f8f\n``#1f8f8f` `rgb(31,143,143)``\n• #199595\n``#199595` `rgb(25,149,149)``\n• #129c9c\n``#129c9c` `rgb(18,156,156)``\n• #0ba3a3\n``#0ba3a3` `rgb(11,163,163)``\n• #04aaaa\n``#04aaaa` `rgb(4,170,170)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #3a7474 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":0.590495,"math_prob":0.59792835,"size":3699,"snap":"2021-04-2021-17","text_gpt3_token_len":1663,"char_repetition_ratio":0.12016238,"word_repetition_ratio":0.011090573,"special_character_ratio":0.56420654,"punctuation_ratio":0.2331081,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99185365,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-01-23T03:30:06Z\",\"WARC-Record-ID\":\"<urn:uuid:75b40cf4-3698-47f2-8cc7-0c84c08fce53>\",\"Content-Length\":\"36291\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e7c6ffd0-da31-4a34-bafc-f56f9ab2addc>\",\"WARC-Concurrent-To\":\"<urn:uuid:68b41931-db8b-4635-bff6-efc8d2eb5fff>\",\"WARC-IP-Address\":\"178.32.117.56\",\"WARC-Target-URI\":\"https://www.colorhexa.com/3a7474\",\"WARC-Payload-Digest\":\"sha1:HNRL5KOYT6BBZZKDLE32PYDX3M5ONQQN\",\"WARC-Block-Digest\":\"sha1:ZVZCUY25C7CTGXL2T25GIDTUI7P35AC4\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-04/CC-MAIN-2021-04_segments_1610703533863.67_warc_CC-MAIN-20210123032629-20210123062629-00706.warc.gz\"}"}
https://www.journalduparanormal.com/id/cos-graph-equation-90d75d	[ "For example. Like with sine graphs, the domain of cosine is all real numbers, and its range is Calculate the graph’s x- intercepts. This means that it repeats itself every 360 . Thus, the two graphs are shifts of 1/4 of the period from each other. Even though the input sign changed, the output sign for cosine stayed the same, and it always does for any theta value and its opposite. The graph of y = cos θ The graph of $$y = \\cos{\\theta}$$ has a maximum value of 1 and a minimum value of -1. This article has been viewed 13,641 times. Here are the steps: Like with sine graphs, the domain of cosine is all real numbers, and its range is, Referring to the unit circle, find where the graph f(x)=cos x crosses the x-axis by finding the angles on the unit circle where the cosine is 0. Unlike the sine function, which has 180-degree symmetry, cosine has y-axis symmetry. Connection between period of graph, equation and formula Table of contents top Formula Practice What is the period of a sine cosine curve? It’s symmetrical about the y-axis (in mathematical dialect, it’s an even function). For each y-value, move the y-value up by D if D is positive, or move the y-value down if D is negative. In other words, you can fold the graph in half at the y-axis and it matches exactly. {\"smallUrl\":\"https:\\/\\/www.wikihow.com\\/images\\/thumb\\/c\\/c7\\/Graphingtemplate.jpg\\/460px-Graphingtemplate.jpg\",\"bigUrl\":\"\\/images\\/thumb\\/c\\/c7\\/Graphingtemplate.jpg\\/552px-Graphingtemplate.jpg\",\"smallWidth\":460,\"smallHeight\":329,\"bigWidth\":\"552\",\"bigHeight\":\"395\",\"licensing\":\"" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.5397814,"math_prob":0.99533665,"size":4788,"snap":"2021-04-2021-17","text_gpt3_token_len":1495,"char_repetition_ratio":0.25167224,"word_repetition_ratio":0.87763715,"special_character_ratio":0.3565163,"punctuation_ratio":0.29895714,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9992165,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-01-15T15:18:07Z\",\"WARC-Record-ID\":\"<urn:uuid:b871e347-84fe-458d-aca5-18c5e4edf40c>\",\"Content-Length\":\"17101\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c4e6a3a0-5fb3-4c9c-9ffb-16718f040fc1>\",\"WARC-Concurrent-To\":\"<urn:uuid:615cafd2-5e9f-4693-a4c0-d5d84c7ca051>\",\"WARC-IP-Address\":\"213.186.33.16\",\"WARC-Target-URI\":\"https://www.journalduparanormal.com/id/cos-graph-equation-90d75d\",\"WARC-Payload-Digest\":\"sha1:OUHG7CAD66AHAKLFEG26YMWODC3YU5CX\",\"WARC-Block-Digest\":\"sha1:76FTP7GZEJFU65STBYJTYJJZHCFINZ7C\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-04/CC-MAIN-2021-04_segments_1610703495901.0_warc_CC-MAIN-20210115134101-20210115164101-00549.warc.gz\"}"}
https://mycbseguide.com/blog/ncert-solutions-class-10-maths-exercise-14-4/	[ "# NCERT Solutions for Class 10 Maths Exercise 14.4", null, "## myCBSEguide App\n\nComplete Guide for CBSE Students\n\nNCERT Solutions, NCERT Exemplars, Revison Notes, Free Videos, CBSE Papers, MCQ Tests & more.\n\nNCERT solutions for Maths Statistics", null, "## NCERT Solutions for Class 10 Maths Statistics\n\n1. The following distribution gives the daily income of 50 workers of a factory:", null, "Convert the distribution above to a less than type cumulative frequency distribution and draw its ogive.\n\nAns.", null, "Now, by drawing the points on the graph,\n\ni.e., (120, 12); (140, 26); 160, 34); (180, 40); (200, 50)\n\nScale: On", null, "axis 10 units = Rs. 10 and on", null, "axis 10 units = 5 workers", null, "NCERT Solutions for Class 10 Maths Exercise 14.4\n\n2. During the medical checkup of 35 students of a class, their weights were recorded as follows:", null, "Draw a less than type ogive for the given data. Hence obtain the median weight from the graph and verify the result by using the formula.\n\nAns.", null, "Hence, the points for graph are:\n\n(38, 0), (40, 3), (42, 5), (44, 9), (46, 14), (48, 28), (50, 32), (52, 35)\n\nScale: On", null, "axis, 10 units = 2 kg and on", null, "axis, 10 units = 5 students", null, "From the above graph, Median = 46.5 kg, which lies in class interval 46 – 48.\n\nHere,", null, ", then", null, ", which lies in interval 46 – 48.", null, "Median class = 46 – 48\n\nSo,", null, "and", null, "Now, Median =", null, "=", null, "=", null, "= 46 + 0.5\n\n= 46.5\n\nHence median weight of students is 46.5 kg.\n\n3. The following table gives production yield per hectare of wheat of 100 farms of a village.", null, "Change the distribution to a more than type distribution and draw its ogive.\n\nAns.", null, "The points for the graph are:\n\n(50, 100), (55, 98), (60, 90), (65, 78), (70, 54), (75, 16)\n\nScale: On", null, "axis, 10 units = 5 kg/ha and on", null, "axis, 10 units = 10 forms.", null, "## NCERT Solutions for Class 10 Maths Exercise 14.4\n\nNCERT Solutions Class 10 Maths PDF (Download) Free from myCBSEguide app and myCBSEguide website. Ncert solution class 10 Maths includes text book solutions from Mathematics Book. NCERT Solutions for CBSE Class 10 Maths have total 15 chapters. 10 Maths NCERT Solutions in PDF for free Download on our website. Ncert Maths class 10 solutions PDF and Maths ncert class 10 PDF solutions with latest modifications and as per the latest CBSE syllabus are only available in myCBSEguide.\n\n## CBSE app for Class 10\n\nTo download NCERT Solutions for Class 10 Maths, Computer Science, Home Science,Hindi ,English, Social Science do check myCBSEguide app or website. myCBSEguide provides sample papers with solution, test papers for chapter-wise practice, NCERT solutions, NCERT Exemplar solutions, quick revision notes for ready reference, CBSE guess papers and CBSE important question papers. Sample Paper all are made available through the best app for CBSE", null, "## Test Generator\n\nCreate Papers with your Name & Logo\n\n### 8 thoughts on “NCERT Solutions for Class 10 Maths Exercise 14.4”\n\n1. My cbse guide is the best\n\n2. Every thing is great but graphs are wrong. We can’t put origin’s value anything except zero." ]	[ null, "https://media-mycbseguide.s3.amazonaws.com/images/logo.png", null, "https://media-mycbseguide.s3.ap-south-1.amazonaws.com/images/blog/10%20Maths%20Book.jpg", null, "http://media.mycbseguide.com/images/static/ncert/10/mathematics/ch14/Ex14.4/image001.png", null, "http://media.mycbseguide.com/images/static/ncert/10/mathematics/ch14/Ex14.4/image002.png", null, "http://media.mycbseguide.com/images/static/ncert/10/mathematics/ch14/Ex14.4/image003.png", null, "http://media.mycbseguide.com/images/static/ncert/10/mathematics/ch14/Ex14.4/image004.png", null, "http://media.mycbseguide.com/images/static/ncert/10/mathematics/ch14/Ex14.4/image005.jpg", null, "http://media.mycbseguide.com/images/static/ncert/10/mathematics/ch14/Ex14.4/image006.png", null, "http://media.mycbseguide.com/images/static/ncert/10/mathematics/ch14/Ex14.4/image007.png", null, "http://media.mycbseguide.com/images/static/ncert/10/mathematics/ch14/Ex14.4/image003.png", null, "http://media.mycbseguide.com/images/static/ncert/10/mathematics/ch14/Ex14.4/image004.png", null, "http://media.mycbseguide.com/images/static/ncert/10/mathematics/ch14/Ex14.4/image008.jpg", null, "http://media.mycbseguide.com/images/static/ncert/10/mathematics/ch14/Ex14.4/image009.png", null, "http://media.mycbseguide.com/images/static/ncert/10/mathematics/ch14/Ex14.4/image010.png", null, "http://media.mycbseguide.com/images/static/ncert/10/mathematics/ch14/Ex14.4/image011.png", null, "http://media.mycbseguide.com/images/static/ncert/10/mathematics/ch14/Ex14.4/image012.png", null, "http://media.mycbseguide.com/images/static/ncert/10/mathematics/ch14/Ex14.4/image013.png", null, "http://media.mycbseguide.com/images/static/ncert/10/mathematics/ch14/Ex14.4/image014.png", null, "http://media.mycbseguide.com/images/static/ncert/10/mathematics/ch14/Ex14.4/image015.png", null, "http://media.mycbseguide.com/images/static/ncert/10/mathematics/ch14/Ex14.4/image016.png", null, "http://media.mycbseguide.com/images/static/ncert/10/mathematics/ch14/Ex14.4/image017.png", null, "http://media.mycbseguide.com/images/static/ncert/10/mathematics/ch14/Ex14.4/image018.png", null, "http://media.mycbseguide.com/images/static/ncert/10/mathematics/ch14/Ex14.4/image003.png", null, "http://media.mycbseguide.com/images/static/ncert/10/mathematics/ch14/Ex14.4/image004.png", null, "http://media.mycbseguide.com/images/static/ncert/10/mathematics/ch14/Ex14.4/image019.jpg", null, "https://media-mycbseguide.s3.amazonaws.com/images/category/flat64/tgnr1.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.864051,"math_prob":0.9178808,"size":3466,"snap":"2020-45-2020-50","text_gpt3_token_len":925,"char_repetition_ratio":0.16262276,"word_repetition_ratio":0.05098684,"special_character_ratio":0.2798615,"punctuation_ratio":0.15694444,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98676366,"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,null,null,null,null,1,null,2,null,3,null,3,null,1,null,1,null,1,null,3,null,3,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,3,null,3,null,1,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-10-21T23:59:39Z\",\"WARC-Record-ID\":\"<urn:uuid:53c3e672-cc31-49df-b779-592358fc61e5>\",\"Content-Length\":\"266562\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:71c5a5d3-bcbe-46d1-af72-cd176b7ec9ca>\",\"WARC-Concurrent-To\":\"<urn:uuid:a4895831-6783-41ca-874d-96a73ec4a9bb>\",\"WARC-IP-Address\":\"164.52.201.197\",\"WARC-Target-URI\":\"https://mycbseguide.com/blog/ncert-solutions-class-10-maths-exercise-14-4/\",\"WARC-Payload-Digest\":\"sha1:JIYLGWQJ3UIHZJYVIXXI2PH4X6O2GP2J\",\"WARC-Block-Digest\":\"sha1:KEQTWWAAEE2A7WJERRVVPNHAPAT7RHVJ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-45/CC-MAIN-2020-45_segments_1603107878662.15_warc_CC-MAIN-20201021235030-20201022025030-00045.warc.gz\"}"}
https://www.math-only-math.com/high-school-math-puzzles.html	[ "# High School Math Puzzles\n\nIn high school math puzzles and games we will solve different types of math logical questions. Numerous collections of high school and college grade math puzzles ranging over algebra, geometry, probability, number theory etc. These printable math puzzles are great to get prepared for exams.\n\n1. Which one of the answer figures should come after the problem figure if the sequence is continued?\n\n2. 5 + 3 = 16, 7 + 3 = 40 then 9 + 3 = ……….. .\n\n(a) 70\n\n(b) 72\n\n(c) 75\n\n(d) 80\n\n[Hints: 5² – 3² = 5 × 5 – 3 × 3 = 25 – 9 = 16]\n\n3. If a = 11 (242) 121, b = 14 (392) 196 then, find c = 13 (?) 169.\n\n(a) 338\n\n(b) 225\n\n(c) 337\n\n(d) 119\n\n4. Which one of the answer figures should come after the problem figure if the sequence is continued?\n\n5. How many triangles are there in the given figure?\n\n(a) 15\n\n(b) 14\n\n(c) 12\n\n(d) 13\n\n6. If GUN = 42 and ME = 18 then, HOME =?\n\n(a) 51\n\n(b) 31\n\n(c) 41\n\n(d) 21\n\n7. If X = 15 (104) 11 and Y = 13 (88) 9 then, Z = 14 (?) 10; find the number in place of the question mark (?).\n\n(a) 96\n\n(b) 95\n\n(c) 85\n\n(d) 75\n\n[Hints: 15² – 11² = 104]\n\n8. If A = 1, PAT = 37 then find TAP = …………… .\n\n(a) 73\n\n(b) 37\n\n(c) 36\n\n(d) 38\n\n9. 5, 16, 49, 148 then, find the next number.\n\n(a) 445\n\n(b) 449\n\n(c) 440\n\n(d) 442\n\n[Hints: Multiply 3 with each number and then add 1]\n\n10. If E = 5 and AMENDMENT = 89 then find SECRETARY =?\n\n(a) 114\n\n(b) 113\n\n(c) 123\n\n(d) 115\n\nFree printable math puzzles are available for everyone and even parents and teachers can encourage and suggest the child to practice the cool math puzzles for increasing the joy of thinking." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8045364,"math_prob":0.9991868,"size":2116,"snap":"2021-31-2021-39","text_gpt3_token_len":705,"char_repetition_ratio":0.12357955,"word_repetition_ratio":0.062921345,"special_character_ratio":0.38657844,"punctuation_ratio":0.12200436,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99421567,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-23T06:03:36Z\",\"WARC-Record-ID\":\"<urn:uuid:b936c61e-3afa-4a2b-93db-a8c704b43d79>\",\"Content-Length\":\"39398\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f6ea8c08-9e55-46b7-aee5-e8d9632fbd7a>\",\"WARC-Concurrent-To\":\"<urn:uuid:c45751a4-86c0-4892-a015-d33f6ab0ee32>\",\"WARC-IP-Address\":\"173.247.219.53\",\"WARC-Target-URI\":\"https://www.math-only-math.com/high-school-math-puzzles.html\",\"WARC-Payload-Digest\":\"sha1:6GRPVO3MLZ25VHOTQLJF23SRXL3MQAX7\",\"WARC-Block-Digest\":\"sha1:CWUIJYNLBEMTTJVSNH6DC5GVCLGOAZE6\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780057417.10_warc_CC-MAIN-20210923044248-20210923074248-00154.warc.gz\"}"}
https://www.studystack.com/flashcard-173159	[ "", null, "or", null, "or", null, "taken", null, "why\n\nMake sure to remember your password. If you forget it there is no way for StudyStack to send you a reset link. You would need to create a new account.\n\nDon't know\nKnow\nremaining cards\nSave\n0:01\nEmbed Code - If you would like this activity on your web page, copy the script below and paste it into your web page.\n\nNormal Size Small Size show me how\n\n# QDM Exam #2\n\n### Ch. 8\n\nQuantitative Forecasting Methods Forecast is based on mathematical modeling.\nExecutive Opinion Forecasting method in which a group of managers collectively develop a forecast.\nMarket Research Approach to forecasting that relies on surveys and interviews to determine customer preferences.\nDelphi Method Approach to forecasting in which a forecast is the product of a consensus among a group of experts.\nTime Series Models Based on the assumption that a forecast can be generated from the information contained in a time series of data.\nTime Series A series of observations taken over time.\nCausal Models Based on the assumption that the variable being forecast is related to other variables in the environment.\nLevel or Horizontal Pattern Pattern in which data values fluctuate around a constant mean.\nTrend Pattern in which data exhibit increasing or decreasing values over time.\nSeasonality Any pattern that regularly repeats itself and is constant in length.\nCycles Data patterns created by economic fluctuations.\nRandom Variation Unexplained variation that cannot be predicted.\nNaive Method Forecasting method that assumes next period's forecast is equal to the current period's actual value.\nSimple Mean or Average The average of a set of data.\nSimple Moving Average A forecasting method in which only n of the most recent observations are averaged.\nWeighted Moving Average A forecasting method in which n of the most recent observations are averaged and past observations may be weighted differently.\nExponential Smoothing Model Uses a sophisticated weight average procedure to generate a forecast.\nTrend-Adjusted Exponential Smoothing Exponential smoothing model that is suited to data that exhibit a trend.\nSeasonal Index Percentage amount by which data for each season are above or below the mean.\nLinear Regression Procedure that models a straight-line relationship between two variables.\nCorrelation Coefficient Statistic that measures the direction and strength of the linear relationship between two variables.\nForecast Error Difference between forecast and actual value for a given period.\nMean Absolute Deviation (MAD) Measure of forecast error that computes error as the average of the sum of the absolute errors.\nMean Squared Error (MSE) Measure of forecast error that computes error as the average of the squared error.\nForecast Bias A persistent tendency for a forecast to be over or under the actual value of the data.