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[ "## Monday 23 October 2017\n\n### Loop de loop patterns\n\nDuring The Man Who Knew Infinity, Thomas produced a picture for Liz based on a random set of numbers taken from Donald Duck in Mathemagic Land.\n\nThe picture was a Loop de Loop pattern.\n\nThe idea is that you consider a moving pointer, known as a \"turtle\". The turtle is given instructions to move around and the picture is made up of the turtle's track history. The set of rules that the turtle follows is:\n• the turtle moves along in a straight line a given distance. Here, the random numbers where used as the distances;\n• after moving each distance the turtle is turned 90 degrees to the right;\n• iterate the pattern until you get back to the beginning, or until you want to stop.\nThere are many websites that will allow you to produce such a picture using a programming language known as Logo.\n\nFor a simple example consider the list of numbers {50, 100, 150}. A single iteration of the Loop de Loop algorithm is given by\n\nfd 50 rt 90 fd 100 rt 90 fd 150,\n\nwhere \"fd\" means \"go forward\" the given amount and \"rt\" means \"turn right\" the given number of degrees. Following this set of instruction would produce a \"J\" type pattern, as shown below. Iterating the pattern can be done simply through\n\nrepeat 4[fd 50 rt 90 fd 100 rt 90 fd 150 rt 90].\n\nThis code will repeat the instructions 4 times and produce a \"pin wheel\" type pattern, shown below.", null, "Iterations of the code using the set of numbers {50, 100, 150}. Top left - one iteration, top right - two iterations, bottom left - three iterations, bottom right - four iterations.\nThe numbers Thomas used (although apparently not the numbers that Liz gave) were\n\n{7, 9, 4, 8, 5, 7, 4, 3, 9, 2, 5, 7, 3, 8, 4, 9, 7, 5, 2, 6, 3}.\n\nFour iterations take you back to the beginning point; the four iterations are illustrated below. Once, you have made your pattern, you can colour it in and make it funky.", null, "The four iterations of the Loop de Loop pattern, using the Donald Duck random number list.\nThomas generated his pictures using MatLab. The code is given below.\n\nWhy not produce your own Loop de Loop pattern and tweet it to us @PodcastMathsAt?\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\n% Clear all information\nclear\nclose all\nclc\nN=[7 9 4 8 5 7 4 3 9 2 5 7 3 8 4 9 7 5 2 6 3]; %The list of numbers.\n\nfigure('position',[0 0 1 1]) %Set the figure space to fill the screen.\n\n% Iteration 1\nM=N;\ntheta=0;\nx(1)=0;\n\nfor i=1:length(M)\n\nx(i+1)=x(i)+M(i)*exp(theta*1i);\ntheta=theta+pi/2;\nend\nsubplot(2,2,1)\nplot(real(x),imag(x))\naxis equal\naxis off\n\n% Iteration 2\nM=[N N]\ntheta=0;\nx(1)=0;\nfor i=1:length(M)\n\nx(i+1)=x(i)+M(i)*exp(theta*1i);\ntheta=theta+pi/2;\nend\nsubplot(2,2,2)\nplot(real(x),imag(x))\naxis equal\naxis off\n\n% Iteration 3\nM=[N N N];\ntheta=0;\nx(1)=0;\n\nfor i=1:length(M)\n\nx(i+1)=x(i)+M(i)*exp(theta*1i);\ntheta=theta+pi/2;\nend\nsubplot(2,2,3)\nplot(real(x),imag(x))\naxis equal\naxis off\n\n% Iteration 4\nM=[N N N N];\ntheta=0;\nx(1)=0;\n\nfor i=1:length(M)\n\nx(i+1)=x(i)+M(i)*exp(theta*1i);\ntheta=theta+pi/2;\nend\nsubplot(2,2,4)\nplot(real(x),imag(x))\naxis equal\naxis off" ]

[ null, "https://1.bp.blogspot.com/-YOxyy__VKqQ/WeyH1kgLAcI/AAAAAAAAI4Y/9V_0LMw7UtUc79sDYMgWHk1njKlAK36yACLcBGAs/s400/Loop.png", null, "https://2.bp.blogspot.com/-l8ItRxLhBHA/WeyAKDSasTI/AAAAAAAAI4I/C02GjQaONME1Bb7U9FL2via3fFU23OgfgCLcBGAs/s400/Loop.png", null ]

[ "# #9ecdea Color Information\n\nIn a RGB color space, hex #9ecdea is composed of 62% red, 80.4% green and 91.8% blue. Whereas in a CMYK color space, it is composed of 32.5% cyan, 12.4% magenta, 0% yellow and 8.2% black. It has a hue angle of 202.9 degrees, a saturation of 64.4% and a lightness of 76.9%. #9ecdea color hex could be obtained by blending #ffffff with #3d9bd5. Closest websafe color is: #99ccff.\n\n• R 62\n• G 80\n• B 92\nRGB color chart\n• C 32\n• M 12\n• Y 0\n• K 8\nCMYK color chart\n\n#9ecdea color description : Very soft blue.\n\n# #9ecdea Color Conversion\n\nThe hexadecimal color #9ecdea has RGB values of R:158, G:205, B:234 and CMYK values of C:0.32, M:0.12, Y:0, K:0.08. Its decimal value is 10407402.\n\nHex triplet RGB Decimal 9ecdea `#9ecdea` 158, 205, 234 `rgb(158,205,234)` 62, 80.4, 91.8 `rgb(62%,80.4%,91.8%)` 32, 12, 0, 8 202.9°, 64.4, 76.9 `hsl(202.9,64.4%,76.9%)` 202.9°, 32.5, 91.8 99ccff `#99ccff`\nCIE-LAB 80.107, -8.538, -19.272 50.78, 56.87, 86.139 0.262, 0.293, 56.87 80.107, 21.079, 246.105 80.107, -24.032, -29.112 75.412, -11.777, -14.934 10011110, 11001101, 11101010\n\n# Color Schemes with #9ecdea\n\n• #9ecdea\n``#9ecdea` `rgb(158,205,234)``\n• #eabb9e\n``#eabb9e` `rgb(234,187,158)``\nComplementary Color\n• #9eeae1\n``#9eeae1` `rgb(158,234,225)``\n• #9ecdea\n``#9ecdea` `rgb(158,205,234)``\n• #9ea7ea\n``#9ea7ea` `rgb(158,167,234)``\nAnalogous Color\n• #eae19e\n``#eae19e` `rgb(234,225,158)``\n• #9ecdea\n``#9ecdea` `rgb(158,205,234)``\n• #ea9ea7\n``#ea9ea7` `rgb(234,158,167)``\nSplit Complementary Color\n• #cdea9e\n``#cdea9e` `rgb(205,234,158)``\n• #9ecdea\n``#9ecdea` `rgb(158,205,234)``\n• #ea9ecd\n``#ea9ecd` `rgb(234,158,205)``\n• #9eeabb\n``#9eeabb` `rgb(158,234,187)``\n• #9ecdea\n``#9ecdea` `rgb(158,205,234)``\n• #ea9ecd\n``#ea9ecd` `rgb(234,158,205)``\n• #eabb9e\n``#eabb9e` `rgb(234,187,158)``\n``#5faddc` `rgb(95,173,220)``\n• #74b7e1\n``#74b7e1` `rgb(116,183,225)``\n• #89c2e5\n``#89c2e5` `rgb(137,194,229)``\n• #9ecdea\n``#9ecdea` `rgb(158,205,234)``\n• #b3d8ef\n``#b3d8ef` `rgb(179,216,239)``\n• #c8e3f3\n``#c8e3f3` `rgb(200,227,243)``\n• #ddedf8\n``#ddedf8` `rgb(221,237,248)``\nMonochromatic Color\n\n# Alternatives to #9ecdea\n\nBelow, you can see some colors close to #9ecdea. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #9ee0ea\n``#9ee0ea` `rgb(158,224,234)``\n• #9edaea\n``#9edaea` `rgb(158,218,234)``\n• #9ed3ea\n``#9ed3ea` `rgb(158,211,234)``\n• #9ecdea\n``#9ecdea` `rgb(158,205,234)``\n• #9ec7ea\n``#9ec7ea` `rgb(158,199,234)``\n• #9ec0ea\n``#9ec0ea` `rgb(158,192,234)``\n• #9ebaea\n``#9ebaea` `rgb(158,186,234)``\nSimilar Colors\n\n# #9ecdea Preview\n\nThis text has a font color of #9ecdea.\n\n``<span style=\"color:#9ecdea;\">Text here</span>``\n#9ecdea background color\n\nThis paragraph has a background color of #9ecdea.\n\n``<p style=\"background-color:#9ecdea;\">Content here</p>``\n#9ecdea border color\n\nThis element has a border color of #9ecdea.\n\n``<div style=\"border:1px solid #9ecdea;\">Content here</div>``\nCSS codes\n``.text {color:#9ecdea;}``\n``.background {background-color:#9ecdea;}``\n``.border {border:1px solid #9ecdea;}``\n\n# Shades and Tints of #9ecdea\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, #030b10 is the darkest color, while #ffffff is the lightest one.\n\n• #030b10\n``#030b10` `rgb(3,11,16)``\n• #071620\n``#071620` `rgb(7,22,32)``\n• #0a2230\n``#0a2230` `rgb(10,34,48)``\n• #0e2d40\n``#0e2d40` `rgb(14,45,64)``\n• #113850\n``#113850` `rgb(17,56,80)``\n• #154460\n``#154460` `rgb(21,68,96)``\n• #184f71\n``#184f71` `rgb(24,79,113)``\n• #1c5a81\n``#1c5a81` `rgb(28,90,129)``\n• #1f6691\n``#1f6691` `rgb(31,102,145)``\n• #2371a1\n``#2371a1` `rgb(35,113,161)``\n• #267cb1\n``#267cb1` `rgb(38,124,177)``\n• #2a87c1\n``#2a87c1` `rgb(42,135,193)``\n• #2d93d1\n``#2d93d1` `rgb(45,147,209)``\n• #3d9bd5\n``#3d9bd5` `rgb(61,155,213)``\n• #4da3d9\n``#4da3d9` `rgb(77,163,217)``\n• #5eacdc\n``#5eacdc` `rgb(94,172,220)``\n• #6eb4e0\n``#6eb4e0` `rgb(110,180,224)``\n• #7ebce3\n``#7ebce3` `rgb(126,188,227)``\n• #8ec5e7\n``#8ec5e7` `rgb(142,197,231)``\n• #9ecdea\n``#9ecdea` `rgb(158,205,234)``\n• #aed5ed\n``#aed5ed` `rgb(174,213,237)``\n• #bedef1\n``#bedef1` `rgb(190,222,241)``\n• #cee6f4\n``#cee6f4` `rgb(206,230,244)``\n• #deeef8\n``#deeef8` `rgb(222,238,248)``\n• #eff7fb\n``#eff7fb` `rgb(239,247,251)``\n• #ffffff\n``#ffffff` `rgb(255,255,255)``\nTint Color Variation\n\n# Tones of #9ecdea\n\nA tone is produced by adding gray to any pure hue. In this case, #c2c4c6 is the less saturated color, while #8cd1fc is the most saturated one.\n\n• #c2c4c6\n``#c2c4c6` `rgb(194,196,198)``\n• #bec5ca\n``#bec5ca` `rgb(190,197,202)``\n• #b9c7cf\n``#b9c7cf` `rgb(185,199,207)``\n• #b5c8d3\n``#b5c8d3` `rgb(181,200,211)``\n• #b0c9d8\n``#b0c9d8` `rgb(176,201,216)``\n``#accadc` `rgb(172,202,220)``\n• #a7cbe1\n``#a7cbe1` `rgb(167,203,225)``\n• #a3cce5\n``#a3cce5` `rgb(163,204,229)``\n• #9ecdea\n``#9ecdea` `rgb(158,205,234)``\n• #99ceef\n``#99ceef` `rgb(153,206,239)``\n• #95cff3\n``#95cff3` `rgb(149,207,243)``\n• #90d0f8\n``#90d0f8` `rgb(144,208,248)``\n• #8cd1fc\n``#8cd1fc` `rgb(140,209,252)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #9ecdea 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" ]

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[ "The following document demonstrates use of the LAGOSNE package with two analyses:\n\nBefore we can begin, we must load LAGOSNE data with lagosne_load:\n\nlibrary(LAGOSNE)\ndt <- lagosne_load()\n\n## Boxplot of average chlorophyll by state\n\nThe first analysis involves constructing a boxplot of average lake chlorophyll by state. To begin, we join the nutrient water quality table (epi.nutr) to the location information for each lake in the locus table. Next, we join the output to the state table to obtain state abbreviations that correspond to each lake. Finally, we filter the dataset to exclude any lakes that do not fall within a state.\n\nlibrary(dplyr)\nlibrary(ggplot2)\n\nlg <- left_join(dt$epi_nutr, dt$locus)\nlg <- left_join(lg, dt$state) lg <- group_by(lg, state) lg <- filter(lg, !is.na(state)) Now, the state column is of character type so will be in alphabetical order by default. We need to cast this column as a factor and reorder its levels from East to West in preparation for plotting. state_easting <- summarize(lg, mean_easting = mean(nhd_long, na.rm = TRUE)) ## summarise() ungrouping output (override with .groups argument) state_easting <- arrange(state_easting, mean_easting) lg$state <- factor(lg$state, levels = state_easting$state)\n\nFinally, we construct a boxplot with ggplot:\n\nggplot(lg) +\ngeom_boxplot(aes(x = state, y = chla), outlier.shape = NA) +\nylim(c(0, 80)) + ylab(\"Chlorophyll a (ug/L)\")", null, "## Map of average chlorophyll in Pennsylvania\n\nThe second analysis involves constructing a map of average lake chlorophyll concentration in Pennsylvania lakes. To begin, we use the maps and sf package to obtain a geographic outline of the state.\n\nlibrary(maps)\nlibrary(sf)\n\nstates <- st_as_sf(map(\"state\", plot = FALSE, fill = TRUE))\npa <- filter(states, ID == \"pennsylvania\")\n\nNext, we filter the output from the previous analysis to only include PA lakes and calculate mean chlorophyll for each lake:\n\npa_chl <- filter(lg, state == \"PA\")\npa_chl <- group_by(pa_chl, lagoslakeid)\npa_chl <- summarize(pa_chl,\nmean_chl = mean(chla, na.rm = TRUE),\nmean_long = mean(nhd_long, na.rm = TRUE),\nmean_lat = mean(nhd_lat, na.rm = TRUE))\npa_chl <- filter(pa_chl, !is.na(mean_chl))\n\nFinally, we turn our data.frame into an sf geospatial object using st_as_sf before constructing a map with ggplot and geom_sf.\n\npa_chl <- st_as_sf(pa_chl, coords = c(\"mean_long\", \"mean_lat\"), crs = 4326)\n\nggplot() +\ngeom_sf(data = pa) +\ngeom_sf(data = pa_chl, aes(color = log(mean_chl))) +\ntheme_bw() + scale_color_gradient(low = \"white\", high = \"darkgreen\") +\nlabs(color = \"log(Chlorophyll a (ug/L))\")" ]

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JVvbg2uyHM3fBQ3kby4iQQRN5H8HAGdM/AM9O7mfaN5+fzWYFb+PgHtJQIKEQH1cwT0N/nYNc3tec23fi7Mk1Y29pfsP37IDSoC6kVAISKgfo6AFifkM5YUq7815oAnGoPp+ay+YuXMfN6QBxFQLwIKEQH1swT04Y/nMZ8amw+4qTl4bGI+8Pjx+TA+jam3CChEBNTPEtBi0Zc/Nf7I2Y8PDOZMHjdl3vKhjyGgXgQUIgLq5wloGQTUi4BCRED9CGhUBBQiAupHQKMioBARUD8CGhUBhYiA+hHQqAgoRATUj4BGRUAhIqB+BDQqAgoRAfUjoFERUIgIqB8BjYqAQkRA/QhoVAQUIgLqR0CjIqAQEVA/AhoVAYWIgPoR0KgIKEQE1I+ARkVAISKgfgQ0KgIKEQH1I6BREVCICKgfAY2KgEJEQP0IaFQEFCIC6kdAoyKgEBFQPwIaFQGFiID6EdCoCChEBNSPgEZFQCEioH4ENCoCChEB9SOgURFQiAioHwGNioBCRED9CGhUBBQiAupHQKMioBARUD8CGhUBhYiA+hHQqAgoRATUj4BGRUAhIqB+BDQqAgoRAfUjoFERUIgIqB8BjYqAQkRA/QhoVAQUIgLqR0CjIqAQEVA/AhoVAYWIgPoR0KgIKEQE1I+ARkVAISKgfgQ0KgIKEQH1I6BREVCICKgfAY2KgEJEQP0IaFQEFCIC6kdAoyKgEBFQPwIaFQGFiID6EdCoCChEBNSPgEZFQCEioH4ENCoCChEB9SOgURFQiAioHwGNioBCRED9CGhUBBQiAupHQKMioBARUD8CGhUBhYiA+hHQqAgoRATUj4BGRUAhIqB+BDQqAgoRAfUjoFERUIgIqB8BjYqAQkRA/QhoVAQUIgLqR0CjIqAQEVA/AhoVAYWIgPoR0KgIKEQE1I+ARkVAISKgfmUD+vzPr/+nM8775x//PwJaEwQUIgLqVyqgv5r94W3TgLdM/eelBLQGCChEBNSvc0D7/2WfdjpH/v72I1o7rzzufgJaOQIKEQH16xjQq9+W0jZ/OWP+b59tDNY8/fMLpvxRSiMmPEJAK0ZAISKgfh0C+tiENGKfby7bcPK+WW9J285dQ0ArRUAhIqB+HQL6qpFHbvS55g17pekEtFIEFCIC6tchoJ984MV6eNW5BLRSBBQiAurXIaAb8ePJk8V0EtBeIKAQEVA/PaD/mLp7tz0B9SKgEBFQPwIaFQGFiID6EdCoCChEBNSPgEZFQCEioH4ENCoCChEB9SOgURFQiAioHwGNioBCRED9CGhUBBQiAupHQKMioBARUL8OAV38QucT0FogoBARUL8OAU0bRUBrgIBCRED9CGhUBBQiAurXIaAf2ygCWgMEFCIC6tchoEYE1IuAQkRA/QhoVAQUIgLq1yGgdxHQuiKgEBFQvw4BTW874d/6CWgdEVCICKhfh4B+8X+MSG844l9XE9DaIaAQEVC/DgEtise/+qFXpO0mfm/ZC46IFj0Np2cbAa16DQhlUSOgVa9huOkY0IbFlx4wOm2934VDH6wxvRIAADX1YnfhV1x9yA5p873Pf6j7X70UViuKYmXVa0AsRbGm6iUMNyUD2tB307FvTmnP0+/pMqC8BurFa6AQ8RqoX/mANv3nybultOsMAloDBBQiAuqnBbTh12futRkBrQECChEB9ZMD2vAYAa0BAgoRAfUrFdAn1vfMyq76SUDNCChEBNSvVECHfJrd1rtPukp/UxIB9SKgEBFQv24C2rTnbwlotQgoRATUr1RAf3ndVmmzDx77xdmf2XdEeu95J0/dMaVd1Ct5AupFQCEioH6lAvrsO9M+v27v3v+hdHRR9F04In2FgFaKgEJEQP1KBfSE9M61fwq//F3p0sZmRvoQAa0UAYWIgPqVCuhb0xfWDc5M7238vCW9noBWioBCRED9SgV0y/S9dYOr0ujGzyfTlgS0UgQUIgLqVyqgr0t/t25wStqh8fPetDMBrRQBhYiA+pUK6AFpp0cH9x9/c9qvsTktfZCAVoqAQkRA/UoF9N9GpHdcs6a51//93dJmPy0ePXNk+icCWikCChEB9SsV0OKslNJO//tTn/rQmxo7ZxXF7im9ew0BrRQBhYiA+pULaDF/98E/QXpV801Mu6ePLBb7SUDNCChEBNSvZECL/u9M+cCO2//ZxNmtcM6/X80nAXUjoBARUL+yAX3pCKgXAYWIgPoR0KgIKEQE1K9UQB8cgoDWAAGFiID6lQro0M+yI6A1QEAhIqB+BDQqAgoRAfUrFdD5g75x4h+kEV98moDWAAGFiID6lQroelZOT+mbBLQGCChEBNRPDWjR/6H0qtUEtHoEFCIC6icHtPhWSj8noNUjoBARUD89oHem9AgBrR4BhYiA+ukBnZ9GddNPAmpGQCEioH5yQJ9/f3oPAa0BAgoRAfUrFdDvrnX57D9JaToBrQECChEB9SsV0A3fRv/aRQS0BggoRATUTw7o6A8/1FU/CagZAYWIgPqVCugT61E/iJ6A9ggBhYiA+pUKqAUB9SKgEBFQPwIaFQGFiID6dQjoD1+0h8tvJaCVIqAQEVC/DgFNf3XbRmu4au7rDyWglSKgEBFQvw4B/fIr0z6XLB/awl9/7k3pzxYS0EoRUIgIqF+HgBaP7p/S6IkX3ts/OLHspi+8L6Vtz5fvxhNQLwIKEQH16xTQorhlwhYppe3+fNyhxx87cd93bt4YvP6UJ9V8ElA3AgoRAfXrHNCieOjkd2+27p302374klV6PgmoGwGFiID6lQlow1Pf+ezH/9due37k0NNueb6behJQOwIKEQH1KxlQAwLqRUAhIqB+BDQqAgoRAfUjoFERUIgIqB8BjYqAQkRA/QhoVAQUIgLqR0CjIqAQEVA/AhoVAYWIgPoR0KgIKEQE1I+ARkVAISKgfgQ0KgIKEQH10wO6dOyxBLQGCChEBNRPD+h1aRsCWgMEFCIC6lcuoH3njtl9wDtGpFcT0BogoBARUL9SAV2z7/rfDJ8OJ6A1QEAhIqB+pQJ6RUo7T/5wSp86/NB3p9GX92/0QQT05UVAISKgfqUC+uG0y5Ki2D/9oCj6j0wndtVPAmpGQCEioH6lArprOqvx82vpc42fK96yufp1cgS0FwgoRATUr1RAR6UrGz9vTh9rDs5LUwloDRBQiAioX6mA/l4roA+nPZqDH6U/IKA1QEAhIqB+pQL6F+nsxs81W41sfpvcnWkrAloDBBQiAupXKqCfTrs0/nMt/jj9S+PnpelNBLQGCChEBNSvVEBvSOnt5xfFkemdTxZP7Jb2IaA1QEAhIqB+pQJaHJTSG4virs3T1rttldJVBLQGCChEBNSvXEBXn/HmnRqbM1p/iDSlq34SUDMCChEB9SsX0IaVzR8/PHjvqfO76ycBNSOgEBFQv9IBfckIqBcBhYiA+hHQqAgoRATUj4BGRUAhIqB+BDQqAgoRAfUjoFERUIgIqB8BjYqAQkRA/QhoVAQUIgLqR0CjIqAQEVA/AhoVAYWIgPoR0KgIKEQE1I+ARkVAISKgfh0CevZGEdAaIKAQEVC/DgFNG0VAa4CAQkRA/QhoVAQUIgLq1yGgfRtFQGuAgEJEQP06BNSIgHoRUIgIqB8BjYqAQkRA/QhoVAQUIgLq1yGgvI2ptggoRATUr0NAuQtfWwQUIgLq5wvoA/uf3N5ZccGUCTMue8GtegLqRUAhIqB+HQJa/m1Mq47K7YA+dUzOE3OeuZSA9hQBhYiA+nUIaHlfzwMBPSnPeKp4dFqeTUB7ioBCRED9XAH9zzET2wG9Nx/UfO75xNgxvyOgvURAISKgfqUC+kenP9Chn0smTbq2HdB5+ZzWzMx8NQHtJQIKEQH1KxXQlDb7n19++r8L6On59p+2A3pSXtCauXzoNTwB9SKgEBFQv1IB3aN56/0V+bLnXqyf1+cvFwMBnZwXtqYW5OntY6seaXv6GTgtK4rnql4DQvmvouireg3DTamAFveduluzodtO+sFG78A/OuHwlYMBnZAfbM3dnqe1D96xR9viF6svAAwLL34TaeHf/mGzoa877vYXHOr7zNhfFYMBHZufbE3enae0jxJQAJuG//Yu/O2ffWOzobucct+G8xfnbxVrA3rw2megx7WP/vbEtmdWwml1UTxf9RoQS1H0V72E4UYJaONf/0+mbdNs6HvnrHdL6Tf7fbZvXUCPzXe1ZhfkUzb8Z7mJ5MVNJIi4ieQnBbT/thm7DPw156gZa18NvTmvc18xK9/cmr0izyWgvURAISKgfuUD2vejo1tX8Ft++BvnvquxPXTwwO2fbpmUx3/60w8W8/L5rdlZ+fsEtJcIKEQE1K9kQFdcc8hrmvXcetylS5rjn74rbfarDR8ycAm/ME9qvi6wZP/xywhoLxFQiAioX6mAfvujo5v1HH3g5WujeGdKl240oMX0fFZfsXJmPm/ILyGgXgQUIgLqVyqgzXpud/BVK9abWprSzRsP6GMT84HHj8+H8WlMvUVAISKgfuUCusOU61ZvOLV47twVG84MBrRYNGfyuCnzlg/9JQTUi4BCRED9SgX0xu6+yJiA9hIBhYiA+pUKqAUB9SKgEBFQv7IBfeTHN61DQGuAgEJEQP3KBfTWt/KlcnVDQCEioH6lAnr/SL6Vs3YIKEQE1K9UQD+Z0v7z77hzLQJaAwQUIgLqVyqgb0/jumomAe0hAgoRAfUrFdCt0rUEtG4IKEQE1K9UQF+b/p2A1g0BhYiA+pUK6L7pawS0bggoRATUr1RAr027PUtAa4aAQkRA/UoFtDgx7X03Aa0XAgoRAfXrENDT2k59axrxF4eeMjA6jYDWAAGFiID6dQho2igCWgMEFCIC6tchoKM3ioDWAAGFiID6dQioEQH1IqAQEVA/AhoVAYWIgPqVD2j/Yz/5wQNrCGhdEFCICKhf2YDetu/WzdtHI9/f3YeBElA7AgoRAfUrF9Al+627Bf/XL/i2IwJaBQIKEQH1KxXQNe9PafOPnXfN9XMnbZHSwQS0DggoRATUr1RA56W0533t3fvfm9LFBLQGCChEBNSvVED/Mv3+Y4P7j78h7U1Aa4CAQkRA/UoF9DXpb9YNpqcdCGgNEFCICKhfmYD2pXTJutG301YEtAYIKEQE1K9MQJemtN7nh3whvZWA1gABhYiA+pUJaPHW9KfPD+737ZWOIaA1QEAhIqB+pQJ6YUqHrm7vPv/ptN1jG30QAX15EVCICKhfqYAWszZLbzr3Jw8/8pMv7Zy2vm5ZywoCWikCChEB9SsV0N133/qFHwr6QQJaKQIKEQH1KxXQjX6qMgGtFgGFiID6lQro3I2ZT0ArRUAhIqB+pQJqQUC9CChEBNSPgEZFQCEioH5lA9p38eH7vOUNH5g67/kXewQBfXkRUIgIqF/JgF75jsF7R7uor30S0N4goBARUL9yAZ3dKOfo94ybsOcrGztXEtA6IKAQEVC/UgG9d4u0/TnPNvee+9Jr0mseJaA1QEAhIqB+pQI6LY28ZXD/1pHpaAJaAwQUIgLqVyqgu6bD16/p2wloDRBQiAioX6mAjkoXrRtckkYT0BogoBARUL9SAd1m/YBenEYR0BogoBARUL9SAX0bl/D1Q0AhIqB+pQI6LW156+D+z7ZMRxHQGiCgEBFQv1IBfejVaYc5rc//XDH3tekVvyKgNUBAISKgfqUCWlyZUtpur4997H3bNXa+1FU/CagZAYWIgPqVC2gxf9fBP+Xctcs/RCKgZgQUIgLqVzKgRd+FU/feaae9p36DDxOpCQIKEQH1KxvQl46AehFQiAioX5mAPvnVr3b3RZwEtIcIKEQE1K9MQPu2SN8loHVDQCEioH5lAlq8K51BQOuGgEJEQP1KBfS7acclBLRmCChEBNSvVECLOZv/1d0EtF4IKEQE1K9UQK+//nNbjHj/J08+bQABrQECChEB9SsV0DQEAa0BAgoRAfUrFdAthiCgNUBAISKgfqUCakFAvQgoRATUj4BGRUAhIqB+BDQqAgoRAfUrG9C+iw/f5y1v+MDUeXyYSE0QUIgIqF/JgF75jsE78LvMJ6C1QEAhIqB+5QI6u1HO0e8ZN2HPVzZ2uvxAUALqRUAhIqB+pQJ67xZp+3Oebe4996XXpNc8SkBrgIBCRED9SgV0Whp5y+D+rSPT0QS0BggoRATUr1RAd+VrjeuHgEJEQP1KBXRUumjd4JI0moDWAAGFiID6lQroNusH9OI0ioDWAAGFiID6lQro27iErx8CChEB9SsV0Glpy1sH93+2ZTqKgNYAAYWIgPqVCuhDr047zFnR3Fsx97XpFb8ioDVAQCEioH6lAlpcmVLabq+Pfex92zV2vtRVPwmoGQGFiID6lQtoMX/XwT/l3LXLP0QioGYEFCIC6lcyoEXfhVP33mmnvad+gw8TqQkCChEB9Ssb0JeOgHoRUIgIqB8BjYqAQkRA/QhoVAQUIgLq1yGgZ28UAa0BAgoRAfXrENChX2jM1xrXBgGFiID6bSoBffTW4eY/7rnnjqrXYPdApSfJcEdA/ToEtG+jAgb0tj0QwFWVniTDHQH16xBQo0oDes+hVacBZRz071WeJcMdAfXbRAL6+arLgHKOrfIsGe4IqF/ZgD7y45vWiRfQ695bdRlQxp7frPIsGe4IqF+5gN761ug3kW6pOg0o47uVniTDHQH1KxXQ+0eGvwv/qzOHm7PPP/+cqtdg9x+VniTDHQH1KxXQT6a0//w77lwrYECHH94HChEB9SsV0LencV01k4D2EAGFiID6lQroVunalx7Q1c/Dqa8o1lS9BsRSFP1VL2G4KRXQ16Z/f+kBXQGr1Y2nE1WvAbE0/p9b9RKGm1IB3Td97aUHlEt4Ly7hIeIS3q9UQK9Nuz1LQGuGgEJEQP1KBbQ4Me19NwGtFwIKEQH16xDQ09pOfWsa8ReHnjIwOo2A1gABhYiA+nUI6LD5OLvhh4BCRED9OgR09EYR0BogoBARUL8OATUioF4EFCIC6kdAoyKgEBFQPwIaFQGFiID6lQno6p+2tzfsN+sOAloXBBQiAurXOaALP/nqndt734bcbiYAABR/SURBVE0pvftfCWg9EFCICKhfx4B+e5uU1gtoSrMIaC0QUIgIqF+ngF7YSOYfn93e7//l19/XGH6VgNYBAYWIgPp1COjy30+jvrX+xEVbpVGLCGgNEFCICKhfh4B+MaU5G858NaWzCGgNEFCICKhfh4COTTuv3nCm/w/ShwhoDRBQiAioX4eAvjNNHBrCY9OOBLQGCChEBNSvQ0BHpb8fGsLZaSsCWgMEFCIC6tchoNumc4eG8Ow0ioDWAAGFiID6dQjoH6bPDA3hEelNBLQGCChEBNSvQ0APTG8fMrPmjWkMAa0BAgoRAfXrENCLUvrOhjNzU5pLQGuAgEJEQP06BHTVm9P2G3yAyI1bpx2WEdAaIKAQEVC/DgEtrhmRRp3w5ODo4cNekdKlXfWTgJoRUIgIqF+ngBZf3Tylbaac87077vzeWZO2TGmz07vrJwE1I6AQEVC/jgEtbtpx/e+T2+6KLvtJQM0IKEQE1K9zQIvn5u46mM83nra4234SUDMCChEB9SsR0IaHLvnCZ487bd4vu64nAbUjoBARUL9yAXUgoF4EFCIC6kdAoyKgEBFQPwIaFQGFiID6EdCoCChEBNSPgEZFQCEioH4ENCoCChEB9SOgURFQiAioHwGNioBCRED9CGhUBBQiAupHQKMioBARUD8CGhUBhYiA+hHQqAgoRATUj4BGRUAhIqB+BDQqAgoRAfUjoFERUIgIqB8BjYqAQkRA/QhoVAQUIgLqR0CjIqAQEVA/AhoVAYWIgPoR0KgIKEQE1I+ARkVAISKgfgQ0KgIKEQH1I6BREVCICKgfAY2KgEJEQP0IaFQEFCIC6kdAoyKgEBFQPwIaFQGFiID6EdCoCChEBNSPgEZFQCEioH4ENCoCChEB9SOgURFQiAioHwGNioBCRED9CGhUBBQiAupHQKMioBARUD8CGhUBhYiA+hHQqAgoRATUj4BGRUAhIqB+BDQqAgoRAfUjoFERUIgIqB8BjYqAQkRA/QhoVAQUIgLqR0CjIqAQEVA/AhoVAYWIgPoR0KgIKEQE1I+ARkVAISKgfgQ0KgIKEQH1I6BREVCICKgfAY2KgEJEQP0IaFQEFCIC6kdAoyKgEBFQPwIaFQGFiID6EdCoCChEBNSPgEZFQCEioH4ENCoCChEB9SOgURFQiAioHwGNioBCRED9CGhUBBQiAupHQKMioBARUD8CGhUBhYiA+hHQqAgoRATUzxPQB2Z/av+pp97THqy4YMqEGZf1EdDeIqAQEVA/S0B/MjbnKePzmEubg6eOyXlizjOXEtCeIqAQEVA/R0AXj89nLSr6rhwz5ueN0Ul5xlPFo9PybALaUwQUIgLq5wjo5fmE1gX7vHx6UdybD2o+93xi7JjfEdBeIqAQEVA/R0C/kK9tbe/NhzQrek5rMDNfTUB7iYBCRED9HAE99ZBftLa/yZOaV/ALWoPLh17DE1AvAgoRAfVzBHTQpfmUopicF7YGC/J0AtpLBBQiAupnDOid++fGM9EJ+cHW6PY8rT1995i2Z/rgtKYo1lS9BsRSFP1VL2G4sQV05UX75csa27H5yXY385T2gTv2aFvc9a8GgAi6Dugth+QJ1zd3Dl77DPS49hECCmDT0GVAl5yR85ntZ57H5rta2wXNF0TXx2ugXrwGChGvgfpZAvrk5HzkvQP7s/LNre0VeS4B7SUCChEB9XMEdPnh+bzVg4N5+fyBkH6fgPYSAYWIgPo5AnrV+lfrC/OklY3Nkv3HLyOgvURAISKgfo6AHpl/vmZAYzQ9n9VXrJyZzxvyKALqRUAhIqB+hoD27ZcHfboxfGxiPvD48fkwPo2ptwgoRATUzxDQx/MGAS0WzZk8bsq85UMfRkC9CChEBNTPENCSCKgXAYWIgPoR0KgIKEQE1I+ARkVAISKgfgQ0KgIKEQH1I6BREVCICKgfAY2KgEJEQP0IaFQEFCIC6kdAoyKgEBFQPwIaFQGFiID6EdCoCChEBNSPgEZFQCEioH4ENCoCChEB9SOgURFQiAioHwGNioBCRED9CGhUBBQiAupHQKMioBARUD8CGhUBhYiA+hHQqAgoRATUj4BGRUAhIqB+BDQqAgoRAfUjoFERUIgIqB8BjYqA9tqDvx1mHliyZFHVa7B7uNqThIBGRUB7bcweqL8Tqj1JCGhUBLTHLvjzqtuAEv76mkrPEgIaFQHtrZ9WXQaUs9fjVZ4mBDQqAtpb9+1ddRlQykGVniYENCoC2mMP7Ft1GlDCEY9VepYQ0KgIaK+d/XfDzemnn1r1EuwurfYkIaBREVCIeB+oHwGNioBCRED9CGhUBBQiAupHQKMioBARUD8CGhUBhYiA+hHQqAgoRATUj4BGRUAhIqB+BDQqAgoRAfUjoFERUIgIqB8BjYqAQkRA/QhoVAQUIgLqR0CjIqAQEVA/AhoVAYWIgPoR0KgIKEQE1I+ARkVAISKgfgQ0KgIKEQH1I6BREVCICKgfAY2KgEJEQP0IaFQEFCIC6kdAoyKgEBFQPwIaFQGFiID6EdCoCChEBNSPgEZFQCEioH4ENCoCChEB9SOgURFQiAioHwGNioBCRED9CGhUBBQiAupHQKMioBARUD8CGhUBhYiA+hHQqAgoRATUj4BGRUAhIqB+BDQqAgoRAfUjoFERUIgIqB8BjYqAQkRA/QhoVAQUIgLqR0CjIqAQEVA/AhoVAYWIgPoR0KgIKEQE1I+ARkVAISKgfgQ0KgIKEQH1I6BREVCICKgfAY2KgEJEQP0IaFQEFCIC6kdAoyKgEBFQPwIaFQGFiID6EdCoCChEBNSPgEZFQCEioH4ENCoCChEB9SOgURFQiAioHwGNioBCRED9CGhUBBQiAupHQKMioBARUD8CGhUBhYiA+hHQqAgoRATUj4BGRUAhIqB+BDQqAgoRAfUjoFERUIgIqB8BjYqAQkRA/QhoVAQUIgLqR0CjIqAQEVC/ly+gy2C1sihWVb0GxFIUa6pewnDz8gW0rx9WRVH1ChBMwTnj9vIFlEt4Ly7hIeIS3o+ARkVAISKgfgQ0KgIKEQH1I6BREVCICKgfAY2KgEJEQP0IaFQEFCIC6kdAoyKgEBFQPwIaFQGFiID6EdCoCChEBNSPgEZFQCEioH4ENCoCChEB9SOgURFQiAioHwGNioBCRED9CGhUBBQiAupHQKMioBARUD8CGhUBhYiA+hHQqAgoRATUj4BGRUAhIqB+BDQqAgoRAfUjoFERUIgIqB8BjYqAQkRA/QhoVAQUIgLqR0CjIqAQEVA/AhoVAYWIgPoR0KgIKEQE1I+ARkVAISKgfgQ0KgIKEQH1I6BREVCICKgfAY2KgEJEQP0IaFQEFCIC6kdAoyKgEBFQPwIaFQGFiID6EdCoCChEBNSPgEZFQCEioH4ENCoCChEB9SOgURFQiAioHwGNioBCRED9CGhUBBQiAupHQKMioBARUD8CGhUBhYiA+hHQqAgoRATUj4BGRUAhIqB+BDQqAgoRAfUjoFERUIgIqB8BjYqAQkRA/QhoVAQUIgLqR0CjIqAQEVA/AhoVAYWIgPoR0KgIKEQE1I+ARkVAISKgfgQ0KgIKEQH1I6BREVCICKgfAY2KgEJEQP0IaFQEFCIC6kdAoyKgEBFQPwIaFQGFiID6EdCoCChEBNSPgEZFQCEioH4ENCoCChEB9SOgURFQiAioHwGNioBCRED9CGhUBBQiAupHQKMioBARUD8CGhUBhYiA+hHQqAgoRATUj4BGRUAhIqB+BDQqAgoRAfUjoFERUIgIqB8BjYqAQkRA/QhoVAQUIgLqR0CjIqAQEVA/AhoVAYWIgPoR0KgIKEQE1I+ARkVAISKgfgQ0KgIKEQH1I6BREVCICKgfAY2KgEJEQP0IaFQEFCIC6kdAoyKgEBFQPwIaFQGFiID6EdCoCChEBNSPgEZFQCEioH4ENCoCChEB9SOgURFQiAioHwGNioBCRED9CGhUBBQiAupHQKMioBARUD8CGhUBhYiA+vUgoCsumDJhxmV9BLS3CChEBNTPH9Cnjsl5Ys4zlxLQniKgEBFQP39AT8oznioenZZnE9CeIqAQEVA/e0DvzQc1n3s+MXbM7whoLxFQiAionz2g8/I5re3MfDUB7SUCChEB9bMH9KS8oLW9fOg1PAH1IqAQEVA/e0An54Wt7YI8nYD2EgGFiID62QM6IT/Y2t6ep7UnHj2/bdFzcFpVFKurXgNiKYo1VS9huLEHdGx+srW9O09pT9yxR9vil/yrAaDOXnpAD177DPS49gQBBbBpeOkBPTbf1douyKe0J5be1rZoMZyWF8WKqteAWIqir+olDDf2gM7KN7e2V+S5Gx7gJpIXN5Eg4iaSnz2g8/L5AyH9PgHtJQIKEQH1swd0YZ60srFZsv/4ZQS0lwgoRATUzx7QYno+q69YOTOfN2SegHoRUIgIqJ8/oI9NzAcePz4fxqcx9RYBhYiA+vkDWiyaM3nclHnLh04TUC8CChEB9etBQF8EAfUioBARUD8CGhUBhYiA+hHQqAgoRATUj4BGRUAhIqB+BDQqAgoRAfUjoFERUIgIqB8BjYqAQkRA/QhoVAQUIgLqR0CjIqAQEVA/AhoVAYWIgPoR0KgIKEQE1I+ARkVAISKgfgQ0KgIKEQH1I6BREVCICKgfAY2KgEJEQP0IaFQEFCIC6kdAoyKgEBFQPwIaFQGFiID6EdCoCChEBNSPgEZFQCEioH4ENCoCChEB9SOgURFQiAioHwGNioBCRED9CGhUBBQiAupHQKMioBARUD8CGhUBhYiA+hHQqAgoRATUj4BGRUAhIqB+BDQqAgoRAfUjoFERUIgIqN/LF9BnnobTosWL/6vqNSAWThm7ly+g8Lp2jz0uqHoNCGXVHnscVPUahjkCGgYBhYiA9hwBDYOAQkRAe46AhkFAISKgPUdAwyCgEBHQniOgYRBQiAhozxHQMAgoRAS05whoGE/fdtujVa8Boay57baFVa9hmCOgANAlAgoAXSKgANAlAgoAXSKgANAlAlp/Z+VvDOx9PufLKl0Kam5hzh/vGxzcnPMRA7ucOT1CQOtvbUCfHbfuvwhgIxoBzbcPDj6/9nThzOkVAlp/awN6fT5sv3x/tYtBrS3MY/Lsgf0V48cMVpMzp1cIaP2tDeiMPP9zay/ngRdamI/8xIGr2vs/zicNBpQzp1cIaP0NBvTJMfs9c0Oe3F/xclBjC/Ox/5D/rb1/er52IKCcOT1DQOtvMKDfyacUy8blX1S8HNRYI6C/yGe1dp8b/+lfDgSUM6dnCGj9DQZ0Wr65KE7NcyteDmqsEdD+SQesaO4uyBf/aiCgnDk9Q0DrbyCgv8nNF7cW5IOer3pBqK1GQIuvNXPZDOaDAwHlzOkdAlp/AwH9xzynaN5azT+reD2or2ZA78mfb+wtH3dEMRBQzpzeIaD11w7omkn5rubojHxmxetBfTUD2n/I/suL4sb8rYGAcub0EAGtv3ZA78iDxj9X9YpQV82AFvPyj4rilPzQQEA5c3qIgNZfO6Dn5k8c0TI231j1ilBXrYDe17rpflQxEFDOnB4ioPXXCujKCfk/28Mz8qxKl4MaawW0mDp26Q+bf/veCihnTi8R0PprBfSm/ImBD4m4Ne+3uNoFobbaAb0o/2BWfmQgoJw5vURA668V0L/PXx4YrvpovqbS9aC+2gF9IP/N2Oa2FVDOnF4ioPXXDOjise07qU3n5OlVLgc11g5ocUTOlxftgHLm9BQBrb9mQK/Oh6z9S+bbc368yvWgvgYC+q32KdIMKGdOTxFQAOgSAQWALhFQAOgSAQWALhFQAOgSAQWALhFQAOgSAQWALhFQAOgSAQWALhFQAOgSAQWALhFQAOgSAUVED//sP6R5oCcIKCI6Oe0gzQM9QUAREQFFLRBQRERAUQsEFBERUNQCAUUMi//PB3bc5o/2v6W5f3hq+VJz/8bDPvC60e/4yPw1Q+aXnbXX9qPfdfgvK1wyhj8CihBueXU7jukzxfqhXH3QwHTat2+DgP7fN7b3N/t81SvHcEZAEcHjv5dGHfPtq76wY0qtb+YdvFQ/I6VdvnjlFWfsktK568//rtHbA7/2z8e/KqUvVLVmbAIIKCKYl9INze39I9Mhze1gKN+c3r60uV2ycxqz/vzRaYv5ze0Du6atH3z5l4tNBQFFBEenke0v5z158qmtTTuUK3d64/ntB0xOu683/8DI9Nn2/I0p/e3LvFZsQggoIjg9pa+vP37B3fYPbxDQf0jp4YEDb0t/+TKsD5soAooIfj0qpT1Pu2Xl4Hj9gPY/8pOvj08bBPS4tOPg0f14YxN6h4AihBv+sHlPfcsPfmVJa7g2oHcd9cdbt++3rx/Qj6T1jKhoydgEEFDEsPJ7k1/fzOEbftQcDQb0zM1TGvknB37uB5/aIKB/un5A0/OVLRrDHQFFHL++YN+Udni6WBvKGxtPPK9pXdcfvkFAP5reV9kisSkhoIjgwQcXtXfmpHRFsTaUB6etn2jPH7pBQP8uva5/4J/s61vz8i4VmxICigh2TB9p79yb0jeKtaHcO+3Wnu5/9wYBnZ/S/PaBp7blbUzoHQKKCHLaamFrZ0ZK9xTNUL66OZqatnqyuV1xZErvKNbN9/9p2v6m5s6qnDb7TRULxqaBgCKCa1La8uhvXveP+6a0T/Pi/O/TZlfc9VhxZUq7ffvnC857SxqVRly/dO18cetm6RWTvvIvs9/O++jRSwQUIZy42cAt9b0WN4dXD3xoyJSBdyqd8J3Gz7Hr5osf7jhw5OiKF45hjYAihjs+8Z7tR79z3PcHhqe/ecsdL2lsr/zgTlvuPOUXxZrDttv6mPXmi2dOeOcrt/2TyXdVtmBsCggoAHSJgAJAlwgoAHSJgAJAlwgoAHSJgAJAlwgoAHSJgAJAlwgoAHSJgAJAlwgoAHSJgAJAlwgoAHSJgAJAlwgoAHSJgAJAlwgoAHSJgAJAlwgoAHSJgAJAlwgoAHSJgAJAl/4/YGJG1shA23kAAAAASUVORK5CYII=", null ]