\nTracking Signal Tool used to monitor the quality of a forecast." ]	[ null, "https://sstk.biz/images/studystacklogo.svg", null, "https://sstk.biz/images/blackeye.png", null, "https://sstk.biz/images/greenCheckMark.svg", null, "https://sstk.biz/images/blackeye.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8608583,"math_prob":0.87912637,"size":2855,"snap":"2020-45-2020-50","text_gpt3_token_len":580,"char_repetition_ratio":0.13188355,"word_repetition_ratio":0.08695652,"special_character_ratio":0.19124344,"punctuation_ratio":0.06410257,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97460407,"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\":\"2020-10-24T21:15:23Z\",\"WARC-Record-ID\":\"<urn:uuid:cd292e0b-c4e0-48a7-8c2a-d45c29d577d0>\",\"Content-Length\":\"58988\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:942e12c3-bb46-432a-9d71-3a77a00601a5>\",\"WARC-Concurrent-To\":\"<urn:uuid:49ba8eb4-81f5-430c-9258-9e939b80d217>\",\"WARC-IP-Address\":\"69.89.15.53\",\"WARC-Target-URI\":\"https://www.studystack.com/flashcard-173159\",\"WARC-Payload-Digest\":\"sha1:CIEZCADQFJUVTIDI5KR5V2KZPLM4OMAH\",\"WARC-Block-Digest\":\"sha1:RZE3Z5LGUXFLUIVSTIKMKFU5CDPD424L\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-45/CC-MAIN-2020-45_segments_1603107884755.46_warc_CC-MAIN-20201024194049-20201024224049-00155.warc.gz\"}"}
https://researchportal.uc3m.es/display/act532471	[ "# A dirichlet process prior approach for covariate selection Articles", null, "• August 2020\n\n• 948\n\n• 9\n\n• 22\n\n• 1099-4300\n\n### abstract\n\n• The variable selection problem in general, and specifically for the ordinary linear regression model, is considered in the setup in which the number of covariates is large enough to prevent the exploration of all possible models. In this context, Gibbs-sampling is needed to perform stochastic model exploration to estimate, for instance, the model inclusion probability. We show that under a Bayesian non-parametric prior model for analyzing Gibbs-sampling output, the usual empirical estimator is just the asymptotic version of the expected posterior inclusion probability given the simulation output from Gibbs-sampling. Other posterior conditional estimators of inclusion probabilities can also be considered as related to the latent probabilities distributions on the model space which can be sampled given the observed Gibbs-sampling output. This paper will also compare, in this large model space setup the conventional prior approach against the non-local prior approach used to define the Bayes Factors for model selection. The approach is exposed along with simulation samples and also an application of modeling the Travel and Tourism factors all over the world.\n\n### keywords\n\n• conventional priors; covariate inclusion probability; dirichlet process prior; non-local prior; ordinary linear regression; variable selection" ]	[ null, "https://researchportal.uc3m.es/images/individual/uriIcon.gif", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.837677,"math_prob":0.9079986,"size":1614,"snap":"2021-43-2021-49","text_gpt3_token_len":305,"char_repetition_ratio":0.1173913,"word_repetition_ratio":0.0,"special_character_ratio":0.17905824,"punctuation_ratio":0.071428575,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9511032,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-11-29T21:43:54Z\",\"WARC-Record-ID\":\"<urn:uuid:8da520c0-926e-45e0-8772-2be289699644>\",\"Content-Length\":\"17517\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:19c8ad26-b0f6-4c48-bcec-2845da89fb42>\",\"WARC-Concurrent-To\":\"<urn:uuid:7313c58b-d885-4fcf-93c9-99116bea00dc>\",\"WARC-IP-Address\":\"163.117.98.14\",\"WARC-Target-URI\":\"https://researchportal.uc3m.es/display/act532471\",\"WARC-Payload-Digest\":\"sha1:OKURM7T6R4CE2GUTKZCOZ7GNZNSZ5UJ6\",\"WARC-Block-Digest\":\"sha1:7CGAOYVETTWMYQ6QF6XOO5QKXGWFLU4I\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964358842.4_warc_CC-MAIN-20211129194957-20211129224957-00141.warc.gz\"}"}
https://www.arxiv-vanity.com/papers/hep-ph/0006158/	[ "# Space-time description of the hadron interaction at high energies.\n\nV.N.Gribov\n###### Abstract\n\nIn this lecture we consider the strong and electromagnetic interactions of hadrons in a unified way. It is assumed that there exist point-like particles (partons) in the sense of quantum field theory and that a hadron with large momentum consists of partons which have restricted transverse momenta, and longitudinal momenta which range from to zero. The density of partons increases with the increase of the coupling constant. Since the probability of their recombination also increases, an equilibrium may be reached. In the lecture we will consider consequences of the hypothesis that the equilibrium really occurs. We demonstrate that it leads to constant total cross sections at high energies, and to the Bjorken scaling in the deep inelastic ep-scattering. The asymptotic value of the total cross sections of hadron-hadron scattering turns out to be universal, while the cross sections of quasi-elastic scattering processes at zero angle tend to zero.\n\nThe multiplicity of the outgoing hadrons and their distributions in longitudinal momenta (rapidities) are also discussed.\n\n## Introduction\n\nIn this lecture we will try to describe electromagnetic and strong interactions of hadrons in the same framework which follows from general quantum field theory considerations without the introduction of quarks or other exotic objects.\n\nWe will assume that there exist point-like constituents in the sense of quantum field theory which are, however, strongly interacting. It is convenient to refer to these particles as partons. We will not be interested in the quantum numbers of these partons, or the symmetry properties of their interactions. We will assume that, contrary to the perturbation theory, the integrals over the transverse momenta of virtual particles converge like in the theory. It turns out that within this picture a common cause exists for two seemingly very different phenomena: the Bjorken scaling in deep inelastic scattering, and the recent theoretical observation that all hadronic cross sections should approach the same limit (provided that the Pomeranchuk pole exists). The lecture is organized as follows. In the first part we discuss the propagation of the hadrons in the space as a process of creation and absorption of the virtual particles (partons) and formulate the notion of the parton wave function of the hadron. The second part describes momentum and coordinate parton distributions in hadrons. In the third part we consider the process of deep inelastic scattering. It is shown that from the point of view of our approach the deep inelastic scattering satisfies the Bjorken scaling, and, in contrast to the quark model, the multiplicity of the produced hadrons is of the order of . The fourth part is devoted to the strong interactions of hadrons and it is shown that in the same framework the total hadron cross sections have to approach asymptotically the same limiting value. In the last part of the lecture we discuss the processes of elastic and quasi-elastic scattering at high energies. It is demonstrated that the cross sections of the quasi-elastic scattering processes at zero angle tend to zero at asymptotically high energies.\n\nLet us discuss, how one can think of the space-time propagation of a physical particle in terms of virtual particles which are involved in the interaction with photon and other hadrons. It is well known that the propagation of a real particle is described by its Green function, which corresponds to a series of Feynman diagrams of the type\n\n(for simplicity, we will consider identical scalar particles). The Feynman diagrams, having many remarkable properties, have, nevertheless, a disadvantage compared to the old-fashioned perturbation theory. Indeed, they do not show how a system evolves with time in a given coordinate reference frame. For example, depending on the relations between the time coordinates , , and , the graph in Fig.1b corresponds to different processes:\n\nSimilarly, the diagram Fig.1c corresponds to the processes\n\nIn quantum electrodynamics, where explicit calculations can be carried out, this complicated correspondence is of little interest. However, for strong interactions, where explicit calculations are impossible, distinguishing between different space developments will be useful.\n\nObviously, if the interaction is strong (the coupling constant is large), many diagrams are relevant. The first question which arises is which configurations dominate: the ones which correspond to the subsequent decays of the particles - the diagrams Fig.2a and Fig3.a, or those which correspond to the interaction of the initial particle with virtual ”pairs” created in the vacuum. It is clear that if the coupling constant is large and the momentum of the incoming particle is small (see below), configurations with ”pairs” dominate (at least if the theory does not contain infinities). Indeed, if is small, then in the case of configurations without ”pairs” the integration regions corresponding to each correction will tend to zero with an increase of the number of corrections. At the same time, for the configurations containing ”pairs” the region of integration over time will remain infinite. Hence, if the retarded Green function does not have a strong singularity at , the contribution of the configurations without ”pairs” will be relatively small if the coupling constant is large. Even the graphs of the type Fig.1d are determined mainly by configurations like\n\nThis means that if we observe a low energy particle at any particular moment of time (the cut in the diagram in Fig. 4), we will see few partons which are decay products of the particle, and a large number of virtual ”pairs” which will interact with these partons in the future.\n\nWhat happens if a particle has a large momentum in our coordinate reference frame? To analyze the space-time evolution of a fast particle we have to consider a space-time interval such that , and . Here is the mass of the particle, its energy. In this case, , . For such intervals the relation between the configurations with and without ”pairs” changes. Configurations corresponding to a decay of one parton into many others start to dominate, while the role of configurations with ”pairs” decreases.\n\nThe physical origin of this phenomenon is evident. A fast parton can decay, for example, into two fast partons which, due to the energy-time uncertainty relation, will exist for a long time (of the order of ), since\n\n ΔE=√μ2+→p2−√μ2+→p21−√μ2+(→p−→p1)2∼μ22\|→p\|−μ22\|→p1\|−μ22\|→p−→p1\|.\n\nEach of these two partons can again decay into two partons and this will continue up to the point when slow particles, living for a time of the order of , are created. After that the fluctuations must evolve in the reverse direction, i.e. the recombination of the particles begins.\n\nOn the other hand, due to the same uncertainty relation, creation of virtual ”pairs” with large momenta in vacuum is possible only for short time intervals of the order of . Hence, it affects only the region of small momentum partons. The way in which this phenomenon manifests itself can be seen using the simplest graph in Fig.5. as an example. We will observe that it is possible to place here many emissions in spite of the fact that the interval is of the order of unity (), and the Green function depends only on the invariants.\n\nFor the sake of simplicity, let us verify this for one space dimension (). Suppose that and , . Then . Let us choose the variables , in the same way: , , and consider the following region of integration in the integral, corresponding to the diagram in Fig.5:\n\n 1\n 1\n zi∼z′i,ti=zi+y2i2zi,t′i=z′i+y′2i2z′i.\n\nThe integrations over can be substituted by integrations over , and\n\n d2yi=12dy2idzizi,d2y′i=12dy′2idz′iz′i\n\nIt is easy to see that in this region of integration the arguments of all Green functions: , , , are of the order of unity, and the integrals do not contain any small factors. All these conditions for can be satisfied simultaneously for a large number of emissions: . Indeed, if we write in the form , all conditions will be fulfilled for\n\n n∼lntlnC,C≥1.\n\nObviously, one can consider a more complicated diagram than Fig.5 by including interactions of the virtual particles. On the other hand, configurations containing vacuum ”pairs” play a minor role. Moving backwards in time is possible only for short time intervals (Fig.6):\n\nHence, we reach the following picture. A real particle with a large momentum can be described as an ensemble of an indefinite number of partons of the order of with momenta in the range from to zero, and several vacuum pairs with small momenta which in the future can interact with the target.\n\nThe observation of a slow particle during an interval of the order of does not tell us anything about the structure of the particle since we cannot distinguish it from the background of the vacuum fluctuations, and we can speak only about the interaction of particles or about the spectrum of states. On the contrary, in the case of a fast particle we can speak about its structure, i.e. about the fast partons which do not mix with the vacuum fluctuations. As a result, in a certain sense a fast particle becomes an isolated system which is only weakly coupled to the vacuum fluctuations. Hence, it can be described using a quantum mechanical wave function or an ensemble of wave functions, which determine probabilities of finding certain numbers of partons and their momentum distribution. Such a description is not invariant , since the number of partons depends on the momentum of the particle, but it can be considered as covariant. Moreover, it may be even invariant, if the momentum distribution of the partons is homogeneous in the region of momenta much smaller than the maximal one, and much larger than .\n\nIndeed, under the transformation from one reference frame to another in which the particle has, for example, a larger momentum, a new region emerges in the distribution of partons; in the old region, however, the parton distribution remains unchanged. One usually describes hadrons in terms of the quantum mechanics of partons in the reference frame which moves with an infinite momentum, because in this case all partons corresponding to vacuum fluctuations have zero momenta, and such a description is exact. Such a reference frame is convenient for the description of the deep inelastic scattering. However, it is not as good for describing strong interactions, where the slow partons are important. In any case, it appears useful to preserve the freedom in choosing the reference frame and to use the covariant description. This allows a more effective analysis of the accuracy of the derivations.\n\n## 1 Wave function of the hadron. Orthogonality and normalization\n\nThe previous considerations allow us to introduce the hadron wave function in the following way. Let us assume, as usual, that at the hadron can be represented as a bare particle (the parton). After a sufficiently long time the parton will decay into other partons and form a stationary state which we call a hadron. Diagrams corresponding to this process are shown in Fig.7.\n\nLet us exclude from the Feynman diagrams those configurations (in the sense of integrations over intermediate times) which correspond to vacuum pair creation.\n\nFor the theory such a separation of vacuum fluctuations corresponds to decomposing into positive and negative frequency parts and substituting by . The previous discussion shows that the ignored term would mix only partons with small momenta.\n\nIt is natural to consider the set of all possible diagrams with a given number of partons at the given moment of time as a component of the hadron wave function . Similarly, we can determine the wave functions of several hadrons with large momenta provided the energy of their relative motion is small compared to their momenta. The latter condition is necessary to ensure that slow partons are not important in the interaction. The Lagrangian of the interaction remains Hermitian even after the terms corresponding to the vacuum fluctuations are omitted. As a result, the wave functions will be orthogonal, and will be normalized in the usual way:\n\n ∑n∫Ψb∗n(→y1…,→yn,pb)i↔∂Ψan(→y1…,→yn,pa)d3y1…d3ynn!=(2π)3δ(→pa−→pb)δab, (1)\n\nor similarly in the momentum space, after separating\n\n ∑n1n!∫Ψb∗n(→k1…,→kn,→p)Ψan(→k1…,→kn,→p)d3k1…d3kn2k10…2kn0δ(p−∑ki)(2π)3n−1=δab. (2)\n\nFor the momentum range , the wave functions coincide with those calculated in the infinite momentum frame. In this reference frame they do not depend on the momentum of the system (except for a trivial factor). This can be easily proven by expanding the parton momenta\n\n →ki=βi→p+ki⊥, (3)\n\nand writing the parton energy in the form\n\n εi=√→k2i+m2=βip+m2+k2i⊥2pβi. (4)\n\nNote now that the integrals which determine , corresponding to Fig.7, can be represented in the form of the old-fashioned perturbation theory where only the differences between the energies of the intermediate states and the initial state enter, and the momentum is conserved. Hence, the terms linear in cancel in these differences, and concequently\n\n Ek−E=12p(∑im2+k2i⊥βi−m2) (5)\n\nEach consequent intermediate state in Fig.7 in the model differs from the previous one by the appearance or disappearance of one particle. The factor\n\n ∑n1n!