[ "Solutions by everydaycalculation.com\n\n## Divide 9/18 with 49/24\n\n1st number: 9/18, 2nd number: 2 1/24\n\n9/18 ÷ 49/24 is 12/49.\n\n#### Steps for dividing fractions\n\n1. Find the reciprocal of the divisor\nReciprocal of 49/24: 24/49\n2. Now, multiply it with the dividend\nSo, 9/18 ÷ 49/24 = 9/18 × 24/49\n3. = 9 × 24/18 × 49 = 216/882\n4. After reducing the fraction, the answer is 12/49\n\nMathStep (Works offline)", null, "Download our mobile app and learn to work with fractions in your own time:" ]

[ null, "https://answers.everydaycalculation.com/mathstep-app-icon.png", null ]

[ "💬 👋 We’re always here. Join our Discord to connect with other students 24/7, any time, night or day.Join Here!", null, "# (a) Find the average value of $f$ on the given interval.(b) Find $c$ such that $f_{ave} = f(c)$.(c) Sketch the graph of $f$ and a rectangle whose area is the same as the area under the graph of $f$.$f(x) = \\dfrac{1}{x}$ , $[1, 3]$\n\n## (A). $f_{a v g}=\\frac{1}{2} \\ln 3$(B). $\\frac{2}{\\ln (3)}$(C). SEE SOLUTION\n\n#### Topics\n\nApplications of Integration\n\n### Discussion\n\nYou must be signed in to discuss.\n##### Top Calculus 2 / BC Educators", null, "##### Catherine R.\n\nMissouri State University", null, "", null, "##### Kristen K.\n\nUniversity of Michigan - Ann Arbor", null, "##### Michael J.\n\nIdaho State University\n\nLectures\n\nJoin Bootcamp\n\n### Video Transcript\n\nand this problem, we are going to be using a specific application of the integral, which is finding the average value of our function. So let's just review how we would find that. Remember, on the closed interval, a B theatric value of our function is equivalent to one over B minus a times the integral from A to B of f of x indie X. And in this problem, we're told that f of X equals one over acts and our interval is 123 So we can simply plug these into this general format. So the average value of our function would be equivalent to 1/3 minus one, and we'll multiply that value by the integral from 1 to 3 of one over X in D X. So then we get one half will take the anti derivative, which is the natural log of X, and we'll evaluate that for 123 And you might be asking, why isn't this the natural? The natural log of the absolute value of X and that is you are correct. Um, the only thing that we have to consider here is that our intervals from 1 to 3 were in the positive number, so we can't have a negative value in the natural log. So then, when we evaluate this will do B minus a will take one half times the natural log of three minus the natural log of one. But remember, the actual log of one is zero. So we'll find that the average value of our function is one half times the natural log of three. So now we were asked to find some value. See, and remember, by the mean value the're, um we're told that there exists number C in our average value of our function. FFC So we'll set one oversee equal to one half times the natural log of three will get see moved on to the other side. So I'll get two equals. A natural log of three times see, and then to find C by itself will have two divided by the natural log of three. How and now Finally, we're asked, Let's put everything together that we've solved and really analyze it in a visual format in the graph of our function. So this is this portion is that portion of our function that we just analyzed this is one over X and we are all the wayto one, which is basically the height of our function. And we're evaluating it from 1 to 3 because remember, that's our interval. And this is the portion we just analyzing using the the average value of the function. So I hope that this helps you understand how we can find the average value of a function using integration and then use an application of the mean value for him to find the number see in our in our average of the function. And then finally just thinking about this visually by graphing our function and the portion of our interval that we were looking at.", null, "University of Denver\n\n#### Topics\n\nApplications of Integration\n\n##### Top Calculus 2 / BC Educators", null, "##### Catherine R.\n\nMissouri State University", null, "", null, "##### Kristen K.\n\nUniversity of Michigan - Ann Arbor", null, "##### Michael J.\n\nIdaho State University\n\nLectures\n\nJoin Bootcamp" ]

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[ "# Answer to Question #54614 in Inorganic Chemistry for alex\n\nQuestion #54614\nIn nature, the element X consists of two naturally occurring isotopes. 107X with abundance 55.84% and isotopic mass 106.9051 amu and 109X with isotopic mass 108.9048 amu. Use the given information to calculate the atomic mass of the element X to an accuracy of .001% (Report your answer like this yyy.yyyy)\n1\n2015-09-13T04:19:33-0400\n«W» is the mass fraction (abundance) of the isotope (%);\nW(107X) = 55.84%;\nW(109X) = 100 – 55.84 = 44.16%;\nAr – the relative atomic mass (a.m.u.) of the element. Symbol: u;\nAr107X = 106.9051 amu (u);\nAr109X = 108.9048 amu (u);\nArX=W(107X)*Ar107X+W(109X)*Ar109X\nArX = 106.9051*0.5584 + 108.9048*0.4416 = 59.6958 + 48.0924 = 107.7882 amu (u);\n\nNeed a fast expert's response?\n\nSubmit order\n\nand get a quick answer at the best price\n\nfor any assignment or question with DETAILED EXPLANATIONS!" ]

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[ "# Testing the Accuracy of the FLIR One\n\nIn my review of the FLIR One, I mentioned that this would be an excellent device for a physics lab. So, here is my first \"lab\", hopefully there will be more posts like this. The FLIR One Tech Specs states the camera can detect temperature differences down to 0.1°C. However, detecting temperature differences and measuring temperatures are two very different things.\n\nHow Do You Measure Temperature? ——————————-\n\nThis might seem like an easy thing to measure, but it isn't. Here are three common methods for measuring temperature.\n\nThermal Expansion of a Liquid. When a liquid increases in temperature, it expands (true for solids and gases too). If you put this liquid in a tube, you can correlate the increase in volume with an increase in temperature. In the past, these thermometers were filled with mercury. However, mercury isn't the best thing to have around if the thermometer breaks. Newer liquid thermometers have alcohol in them. How do you calibrate these thermometers? The typical method is to use at least two reference temperatures (freezing and boiling point of water) and then divide the scale into 100 pieces (for Celsius).\n\nElectric Potential of Different Metals (Thermocouple). The basic idea is that two different metals can produce a change in electric potential when in contact with each other. This small voltage depends on the temperature of the two metals. That's it. Just record the voltages for some known temperatures and use that to calibrate the device.\n\nMeasuring the Infrared Light. For any object, there are two sources of light. First, light (of any wavelength) could hit the object and then reflect off. This isn't very useful for measuring temperature. Second, light is also emitted from objects. The wavelength and intensity of this light depend on the temperature of the object. We call this blackbody radiation (since a black body doesn't reflect light). For most objects around you, the blackbody radiation is in the infrared region so that you can't see it - but an IR camera can. By measuring this IR light, the camera can estimate the temperature.\n\nOf course with all of these methods, there are problems. The thermocouple and liquid thermometer have to be in contact with the object to measure its temperature. This means that the location of the probe might be a different temperature than other parts of the object. Also, there is the calibration issue. In the end, measuring temperature isn't as simple as you would like.\n\nComparing Temperatures ———————-\n\nAs a means for estimating the accuracy of the FLIR One, I am just going to set up a simple experiment. I will use a few different temperature measuring devices and compare the temperature values for water on a hot plate.\n\nI have three thermometers to measure the water temperature. There is a hand held digital thermometer, a digital thermometer with a probe on a wire (I assume both are thermocouples) and then a liquid thermometer. My plan was to measure the temperature from 0°C to 100°C (the stated range of the FLIR One), but those two end points didn't give the best data. Also, the liquid thermometer only goes up to 50°C.\n\nFor the FLIR One measurements, the device needs to recalibrate to obtain a temperature measurement. For each temperature reading, I would recalibrate at least 5 times and get 5 different values for temperature. This way I could get both an average and standard deviation for the temperature.\n\nLet's just get to the data. Here is a plot of difference in temperature of different devices vs. a reference temp. For the reference temperature, I am going to use the readings from the digital thermometer with the remote probe. The liquid thermometer is probably the most accurate, but it only goes up to 50°C.\n\nOk, how about the average deviation from the reference temperature for each device?\n\nClearly there are some problems with this data. Here are a few of the things I can think of.\n\n• It's possible that the FLIR One is measuring reflected IR instead of IR from the water. I tried putting a box over the water to shield it and it didn't seem to make a difference.\n• Both the digital thermometers and the mercury thermometer take some time to reach an equilibrium temperature. The reading might not yet be from the equilibrium temp.\n• The water isn't all just at one temperature. I think the contact thermometers sort of average out any difference in water temperature by \"touching\" different parts. However, the FLIR one just looks at one spot.\n\nHere is a quick video where you can see the temperature differences at the surface of the water.\n\nYou can see the problem of measuring temperature. However, notice that even if the FLIR One produces uncertainty in the temperature, you can still see relatively small temperature differences.\n\nBut what is the accuracy of the FLIR One? Based on the bar chart above, I can give the following conservative estimates for the accuracy.\n\n• FLIR One = +/- 2°C\n• Mercury thermometer = +/- 0.5°C\n• Digital thermometer = +/- 1°C\n\nThat's my best guess on the accuracy of the FLIR One." ]

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[ "# Spatial Regression Models\n\n2.1 Spatial lag and spatial error models\n\nIn the standard linear regression model, spatial dependence can be incorporated in two distinct ways: as an additional regressor in the form of a spatially lagged dependent variable (Wy), or in the error structure (Е[є;є;] Ф 0). The former is referred to as a spatial lag model and is appropriate when the focus of interest is the assessment of the existence and strength of spatial interaction. This is inter­preted as substantive spatial dependence in the sense of being directly related to a spatial model (e. g. a model that incorporates spatial interaction, yardstick com­petition, etc.). Spatial dependence in the regression disturbance term, or a spatial error model is referred to as nuisance dependence. This is appropriate when the concern is with correcting for the potentially biasing influence of the spatial autocorrelation, due to the use of spatial data (irrespective of whether the model of interest is spatial or not).\n\nFormally, a spatial lag model, or a mixed regressive, spatial autoregressive model is expressed as\n\ny = pWy + Xp + e, (14.9)\n\nwhere p is a spatial autoregressive coefficient, e is a vector of error terms, and the other notation is as before.16 Unlike what holds for the time series counterpart of this model, the spatial lag term Wy is correlated with the disturbances, even when the latter are iid. This can be seen from the reduced form of (14.9),\n\ny = (I – pW)-1Xp + (I – pW)-1E, (14.10)\n\nin which each inverse can be expanded into an infinite series, including both the explanatory variables and the error terms at all locations (the spatial multiplier). Consequently, the spatial lag term must be treated as an endogenous variable and proper estimation methods must account for this endogeneity (OLS will be biased and inconsistent due to the simultaneity bias).\n\nA spatial error model is a special case of a regression with a non-spherical error term, in which the off-diagonal elements of the covariance matrix express the structure of spatial dependence. Consequently, OLS remains unbiased, but it is no longer efficient and the classical estimators for standard errors will be biased. The spatial structure can be specified in a number of different ways, and (except for the non-parametric approaches) results in a error variance-covariance matrix of the form", null, "E[ee’] = Q(0),\n\nwhere 0 is a vector of parameters, such as the coefficients in an SAR error\n\nprocess." ]

[ null, "http://cacasa2.info/img/1149/image365.png", null ]

[ "It’s best to think of iterators as pointers:\n\n## Check if element is in vector\n\nstd::find, std::find_if, std::find_if_not - cppreference.com\n\n``````#include <iostream>\n#include <algorithm>\n#include <vector>\n#include <iterator>\n\nint main()\n{\nint n1 = 3;\nint n2 = 5;\n\nstd::vector<int> v{0, 1, 2, 3, 4};\n\nauto result1 = std::find(std::begin(v), std::end(v), n1);\nauto result2 = std::find(std::begin(v), std::end(v), n2);\n\nif (result1 != std::end(v)) {\nstd::cout << \"v contains: \" << n1 << '\\n';\n} else {\nstd::cout << \"v does not contain: \" << n1 << '\\n';\n}\n\nif (result2 != std::end(v)) {\nstd::cout << \"v contains: \" << n2 << '\\n';\n} else {\nstd::cout << \"v does not contain: \" << n2 << '\\n';\n}\n}\n\n``````\n\n## Finding the index of an item\n\nC++ : How to find an element in vector and get its index ? – thispointer.com\n\n``````int index = std::distance(std::begin(v), result1);\n\n``````\n\n### Generic function to find the index in a vector\n\n``````Generic function to find an element in vector and also its position.\nIt returns a pair of bool & int i.e.\n\nbool : Represents if element is present in vector or not.\nint : Represents the index of element in vector if its found else -1\n\n*/\ntemplate < typename T>\nstd::pair<bool, int > findInVector(const std::vector<T> & vecOfElements, const T & element)\n{\nstd::pair<bool, int > result;\n\n// Find given element in vector\nauto it = std::find(vecOfElements.begin(), vecOfElements.end(), element);\n\nif (it != vecOfElements.end())\n{\nresult.second = distance(vecOfElements.begin(), it);\nresult.first = true;\n}\nelse\n{\nresult.first = false;\nresult.second = -1;\n}\n\nreturn result;\n}\n\n``````\n\nTags:\n\nCategories:" ]

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[ "", null, "Caffe2 - C++ API A deep learning, cross platform ML framework\ntorch::ExpandingArray< D, T > Class Template Reference\n\nA utility class that accepts either a container of `D`-many values, or a single value, which is internally repeated `D` times. More...\n\n`#include <expanding_array.h>`\n\n## Public Member Functions\n\nExpandingArray (std::initializer_list< T > list)\nConstructs an `ExpandingArray` from an `initializer_list`. More...\n\nExpandingArray (at::ArrayRef< T > values)\nConstructs an `ExpandingArray` from an `initializer_list`. More...\n\nExpandingArray (T single_size)\nConstructs an `ExpandingArray` from a single value, which is repeated `D` times (where `D` is the extent parameter of the `ExpandingArray`). More...\n\nExpandingArray (const std::array< T, D > &values)\nConstructs an `ExpandingArray` from a correctly sized `std::array`.\n\nstd::array< T, D > & operator* ()\nAccesses the underlying `std::array`.\n\nconst std::array< T, D > & operator* () const\nAccesses the underlying `std::array`.\n\nstd::array< T, D > * operator-> ()\nAccesses the underlying `std::array`.\n\nconst std::array< T, D > * operator-> () const\nAccesses the underlying `std::array`.\n\noperator at::ArrayRef< T > () const\nReturns an `ArrayRef` to the underlying `std::array`.\n\nsize_t size () const noexcept\nReturns the extent of the `ExpandingArray`.\n\n## Detailed Description\n\n### template<size_t D, typename T = int64_t> class torch::ExpandingArray< D, T >\n\nA utility class that accepts either a container of `D`-many values, or a single value, which is internally repeated `D` times.\n\nThis is useful to represent parameters that are multidimensional, but often equally sized in all dimensions. For example, the kernel size of a 2D convolution has an `x` and `y` length, but `x` and `y` are often equal. In such a case you could just pass `3` to an `ExpandingArray<2>` and it would \"expand\" to `{3, 3}`.\n\nDefinition at line 21 of file expanding_array.h.\n\n## Constructor & Destructor Documentation\n\ntemplate<size_t D, typename T = int64_t>\n torch::ExpandingArray< D, T >::ExpandingArray ( std::initializer_list< T > list )\ninline\n\nConstructs an `ExpandingArray` from an `initializer_list`.\n\nThe extent of the length is checked against the `ExpandingArray`'s extent parameter `D` at runtime.\n\nDefinition at line 26 of file expanding_array.h.\n\ntemplate<size_t D, typename T = int64_t>\n torch::ExpandingArray< D, T >::ExpandingArray ( at::ArrayRef< T > values )\ninline\n\nConstructs an `ExpandingArray` from an `initializer_list`.\n\nThe extent of the length is checked against the `ExpandingArray`'s extent parameter `D` at runtime.\n\nDefinition at line 32 of file expanding_array.h.\n\ntemplate<size_t D, typename T = int64_t>\n torch::ExpandingArray< D, T >::ExpandingArray ( T single_size )\ninline\n\nConstructs an `ExpandingArray` from a single value, which is repeated `D` times (where `D` is the extent parameter of the `ExpandingArray`).\n\nDefinition at line 43 of file expanding_array.h.\n\nThe documentation for this class was generated from the following file:" ]

[ null, "https://caffe2.ai/doxygen-c/html/Caffe2-with-name-55-tall.png", null ]

[ "## Sum Embedded Numbers\n\n### May 15, 2018\n\nIt’s mid-May, so at most universities the semester has ended and it is safe for me to solve some of the exercises that students sent to me over the last few months. Here’s a fun little one:\n\nGiven a string containing embedded numbers, find the sum of the embedded numbers. For instance, given the string “11aa22bb33cc44”, the desired sum is 11 + 22 + 33 + 44 = 110. You may not use regular expressions.\n\nAlthough the problem statement doesn’t say so, you may assume that the numbers of interest are non-negative integers. Thus, the purpose of the exercise is for students to iterate through a string, identify the digits in the string, and manipulate them numerically.\n\nYour task is to write a program that sums the numbers embedded in a string. When you are finished, you are welcome to read or run a suggested solution, or to post your own solution or discuss the exercise in the comments below.\n\nPages: 1 2\n\n### 35 Responses to “Sum Embedded Numbers”\n\n1. Milbrae said\n\nPython\n\n```def embeddednumbers(s):\nsd, cd = 0, 0\n\n# single digit numbers\nfor d in s:\nv = ord(d)\nif v >= 0x30 and v <= 0x39:\nsd += v - 0x30\n\n# consecutive numbers\nl = len(s)\ni = 0\nwhile i < l:\nv = ord(s[i])\nif v >= 0x30 and v <= 0x39:\nj = 1\nwhile i + j < l:\nv = ord(s[i+j])\nif v >= 0x30 and v <= 0x39:\nj += 1\nelse:\nbreak\ncd += int(s[i:i+j])\ni += j\n\nreturn sd, cd\n\nif __name__ == \"__main__\":\n''\ntext = \"11aa22bb33cc44\"\nprint (text + \" = \" + str(embeddednumbers(text)))\n\ntext = \"1k23jk34jk56jk3454\"\nprint (text + \" = \" + str(embeddednumbers(text)))\n```\n\nSample output:\n11aa22bb33cc44 = (20, 110)\n1k23jk34jk56jk3454 = (40, 3568)\n\n2. ```(defun sum-embedded-cardinals (string)\n(loop\n:for start := (position-if (function digit-char-p) string)\n:then (position-if (function digit-char-p) string :start end)\n:for (cardinal end) := (when start\n(multiple-value-list\n(parse-integer string :start start :junk-allowed t)))\n:while cardinal\n:sum cardinal))\n\n(defun test/sum-embedded-cardinals ()\n(assert (= 0 (sum-embedded-cardinals \"\")))\n(assert (= 0 (sum-embedded-cardinals \"foo\")))\n(assert (= 42 (sum-embedded-cardinals \"42\")))\n(assert (= 42 (sum-embedded-cardinals \"-42\")))\n(assert (= 42 (sum-embedded-cardinals \"-42-\")))\n(assert (= 10 (sum-embedded-cardinals \"1-2-3-4\")))\n(assert (= 10 (sum-embedded-cardinals \" 1 2, 3; 4. \")))\n(assert (= 110 (sum-embedded-cardinals \"11aa22bb33cc44\")))\n:success)\n\n(test/sum-embedded-cardinals)\n;; --> :success\n```\n3. Milbrae said\n\nTested the samples of Pascal and found a misplacement of a variable. Should be fixed now.\n\n```def embeddednumbers(s):\nsd, cd = 0, 0\n\n# single digit numbers\nfor d in s:\nv = ord(d)\nif v >= 0x30 and v <= 0x39:\nsd += v - 0x30\n\n# consecutive numbers\nl = len(s)\nif l == 0: return sd, 0\n\ni = 0\nwhile i < l:\nv = ord(s[i])\nj = 1\nif v >= 0x30 and v <= 0x39:\nwhile i + j < l:\nv = ord(s[i+j])\nif v >= 0x30 and v <= 0x39:\nj += 1\nelse:\nbreak\ncd += int(s[i:i+j])\ni += j\n\nreturn sd, cd\n\nif __name__ == \"__main__\":\n''\ndata = [\"\", \"42\", \"-42\", \"-42-\", \"1-2-3-4\", \" 1 2, 3; 4. \", \"11aa22bb33cc44\"]\nfor text in data:\nprint (str(embeddednumbers(text)) + \" = \" + text)\n```\n\nOutput is:\n(0, 0) =\n(6, 42) = 42\n(6, 42) = -42\n(6, 42) = -42-\n(10, 10) = 1-2-3-4\n(10, 10) = 1 2, 3; 4.\n(20, 110) = 11aa22bb33cc44\n\n4. Phil Runninger said\n\nHere’s a one-line Erlang solution\n\n```lists:sum(lists:foldl(fun(X,[H|T]) when X >= \\$0 andalso X =< \\$9 -> [H*10+X-\\$0 | T]; (X,L) -> [0|L] end, , \"11aa22bb33cc44\")).\n```\n\nand an attempt to explain it:\n\n```lists:sum(\nlists:foldl(\nfun(X,[H|T]) when X >= \\$0 andalso X =< \\$9 -> [H*10+X-\\$0 | T];\n(X,L) -> [0|L]\nend,\n,\n\"11aa22bb33cc44\")\n).\n```\n\nStart with a list containing the value 0 (line 6). For each character X (lines 3,4) in the string (line 7), if it’s between “0” and “9”, multiply the head of the list by 10, add X to it, and put that back at the head of the list (line 3); and if X is not numeric, tack on another 0 to the list (line 4). When lists:foldl is done, you have the list of numbers [44,0,33,0,22,0,11], which lists:sums to 110 (line 1).\n\n5. Steve said\n\nCache/Mumps version\n\n```SUMEMBNUM(STR) ;\nN SUBTOTAL,TOTAL\nS TOTAL=0\nF S SUBTOTAL=\"\" D GETNUMS(.STR) S TOTAL=TOTAL+SUBTOTAL D FINDNEXTNUMS(.STR) Q:STR=\"\"\nQ TOTAL\n;\nGETNUMS(STR) ;\nN CHAR\nF Q:STR=\"\" S CHAR=\\$A(STR) Q:CHAR<48!(CHAR>57) S SUBTOTAL=SUBTOTAL_\\$C(CHAR),STR=\\$E(STR,2,\\$L(STR))\nQ\n;\nFINDNEXTNUMS(STR) ;\nN CHAR\nF Q:STR=\"\" S CHAR=\\$A(STR) Q:CHAR'<48&(CHAR'>57) S STR=\\$E(STR,2,\\$L(STR))\nQ\n```\n\nf i=””,”foo”,”42″,”-42″,”-42-“,”1-2-3-4″,” 1 2, 3; 4. “,”11aa22bb33cc44″ w !,””””,i,””” –>”,?40,””””,\\$\\$SUMEMBNUM(i),””””\n\n“” –> “0”\n“foo” –> “0”\n“42” –> “42”\n“-42” –> “42”\n“-42-” –> “42”\n“1-2-3-4” –> “10”\n” 1 2, 3; 4. ” –> “10”\n“11aa22bb33cc44” –> “110”\n\n6. sbocq said\n\nBash.\n\n```function sum_embedded {\nlocal tot=0 acc=\"\" c i\n\nwhile read -r -n1 c; do\ni=\\$(printf \"%d\\n\" \"'\\$c\");\nif (( i >= 0x30 && i <= 0x39 )); then\nacc+=\\$c;\nelif [[ -n \\${acc} ]]; then\n(( tot+=acc ))\nacc=\"\"\nfi\ndone <<< \"\\$1\"\n\necho \\$tot\n}\n```\n\nTest:\n\n```\\$ for assert in \"foo=0\" \\\n\"42=42\" \\\n\"-42=42\" \\\n\"-42-=42\" \\\n\"1-2-3-4=10\" \\\n\" 1 2, 3; 4. =10\" \\\n\"11aa22bb33cc44=110\"; do\necho -en \"\\${assert}=\"\nIFS='=' read s expect <<< \"\\${assert}\"\n(( \\$(sum_embedded \"\\$s\") == expect )) && echo -e \"OK\" || echo -e \"FAIL!\"\ndone|column -s\\$'=' -t\n\nfoo=0 OK\n42=42 OK\n-42=42 OK\n-42-=42 OK\n1-2-3-4=10 OK\n1 2, 3; 4. =10 OK\n11aa22bb33cc44=110 OK\n```\n7. sbocq said\n\nClojureScript.\n\n```(defn sum-embedded [s]\n(first (reduce (fn [[tot acc] c] (cond (<= 0x30 (.charCodeAt c 0) 0x39) [tot (str acc c)]\n(not (= acc \"\")) [(+ tot (js/parseInt acc)) \"\"]\n:else [tot acc]))\n[0 \"\"]\n(str s \"_\"))))\n```\n\nTest:\n\n```(doseq ([s expect] [[\"foo\" 0]\n[\"42\" 42]\n[\"-42\" 42]\n[\"-42-\" 42]\n[\"1-2-3-4\" 10]\n[\"1 2, 3; 4. \" 10]\n[\"11aa22bb33cc44\" 110]])\n(print s \"=\" expect \" \" (if (= (sum-embedded s) expect) \"OK\" \"FAIL\")))\n\nfoo = 0 OK\n42 = 42 OK\n-42 = 42 OK\n-42- = 42 OK\n1-2-3-4 = 10 OK\n1 2, 3; 4. = 10 OK\n11aa22bb33cc44 = 110 OK\nnil\n```\n8. Milbrae said\n\nRe-implementation of my Python solution in D\n\n```import std.stdio;\nimport std.range;\nimport std.string;\nimport std.conv;\n\nulong embedded(string s)\n{\nulong l = s.length;\nulong sum = 0;\n\nif (l > 0) {\nulong i = 0;\nwhile (i < l) {\nbyte v = cast(byte)s[i];\nulong j = 1;\nif (v >= 0x30 && v <= 0x39) {\nwhile (i + j < l) {\nv = cast(byte)s[i+j];\nif (v >= 0x30 && v <= 0x39) {\nj++;\n}\nelse {\nbreak;\n}\n}\nulong t = to!ulong(s[i..(i+j)]);\nsum += t;\n}\ni += j;\n}\n}\n\nreturn sum;\n}\n\nvoid main()\n{\nstring[] data = [\"\", \"42\", \"-42\", \"-42-\", \"1-2-3-4\", \" 1 2, 3; 4. \", \"11aa22bb33cc44\"];\n\nforeach (text; data) {\ntext.embedded.writeln;\n}\n}\n```\n9. Globules said\n```import Data.Char (isDigit)\nimport Data.List.Split (wordsBy)\n\nembeddedSum :: (Read a, Integral a) => String -> a\nembeddedSum = sum . map read . wordsBy (not . isDigit)\n\nmain :: IO ()\nmain = do\nlet strs = [\"\", \"foo\", \"42\", \"-42\", \"-42-\", \"1-2-3-4\",\n\"1 2, 3; 4. \", \"11aa22bb33cc44\"]\nmapM_ (print . embeddedSum) strs\n```\n```\\$ ./embedsum\n0\n0\n42\n42\n42\n10\n10\n110\n```\n10. Daniel said\n\nHere’s a solution in C.\n\n```#include <stdio.h>\n#include <stdlib.h>\n#include <string.h>\n\nint sum_embedded(char* str) {\nsize_t i = strlen(str);\nint sum = 0;\nint place = 1;\nwhile (1) {\nif (i == 0) break;\n--i;\nchar c = str[i];\nif (c < '0' || c > '9') {\nplace = 1;\ncontinue;\n}\nsum += place * (c - '0');\nplace *= 10;\n}\nreturn sum;\n}\n\nint main(int argc, char* argv[]) {\nif (argc != 2) {\nfprintf(stderr, \"Usage: %s <STRING>\\n\", argv);\nreturn EXIT_FAILURE;\n}\nint sum = sum_embedded(argv);\nprintf(\"%d\\n\", sum);\nreturn EXIT_SUCCESS;\n}\n```\n\nExample:\n\n```\\$ ./sum_embedded \"11aa22bb33cc44\"\n110\n```\n11. sbocq said\n\nI really like the Haskell solution of Globules. The closest I could come with in ClojureScript, without using regex, is with partition-by:\n\n```(defn sum-embedded [s]\n(->> (partition-by #(<= 0x30 (.charCodeAt % 0) 0x39) s)\n(filter integer?)\n(reduce +)))\n```\n\nTest:\n\n```(doseq ([s expect] [[\"\" 0] [\"foo\" 0] [\"42\" 42] [\"-42\" 42] [\"-42-\" 42] [\"1-2-3-4\" 10]\n[\"1 2, 3; 4. \" 10] [\"11aa22bb33cc44\" 110]])\n(print s \"=\" expect \" \" (if (= (sum-embedded s) expect) \"OK\" \"FAIL\")))\n```\n\n= 0 OK\nfoo = 0 OK\n42 = 42 OK\n-42 = 42 OK\n-42- = 42 OK\n1-2-3-4 = 10 OK\n1 2, 3; 4. = 10 OK\n11aa22bb33cc44 = 110 OK\n\n12. Daniel said\n\nHere’s a solution in Python.\n\n```def sum_embedded(s):\ns = ''.join(c if c.isdigit() else ' ' for c in s)\nreturn sum(map(int, s.split()))\n\ns = \"11aa22bb33cc44\"\nprint sum_embedded(s)\n```\n\nOutput:\n\n```110\n```\n13. Daniel said\n\nHere’s the same solution using double quotes, to attempt to improve the formatting.\n\n```def sum_embedded(s):\ns = \"\".join(c if c.isdigit() else \" \" for c in s)\nreturn sum(map(int, s.split()))\n```\n14. Daniel said\n\nHere’s another solution in C.\n\n```#include <stdio.h>\n#include <stdlib.h>\n\nint sum_embedded(char* str) {\nint sum = 0;\nchar* endptr;\nwhile (1) {\nsum += strtol(str, &endptr, 10);\nif (*endptr == '\\0') break;\nstr = endptr + 1;\n}\nreturn sum;\n}\n\nint main(int argc, char* argv[]) {\nif (argc != 2) {\nfprintf(stderr, \"Usage: %s <STRING>\\n\", argv);\nreturn EXIT_FAILURE;\n}\nint sum = sum_embedded(argv);\nprintf(\"%d\\n\", sum);\nreturn EXIT_SUCCESS;\n}\n```\n\nExample:\n\n```\\$ ./sum_embedded \"11aa22bb33cc44\"\n110\n```\n15. programmingpraxis said\n\n@sbocq: It’s not hard to adapt Globules’ Haskell solution to Scheme. Start with a variant of the Standard Prelude’s `string-split` function that takes a predicate instead of just comparing to a character:\n\n```(define (string-split-by pred? str)\n(define (f cs xs) (cons (list->string (reverse cs)) xs))\n(let loop ((ss (string->list str)) (cs (list)) (xs (list)))\n(cond ((null? ss) (reverse (if (null? cs) xs (f cs xs))))\n((pred? (car ss)) (loop (cdr ss) '() (f cs xs)))\n(else (loop (cdr ss) (cons (car ss) cs) xs)))))```\n\nThen just build a pipeline of Scheme functions:\n\n```(define (sum-embedded-numbers str)\n(sum\n(filter integer?\n(map string->number\n(string-split-by (complement char-numeric?)\nstr)))))```\n\nAnd here’s the result:\n\n```> (sum-embedded-numbers \"11aa22bb33cc44\")\n110```\n\nBe sure you understand the output of `string-split-by`, which is slightly different than the Haskell `wordsBy` function.\n\n16. sbocq said\n\nOh sure, I could have reimplemented split-by myself. But for these exercises, I’m interested in how much bang you get for the bucks sticking to the language and its standard library. In my eyes, you give up a bit of that if you have to resort to custom loops or recursion to implement split-by. This is why I like the Haskell solution.\n\n17. V said\n\nIn Ruby. An imperative and a functional version.\n\n```def sum_embedded_numbers_imperatibely(str)\nsum = 0\ncurrent_val = nil\n\nstr.chars do |c|\nif (Integer(c) rescue false)\ncurrent_val = current_val ? current_val << c : c\nelsif current_val\nsum += current_val.to_i\ncurrent_val = nil\nend\nend\nsum += current_val.to_i\nsum\nend\n\ndef sum_embedded_numbers_functionaly(str)\nstr\n.chars\n.chunk { |char| (Integer(char) rescue false) }\n.select { |is_match, _| is_match }\n.sum { |_,v| v.join.to_i }\nend\n\n# Test\n\ndef test(output, target)\nputs \"target: #{target}\"\nputs \"output: #{output}\"\nputs \"pass: #{output == target}\"\nputs\nend\n\ntest sum_embedded_numbers_imperatibely(\"11aa22bb33cc44\"), 110\ntest sum_embedded_numbers_functionaly(\"11aa22bb33cc44\"), 110\n\n```\n\nOutput:\n\n```target: 110\noutput: 110\npass: true\n\ntarget: 110\noutput: 110\npass: true\n```\n18. sbocq said\n\nAnother Bash solution inspired from Daniel’s nice use of strtol in his C version:\n\n```function sum_embedded {\nlocal tot=0 n\nwhile read -d' ' n; do\n(( tot+=n ))\ndone < <(tr -c '[:digit:]' \\$' ' <<< \"\\$1\")\necho \\${tot}\n}\n```\n\nExample:\n\n```\\$ sum_embedded 11aa22bb33cc44\n110\n```\n19. Mike said\n\nPython\n\n```import operator as op\nimport itertools as it\n\nisdigit = op.methodcaller('isdigit')\n\ndef sum_embedded_numbers(s):\nreturn sum(int(''.join(vs)) for k,vs in it.groupby(s, key=isdigit) if k)\n```\n\nUses ‘groupby’ to break the input string into groups of characters according to the function provided as the ‘key’ parameter, e.g., ‘isdigit’. If the key, k, is True, the characters in the group are concatenated and converted to an integer. The sum of the integers is returned.\n\nHere is a link to a pastebin that lets you execute the code.\nTry it online!\n\n20. WWW.ХХХ.SGDLСQ.ХУZ said\n\nWhat ?\n\n21. WWW.ХХХ.ХQТSРN.ХУZ said\n\nWhat ?\n\n22. WWW.ХХХ.SРNСР.ХУZ said\n\nWhat ?\n\n23. WWW.ХХХ.GЕРBJZ.ХУZ said\n\nWhat ?\n\n24. WWW.ХХХ.SРNСР.ХУZ said\n\nWhat ?\n\n25. WWW.ХХХ.ТАJКХР.ХУZ said\n\nWhat ?\n\n26. WWW.ХХХ.UЕLЕЕS.ХУZ said\n\nWhat ?\n\n27. WWW.ХХХ.ЕWZJJ.ХУZ said\n\nWhat ?\n\n28. WWW.ХХХ.LЕАХ.ХУZ said\n\nWhat ?\n\n29. WWW.ХХХ.VIDIОХХХ.ХУZ said\n\nWhat ?\n\n30. sealfin said\n\nMay 15th, 2018.c:\n\n```#include \"seal_bool.h\" /* <http://GitHub.com/sealfin/C-and-C-Plus-Plus/blob/master/seal_bool.h> */\n#include <string.h>\n#include <limits.h>\n#include <stdio.h>\n#include <stdlib.h>\n\nint main( const int argc, const char * const argv[] )\n{\n#ifdef LEONARDO\nconst size_t length = 1;\n#else\nconst size_t length = 2;\n#endif\n\nif( argc == length )\n{\nsize_t i = 0;\n#ifdef LEONARDO\nconst size_t index = 0;\n#else\nconst size_t index = 1;\n#endif\nbool any_digits_encountered = false, number_begun = false;\nunsigned long\n#ifndef LEONARDO\nlong\n#endif\nnumber = 0;\nsize_t number_of_numbers_encountered = 0;\nunsigned long\n#ifndef LEONARDO\nlong\n#endif\nsum = 0;\n\nfor( ; i < strlen( argv[ index ] ); i ++ )\n{\nconst char c = argv[ index ][ i ];\n\nif(( c >= '0' ) && ( c <= '9' ))\n{\nany_digits_encountered = true;\nnumber_begun = true;\nnumber *= 10;\nnumber += ( c - '0' );\nif( number > UINT_MAX )\ngoto l_Error;\n}\nelse\nif( number_begun )\n{\nnumber_begun = false;\nnumber_of_numbers_encountered ++;\nsum += number;\nif( sum > UINT_MAX )\ngoto l_Error;\nnumber = 0;\n}\n}\nif( !any_digits_encountered )\ngoto l_Error;\nif( number_begun )\n{\nnumber_of_numbers_encountered ++;\nsum += number;\nif( sum > UINT_MAX )\ngoto l_Error;\n}\nprintf( \"\\nThe %snumber%s embedded in the string \\\"%s\\\" is \"\n#ifdef LEONARDO\n\"%lu\"\n#else\n\"%llu\"\n#endif\n\".\\n\", ( number_of_numbers_encountered == 1 )?\"\":\"sum of the \", ( number_of_numbers_encountered == 1 )?\"\":\"s\", argv[ index ], sum );\n#ifndef LEONARDO\nprintf( \"\\n\" );\n#endif\nexit( EXIT_SUCCESS );\n}\nl_Error:;\nprintf( \"\\nThis program must be passed, via the command line, a string containing at least one digit and – optionally – one or more letters; this program will then sum the number(s) embedded in that string.\\n\" );\nprintf( \"\\n(Furthermore, the number(s) embedded in that string – and the sum of those number(s) – must be in the range [ 0, %lu ].)\\n\", UINT_MAX );\n#ifndef LEONARDO\nprintf( \"\\n\" );\n#endif\nexit( EXIT_FAILURE );\n}```\n\nThe solution is known to run on an Apple Power Mac G4 (AGP Graphics) (450MHz processor, 1GB memory) on both Mac OS 9.2.2 (International English) (the solution interpreted using Leonardo IDE 3.4.1) and Mac OS X 10.4.11 (the solution compiled using Xcode 2.2.1).\n\n(I’m just trying to solve the problems posed by this ‘site whilst I try to get a job; I’m well-aware that my solutions are far from the best – but, in my defence, I don’t have any traditional qualifications in computer science :/ )" ]