∫Ψb∗n(ki⊥,βi)Ψan(ki⊥,βi)∏d2ki⊥2(2π)2dβiβi(2π)3δ(1−∑βi)=δab. (6)\n\nFor slow partons, where the expansion (4) is not correct, the dependence on momentum does not disappear, and contrary to the case of the system moving with , this dependence cuts off the sum over the number of partons.\n\n## 2 Distribution of the partons in space and momentum\n\nThe distribution of partons in longitudinal momenta can be characterized by the rapidity:\n\n ηi=12lnεi+kizεi−kiz, (7)\n\nwhere is the component of the parton momentum along the hadron momentum.\n\n ηi≈ln2βip√m2+k2i⊥. (8)\n\nAs it is well known, this quantity is convenient since it simply transforms under the Lorentz transformations along the direction: , where is the rapidity of the coordinate system.\n\nThe determination of the parton distribution over is based on the observation that in each decay process shown in Fig.7 the momenta and are, in the average, of the same order. This means that in the process of subsequent parton emission and absorption the rapidities of the partons change by a factor of the order of unity. At the same time the overall range of parton rapidities is large, of the order of . This implies that in the rapidity space we have short range forces.\n\nLet us consider the density of the distribution in rapidity\n\n φ(η,k⊥,p)= (9) ∑n1n!∫\|Ψ(k⊥,η,k⊥1,η1,…,k⊥n,ηn,)\|2(2π)3δ(→p−→k−∑→ki)∏dkidηi2(2π)3\n\nin the interval (see Fig.8).\n\nThe independence of on for these values of means that depends only on the differences . If decreases with the increase of , this corresponds to a weak coupling, i.e. to a small probability of the decay of the initial parton. If the coupling constant grows, the number of partons increases and at a certain value of the coupling constant an equilibrium is reached, since the probability of recombination also increases. The value of this critical coupling constant has to be such that the recombination probability due to the interaction should be larger than the recombination probability related to the uncertainty principle.\n\nThe basic hypothesis is that such an equilibrium does occur and that due to the short-range character of interaction it is local. This is equivalent to the hypothesis of the constant total cross sections of interaction at . Hence we assume that the equilibrium is determined by the vicinity of the point of the order of unity and it does not depend on . Obviously, this can be satisfied only if does not depend on and at . According to the idea of Feynman, this situation resembles the case of a sufficiently long one-dimensional matter in which, due to the homogeneity of the space, far from the boundaries the density is either constant or oscillating (for a crystal). In our case the analogue of the homogeneity of space is the relativistic invariance (the shift in the space of rapidities). For the time being we will not consider the case of the crystal. According to (9), the integral of over and has the meaning of the average parton density which is, obviously, of the order of .\n\nGenerally speaking, we cannot say anything about the parton distribution in the transverse momenta except for one statement: it is absolutely crucial for the whole concept that it must be restricted to the region of the order of parton masses, like in the theory.\n\nConsider now the spatial distribution of the partons. First, let us discuss parton distribution in the plane perpendicular to the momentum . For that purpose it is convenient to transform from to the impact parameter representation :\n\n Ψn(→ρn,ηn)=∫ei∑ki⊥ρiΨ(ki⊥,ηi)δ(∑k⊥i)(2π)2∏d2ki(2π)2. (10)\n\nLet us rank the partons in the order of their decreasing rapidities. Consider a parton with the rapidity and let us follow its history from the initial parton. Initially, we will assume that it was produced solely via parton emissions (Fig.9).\n\nIn this case it is clear that if the transversal momenta of all partons are of the order of , than each parton emission leads to a change of the impact parameter by . If emissions are necessary to reduce the rapidity from to , and they are independent and random, . If every emission changes the rapidity of the parton by about one unit, then\n\n ¯¯¯¯¯¯¯¯¯¯¯¯¯¯(Δρ)2=γ(ηp−η). (11)\n\nHence, the process of the subsequent parton emissions results in a kind of diffusion in the impact parameter plane. The parton distribution in for the rapidity has the Gaussian form:\n\n φ(ρ,η)=C(η)πγ(ηp−η)e−ρ2γ(ηp−η), (12)\n\nif the impact parameter of the initial parton is considered as the origin. Consequently, the partons with have the broadest distribution, and, hence, the fast hadron is of the size\n\n R=√γηp≈√γln2pm. (13)\n\nThe account of the recombination and the scattering of the partons affects only densities of partons and fluctuations, but does not change the radius of the distribution which can be viewed as the front of the diffusion wave.\n\nLet us discuss the parton distribution over the longitudinal coordinate. A relativistic particle with a momentum is commonly considered as a disk of the thickness . In fact, this is true only in the first approximation of the perturbation theory. Really, a hadron is a disk with radius and the thickness of the order of . Indeed, each parton with a longitudinal momentum is distributed in the longitudinal direction in an interval . Since the parton spectrum exists in the range of momenta from down to , the longitudinal projection of the hadron wave function has the structure depicted in Fig.11.\n\nFinally, let us consider what is the lifetime of a particular parton. As we have discussed in the Introduction, in a theory which is not singular at short distances, the intervals between two events represented by a Feynman diagram are of the order of unity. For a fast particle moving along the axis, and . Consequently, the lifetime of a fast parton with a momentum is of the order of . The presented arguments were based on the theory which is the only theory which provides a cutoff in transverse momenta. Still, the argument should hold for other theories and for particles with spins, if one assumes that in these theories the cutoff of transverse momenta occures in some way. On the other hand, the theory cannot be considered as a self-consistent example. Indeed, due to the absence of a vacuum state, the series of perturbation theory do not make sense (series with positive coefficients are increasing as factorials). Hence, the picture we have presented here does not correspond literally to any particular field theory. At the same time, it corresponds fully to the main ideas of the quantum field theory and to its basic space-time relations.\n\n## 3 Deep inelastic scattering\n\nIt is convenient to consider the deep inelastic scattering of electrons in the frame where the time component of the virtual photon momentum is . In this reference frame the momentum of the photon is equal to (), while the momentum of the hadron is (). Suppose that is large and . According to our previous considerations, a fast hadron can be viewed as an ensemble of partons. In this system a photon looks as a static field with the wavelength .\n\nThe main question is, with which partons can the photon interact. We can consider the static field of a photon as a packet with a longitudinal size of the order of . The interaction time between a hadron with the size and such a packet is of the order of . However, due to the big difference between the parton and photon wave lengths, the interaction with a slow parton is small. Hence, the photon interacts with partons which have momenta of the order of . Partons with such momenta are distributed in the longitudinal direction in the region . Because of this, the time of the hadron-photon interaction is in fact of the order of , i.e. much shorter than the lifetime of a parton. This means that the photon interacts with a parton as with a free particle, and so not only the momentum but also the energy is conserved. As a result, the energy-momentum conservation laws select the parton with momentum , which can absorb a photon\n\n kiz−qz=k′iz,\|kiz−qz\|=kiz.\n\nThis gives\n\n kiz=qz2,k′iz=−qz2.\n\nThe cross section of such a process is, obviously, equal to the cross section of the absorption of a photon by a free particle, multiplied by the probability to find a parton with a longitudinal momentum inside the hadron, i.e. by the value (9), integrated over . (The necessary accuracy of fulfilment of the conservation laws allows any ).\n\nAs it was already discussed, . Hence, using the known cross section for the interaction of the photon with a charged spinless particle, we obtain for the cross section of the deep inelastic scattering\n\n d2σdq2dω=4πα2q4(1−pqppe)φ(ω), (14)\n\nwhere is the electron momentum. If the partons have spins, the situation becomes more complicated, since the cross sections of the interactions between photons and partons with different spins are different. The parton distributions in rapidities for different spins may also be different, leading to the form:\n\n d2σdq2dω=4πα2q4{(1−(pq)(ppe))φ0(ω)+[1−pqppe+12(pqppe)2]φ12(ω)}. (15)\n\nLet us discuss now a very important question, namely: what physical processes take place in deep inelastic scatterings. To clarify this, we go back to Fig.7 determining the hadron wave function. We will neglect the parton recombinations in the process of their creation from the initial parton, i.e. we consider fluctuations of the type shown in Fig.9. Suppose that the photon was absorbed by a parton with a large momentum . As a result, this parton obtained a momentum and moves in the opposite direction with momentum . The process is depicted in Fig.12. What will now happen to this parton and to the remaining partons? Within the framework we are using it is highly unlikely that the parton with the momentum will have time to interact with the other partons. The probability to interact directly with residual partons will be small, because the relative momentum of the parton with and the rest of the partons is large. It could interact with other partons after many subsequent decays which, in the end, could create a slow parton. However, the time needed for these decays is large, and during this time the parton and its decay products will move far away from the remaining partons, thus the interaction will not take place.\n\nHence, we come to the conclusion that one free parton is moving in the direction . What will we observe experimentally, if we investigate particles moving in this direction? To answer this question, it is sufficient to note that, in average, a hadron with a momentum consists of partons; at .\n\nIn a sense there should exist an uncertainty relation between the number of partons in a hadron () and the number of hadrons in a parton ()\n\n npn>∼clnkzμ, (16)\n\nwhere is the momentum of the state.\n\nWe came to the conclusion that the parton decays into a large number of hadrons i.e. in fact the parton is very short-lived, highly virtual. Hence, we have to discuss whether this conclusion is consistent with the assumption that the photon-parton interaction satisfies the energy conservation. To answer this question, let us calculate the mass of a virtual parton with momentum , decaying into hadrons with momenta and masses .\n\n M2=(∑√m2i+k2i)2−k2z=(kz+∑im2i+k2i⊥2kiz)2−k2z≈kz∑im2i+k2i⊥kiz.\n\nIf the hadrons are distributed almost homogeneously in rapidities, their longitudinal momenta decrease exponentially with their number, and in the sum only a few terms, corresponding to slow hadrons, are relevant. As a result, , i.e. the time of the existence of the parton is of the order of , much larger than the time of interaction with a photon .\n\nLet us discuss now, what happens to the remaining partons. Little can be determined using only the uncertainty relation eq.(16). This is because the number of partons before the photon absorption was , after the photon absorption it became and, consequently, according to the uncertainty relation, the number of hadrons corresponding to this state can range from 1 to . Hence, everything depends on the real perturbation of the hadron wave function due to the photon absorption.\n\nConsider now the fluctuation shown in Fig.12. The photon absorption will not have any influence on partons created after the parton ”” which absorbed the photon was produced, and and which have momenta smaller than “b”. These fluctuations will continue, and the partons can, in particular, recombine back into the parton ””. The situation is different for partons which occured earlier and have large momenta (””, ””). In this case the fluctuation cannot evolve further the same way, since the parton ”” has moved in the opposite direction. As a result, it is highly probable that partons ”” and ”” will move apart and lose coherence. On the other hand, slow partons which were emitted by ”” and ”” earlier and which are not connected with the parton ””, will be correlated, as before, with each of them. Thus ”” and ”” will move in space together with their slow partons, i.e. in the form of hadrons. Hence, it appears that partons flying in the initial direction lead to the production of the order of\n\nContrary to the case of rapidities of partons, we will count the rapidity of the hole not from zero rapidity but from the rapidity . In this case the rapidity of the hole is . If we now represent the parton hole with rapidity as a superposition of the hadron states, this superposition will contain hadron states.\n\nLet us represent the whole process by a diagram describing rapidity distributions of partons and hadrons. Before the photon absorption the partons in the hadrons are distributed at rapidities between zero and , while after the photon absorption a parton distribution is produced which is shown in Fig. 13.\n\nThis parton distribution leads to the hadron distribution shown in Fig. 14. The total multiplicity corresponding to this distribution is\n\n ¯n=clnqzμ+clnω=clnνμ√−q2.\n\nThis hadron distribution in rapidities in the deep inelastic scattering differs qualitatively from those previously discussed in the literature. It corresponds to hadrons moving in the photon momentum direction, and hadrons are moving in the nucleon momentum direction, with a gap in rapidity between these distributions. The hadron distribution which was obtained in the framework of perturbation theory for superconverging theories like (Drell, Yan) differs qualitatively from the distribution in Fig. 14.\n\nIn conclusion of this part, it is necessary to point out that the problem of spin properties of the partons exist in this picture even if the partons do not have quark quantum numbers. If, as the experiment shows, the cross section for the interaction of the transversal photons is larger than the cross section for the interaction of the longitudinal photons, , the charged partons have predominantly spin . This means that at least one fermion, for example a nucleon, has to move in the direction of the photon momentum. In other words, in deep inelastic scattering the distribution of the created hadrons in quantum numbers as the function of their rapidities differs essentially from what we are used to in the strong interactions. Perhaps this is one of the key prediction of the non-quark parton picture for .\n\n## 4 Strong interactions of hadrons\n\nLet us discuss now the strong interactions of hadrons. First, we consider a collision of two hadrons in the laboratory frame. Suppose that a hadron ”1” with momentum hits hadron ”2” which is at rest. Obviously, the parton wave function makes no sense for the hadron at rest, since for the latter the vacuum fluctuations are absolutely essential. However, the hadron at rest can also be understood as an ensemble of slow partons distributed in a volume of the order of , independent of the origin of the partons. Indeed, it does not matter whether these partons are decay products of the initial parton or the result of the vacuum fluctuations. How can a fast hadron, consisting of partons with rapidities from to zero, interact with the target which consists of slow partons? Obviously, the cross section of the interaction of two point-like particles with a large relative energy is not larger than (where is the wave length in the c.m. frame, is the relative rapidity). That is why only slow partons of the incident hadron can interact with the target with a cross section which is not too small. This process is shown in Fig. 15.