[ null ]

[ "# Algebraic number theory\n\nAlgebraic number theory is a major branch of number theory that studies algebraic structures related to algebraic integers. This is generally accomplished by considering a ring of algebraic integers O in an algebraic number field K/Q, and studying their algebraic properties such as factorization, the behaviour of ideals, and field extensions. In this setting, the familiar features of the integers—such as unique factorization—need not hold. The virtue of the primary machinery employed—Galois theory, group cohomology, group representations, and L-functions—is that it allows one to deal with new phenomena and yet partially recover the behaviour of the usual integers.\n\n## History of algebraic number theory\n\n### Diophantus\n\nThe beginnings of algebraic number theory can be traced to Diophantine equations, named after the 3rd-century Alexandrian mathematician, Diophantus, who studied them and developed methods for the solution of some kinds of Diophantine equations. A typical Diophantine problem is to find two integers x and y such that their sum, and the sum of their squares, equal two given numbers A and B, respectively:\n\n$A=x+y\\$", null, "$B=x^{2}+y^{2}.\\$", null, "Diophantine equations have been studied for thousands of years. For example, the solutions to the quadratic Diophantine equation x2 + y2 = z2 are given by the Pythagorean triples, originally solved by the Babylonians (c. 1800 BC). Solutions to linear Diophantine equations, such as 26x + 65y = 13, may be found using the Euclidean algorithm (c. 5th century BC).\n\nDiophantus's major work was the Arithmetica, of which only a portion has survived.\n\n### Fermat\n\nFermat's last theorem was first conjectured by Pierre de Fermat in 1637, famously in the margin of a copy of Arithmetica where he claimed he had a proof that was too large to fit in the margin. No successful proof was published until 1995 despite the efforts of countless mathematicians during the 358 intervening years. The unsolved problem stimulated the development of algebraic number theory in the 19th century and the proof of the modularity theorem in the 20th century.\n\n### Gauss\n\nOne of the founding works of algebraic number theory, the Disquisitiones Arithmeticae (Latin: Arithmetical Investigations) is a textbook of number theory written in Latin by Carl Friedrich Gauss in 1798 when Gauss was 21 and first published in 1801 when he was 24. In this book Gauss brings together results in number theory obtained by mathematicians such as Fermat, Euler, Lagrange and Legendre and adds important new results of his own. Before the Disquisitiones was published, number theory consisted of a collection of isolated theorems and conjectures. Gauss brought the work of his predecessors together with his own original work into a systematic framework, filled in gaps, corrected unsound proofs, and extended the subject in numerous ways.\n\nThe Disquisitiones was the starting point for the work of other nineteenth century European mathematicians including Ernst Kummer, Peter Gustav Lejeune Dirichlet and Richard Dedekind. Many of the annotations given by Gauss are in effect announcements of further research of his own, some of which remained unpublished. They must have appeared particularly cryptic to his contemporaries; we can now read them as containing the germs of the theories of L-functions and complex multiplication, in particular.\n\n### Dirichlet\n\nIn a couple of papers in 1838 and 1839 Peter Gustav Lejeune Dirichlet proved the first class number formula, for quadratic forms (later refined by his student Kronecker). The formula, which Jacobi called a result \"touching the utmost of human acumen\", opened the way for similar results regarding more general number fields. Based on his research of the structure of the unit group of quadratic fields, he proved the Dirichlet unit theorem, a fundamental result in algebraic number theory.\n\nHe first used the pigeonhole principle, a basic counting argument, in the proof of a theorem in diophantine approximation, later named after him Dirichlet's approximation theorem. He published important contributions to Fermat's last theorem, for which he proved the cases n = 5 and n = 14, and to the biquadratic reciprocity law. The Dirichlet divisor problem, for which he found the first results, is still an unsolved problem in number theory despite later contributions by other researchers.\n\n### Dedekind\n\nRichard Dedekind's study of Lejeune Dirichlet's work was what led him to his later study of algebraic number fields and ideals. In 1863, he published Lejeune Dirichlet's lectures on number theory as Vorlesungen über Zahlentheorie (\"Lectures on Number Theory\") about which it has been written that:\n\n\"Although the book is assuredly based on Dirichlet's lectures, and although Dedekind himself referred to the book throughout his life as Dirichlet's, the book itself was entirely written by Dedekind, for the most part after Dirichlet's death.\" (Edwards 1983)\n\n1879 and 1894 editions of the Vorlesungen included supplements introducing the notion of an ideal, fundamental to ring theory. (The word \"Ring\", introduced later by Hilbert, does not appear in Dedekind's work.) Dedekind defined an ideal as a subset of a set of numbers, composed of algebraic integers that satisfy polynomial equations with integer coefficients. The concept underwent further development in the hands of Hilbert and, especially, of Emmy Noether. Ideals generalize Ernst Eduard Kummer's ideal numbers, devised as part of Kummer's 1843 attempt to prove Fermat's Last Theorem.\n\n### Hilbert\n\nDavid Hilbert unified the field of algebraic number theory with his 1897 treatise Zahlbericht (literally \"report on numbers\"). He also resolved a significant number-theory problem formulated by Waring in 1770. As with the finiteness theorem, he used an existence proof that shows there must be solutions for the problem rather than providing a mechanism to produce the answers. He then had little more to publish on the subject; but the emergence of Hilbert modular forms in the dissertation of a student means his name is further attached to a major area.\n\nHe made a series of conjectures on class field theory. The concepts were highly influential, and his own contribution lives on in the names of the Hilbert class field and of the Hilbert symbol of local class field theory. Results were mostly proved by 1930, after work by Teiji Takagi.\n\n### Artin\n\nEmil Artin established the Artin reciprocity law in a series of papers (1924; 1927; 1930). This law is a general theorem in number theory that forms a central part of global class field theory. The term \"reciprocity law\" refers to a long line of more concrete number theoretic statements which it generalized, from the quadratic reciprocity law and the reciprocity laws of Eisenstein and Kummer to Hilbert's product formula for the norm symbol. Artin's result provided a partial solution to Hilbert's ninth problem.\n\n### Modern theory\n\nAround 1955, Japanese mathematicians Goro Shimura and Yutaka Taniyama observed a possible link between two apparently completely distinct, branches of mathematics, elliptic curves and modular forms. The resulting modularity theorem (at the time known as the Taniyama–Shimura conjecture) states that every elliptic curve is modular, meaning that it can be associated with a unique modular form.\n\nIt was initially dismissed as unlikely or highly speculative, and was taken more seriously when number theorist André Weil found evidence supporting it, but no proof; as a result the \"astounding\" conjecture was often known as the Taniyama–Shimura-Weil conjecture. It became a part of the Langlands programme, a list of important conjectures needing proof or disproof.\n\nFrom 1993 to 1994, Andrew Wiles provided a proof of the modularity theorem for semistable elliptic curves, which, together with Ribet's theorem, provides a proof for Fermat's Last Theorem. Both Fermat's Last Theorem and the Modularity Theorem were almost universally considered inaccessible to proof by contemporaneous mathematicians (meaning, impossible or virtually impossible to prove using current knowledge). Wiles first announced his proof in June 1993 in a version that was soon recognized as having a serious gap in a key point. The proof was corrected by Wiles, in part via collaboration with Richard Taylor, and the final, widely accepted, version was released in September 1994, and formally published in 1995. The proof uses many techniques from algebraic geometry and number theory, and has many ramifications in these branches of mathematics. It also uses standard constructions of modern algebraic geometry, such as the category of schemes and Iwasawa theory, and other 20th-century techniques not available to Fermat.\n\n## Basic notions\n\n### Unique factorization and the ideal class group\n\nOne of the first properties of Z that can fail in the ring of integers O of an algebraic number field K is that of the unique factorization of integers into prime numbers. The prime numbers in Z are generalized to irreducible elements in O, and though the unique factorization of elements of O into irreducible elements may hold in some cases (such as for the Gaussian integers Z[i]), it may also fail, as in the case of Z[√Template:Overline] where\n\n$6=2\\cdot 3=(1+{\\sqrt {-5}})\\cdot (1-{\\sqrt {-5}}).$", null, "The ideal class group of O is a measure of how much unique factorization of elements fails; in particular, the ideal class group is trivial if, and only if, O is a unique factorization domain.\n\n### Factoring prime ideals in extensions\n\nUnique factorization can be partially recovered for O in that it has the property of unique factorization of ideals into prime ideals (i.e. it is a Dedekind domain). This makes the study of the prime ideals in O particularly important. This is another area where things change from Z to O: the prime numbers, which generate prime ideals of Z (in fact, every single prime ideal of Z is of the form (p):=pZ for some prime number p,) may no longer generate prime ideals in O. For example, in the ring of Gaussian integers, the ideal 2Z[i] is no longer a prime ideal; in fact\n\n$2{\\mathbf {Z} }[i]=\\left((1+i){\\mathbf {Z} }[i]\\right)^{2}.$", null, "On the other hand, the ideal 3Z[i] is a prime ideal. The complete answer for the Gaussian integers is obtained by using a theorem of Fermat's, with the result being that for an odd prime number p\n\n$p{\\mathbf {Z} }[i]{\\mbox{ is a prime ideal if }}p\\equiv 3\\,(\\operatorname {mod} \\,4)$", null, "$p{\\mathbf {Z} }[i]{\\mbox{ is not a prime ideal if }}p\\equiv 1\\,(\\operatorname {mod} \\,4).$", null, "Generalizing this simple result to more general rings of integers is a basic problem in algebraic number theory. Class field theory accomplishes this goal when K is an abelian extension of Q (i.e. a Galois extension with abelian Galois group).\n\n### Primes and places\n\nAn important generalization of the notion of prime ideal in O is obtained by passing from the so-called ideal-theoretic approach to the so-called valuation-theoretic approach. The relation between the two approaches arises as follows. In addition to the usual absolute value function |·| : QR, there are absolute value functions |·|p : QR defined for each prime number p in Z, called p-adic absolute values. Ostrowski's theorem states that these are all possible absolute value functions on Q (up to equivalence). This suggests that the usual absolute value could be considered as another prime. More generally, a prime of an algebraic number field K (also called a place) is an equivalence class of absolute values on K. The primes in K are of two sorts: ${\\mathfrak {p}}$", null, "-adic absolute values like |·|p, one for each prime ideal ${\\mathfrak {p}}$", null, "of O, and absolute values like |·| obtained by considering K as a subset of the complex numbers in various possible ways and using the absolute value |·| : CR. A prime of the first kind is called a finite prime (or finite place) and one of the second kind is called an infinite prime (or infinite place). Thus, the set of primes of Q is generally denoted { 2, 3, 5, 7, ..., ∞ }, and the usual absolute value on Q is often denoted |·| in this context.\n\nThe set of infinite primes of K can be described explicitly in terms of the embeddings KC (i.e. the non-zero ring homomorphisms from K to C). Specifically, the set of embeddings can be split up into two disjoint subsets, those whose image is contained in R, and the rest. To each embedding σ : KR, there corresponds a unique prime of K coming from the absolute value obtained by composing σ with the usual absolute value on R; a prime arising in this fashion is called a real prime (or real place). To an embedding τ : KC whose image is not contained in R, one can construct a distinct embedding Template:Overline, called the conjugate embedding, by composing τ with the complex conjugation map CC. Given such a pair of embeddings τ and Template:Overline, there corresponds a unique prime of K again obtained by composing τ with the usual absolute value (composing Template:Overline instead gives the same absolute value function since |z| = |Template:Overline| for any complex number z, where Template:Overline denotes the complex conjugate of z). Such a prime is called a complex prime (or complex place). The description of the set of infinite primes is then as follows: each infinite prime corresponds either to a unique embedding σ : KR, or a pair of conjugate embeddings τ, Template:Overline : KC. The number of real (respectively, complex) primes is often denoted r1 (respectively, r2). Then, the total number of embeddings KC is r1+2r2 (which, in fact, equals the degree of the extension K/Q).\n\n### Units\n\nThe fundamental theorem of arithmetic describes the multiplicative structure of Z. It states that every non-zero integer can be written (essentially) uniquely as a product of prime powers and ±1. The unique factorization of ideals in the ring O recovers part of this description, but fails to address the factor ±1. The integers 1 and -1 are the invertible elements (i.e. units) of Z. More generally, the invertible elements in O form a group under multiplication called the unit group of O, denoted O×. This group can be much larger than the cyclic group of order 2 formed by the units of Z. Dirichlet's unit theorem describes the abstract structure of the unit group as an abelian group. A more precise statement giving the structure of O×Z Q as a Galois module for the Galois group of K/Q is also possible. The size of the unit group, and its lattice structure give important numerical information about O, as can be seen in the class number formula.\n\n### Local fields\n\n{{#invoke:main|main}} Completing a number field K at a place w gives a complete field. If the valuation is archimedean, one gets R or C, if it is non-archimedean and lies over a prime p of the rationals, one gets a finite extension Kw / Qp: a complete, discrete valued field with finite residue field. This process simplifies the arithmetic of the field and allows the local study of problems. For example the Kronecker–Weber theorem can be deduced easily from the analogous local statement. The philosophy behind the study of local fields is largely motivated by geometric methods. In algebraic geometry, it is common to study varieties locally at a point by localizing to a maximal ideal. Global information can then be recovered by gluing together local data. This spirit is adopted in algebraic number theory. Given a prime in the ring of algebraic integers in a number field, it is desirable to study the field locally at that prime. Therefore one localizes the ring of algebraic integers to that prime and then completes the fraction field much in the spirit of geometry.\n\n## Major results\n\n### Finiteness of the class group\n\nOne of the classical results in algebraic number theory is that the ideal class group of an algebraic number field K is finite. The order of the class group is called the class number, and is often denoted by the letter h.\n\n### Dirichlet's unit theorem\n\n{{#invoke:main|main}} Dirichlet's unit theorem provides a description of the structure of the multiplicative group of units O× of the ring of integers O. Specifically, it states that O× is isomorphic to G × Zr, where G is the finite cyclic group consisting of all the roots of unity in O, and r = r1 + r2 − 1 (where r1 (respectively, r2) denotes the number of real embeddings (respectively, pairs of conjugate non-real embeddings) of K). In other words, O× is a finitely generated abelian group of rank r1 + r2 − 1 whose torsion consists of the roots of unity in O.\n\n### Reciprocity laws\n\n{{#invoke:main|main}} In terms of the Legendre symbol, the law of quadratic reciprocity for positive odd primes states\n\n$\\left({\\frac {p}{q}}\\right)\\left({\\frac {q}{p}}\\right)=(-1)^{{\\frac {p-1}{2}}{\\frac {q-1}{2}}}.$", null, "A reciprocity law is a generalization of the law of quadratic reciprocity.\n\nThere are several different ways to express reciprocity laws. The early reciprocity laws found in the 19th century were usually expressed in terms of a power residue symbol (p/q) generalizing the quadratic reciprocity symbol, that describes when a prime number is an nth power residue modulo another prime, and gave a relation between (p/q) and (q/p). Hilbert reformulated the reciprocity laws as saying that a product over p of Hilbert symbols (a,b/p), taking values in roots of unity, is equal to 1. Artin reformulated the reciprocity laws as a statement that the Artin symbol from ideals (or ideles) to elements of a Galois group is trivial on a certain subgroup. Several more recent generalizations express reciprocity laws using cohomology of groups or representations of adelic groups or algebraic K-groups, and their relationship with the original quadratic reciprocity law can be hard to see." ]

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[ "The Excel Function can be allocated to a particular Cell (Absolute), or relative to the current position (Relative). If relative, then the function, called by name in one cell, is applied based on taking the current position, and  the applying the cell references in the equation from there. If Absolute, then the actual cell references are used.\n\nUse a Relative Function Reference by name:\n\n1.       Place cursor anywhere in the spreadsheet.\n\n2.       Choose Insert | Name | Define.\n\n3.       Enter a Name for the function. This will be the name which will be called in cells as though it is a value.\n\n4.       Enter an equation in the “Refers To” box. This needs to be worked out so that where you are going to want to use it, it will reference the required places based on the current cell, and the relative reference of cells being used.\n\n5.       Choose OK.\n\nEg if the function is always going to multiply the next adjacent cell to the right by two, then do the following:\n\na)       Click in A1\n\nb)       Choose Insert | Name | Define\n\nc)       Enter “Test” for a name.\n\nd)       Enter “=B1*2” in the Refers to box. (This will mean any use of =Test will work out to be the quantity in the Cell to the right times 2, as though the current cell is always A1, then the cell one to the right will always be B1).\n\ne)       Choose OK.\n\nf)        Click on any cell and type “=Test”.\n\ng)       To test, enter a value in the cell one to the Right.\n\nh)       The result will be where you typed “=Test” will show the value of the cell to the right multiplied by 2.\n\nUse an Absolute Function Reference by name:\n\n1.       Place cursor anywhere in the spreadsheet.\n\n2.       Choose Insert | Name | Define.\n\n3.       Enter a Name for the function. This will be the name which will be called in cells as though it is a value.\n\n4.       Enter an equation in the “Refers To” box. This needs to be worked out so that the actual cell reference is entered with \\$ before the Column and Row. Eg =\\$B\\$1*2\n\n5.       Choose OK.\n\nEg if the function is always going to multiply the contents of B1, then do the following:\n\na)       Click in A1\n\nb)       Choose Insert | Name | Define\n\nc)       Enter “Test” for a name.\n\nd)       Enter “=\\$B\\$1*2” in the Refers to box. (This will mean any use of =Test will work out to be the quantity in the Cell B1 times 2.\n\ne)       Choose OK.\n\nf)        Click on any cell and type “=Test”.\n\ng)       To test, enter a value in the cell B1.\n\nh)       The result will be where you typed “=Test” will show the value of the cell B1 multiplied by 2." ]

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[ "# Real Number Math Projects", null, "••• Mathematik image by bbroianigo from Fotolia.com\n\nThe real number is a difficult concept to grasp for many introductory math students because it's abstract. The simplest way to define a real number is a number with real value. For example, the number 14 has real value, and so does the number -8. We understand what those numbers mean and can conceptualize them. Infinity, on the other hand, is a math concept with no real value. Infinity is not a real number then. The best way to solidify this point is with math projects that clearly explain the types of real numbers and their characteristics.\n\n## Real Number Relationship Box\n\nOne of the best ways to understand real numbers is to see how they are related to other categories of numbers. In short, \"real numbers\" is an extremely broad term that encompasses just about every other number category. It may be helpful for children to see just how all-encompassing the definition is. Start by drawing a large box that represents real numbers. Then, draw the next largest category of numbers that fit into the the real number box: rational numbers (numbers that have a repeating pattern, such as 2/3 or 5). The next box will be integers, or all whole numbers, either positive or negative (for example, -2, -1, 0, 1, and 2). Integers will contain two smaller boxes: negative numbers and whole numbers. Finally, whole numbers will contain two boxes, one for the number zero and another for positive natural numbers (such as 1, 2 and 3).\n\nThis completes all the rational numbers that represent all real numbers. Now, draw a second large box next to the rational number box and label it \"irrational numbers.\" This is the final category of real numbers you have not covered with this project. An irrational number is a number that does not have a repeating pattern, such as Pi. These numbers are real but fit in no other category.\n\nOnce the boxes have been drawn out, it will be easier for students to visualize the different types of real numbers and how they relate to each other.\n\n## Real Number Line\n\nA real number line is a simple project that will help children understand the different values a real number can have. First, draw a line and, in the center of the line, draw a hash mark that indicates the number zero. Next, draw other hash marks on either side of the zero to represent other numbers, either negative or positive. No matter what number is written down on the number line, it will be real. This project will help demonstrate that real numbers exist in a continuum. As long as the number can exist on the number line, it is a real number.\n\n## Real Numbers in Real Life\n\nAn out-of-classroom project that will help demonstrate that real numbers have real value is the \"real numbers in real life\" project. A student will identify all the numbers (or as many as possible) that they encounter in real life. This will include volume measurements on grocery items (e.g., ounces, liters) and speed limit signs. Then, the students will identify what the real number is measuring. For example, a student may show that a gallon of milk is 128 ounces. The student must explain that 128 is a real number that values how much milk is contained in a milk jug.\n\n## Real Number Characteristics\n\nAn important way to fully comprehend real numbers is to demonstrate their characteristics. A project that shows as many real number characteristics as possible will demonstrate actual mechanics. First, the basic types of real numbers should be identified: zero, whole numbers, negative numbers, fractions, decimals, integers and rational numbers. Next, general math characteristics of real numbers should be examined. For example, a real number squared (i.e., multiplied by itself) will always yield a positive number. So 2 x 2 will equal 4. Similarly, -2 x -2 also equals 4.\n\nDont Go!\n\nWe Have More Great Sciencing Articles!" ]

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[ "Welcome, guest | Sign In | My Account | Store | Cart\n\nThis calculates the midpoint between two GPS coordinates along the Earth's surface. Based on formula from http://www.movable-type.co.uk/scripts/latlong.html\n\nPython, 20 lines\n 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 import math def midpoint(x1, y1, x2, y2): #Input values as degrees #Convert to radians lat1 = math.radians(x1) lon1 = math.radians(x2) lat2 = math.radians(y1) lon2 = math.radians(y2) bx = math.cos(lat2) * math.cos(lon2 - lon1) by = math.cos(lat2) * math.sin(lon2 - lon1) lat3 = math.atan2(math.sin(lat1) + math.sin(lat2), \\ math.sqrt((math.cos(lat1) + bx) * (math.cos(lat1) \\ + bx) + by**2)) lon3 = lon1 + math.atan2(by, math.cos(lat1) + Bx) return [round(math.degrees(lat3), 2), round(math.degrees(lon3), 2)]", null, "Created by Stijn de Graaf on Sat, 21 May 2011 (MIT)" ]

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[ "## Root extraction, part II: cube roots\n\nby", null, "As you might guess, this post builds on “Root extraction, part I“, which gave a way to visualize the traditional square root algorithm geometrically, an approach that has the advantage that each step appears natural and easily motivated.\n\nOur goal herein is to do much the same for cube roots. The point is to find a geometric construction, ideally one well-suited to physical manipulatives, in which the steps in building the successive digits of the cube root of a number are transparent.  As with the post on square roots, I make no claims to originality in what follows.\n\nExample:  Find", null, "$\\sqrt{22665187}$\n\nOh, nice.\n\nWe’re looking for a cube whose volume is 22665187.  Comparing 100³=1,000,000 with 1000³=1,000,000,000, we see our answer is in the hundreds; further, the cube root lies between 200 and 300, since 200³=8,000,000 < 22,665,187 < 300³=27,000,000.\n\nWe know the hundreds digit of our root is “2”; we create a cube, where each edge has its length split into two segments, one of length 200, the other of length 10d, where “d” is the tens digit of the root.  Our goal is to determine an appropriate choice for d so that the volume of this cube is close to but no more than 22,665,187.", null, "In the diagram above, the red core cube is (200)(200)(200); each of the three green faces has dimensions (200)(200)(10d); each of the three yellow seams has dimensions (200)(10d)(10d); the final little cube has dimensions (10d)(10d)(10d).  The outer fringes of the big cube (the non-red portions) are meant to have volume nearly equal to 22,665,187 – 8,000,000= 14,665,187.  But since d is small, we expect the yellow and blue volumes to be negligible relative to the green faces, and so we compare 3x400000d to 14,665,187, which leads to a guess of d = 12.  (!)  Apparently, those negligible seams are bigger than we thought!\n\nIf we guess d=9, we find the volume of the big cube to be 8,000,000 [red]+ 3x(40000)(90) [green] + 3x(200)(90)(90)[yellow] + (90)(90)(90)[blue] = 24,389,000, a bit too big.\n\nRevising our guess to be d=8, we find the volume of the big cube to be 8,000,000 [red]+ 3x(40000)(80) [green] + 3x(200)(80)(80)[yellow] + (80)(80)(80)[blue]=21,952,000, a bit smaller than our target.  Thus we’ve determined that the tens place of our cube root is an “8”.\n\nWe repeat the process, now with a cube whose sides have length 280+d, and where d is the ones place of our root.", null, "The red inner cube has volume 280³=21,952,000 from our earlier work, and so the remaining regions seek to have volume 713,817.  As before, we’ll focus on only the three green regions initially, as the others are quite small by comparison.  The green flats each have dimensions 280 by 280 by d, and so the total volume is (3)(280)²(d)=235200d.  Comparing 235,200d to 713817, we find that d can be no more than 3.  Setting d=3, we find a total volume of the big cube of 280³ [red] + (3)(280)²(3) [green]+3(280)(3)²[yellow]+3³[blue], which yields 22,665,187.\n\nVoila!  We’ve determined that", null, "$\\sqrt{22665187} = 283$.\n\nNow it is indeed possible to apply this very same technique to finding cube roots of any number to an arbitrary number of decimal places; one is only constrained by one’s willingness to compute the volumes of the green faces (which forces you to triple the square of the digits found so far for the root), the volumes of the yellow strips (multiplying the digits found so far for the root by 3 x the next digit squared), and the volume of the blue cube (the next digit cubed).    Naturally, this is precisely the traditional cube root algorithm (and precisely the ancient Chinese algorithm for root extraction with number rods or with an abacus).\n\nI gather that classroom manipulatives were in vogue at the beginning of the 20th century, consisting essentially of a large block, three flats, three strips, and a small block (much as in our diagram above); the intent was that pupils would use the manipulative as a support as they extracted cube roots by hand of various numbers.\n\nIf one wished to extend this idea to higher order roots, one could, but one loses the ability to do the calculation geometrically, at least with a tangible object; the binomial theorem and algebraic reasoning become necessary tools for computing fifth and higher order roots by these means.\n\nAdvertisements\n\n### 7 Responses to “Root extraction, part II: cube roots”\n\n1.", null, "jd2718 Says:\n\nHi! Thanks for this.\n\nWhere do the diagrams come from? (very nice visual for centered hexagonal numbers, believe it or not)\n\nJonathan\n\n2.", null, "TwoPi Says:\n\nI built them in Microsoft Word, trying to capture the essence what I had done in class using base ten blocks.\n\n3.", null, "jd2718 Says:\n\nvery nice. I know it’s not your problem, but look how easy it is to see that", null, "$(n+1)^3 - n^3 = 3(n-1)^2 + 3(n-1) + 1$\n\nOh, yeah, the post is impressive, too. I’m just stuck on that diagram.\n\n4. Carnival of Mathematics 1000 « JD2718 Says:\n\n[…] Borovik discusses at Math Under the Microscope. 00 – A geometric interpretation of how to extract cube roots (for the brave) over at Blog 360. (Also, for the brave and non-brave alike, discussion of how to […]\n\n5.", null, "Michael Says:\n\nYou might be interested to know that this basic approach was actually used in schools in the late 19th and early 20th centuries. Three dimensional wooden models called Cube Root Blocks were available for teachers to use as visualization aids. You can find an example of a “Double” cube root block on my website at sawbonesantiques.com under the “Technical” link. The double block allowed calculation of roots to three significant digits.\n\n6.", null, "jovelyn Says:\n\nhey! thank you verry much for the illustration and discussion on how to extract the cube root of a number……………..\n\n7.", null, "Auroran Says:\n\nmany higher order roots can be obtained by multiplying by other roots, the difficulty seems to be in dealing with…..prime roots-excepting the number 3, of course. 🙂" ]

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[ "###", null, "Author Topic: Term Test 2 Question 1 Solve Integral  (Read 3145 times)\n\n#### Qingyang Wei\n\n• Jr. Member\n•", null, "", null, "• Posts: 6\n• Karma: 0", null, "##### Term Test 2 Question 1 Solve Integral\n« on: December 05, 2018, 04:39:52 PM »\nFor question 1 a) of Term Test 2, i.e. Solve equation $y'' + 4y = 2\\tan (t)$, how do you solve the integral to arrive at the particular solution of the equation? I see the solution to the question being posted in the Term test section of the forum but I am not sure how the particular solution $$y_p(t)=-\\cos 2t(t-\\sin (t)\\cos (t))+\\sin 2t(\\log (\\cos (t)) - 1/2 \\cos 2t)$$ is calculated, specifically, how the integral is solved to arrive at this answer.\n\n#### Tzu-Ching Yen\n\n• Full Member\n•", null, "", null, "", null, "• Posts: 31\n• Karma: 22", null, "##### Re: Term Test 2 Question 1 Solve Integral\n« Reply #1 on: December 05, 2018, 05:25:18 PM »\nThere is this method call variation of parameter that give particular solution as linear combination of general solution $y_i$ where coefficients are determined by integrals.\n\n#### Shlok Somani\n\n• Newbie\n•", null, "• Posts: 3\n• Karma: 0", null, "##### Re: Term Test 2 Question 1 Solve Integral\n« Reply #2 on: December 05, 2018, 05:55:15 PM »\nIn these type of question, I just find the wronskian and W1 and W2 as this equation was second order and then find their respective U's, So for U1 integrate the division of W1 and W and for the U2 integrate the division of W2 and W. Then you multiply respective U's to your Y's in this case y1 and y2 then add it to your general solution. For a better understanding of this process, I would recommend seeing the example 1 of section 4.4 in the textbook. I think that will clarify how the integral was solved in this question.\n\n#### Qingyang Wei\n\n• Jr. Member\n•", null, "", null, "• Posts: 6\n• Karma: 0", null, "##### Re: Term Test 2 Question 1 Solve Integral\n« Reply #3 on: December 05, 2018, 08:49:38 PM »\nSo I used variation of parameter and concluded that the particular solution follows:\n$$y_p = 2\\cos (2t) \\int{\\tan (t) \\sin (2t) dt} + 2 \\sin (2t) \\int{\\tan(t) \\cos(2t) dt}$$\nI am not sure how to solve the two integrals $\\int{\\tan (t) \\sin (2t) dt}$ and $\\int{\\tan(t) \\cos(2t) dt}$ since it involves the $2t$ term inside the sine and cosine function. Can anyone help me solve this?\n\n#### Monika Dydynski\n\n• Full Member\n•", null, "", null, "", null, "• Posts: 26\n• Karma: 30", null, "##### Re: Term Test 2 Question 1 Solve Integral\n« Reply #4 on: December 05, 2018, 09:30:19 PM »\nSo I used variation of parameter and concluded that the particular solution follows:\n$$y_p = 2\\cos (2t) \\int{\\tan (t) \\sin (2t) dt} + 2 \\sin (2t) \\int{\\tan(t) \\cos(2t) dt}$$\nI am not sure how to solve the two integrals $\\int{\\tan (t) \\sin (2t) dt}$ and $\\int{\\tan(t) \\cos(2t) dt}$ since it involves the $2t$ term inside the sine and cosine function. Can anyone help me solve this?\n\nFirst, recall the double-angle trig identity, $\\sin(2t)=2\\sin(t)\\cos(t)$. Thus, the integrand $\\tan(t)\\sin(2t)=\\tan(t)(2\\sin(t)\\cos(t))=2\\sin^2(t)$.\n\nThe half-angle trig identity allows us to rewrite $2\\sin^2(t)$ as $1-\\cos(2t)$.\n\nNow, we have\n$$\\int{1}dt-\\int{\\cos(2t)}dt$$\n\nLet $u=2t \\Rightarrow du=2dt$\n$$\\int{1}dt-\\int{\\cos(2t)}dt=t-\\frac{1}{2}\\int{\\cos(u)}du=t-\\frac{1}{2}\\sin(u)=t-\\frac{1}{2}\\sin(2t)=t-\\sin(t)\\cos(t)$$\n\nThe process is similar for $\\int{\\tan(t)\\cos(2t)}dt$. Hope that helps!\n« Last Edit: December 06, 2018, 07:36:41 AM by Monika Dydynski »" ]

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[ "By | 23.12.2016\n\n# Multiplication of two Numbers\n\nWrite a C++ Program to Multiply two Numbers. Here’s simple C++ Program to Calculate Multiplication of two Numbers in C++ Programming Language.\n\nIn this program, user is asked to enter two numbers (floating point numbers). Then, the product of those two numbers is stored in a variable and displayed on the screen.\n\nHere is source code of the C++ Program to Calculate Multiplication of two Numbers. The C++ program is successfully compiled and run(on Codeblocks) on a Windows system. The program output is also shown in below.\n\n## SOURCE CODE : :\n\n```/* C++ Program to Calculate Multiplication of two Numbers */\n\n#include <iostream>\nusing namespace std;\n\nint main()\n{\ndouble first, second, product;\n\ncout << \"Enter 1st number :: \";\ncin >> first;\ncout << \"\\nEnter 2nd number :: \";\ncin >> second;\n\nproduct = first * second;\n\ncout << \"\\nProduct of Two Numbers [ \"<<first<<\" * \"<<second<<\" ] = \" << product<<\"\\n\";\n\nreturn 0;\n}```\n\n### Output : :\n\n```/* C++ Program to Calculate Multiplication of two Numbers */\n\nEnter 1st number :: 5\n\nEnter 2nd number :: 8\n\nProduct of Two Numbers [ 5 * 8 ] = 40\n\nProcess returned 0```\n\n• In this program, user is asked to enter two numbers. These two numbers entered by the user is stored in variable first and second respectively.\n• Then, the product of first and second is evaluated and the result is stored in variable product.\n• Finally, the product is displayed on the screen.\n\nAbove is the source code for C++ Program to Calculate Multiplication of two Numbers which is successfully compiled and run on Windows System.The Output of the program is shown above .\n\nIf you found any error or any queries related to the above program or any questions or reviews , you wanna to ask from us ,you may Contact Us through our contact Page or you can also comment below in the comment section.We will try our best to reach up to you in short interval.\n\nThanks for reading the post….", null, "" ]

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[ "# Table 2 Graph theoretical features of functional connectivity networks\n\nFeature Feature description Feature calculation\nldg Link density of the graph (2 × ne)/(nn × (nn - 1))\nacc Average of closeness centrality (1/nn) ∑ nn (sum of reciprocal distances from a node to all other nodes)\ngcc Graph clustering coefficient (3 × number of triangles)/(number of connected triples of nodes)\nrcc Rich club coefficient (ne_k)/(nn_k × (nn k - 1))\nsmg S-metric of graph Sum of the nodal degree products for every edge.\nacg Algebraic connectivity of graph Second smallest eigenvalue of the Laplacian of adjacency matrix.\neng Energy of network graph Sum of absolute values of the real components of eigenvalues of adjacency matrix.\n1. ne: Number of graph edges; nn: number of graph nodes;\n2. nn_k: Number of nodes with degree larger thank; ne_k: number of edges among nodes with degree larger than k.", null, "" ]

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[ "# Is It Possible to Obtain Multi-Dimensional Energy-Momentum Equation by Using Minkowski Spacetime?\n\n14 Pages Posted: 21 Jul 2020\n\nSee all articles by Bahram Kalhor\n\n## Bahram Kalhor\n\nIslamic Azad University (IAU) - Karaj Branch\n\n## Farzaneh Mehrparvar\n\nIslamic Azad University (IAU) - Karaj Branch\n\n## Behnam Kalhor\n\naffiliation not provided to SSRN\n\nDate Written: June 28, 2020\n\n### Abstract\n\nWe use Generalized Minkowski spacetime and obtain a multi-dimensional energy equation in the spacetimes. We use potential energy and total energy as two components in the Makowski spacetime. We present a new multi-dimensional energy equation that relates the total energy of the multi-dimensional object to its rest mass. The multi-dimensional energy equation shows that at the speed of the light, the kinetic energy of the particle reaches the potential energy multiplied by the speed of the light. At the speed of the light, the kinetic energy will be converted to the potential energy, hence the potential energy in the higher dimension is c times of the potential energy in the previous dimension.\n\nKeywords: multi-dimensional energy-momentum equation; Minkowski spacetime\n\nSuggested Citation\n\nKalhor, Bahram and Mehrparvar, Farzaneh and Kalhor, Behnam, Is It Possible to Obtain Multi-Dimensional Energy-Momentum Equation by Using Minkowski Spacetime? (June 28, 2020). Available at SSRN: https://ssrn.com/abstract=3637202 or http://dx.doi.org/10.2139/ssrn.3637202\n\n## Paper statistics", null, "" ]

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[ "# 自减（--）\n\n## 语法\n\njs\n\n``````x--\n--x\n``````\n\n## 描述\n\njs\n\n``````--(--x); // SyntaxError: Invalid left-hand side expression in prefix operation\n``````\n\n## 示例\n\n### 后缀式\n\njs\n\n``````let x = 3;\nconst y = x--;\n\n// x = 2\n// y = 3\n``````\n\n### 前缀式\n\njs\n\n``````let x = 3;\nconst y = --x;\n\n// x = 2\n// y = 2\n``````\n\n## 浏览器兼容性\n\nBCD tables only load in the browser" ]

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[ "# SSC Combined Graduate Level (Tier-I) Examination 1st Shift Held on 9-8-2015 General Intelligence and Reasoning Question Paper With Answer Key\n\nSSC Combined Graduate Level (Tier-I) Examination 1st Shift Held on 9-8-2015\n\nGeneral Intelligence and Reasoning\n\nDirections – (Q. 1-6) In the following Six Questions, find the odd word/number/letters/number pair from t he given alternatives\n\n1.\n\n(A)  Morning\n\n(B)  Noon\n\n(C)  Evening\n\n(D)  Night\n\n2.\n\n(A)  Liberty\n\n(B)  Society\n\n(C)  Equality\n\n(D)  Fraternity\n\n3.\n\n(A)  DWFU\n\n(B)  EVHS\n\n(C)  HSKP\n\n(D)  KQNN\n\n4.\n\n(A)  CBEF\n\n(B)  EDGH\n\n(C)  IHKL\n\n(D)  GFHJ\n\n5.\n\n(A)  4025\n\n(B)  7202\n\n(C)  6023\n\n(D)  5061\n\n6.\n\n(A)  96 : 80\n\n(B)  64 : 48\n\n(C)  80 : 60\n\n(D)  104 : 78\n\n7. Choose the correct alternative to complete the series.\n\nLily, Daisy, Datura, ?\n\n(A)  Sun Flower\n\n(B)  Hibiscus\n\n(C)  Marigold\n\n(D)  Jasmine\n\n8. Which one of the given responses would be a meaningful order of the following?\n\n1. Elephant 2. Cat\n\n3. Mosquito 4. Tiger\n\n5. Whale\n\n(A)  5, 3, 1, 2, 4\n\n(B)  1, 3, 5, 4, 2\n\n(C)  3, 2, 4, 1, 5\n\n(D)  2, 5, 1, 4, 3\n\nDirections- (Q. 9-10) In the following Two Questions, which one set of letters when sequentially placed at the gaps in the given letter series shall complete it?\n\n9. ccbab_caa_bccc_a_\n\n(A)  babb\n\n(B)  bbba\n\n(C)  baab\n\n(D)  babc\n\n10. a_ _ dba_ _ bacad _ _ da _ _ cd\n\n(A)  bccdbcab\n\n(B)  abcddcba\n\n(C)  cbcddcba\n\n(D)  aabbccdd\n\nDirections – (Q. 11-12) In the following Two Questions, a series is given, with one term missing. Choose the correct alternative from the given ones that will complete the series.\n\n11. 4, 6, 10, 16, 24, ?\n\n(A)  28\n\n(B)  30\n\n(C)  34\n\n(D)  40\n\n12. 3, 5, 9, 17, ?\n\n(A)  26\n\n(B)  65\n\n(C)  33\n\n(D)  42\n\n13. A train starts from station A and reaches B 15 minutes late when it moves with 40 km/hr and 24 minutes late when it goes 30 km/hr. The distance between the two stations is-\n\n(A)  16 km\n\n(B)  18 km\n\n(C)  21 km\n\n(D)  24 km\n\n14. In a row men, Manoj is 30th from the right and Kiran is 20th from the left. When they interchange their position, Manoj becomes 35th from the right. What is the total number of men in the row?\n\n(A)  45\n\n(B)  44\n\n(C)  54\n\n(D)  34\n\n15. Seven persons A, B, C, D, E, F and G are standing in a straight line.\n\nD is to the right of G.\n\nC is between A and B.\n\nE is between F and D.\n\nThere are three persons between G and B. Who is on the extreme left?\n\n(A)  A\n\n(B)  B\n\n(C)  D\n\n(D)  G\n\n16. From the given alternative words, select the word which cannot be formed using the letters of the given word :\n\nCUMBERSOME\n\n(A)  MOUSE\n\n(B)  SOBER\n\n(C)  ROME\n\n(D)  MERCY\n\n17. Name a single letter, which can be prefixed to the following words in order to obtain entirely new words?\n\nTILL TABLE PILE TAB PRING\n\n(A)  S\n\n(B)  B\n\n(C)  H\n\n(D)  C\n\n18. Unscramble the following letters to frame a meaningful word. Then find out the correct numerical position of the letters :\n\nB  C  U  S  M  E  L  R  N  A\n\n1   2  3   4   5   6   7  8   9  10\n\n(A)  6 1 4 3 2 5 8 7 9 10\n\n(B)  3 1 5 7 10 4 2 6 9 8\n\n(C)  3 9 4 2 8 10 5 1 7 6\n\n(D)  2 1 3 4 6 8 9 7 5 10\n\n19. Using the following code and key decode the given coded word :\n\nCode : L  X  P  Z  J  Y   Q   M   N   B\n\nKey :   b   a   e   s   p  r   h    i     g    t\n\nCode word : ZBYXMNQB\n\n(A)  height\n\n(B)  struggle\n\n(C)  straight\n\n(D)  strength\n\n20. In a certain code ‘MOUSE’ is written as ‘PRUQC’. How is ‘SHIFT’ written in that code?\n\n(A)  VJIDR\n\n(B)  VIKRD\n\n(C)  RKIVD\n\n(D)  VKIDR\n\n21. In a certain code, 253’ means ‘books are old’ ; ‘546’ means ‘mean is old’ and ‘378’ means ‘buy good books’. What stands for “are “ in that code ?\n\n(A)  2\n\n(B)  4\n\n(C)  5\n\n(D)  6\n\n22. If, + stands for division; x stands for addition; − stands for multiplication; ÷ stands for subtraction, which of the following is correct ?\n\n(1)   46 × 6 ÷ 4 – 5 + 3 = 74\n\n(2)   46 – 6 + 4 × 5 ÷ 3 = 71\n\n(3)   46 ÷ 6 × 4 – 5 + 3 = 75.5\n\n(4)   46 × 6 – 4 + 5 ÷ 3 = 70.1\n\n(A)  4\n\n(B)  2\n\n(C)  1\n\n(D)  3\n\n23. If + = ×, − = ÷, × = +, ÷ = −, then which is the correct equation out of the following ?\n\n(A)  18 ÷ 6 + 4 – 2 ÷ 3 = 22\n\n(B)  18 + 6 – 4 × 2 ÷ 3 = 26\n\n(C)  18 × 6 – 4 + 7 × 8 = 47\n\n(D)  18 – 6 × 7 ÷ 2 + 8 = 63\n\nDirections – (Q. 24-28) In the following Five Questions, select the missing number from the given responses.\n\n24.", null, "(A)  97\n\n(B)  907\n\n(C)  1097\n\n(D)  9107\n\n25.", null, "(A)  12\n\n(B)  17\n\n(C)  18\n\n(D)  16\n\n26. 9  11   13\n\n13   15    17\n\n10    12    14\n\n14    16    18\n\n11    13     ?\n\n(A)  22\n\n(B)  14\n\n(C)  156\n\n(D)  21\n\n27.", null, "(A)  30\n\n(B)  11\n\n(C)  0\n\n(D)  1\n\n28. 7     5                 3\n\n8      4                 9\n\n2          8               ?\n\n112      60               162\n\n(A)  4\n\n(B)  6\n\n(C)  8\n\n(D)  12\n\n29. Pinky walks a distance of 600 m towards east, turns left and moves 500 m then turns left and walks 600 m and then turns left again and moves 500 m and halts At what distance in metres is she from the starting point ?\n\n(A)  2200\n\n(B)  500\n\n(C)  0\n\n(D)  600\n\n30. Sunita rode her scooty northwards, then turned left and then again rode to her left 4 km. She found herself exactly 2 km west of her starting point. How far did she ride northwards initially?\n\n(A)  2 km\n\n(B)  4 km\n\n(C)  5 km\n\n(D)  6 km\n\nDirections – (Q. 31-32) In the following Two Questions, one statement is given followed by two Conclusions, I and II. You have to consider the statement to be true, even if it seems to be at variance from commonly known facts. You are to decide which of the given conclusions can definitely by drawn from the give statement. Indicate your answer.\n\n31. Statement : Every school should promote partnerships that will increase parental involvement and participation for promoting the growth of children.\n\nConclusion :\n\nI. For the growth of the children, parents should be involved in various school activities.\n\nII. Involvement of parents in school activities has no influence on the growth of the children.\n\nCodes :\n\n(A)  Only I follows.\n\n(B)  Only II follows.\n\n(C)  Neither I nor II follows.\n\n(D)  Both I and II follow.\n\n32. Statement : Aggressive animals can be trained with care and affection to behave as the occasion demands.\n\nConclusions :\n\nI. Trained dogs cannot be aggressive.\n\nII. Animals are always aggressive unless care and affection is given to them.\n\nCodes :\n\n(A)  Only I follows.\n\n(B)  Only II follows.\n\n(C)  Neither I nor II follows.\n\n(D)  Both I and II follow\n\nDirections – (Q. 33-38) In the following Six Questions, select the related word / letters / number from the given alternatives\n\n33. Haematology : Blood : : Phycology : ?\n\n(A)  Fungi\n\n(B)  Fishes\n\n(C)  Algae\n\n(D)  Diseases\n\n34. Pri9de of Lions : : ………. of cats.\n\n(A)  Herd\n\n(B)  School\n\n(C)  Clowder\n\n(D)  Bunch\n\n35. MAN : PDQ : : WAN : ?\n\n(A)  ZDQ\n\n(B)  NAW\n\n(C)  YQD\n\n(D)  YDQ\n\n36. AEFJ : KOPT : : ? : QUVZ\n\n(A)  GLKP\n\n(B)  GKLP\n\n(C)  HLKP\n\n(D)  HKQL\n\n37. 2 : 32 : : 3 : ?\n\n(A)  243\n\n(B)  293\n\n(C)  183\n\n(D)  143\n\n38. D × H : 4 × 8 as M × Q : ?\n\n(A)  12 × 17\n\n(B)  12 × 16\n\n(C)  13 × 17\n\n(D)  14 × 18\n\n39. How many triangles are there in the figure ?", null, "(A)  24\n\n(B)  14\n\n(C)  28\n\n(D)  20\n\n40. The figure given on the left hand side is folded to form a box. Choose from the alternatives (1), (2), (3) and (4) the boxes that is similar to the box formed.", null, "(A)  (2) and (3) only\n\n(B)  (1), (3) and (4) only\n\n(C)  (2) and (4) only\n\n(D)  (1) and (4) only\n\n41. Find out the number of circles in the given figure :", null, "(A)  14\n\n(B)  16\n\n(C)  17\n\n(D)  18\n\n42. Identify the diagram that best represents the relationship among the classes given below : Animals, land animals, sea animals", null, "43.", null, "Which number indicates doctors who are not married ?\n\n(A)  6\n\n(B)  4\n\n(C)  2\n\n(D)  1\n\n44.", null, "In the fig :\n\nK represents all Kites\n\nR represents all Rhombus\n\nP represents all Parallelogram\n\nThe statement ‘Rhombus is also a Kite’ can be described as-\n\n(A)  P and K is nothing but R\n\n(B)  P or K is nothing but R\n\n(C)  P and R is nothing but K\n\n(D)  P or R is nothing but K\n\nDirections – (Q. 45-46) In the following Two Questions, which answer figure will complete the pattern in the question figure ?\n\n45.", null, "46.", null, "47. From the given answer figures, select the one which is hidden/ embedded in the question figure :", null, "48. A piece of paper is folded and cut as shown below in the question figures. From the given answer figures, indicate how it will appear when opened.", null, "49. If a mirror is places on the line MN, then which of the answer figures is the right image of the given figure?", null, "50. Directions- A word is represented by only one set of numbers as given in any one of the alternatives. The sets of numbers given in the alternatives are represented by two classes of alphabets as in two matrices a given below. The columns and rows of Matrix I are numbered from 0 to 4 and that of numbered from 0 to 4 and that of Matrix II are numbered from 5 to 9. A letter from these matrices can be represented first by its row and next by its column, e.g., ‘A’ can be represented by 01, 14 etc., and ‘O’ can be represented by 59 , 67 etc. similarly, you have to identify the set for the word ‘PEARL’.", null, "(A)  00, 55, 22, 11, 96\n\n(B)  00, 66, 14, 32, 56\n\n(C)  13, 77, 30, 14, 88\n\n(D)  12, 88, 43, 89" ]

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[ "+0\n\n# maths help\n\n+1\n136\n2\n\n( x + sqrt {y} ) ^2 = 50 +  (14 sqrt {y})\n\nhelp...\n\nMay 28, 2019\n\n#1\n+2\n\nx^2 + 2xsqrty + y = 50 + 14 sqrt y             You have one equation and two variables.   DO you have another eqaution relating x and y?\n\nOne possible solution would be\n\nx^2 +y = 50\n\nand    2x sqrty = 14 sqrty      so x = 7      then y = 1 (from first equation)\n\nMay 28, 2019\n#2\n+1\n\nHere is a graph of all of the possible solutions:", null, "May 28, 2019" ]

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[ "# Aptitude - Simplification\n\nWhile simplifying an expression, the VBODMAS must be learnt as V,B,O,D,M,A,S where every alphabet have a special stand which are as follows:\n\n1. Virnaculum\n\n2. Bracket\n\n3. Of\n\n4. Division\n\n5. Multiplication\n\n7. Subtraction\n\nWe should follow the order strictly.\n\nModulus of a real number x is a positive value.Its modulus is denoted by |x|. Thus , |5 |= 5 and |-5| =5.\n\nImportant Formulaes\n\n```(a+b)2 =(a2+b2+2ab)\n(a-b)2 = (a2+b2-2ab)\n(a+b)2-(a-b)2 = 4ab\n(a+b)2+(a-b)2 = 2 (a2+b2)\n(a+b)(a-b) = (a2-b2)\n(a+b+c)2 = (a2+b2+c2)+2(ab+bc+ca )\n(a+b)3 = a3+b3+3ab(a+b)\n(a-b)3 = a3-b3-3ab (a-b)\n(a3+b3) = (a+b)(a2+b2-ab)\n(a3-b3 ) = (a-b)(a2+b2+ab)\n```\n\n## Solved Examples\n\nSolved Examples\naptitude_simplification.htm", null, "" ]

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[ "Home | Menu | Get Involved | Contact webmaster", null, "", null, "", null, "", null, "", null, "# Number 3804\n\nthree thousand eight hundred four\n\n### Properties of the number 3804\n\n Factorization 2 * 2 * 3 * 317 Divisors 1, 2, 3, 4, 6, 12, 317, 634, 951, 1268, 1902, 3804 Count of divisors 12 Sum of divisors 8904 Previous integer 3803 Next integer 3805 Is prime? NO Previous prime 3803 Next prime 3821 3804th prime 35803 Is a Fibonacci number? NO Is a Bell number? NO Is a Catalan number? NO Is a factorial? NO Is a regular number? NO Is a perfect number? NO Polygonal number (s < 11)? NO Binary 111011011100 Octal 7334 Duodecimal 2250 Hexadecimal edc Square 14470416 Square root 61.676575780437 Natural logarithm 8.2438084236653 Decimal logarithm 3.5802405082654 Sine 0.45172998432356 Cosine -0.89215470702285 Tangent -0.5063359311649\nNumber 3804 is pronounced three thousand eight hundred four. Number 3804 is a composite number. Factors of 3804 are 2 * 2 * 3 * 317. Number 3804 has 12 divisors: 1, 2, 3, 4, 6, 12, 317, 634, 951, 1268, 1902, 3804. Sum of the divisors is 8904. Number 3804 is not a Fibonacci number. It is not a Bell number. Number 3804 is not a Catalan number. Number 3804 is not a regular number (Hamming number). It is a not factorial of any number. Number 3804 is an abundant number and therefore is not a perfect number. Binary numeral for number 3804 is 111011011100. Octal numeral is 7334. Duodecimal value is 2250. Hexadecimal representation is edc. Square of the number 3804 is 14470416. Square root of the number 3804 is 61.676575780437. Natural logarithm of 3804 is 8.2438084236653 Decimal logarithm of the number 3804 is 3.5802405082654 Sine of 3804 is 0.45172998432356. Cosine of the number 3804 is -0.89215470702285. Tangent of the number 3804 is -0.5063359311649" ]

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[ "We have two groups of adding fractions worksheets.\n\n### How to add like fractions?\n\nFree printable and online worksheets to help Grade 5 students practice adding like fractions or fractions with the same denominators.\n\nThere are three printable worksheets. The first will introduce adding like fractions with sums not more than one and reducing is not required. The second will have sums not more than one and reducing may be necessary. The third may have sums that are more than one and will require converting to mixed numbers.\n\nThere are also some online worksheets and word problems for further practice.\n\nTo add like fractions, which are fractions with the same denominator, you can follow these steps:\nStep 1: Make sure the fractions have the same denominator. If the denominators are different then check here to see how to make the denominators the same.\nStep 2: Add the numerators of the fractions together. Keep the denominator unchanged.\nStep 3: Simplify the fraction, if necessary. Check here if you need to review how to simplify fractions.\n\nHere’s an example:\nFraction 1: 3/5\nFraction 2: 1/5\nStep 1: The fractions already have the same denominator, which is 5.\nStep 2: Add the numerators: 3 + 1 = 4\nThe sum of the numerators is 4.\nStep 3: Keep the denominator unchanged, which is 5.\nSo, the sum of the fractions is 4/5.\n\nExample:", null, "Have a look at this video if you need to review how to add fractions.\n\nClick on the following worksheet to get a printable pdf document.\nScroll down the page for more Add Like Fractions Worksheets.\n\n### More Add Like Fractions Worksheets\n\nPrintable\nAdd Like Fractions Worksheet #1 (sums ≤ 1)\nAdd Like Fractions Worksheet #2 (sums ≤ 1, simplify)\nAdd Like Fractions Worksheet #3 (sums can be mixed numbers)\n\nOnline\nSubtract Like Fractions (no reducing)\n\nLeast Common Denominator\n\n### Add Like Fractions Word Problems\n\n1. Tim has 2/8 of a pizza left, and his friend Lisa has 5/8 of a pizza left. How much pizza do they have altogether?\n\n2. A recipe requires 3/5 cup of milk and 4/5 cup of sugar. How much liquid do you need to measure altogether?\n\n3. Emma and Ryan are painting a room together. Emma has already painted 3/8 of the room, and Ryan has painted 1/8 of the room. What fraction of the room have they painted together?\n\n4. Jessica has a garden with 1/8 of it planted with roses, and Sarah has a garden with 5/8 of it planted with roses. If they combine their gardens, what fraction of the combined garden will be planted with roses?\n\n5. A fruit salad recipe calls for 3/4 cup of strawberries, 1/4 cup of blueberries, and 3/4 cup of raspberries. What is the total amount of fruit in the salad?\n\nMore Printable Worksheets\n\nTry the free Mathway calculator and problem solver below to practice various math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations.", null, "" ]

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[ "Limited Time Offer: Save 10% on all 2022 Premium Study Packages with promo code: BLOG10", null, "# Foreign Exchange Markets\n\nAfter completing this reading, you should be able to:\n\n• Explain and describe the mechanics of spot quotes, forward quotes, and future quotes in the foreign exchange market and distinguish between the bid and ask exchange rates.\n• Compare outright (forward) and swap transactions.\n• Define, compare and contrast transaction risk, translation risk, and economic risk\n• Describe the examples of the transaction, translation, and economic risk and explain how to hedge these risks.\n• Describe the rationale for multi-currency hedging using options.\n• Identify and explain the factors that determine the exchange rates.\n• Calculate and explain the effect of an appreciation/depreciation of a currency relative to a foreign currency.\n• Explain the purchasing power parity theorem and use this theorem to calculate the appreciation or depreciation of a foreign currency.\n• Explain how the no-arbitrage assumption in the foreign exchange markets leads to the interest rate parity theorem and use this theorem to calculate forward foreign exchange rates.\n• Distinguish between covered and uncovered interest rate parity conditions.\n\n## The Foreign Exchange Market\n\nThe foreign exchange market is a type of market where the parties involved (collection of hedgers and speculators) exchange one currency for another at the specified rates. The market is also called Forex, Fx, or currency market.\n\n### Currency Quotes\n\nThe exchange rate can be defined as the number of units of one currency (the quote currency) that are needed to purchase one unit of another currency (base currency). The exchange market is the world’s largest market, where all forms of exchange transactions are carried out.\n\nGenerally, currency quotes always appear as AAA/BBB or AAABBB, where AAA and BBB are different currencies. The currency to the left of the slash is the base currency, while the currency to the right of the slash is the quote currency.\n\nThe base currency (in this case, AAA) is always equal to one unit, and the quoted currency (in this case, BBB) is what that one base unit is equivalent to in the other currency.\n\nThe general convention is to quote base currency/ quoted currency. Currency quotes between USD and another currency is the most common exchange rate. Other quotes, for instance, between GPB and CAD, are called cross-currency. In most cases, the USD is the base currency, while the other currency is the quote currency. However, when the US dollar is quoted with the British Pound, the Euro, the Newzealand dollar, and the Australian dollar, then the USD becomes the quote currency.\n\n#### Example: Interpreting the Currency Quotes\n\nThe EUR/USD is quoted as 1.2563. How do we interpret this quote? This quote implies that we need 1.2563 USD to buy one euro.\n\nThe bid price is the price at which the counterparty is willing to buy one unit of the base currency, expressed in terms of the price currency. On the other hand, the ask price is the price at which a counterparty is willing to sell one unit of the base currency, expressed in terms of price currency. For instance, a dealer might quote EUR/USD exchange rate of 1.3849. This quote implies that the dealer is willing to pay 1.3849 USD to buy 1 Euro. Intuitively, we expect the bid price to be slightly less than the offer price because the dealer’s goal is to make some cash in every transaction. With that in mind, it is easier to single out the bid price or the offer price given a quote.\n\n1. The ask price should always be higher than the bid price.\n2. A market participant requesting the two-sided price quote has the option but not the obligation to transact at either the bid or the ask quoted by the dealer. If a party chooses to transact at the quoted prices, they are said to have “hit the bid” or “paid the offer.”\n\n## Spot and Forward Exchange Rates\n\n### Spot Exchange Rates\n\nA spot exchange rate is the prevailing price level in the market used to directly trade one currency for another, which is delivered at the earliest possible time. The standard delivery time for spot currency transactions is no longer than T+2, after which it will be deemed a forward contract. Spot exchange rates are usually quoted with four decimal places. For instance, the spot-bid for EUR/USD could be stated as 1.1745 and the spot-ask as 1.1747.\n\n### Forward Exchange Rates\n\nA forward exchange is a price at which one currency is traded against another at some specified time in the future. The forward exchange rate must respect the arbitrage relationship – the returns from two alternatives, but equivalent investments must be equal (as we will see in covered rate parity).\n\nForward exchange rates are not quoted with the same base as the spot exchange rates but rather as points that are multiplied by $$\\frac {1}{10,000}$$ then added to the spot exchange rate.\n\n#### Example: Calculating the Forward Exchange Rates\n\nThe following table gives the forward rates as of June 16, 2019. The spot bid and ask rates are 1.1745 and 1.1749, respectively.\n\n$$\\begin{array}{c|c|c} \\textbf{Maturity} & \\textbf{Bid} & \\textbf{Ask} \\\\ \\hline \\text{1 month} & {27.12} & {28.60} \\\\ \\hline \\text{2 months} & {53.15} & {54.15} \\\\ \\hline \\text{3 months} & {81.87} & {83.07} \\\\ \\hline \\text{4 months} & {113.59} & {114.99} \\\\ \\hline \\text{5 months} & {139.07} & {140.47} \\\\ \\end{array}$$\n\nWhat is the three-month forward bid and ask quotes?\n\n#### Solution\n\nRecall that forward exchange rates are not quoted with the same base as the spot exchange rates but rather as points that are multiplied by $$\\frac {1}{10,000}$$ then added to the spot exchange rate.\n\nSince we are given the spot bid rate as 1.1745, then the 3-month forward bid rate is:\n\n$$1.1745+\\cfrac {1}{10,000}×81.87=1.1745+0.008187=1.182687$$\n\nAnalogously, the 3-month forward ask quote is\n\n$$1.1749+\\cfrac {1}{10,000}×83.07=1.1749+0.008307=1.183207$$\n\nThe bid-ask spread is the amount by which the offer price exceeds the bid price of a currency in a market. Basically, it is the difference between the highest price that the purchaser is willing to pay and the lowest price that a seller is willing to accept.\n\nContinuing with the example above, the bid-ask spread for the 3-month forward rate is calculated as:\n\n$$1.183207-1.182687=0.00052$$\n\nNote that this can also be calculated as:\n\n$$(1.1749-1.1745)+(0.008307-0.008187)=0.0004+0.00012=0.00052$$\n\nWhen the forward exchange rate is less than the spot rate, the points are negative. However, it should be apparent that the magnitude of the negative ask price is less than that of the bid price. This makes sure that the ask price is larger than the bid price. Consider the following example.\n\n#### Example: Calculating the Forward Exchange Rate\n\nThe following table gives the forward rates as of July 8, 2019. The spot bid and ask rates are 1.3184 and 1.3185 respectively.\n\n$$\\begin{array}{c|c|c} \\textbf{Maturity} & \\textbf{Bid} & \\textbf{Ask} \\\\ \\hline \\text{1 Month} & {-9.39} & {-7.67} \\\\ \\hline \\text{2 months} & {-18.20} & {-17.29} \\\\ \\hline \\text{3 months} & {-32.10} & {-30.91} \\\\ \\hline \\text{4 months} & {-40.91} & {-39.42} \\\\ \\hline \\text{5 months} & {-45.90} & {-43.95} \\\\ \\end{array}$$\n\nWhat is the five-month forward bid and ask quotes?\n\n#### Solution\n\nThe 5-month forward bid quote is:\n\n$$1.3184+\\cfrac {1}{10,000} ×-45.90=1.3184-0.004590=1.31381$$\n\nAnalogously, the forward ask quote is:\n\n$$1.3185 +\\cfrac {1}{10,000} ×-43.95=1.3185-0.004395=1.314105$$\n\n$$1.314105-1.31381=0.000295$$\n\n### Outrights and Swaps\n\nRecall that a forward exchange is a price at which two parties agree to trade one currency against another at some specified time in the future. This is termed as an outright transaction or outright forward transaction.\n\nOn the other hand, a foreign exchange swap is a type of exchange rate transaction where a currency is bought (sold) in a spot market and then sold (bought) in the forward market. From a different angle, an FX swap is a method of funding an asset transacted in foreign currency by paying the interest due in terms of the domestic currency. An example of an FX swap is where a US-based company funds its Chinese investment by borrowing in USD and buying the Chinese Yuan, and after some time, the company exchanges the money back to USD. By doing this, the company can fund its operation in the Chinese Yuan.\n\nSwap transactions are usually profitable if the foreign currency used will be of more value in the forward market, i.e., more of the local currency will be received for less of the foreign currency at the agreed-upon future date.\n\nFor instance, if we use the following table:\n\n$$\\begin{array}{c|c|c} \\textbf{Maturity} & \\textbf{Bid} & \\textbf{Ask} \\\\ \\hline \\text{1 month} & {27.12} & {28.60} \\\\ \\hline \\text{2 months} & {53.15} & {54.15} \\\\ \\hline \\text{3 months} & {81.87} & {83.07} \\\\ \\hline \\text{4 months} & {113.59} & {114.99} \\\\ \\hline \\text{5 months} & {139.07} & {140.47} \\\\ \\end{array}$$\n\nAssume that a US company borrows in USD and buys 1 million EUR today to fund its European operations. At the same time, the company also agrees to sell 1 million EUR for USD within one-month time. In the table above, the points in one month’s time are 27.10. This implies that EUR is valuable in the forward market, and thus, the points reduce the net funding cost in USD since more USD is going to be received in a one-month time relative to the amount that would have been received today.\n\nCurrency swaps will be discussed in details in chapter 20\n\n### Future Quotes\n\nFuture quotes are the exchange-traded futures legal contract that stipulates the price in one currency at which another currency can be bought or sold at a future date. A good example is the Chicago Mercantile Exchange (CME) in the US, where diverse futures contracts on exchange rates between USD and other currencies are made.\n\nIn the CME setting, USD is always the base currency since investors treat foreign currency as assets value in USD. Assume that the 6-month forward quote for the USD/CAD is 1.2900. The future quote is found by finding the reciprocal of the forward quote. That is:\n\n6-month futures quote is equivalent to the 6-month forward quote is: $$\\frac {1}{1.2900}$$=0.7752 USD per CAD\n\n## Estimation of the Foreign Exchange Market (FX) Risk\n\nFirms in the foreign exchange market are exposed to risks. They should, therefore, be keen on the extent to which they could accept risk. Once this is known, the firms should decide whether the risk levels are acceptable and if not, and if not, they should apply appropriate hedging strategies.\n\nThree categories of risk are examined, and these include:\n\n1. Transaction risk\n2. Translation risk\n3. Economic risk\n\n### Transaction Risk\n\nThis kind of risk is associated with received and paid capital; it affects the cash flows of a company. Let us look at a simple example:\n\nAssume that an American company imports goods from Kenya, for which it pays in Kenyan shillings. By doing this, the company is faced with USD/KSH risk. That is, if KSH gains strength, then the company will experience little profits if it is required to buy KSH to pay for its services.\n\nIn summary, a company buying from a foreign company will be exposed to losses (profits) if: the currency of the foreign company strengthens (loses value), implying that more (less) of the local currency will be needed to purchase one unit of the foreign currency.\n\nConversely, a company selling to international clients will suffer losses (profits) if the currency of the foreign company weakens (strengthens), implying that more (less) of the received revenue will be needed so as to convert it to the local currency.\n\n### Hedging Transaction Risk\n\nTransaction risk is hedged using outright forward transactions and swaps.\n\nIn the case of hedging using outright transactions, consider the case of an American company investing in Kenyan Shillings. The company could hedge its position by buying KSH forward, which would hedge the exchange rate paid to Kenyan suppliers while selling USD forward would lock in the FX rate applied to the revenues of USD.\n\nOn the other hand, the FX swap is applied when the company owns a foreign company that owns foreign currency that will be used for purchases at a future time to earn interest in its domestic currency. The company would sell the foreign currency in exchange for its domestic currency in the spot market and repurchase it at a stated future time in the forward market.\n\n### Translation Risk\n\nThis type of risk comes up when an institution’s assets and liabilities are in a foreign currency, which must be valued in the institution’s domestic currency when the financial statements are made. Thus, the institution can experience foreign exchange gains or losses.\n\nAs compared to transaction risk, translation risk does not affect the cash flows of a company.\n\n#### Example: Demonstrating Translation Risk (Investment)\n\nA Canadian company has its investments in the US. At the end of the first year, the company netted USD 20 million, and the USD/CAD at that time was 1.3200. At the end of the second year, the company’s net value remained the same as that of the first year. However, the USD/CAD has changed to 1.2700. At the end of the two years, the company wishes to produce its 2-year financial statements. Intuitively, the company experienced a loss of:\n\n$$(1.32000-1.2700)×20,000,000=\\text {CAD } 1,000,000$$\n\nForeign gains or losses can also arise from borrowing in foreign currency. To illustrate this, let us look at an example.\n\n#### Example: Demonstrating Translation Risk (Borrowing)\n\nSuppose that the Canadian company has a loan of USD 10 million that is supposed to be returned in 10 years (assume that it is paid at par). The interests are paid in USD, implying that the company is prone to transaction risk. Additionally, the company is exposed to translation risk, which is embedded in the repayment of the loan principal.\n\nGiven that the USD/CAD at the end of 1st year was 1.3100 and at the end of the 2nd year, USD/CAD had reduced to 1.2800, the value of the loan at the end of the 1st year is:\n\n$$10,000,000×1.3100=\\text{CAD } 13,100,000$$\n\nAnd at the end of the second year:\n\n$$10,000,000×1.2800=\\text{CAD } 12,800,000$$\n\nIntuitively, the company earns a foreign gain of\n\n$$13,100,000-12,800,000=\\text{CAD } 300,000$$\n\nThe above results can be attributed to the fact that the USD has weakened over the CAD. On the other hand, had the USD strengthened, the company would have experienced a loss.\n\n### Hedging Translation Risk\n\nTranslation risk is hedged using the forward contracts on the reporting date to decrease the volatility of profits. The forward contract involves a plan by the concerned firm to sell foreign currency assets or retire foreign currency liabilities at a later time.\n\nAnother method of hedging translation is by financing the assets in a country with the borrowed funds in that country. By doing this, the gains (losses) on assets are offset by the losses (gains) on the liabilities.\n\nTo see this, suppose that in our example above. Suppose the Canadian company has a USD 20 Million worth of investment. If the company analyzes its position and realizes that the translation is imminent, it could finance its investment with a USD 20 million loan.\n\nIt is worth noting the difference between transaction risk and translation risk. We can see that transaction risk has direct effects on the cashflows of a company which is not the case for translation risk. However, the effects of translation risk on the reported earnings of a company can be big.\n\nAlso, note that it is only reasonable that we hedge translation risks on one future date. Doing otherwise will be overhedging.\n\n### Economic Risk\n\nEconomic risk is the risk that the future cash flows of a firm will be affected by the movements in exchange rates.\n\nEconomic risk arises from the exchange rate movements and thus is difficult to quantify. Consider a Canadian sales firm in Brazil. If the BRL (Brazilian real) weakens in value relative to the Canadian dollar, the Brazilian customers will see the firm’s products as expensive. This will decline the demand for the product or prompts the company’s management to reduce the CAD price of its products.\n\nMoreover, economic risk alters with the competitive nature of a company. Consider a company in a given country, Kenya, which has no investment operations overseas. Exchange rate movements might be favorable for a foreign investor who sees Kenya as a desirable place for business. The foreigner’s increased business activities could negatively impact the domestic firm’s competitive position.\n\nCompared with translation and transaction risks, economic risk is not easily quantified. However, possible exchange rate movements should be taken into consideration before essential strategic business decisions are made. For instance, a production firm might decide to move production overseas due to favorable exchange rate movements.\n\n### Multicurrency Hedging Using Options\n\nMultinational companies are exposed to many different currencies. Just like any other portfolio, multiple exposures to multiple currencies reduce the FX risk due to diversification. That is, volatility from multiple currency exposures is less than exposure to a single currency.\n\nCompanies often prefer options to forward contracts since the options provide downside protection against unfavorable exchange rate movement while allowing a firm to benefit from desirable movements. Therefore, hedging using options involves buying options on individual currencies to cover each adverse exchange rate movement.\n\nAlternatively, a firm might buy an option on a portfolio of currencies to which it is exposed in the over-the-counter market. Such options are basket and Asian options.\n\n## Determination of Exchange Rates\n\nLike any other financial asset, currency exchange rates cannot be determined with ultimate precision because they are influenced by supply and demand, which are also affected by other factors. Some of the factors affecting the exchange rates include:\n\n1. Balance of payments and trade flows\n2. Monetary policies\n3. Inflation\n\n### Balance of Payments and Trade Flows\n\nRecall that the balance of payments between the two countries is the difference between the value of exports and imports. We shall demonstrate the effect of balance of payments using an example:\n\n### Example\n\nSuppose that the exports from country X to country Y increases. If the exporters exchange the foreign currency (which is their revenues) to domestic currency, the demand for country X currency will increase, strengthening its currency relative to country Y’s currency. This causes exports to be more expensive in country Y.\n\nOn the other hand, if the imports from country X to Y increase, the currency of country X will weaken relative or that of country Y since the importers will be forced to buy country Y’s currency to pay for the goods they are importing. Consequently, the imported goods to country Y become more expensive, lowering the demand for imported goods.\n\n## Monetary Policies\n\nThe central bank’s monetary policy also influences the value of a country’s currency. With all other factors held equal, if Country X, say, raises its money supply by 15% while Country Y keeps its money supply at constant, the value of Country X’s currency would begin to fall by 15% compared to Country Y’s currency. This is due to the fact that 15% more of Country X’s currency is being used to buy the same quantity of goods.\n\n## Inflation\n\nInflation gives rise to purchasing power parity; a relationship that allows for theoretical arbitrage opportunities, i.e., a trader can buy goods cheaply from a country with a lower inflation rate and sell them at a higher price in a different country with a higher inflation rate since inflation has negative effects on the exchange rates.\n\nThis condition reflects the link between the exchange rates and the difference in countries’ inflation rates. The laws of one price state the price of a foreign good x denoted as $${\\text P}_{\\text f}^{\\text x}$$ must be the equal price of the similar good in a domestic country, $${\\text P}_{\\text d}^{\\text x}$$, using the spot rate $${\\text S}_{\\frac {\\text f}{\\text d}}$$ (We have used the ($${\\frac {\\text f}{\\text d}}$$) notation for simplicity). Put mathematically,\n\n$${\\text P}_{\\text f}^{\\text x}={\\text S}_{\\frac {\\text f}{\\text d}}×{\\text P}_{\\text d}^{\\text x}$$\n\nFor instance, a product in Canada costs CAD 100. The nominal exchange rate for USD/CAD is 0.76. So, the same product will cost 0.76×100 = USD 76 in the US.\n\nThe purchasing power parity amplifies the law of one price to include a broader range of goods and services and not just good x. The law of one price equation transforms into:\n\n$${\\text P}_{\\text f}={\\text S}_{\\frac {\\text f}{\\text d}}×{\\text P}_{\\text d}$$\n\nwhere:\n\n$${\\text P}_{\\text f}$$-the price level of the foreign country.\n\n$${\\text P}_{\\text d}$$-the price level of the domestic country\n\n$${\\text S}_{\\frac {\\text f}{\\text d}}$$-nominal exchange rate\n\nMaking the $${\\text S}_{\\frac {\\text f}{\\text d}}$$ the subject of the formula, we get:\n\n$${\\text S}_{\\frac {\\text f}{\\text d}}=\\cfrac {{\\text P}_{\\text f}}{{\\text P}_{\\text d}}$$\n\n#### Example: Calculating the Spot Exchange Rate Using the Purchasing Power Parity\n\nThe inflation rate in the US is 3% per year and 1% per year in Canada. You are also given that the USD/CAD exchange rate is 1.0500. A basket of goods in the US costs USD 100. Assuming that the purchasing power parity holds, what is the new USD/CAD after one year.\n\n#### Solution\n\nThe inflation rate in the US is 3% per year, implying that the price of a basket of goods increases by 3% each. This is analogous to Canada. So, after one year, the basket of goods in the US is:\n\n$${\\text P}_{\\text d}=1.03×100=\\text{USD } 103$$\n\n$${\\text P}_{\\text f}=1.05×1.01×100=\\text{CAD } 106.05 .$$\n\n$${\\text S}_{\\frac {\\text f}{\\text d}}=\\cfrac {{\\text P}_{\\text f}}{{\\text P}_{\\text d}} =\\cfrac {106.05}{103}=1.02961$$\n\nTherefore, the equilibrium in the exchange rates is determined by the ratio of the national price level of the two countries. However, if the transaction cost is largely coupled with the non-tradable nature of some goods, this condition might not hold.\n\nMoreover, if the transaction costs and other trading difficulties are constant, the deviation of the exchange rate is entirely determined by the difference between the inflation rates of the foreign and domestic countries. Mathematically,\n\n$$\\% \\Delta {\\text S}_{\\frac {\\text f}{\\text d}} \\approx \\pi_{\\text f}-\\pi_{\\text d}$$\n\nWhere:\n\n$$\\% \\Delta {\\text S}_{\\frac {\\text f}{\\text d}}$$=change in the spot exchange rate\n\n$$\\pi_{\\text f}$$=foreign inflation rate\n\n$$\\pi_{\\text d}$$=domestic inflation rate\n\nThat is:\n\n$$\\text{Percentage Strengthening of Domestic Spot Rate}=\\text{Foreign Inflation Rate}- \\text{ Domestic Inflation Rate.}$$\n\nIf we go back to our example, the domestic currency weakens by:\n\n$$\\% \\Delta {\\text S}_{\\frac {\\text f}{\\text d}} \\approx \\pi_{\\text f}-\\pi_{\\text d}=1\\%-3\\%=-2\\%$$\n\nThis implies the domestic spot rate weakened by 2%\n\n### Calculating Percentage Appreciation / Depreciation\n\nTo calculate the percentage change, one needs to have a clear understanding of the base currency and the quote currency. Let us take an example of the Chinese Yuan (CNY) and South African Rand (ZAR). Suppose that the exchange rate of ZAR/CNY increased from 1.6459 to 1.8356. Therefore, the percentage of appreciation/depreciation of the Chinese Yuan will be:\n\n$$\\cfrac {1.6459}{1.8356} – 1 = -10.33\\%$$\n\nThe Chinese Yuan depreciated by 10.33% because it used to take 1.6459 CNY to buy one ZAR, but now it has increased to 1.8356 CNY for one ZAR.\n\n#### Interpretation of the Percentage Appreciation/Depreciation\n\nThe depreciation of the Chinese Yuan against the South African Rand can also be expressed as an appreciation of the South African Rand against the Chinese Yuan. The appreciation percentage, in this case, will not be equal to the previous depreciation percentage of -10.33%.\n\nTo calculate the percentage appreciation of the South African Rand, we have to invert the exchange rate. To invert a currency exchange rate, we have to divide 1 by the exchange rate. For example, if\n\n$$\\text {ZAR/CNY} = 1.6459$$\n\nThen,\n\n$$\\text {CNY/ZAR} =\\cfrac {1}{1.6459} = 0.6076$$\n\nTherefore, the appreciation percentage of South African Rand if the exchange rate ZAR/CNY increased from 1.6459 to 1.8356. We have to invert this exchange to CNY/ZAR so that the Chinese Yuan is now the base currency and the South African Rand is the quote currency. That is:\n\n$$\\frac{\\left(\\frac{1}{1.6459}\\right)}{\\left(\\frac{1}{1.8356}\\right)}-1=\\cfrac {1.8356}{1.6459} – 1 = 11.53\\%$$\n\nThis represents an 11.53 percent appreciation in the South African Rand against the Chinese Yuan because the CNY/ZAR (as a result of inversion) exchange rate is expressed with the Chinese Yuan as the base currency and the South African Rand as the quote currency. In other words, you now need fewer South African Rands to buy one Chinese Yuan.\n\nThus, we can see that the appreciation percentage of the South African Rand is different from the Chinese Yuan’s depreciation.\n\n#### Example: Appreciation/Depreciation\n\nA forex trader noticed that the USD/EUR spot rate was 1.2960 and expected to be 1.2863 after one year. Similarly, the CHF/USD spot rate is 0.9587 and is expected to drop to 0.8885. Calculate the euro (EUR) expected appreciation/depreciation against the US dollar over the next year.\n\n#### Solution\n\nWe know that we are dealing with USD/EUR quotes. So, we calculate as: $$\\cfrac {1.2960}{1.2863}-1=0.007541=0.7541\\%$$\n\nThis was expected because clearly, there was a decrease in USD/EUR, indicating that EUR is appreciating.\n\n## Real and Nominal Interest Rates\n\n### Nominal Interest Rates\n\nNominal interest rates are those rates that are listed in the market and show the return that will be earned on a currency. For instance, 5% per year for a given currency of a country implies that 100 units of a currency are anticipated to grow to 105 in one year.\n\n### Real Interest Rates\n\nReal interest rate is those rates that are adjusted to accommodate the effects of inflation. The real interest is given by:\n\n$${\\text r}_{\\text{real}}=\\cfrac {1+{\\text r}_{\\text{nominal}}}{1+{\\text r}_{\\text {inflation}}}-1$$\n\nwhere\n\n$${\\text r}_{\\text{real}}$$=real interest rate\n\n$${\\text r}_{\\text{nominal}}$$=nominal interest rate\n\n$${\\text r}_{\\text {inflation}}$$=rate of inflation\n\nThe above equation is usually approximated as:\n\n$${\\text r}_{\\text{real}} \\approx {\\text r}_{\\text{nominal}}-{\\text r}_{\\text {inflation}}$$\n\nFor instance, if the nominal interest rate is 5% and the inflation rate is assumed to be 2%, then the real interest rate is approximated as:\n\n$${\\text r}_{\\text{real}} \\approx 5\\%-2\\%=3\\%$$\n\nOr\n\n$${\\text r}_{\\text{real}} =\\cfrac {1.05}{1.02}-1=0.02941 \\approx 2.941\\%$$\n\nNote that this is pretty close to the approximation.\n\nNote also that the real and nominal interest rates can both be negative.\n\n## The Covered Interest Parity\n\nThis is a no-arbitrage condition which states that an investment in a foreign market that is entirely hedged against exchange rate risk should give the same return as a similar investment in a domestic market.\n\n### Derivation of Covered Interest Rate Parity\n\nConsider an investor who wishes to start with 1 unit of domestic currency so that he ends up with an amount in foreign currency terms. To achieve this, there are two ways to accomplish this:\n\n1. The trader can invest in the funds in the risk-free foreign rate of interest $$({\\text i}_{\\text f})$$ so that the funds grows to $$(1+{\\text i}_{\\text f} )^{\\text T}$$ at time T. At time T, the trader can enter into a forward contract to exchange $$(1+{\\text i}_{\\text f} )^{\\text T}$$ for foreign currency at a foreign forward rate of exchange $${\\text F}_{\\frac {\\text f}{\\text d}}$$ to get: $$(1+{\\text i}_{\\text f} )^{\\text T} {\\text F}_{\\frac {\\text f}{\\text d}}$$\n2. The trader immediately exchange the proceeds to USD and invest in risk-free domestic rate of interest to get $${\\text S}_{\\frac {\\text f}{\\text d}} (1+{\\text i}_{\\text d} )^{\\text T}$$\n\nSince we are assuming there is a no-arbitrage condition, then these two investments should give the same result. That is:\n\n$$(1+{\\text i}_{\\text f} )^{\\text T} {\\text F}_{\\frac {\\text f}{\\text d}}={\\text S}_{\\frac {\\text f}{\\text d}} (1+{\\text i}_{\\text d} )^{\\text T}$$\n\nRearranging we get:\n\n$${\\text F}_{\\frac {\\text f}{\\text d}} ={\\text S}_{\\frac {\\text f}{\\text d}} \\left( \\cfrac { (1+{\\text i}_{\\text d} )^{\\text T}}{ (1+{\\text i}_{\\text f} )^{\\text T} } \\right)$$\n\nWhere\n\n$${\\text i}_{\\text d}$$ =The interest rate in the domestic currency or the quoted currency\n\n$${\\text i}_{\\text f}$$=interest rate in the foreign currency or the base currency\n\n$${\\text S}_{\\frac {\\text f}{\\text d}}$$=current spot exchange rate\n\n$${\\text F}_{\\frac {\\text f}{\\text d}}$$=forward foreign exchange rate\n\nIn other words, the forward exchange rate should give the same rate involving the spot exchange rate, domestic and foreign risk-free yields either in domestic market instruments or currency-hedged foreign market instruments within the same investment horizon.\n\nFor this condition to hold, it is assumed that:\n\n1. There is no transaction cost and\n2. The domestic and foreign market instruments should be identical in liquidity, time to maturity, and the default risk.\n\nNote that we have to use the notation f/d, where f stands for the foreign currency and d for the domestic currency. So, f/d can be anything else such as CAD/USD, in which USD is the domestic currency.\n\nGenerally given the exchange rate XXXYYY or XXX/YYY then:\n\n$${\\text F}_{\\frac {\\text{XXX}}{\\text{YYY}}}={\\text S}_{\\frac {\\text{XXX}}{\\text{YYY}}} \\left(\\cfrac { (1+{\\text i}_{\\text{YYY}} )^{\\text T}}{(1+{\\text i}_{\\text {XXX}} )^{\\text T} } \\right)$$\n\nWhere the variables are defined as above.\n\n### Example: Covered Interest Rate Parity\n\nThe following are interest rates as listed in an interbank market:\n\n$$\\begin{array}{c|c|c|c} \\textbf{Currency} & \\textbf{Libor (annualized)} & \\textbf{Currency} & \\textbf{Spot Rate} \\\\ {} & {} & \\textbf{Combinations} & {} \\\\ \\hline \\text{USD} & {0.30\\%} & \\text{USD/EUR} & {1.6975} \\\\ \\hline \\text{EUR} & {5.00\\%} & \\text{JPY/EUR} & {0.0085} \\\\ \\hline \\text{JPY} & {0.30\\%} & \\text{JPY/USD} & {82.25} \\\\ \\end{array}$$\n\nA Japanese company investment manager wants to estimate all-in investment returns on a hedged EUR currency exposure if the covered interest parity holds.\n\n#### Solution\n\nIn this question, we do not require any calculations because it is just a matter of intuition. If the covered interest rate parity holds, then the all-in investment for the Japanese company is the fully-hedged EUR Libor, which is equal to a one-year JPY Libor of 0.30%. So, the investment return is simply 0.30% since, according to covered interest rate parity, an investment in a foreign market that is entirely hedged against exchange rate risk should give the same return as a similar investment in a domestic market.\n\n### Arbitrage Opportunities in Case the Covered Interest Rate Fails\n\nAssume that a US investor starts with 1 unit of USD to end up with Canadian dollars (CAD). So, the spot rate is quoted as CAD/USD. If the covered rate parity holds then:\n\n\\begin{align*}{\\text F}_{\\frac {\\text{CAD}}{\\text{USD}}}&={\\text S}_{\\frac {\\text{CAD}}{\\text{USD}}} \\left( \\cfrac {\\left(1+{\\text i}_{\\text{USD}} \\right)^{\\text T}}{ \\left(1+{\\text i}_{\\text{CAD}} \\right)^{\\text T}} \\right)\\\\ \\end{align*}\n\nIf\n\n\\begin{align*} {\\text F}_{\\frac {\\text{CAD}}{\\text{USD}}} &< {\\text S}_{\\frac {\\text{CAD}}{\\text{USD}}} \\left( \\cfrac {\\left(1+{\\text i}_{\\text{USD}} \\right)^{\\text T}}{ \\left(1+{\\text i}_{\\text{CAD}} \\right)^{\\text T}} \\right) \\\\ \\end{align*}\n\nThen:\n\n\\begin{align*}{\\text F}_{\\frac {\\text{CAD}}{\\text{USD}}} \\left(1+{\\text i}_{\\text{CAD}} \\right)^{\\text T} &< {\\text S}_{\\frac {\\text{CAD}}{\\text{USD}}} \\left(1+{\\text i}_{\\text{USD}} \\right)^{\\text T}\\\\ \\Rightarrow \\cfrac { {\\text S}_{\\frac {\\text{CAD}}{\\text{USD}}} \\left(1+{\\text i}_{\\text{USD}} \\right)^{\\text T} }{ {\\text F}_{\\frac {\\text{CAD}}{\\text{USD}}} } &>\\left(1+{\\text i}_{\\text{CAD}} \\right)^{\\text T}\\\\ \\end{align*}\n\nThis implies that the investor has more CAD than required to pay the borrowed amount; hence he can make a riskless profit.\n\nOn the other hand, if\n\n\\begin{align*}{\\text F}_{\\frac {\\text{CAD}}{\\text{USD}}}&>{\\text S}_{\\frac {\\text{CAD}}{\\text{USD}}} \\left( \\cfrac {\\left(1+{\\text i}_{\\text{USD}} \\right)^{\\text T}}{ \\left(1+{\\text i}_{\\text{CAD}} \\right)^{\\text T}} \\right) \\\\ \\end{align*}\n\nThen:\n\n\\begin{align*}{\\text F}_{\\frac {\\text{CAD}}{\\text{USD}}} \\left(1+{\\text i}_{\\text{CAD}} \\right)^{\\text T} &> {\\text S}_{\\frac {\\text{CAD}}{\\text{USD}}} {\\left(1+{\\text i}_{\\text{USD}} \\right)^{\\text T}} \\\\ \\end{align*}\n\nTherefore, the investor would have more of USD than the borrowed amount in USD and make a riskless profit.\n\n#### Example: Calculating the Forward Rate Using Covered Interest Parity\n\nSuppose that the risk-free rate of interest in USD and EUR are 3% and 6% per year, respectively. Given that the USD/EUR spot rate is 0.90. What is the 6-month USD/EUR forward rate?\n\n#### Solution\n\nUsing the formula:\n\n\\begin{align*} {\\text F}_{\\frac {\\text{USD}}{\\text{EUR}}} & ={\\text S}_{\\frac {\\text{USD}}{\\text{EUR}}} \\left( \\cfrac {\\left(1+{\\text i}_{\\text{EUR}} \\right)^{\\text T}}{ \\left(1+{\\text i}_{\\text{USD}} \\right)^{\\text T}} \\right) \\\\ & =0.90× \\cfrac {1.06^{0.5}}{1.03^{0.5}}=0.9130 \\\\ \\end{align*}\n\nNote that the spot rate (0.90) is less than the forward rate (0.9130), implying that USD is (theoretically) stronger than the EUR.\n\n### Interpretation of Points\n\nWhen T<1 $${\\text F}_{\\frac {\\text f}{\\text d}}={\\text S}_{\\frac {\\text f}{\\text d}} \\left( \\cfrac { (1+{\\text i}_{\\text d} )^{\\text T}}{(1+{\\text i}_{\\text f} )^{\\text T} } \\right)$$\n\nFor simplicity let $${\\text F}_{\\frac {\\text f}{\\text d}}=\\text F$$ and $${\\text S}_{\\frac {\\text f}{\\text d}}=\\text S$$\n\nThe above equation can be approximated as:\n\n$$\\cfrac {\\text F}{\\text S}=\\cfrac {1+{\\text i}_{\\text d} {\\text T}}{1+{\\text i}_{\\text f} {\\text T}}$$\n\nIf we subtract 1 from both sides,\n\n$$\\cfrac {\\text F}{\\text S} -1 =\\cfrac {1+{\\text i}_{\\text d} {\\text T}}{1+{\\text i}_{\\text f} {\\text T}} -1$$\n\nWe get:\n\n$$\\cfrac {{\\text F} – {\\text S} }{\\text S}=\\cfrac {1+{\\text i}_{\\text d} {\\text T}-1-{\\text i}_{\\text f} {\\text T}}{1+{\\text i}_{\\text f} {\\text T}}=\\cfrac {{\\text i}_{\\text d} {\\text T}-{\\text i}_{\\text f} {\\text T}}{1+{\\text i}_{\\text f} {\\text T}}$$\n\nSo,\n\n$$\\cfrac {{\\text F} – {\\text S} }{\\text S}=\\cfrac {{\\text i}_{\\text d} {\\text T}-{\\text i}_{\\text f} {\\text T}}{1+{\\text i}_{\\text f} {\\text T}}$$\n\nThis can be approximated as:\n\n$$\\cfrac {{\\text F} – {\\text S} }{\\text S} \\approx({{\\text i}_{\\text d} {\\text T}-{\\text i}_{\\text f} {\\text T}}){\\text T}$$\n\nIn the last expression, F-S expressed as a percentage of the spot rate is equivalent to the number of points divided by 10,000 and it is an approximate value of the interest rate differential at time T.\n\n#### Example: Calculating the Forward Rate as Points\n\nSuppose that the risk-free rate of interest in USD and EUR are 3% and 6% per year, respectively. Given that the USD/EUR spot rate is 0.90. What is the 6-month USD/EUR forward rate expressed as points?\n\n#### Solution\n\nIn the previous calculation, we had calculated the forward rate as 0.9130. So, the 6-month forward points are:\n\n$$(0.9130-0.9000)×10,000=130$$\n\nIf we express in terms of percentage of spot rate, we have:\n\n$$\\cfrac {0.9130-0.9000}{0.9000}=0.014444=1.444\\%$$\n\n## The Uncovered Interest Rate Parity\n\nWhile covered interest parity is concerned with forward rates and depends on arbitrage arguments, Uncovered interest parity is concerned with exchange rates themselves.\n\nThe uncovered interest parity condition postulates that the expected yield from a risky foreign investment must be equal to that of an equivalent domestic currency investment.\n\nIt states that the change in spot rate over the investment period should be averagely equal to the difference between the interest rates in two different countries, or simply, the expected appreciation or depreciation should approximately offset the difference in the interest rates.\n\nWhile using the (f/d) notation (domestic (d) currency as the base currency), assume that an investor has a choice of venturing in one-year domestic market investment and a risky (unhedged) foreign market investment. The uncovered parity condition compels the investor to weigh between the certain return from domestic investment and expected return from the risky foreign investment (in terms of foreign currency).\n\nThe foreign investment return in domestic currency will be given by:\n\n$$(1+{\\text i}_{\\text f} )\\left(1-\\%\\Delta {\\text S}_{\\frac {f}{d}} \\right)-1$$\n\nThis can also be represented as:\n\n$$\\approx {\\text i}_{\\text f}-\\%\\Delta {\\text S}_{\\frac {f}{d}}$$\n\nAlso, the uncovered interest rate parity implies that the anticipated change in the spot rate over the investment period should show the difference between the foreign and domestic interest rates. This is mathematically represented as:\n\n$$\\%\\Delta {\\text S}_{\\frac {f}{d}}^{\\text e}={\\text i}_{\\text f}-{\\text i}_{\\text d}$$\n\nWhere $$\\Delta {\\text S}^{\\text e}$$ is the future change in the spot rate.\n\n#### Example: Uncovered Interest Parity\n\nCurrencies A (domestic) and B (foreign) have risk-free rates of interest of 2% and 5% respectively. Assuming that the uncovered interest rate parity holds, what percentage would B weaken (strengthen) relative to A?\n\n#### Solution\n\nAccording to uncovered interest rate parity\n\n$$\\%\\Delta {\\text S}_{\\frac {f}{d}}^{\\text e}={\\text i}_{\\text f}-{\\text i}_{\\text d}=5\\%-2\\%=3\\%$$\n\nSo, we would expect currency B to weaken by 3% relative to the value of currency A.\n\nTherefore, the assumption brought forward by the uncovered interest rate is that when a country has higher interest rates, its currency will depreciate, which offsets the high yields, bringing the return of the two investments to the same level.\n\n## Question\n\nThe following are interest rates as listed in an interbank market:\n\n$$\\begin{array}{c|c|c|c} \\textbf{Currency} & \\textbf{Libor (annualized)} & \\textbf{Currency} & \\textbf{Spot Rate} \\\\ {} & {} & \\textbf{Combinations} & {} \\\\ \\hline \\text{USD} & {0.70\\%} & \\text{EUR/USD} & {1.8975} \\\\ \\hline \\text{EUR} & {7.00\\%} & \\text{EUR/USD} & {0.0075} \\\\ \\hline \\text{JPY} & {0.50\\%} & \\text{USD/JPY} & {82.25} \\\\ \\end{array}$$\n\nAssuming that the Uncovered Interest Parity holds, by how much is the JPY currency expected to change relative to USD over one year?\n\nA. -0.2%\n\nB. 0.2%\n\nC. 0%\n\nD. 0.4%\n\nSolution\n\nAccording to uncovered interest rate parity, the expected change in a spot exchange rate is equivalent to the difference between the interest rates corresponding to each currency (Libors). That is,\n\n$$\\%\\Delta {\\text S}_{\\frac {f}{d}}^{\\text e}={\\text i}_{\\text f}-{\\text i}_{\\text d}=(0.5-0.7)\\%=-0.2\\%$$ Therefore, JPY currency has increased by 0.2% relative to the USD currency.\n\nShop CFA® Exam Prep\n\nOffered by AnalystPrep", null, "Level I\nLevel II\nLevel III\nAll Three Levels\nFeatured Shop FRM® Exam Prep", null, "FRM Part I\nFRM Part II\nFRM Part I & Part II\nLearn with Us\n\nSubscribe to our newsletter and keep up with the latest and greatest tips for success\nShop Actuarial Exams Prep", null, "Exam P (Probability)\nExam FM (Financial Mathematics)\nExams P & FM\nShop GMAT® Exam Prep", null, "Complete Course", null, "Daniel Glyn\n2021-03-24\nI have finished my FRM1 thanks to AnalystPrep. And now using AnalystPrep for my FRM2 preparation. Professor Forjan is brilliant. He gives such good explanations and analogies. And more than anything makes learning fun. A big thank you to Analystprep and Professor Forjan. 5 stars all the way!", null, "michael walshe\n2021-03-18\nProfessor James' videos are excellent for understanding the underlying theories behind financial engineering / financial analysis. The AnalystPrep videos were better than any of the others that I searched through on YouTube for providing a clear explanation of some concepts, such as Portfolio theory, CAPM, and Arbitrage Pricing theory. Watching these cleared up many of the unclarities I had in my head. Highly recommended.", null, "Nyka Smith\n2021-02-18\nEvery concept is very well explained by Nilay Arun. kudos to you man!", null, "2021-02-13", null, "", null, "", null, "" ]