\n\nIf the slow parton which initiated the interaction was absorbed in this interaction, the fluctuation which lead to its creation from a fast parton was interrupted. Hence, all partons which were emitted by the fast parton in the process of fluctuation cannot recombine any more. They disperse in space and ultimately decay into hadrons leading to the creation of hadrons with rapidities from zero to . The interaction between the partons is short-range in rapidities. Hence, the hadron distribution in rapidities will reproduce the parton distribution in rapidities. In particular, the inclusive spectrum of hadrons will have the form shown in Fig. 8, with an unknown distribution near the boundaries. The total hadron multiplicity will be of the order of . If the probability of finding a slow parton in the hadron does not depend on the hadron momentum (this would be quite natural, since with the increase of the momentum the life-time of the fluctuation is also growing), the total cross section of the interaction will not depend on the energy at high energies.\n\nBefore continuing the analysis of inelastic processes, let us discuss, how to reconcile the energy independence of the total interaction cross section at high energies with the observation discussed above that the transverse hadron sizes increase with the increase of the energy as . The answer is that slow partons are distributed almost homogeneously over the disk of the radius (Eq.(11)) , while their overall multiplicity during the time of is of order of unity.\n\nLet us see now how the same process will look, for example, in the c.m. frame. In this reference frame the interaction will have the form as shown in Fig. 16.\n\nEach of the hadrons consists of partons with rapidities ranging from to zero and from zero to , respectively. The slow partons interact with cross sections which are not small. As a result, the fluctuations will be interrupted in both hadrons, and the partons will fly away in the opposite directions, leading to the creation of hadrons with rapidities from to . From the point of view of this reference frame the inclusive spectrum must have the form shown in Fig. 17, with unknown distributions not only at the boundaries but also in the centre, since the distribution of the slow partons in the hadrons and in vacuum fluctuations is unknown. The hadron inclusive spectrum, however, should not depend on the reference frame. Thus the inclusive spectrum in Fig. 17 should coincide with the inclusive spectrum in Fig. 8, and they should differ only by a trivial shift along the rapidity axis, i.e. due to relativistic invariance we know something about the spectra of slow partons and vacuum fluctuations.\n\nLet us demonstrate that this comparison of processes in two reference frames leads to a very important statement, namely that at ultra-high energies the total cross sections for the interactions of arbitrary hadrons should be equal. Indeed, we have assumed that the distribution of hadrons reproduces the parton distribution.\n\nFrom the point of view of the laboratory frame the distribution of partons and, consequently, distribution of hadrons in the central region of the spectrum is completely determined by the properties (quantum numbers, mass, etc.) of particle 1, and does not depend on the properties of particle 2. On the other hand, from the point of view of the antilaboratory frame (where the particle 1 is at rest) everything is determined by the properties of particle 2. This is possible only if the distribution of partons in the hadrons with rapidities much smaller than the hadron rapidity does not depend on the quantum numbers and the mass of the hadron, that is the parton distribution with should be universal. From the point of view of the c.m. system the same region is determined by slow partons of both hadrons and by vacuum fluctuations (which are universal), and, consequently, the distribution of slow partons is also universal.\n\nIt is natural to assume that the probability of finding a hadron in a sterile state without slow partons tends to zero with the increase of its momentum, in other words assume that slow partons are always present in a hadron (compare to the decrease of the cross section of the elastic electron scattering at large ). In this case considering the process in the c.m. system, we see that the total cross section of the hadron interaction is determined by the cross section of the interaction of slow partons and by their transverse distribution which is universal. Consequently, the total hadron interaction cross section is also universal, i.e. equal for any hadrons.\n\nThis statement looks rather strange if we regard it, for instance, from the following point of view. Let us consider the scattering of a complicated system with a large radius, for example, deuteron-nucleon scattering. As we know, the cross section of the deuteron-nucleon interaction equals the sum of the nucleon-nucleon cross sections, thus it is twice as large as the nucleon-nucleon cross section. How and at what energies can the deuteron-nucleon cross section become equal to the nucleon-nucleon cross section? How is it possible that the density of slow partons in the deuteron turns out to be equal to the density of slow partons in the nucleon? To answer this question, let us discuss the parton structure of two hadrons which are separated in the plane transverse to their longitudinal momenta by a distance much larger than their Compton wave length . Suppose that at the initial moment they were point-like particles. Next, independently of each other, they begin to emit partons with decreasing longitudinal momenta. At the same time the diffusion takes place in the transverse plane so that the partons will be distributed in a growing region. The basic observation which we shall prove and which answers our question is that if the momenta of the initial partons are sufficiently large, then during one fluctuation the partons coming from different initial partons will inevitably meet in space (Fig. 18) in the region of the order of . They will have similar large rapidities and, hence, will be able to interact with a probability of the order of unity. If such “meetings” take place sufficiently frequently, the probability of the parton interaction will be unity. Consequently, the further evolution and the density of the slow partons which are created after the meeting may not depend on the fact that initially the transverse distance between two partons was large.\n\nIn terms of the diffusion in the impact parameter plane this statement corresponds to the following picture. Suppose that initial partons were placed at points and in Fig. 19 and that their longitudinal momenta are of the same order of magnitude, i.e. difference of their rapidities is of the order of unity, while each of the rapidities is large. We will follow the parton starting from point , which decelerates via emission of other partons. As we have seen, its propagation in the perpendicular plane corresponds to diffusion. The difference of rapidities at the initial and the considered moments serves the role of time in this diffusion process.\n\nThe diffusion character of the process means that the probability density of finding a parton with rapidity at the point if it started from the point with rapidity is\n\n ω(→ρ,→ρ1,ηp−η)=1πγ(ηp−η)e−(→ρ−→ρ1))2γ(ηp−η). (17)\n\nThe situation is exactly the same for a decelerating parton which started from the point . Thus, the probability of finding both partons at the same point with equal rapidities is proportional to\n\n ω(ρ12,ηp−η)= (18) ∫ω(→ρ,→ρ1,ηp−η)ω(→ρ,→ρ2,ηp−η)d2ρ= 12πγ(ηp−η)exp[−(→ρ1−→ρ2))22γ(ηp−η)]\n\nIf we now integrate this expression over , i.e. estimate the probability for the partons to meet at some rapidities, we obtain\n\n ∫ηp0ω(ρ12,ηp−η)dη≈1πlog2γηpρ212\|ηp→∞⟶∞. (19)\n\nThis means that if , the partons will inevitably meet. According to (19) we get a probability much larger than unity. The reason is that under these conditions the meeting of partons at different values of are not independent events and therefore it does not make sense to add the probabilities. It is easy to prove this statement directly, for example with the help of the diffusion equation. We will not do this, however. According to a nice analogy suggested by A. Larkin, this theorem is equivalent to the statement that if you are in an infinite forest in which there is a house on a finite distance from you, then, randomly wandering in the forest, you sooner or later arrive at this house. Essentially, the reason is that in the two-dimensional space the region inside of which the diffusion takes place and the length of the path travelled during the diffusion increase with time in the same way. ¿From the point of view of the reference frame in which the deuteron is at rest and is hit by a nucleon in the form of a disk, the radius of which is much larger than that of the deuteron, the statement of the equality of cross sections means that the parton states inside the disk are highly coherent.