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[ "", null, "# Trying to make a \"three-level nested linear model\" run faster\n\nI am trying to fit a 3 level hierarchical random intercept model.\nTo understand how to tame such a beast I looted this code by Kazuki Yoshida.\n\nSo i tried his un-vectorized code (sorry no simulated data, but real data here)\n\n``````classroom <- read.csv(\"http://www-personal.umich.edu/~bwest/classroom.csv\")\nschoolLookupVec <- unique(classroom[c(\"classid\",\"schoolid\")])[,\"schoolid\"]\ndat <- with(classroom,\nlist(Ni = length(unique(childid)),\nNj = length(unique(classid)),\nNk = length(unique(schoolid)),\nclassid = classid,\nschoolid = schoolid,\nschoolLookup = schoolLookupVec,\nmathgain = mathgain))\nresStan <- stan(model_code = stan_code, data = dat, chains = 4, iter = 10000, warmup = 1000, thin = 10)\n``````\n\nwith the following stan code:\n\n``````data {\n// Define variables in data\n// Number of level-1 observations (an integer)\nint<lower=0> Ni;\n// Number of level-2 clusters\nint<lower=0> Nj;\n// Number of level-3 clusters\nint<lower=0> Nk;\n\n// Cluster IDs\nint<lower=1> classid[Ni];\nint<lower=1> schoolid[Ni];\n\n// Level 3 look up vector for level 2\nint<lower=1> schoolLookup[Nj];\n\n// Continuous outcome\nreal mathgain[Ni];\n\n// Continuous predictor\n// real X_1ijk[Ni];\n}\n\nparameters {\n// Define parameters to estimate\n// Population intercept (a real number)\nreal beta_0;\n// Population slope\n// real beta_1;\n\n// Level-1 errors\nreal<lower=0> sigma_e0;\n\n// Level-2 random effect\nreal u_0jk[Nj];\nreal<lower=0> sigma_u0jk;\n\n// Level-3 random effect\nreal u_0k[Nk];\nreal<lower=0> sigma_u0k;\n}\n\ntransformed parameters {\n// Varying intercepts\nreal beta_0jk[Nj];\nreal beta_0k[Nk];\n\n// Individual mean\nreal mu[Ni];\n\n// Varying intercepts definition\n// Level-3 (10 level-3 random intercepts)\nfor (k in 1:Nk) {\nbeta_0k[k] <- beta_0 + u_0k[k];\n}\n// Level-2 (100 level-2 random intercepts)\nfor (j in 1:Nj) {\nbeta_0jk[j] <- beta_0k[schoolLookup[j]] + u_0jk[j];\n}\n// Individual mean\nfor (i in 1:Ni) {\nmu[i] <- beta_0jk[classid[i]];\n}\n}\n\nmodel {\n// Prior part of Bayesian inference\n// Flat prior for mu (no need to specify if non-informative)\n\n// Random effects distribution\nu_0k ~ normal(0, sigma_u0k);\nu_0jk ~ normal(0, sigma_u0jk);\n\n// Likelihood part of Bayesian inference\n// Outcome model N(mu, sigma^2) (use SD rather than Var)\nfor (i in 1:Ni) {\nmathgain[i] ~ normal(mu[i], sigma_e0);\n}\n}\n``````\n\nThis model takes 574.467 seconds (Total) to run with all 4 chains where the estimated Bayesian Fraction of Missing Information was low. It is impossible to use the pairs() command to identify problems. And:\n\n``````print(fit, pars = c(\"beta_0\",\"sigma_e0\",\"sigma_u0jk\",\"sigma_u0k\"))\nmean se_mean sd 2.5% 25% 50% 75% 97.5% n_eff Rhat\nbeta_0 57.40 0.02 1.47 54.57 56.43 57.42 58.37 60.35 3494 1.00\nsigma_e0 32.14 0.02 0.76 30.67 31.62 32.14 32.65 33.66 1854 1.00\nsigma_u0jk 9.99 0.11 2.31 4.90 8.61 10.10 11.52 14.20 452 1.01\nsigma_u0k 8.62 0.07 2.10 3.97 7.33 8.76 10.01 12.46 821 1.00\n\n``````\n\nI tried a vectorized version of the same model (hoping in speeding up a little the model)\n\n``````data {\n// Define variables in data\n// Number of level-1 observations (an integer)\nint<lower=0> Ni;\n// Number of level-2 clusters\nint<lower=0> Nj;\n// Number of level-3 clusters\nint<lower=0> Nk;\n\n// Cluster IDs\nint<lower=1, upper=Nj> classid[Ni];\nint<lower=1, upper=Nk> schoolid[Ni];\n\n// Level 3 look up vector for level 2\nint<lower=1, upper=Nk> schoolLookup[Nj];\n\n// Continuous outcome\nvector[Ni] mathgain;\n}\n\nparameters {\n// Population intercept\nreal beta_0;\n\n// Level-1 errors\nreal<lower=0> sigma_e0;\n\n// Level-2 random effect\nvector[Nj] u_0jk;\nreal<lower=0> sigma_u0jk;\n\n// Level-3 random effect\nvector[Nk] u_0k;\nreal<lower=0> sigma_u0k;\n}\n\ntransformed parameters {\n\n// Varying intercepts\nvector[Nj] beta_0jk;\nvector[Nk] beta_0k;\n\n// Varying intercepts definition\n// Level-3\nbeta_0k = beta_0 + u_0k;\n\n// Level-2\nbeta_0jk = beta_0k[schoolLookup] + u_0jk;\n}\n\nmodel {\n// Prior part of Bayesian inference\n\n// Random effects distribution\nu_0k ~ normal(0, sigma_u0k);\nu_0jk ~ normal(0, sigma_u0jk);\n\n// Likelihood part of Bayesian inference\nmathgain ~ normal(beta_0jk[classid], sigma_e0);\n}\n``````\n\nFitted values are similar (with better Rhat), but this vectorized model takes longer to sample (more or less it takes alway double of the unvectorized code:837.244 seconds (Total) while I expected an improvement) with the same warnings on low estimated Bayesian Fraction of Missing Information (which still I am not able to explore )\n\n`````` mean se_mean sd 2.5% 25% 50% 75% 97.5% n_eff Rhat\nbeta_0 57.37 0.02 1.44 54.56 56.42 57.35 58.35 60.14 3571 1\nsigma_e0 32.15 0.02 0.78 30.68 31.60 32.14 32.68 33.67 1389 1\nsigma_u0jk 9.95 0.13 2.47 4.56 8.37 10.15 11.62 14.38 354 1\nsigma_u0k 8.47 0.09 2.19 3.45 7.19 8.60 9.89 12.54 593 1\n``````\n\nAny advice to make my model run faster?\nThe QR reparametrization could be applied to this kind of models? Or should I try something else like adding some weakly informative priors on parameters?\n\nHave u tried the non-centered parametrization? This likely very relevant for such a model. So in your formulation you replace\n\n``````beta_0k = beta_0 + u_0k;\n``````\n\nwith\n\n``````beta_0k = beta_0 + u_0k * sigma_u0k;\n``````\n\nand then place a normal(0,1) prior on u_0k; same for the others. The manual has a ton of material on this.\n\n3 Likes\n\nThanks @wds15!\nI have read of non-centerd reparametrization in manual and in the forum (many users asking unceasingly about it).\nStill it is -for me- a little “obscure”. I will study more about it and trying it. I will try it but i I still feel it like “a hand grenade in my pocket”.\n\nWell, you are just reparametrizing the very same density. In a nutshell it works so well because:\n\n• NUTS “likes” to sample unit normals\n\n• hierarchical models are typically used in data sparse situations to borrow the little information there is on the data across the units (its a data driven prior)\n\n• sparse data loosely means that the impact of your data is weak on the prior. So the prior does not change much when compared to the posterior.\n\n• so if you represent your model in a way such that your prior is essentially a bunch of N(0,1), then that won’t change much for sparse data and hence NUTS runs insanely better.\n\nThis is the poor mans explanation. The more involved explanation makes the argument of easier to sample from geometry in the posterior (funnel geometry). @betanalpha has good papers on this which are referenced to in the manual.\n\nDon’t be shy with non-centered! It works quite amazing.\n\nLooking at the sigma estimates which you are getting makes me think you could rescale the data such that the sigma estimates land on a unit scale (the betas as well).\n\n3 Likes\n\nThanks for the ‘poor man’ explanation! I started to read Betancourt & Girolami paper but your explanation helps me to procede with my work.\n\nFrom what I read, the non-centered parametrization works well only when with the parameters.\n\nWith the data it is better to use a centered parametrization. So, in my case:\n\n``````mathgain ~ normal(beta_0jk[classid], sigma_e0);\n``````\n\nwill be rescaled to have unit scale parameters (and put good priors on them)\n\n``````transformed data {\nvector[Ni] Y_ijk;\nY_ijk = (mathgain- mean(mathgain))/sd(mathgain);\n}\n\nmodel {\n...\nbeta_0jk ~ normal(0, 1);\nsigma_e0 ~ normal(0, 1);\nY_ijk ~ normal(beta_0jk[classid], sigma_e0);\n}\n\n``````\n\nhave I understood well?\n\nI think that’s correct, yes.\n\nSo, with the help of @wds15, I post here the vectorized code of the model, with the non-centered parametrization.\nI hope someone else will find it useful in future. Many thanks to Kazuki Yoshida that initially shared his code.\n\n``````data {\n// Define variables in data\n// Number of level-1 observations\nint<lower=0> Ni;\n// Number of level-2 clusters\nint<lower=0> Nj;\n// Number of level-3 clusters\nint<lower=0> Nk;\n\n// Cluster IDs\nint<lower=1, upper=Nj> classid[Ni];\nint<lower=1, upper=Nk> schoolid[Ni];\n\n// Level 3 look up vector for level 2\nint<lower=1, upper=Nk> schoolLookup[Nj];\n\n// Continuous outcome\nvector[Ni] mathgain;\n}\n\ntransformed data {\nvector[Ni] Y_ijk;\nY_ijk = (mathgain - mean(mathgain))/sd(mathgain);\n}\n\nparameters {\n// Population intercept\nreal beta_0;\n\n// Level-1 errors\nreal<lower=0> sigma_e0;\n\n// Level-2 random effect\nvector[Nj] u_0jk;\nreal<lower=0> sigma_u0jk;\n\n// Level-3 random effect\nvector[Nk] u_0k;\nreal<lower=0> sigma_u0k;\n}\n\ntransformed parameters {\n\n// Varying intercepts\nvector[Nj] beta_0jk;\nvector[Nk] beta_0k;\n\n// Varying intercepts definition\n// Level-3 with centered parametrization\nbeta_0k = beta_0 + u_0k * sigma_u0k;\n\n// Level-2\nbeta_0jk = beta_0k[schoolLookup] + u_0jk * sigma_u0jk;\n}\n\nmodel {\n// Prior part of Bayesian inference\n\nbeta_0 ~ normal(0, 1);\nsigma_e0 ~ normal(0, 1);\n\nsigma_u0k ~ normal(0,1);\nsigma_u0jk ~ normal(0,1);\n\n// Random effects distribution\nu_0k ~ normal(0, 1);\nu_0jk ~ normal(0, 1);\n\n// Likelihood\nY_ijk ~ normal(beta_0jk[classid], sigma_e0);\n}\n``````\n\nIt runs in 281.321 seconds (Total) which is a great improvement form the initial version, with higher effective sample size for the estimated model parameters.\n\n``````4 chains, each with iter=10000; warmup=1000; thin=10;\npost-warmup draws per chain=900, total post-warmup draws=3600.\n\nmean se_mean sd 2.5% 25% 50% 75% 97.5% n_eff Rhat\nbeta_0 0.00 0 0.04 -0.09 -0.03 0.00 0.02 0.08 3573 1\nsigma_e0 0.93 0 0.02 0.89 0.91 0.93 0.94 0.97 3436 1\nsigma_u0jk 0.28 0 0.07 0.14 0.24 0.29 0.33 0.41 2456 1\nsigma_u0k 0.25 0 0.06 0.11 0.21 0.25 0.29 0.36 2528 1\n``````\n5 Likes\n\nMore technically centering or non-centered affects the joint posterior geometry only for latent parameters. For terms that involve observed data non-centered can affect the marginally posteriors but the effect if very different and typically not as impactful.\n\n1 Like\n\nThank you @betanalpha. I am reading the paper you co-authored with Girolami just right now!\n\nYou might also find some of the discussion in https://betanalpha.github.io/assets/case_studies/divergences_and_bias.html useful as a compliment to the paper.\n\n1 Like\n\n@betanalpha WOW. I totally missed “Diagnosing Biased Inference with Divergences”. And actually it is available in the stan’s Case Studies. Adding this to the help that @wds15 gave me… this is pure gold!.\n\nThis post has been really helpful for me as well. One comment on the latest stan model- shouldn’t the sigma’s have a half cauchy or exponential distribution? It is not possible to have a negative SD, so a standard normal prior doesn’t make a lot of sense.\n\nThe variable declarations for the sigmas include the lower bound of zero, so those normal priors get implemented as half-normal automatically.\n\n1 Like" ]

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[ "# Backward Induction.\n\nBackward induction is a process of reasoning in which one starts from the end of a problem or situation, and then works backwards to the beginning. It is often used in game theory to analyze situations in which players have perfect information (meaning they know all relevant information about the situation, including what other players are thinking and what the possible outcomes of their actions might be).\n\nIn game theory, backward induction is often used to find what is called a Nash equilibrium, which is a situation in which no player has an incentive to change their strategy, because doing so would not improve their situation.\n\nTo use backward induction to find a Nash equilibrium, one starts by looking at the end of the game, and then works backwards to see what each player would need to do in order to achieve that outcome. For example, in a game of chicken, each player knows that the best outcome for them is if the other player chicken out first. So, if each player knows that the other player is reasoning backwards from the end of the game, they will both chicken out, and the Nash equilibrium is achieved.\n\n### Can backward induction be applied in this game to find a solution?\n\nIn game theory, backward induction is a solution concept used to determine the optimal course of action in a sequential game, where the player has perfect information. In a sequential game, each player has a turn to choose their action, and the players can anticipate the actions of their opponents.\n\nIn this game, there are two players, Alice and Bob. Alice goes first, and can either cooperate or defect. If Alice defects, then Bob can either cooperate or defect. If Bob defects, then Alice gets 1 point and Bob gets 0 points. If Bob cooperates, then Alice gets 2 points and Bob gets 1 point.\n\nIf Alice cooperates, then Bob can either cooperate or defect. If Bob defects, then Alice gets 0 points and Bob gets 2 points. If Bob cooperates, then Alice gets 3 points and Bob gets 1 point.\n\nThe game can be represented by the following payoff matrix:\n\nAlice Bob\n\nC C 3,1\n\nC D 2,0\n\nD C 0,2\n\nD D 1,0\n\nUsing backward induction, we can see that the optimal course of action for Alice is to defect, since she can get more points by doing so. Bob's optimal course of action is then to cooperate, since he can get more points by doing so.\n\n#### What is backward induction quizlet?\n\nBackward induction is a decision-making process in which one starts with the terminal (end) stage of a problem or process, and then works backwards to the present. It is also sometimes called \"reverse induction\".\n\nThe main idea behind backward induction is that it allows one to focus on the future consequences of present choices, and make decisions accordingly. This can be contrasted with other decision-making processes, such as forward induction, in which one starts with the present and works forwards into the future.\n\nThere are a number of different applications of backward induction. One well-known example is in game theory, where it can be used to analyze situations in which two or more people are making decisions that affect each other's payoffs.\n\nBackward induction can also be used in other settings, such as when making investment decisions or when trying to optimize a process. In each case, the goal is to take into account the future consequences of present choices, in order to make the best possible decision. What are the four types of games in game theory? There are four types of games in game theory:\n\n1. Zero-sum games\n2. Non-zero-sum games\n3. Constant sum games\n4. Sequential games What are the two basic types of game in game theory? There are two types of games in game theory: static games and dynamic games. Static games are those in which the player's strategies do not change over time, while dynamic games are those in which player's strategies may change over time. Is game theory math or economics? Game theory is a branch of mathematics that deals with the analysis of strategic interactions between different agents. In game theory, an agent is any decision-making entity, whether it is an individual, a firm, or a government.\n\nGame theory has been used to study a wide variety of phenomena in a variety of fields, such as economics, political science, biology, and psychology. In recent years, game theory has also been used to study problems in a variety of other fields, such as computer science, engineering, and operations research." ]

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[ "## Conversion formula\n\nThe conversion factor from kilometers per hour to knots is 0.53995680345662, which means that 1 kilometer per hour is equal to 0.53995680345662 knots:\n\n1 km/h = 0.53995680345662 kt\n\nTo convert 447 kilometers per hour into knots we have to multiply 447 by the conversion factor in order to get the velocity amount from kilometers per hour to knots. We can also form a simple proportion to calculate the result:\n\n1 km/h → 0.53995680345662 kt\n\n447 km/h → V(kt)\n\nSolve the above proportion to obtain the velocity V in knots:\n\nV(kt) = 447 km/h × 0.53995680345662 kt\n\nV(kt) = 241.36069114511 kt\n\nThe final result is:\n\n447 km/h → 241.36069114511 kt\n\nWe conclude that 447 kilometers per hour is equivalent to 241.36069114511 knots:\n\n447 kilometers per hour = 241.36069114511 knots\n\n## Alternative conversion\n\nWe can also convert by utilizing the inverse value of the conversion factor. In this case 1 knot is equal to 0.0041431767337739 × 447 kilometers per hour.\n\nAnother way is saying that 447 kilometers per hour is equal to 1 ÷ 0.0041431767337739 knots.\n\n## Approximate result\n\nFor practical purposes we can round our final result to an approximate numerical value. We can say that four hundred forty-seven kilometers per hour is approximately two hundred forty-one point three six one knots:\n\n447 km/h ≅ 241.361 kt\n\nAn alternative is also that one knot is approximately zero point zero zero four times four hundred forty-seven kilometers per hour.\n\n## Conversion table\n\n### kilometers per hour to knots chart\n\nFor quick reference purposes, below is the conversion table you can use to convert from kilometers per hour to knots\n\nkilometers per hour (km/h) knots (kt)\n448 kilometers per hour 241.901 knots\n449 kilometers per hour 242.441 knots\n450 kilometers per hour 242.981 knots\n451 kilometers per hour 243.521 knots\n452 kilometers per hour 244.06 knots\n453 kilometers per hour 244.6 knots\n454 kilometers per hour 245.14 knots\n455 kilometers per hour 245.68 knots\n456 kilometers per hour 246.22 knots\n457 kilometers per hour 246.76 knots" ]

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[ "# Journal of Nano- and Electronic Physics\n\nYüklə 239,46 Kb.\n səhifə 1/3 tarix 12.01.2019 ölçüsü 239,46 Kb. #95514\n1   2   3\n\nJournal of Nano- and Electronic Physics Журнал нано- та електронної фізики\n\nVol. 5 No 4, 04047(pp) (2012) Том 5 № 4, 04047(cc) (2012)\n\nMicrofluidic Mechanics and Applications: a Review\nSandeep Arya1, Saleem Khan1, Akhil Vaid2, Harneet Kour2, Parveen Lehana1\n1 Department of Physics & Electronics, University of Jammu, 180006 Jammu, India\n\n2 Department of E&C, SSCET, Badhani, Pathankot, Punjab, India\n(Received 26 July 2013; published online 31 January 2014)\nMicrofluidics involves the transportation, splitting and mixing of minute fluids to perform several chemical and biological reactions including drug screening, heating, cooling or dissolution of reagents.\nEfforts have been made to develop different microfluidic devices, droplets and valves that can stop and resume flow of liquids inside a microchannel. This paper provides the review related to the theory and mechanics of microfluidic devices and fluid flow. Different materials and techniques for fabricating microfluidic devices are discussed. Subsequently, the microfluidic components that are responsible for successful micrfluidic device formation are presented. Finally, recent applications related to the microfluidics are highlighted.\nKeywords: Microfluidics; Reynold’s number; Capillary; Surface tension; Fabrication.\n\n PACS numbers: 47.63. – b, 47.85.L –\n\n1. iNTRODUCTION\n\nMicrofluidics is one of the potential areas of research and is at its initial step of developments to influence other research fields such as chemical, biological, physical, optics and information technology. Microfluidics, technically known as the science of fluid mechanics studied at micro-scale, provide a solution to overcome the problems related to development of an analytical method with high resolution and sensitivity to accomplish microanalysis purposes. Microfluidics deals with the study related to behavior of fluids, geometric manipulation of its micro domain and precise control of small volumes of fluid flow in microlitres (L), nanolitres (nL), or even picolitres (pL). Microfluidics possesses the aptitude to carry out high resolution and sensitive separations to detect very small quantities of samples and reagents which makes this system relatively inexpensive and requires short time analysis . Microfluidic detection includes some common fluids such as bacterial cell suspensions; blood samples; proteins or antibody solutions. Microanalysis of such fluids is successfully being used in several micro-scale applications and measurements that include fluid viscosity; capillary electrophoresis; DNA sequencing; diffusion coefficient detections; chemical binding; immunoassays; cell separation and many more [2, 3]. These factors are analyzed and measured using a microfluidic device that contains either one or more channel of small dimensions nearly less than 1 mm.\n\nThe focus of this paper is to present the recent research work on microfluidic mechanics and its applications that has been comprehensively investigated. The miniaturization and integration of microfluidic components has been exploited immensely resulting into development of several microfluidic devices. Different fabrication methods were reported by the researchers to show some specific application particularly keeping in view an idea to fit ‘entire lab on a chip’. Microfluidic components like channels, valves and diaphragms can be made using different materials such as elastomers, PMMA, PDMS and solution gels .", null, "a", null, "b\n\nFig. 1 – Fabricated microfluidics (a) and natural microfluidics (b)\n\n1. MECHANICS of MICROFLUIDICS\n\nThe numerical simulation of microfluidics is essentially valuable to know the behavior of a particular system which is difficult to predict at the outset. It can be analyzed and studied that further requires the incorporation of the complexities of different factors such as channel geometry, fluids flow rate, diffusion coefficients and possible chemical interactions altogether into a numerical model. This numerical modeling can be used to visualize the complex flow phenomenon which otherwise is difficult to obtain experimentally. Several important parameters have to be studied to model it successfully. Some of them are mentioned in subsequent sections.\n\n1.1Reynolds Number\nIn fluid dynamics, when fluid flows through a micro channel, it is important to observe whether it is turbulent, that is, lesser orderly regime or laminar (stream lined flow). Table 1 shows the nature of flow for different Reynolds number . The type of flow can be determined by a dimensionless parameter known as Reynolds number depending on its uncharacteristic flow geometry.\nTable 1 – Fluid Flow on the basis of Reynolds Number\n\n Range of Reynolds Number (Re) Nature of fluid flow Re  2300 Laminar Flow 2300  Re  4000 Transient Flow Re  4000 Turbulent Flow\n\nReynolds number is used to characterize the flow of a fluid through micro channel and is defined as\n\nRe  LVavg/,\nwhere is the viscosity, is the fluid density, Vavg is the average velocity of flow and ‘L’ is most relevant length scale, and\nL  4A/P,\nA is crossectional area of the channel and P is wetted perimeter of the channel. Re depends on three factors, i.e. material properties (density, viscosity), boundary conditions and critical velocity.\nTable 2 – Properties of different Fluid Flows \n\n Laminar Turbulent Periodic flow Small flow velocities Curling of field lines 3rd flow regime Even sliding of adjacent layers Mixing between adjacent layers Surface waves Field of velocity vectors constant in time “Random\" development of field of velocity vectors Acoustic waves Shear stress depends on viscosity () and independent of density(). Flow patterns increasingly turbulent towards high velocities Sometimes laminar flow preserved up to higher velocities Shear stress :function of density () of density\n\nTable 2 shows different fluid flow and its properties . The lower values of Reynolds numbers represents the laminar flow as the viscous forces are dominant and shows smooth fluid motion where as high Reynolds numbers represents the high inertial forces representing flow variations and the turbulence. This indicates that on the basis of Reynolds number value and channel crossectional geometry, the variation in critical transition point of fluid flow can be determined, i.e. whether the flow is laminar, transient or turbulent as tabulated below. It has also been reported that momentum based phenomenon’s such as flow separation can be achieved at Reynolds numbers below 100 .\n\n1.2Flow Types in Microfluidics\nThe different types of flow are as:\n\nBubbly flow: With a high focus of bubbles in the upper half of the tube the gas bubbles are isolated in the liquid due to their buoyancy. The bubbles tend to disperse uniformly in the tube when shear forces are dominant. The regime only occurs at high mass flow rates in horizontal flows.\n\nSlug flow: The diameters of out such elongated bubbles can also be termed as large amplitude waves.\n\nStratified-wavy flow: Waves are formed on the interface when the gas velocity is increased which travel in the direction of flow and amplitude of the waves is noteworthy and depends stretched out bubbles become similar in size to the channel height at higher gas velocities.\n\nAnnular flow: A continuous annular film around the perimeter of the tube is formed by the liquid at large gas flow rates, same as in vertical flow but differs only in the liquid film which is thicker at the bottom than the top.\n\nMist flow: The liquid may be stripped from the wall and entrained as small droplets at very high gas velocities as in case of vertical flows in the now continuous gas phase.", null, "Fig 2 – Flow regimes in microchannel . a) bubbly flow, b) slug / taylor flow, c) churn flow, d) slug / annular flow, and e) annular flow\nSometimes, transition of fluid flow from laminar to turbulent can also occur due to its sensitivity to flow disturbances and channel imperfections. The extreme case of laminar flow is the Stokes flow which involves the creeping motion of fluid through channels at Reynolds number lesser than 1. This is due to greater effects of viscous forces acting relative to the inertial forces at low Reynolds number values.\n1.3Micromixing\nMicromixing is a phenomenon in microfluids in which both stirring and diffusion occurs simultaneously. During micromixing, the molecules of different liquids exhibit different properties there by resulting in the random molecular motion at the interface which allows permeability of molecules from one liquid to\nother liquid apparently. This permeation is called diffusion which is quite apparent and referred as final stage of micromixing. Micromixing is an important parameter in microfluidic applications especially for bio and chemical analysis that using lab on chips. The low diffusivity of fluids extends chemical reaction time without any chaotic advection. In micromixing, mixing quantities or mixing index, and residence time are used to evaluate the mixing performance and are developed from nonlinear dynamical systems. Since the microfluids deal with fluids flowing at low Reynolds numbers mostly less than 1, the main concern is confined to miscible liquids considering the diffusion as the final stage of micro mixing [13-16].\n1.4Diffusion\nDiffusion implies the spreading of particles due to their Brownian motion. This occurs mostly due to thermal energy. The easiest and simplest way of modeling diffusion is in one dimension . This can be represented as", null, "where is the root mean square displacement of a particle in a time t. D is the diffusion coefficient of the particle in surrounding medium, typically 10 – 5 cm/s for a small molecule in water at room temperature. According to Einstein-Stokes equation , D is described as", null, "When two different fluid flows from different reservoirs across a micro channel that has no specific mixing element, the flow is parallel and there occurs no stirring and mixing is completely diffusive.\n1.5Poiseuille’s Law\nThe Poiseuille’s law helps in obtaining the flow rate, pressure drop as well as the effective resistance of viscous and incompressible fluids which has laminar flow nature. Mathematically, the Poiseuille’s law is given as", null, "where ∆P is the pressure drop, Q is volume flow rate, L is length of the channel, η is the dynamic fluid viscosity, and r is radius of the channel. The resistance to flow denoted as R and is given by,", null, "where x is the distance in direction of flow. Microfluidics is characterized by microchannels that may vary in diameter from 100 nm to 100 micron range or even from 10 nm to 10 microns. At these length scales of micro channels the mass transfer peclet number is large which in turn leads to microfluidic mixing regimes . Peclet number is dimensionless and is used in calculations involving convective heat transfer. It is defines as the ratio of the thermal energy convected to the fluid to the thermal energy conducted within the fluid. Mathematically, Peclet number is represented as", null, "1.6Other Mechanical Parameters\nThere are some other parameters which are equally important to understand the mechanics of microfluidics. Some of them are described in this sub-section.\n2.6.1 Surface Tension\nIt is a contractive tendency of the liquid surface that allows it to resist an external acting force as the behavior of the liquid depends on cohesion forces acting between similar molecules. Its dimension is force per unit length or energy per unit area. Surface tension is due to imbalance of intermolecular attractive forces that one molecule experiences in vicinity of other molecules in a liquid. These attractive forces may be hydrogen bonds in case of polar molecules or vander waals forces for other molecules. There exist three system interfaces such as solid surface – gas interface; solid-liquid interface, liquid-gas interface that inturn give rise to surface tension forces in order to maintain a local equilibrium at specific contact angles forming a meniscus. Meniscus is a curve in the upper surface of a liquid close to the surface of the container or another object, caused by surface tension. The fluid attracts to surface due to maintenance of this local equilibrium carried by three system interfaces. This further propels the fluid along the surface . When a single layer of fluid molecules are absorbed by the channel, then dynamic contact angles can be observed which causes the flow through subsequent layers more easily. When the gas pressure of fluid nearly equals the atmospheric pressure acting on the open end of channel, the capillary action achieved is termed as perfect capillary action . Capillary action is defined as the movement of liquid through thin tubes, not a specific force. Capillary flow is due to the balance of surface energy and gravitational potential energy. Table 3 surface tension and thermal coefficient of Fluids at liquid-air interface at 200 C. The increase in capillary forces with time depends on channel aspect ratio. Higher the aspect ratio of the channel, faster is the displacement and this leads to increase of capillary forces with time. The surface tension is determined by equation", null, "where is surface tension, U is the average cohesive energy of a molecule,  is characteristic dimension of a molecule and 2 represents the effective surface area of molecule. Total energy E stored in the interface is", null, ",\nwhere S is total surface area of interface . Table 3 shows the values of surface tension and the thermal coefficients of some fluids used in the microchannels.\nTable 3 – Surface Tension and Thermal Coefficient of Fluids at Liquid-Air Interface at 200 C \n\n Liquid Surface tension () Thermal coefficient (Α) Acetone 25.2 – 0.112 Benzene 28.9 – 0.129 Ethanol 22.1 – 0.0832 Glycerol 64.0 – 0.060 Mercury 425.4 – 0.205 Water 72.8 – 0.1514\n\n2.6.2 Contact Angles\nCapillary flow is required to maintain the surface tension equilibrium that can be achieved by varying the contact angles. Contact angles can be calculated using young’s modulus and is an essential parameter in the process of slug formation. It is used to determine the feature when gas and liquid slugs interacts with the channel wall. The wall adhesion equation helps to determine the shape of fluid interface by specifying the value of static contact angle.", null, "Fig. 3 – Different forces acting at contact angles (θ). From article \nThe contact angle at any point of intersection is used to organize surfaces as wetting, hydrophilic, non wetting or hydrophobic. The wetting surfaces includes the contact angle range as 0 ≤ θ  900 where as the range of 900 ≤ θ ≤ 1800 includes the non wetting surfaces . Fig. 3 shows the formation of contact angle at the surface of a fluid. The contact angle is given as", null, "The wall adhesion equation can be cast into the better known form", null, "where,", null, "is a measure of the attraction between the particles, and, is the volume excluded by a mole of particles. p is the pressure of the fluid, V is the total volume of the container containing the fluid, n is the number of moles, and T is the absolute temperature of the system .\n\n1. Dynamics of Fluidic Flow\n\nThe dynamics of microfluids includes the parameters such as viscosity of fluids, resistance of fluids and capillary pressure. These parameters are illustrated below.\n\n1.7Viscosity of Fluids\nThe fluid – fluid interface gets affected due to variations in surface tensions that lead to marangoni flow instabilities. For example, surfactant laden flows exhibits surface tension variations at either gas-liquid or liquid- liquid interface causing instabilities due to accumulation of surfactants close to stagnation points . The marangoni effects for gas-liquid interface cause hardening of the gas bubbles that attains by surrogate no-shear boundary condition or with a no slip boundary condition. These effects alter the pressure drop and theoretical calculation based on no shear at new interfaces in microfluidic network which require intense care for its use in practical applications . The flows that are driven by surface tension gradients are Marangoni flows. Marangoni flow is generated by gradients in temperature or chemical concentration at an interface as it depends on surface tension. If the surface tension of a interface is varied, there would be disproportion of the forces which would result in flow. This flow is called the Marangoni effect.\n1.8Pressure Driven Micro-Flows\nNavier stocks equations simplifies the flow in nano and microfluidic systems and is the ratio of the inertia terms to the viscous term that is characterized by Reynolds number Re attaining negligible value much lesser than one . According to Navier Stockes equation", null, "where p is the pressure, u is the fluid velocity and is the dynamic viscosity of the fluid. This equation helps in determining various shapes of micro channels through which the fluid flows . The relation between the pressure and flow rate of system is obtained by the Hagen-Poiseuille equation as described above .\n1.9Fluidic Resistance\nThe fluidic resistance comprises of fluidic analogy to electrical circuit. The fluidic resistance depends on the crossectional geometry and can be obtained by", null, "where h is the smallest dimension of crossection of channels and w is the widest dimension of channel cross sections . The average fluidic flow rate in a microchannel depends on pressure gradient imposed at capillary ends on both sides proportionally and it justifies haven Poiseuille’s equation as classical ohms law given as", null, "where Rfluid depends on geometry of the channel crossection. In any micro or nano networks, the fluidic resistance can be calculated in same way as for electrical circuits i.e. by Kirchhoff’s laws . Fluidic resistance also helps to calculate the effective section Seffect that helps to calculate pressure drop in microchannels. The effective section is calculated as", null, "1.10Fluid Flow Control\nFluid flow can be controlled externally by three different ways. Hydrostatic generator is one such system by which flow is controlled using pressure difference by varying the altitude of fluid to atmosphere interface in different reservoirs . Another system called pressure generator controls the flow rate by varying pressure drops. It comprises of a compressor, a static pressure regulator and a manometer to keep eye on pressure values with respect to atmospheric pressure. The compatibility of all these components must be essentially good to achieve the precise and robust results. Pressure control can also be achieved using a set of electro valves enslaved electronically to a pressure sensor. Syringe pumps are used to control the flow rate directly. The main advantage of such system is that the flow is independent of fluid resistance across the micro channels. The limitation of syringe pumps is that at low flow rates there occurs development of pulsate flow rates and the time required to stabilize them is in negligible. There are certain other pumps that are not perfect flow generators due to low flow rate by back pressure. This includes peristaltic, piezoelectric etc. In contrary Electro osmotic pumps are based on electrical pumping of fluid through nonporous materials which can bear the back pressures but it do not reveal flow fluctuation problems and requires low conductive fluids there by suffering from lack of reproducibility .\n\nYüklə 239,46 Kb.\n\nDostları ilə paylaş:\n1   2   3\n\nVerilənlər bazası müəlliflik hüququ ilə müdafiə olunur ©muhaz.org 2023\nrəhbərliyinə müraciət" ]

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[ "Search a number\nBaseRepresentation\nbin111010001\n3122020\n413101\n53330\n62053\n71233\noct721\n9566\n10465\n11393\n12329\n1329a\n14253\n15210\nhex1d1\n\n465 has 8 divisors (see below), whose sum is σ = 768. Its totient is φ = 240.\n\nThe previous prime is 463. The next prime is 467. The reversal of 465 is 564.\n\n465 is nontrivially palindromic in base 11 and base 16.\n\n465 is an esthetic number in base 15, because in such base its adjacent digits differ by 1.\n\n465 is a nontrivial binomial coefficient, being equal to C(31, 2).\n\nIt is an interprime number because it is at equal distance from previous prime (463) and next prime (467).\n\nIt is a sphenic number, since it is the product of 3 distinct primes.\n\nIt is not a de Polignac number, because 465 - 21 = 463 is a prime.\n\nIt is a Harshad number since it is a multiple of its sum of digits (15), and also a Moran number because the ratio is a prime number: 31 = 465 / (4 + 6 + 5).\n\n465 is an undulating number in base 11 and base 16.\n\nIt is a plaindrome in base 7, base 9 and base 13.\n\nIt is a nialpdrome in base 5, base 8 and base 15.\n\nIt is a congruent number.\n\nIt is not an unprimeable number, because it can be changed into a prime (461) by changing a digit.\n\nIt is a pernicious number, because its binary representation contains a prime number (5) of ones.\n\nIt is a polite number, since it can be written in 7 ways as a sum of consecutive naturals, for example, 1 + ... + 30.\n\nIt is an arithmetic number, because the mean of its divisors is an integer number (96).\n\n465 is the 30-th triangular number.\n\nIt is an amenable number.\n\n465 is a deficient number, since it is larger than the sum of its proper divisors (303).\n\n465 is a wasteful number, since it uses less digits than its factorization.\n\n465 is an odious number, because the sum of its binary digits is odd.\n\nThe sum of its prime factors is 39.\n\nThe product of its digits is 120, while the sum is 15.\n\nThe square root of 465 is about 21.5638586528. The cubic root of 465 is about 7.7473108950.\n\nAdding to 465 its product of digits (120), we get a palindrome (585).\n\nSubtracting 465 from its reverse (564), we obtain a palindrome (99).\n\nThe spelling of 465 in words is \"four hundred sixty-five\", and thus it is an aban number.\n\nDivisors: 1 3 5 15 31 93 155 465" ]

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[ "# Tag Info\n\n2\n\nAside from the points raised in nbro's answer, I'd like to point out that for a single MDP (a single instance of a \"problem\"), it may be sensible to study it from perspectives that include no policy at all, or multiple different policies. For instance, if I have an MDP, I may be interested in studying it by looking at various inherent properties of the ...\n\n6\n\nThe MDP defines the environment (which corresponds to the task that you need to solve), so it defines e.g. the states of the environment, the actions that you can take in those states, the probabilities of transitioning from one state to the other and the probabilities of getting a reward when you take a certain action in a certain state. The policy ...\n\n2\n\nIsn't the environment constantly changing in this game? The current state of the agent and the environment is constantly changing as you play, but not necessarily the transition probabilities. For simplicity, you may assume that the transition probabilities do not change (e.g. if the dealer and the deck are the same every time you play). How would the ...\n\n0\n\nTo my knowledge you can't compute or solve an uncountably large MDP numerically. It will need to be discretized in some capacity. The same applies for classic control: you can't optimize over the true functional so you use a discrete approximation to the system and solve that.\n\n1\n\nNote: I assume you mean, countable Action and State Sets by 'Finite'. MDP(s) are not exclusive to finite spaces only. They can be used in Continuous/uncountable sets of Action and States too. Markov Decision Process (MDP) is a tuple $(\\mathcal S, \\mathcal A, \\mathcal P^a_s, \\mathcal R^a_{ss'}, \\gamma, \\mathcal S_o)$ where $\\mathcal S$ is a set of States, $\\... 1 In addition to the reason outlined in the comment, also note that if the state-space and action-space are both finite and of feasible size, tabular methods can be used, and there are some advantages to them (like the existence of convergence guarantees and generally a smaller number of hyperparameters to tune). 0 Transition Probabilities: Consider that you are at state$s$and from that state take an action$a$.Then there are some probability you will land up at state$s_{1}'$or$s_{2}'$($s'\\$ indicate the next states). Those probabilities are called transition probabilities. In this example, the transition matrix is just a 3D array since it depends on your state ...\n\nTop 50 recent answers are included" ]

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[ "show · character.dirichlet.value_field all knowls · up · search:\n\nThe field of values of a Dirichlet character $\\chi\\colon\\Z\\to\\C$ is the field $\\Q(\\chi(\\Z))$ generated by its values; it is equal to the cyclotomic field $\\Q(\\zeta_n)$, where $n$ is the order of $\\chi$.\n\nAuthors:\nKnowl status:\n• Review status: reviewed\n• Last edited by Pascal Molin on 2019-04-29 05:45:32\nReferred to by:\nHistory:\nDifferences" ]

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[ "# Maximization of capacity and lp norms for some product channels.\n\nChristopher King\nDepartment of Mathematics\nNortheastern University\nBoston, MA 02115\n###### Abstract\n\nIt is conjectured that the Holevo capacity of a product channel is achieved when product states are used as input. Amosov, Holevo and Werner have also conjectured that the maximal norm of a product channel is achieved with product input states. In this paper we establish both of these conjectures in the case that is arbitrary and is a CQ or QC channel (as defined by Holevo). We also establish the Amosov, Holevo and Werner conjecture when is arbitrary and either is a qubit channel and , or is a unital qubit channel and is integer. Our proofs involve a new conjecture for the norm of an output state of the half-noisy channel , when is a qubit channel. We show that this conjecture in some cases also implies additivity of the Holevo capacity.\n\n## 1 Introduction\n\nA quantum channel is the mathematical description of a device which stores and transmits quantum states. Much work has been devoted to the study of particular quantum channels with highly non-classical properties, and also to general questions such as the information capacity of classes of channels. In this paper we will consider some problems of the second type, concerning additivity and multiplicativity properties that are believed to hold for all product channels.\n\nThe basic components of a quantum channel are a Hilbert space and a noise operator . The quantum states are positive operators on , with trace equal to one. The noise operator is a completely positive, trace-preserving map which acts on the set of states. Positivity means that is a positive operator on (the algebra of bounded operators on ). Complete positivity means that the map is also a positive operator on for every .\n\nWhen the channel is used to store or transmit information, it is assumed that the information is encoded as a state on the product space for some , and that the noise acts on this state through the product operator , thereby mimicking the action of a memoryless channel in classical information theory. The basic properties of such quantum memoryless channels have been studied by many authors , , , , . One outstanding problem is to determine the ultimate rate at which classical information can be transmitted through this channel, when no prior entanglement is available between sender and receiver. The protocol that achieves this capacity may require messages to be encoded using entangled states and/or decoded using collective measurements. It is conjectured that this ultimate capacity is given by the well-known Holevo bound \n\n CHolv(Φ)=supπ,ρ[S(∑πiΦ(ρi))−∑πiS(Φ(ρi))], (1)\n\nwhere is the von Neumann entropy, and the runs over all probability distributions and collections of states on . This capacity conjecture is equivalent to the statement that there is no benefit gained when entangled states are used to encode messages for transmission through a quantum channel. As shown by Holevo and Schumacher-Westmoreland , the ultimate rate for information transmission using non-entangled coding states is exactly . Thus the capacity conjecture is implied by the additivity conjecture for , which states that for any channels and\n\n CHolv(Ω⊗Φ)=CHolv(Ω)+CHolv(Φ) (2)\n\nAlthough the equality (2) has been shown in some special cases , , , , , it remains a challenging problem to prove this result for a general pair of channels . Amosov, Holevo and Werner introduced a related conjecture, concerning the noncommutative norm of output states from a product channel (this norm is defined below). In this paper we report progress toward establishing these conjectures for some special product channels, namely the cases when is arbitrary and either (i) is a CQ or QC channel (these are defined below), or (ii) is a qubit channel. In the first case we establish both conjectures. In the second case we establish the Amosov, Holevo and Werner conjecture for integer values of . A principal ingredient in our proof in the second case is a new bound concerning the norm of the output from a “half-noisy” channel , for integer values of . We conjecture that this bound holds for all , and we show that in some cases this conjecture implies additivity of the Holevo bound (2).\n\nThe paper is organised as follows. Section 2 contains a precise statement of the results, and the conjectured bound for half-noisy channels. In section 3 we review the relation of relative entropy and the Holevo bound. In sections 4 and 5 we prove the results for CQ and QC channels. Then in section 6 we prove the results for qubit channels, and in section 7 we prove the Corollaries of our new conjecture. In section 8 we give a summary and overview of the results in the paper. Finally the Appendix contains a proof by Lieb and Ruskai of a special case of the conjecture.\n\n## 2 Statement of results\n\nThe noncommutative norm of a matrix is defined by\n\n ||A||p=(Tr|A|p)1p=[Tr(A∗A)p2]1p (3)\n\nThe corresponding maximal norm for a positive map on is\n\n νp(Φ)=supρ||Φ(ρ)||p (4)\n\nwhere the runs over states in (this quantity was introduced in , where it was called the ‘maximal output purity’ of the channel). It is always true that for any maps and , and any\n\n νp(Ω⊗Φ)≥νp(Ω)νp(Φ) (5)\n\nThe multiplicativity conjecture of states that for any completely positive trace-preserving maps and , and for all ,\n\n νp(Ω⊗Φ)=νp(Ω)νp(Φ) (6)\n\nEquality always holds in (6) for . It has been shown in several different ways that (6) holds for all and all when , , . Recently, it has been shown that (6) holds when both and are depolarizing channels, and is integer . In this paper we provide some further examples where it holds.\n\nThe first case we consider involves the CQ and QC channels introduced by Holevo , so we recall their definitions now. Let be a POVM on (so and ) and let be any collection of states. Then we can define a channel by the formula\n\n Φ(ρ)=∑Tr(ρXb)Qb (7)\n\nHolevo considered two special cases of (7). First, if are projections onto an orthonormal basis in , then (7) is called a CQ channel. Second, if , then (7) is called a QC channel. Holevo proved the additivity result (2) when is either a CQ or QC channel. Our first result generalises this by allowing an arbitrary channel .\n\n###### Theorem 1\n\nLet be a CQ or QC channel. Then for any completely positive trace-preserving map , -multiplicativity (6) holds for all , and Holevo additivity (2) holds.\n\nFor our second set of results we restrict to channels on a two-dimensional Hilbert space. For brevity of notation we will say that is a qubit map if it is a completely positive trace-preserving map on .\n\n###### Theorem 2\n\nLet be a qubit channel. Then the equality (6) holds for , that is for all channels .\n\nIn order to state the next result we need to recall the classification of qubit maps. Any qubit map can be represented by a real matrix with respect to the basis , where are the Pauli matrices. In it was explained that by using independent unitary transformations in its domain and range, this matrix can be put into the following form:\n\n Φ=⎛⎜ ⎜ ⎜⎝1000t1λ100t20λ20t300λ3⎞⎟ ⎟ ⎟⎠ (8)\n\nThis form makes it easy to see how acts on the Bloch sphere. The sphere is first compressed to an ellipsoid with semi-major axes , and is then translated by the vector . There are constraints on the allowed values of these six parameters (coming from the requirements that be completely positive and trace-preserving), and these constraints have been fully worked out in . If for then , in which case is a unital qubit map.\n\nOur next result requires a slightly stronger condition on the map , which we now state in terms of these parameters:\n\n if|λi|<|λj|<|λk|thentitj=0 (9)\n\nThis condition can be stated in words as follows: the ellipsoid may be translated only in directions lying in the two planes that are perpendicular to its two smaller axes (if any two axes have equal length, there is no restriction).\n\n###### Theorem 3\n\nLet be a qubit channel satisfying the condition (9). Then -multiplicativity (6) holds for all integer , that is for all channels and all integers .\n\nThe proofs of Theorem 2 and Theorem 3 make use of a bound for the norm of the output state from the half-noisy channel . We believe that this bound holds for all , however we can prove it only for the cases listed in the Theorems. So we state the general bound as a conjecture.\n\n###### Conjecture 4\n\nLet be a qubit channel, and let be a matrix. Write in the form\n\n M=(XYY∗Z), (10)\n\nwhere , and are matrices. Then for all\n\n ||(I⊗Φ)(M)||p≤νp(Φ)(||X||p+||Z||p) (11)\n\nThis conjecture has several important consequences, which we list in the next three Corollaries. In particular, the first Corollary shows that Conjecture 4 implies Theorems 2 and 3.\n\n###### Corollary 5\n\nLet be a qubit channel, and suppose that (11) holds for all positive matrices , for some . Then for any completely positive map on , -multiplicativity (6) holds for the same value of .\n\nIn Section 5 we will prove that (11) holds for all qubit maps when , and also for the cases listed in Theorem 3. Combining this with Corollary 5 will prove Theorems 2 and 3.\n\nOur next result concerns the additivity of minimal entropy. The minimal entropy of a completely positive trace-preserving map is defined by\n\n Smin(Φ)=infρS(Φ(ρ)) (12)\n\nThe additivity of minimal entropy is the statement that\n\n Smin(Ω⊗Φ)=Smin(Ω)+Smin(Φ) (13)\n\n###### Corollary 6\n\nLet be a qubit channel, and suppose that (11) holds for all positive matrices , and for all for some . Then for any completely positive map on , additivity of minimal entropy (13) holds.\n\nFor our last corollary, recall that a map is unital if , which means roughly that leaves unchanged the “noisiest” state through the channel.\n\n###### Corollary 7\n\nLet be a unital qubit channel, and suppose that (11) holds for all positive matrices , and for all for some . Then for any completely positive trace-preserving map on , Holevo additivity (2) holds.\n\nRemarks.\n\n1) There are two special cases where it is easy to verify Conjecture 4. First, if is a one-dimensional projection then the right side of (11) becomes , and then the result follows immediately from the definition (4). Second, suppose that is the identity map, so . Define the projections\n\n P0=(I000),P1=(000I) (14)\n\nThen convexity of the norm for implies that\n\n ||M||p=||M1/2(P0+P1)M1/2||p≤||M1/2P0M1/2||p+||M1/2P1M1/2||p (15)\n\nFurthermore for any matrix , the matrices and have the same spectrum, so we deduce that\n\n ||M||p≤||P0MP0||p+||P1MP1||p=||X||p+||Z||p (16)\n\n(this derivation is a special case of a more general result for POVM’s which is described in ).\n\n2) Lieb and Ruskai have recently established Conjecture 4, eq. (11) for a depolarizing channel in the special case , for all . Recall that the depolarizing channel is described by the parameter values , and , so that in this case the bound (11) becomes\n\n ∣∣∣∣∣∣(XλYλY∗X)∣∣∣∣∣∣p≤νp(Φ)(2||X||p) (17)\n\nwhere\n\n νp(Φ)=[(1+λ2)p+(1−λ2)p]1/p (18)\n\nTheir proof appears as an Appendix to this paper.\n\n3) Theorem 3 was proved in for unital maps in the case , and our proof here extends this result to all integer values of (and to a larger class of maps). The class of qubit maps which satisfy (9) includes all unital qubit maps and many non-unital maps. In particular, our proof applies to any extreme point in the set of qubit maps (this refers to recent work in , and we discuss it more fully in section 3).\n\n4) To prove Corollaries 6 and 7 we need only the derivative of (11) at , which we now state as a separate bound. Assume that has the form (10) with , and define the states\n\n ξ=1TrXX,ζ=1TrZZ (19)\n\nThen taking the derivative of (11) at gives\n\n S((I⊗Φ)(M))≥Smin(Φ)+Tr(X)S(ξ)+Tr(Z)S(ζ) (20)\n\n5) When and are both unital qubit maps, the additivity result (2) follows immediately from the additivity of minimal entropy (13), as was discussed in . This is also true if is a product of unital qubit maps. Additivity of Holevo capacity (2) for the ‘half-noisy’ case was proved by Schumacher and Westmoreland , and their analysis underlies our proof of Corollary 7.\n\n## 3 Relative entropy and the Holevo bound\n\nThe Holevo bound (1) can be re-expressed in terms of relative entropy in several ways (see for example the discussion in ). Here we will follow the approach of Ohya, Petz and Watanabe and Schumacher and Westmoreland , who express (1) as an optimization of relative entropy.\n\nLet be a channel, and let be an ensemble of input states for the channel. Define\n\n (21)\n\nFollowing the notation of , the Holevo capacity of the channel is denoted\n\n χ∗(Φ)=CHolv(Φ)=supEχ(Φ;E) (22)\n\nAs shown in there is an ensemble which achieves this supremum. The ensemble may not be unique, however its average input state is unique. We let denote this optimal average input state.\n\nThe relative entropy of a state with respect to a state is defined by\n\n S(ω|ρ)=Trω(logω−logρ) (23)\n\nRelative entropy is non-negative: , with equality if and only if . There is a useful characterization of the capacity in terms of relative entropy, namely\n\n χ∗(Φ)=infρsupωS(Φ(ω)|Φ(ρ)) (24)\n\nThis result was derived in and also in . For our purposes it is convenient to restate it as follows:\n\nfor any state ,\n\n χ∗(Φ)≤supωS(Φ(ω)|Φ(ρ)) (25)\n\nand equality holds in (25) if and only if .\n\nOur goal is the additivity result (2). By restricting to product states it is clear that\n\n χ∗(Ω)+χ∗(Φ)≤χ∗(Ω⊗Φ) (26)\n\nSo to establish (2) it is sufficient to prove the bound\n\n χ∗(Ω⊗Φ)≤χ∗(Ω)+χ∗(Φ) (27)\n\nFor a channel , denote the optimal average output state by\n\n ρΦ:=Φ(ρ∗) (28)\n\nThen (25) implies that\n\n χ∗(Ω⊗Φ)≤supτS((Ω⊗Φ)(τ)|ρΩ⊗ρΦ) (29)\n\nTherefore in order to prove (27), and hence (2), it is sufficient to show that for any state ,\n\n S((Ω⊗Φ)(τ)|ρΩ⊗ρΦ)≤χ∗(Ω)+χ∗(Φ) (30)\n\n## 4 Proof for CQ channel\n\nLet be a CQ channel on , so that\n\n Φ(ρ)=∑Tr(ρXb)Qb, (31)\n\nwhere are one-dimensional orthogonal projections. It follows that for all ,\n\n Qb=Φ(Xb) (32)\n\nLet be a completely positive map on . Then for any state in ,\n\n (Ω⊗Φ)(τ)=∑Ω(Tr2((I⊗Xb)τ))⊗Qb (33)\n\nwhere is the trace over the second factor. For each let\n\n nb=Tr((I⊗Xb)τ), (34)\n\nand define the state\n\n τb=1nbTr2((I⊗Xb)τ) (35)\n\nThen (33) can be written\n\n (Ω⊗Φ)(τ)=∑nbΩ(τb)⊗Qb=∑nbΩ(τb)⊗Φ(Xb) (36)\n\nwhere in the second equality we used (32).\n\nTurning first to the norm result, it follows from (36) and the definition (4) that\n\n ||(Ω⊗Φ)(τ)||p≤∑nbνp(Ω)νp(Φ)=νp(Ω)νp(Φ) (37)\n\nand this proves (6).\n\nTurning next to the channel capacity result, we will prove that (30) holds. Indeed (36) implies that\n\n S((Ω⊗Φ)(τ)|ρΩ⊗ρΦ)≤∑nb[S(Ω(τb)|ρΩ)+S(Φ(Xb)|ρΦ)] (38)\n\nwhere we used the additivity of relative entropy for product states. Now (24) implies\n\n S(Ω(τb)|ρΩ)≤χ∗(Ω),S(Φ(Xb)|ρΦ)≤χ∗(Φ) (39)\n\nwhich proves the result.\n\n## 5 Proof for QC channel\n\nLet be a QC channel, so that\n\n Φ(ρ)=∑Tr(ρXb)Qb, (40)\n\nwhere are one-dimensional orthogonal projections. For any state ,\n\n (Ω⊗Φ)(τ) = ∑Ω(Tr2(I⊗Xb)τ)⊗Qb = ∑nbΩ(τb)⊗Qb\n\nwhere we use the definitions (34) and (35). Now define\n\n θ=Tr1(τ), (42)\n\nthen it follows that\n\n nb=Tr(θXb) (43)\n\nand (5) can be written as\n\n (Ω⊗Φ)(τ)=∑Ω(τb)⊗(Tr(θXb)Qb) (44)\n\nFirst we prove the bound for the norm. Using the fact that are orthogonal projections, we get\n\n Tr|(Ω⊗Φ)(τ)|p=∑Tr|Ω(τb)|p(Tr(θXb))p (45)\n\nThe definition of the norm implies that for any positive matrix ,\n\n ||Ω(A)||p≤νp(Ω)Tr(A) (46)\n\nand hence (45) implies that\n\n Tr|(Ω⊗Φ)(τ)|p≤(νp(Ω))p∑(Tr(θXb))p (47)\n\nFurthermore, from (40) it follows that\n\n Tr|Φ(θ)|p=∑[Tr(θXb)]p (48)\n\nCombining (47) and (48) and taking the root gives\n\n ||Ω⊗Φ(τ)||p≤νp(Ω)||Φ(θ)||p≤νp(Ω)νp(Φ) (49)\n\nwhich then proves the result.\n\nTurning now to the additivity of the channel capacity, we will again establish the bound (30). We claim that the following identity holds:\n\n S((Ω⊗Φ)(τ)|ρΩ⊗ρΦ)=∑Tr(θXb)S(Ω(τb)|ρΩ)+S(Φ(θ)|ρΦ) (50)\n\nFrom the result (25) it follows that\n\n S(Φ(θ)|ρΦ)≤χ∗(Φ),S(Ω(τb)|ρΩ)≤χ∗(Ω) (51)\n\nTherefore (50) implies\n\n S((Ω⊗Φ)(τ)|ρΩ⊗ρΦ)≤∑Tr(θXb)χ∗(Ω)+χ∗(Φ)=χ∗(Ω)+χ∗(Φ) (52)\n\nand this proves the result.\n\nSo it remains to verify the identity (50). This follows easily from the definition of relative entropy, and the fact that are orthogonal projections.\n\n## 6 Proofs for qubit channels\n\nIn this section we prove Theorems 2 and 3. We do this by establishing the bound (11), and then using Corollary 5, which will be proved in the next section.\n\nLet be a qubit map, and assume that bases have been chosen in its domain and range so that it has the form (8). Clearly, the maximal norm of is invariant under permutations of the three coordinates. It is also invariant under the following symmetry operations.\n\n###### Lemma 8\n\nFor every , is invariant if the signs of any two of are reversed, or if the signs of any two of are reversed.\n\nThe proof is easy: first notice that conjugation by in the domain of switches the signs of without any other changes, and similarly for conjugation by and . Then notice that simultaneous conjugation by in both the domain and range of switches the signs of without any other changes, and similarly for and .\n\nAs a consequence, we will assume henceforth without loss of generality that\n\n t1≥0,t2≥0,andλ1≥λ2≥0 (53)\n\nOur first goal is to establish Conjecture 4 for , for any map . We rewrite (10) more fully as\n\n M=(XY1−iY2Y1+iY2Z) (54)\n\nwhere , and , are hermitian. Let . Then using the special form (8) we get\n\n (I⊗Φ)(M)= (55) ⎛⎜⎝c++X+c−+Z(t1W+λ1Y1)−i(t2W+λ2Y2)(t1W+λ1Y1)+i(t2W+λ2Y2)c−−X+c+−Z⎞⎟⎠\n\nwhere\n\n c++=(1+λ3+t3)/2, c−+=(1−λ3+t3)/2 (56) c+−=(1+λ3−t3)/2, c−−=(1−λ3−t3)/2\n\nNote that since and is a qubit map, it follows that for all choices of and . Hence the four coefficients in (56) are positive, for all allowed values of and .\n\nWe consider first the case that , and is any qubit map. Taking the trace of the square of (55) gives\n\n Tr|(I⊗Φ)(M)|2 = Tr(c++X+c−+Z)2+Tr(c+−X+c−−Z)2 + 2Tr(t1W+λ1Y1)2+2Tr(t2W+λ2Y2)2\n\nDefine\n\n x=||X||2,z=||Z||2,y1=||Y1||2,y2=||Y2||2 (58)\n\nThen using the Cauchy-Schwarz inequality for the Hilbert-Schmidt norm, and our positivity condition (53) we get\n\n Tr|(I⊗Φ)(M)|2 ≤ (c++x+c−+z)2+(c+−x+c−−z)2 + 2(t1(x+z)2+λ1y1)2+2(t2(x+z)2+λ2y1)2\n\nDefine the matrix\n\n m=(xy1−iy2y1+iy2z) (60)\n\nThen (6) can be re-written as\n\n ||I⊗Φ(M)||2≤||Φ(m)||2 (61)\n\nThe positivity of implies that\n\n Tr|Y1−iY2|2=y21+y22≤xz, (62)\n\nand hence that is positive. Therefore\n\n ||(I⊗Φ)(M)||2≤ν2(Φ)Tr(m)=ν2(Φ)(x+z)=ν2(Φ)(||X||2+||Z||2) (63)\n\nwhich establishes (11) for , and hence by Corollary 5 proves Theorem 2.\n\nIn order to prove Theorem 3 we will assume that the condition (9) is satisfied. Without loss of generality, this condition can be rewritten as follows:\n\n t1≥0andt2=0andλ1≥λ2≥0. (64)\n\nTo see this, suppose first that for any . Then the condition (9) implies that at least one of the is zero, and also that the corresponding is not the largest. Hence by permuting coordinates we can arrange that and that . By switching signs of pairs of parameters we can then re-state (9) as (64). Suppose now that for some . By permuting coordinates we can assume that , and by changing signs that . This allows a further symmetry transformation, namely we can conjugate by a unitary matrix in the range of without changing . With such a conjugation we can set , and then the condition (64) again holds.\n\nThe condition (64) is clearly satisfied for all unital maps, since in that case for all . It is also satisfied by all maps in the closure of the set of extreme points of the (convex) set of qubit maps. This fact follows from Theorem 4 in , where it was shown that all such maps have only one of the parameters being non-zero.\n\nIn order to prove (11), we re-write (55) as\n\n (I⊗Φ)(M) = (R11R12R21R22) = R11⊗E11+R12⊗E12+R21⊗E21+R22⊗E22\n\nwhere is the matrix with in position and elsewhere, and where\n\n R11 = c++X+c−+Z, (66) R12 = (t1W+λ1Y1)−iλ2Y2, R21 = (t1W+λ1Y1)+iλ2Y2, R22 = c−−X+c+−Z\n\n(we have used the condition (64) to set ).\n\nFor integer we can evaluate by multiplying the right side of (6) with itself times, and taking the trace with respect to a product basis where span and span . The result is\n\n Tr|(I⊗Φ)(M)|p=∑Tr[Ei1j1Ei2j2…Eipjp]Tr[Ri1j1Ri2j2…Ripjp], (67)\n\nwhere the sum runs over all indices . The coefficient in each of these terms is non-negative, since the matrices are all non-negative. Furthermore, repeated application of Hölder’s inequality shows that\n\n |TrA1A2…Ap|≤||A1||p||A2||p…||Ap||p (68)\n\nfor any product of matrices. Hence the sum in (67) is bounded above by\n\n Tr|(I⊗Φ)(M)|p≤∑Tr[Ei1j1Ei2j2…Eipjp]||Ri1j1||p||Ri2j2||p…||Ripjp||p (69)\n\nWe define the matrix\n\n m′=(x′y′y′z′) (70)\n\nwhere now\n\n x′=||X||p,z′=||Z||p,y′=||Y1−iY2||p (71)\n\nThe matrix is positive. This can be seen most easily by noting that the positivity of implies that where is a contraction , and hence by Hölder’s inequality . Applying the map gives\n\n Φ(m′) = [c++x′+c−+z′]⊗E11+[t1(x′+z′)/2+λ1y′]⊗E12 +[t1(x′+z′)/2+λ1y′]⊗E21+[c−−x′+c+−z′]⊗E22\n\nApplying the same method to evaluate gives\n\n Tr|Φ(m′)|p=∑Tr[Ei1j1Ei2j2…Eipjp]ri1j1ri2j2…ripjp (73)\n\nwhere\n\n r11 = c++x′+c−+z′, (74) r12 = r21=t1(x′+z′)/2+λ1y′, r22 = c−−x′+c+−z′\n\nWe now claim that\n\n Tr|(I⊗Φ)(M)|p≤Tr|Φ(m′)|p (75)\n\nIf we assume for the moment that (75) is valid, then it implies\n\n ||(I⊗Φ)(M)||p≤||Φ(m′)||p≤νp(Φ)Tr(m′)≤νp(Φ)(x′+z′) (76)\n\nThis proves (11), which by Corollary 5 implies Theorem 3.\n\nSo it sufficient to demonstrate (75). From (69) and (73) it is sufficient to show that\n\n ||Rij||p≤rij (77)\n\nfor all . First, using the positivity of etc we have\n\n ||R11||p = ||c++X+c−+Z||p≤c++x′+c−+z′=r11 ||R22||p = ||c+−X+c−−Z||p≤c+−x′+c−−z′=r22\n\nThe remaining bound also follows easily, since\n\n ||R12||p = ||t1(X+Z)2+λ1Y1−iλ2Y2||p ≤ ||t1(X+Z)2||p+||(λ1−λ2)Y1+λ2(Y1−iY2)||p ≤ t1(x′+z′)/2+(λ1−λ2)||Y1||p+λ2||Y1−iY2||p\n\nwhere in the last line we used (53). Furthermore\n\n ||Y1||p = ||(Y1−iY2)/2+(Y1+iY2)/2||p ≤ 12||Y1−iY2||p+12||Y1+iY2||p = y′\n\nHence (6) becomes\n\n ||R12||p≤t1(x′+z′)/2+(λ1−λ2)y′+λ2y′=t1(x′+z′)/2+λ1y′=r12 (79)\n\nwhich establishes the result.\n\n## 7 Proofs of Corollaries\n\n### 7.1 Corollary 5\n\nLet be any completely positive map on , and let be a state on of the form\n\n τ=(ABB∗C) (80)\n\nwhere are matrices, with , and . Then has the form (10) with , and . Hence from the definition of the maximal norm it follows that\n\n ||X||p≤νp(Ω)Tr(A),||Z||p≤νp(Ω)Tr(C) (81)\n\nApplying (11) and using the facts that and\nwe immediately deduce Corollary 5.\n\n### 7.2 Corollary 2\n\nRecall that the entropy of a state is defined by\n\n S(ρ)=−Trρlogρ (82)\n\nUsing it follows that\n\n ddp(||ρ||p)p=1=−S(ρ), (83)\n\nand hence that\n\n ddp(νp(Φ))p=1=−Smin(Φ) (84)\n\nTherefore taking the derivative of (6) at yields immediately (13).\n\n### 7.3 Corollary 3\n\nFrom the results of Section 2, it is sufficient to establish the bound (30). For any states and we have\n\n log(ω⊗ρ)=logω⊗I+I" ]