\n\nIt is clear from above that the cross sections of two hadrons can become equal only when the radius of parton distribution which is increasing with the energy becomes much larger than the size of both hadrons. Substituting for the value of ( is the proton mass) 111It will be demonstrated below that , where is the slope of the Pomeron trajectory. The current data give . we see that the deuteron-nucleon cross section will practically never coincide with the nucleon-nucleon cross section, while the tendency for convergence of cross sections for pion-nucleon, kaon-nucleon and nucleon-nucleon scatterings may be manifested already starting at the incident energies GeV.\n\n## 5 Elastic and quasi-elastic processes\n\nSo far we focused on the implications of the considered picture for inelastic processes with multiplicities, growing logarithmically with the energy. However, with a certain probability it can happen that slow partons scatter at very small angles and the fluctuations will not be interrupted in either of the hadrons (for example, if we discuss the process in the c.m. frame). In this case small angle elastic or quasi-elastic scattering will take place (Fig. 20).\n\nFirst, let us calculate the elastic scattering amplitude. It is well known that the imaginary part of the elastic scattering amplitude can be written in the form\n\n A1(s12)=s12∫d2ρ12ei→q→ρ12σ(ρ12,s12), (20)\n\nwhere is the energy squared in the c.m. system, is the relative impact parameter, - the total interaction cross section of particles being at the distance and is the momentum transferred. In order to calculate it is sufficient to notice that, according to (12), the probability of finding a slow parton with rapidity at the impact parameter which originated from the first hadron with an impact parameter is\n\n φ1(→ρ1,→ρ′1,η1,ηpc)C(η1)πγηpcexp[−(→ρ1−→ρ′1)2γηpc]. (21)\n\nThe probaility a parton originating from the second hadron at impact parameter is\n\n φ2(→ρ2,→ρ′2,η2,ηpc)C(η2)πγηpcexp[−(→ρ2−→ρ′2)2γηpc]. (22)\n\nThe total cross section of the hadron interaction which is due to the interaction of slow partons is equal to\n\n σ(ρ12,s12)=∫dη1dη2d2ρ′12C(η1)C(η2)\n ×∫d2ρ(πγηpc)2exp[−(→ρ−→ρ1)2γηpc−(→ρ−→ρ2)2γηpc].\n\nWe have taken into account that , , and that the dependence on can be neglected in the exponential factor.\n\nAfter carrying out the integration over , we obtain\n\n σ(ρ12,s12)=e−(→ρ1−→ρ2)22γηpcσ02πγηpc (23)\n\nInserting (22) into (20), we get\n\n A1=s12σ0e−γ4q2ξ,\n ξ=2ηpc=logs12μ2. (24)\n\nWe obtained the scattering amplitude corresponding to the exchange by the Pomeranchuk pole with the slope , where is the universal coupling constant of the Pomeron and hadron. The amplitude (24) is usually represented by diagram in Fig. 21\n\nwhere a propagator of the form corresponds to the Pomeron. In the impact parameter space this propagator has the form (22).\n\nLet us discuss the physical meaning of in more detail. For this purpose, let us calculate the zero angle scattering amplitude at (), without using the impact parameter representation. The probability of finding a parton with rapidity and a transverse momentum is described by (9). This expression at corresponds to the diagram in Fig.22. The wavy line represents integration over parton rapidities from to zero.\n\nThis figure reflects the hypothesis that the calculation of for suffiently large and leads to an expression for which is factorized in the same way as the Pomeron contribution to the scattering amplitude. This is because the parton distribution in this region is independent of the properties of the hadron as well as the values of . Compared to the diagram in Fig. 7, Fig. 22 indicates that the calculation of is similar to the calculation of the inclusive cross section due to the Pomeron exchange. The only difference is that the coupling of the hadron with the Pomeron should be substituted by unity, since a hadron always exists in a Pomeron state. If , corresponds to the diagram in Fig.22a, which shows that depends on . Similarly, it is possible to determine the probability of finding several slow partons (Fig. 22b), and even the density matrix of slow partons. In this case the amplitude of elastic hadron-hadron scattering in the center of mass frame is determined by the diagram of fig.23 and the value of is determined solely by the interaction of slow partons.\n\nNow let us consider the quasi-elastic scattering, corresponding to the Pomeron exchange (Figs.24,25) at zero transverse momentum. While the probability to find the parton in hadron ”a” is determined in eq.(8) by the integral of the wave function squared, the analogous quantity for the amplitude of the inelastic diffractive process (Fig.25) will lead to the integral of the product of the parton functions of different hadrons. They are orthogonal to each other and it is almost obvious that amplitude for inelastic diffractive process at zero angle should vanish for this reason. Indeed the orthogonality condition of eq.(6) has the same structure as the imaginary part of the amplitude. Thus, if at high energies the amplitude factorizes (as it should be for the Pomeron exchange), than the orthonormality condition should also have factorized form in the sense that the integral over parton rapidities with factors out, and only constants depend on the properties of specific hadrons (see Fig.26).\n\nOrthogonality of the wave functions of different hadrons implies that at . In fact the reason, why the amplitude of inelastic diffractive process vanishes at zero angle is the same as the reason why all cross sections should approach the same value at high energies. Both phenomena are due to the fact that properties of slow partons do not depend on the properties of hadrons to which they belong. We can illustrate this again using the example of quasi-elastic dissociation of the composite system — e.g. deuteron. Let us consider the interaction of a fast nucleon with a deuteron. As we discussed in the previous section, at very large energies partons from different nucleons will always interact with each other independent of the distance between nucleons. This will lead to production of the spectrum of slow partons which does not depend on the relative distance between nucleons in deuteron. This means that the amplitude of the nucleon-nucleon interaction will not depend on the internucleon distance as well. Thus, if nucleons inside the deuteron will remain intact after the interaction, than the deuteron will not dissociate as well, since if the amplitude does not depend on the internucleon distance, the wave function of the deuteron will not change after the interaction.\n\n## References\n\n• R.Feynman, What neutrinos can tell about partons, Neutrino-72, v.II, p.75, Proceedings Conference, Hungary, June 1972. .\n• J.D.Bjorken, Phys.Rev., 179 (1969), p.1547.\n• S.D.Drell and T.M.Yan, Annals of Phys., 66(1971), p.555\n• V.N.Gribov, Proceedings of Batavia Conference, 1972." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.91924936,"math_prob":0.9694566,"size":44614,"snap":"2023-14-2023-23","text_gpt3_token_len":9149,"char_repetition_ratio":0.20076217,"word_repetition_ratio":0.030396773,"special_character_ratio":0.19590263,"punctuation_ratio":0.09835866,"nsfw_num_words":1,"has_unicode_error":false,"math_prob_llama3":0.98393965,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-06-02T08:30:52Z\",\"WARC-Record-ID\":\"<urn:uuid:819cb864-e4ab-4bc5-9bc8-41d2c1252415>\",\"Content-Length\":\"800020\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e42546f9-84c2-4c1c-9bae-b753a5fb0ec5>\",\"WARC-Concurrent-To\":\"<urn:uuid:90403bab-8c27-42bd-9d4b-0c61cc4a2163>\",\"WARC-IP-Address\":\"172.67.158.169\",\"WARC-Target-URI\":\"https://www.arxiv-vanity.com/papers/hep-ph/0006158/\",\"WARC-Payload-Digest\":\"sha1:MDRUQXNV63SBZCWCY4W6GVC7GLMPGENU\",\"WARC-Block-Digest\":\"sha1:BVIRZCNH3E3JZFYOSYVIASLCNOVBTHLD\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224648465.70_warc_CC-MAIN-20230602072202-20230602102202-00103.warc.gz\"}"}