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[ "# Parallel solutions of simple indexed recurrence equations\n\nResearch output: Contribution to journalArticlepeer-review\n\n## Abstract\n\nWe define a new type of recurrence equations called 'Simple Indexed Recurrences' (SIR). In this type of equations, ordinary recurrences are generalized to X[g(i)] = opi(X[f(i)], X[g(i)]), where f,g: {1...n}→{1...m}, opi(x,y) is a binary associative operator and g is distinct. This enables us to model certain sequential loops as a sequence of SIR equations. A parallel algorithm that solves as a set of SIR equations will, in fact, parallelize sequential loops of the above type. Such a parallel SIR algorithm must be efficient enough to compete with the O(n) work complexity of the original loop. We show why efficient parallel algorithms for the related problems of List Ranking and Tree Contraction, which require O(n) work, cannot be applied to solving SIR. A sequence of experiments was performed to test the effect of synchronous and asynchronous executions on the actual performance of the algorithm. An efficient solution is given for the special case where we know how to compute the inverse of opi, and finally, useful applications of SIR to the well-known Livermore Loops benchmark are presented.\n\nOriginal language English 22-37 16 IEEE Transactions on Parallel and Distributed Systems 12 1 https://doi.org/10.1109/71.899937 Published - Jan 2001\n\n## ASJC Scopus subject areas\n\n• Signal Processing\n• Hardware and Architecture\n• Computational Theory and Mathematics\n\n## Fingerprint\n\nDive into the research topics of 'Parallel solutions of simple indexed recurrence equations'. Together they form a unique fingerprint." ]

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[ "# How to calculate element-wise matrix functions in Sage", null, "Anonymous\n\nIn MATLAB one can write:\n\nA = [1,2,3]; B = 2*A.^3;\n\nwhere B gets element-wise result of the function 2*A.^3 given matrix (or vector) A.\n\nIn Sage the notation .^ does not function. How to do this in Sage?\n\nThen how to calculate the element-wise sin(A) in Sage?\n\nedit retag close merge delete\n\nSort by » oldest newest most voted\nsage: A = vector([1,2,3])\n(1, 8, 27)\nsage: A.apply_map(2*x^3)\n(2, 16, 54)\nsage: A.apply_map(sin(x))\n(sin(1), sin(2), sin(3))\n\n\nNote that the first time you do this you might get a warning.\n\nsage: A.apply_map(x^3)\n/Users/.../sage-5.8.beta3/local/lib/python2.7/site-packages/IPython/core/interactiveshell.py:2721: DeprecationWarning: Substitution using function-call syntax and unnamed arguments is deprecated and will be removed from a future release of Sage; you can use named arguments instead, like EXPR(x=..., y=...)\nSee http://trac.sagemath.org/5930 for details.\nexec code_obj in self.user_global_ns, self.user_ns\n\n\nSo you wouldn't want to rely on that syntax long-term. There are several ways around this; here's one.\n\nsage: A = vector([1,2,3])\nsage: A.apply_map(lambda x: sin(x))\n(sin(1), sin(2), sin(3))\n\nmore\n\nI don't think Sage has the elementwise matrix operations as it is used in MATLAB, but numpy has. So, the OP can either use numpy directly (it is vectorized), or use the Sage equivalent for the Schur product. Elementwise matrix products (Schur product) are present in Sage, and can be accessed by using A.elementwise_product(B), where A, B are matrices. However, the elementwise operations in MATLAB are a bit different. When we write A.^3 MATLAB actually considers 3 to be a matrix B = 3*ones(size(A)). And, then it does A.^B. Here, B can be any other matrix of the same dimensions as A. I think this (vectorized) elementwise power is not present in Sage." ]

[ null, "https://ask.sagemath.org/m/default/media/images/anon.png", null ]

[ "# Examples of Falsifiability\n\nI came across the notion of falsifiability quite recently.\nThe wikipedia article on the same states that:\n\nFalsifiability or refutability of a statement, hypothesis, or theory is the inherent possibility that it can be proven false. A statement is called falsifiable if it is possible to conceive of an observation or an argument which negates the statement in question. In this sense, falsify is synonymous with nullify, meaning to invalidate or \"show to be false\".\nFor a statement to be questioned using observation, it needs to be at least theoretically possible that it can come into conflict with observation.\n\nWhile I can understand the general concept - I would like to have a deeper understanding of the same. Popper mentions that this notion differentiates science from pseudo - science.\n\nCan someone please give me some examples for the same? - So that I might understand the idea more intuitively. Specifically if you could provide what would be the falsifiability arguments/observations would be for:\n\n• Newton's theory of gravitation.\n• Heliocentralism\n• Theorem of calculus.\n• Probability theory.\n\nBasically two popular theories from the realm of physics and two popular theories from mathematics (which I might possibly be familiar with), would do. Need not be just these four.\n\n• Newton's theory: a free apple \"falling\" from the floor to the ceiling. – Mauro ALLEGRANZA Jul 2 '16 at 15:29\n• Right - if we observe that theory of gravitation would be falsified. – user12196 Jul 2 '16 at 15:43\n• Of course, the more complex is the theory, more difficult is to found \"simple\" falsifying conditions like that. When many \"factors\" are involved, a falsifying experiment must \"manage\" all of them. Consider the well-known discovery of Neptune by Urbain Le Verrier : a potential \"falsifier\" has been transformed into a brilliant \"verification\". – Mauro ALLEGRANZA Jul 2 '16 at 15:56\n• For mathematical theories, it is not so clear if Popper's criteria applies. In principle, we can say that the only way to \"falsify\" a math theory is proving his inconsistency. – Mauro ALLEGRANZA Jul 2 '16 at 15:57\n• @MauroALLEGRANZA Can you give me some sources to study up Urbain Le Verrier's discovery - specifically how it relates to falsifiability. – user12196 Jul 2 '16 at 17:25\n\nFalsification is an excellent and easy to understand system in principle, but much more nuanced in implementation. The easiest to falsify hypotheses are those famous ones such as \"all swans are white,\" which can be falsified by observing a black swan. Of course, this assumes we all agree on what is a swan is. Hypotheses get murkier from there.\n\nWhen it comes to real meaningful scientific hypotheses, falsification is typically more of an extended process rather than an instantaneous event. A scientific theory which is falsifiable is one where some results could cast substantial doubt on the hypothesis, and that doubt can be compounded by future tests.\n\nFor example, if one believed the hypothesis that light acts as a wave, one would be surprised to see particle like behavior. The photoelectric effect is one such effect that we now know exhibits particle like behavior. The first time one observes particle like behavior from an experiment, one might assume the results were a measurement error. Doing it a second time would begin casting doubt on the theory that light always behaves like a wave. Having dozens of scientists all run such experiments multiple times, and each discovering particle like effects would eventually \"falsify\" the hypothesis.\n\nThis process is even more complicated due to statistics. If I claim there is a gaussian error term on my results, you can never truly prove that my theory is wrong, because there is always a non-zero chance that you simply observed random luck. However, in practice, once the probability of such chance events is low enough, we declare a theory \"falsified.\" How high one has to go is discipline dependent. In sociology, we regularly see error terms permitting 10% or even 20% due to unexplained factors. In particle physics, a hypothesis is not declared \"confirmed\" until those unexplained factors account for no more than 0.00001% of the total observed effects. This is because subatomic particles behave quite regularly, and we're able to generate as many results as needed to attain such high degrees of confidence. In sociology, it is much harder to repeat experiments and there is a great deal of variance between individuals, so the best we can do is lower degrees of confidence.\n\nAs for your list of examples:\n\nNewton's theory of gravitation.\n\nIt is generally accepted that the motion of planets is almost completely governed by gravitational interactions. If we were to observe the motion of the planets, and find substantial deviations not explained by his theory of gravity, this would either falsify his theory or show that there are other forces at work. I believe we actually do see results which would falsify his work: you have to account for relativity to explain some movements (particularly in cases near a black hole)\n\nHeliocentralism\n\nHeliocentralism actually cannot be proven nor disproven because it is merely a model. Its more akin to a coordinate system transform than a theory. However, if one assumes geocentralism, one is forced to admit many strange forces which account for all of the movement we see in the planets. If one assumes heliocentralism, the movement can be explained entirely with simple conservative gravity models like Newton's theory of gravity. It is the simplicity of the heliocentric model that made it so effective.\n\nTheorem of calculus.\n\nConsider if we couldn't go anywhere, because of Xeno's paradox. This would demonstrate that the assumptions we make regarding limits are false. That being said, calculus is a mathematical construct. All that we can really falsify is its usefulness in describing the world around us.\n\nProbability theory.\n\nOnce again this is a mathematical construct, making it difficult to falsify. However, one could argue that it is \"falsified\" by demonstrating that it does not effectively model reality. One major assumption in much of probability is IID: the idea that observations are (I)ndependent (I)denticaly (D)istributed. If there was a reason to argue that this assumption is invalid in the real world, then much of probability would not apply. This actually does occur when exploring the human mind. In many cases the assumption of IID is very poorly founded, so many simplifications that probability would permit are simply invalid when discussing the behavior of the mind.\n\n• So we can't falsify mathematical theories? I thought Popper's method was a way of distinguishing the scientific from the non scientific - does that imply mathematical constructs are not scientific or is there something wrong with Popper's method. Theories in physics use many mathematical constructs as well - so if they are not falsifiable how come the concepts on physics are? – user12196 Jul 3 '16 at 8:33\n• @novice I cannot answer the question \"does that imply mathematical constructs are not scientific\" directly, because it would involve a detailed discussion of exactly what \"not scientific\" means to you. That would be better suited for a chat room, rather than comments. However, I will try my best to answer obliquely. The validity of a mathematical construct is based in the validity of its assumptions and the validity of the rules of inference associated with it, not physical reality. It is only as one seeks to apply said constructs to the real world that the concept of falsification... – Cort Ammon Jul 3 '16 at 16:39\n• ... becomes meaningful. Without that application to physical reality, mathematical theories are subject to far more stringent validity requirements than falsification would ever ask from them. However, when applying such constructs to science, one does make the assumption that those constructs are indeed valid. If the real numbers do not, indeed, form a field, calculus falls apart. On the other hand, it is totally valid to have mathematical constructs which do not have an immediately obvious connection to reality. The concept of complex numbers, for instance, was a mathematical... – Cort Ammon Jul 3 '16 at 16:41\n• ... curiosity until Euler's function connected them to cyclic motion. Now they are used constantly in science. If you are interested in this question of the validity of mathematical construct, I recommend a beautiful video by Vsauce, How to Count past Infinity. He does an excellent job of explaining the subtle distinction between scientific theories and mathematical ones. – Cort Ammon Jul 3 '16 at 16:44\n• Thanks for your comments as well as the resource you have provided. I confess I am not sure how Popper gives arguments/justifications in favor of the notion of falsifiability to distinguish science from pseudo-science. I read the part about observation and argument - and thought that an argument for the falsifiability of mathematical constructs must be possible - instead of an observation. Let me go into this more deeply and come back with more questions. – user12196 Jul 3 '16 at 17:20\n\nIn the comments to Cort Ammon's answer you say:\n\n\"So we can't falsify mathematical theories? I thought Popper's method was a way of distinguishing the scientific from the non scientific - does that imply mathematical constructs are not scientific or is there something wrong with Popper's method.\"\n\nExactly, mathematical theories are not scientific theories. Mathematics is about abstract mathematical objects, Science is about empirically observables phenomena. The truth of mathematical statements are prove using logic and reason alone, while the truth of statements in physics, chemistry, biology, etc...are proven by experiment and observation. This was best described by David Hume, with his distinction known as Hume's Fork:\n\n\"All the objects of human reason or enquiry may naturally be divided into two kinds, to wit, Relations of Ideas, and Matters of fact. Of the first kind are the sciences of Geometry, Algebra, and Arithmetic ... [which are] discoverable by the mere operation of thought ... Matters of fact, which are the second object of human reason, are not ascertained in the same manner; nor is our evidence of their truth, however great, of a like nature with the foregoing.\" - An Enquiry Concerning Human Understanding\n\nSo things like the fundamental theorem of calculus and probability theory can't be falsified because they don't correspond to anything observable. They, like all mathematical truths are proved solely using the rules and axioms of logic.\n\nThis is the whole point of falsification, one has to attempt to show that they empirically observe a phenomena that contradicts their theory. So the Newton's theory of gravity says that apples should fall every time we let go of them in midair. Pre Popper's falsificationism, Newton's theory is falsified if someone raises an apple lets go of it, and instead of it falling it hovers in the air or goes upwards.\n\nSimilarly per Popper, heliocentrism will be falsified the day that Venus or Mars, or one of the other planets is observed in a different orbit then the one predicted by the theory.\n\nThis points to an interesting problem with Popper's theory, that of auxiliary hypotheses (also called the Duhem-Quine thesis, or they idea that all observations are theory laden): Consider that at the beginning of the 19th century the orbit of Uranus was different than what was predicted by Newtoninan mechanics and heliocentrism. But astronomers, instead of abandoning the theory, concluded that there was an unknown planet modifying the orbit of Uranus, which they later confirmed and called Neptune. So the dilemma is: When observation contradicts theory, is the theory falsified? or is there missing data that can explain the mismatch between theory and predictions?\n\nThe issue of how to solve the problem of auxiallry hypotheses is still debated, and hasn't been solved yet. See the ideas of W.V.O Quine, Thomas Kuhn, Imre Lakatos and Paul Feyerabend, all in response to Popper's concept of falsification.\n\n• The thing is that the wiki article also says arguments could be used to falsify-so I thought it might be applicable for mathematics constructs as well. – user12196 Jul 3 '16 at 19:01\n\nThe best way to understand Popper is to read Popper. There are a few commentators who have correctly understood his ideas, but the vast bulk of commentary on Popper is not even able to state his ideas correctly. Lakatos, Feyerabend and Kuhn are especially bad and should be avoided.\n\nTo understand falsification properly, you need to understand Popper's theory of knowledge more broadly. Most philosophers of science who take science seriously and think it is good are inductivists: they believe in a process called induction. Induction supposedly involves (1) taking observations, (2) using them to make theories, and then (3) showing those theories are true or probably true by more observations. People have looked at many phenomena such as the night sky, biology, medicine and so on, without learning much for thousands of years. So just observing stuff doesn't do much good. If you don't know what to look for, just observing will not produce progress, so step (1) is impossible. In addition, explanations don't follow from observations. The theory of stars has implications for many events we will never observe, e.g. - supernovae that took place before there were human observers, and those events don't follow from observations without a theory of how stars change. So steps (2) and (3) are also impossible.\n\nSo if we don't get theories from observation how do we get them? We guess. You look for a problem: some issue that is not explained by current ideas. You guess solutions to that problem. You then criticise the proposed solutions. This criticism may involve experiments, but many theories can be eliminated without doing experiments, e.g. - inconsistent theories.\n\nAn experiment involves looking for a situation in which two or more different ideas about how the world works make different predictions. You then either set up that situation or look for an existing system that realises that situation. Newton's theory of gravity and Einstein's general theory of relativity made different predictions about Mercury, and Newton's theory was refuted.\n\nSome philosophers make a lot of fuss about the possibility that you might do an experiment wrong or misinterpret the results. But as Popper pointed out in Logic of Scientific Discovery, Chapter V (especially Section 29), this problem is solved by his epistemology. If an experiment contradicts an existing theory, that's a problem. This problem can be solved by any guess that explains the difference and is not eliminated by some criticism. The discovery of Neptune was taken as an example above, so let's look at it. An unsolved problem was found in explaining the orbits of some planets. Urbain Le Verrier guessed that there might be another planet. He worked out some constraints on where the planet could be to produce such effects, Johann Gottfried Galle looked for it and found it. If Galle had not found the planet that problem would have remained unsolved. Perhaps some other explanation could have been found to reconcile Newtonian mechanics with observation, perhaps not. Popper recommended that a proposed solution to a scientific problem should be rejected if it was ad hoc: if it had no implications beyond the problem it was invented to solve.\n\nI'm going to skip the heliocentric theory because it is fairly similar to Newtonian mechanics. If you want a long list of examples, see the introduction to \"Realism and the Aim of Science\" by Popper.\n\nMathematical theories are about abstractions. They can be critically discussed, but not experimentally tested. 1+1 = 2 even though it is possible to think of examples of putting two objects together and only getting one object as a result. If you move two piles of sand together, you may only get one pile. So you have to think carefully about what systems you take as models of mathematical operations such as addition. For a discussion see \"Realism and the Aim of Science\" by Popper Chapter III, Section 24.\n\nAs far as probability is concerned, the best existing explanations have been provided by David Deutsch, see\n\nFor explanations of Popper's positions, see \"Objective Knowledge\" by Popper, Chapter 1, \"Realism and the Aim of Science\" by Popper, \"Logic of Scientific Discovery\" by Popper, \"The Fabric of Reality\" by David Deutsch, Chapters 3 and 7, and \"The Beginning of Infinity\" by David Deutsch, Chapters 1,2,4 and 13.\n\nI'm not sure just how useful falsification is in explaining the actual progress of science, in the sense of the revision of its basic concepts; its not for example a source of new ideas, but of pruning out what is given on the basis of what is known. Its a minor mode of progress, and not its major mode. A major mode would tell us how to find new ideas, unfortunately such a philosphical stone is illusionary.\n\nFor example, calculus can be explained by attempting to give meaning to 0/0; this, in terms of the usual arithmetic operations, is nonsensical; however mathematicians like to 'close up' operations; 0/0 can in fact be given meaning by thinking of it as dx/dy; of course this opens up the whole new world of calculus.\n\nSimilarly, no meaning could be given to the square root of -1; eventually one was found that was useful: i -the imaginary; and it again opened up a whole new world of complex geometry.\n\nProbability is a concept with intuitive appeal; yet Quantum Mechanics relies on the notion of the square root of probability, and in fact a great deal of the bizarre beahviour can be explained on the basis of this new concept which still hasn't found a properly ontological basis in the same way that the infinitesimal or the imaginary has.\n\n• -1 This answer does not even attempt to address the question. – Mr. Kennedy Jun 23 '18 at 15:15" ]

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[ "# Constructions of Representations of Rank Two Semisimple Lie Algebras with Distributive Lattices\n\n• L. Wyatt Alverson II\n• Robert G. Donnelly\n• Scott J. Lewis\n• Robert Pervine\n\n### Abstract\n\nWe associate one or two posets (which we call \"semistandard posets\") to any given irreducible representation of a rank two semisimple Lie algebra over ${\\Bbb C}$. Elsewhere we have shown how the distributive lattices of order ideals taken from semistandard posets (we call these \"semistandard lattices\") can be used to obtain certain information about these irreducible representations. Here we show that some of these semistandard lattices can be used to present explicit actions of Lie algebra generators on weight bases (Theorem 5.1), which implies these particular semistandard lattices are supporting graphs. Our descriptions of these actions are explicit in the sense that relative to the bases obtained, the entries for the representing matrices of certain Lie algebra generators are rational coefficients we assign in pairs to the lattice edges. In Theorem 4.4 we show that if such coefficients can be assigned to the edges, then the assignment is unique up to products; we conclude that the associated weight bases enjoy certain uniqueness and extremal properties (the \"solitary\" and \"edge-minimal\" properties respectively). Our proof of this result is uniform and combinatorial in that it depends only on certain properties possessed by all semistandard posets. For certain families of semistandard lattices some of these results were obtained in previous papers; in Proposition 5.6 we explicitly construct new weight bases for a certain family of rank two symplectic representations. These results are used to help obtain in Theorem 5.1 the classification of those semistandard lattices which are supporting graphs.\n\nPublished\n2006-11-23\nIssue\nArticle Number\nR109" ]

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[ "# Yellow Paper eq 221: Why and how the sender of a signed transaction equals to the address of the signer?\n\nEthereum Yellow Paper: Please see equation 220 and 221.", null, "On equation 220, we obtain the transaction that can be sent to the network and will be tracked by a 256 bit transaction-id. Its right most 160-bits is equalled to S(T) which is the defined as the sender function S of the transaction.\n\nMy question is related to Equation 221 which is an assertion about: :\n\nThe assertion that the sender of a signed transaction equals the address of the signer should be self-evident\n\n[Q] Why and how the sender of a signed transaction(`S(T) on the eq. 220`) equals the address of the signer (`A(pr)`) ? Is there any well explained documentation related to this.\n\nCould we conclude the following statement:\n\n``````B96...255(KEC(ECDSARECOVER(h(T),Tw,Tr,Ts))) == B96...255((KEC(ECDSAPUBKEY(pr)))\n``````\n\nThank you for your valuable time and help.\n\nAn ethereum signature has three parameters `r`, `s` and `v`. Using `r`, `s` and the ECDSA equation you have two candidates for the public key, then using `v` you can disambiguate and know exactly which one is the public key of the signer.\nOnce you have the public key, you use can calculate the address with the last 20 bytes of `keccak256(publicKey)`.\n• I get really confused :( what does `S(T)` stands for, is it Ethereum address of the sender, which is actually the 160-bit we obtain on eq. 215? So `KEC( ECDSARECOVER(h(T),Tw,Tr,Ts)) == KEC(ECDSAPUBKEY(pr))` you could see on my updated Q?@Ismael – alper Dec 16 '17 at 13:55" ]

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[ "Calculate Average Accross Multiple Worksheets\n\nG\n\nGuest\n\nI'm trying to calculate an average accross multiple worksheets. I could pull\nall values into the final worksheet, but there must be a way to do this\nwithout having hidden data...\n\nSheets / Cells\n'Q1'!D35:'Q2'!D35:'Q3'!D35:'Q4'!D35\n\nFour Worksheets: Q1, Q2, Q3 & Q4\nFour Cells: D35, D35, D35 & D35\n\nI also want to prevent a #DIV error if the values are = 0\n\nThis is the formula I have, I just can't seem to make it work:\n=IF(ISERROR(SUMPRODUCT('Q1'!D35+'Q2'!D35+'Q3'!D35+'Q4'!D35)/SUMPRODUCT(('Q1'!D35+'Q2'!D35+'Q3'!D35+'Q4'!D35>0)*(ISNUMBER('Q1'!D35+'Q2'!D35+'Q3'!D35+'Q4'!D35)))),\"\",SUMPRODUCT('Q1'!D35+'Q2'!D35+'Q3'!D35+'Q4'!D35)/SUMPRODUCT(('Q1'!D35+'Q2'!D35+'Q3'!D35+'Q4'!D35>0)*(ISNUMBER('Q1'!D35+'Q2'!D35+'Q3'!D35+'Q4'!D35))))\n\nAlso, (not sure if this matters) all four cells in all four sheets are\naverage calculations themselves. I'm calculating an average in each\nworksheet, and then I want to calculate a total average in a summary sheet.\n\nAny help would be appreciated!\n\nJason\n\nG\n\nGuest\n\n=SUM(Q1:Q4'!D35)/4\nthere should be no possibility of divide by zero" ]

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[ "## Essential University Physics: Volume 1 (3rd Edition)\n\nPublished by Pearson\n\n# Chapter 9 - Exercises and Problems - Page 165: 76\n\n#### Answer\n\n$v_f = 645\\hat{i}+132\\hat{j}$\n\n#### Work Step by Step\n\nWe find the sum of the initial momentums: $P_0=32(580)\\hat{i}+16(870)cos27\\hat{i}+16(870)sin27\\hat{j}$ $P_0=30,962\\hat{i}+6319\\hat{j}$ We divide this by the total final mass to find: $v_f = 645\\hat{i}+132\\hat{j}$\n\nAfter you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback." ]

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[ "# Sum in a sudoku with sum total of 15 and 45\n\nI have created a sudoku puzzle with the following restrictions:\n\n• Each row and column sum to $$45$$.\n• Each row and column in the nine $$3$$ by $$3$$ sub-grids sum to $$15$$.\n\nIs such a sudoku unique?\n\n• \"...where each row and column adds to a total of 45...\" is redundant, since it is in the nature of a sudoku puzzle that its rows and columns sum to $45$. This is because each row contains the numbers $1$ through $9$, and $1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45$. Feb 11 '19 at 5:04\n• \"where each (3*3) 9 individual squares of rows and columns total a sum of 15\" >>> do you mean 45? Feb 11 '19 at 5:33\n• @Kryesec I've assumed that that means that each of the 3 by 3 sub-grids is a magic square with \"magic constant\" 15. In other words, each row/column in each 3 by 3 sub-grid sums to 15. Feb 11 '19 at 5:47\n\nNo.\n\nOf course, there's rotation and reflection, but even beyond that you can swap any two rows within its group of three rows (eg swap row 1 and 2 or swap 4 and 6), and you can swap a group of three with one of the other groups of three (eg swap 1,2,3 with 4,5,6), and the same applies to the columns.\n\n• I wonder if there's any kinda of transformation that would work here that doesn't work on sudoku in general. Feb 11 '19 at 12:00\n\nI was beaten to a very similar answer, but you can create such a square based on the magic square below:\n\nabc\ndef\nghi\n\nTurn it into this:\n\nabcdefghi\ndefghiabc\nghiabcdef\nbcaefdhig\nefdhigbca\nhigbcaefd\ncabfdeigh\nfdeighcab\nighcabfde\n\nSince there are multiple possible magic squares, it's not unique. When you change the number at the very center (i), which you can, you get a different sudoku (not even a rotation or reflection).\n\nOne such arrangement of the 3x3 grid could be:\n\n1 5 9\n8 3 4\n6 7 2\n\nThis would then be repeated along to the right, but rotating the order of the rows to create:\n\n1 5 9 8 3 4 6 7 2\n8 3 4 6 7 2 1 5 9\n6 7 2 1 5 9 8 3 4\n\nThe same could be done to repeat downwards but rotating the columns this time:\n\n1 5 9 8 3 4 6 7 2\n8 3 4 6 7 2 1 5 9\n6 7 2 1 5 9 8 3 4\n5 9 1 3 4 8 7 2 6\n3 4 8 7 2 6 5 9 1\n7 2 6 5 9 1 3 4 8\n9 1 5 4 8 3 2 6 7\n4 8 3 2 6 7 9 1 5\n2 6 7 9 1 5 4 8 3\n\nHowever:\n\nThis creates one such possibility of your given solution, as each of the 3x3 sections could be placed in the top left and the rotations occur from there.\n\nThe answer is no. A Sudoku satisfying the puzzle requirement (i.e. each 3x3 sub-grid is a semi-magic square) is not unique even if we do not make a distinction between Sudokus differing just by rotations or reflections. And the simplest reason, as pointed out by @DR Xorile, is that there are natural Sudoku transformations (namely certain rows and/or columns swaps) which are not rotations and reflections but which preserve the integrity of the Sudoku and the semi-magic property of the sub-grids. This brings up a question alluded to by @Rawling: what if we extend a rotations and reflections group of transformations with row and column permutations suggested by @DR Xorile (let's call this group RRRCP for brevity) and not make distinctions between Sudokus if they can be transformed into each other via any RRRCP transformations. Will it make a Sudoku built out of 3x3 sub-grid of semi-magic squares \"unique\"? At first I thought it might. On a smaller scale, a 3x3 semi-magic square is \"unique\" in a sense that all 72 semi-magic squares can be generated by applying transformations from a similar extended group (rotations, reflections, columns/rows permutations) to just one semi-magic square. But it turns out that for a full Sudoku the answer is still no - a Sudoku built out of 3x3 sub-grid of semi-magic squares is not \"unique\" even modulo RRRCP transformations. Here is why.\n\nFor the sake of precision here is the description of an RRRCP group. Let's call each set of three columns (1, 2, 3), (4, 5, 6), and (7, 8, 9) - a \"macro\" column and each set of three rows (1, 2, 3), (4, 5, 6), and (7, 8, 9) - a \"macro\" row. \"Macro\" columns and \"macro\" rows constitute a 3x3 \"macro\" grid. A RRRCP group consists of compositions of the following Sudoku transformations:\n\n• rotations and reflections;\n\n• permutations of the columns within a \"macro\" column (ex. columns 4, 5, 6)\n\n• permutations of the rows within a \"macro\" row (ex. rows 7, 8, 9)\n\n• permutations of whole \"macro\" columns within a 3x3 \"macro\" grid (ex swap columns 1 2 3 with columns 7, 8, 9)\n\n• permutations of whole \"macro\" rows within a 3x3 \"macro\" grid (ex swap rows 1 2 3 with rows 4, 5, 6).\n\nThe following two Sudokus S1 and S2 are built out of semi-magic squares but cannot be transformed into each other by RRRCP transformations.\n\n2 7 6 1 9 5 3 8 4\n9 5 1 8 4 3 7 6 2\n4 3 8 6 2 7 5 1 9\n\n3 8 4 2 7 6 1 9 5\n7 6 2 9 5 1 8 4 3\n5 1 9 4 3 8 6 2 7\n\n1 9 5 3 8 4 2 7 6\n8 4 3 7 6 2 9 5 1\n6 2 7 5 1 9 4 3 8\n\n-------------------------------\n\n2 7 6 5 1 9 8 4 3\n9 5 1 3 8 4 6 2 7\n4 3 8 7 6 2 1 9 5\n\n8 4 3 2 7 6 5 1 9\n6 2 7 9 5 1 3 8 4\n1 9 5 4 3 8 7 6 2\n\n5 1 9 8 4 3 2 7 6\n3 8 4 6 2 7 9 5 1\n7 6 2 1 9 5 4 3 8\n\nS1 and S2 cannot be transformed into each other by rotations or reflections because they have an identical central 3x3 sub-grid.\nS1 and S2 cannot be transformed into each other by permutations within a \"macro\" row/column because they have an identical \"macro\" diagonal.\nS1 and S2 cannot be transformed into each other by permutations of \"macro\" rows/columns because they are built out of different semi-magic squares." ]

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[ "## Skew symmetry property\n\nThe three types of derivatives: vectors-by-matrices, matrices-by-vectors, and matrices-by-matrices. These are not as widely considered and a notation is not widely agreed upon. However the second type (matrices-by-vectors) is widely used in Robotics. If we consider the following state form for a robot:&nbsp;&nbsp;&nbsp;&nbsp; The matrix is skew-symmetric.Here the matrix-by-vector derivative... [Read More]\nTags: Robotics\n\n## EqualX is a great alternative to MathType\n\nEqualX gives you the power and beauty for writing equations TeX/LaTeX in a simple to use editor.Type TeX or LaTeX Type your equation in the best typesetting language for equations, TeX. You can paste in your existing equations and mix TeX code with click-and-click.LaTeXDraw and Inkscape are best tools for... [Read More]\nTags: Tex\n\n## MailIP\n\n&nbsp;If you want to access to a your personal computer from anywhere via \"Windows Remote Desktop Connection\" you have to: firstly, configure your modem. Secondly, you must know your IP. Here I assume that you know how to configure your modem (if don't please see here) then you should use... [Read More]" ]

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[ "# Power Series Calculator\n\nTo evaluate the power series of a function with steps, input the function and other values in the given input box and hit calculate button using power series calculator.\n\n## Power Series Calculator\n\nFind the power series expansion and representation using the power series calculator with steps. It also gives a plot of approximation of x up to certain order.\n\nFor a concept as complex as power series, this tool does a fair job of making its calculation easy. You can convert any function into a power series using this calculator. The built-in keyboard aids in the input of special operations.\n\n## How to use this calculator?\n\n1. Enter the function. (Take help from the keyboard and examples.)\n2. Select the variable.\n3. Input point and order.\n4. Click Calculate.\n\nOrder means the power of the variable till which you want to find the series.\n\nLet’s learn more about the power series, its form, and how to expand and represent functions using power series with examples.\n\n## What is a power series?\n\nPower series is a type of series expansion of a function into polynomials with infinite terms. It is used by calculators and other computing devices for the evaluation of functions like exponential, logarithmic e.t.c\n\nSay you have two functions;\n\nIf you wanted to perform any kind of mathematical operation on them like addition or subtraction, How would you do it?\n\nThe easiest way is to convert these functions into the polynomial form and then apply the operation. But it cannot be done just like that. There are some rules.\n\nThis is where the power series is used. On expanding these functions using the power series formula, they look like this.\n\nNote: The (!) is factorial\n\nIt is now very easy to add, subtract, or divide these values.\n\nPower series are used in limits and approximate evaluation of integrals. Another important use of power series is error estimation.\n\n## Form of power series:\n\nAlso known as the formula of power series or its syntax, It basically tells what a power series looks like.\n\nIt can be written with the summation sign as well.\n\nHere:\n\n• c represents constant terms.\n• x is the variable of function f(x).\n• a is called the center.\n\nThis term a is subtracted from the variable because in general functions can have different centers. For instance, for a quadratic equation y = x2, the parabolic graph is:\n\nThe center is 0. But the function could have a different center. Like in the case of equation y = ( x - 3 )2, the graph is centered at 3.\n\nHence a is used for generalization so that such functions can be included as well.\n\n## How to represent functions in power series?\n\nIt is convenient to represent a function in the series form before expanding it as it serves as a formula. Remember that the domain of this function should be the interval of convergence.\n\nWe basically manipulate the power series representation of an easy and basic function, here the geometric series. It is represented as:\n\nThis series converges when the absolute value of x is less than one (i.e |x| < 1) and at this point its value is 1/(1 - x). So,\n\nThis is the power series representation because the domain is an interval of convergence at this certain value.\n\nNow, we can use this value to represent other functions. Such as:\n\nAs long as the absolute value of -x is less than one, it eventually means the same thing as |x|<1.\n\nFor a little complex function:\n\nSeparate out the x and it is almost already in the required form.\n\nTherefore writing to represent in power series.\n\n## How to expand the power series?\n\nEnter the numbers at the place of n and solve up to the required order. You can use the sum of the power series calculator as an alternative.\n\nExample:\n\nSolution:\n\nStep 1: Data interpretation.\n\nFunction = x / (1 - 5x2)\nPoint = 0\nOrder = 11\n\nStep 2: Represent in the power series form.\n\nStep 3: Enter the numbers in the expression and solve.\n\n= (5)0(x)2(0) + 1 = (1)(x)0+1 = x\n\n= (5)1(x)2(1) + 1 = (5)(x)2+1 = 5x3\n\n= (5)2(x)2(2) + 1 = (25)(x)4+1 = 25x5\n\n= (5)3(x)2(3) + 1 = (125)(x)6+1 = 125x7\n\n= (5)4(x)2(4) + 1 = (625)(x)8+1 = 625x9\n\n= (5)5(x)2(5) + 1 = (3125)(x)10+1 = 3125x11\n\n(the order 11 is computed at last)\n\nStep 4: Write in the series form.\n\n= x + 5x3 + 25x5 + 125x7 + 625x9 + 3125x11 + 0(x13)", null, "" ]

[ null, "https://www.meracalculator.com/images/push-icon.png", null ]

[ "# Visualizing Portfolio Volatility\n\nby Jonathan Regenstein\n\nThis is the third post in our series on portfolio volatility, variance and standard deviation. If you want to start at the beginning with calculating portfolio volatility, have a look at the first post here - Intro to Volatility. The second post on calculating rolling standard deviations is here: Intro to Rolling Volatility.\n\nToday we will visualize rolling standard deviations with highcharter using two objects from that second post.\n\nThe charts, which are the fun payoff after all of the equations and functions that we have ground out in the previous posts, should highlight any unusual occurrences or volatility spikes/dips that we might want to investigate.\n\nFirst, load the .RDat file saved from our previous Notebook (or you can run the scripts from the previous posts).\n\nload('rolling-sd.RDat')\n\nWe now have 2 objects in our Global Environment - spy_rolling_sd - an xts object of rolling SPY standard deviations - roll_portfolio_result - an xts object of rolling portfolio standard deviations. Because both of those are xts objects, we can pass them straight to highcharter with the hc_add_series() function and set a name and a color with the name and color arguments. Nothing too complicated here - we did the hard work in our previous Notebooks.\n\nhighchart(type = \"stock\") %>%\nhc_title(text = \"SPY v. Portfolio Rolling Volatility\") %>%\nhc_add_series(spy_rolling_sd, name = \"SPY Volatility\", color = \"blue\") %>%\nhc_add_series(roll_portfolio_result, name = \"Port Volatility\", color = \"green\") %>%\nhc_navigator(enabled = FALSE) %>%\nhc_scrollbar(enabled = FALSE)\n\nIt is interesting to note that from late April 2016 to late October 2016, SPY’s rolling standard deviation dipped below that of the diversified portfolio. The portfolio volatility was plunging at the same time, but SPY’s was falling faster. What happened over the 6 preceding months to explain this?\n\nMaybe we should add a flag to highlight this event. We can also add flags for the maximum SPY volatility, maximum and minimum portfolio rolling volatility and might as well include a line for the mean rolling volatility of SPY to practice adding horizontal lines.\n\nWe will use two methods for adding flags. First, we’ll hard code the date for the flag as “2016-04-29” using the date when rolling SPY volatility dipped below the portfolio. Second, we’ll set a flag with the date\nas.Date(index(roll_portfolio_result[which.max(roll_portfolio_result)]),format = \"%Y-%m-%d\") which looks like a convoluted mess but is adding a date for whenever the rolling portfolio standard deviation hit its maximum.\n\nThis is a bit more ‘dynamic’ because we can change our assets but keep this code the same and it will find the date with the maximum rolling standard deviation. Our first flag is not dynamic in the sense that it is specific to the comparison between SPY and this exact portfolio.\n\nspy_important_date <- as.Date(c(\"2016-04-29\"), format = \"%Y-%m-%d\")\n\nport_max_date <- as.Date(index(roll_portfolio_result[which.max(roll_portfolio_result)]),\nformat = \"%Y-%m-%d\")\nport_min_date <- as.Date(index(roll_portfolio_result[which.min(roll_portfolio_result)]),\nformat = \"%Y-%m-%d\")\nspy_max_date <- as.Date(index(spy_rolling_sd[which.max(spy_rolling_sd)]),\nformat = \"%Y-%m-%d\")\n\nhighchart(type = \"stock\") %>%\nhc_title(text = \"SPY v. Portfolio Rolling Volatility\") %>%\nhc_add_series(spy_rolling_sd, name = \"SPY Volatility\", color = \"blue\", id = \"SPY\") %>%\nhc_add_series(roll_portfolio_result, name = \"Portf Volatility\", color = \"green\", id = \"Port\") %>%\ntitle = c(\"SPY Vol Dips\"),\ntext = c(\"SPY rolling sd dips below portfolio.\"),\nid = \"SPY\") %>%\ntitle = c(\"SPY Max \"),\ntext = c(\"SPY max rolling volatility.\"),\nid = \"SPY\") %>%\ntitle = c(\"Portf Max\"),\ntext = c(\"Portfolio maximum rolling volatility.\"),\nid = \"Port\") %>%\ntitle = c(\"Portf Min\"),\ntext = c(\"Portfolio min rolling volatility.\"),\nid = \"Port\") %>%\nhc_yAxis(title = list(text = \"Mean SPY rolling Vol\"),\nshowFirstLabel = FALSE,\nshowLastLabel = FALSE,\nplotLines = list(\nlist(value = mean(spy_rolling_sd), color = \"#2b908f\", width = 2))) %>%\nhc_navigator(enabled = FALSE) %>%\nhc_scrollbar(enabled = FALSE)\n\nHover on the flags and you can see the text we added for explanation.\n\nIt’s remarkable how rolling volatility has absolutely plunged since early-to-mid 2016. Since August of 2016, both the portfolio and SPY rolling standard deviations have been well below the SPY mean.\n\nThanks for sticking with this three-part introduction to volatility. Next time, we’ll port our work to Shiny and play with different assets and allocations.\n\nShare Comments · · · ·" ]

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[ "424,984 Members | 1,461 Online", null, "Need help? Post your question and get tips & solutions from a community of 424,984 IT Pros & Developers. It's quick & easy.\n\nObtaining digits from bit string\n\n P: n/a Im working on an embedded project and i have to obtain individual digits from a byte thats the output of a device. Im trying to output this info to an LCD and need to convert it to ASCII first. I have the following which only works for single digits but not double digits. (Data from chip is stored in current[] array and needs to be converted and stored in array hours[]). hours = 0x30 | ((current & 0x70) >> 4); //obtain the tens minute digit //current stores the byte i want to convert, hours = 0x30 | (current & 0x0F) //obtain the ones minute digit. ; This only works if i have single digit output but i a value like 36 minutes was read from the device, the output goes crazy. I appreciate any suggestions/ideas. zion. Apr 15 '06 #1\n5 Replies\n\n P: n/a On 14 Apr 2006 19:17:14 -0700, \"zion_zii\" wrote in comp.lang.c: Im working on an embedded project and i have to obtain individual digits from a byte thats the output of a device. Im trying to output this info to an LCD and need to convert it to ASCII first. I have the following which only works for single digits but not double digits. (Data from chip is stored in current[] array and needs to be converted and stored in array hours[]). hours = 0x30 | ((current & 0x70) >> 4); //obtain the tens minute digit //current stores the byte i want to convert, hours = 0x30 | (current & 0x0F) //obtain the ones minute digit. ; This only works if i have single digit output but i a value like 36 minutes was read from the device, the output goes crazy. I appreciate any suggestions/ideas. zion. What are the types of \"current\" and \"hours\"? What are the values in \"current\" when you get what you consider correct output, and what is that correct output? What are the values in \"current\" when you get what you consider \"crazy\" output, and what is that output? Exactly how is the format of the data in \"current\" defined? -- Jack Klein Home: http://JK-Technology.Com FAQs for comp.lang.c http://c-faq.com/ comp.lang.c++ http://www.parashift.com/c++-faq-lite/ alt.comp.lang.learn.c-c++ http://www.contrib.andrew.cmu.edu/~a...FAQ-acllc.html Apr 15 '06 #2\n\n P: n/a I apologize for being so vague. current and hours are both int ( 8 - bit ints) arrays. The correct output appears when i have single digits. Prior to what is shown above, the data is gathered from a terminal and fed into a chip (ds1305), where a few minuites later, it is read from the same chip and stored in current[]. The correct output appears when i have something like 8 minutes for example (actually, any single digit). When i have a double digit (> 9), it doesnt work. I should also mention that the most sign bit of the byte just read is immaterial and doesnt do anything. To display this value to the lcd, it has to be in ASCII (thats where the 0x30 comes in). So for single digits that are less than 10, i have no problem (ds1305 outputs in bcd). Its when i need to determine the 10's digit that i fail. This is also happening with other bytes but if i get a soln for one i can change the others accordingly. Thanks Jack. Apr 15 '06 #3\n\n P: n/a zion_zii wrote: I apologize for being so vague. current and hours are both int ( 8 - bit ints) arrays. The correct output appears when i have single digits. Prior to what is shown above, the data is gathered from a terminal and fed into a chip (ds1305), where a few minuites later, it is read from the same chip and stored in current[]. The correct output appears when i have something like 8 minutes for example (actually, any single digit). When i have a double digit (> 9), it doesnt work. I should also mention that the most sign bit of the byte just read is immaterial and doesnt do anything. To display this value to the lcd, it has to be in ASCII (thats where the 0x30 comes in). So for single digits that are less than 10, i have no problem (ds1305 outputs in bcd). Its when i need to determine the 10's digit that i fail. This is also happening with other bytes but if i get a soln for one i can change the others accordingly. Thanks Jack. The request way for sample data (since your code does not work) //Looks like packed BCD No? // 0x30 = '0' it reads better //current stores the byte i want to convert, hours = '0' + (current >> 4); //obtain the tens minute digit hours = '0' + (current & 0x0F) //obtain the ones minute digit. Apr 15 '06 #4\n\n P: n/a Never mind guys. I figured it out. Just divide the num by 10 to get the tens and subtract the tens * 10 from the initial val to get the ones. Thanks though. zion. Apr 15 '06 #5\n\n P: n/a zion_zii wrote: Never mind guys. I figured it out. Just divide the num by 10 to get the tens and subtract the tens * 10 from the initial val to get the ones. Thanks though. Earlier you said that the data was obtained from a chip in BCD. Are you now saying that the number has already been converted to binary, which you didn't realize before, before your code converts it to ASCII decimal? Otherwise, your explanation doesn't make sense. -- Thad Apr 16 '06 #6", null, "" ]

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[ "Risk of Ruin Calculator\n\n# Risk of Ruin Calculator", null, "Moneyzine Editor\nLast updated 4th Oct 2022\nDisclosure\n\nThis online tool is able to calculate the risk of ruin, which is also referred to as a drawdown calculation. The calculator takes into consideration the funds placed at risk, the dollars per transaction, the investor's threshold for ruin, as well as their probability of a positive outcome.\n\n## Calculator Definitions\n\nThe variables used in our online calculator are defined in detail below, including how to interpret the results.\n\n### Total Funds at Risk (\\$)\n\nThis is the total dollars the investor is willing to place at risk, which may be a dedicated portfolio of securities.\n\n### Funds at Risk per Transaction (\\$)\n\nThis is the dollars placed at risk during each transaction. For example, if the investor is able to lose \\$2,000 per transaction, then that is the value to enter in this cell.\n\n### Threshold for Ruin (\\$)\n\nThe threshold for ruin is the value at which the investor would consider the portfolio lost. For example, the investor might have \\$100,000 in the portfolio and is willing to lose up to \\$40,000 (drawdown) before the portfolio is considered failed. In this example, the Threshold for Ruin is \\$60,000.\n\n### The Probability of a Win (%) / The Probability of a Loss (%)\n\nThis value represents the likelihood of a good outcome for each transaction. If a trader is successful 52% of the time when conducting a transaction, then the probability of a win is 52%. Conversely, the probability of a loss would be 100% - 52%, or 48%.\n\n### Risk of Ruin (%)\n\nGiven the variables entered into this calculator, this is the likelihood the portfolio would reach a state of ruin. For example, if the Total Funds at Risk were \\$100,000, the Funds at Risk per Transaction were \\$2,000, and the Probability of a Win is 52%, the investor has a 20% chance the portfolio would be considered ruined.\n\nRisk of Ruin Calculator - Money-Zine.com\n\nDisclaimer: These calculators are made available and meant to be used as a screening tool for the investor. The accuracy of these calculator results is not guaranteed nor is its applicability to your individual circumstances. You should always obtain personal advice from qualified professionals.", null, "" ]

[ null, "https://cdn.moneyzine.com/tr:fo-auto,f-auto,w-30,h-30/uploads/2022/04/1650468271-Moneyzine%20app%20icon.png", null, "https://cdn.moneyzine.com/tr:fo-auto,f-auto,w-135,h-135/uploads/2022/04/1650468271-Moneyzine%20app%20icon.png", null ]

[ "Scilab Home page | Wiki | Bug tracker | Forge | Mailing list archives | ATOMS | File exchange\nChange language to: Français - Português - 日本語 - Русский\n\n# zgrid\n\nzgrid plot\n\n### Calling Sequence\n\n```zgrid()\nzgrid(zeta,wn [,colors])\nzgrid(['new',] zeta,wn [,colors])\nzgrid(zeta,wn [,'new'] [,colors])```\n\n### Arguments\n\nzeta\n\narray of damping factors. Only values in ```[0 1]``` are taken into account. The default value is `0:0.1:1`.\n\nwn\n\narray of normalized natural frequencies (factors of π/dt). Only values in `[0 1]` are taken into account. The default value is `0:0.1:1`.\n\ncolors\n\na scalar or an 2 element array with integer values (color index).\n\n### Description\n\nplots z-plane grid lines: lines of constant damping factor (selection given by `zeta`) and natural frequency (selection given by `wn`) are drawn in within the unit Z-plane circle.\n\nIso-frequency curves are shown in on the interval [0,%pi/dt].\n\nThe `colors` argument may be used to assign a color for constant damping ratio curves (`colors(2)`) and for frequency curves (`colors(1)`).\n\nThe `zgrid` function is often used to draw a grid for evens root locus of continuous time linear systems. In such a case the `zgrid` function should be called after the call to evans. For continuous time linear systems one should use sgrid function instead.\n\nThe optional argument `'new'` can be used to erase the graphic window before plotting the grid.\n\n### Examples\n\n```//zgrid\nclf();zgrid(0:0.2:1,[0.2 0.6 0.8 1])```", null, "```//zgrid with discrete time system root locus\nz=poly(0,'z')\nH=syslin(0.01,(0.54-1.8*z+2.9*z^2-2.6*z^3+z^4)/(0.8+0.78*z-0.1*z^2+0.9*z^3+z^4))\nclf();evans(H,1000);zgrid(0:0.1:0.5)```", null, "• evans — Evans root locus\n• sgrid — s-plane grid lines.\n• freson — peak frequencies\n• datatips — Tool for placing and editing tips along the plotted curves." ]

[ null, "https://help.scilab.org/docs/5.5.0/en_US/zgrid_1.png", null, "https://help.scilab.org/docs/5.5.0/en_US/zgrid_en_US_2.png", null ]

[ "# chain rule in complex analysis\n\nConsider the function f : C → R given by f(z) = |z|2. For the usual real derivative, there are several rules such as the product rule, the chain rule, the quotient rule and the inverse rule. Worked example: Derivative of ∜(x³+4x²+7) using the chain rule. Having inspired from this discussion, I want to share my understanding of the subject and eventually present a chain rule for complex derivatives. c FW Math 321, 2012/12/11 Elements of Complex Calculus 1 Basics of Series and Complex Numbers 1.1 Algebra of Complex numbers ... all integer n6= 1. De nition 1.1 (Chain Complex). 3.2 Cauchy’s theorem Worked example: Derivative of sec(3π/2-x) using the chain rule. (3) (Chain Rule) d dz f(g(z)) = f0(g(z))g0(z) whenever all the terms make sense. So much for similarity. This line passes through the point . Two days ago in Julia Lab, Jarrett, Spencer, Alan and I discussed the best ways of expressing derivatives for automatic differentiation in complex-valued programs. Since z = To see the difference of complex derivatives and the derivatives of functions of two real variables we look at the following example. The chain rule is a method for finding the derivative of composite functions, or functions that are made by combining one or more functions.An example of one of these types of functions is \\(f(x) = (1 + x)^2\\) which is formed by taking the function \\(1+x\\) and plugging it into the function \\(x^2\\). Practice: Chain rule capstone. Suppose is differentiable at , and is differentiable at , and that is a limit point of . All the usual rules of di erentiation: product rule, quotient rule, chain rule,..., still apply for complex di erentiation … 10.13 Theorem (Chain Rule.) The chain rule gives us that the derivative of h is . Then the composition is differentiable at , and It is then not possible to differentiate them directly as we do with simple functions.In this topic, we shall discuss the differentiation of such composite functions using the Chain Rule. Example 2. Contents Preface 6 1 Algebraic properties of complex numbers 8 2 Topological properties of C 18 3 Di erentiation 26 4 Path integrals 38 5 Power series 43 Complex Analysis Mario Bonk Course notes for Math 246A and 246B University of California, Los Angeles Fall 2011 and Winter 2012. Thus, the slope of the line tangent to the graph of h at x=0 is . Click HERE to return to the list of problems. Next lesson. This is the currently selected item. For example, if = a+ biis a complex number, then applying the chain rule to the analytic function f(z) = ez and z(t) = t= at+ (bt)i, we see that d dt e t= e t: 3. Sort by: Top Voted. Fortunately, these carry over verbatim to the complex derivative, and even the proofs remain the same (although … Let be complex functions, and let . Using the point-slope form of a line, an equation of this tangent line is or . Derivative rules review. Proving the chain rule. Chain rule capstone. Sometimes complex looking functions can be greatly simplified by expressing them as a composition of two or more different functions. A chain complex is a set of objects fC ngin a category like vector spaces, abelian groups, R-mod, and graded R-mod, with d n: C n!C n 1 maps such that the kernel of d n is Z n, the n-cycles of C, the image of d n+1 is B n, the n-boundaries of C and H n(C) = Z n=B nis the kernel Implicit differentiation. This is sometimes called the chain rule for analytic functions." ]

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[ "Vibrations and Waves - Lesson 2 - Properties of a Wave\n\n# The Wave Equation\n\nAs was discussed in Lesson 1, a wave is produced when a vibrating source periodically disturbs the first particle of a medium. This creates a wave pattern that begins to travel along the medium from particle to particle. The frequency at which each individual particle vibrates is equal to the frequency at which the source vibrates. Similarly, the period of vibration of each individual particle in the medium is equal to the period of vibration of the source. In one period, the source is able to displace the first particle upwards from rest, back to rest, downwards from rest, and finally back to rest. This complete back-and-forth movement constitutes one complete wave cycle.", null, "The diagrams at the right show several \"snapshots\" of the production of a wave within a rope. The motion of the disturbance along the medium after every one-fourth of a period is depicted. Observe that in the time it takes from the first to the last snapshot, the hand has made one complete back-and-forth motion. A period has elapsed. Observe that during this same amount of time, the leading edge of the disturbance has moved a distance equal to one complete wavelength. So in a time of one period, the wave has moved a distance of one wavelength. Combining this information with the equation for speed (speed = distance/time), it can be said that the speed of a wave is also the wavelength/period.", null, "Since the period is the reciprocal of the frequency, the expression 1/f can be substituted into the above equation for period. Rearranging the equation yields a new equation of the form:\n\nSpeed = Wavelength • Frequency\n\nThe above equation is known as the wave equation. It states the mathematical relationship between the speed (v) of a wave and its wavelength (λ) and frequency (f). Using the symbols v, λ, and f, the equation can be rewritten as\n\nv = f • λ\n\nAs a test of your understanding of the wave equation and its mathematical use in analyzing wave motion, consider the following three-part question:\n\nStan and Anna are conducting a slinky experiment. They are studying the possible effect of several variables upon the speed of a wave in a slinky. Their data table is shown below. Fill in the blanks in the table, analyze the data, and answer the following questions.\n\n Medium Wavelength Frequency Speed Zinc, 1-in. dia. coils 1.75 m 2.0 Hz ______ Zinc, 1-in. dia. coils 0.90 m 3.9 Hz ______ Copper, 1-in. dia. coils 1.19 m 2.1 Hz ______ Copper, 1-in. dia. coils 0.60 m 4.2 Hz ______ Zinc, 3-in. dia. coils 0.95 m 2.2 Hz ______ Zinc, 3-in. dia. coils 1.82 m 1.2 Hz ______\n\n1.  As the wavelength of a wave in a uniform medium increases, its speed will _____.\n\n a. decrease b. increase c. remain the same\n\n2. As the wavelength of a wave in a uniform medium increases, its frequency will _____.\n\n a. decrease b. increase c. remain the same\n\n3. The speed of a wave depends upon (i.e., is causally affected by) ...\n\na. the properties of the medium through which the wave travels\n\nb. the wavelength of the wave.\n\nc. the frequency of the wave.\n\nd. both the wavelength and the frequency of the wave.\n\nThe above example illustrates how to use the wave equation to solve mathematical problems. It also illustrates the principle that wave speed is dependent upon medium properties and independent of wave properties. Even though the wave speed is calculated by multiplying wavelength by frequency, an alteration in wavelength does not affect wave speed. Rather, an alteration in wavelength affects the frequency in an inverse manner. A doubling of the wavelength results in a halving of the frequency; yet the wave speed is not changed.\n\n1. Two waves on identical strings have frequencies in a ratio of 2 to 1. If their wave speeds are the same, then how do their wavelengths compare?\n\n a. 2:1 b. 1:2 c. 4:1 d. 1:4\n\n2. Mac and Tosh stand 8 meters apart and demonstrate the motion of a transverse wave on a snakey. The wave e can be described as having a vertical distance of 32 cm from a trough to a crest, a frequency of 2.4 Hz, and a horizontal distance of 48 cm from a crest to the nearest trough. Determine the amplitude, period, and wavelength and speed of such a wave.\n\n3. Dawn and Aram have stretched a slinky between them and begin experimenting with waves. As the frequency of the waves is doubled,\n\na. the wavelength is halved and the speed remains constant\n\nb. the wavelength remains constant and the speed is doubled\n\nc. both the wavelength and the speed are halved.\n\nd. both the wavelength and the speed remain constant.\n\n4. A ruby-throated hummingbird beats its wings at a rate of about 70 wing beats per second.\n\na. What is the frequency in Hertz of the sound wave?\n\nb. Assuming the sound wave moves with a velocity of 350 m/s, what is the wavelength of the wave?\n\n5. Ocean waves are observed to travel along the water surface during a developing storm. A Coast Guard weather station observes that there is a vertical distance from high point to low point of 4.6 meters and a horizontal distance of 8.6 meters between adjacent crests. The waves splash into the station once every 6.2 seconds. Determine the frequency and the speed of these waves.", null, "6. Two boats are anchored 4 meters apart. They bob up and down, returning to the same up position every 3 seconds. When one is up the other is down. There are never any wave crests between the boats. Calculate the speed of the waves." ]

[ null, "http://www.physicsclassroom.com/Class/waves/u10l2e1.gif", null, "http://www.physicsclassroom.com/Class/waves/u10l2e2.gif", null, "http://www.physicsclassroom.com/Class/waves/u10l2e3.gif", null ]

[ "# murophobia in Scrabble®\n\nThe word murophobia is playable in Scrabble®, no blanks required. Because it is longer than 7 letters, you would have to play off an existing word or do it in several moves.\n\nMUROPHOBIA\n(180)\nMUROPHOBIA\n(180)\n\nbimorph, bohrium\n\nMUROPHOBIA\n(180)\nMUROPHOBIA\n(180)\nMUROPHOBIA\n(120)\nMUROPHOBIA\n(120)\nMUROPHOBIA\n(92)\nMUROPHOBIA\n(88)\nMUROPHOBIA\n(76)\nMUROPHOBIA\n(76)\nMUROPHOBIA\n(76)\nMUROPHOBIA\n(76)\nMUROPHOBIA\n(69)\nMUROPHOBIA\n(66)\nMUROPHOBIA\n(63)\nMUROPHOBIA\n(62)\nMUROPHOBIA\n(60)\nMUROPHOBIA\n(58)\nMUROPHOBIA\n(58)\nMUROPHOBIA\n(54)\nMUROPHOBIA\n(52)\nMUROPHOBIA\n(50)\nMUROPHOBIA\n(48)\nMUROPHOBIA\n(46)\nMUROPHOBIA\n(46)\nMUROPHOBIA\n(46)\nMUROPHOBIA\n(44)\nMUROPHOBIA\n(42)\nMUROPHOBIA\n(42)\nMUROPHOBIA\n(42)\nMUROPHOBIA\n(42)\nMUROPHOBIA\n(40)\nMUROPHOBIA\n(40)\nMUROPHOBIA\n(40)\nMUROPHOBIA\n(38)\nMUROPHOBIA\n(38)\nMUROPHOBIA\n(33)\nMUROPHOBIA\n(31)\nMUROPHOBIA\n(27)\nMUROPHOBIA\n(27)\nMUROPHOBIA\n(26)\nMUROPHOBIA\n(25)\nMUROPHOBIA\n(24)\nMUROPHOBIA\n(24)\nMUROPHOBIA\n(23)\nMUROPHOBIA\n(22)\n\nMUROPHOBIA\n(180)\nMUROPHOBIA\n(180)\nMUROPHOBIA\n(120)\nMUROPHOBIA\n(120)\nBIMORPH\n(114 = 64 + 50)\nBIMORPH\n(110 = 60 + 50)\nBIMORPH\n(107 = 57 + 50)\nBIMORPH\n(107 = 57 + 50)\nBIMORPH\n(107 = 57 + 50)\nBOHRIUM\n(106 = 56 + 50)\nBOHRIUM\n(104 = 54 + 50)\nBIMORPH\n(101 = 51 + 50)\nBOHRIUM\n(101 = 51 + 50)\nBOHRIUM\n(101 = 51 + 50)\nBIMORPH\n(101 = 51 + 50)\nBIMORPH\n(101 = 51 + 50)\nBIMORPH\n(101 = 51 + 50)\nBIMORPH\n(98 = 48 + 50)\nBOHRIUM\n(95 = 45 + 50)\nBOHRIUM\n(95 = 45 + 50)\nBOHRIUM\n(95 = 45 + 50)\nBOHRIUM\n(95 = 45 + 50)\nBOHRIUM\n(95 = 45 + 50)\nBOHRIUM\n(94 = 44 + 50)\nBIMORPH\n(94 = 44 + 50)\nBIMORPH\n(94 = 44 + 50)\nBIMORPH\n(94 = 44 + 50)\nBIMORPH\n(92 = 42 + 50)\nBOHRIUM\n(92 = 42 + 50)\nBOHRIUM\n(92 = 42 + 50)\nMUROPHOBIA\n(92)\nBIMORPH\n(90 = 40 + 50)\nBIMORPH\n(90 = 40 + 50)\nBIMORPH\n(88 = 38 + 50)\nBIMORPH\n(88 = 38 + 50)\nMUROPHOBIA\n(88)\nBIMORPH\n(88 = 38 + 50)\nBIMORPH\n(88 = 38 + 50)\nBOHRIUM\n(86 = 36 + 50)\nBIMORPH\n(86 = 36 + 50)\nBOHRIUM\n(86 = 36 + 50)\nBIMORPH\n(86 = 36 + 50)\nBOHRIUM\n(84 = 34 + 50)\nBIMORPH\n(84 = 34 + 50)\nBOHRIUM\n(84 = 34 + 50)\nBOHRIUM\n(84 = 34 + 50)\nBIMORPH\n(84 = 34 + 50)\nBOHRIUM\n(84 = 34 + 50)\nBOHRIUM\n(82 = 32 + 50)\nBIMORPH\n(82 = 32 + 50)\nBIMORPH\n(82 = 32 + 50)\nBIMORPH\n(82 = 32 + 50)\nBOHRIUM\n(82 = 32 + 50)\nBOHRIUM\n(82 = 32 + 50)\nBIMORPH\n(82 = 32 + 50)\nBIMORPH\n(82 = 32 + 50)\nBOHRIUM\n(80 = 30 + 50)\nBIMORPH\n(80 = 30 + 50)\nBOHRIUM\n(80 = 30 + 50)\nBOHRIUM\n(80 = 30 + 50)\nBOHRIUM\n(78 = 28 + 50)\nBOHRIUM\n(78 = 28 + 50)\nBOHRIUM\n(78 = 28 + 50)\nBOHRIUM\n(78 = 28 + 50)\nBOHRIUM\n(78 = 28 + 50)\nBOHRIUM\n(78 = 28 + 50)\nMUROPHOBIA\n(76)\nBIMORPH\n(76 = 26 + 50)\nMUROPHOBIA\n(76)\nMUROPHOBIA\n(76)\nMUROPHOBIA\n(76)\nBIMORPH\n(74 = 24 + 50)\nBIMORPH\n(74 = 24 + 50)\nBIMORPH\n(74 = 24 + 50)\nBOHRIUM\n(74 = 24 + 50)\nBIMORPH\n(73 = 23 + 50)\nBOHRIUM\n(72 = 22 + 50)\nBOHRIUM\n(71 = 21 + 50)\nBOHRIUM\n(71 = 21 + 50)\nBIMORPH\n(70 = 20 + 50)\nBIMORPH\n(70 = 20 + 50)\nBIMORPH\n(70 = 20 + 50)\nBIMORPH\n(70 = 20 + 50)\nMUROPHOBIA\n(69)\nBOHRIUM\n(69 = 19 + 50)\nBOHRIUM\n(68 = 18 + 50)\nBIMORPH\n(68 = 18 + 50)\nBOHRIUM\n(68 = 18 + 50)\nBIMORPH\n(68 = 18 + 50)\nBIMORPH\n(67 = 17 + 50)\nMUROPHOBIA\n(66)\nBOHRIUM\n(66 = 16 + 50)\nBOHRIUM\n(66 = 16 + 50)\nBOHRIUM\n(66 = 16 + 50)\nBOHRIUM\n(66 = 16 + 50)\nBOHRIUM\n(65 = 15 + 50)\nMUROPHOBIA\n(63)\nMUROPHOBIA\n(62)\nMUROPHOBIA\n(60)\nMUROPHOBIA\n(58)\nMUROPHOBIA\n(58)\nMUROPHOBIA\n(54)\nMUROPHOBIA\n(52)\nBARHOP\n(51)\nPHOBIA\n(51)\nRHOMBI\n(51)\nAMORPH\n(51)\nMUROPHOBIA\n(50)\nMUROPHOBIA\n(48)\nABMHO\n(48)\nAMORPH\n(48)\nRHOMBI\n(48)\nRHOMBI\n(48)\nABOHM\n(48)\nRHOMB\n(48)\nBARHOP\n(48)\nBARHOP\n(48)\nAMORPH\n(48)\nPHOBIA\n(48)\nPHOBIA\n(48)\nMORPH\n(48)\nMUROPHOBIA\n(46)\nMUROPHOBIA\n(46)\nMUROPHOBIA\n(46)\nMOHAIR\n(45)\nABOHM\n(45)\nHUMP\n(45)\nABMHO\n(45)\nMORPH\n(45)\nRUPIAH\n(45)\nMORPH\n(45)\nRHOMB\n(45)\nRHOMB\n(45)\nABOHM\n(45)\nMUROPHOBIA\n(44)\nAMORPH\n(42)\nAMORPH\n(42)\nAMORPH\n(42)\nRUPIAH\n(42)\nRHOMBI\n(42)\nBARHOP\n(42)\nAMORPH\n(42)\nPHUB\n(42)\nBARHOP\n(42)\nMUROPHOBIA\n(42)\nMUROPHOBIA\n(42)\nMUROPHOBIA\n(42)\nMUROPHOBIA\n(42)\nBARHOP\n(42)\nPHOBIA\n(42)\nPHOBIA\n(42)\nPHOBIA\n(42)\nPHUB\n(42)\nMOHAIR\n(42)\nPHOBIA\n(42)\nRHOMBI\n(42)\nRHOMBI\n(42)\nRHOMBI\n(42)\nHUMP\n(42)\nHUMOR\n(42)\nMORPH\n(40)\nMUROPHOBIA\n(40)\nMUROPHOBIA\n(40)\nMUROPHOBIA\n(40)\nBUMP\n(39)\nBUMP\n(39)\nABOHM\n(39)\nPHOBIA\n(39)\nABMHO\n(39)\nPOROMA\n(39)\nRHOMBI\n(39)\nRHOMBI\n(39)\nBARHOP\n(39)\nBARHOP\n(39)\nPOOH\n(39)\nBARIUM\n(39)\nABHOR\n(39)\nABMHO\n(39)\nHOBO\n(39)\nMOHAR\n(39)\nHOOP\n(39)\nBARIUM\n(39)\nAMORPH\n(39)\nHARP\n(39)\nHARM\n(39)\nPOROMA\n(39)\nPHOBIA\n(39)\nAMORPH\n(39)\nMORPH\n(39)\nRHOMB\n(39)\nBARHOP\n(38)\nRHOMBI\n(38)\nBARHOP\n(38)\nAMORPH\n(38)\nAMORPH\n(38)\nMUROPHOBIA\n(38)\nRUPIAH\n(38)\nPHOBIA\n(38)\nMUROPHOBIA\n(38)\nMOHO\n(36)\n\n# murophobia in Words With Friends™\n\nThe word murophobia is playable in Words With Friends™, no blanks required. Because it is longer than 7 letters, you would have to play off an existing word or do it in several moves.\n\nMUROPHOBIA\n(288)\n\nbimorph, bohrium\n\nMUROPHOBIA\n(288)\nMUROPHOBIA\n(270)\nMUROPHOBIA\n(144)\nMUROPHOBIA\n(144)\nMUROPHOBIA\n(108)\nMUROPHOBIA\n(104)\nMUROPHOBIA\n(104)\nMUROPHOBIA\n(96)\nMUROPHOBIA\n(96)\nMUROPHOBIA\n(92)\nMUROPHOBIA\n(92)\nMUROPHOBIA\n(92)\nMUROPHOBIA\n(88)\nMUROPHOBIA\n(88)\nMUROPHOBIA\n(84)\nMUROPHOBIA\n(78)\nMUROPHOBIA\n(76)\nMUROPHOBIA\n(64)\nMUROPHOBIA\n(64)\nMUROPHOBIA\n(64)\nMUROPHOBIA\n(60)\nMUROPHOBIA\n(60)\nMUROPHOBIA\n(56)\nMUROPHOBIA\n(48)\nMUROPHOBIA\n(44)\nMUROPHOBIA\n(44)\nMUROPHOBIA\n(44)\nMUROPHOBIA\n(44)\nMUROPHOBIA\n(34)\nMUROPHOBIA\n(32)\nMUROPHOBIA\n(31)\nMUROPHOBIA\n(31)\nMUROPHOBIA\n(29)\nMUROPHOBIA\n(28)\nMUROPHOBIA\n(28)\nMUROPHOBIA\n(28)\nMUROPHOBIA\n(28)\nMUROPHOBIA\n(28)\nMUROPHOBIA\n(27)\nMUROPHOBIA\n(27)\nMUROPHOBIA\n(27)\nMUROPHOBIA\n(26)\nMUROPHOBIA\n(26)\nMUROPHOBIA\n(25)\nMUROPHOBIA\n(24)\nMUROPHOBIA\n(24)\n\nMUROPHOBIA\n(288)\nMUROPHOBIA\n(270)\nMUROPHOBIA\n(144)\nMUROPHOBIA\n(144)\nBIMORPH\n(137 = 102 + 35)\nBOHRIUM\n(125 = 90 + 35)\nBIMORPH\n(119 = 84 + 35)\nBOHRIUM\n(113 = 78 + 35)\nBIMORPH\n(113 = 78 + 35)\nBIMORPH\n(113 = 78 + 35)\nBIMORPH\n(113 = 78 + 35)\nMUROPHOBIA\n(108)\nBIMORPH\n(107 = 72 + 35)\nBOHRIUM\n(107 = 72 + 35)\nBIMORPH\n(107 = 72 + 35)\nBOHRIUM\n(107 = 72 + 35)\nBIMORPH\n(107 = 72 + 35)\nBIMORPH\n(107 = 72 + 35)\nMUROPHOBIA\n(104)\nMUROPHOBIA\n(104)\nBOHRIUM\n(101 = 66 + 35)\nBIMORPH\n(101 = 66 + 35)\nBOHRIUM\n(99 = 64 + 35)\nBOHRIUM\n(99 = 64 + 35)\nBOHRIUM\n(99 = 64 + 35)\nMUROPHOBIA\n(96)\nMUROPHOBIA\n(96)\nBOHRIUM\n(95 = 60 + 35)\nBIMORPH\n(95 = 60 + 35)\nBIMORPH\n(95 = 60 + 35)\nBOHRIUM\n(95 = 60 + 35)\nBIMORPH\n(95 = 60 + 35)\nMUROPHOBIA\n(92)\nMUROPHOBIA\n(92)\nMUROPHOBIA\n(92)\nBOHRIUM\n(89 = 54 + 35)\nBOHRIUM\n(89 = 54 + 35)\nBOHRIUM\n(89 = 54 + 35)\nMUROPHOBIA\n(88)\nMUROPHOBIA\n(88)\nBIMORPH\n(87 = 52 + 35)\nBIMORPH\n(87 = 52 + 35)\nBIMORPH\n(87 = 52 + 35)\nBARHOP\n(84)\nMUROPHOBIA\n(84)\nBIMORPH\n(83 = 48 + 35)\nBOHRIUM\n(83 = 48 + 35)\nBOHRIUM\n(83 = 48 + 35)\nBOHRIUM\n(79 = 44 + 35)\nBIMORPH\n(79 = 44 + 35)\nBIMORPH\n(79 = 44 + 35)\nBIMORPH\n(79 = 44 + 35)\nMUROPHOBIA\n(78)\nBIMORPH\n(77 = 42 + 35)\nMUROPHOBIA\n(76)\nBOHRIUM\n(75 = 40 + 35)\nMOHAIR\n(75)\nBIMORPH\n(75 = 40 + 35)\nBIMORPH\n(75 = 40 + 35)\nBOHRIUM\n(75 = 40 + 35)\nBOHRIUM\n(75 = 40 + 35)\nBOHRIUM\n(73 = 38 + 35)\nBIMORPH\n(73 = 38 + 35)\nBIMORPH\n(73 = 38 + 35)\nRHOMBI\n(72)\nPHOBIA\n(72)\nBARHOP\n(72)\nPHOBIA\n(72)\nBOHRIUM\n(71 = 36 + 35)\nBIMORPH\n(71 = 36 + 35)\nBIMORPH\n(71 = 36 + 35)\nBIMORPH\n(71 = 36 + 35)\nBIMORPH\n(71 = 36 + 35)\nBOHRIUM\n(71 = 36 + 35)\nBIMORPH\n(71 = 36 + 35)\nBOHRIUM\n(71 = 36 + 35)\nBIMORPH\n(71 = 36 + 35)\nBIMORPH\n(71 = 36 + 35)\nBARIUM\n(69)\nBOHRIUM\n(69 = 34 + 35)\nBARIUM\n(69)\nBOHRIUM\n(69 = 34 + 35)\nBOHRIUM\n(67 = 32 + 35)\nBOHRIUM\n(67 = 32 + 35)\nBOHRIUM\n(67 = 32 + 35)\nBOHRIUM\n(67 = 32 + 35)\nBIMORPH\n(67 = 32 + 35)\nBOHRIUM\n(67 = 32 + 35)\nBOHRIUM\n(67 = 32 + 35)\nBOHRIUM\n(67 = 32 + 35)\nBARHOP\n(66)\nPHOBIA\n(66)\nAMORPH\n(66)\nRUPIAH\n(66)\nBARHOP\n(66)\nPHOBIA\n(66)\nPOROMA\n(66)\nRHOMBI\n(66)\nAMORPH\n(66)\nBUMP\n(66)\nAMORPH\n(66)\nRHOMBI\n(66)\nBUMP\n(66)\nBOHRIUM\n(65 = 30 + 35)\nMUROPHOBIA\n(64)\nMUROPHOBIA\n(64)\nBIMORPH\n(64 = 29 + 35)\nMUROPHOBIA\n(64)\nMORPH\n(63)\nBIMORPH\n(63 = 28 + 35)\nMORPH\n(63)\nRHOMB\n(63)\nRHOMB\n(63)\nBIMORPH\n(63 = 28 + 35)\nBARIUM\n(63)\nHUMP\n(63)\nABMHO\n(63)\nPHUB\n(63)\nBIMORPH\n(63 = 28 + 35)\nBIMORPH\n(63 = 28 + 35)\nABOHM\n(63)\nBARIUM\n(63)\nPHUB\n(63)\nABOHM\n(63)\nBOHRIUM\n(62 = 27 + 35)\nBOHRIUM\n(62 = 27 + 35)\nBIMORPH\n(62 = 27 + 35)\nBIMORPH\n(61 = 26 + 35)\nBOHRIUM\n(61 = 26 + 35)\nBIMORPH\n(61 = 26 + 35)\nBOHRIUM\n(60 = 25 + 35)\nRHOMBI\n(60)\nBIMORPH\n(60 = 25 + 35)\nPHOBIA\n(60)\nBIMORPH\n(60 = 25 + 35)\nBOHRIUM\n(60 = 25 + 35)\nPOROMA\n(60)\nAMORPH\n(60)\nPOROMA\n(60)\nBARHOP\n(60)\nRUPIAH\n(60)\nMUROPHOBIA\n(60)\nMUROPHOBIA\n(60)\nRUMBA\n(60)\nRUPIAH\n(60)\nOPIUM\n(60)\nOPIUM\n(60)\nBOHRIUM\n(59 = 24 + 35)\nBOHRIUM\n(59 = 24 + 35)\nBIMORPH\n(58 = 23 + 35)\nBIMORPH\n(58 = 23 + 35)\nBIMORPH\n(58 = 23 + 35)\nBIMORPH\n(58 = 23 + 35)\nBIMORPH\n(58 = 23 + 35)\nBOHRIUM\n(58 = 23 + 35)\nPRIMO\n(57)\nBIMORPH\n(57 = 22 + 35)\nPUMA\n(57)\nABOHM\n(57)\nBROOM\n(57)\nRHOMB\n(57)\nBROOM\n(57)\nBOHRIUM\n(57 = 22 + 35)\nPRIMO\n(57)\nMORPH\n(57)\nBIMORPH\n(57 = 22 + 35)\nHUMP\n(57)\nBIMORPH\n(57 = 22 + 35)\nMBIRA\n(57)\nMBIRA\n(57)\nBURP\n(57)\nBURP\n(57)\nABMHO\n(57)\nMOHAIR\n(57)\nBOHRIUM\n(56 = 21 + 35)\nRHOMBI\n(56)\nBOHRIUM\n(56 = 21 + 35)\nRHOMBI\n(56)\nBOHRIUM\n(56 = 21 + 35)\nMUROPHOBIA\n(56)\nPHOBIA\n(56)\nBOHRIUM\n(56 = 21 + 35)\nBARHOP\n(56)\nPHOBIA\n(56)\nBARHOP\n(56)\nAMORPH\n(56)\nAMORPH\n(56)\nBIMORPH\n(55 = 20 + 35)\nBOHRIUM\n(55 = 20 + 35)\nBIMORPH\n(55 = 20 + 35)\nBIMORPH\n(55 = 20 + 35)\nBARM\n(54)\nBUHR\n(54)\nBARM\n(54)\nRHOMBI\n(54)\nBOHRIUM\n(54 = 19 + 35)\nBOHRIUM\n(54 = 19 + 35)\nBIMORPH\n(54 = 19 + 35)\nRUPIAH\n(54)\nMOHAR\n(54)\n\n# Word Growth involving murophobia\n\nmu\n\nphobia urophobia\n\n## Longer words containing murophobia\n\n(No longer words found)" ]

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[ "# 已知由实数组成的集合A满足：若x属于A,则1/1-x属于A（1）设A中含有三个元素,且2属于A,求A（2）A能否是仅含一个元素的单元素集,试说明理由\n\n（1）设A中含有三个元素,且2属于A,求A\n（2）A能否是仅含一个元素的单元素集,试说明理由\n\n1.A={2,-1,1/2}\n2.不能.\n\nx=1/1-x\n\n1.①当x=2时 1/(1-x)=-1 把1/(1-x)当作一个整体 令x=-1 1/(1+1)=1/2 此时A={2,-1,1/2}\n②当1/(1-x)=2时 x=1/2 此时A={2,1/2}\n2.不能\n\n1.①当x=2时 1/(1-x)=-1 把1/(1-x)当作一个整体 令x=-1 1/(1+1)=1/2 此时A={2,-1,1/2}\n②当1/(1-x)=2时 x=1/2 此时A={2,1/2}\n2.不能" ]

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[ "# SC-FDMA in LTE system\n\nI am trying to understand how SC-FDMA is used in LTE system. Every description I found so far refers to DFT and IDFT.\n\nHowever, when I checked 3GPP TS 36.211 chapter 5.4 (SC-FDMA baseband signal generation) and 6.12 (OFDM baseband signal generation) I found that there is hardly any difference between given equations (I did not copy the equations here because they are quite long) - OFDMA takes into consideration DC subcarrier, in case of SC-FDMA term \"k + 1/2\" is used wheres OFDMA uses simply term \"k\" (k can be denoted as number of subcarrier). Moreover the document does not say anything about neither DFT nor IDFT.\n\nIn case of OFDMA, I can imagine how the given equation can be considered as IDFT. But what about SC-FDMA? It looks like that modulated symbol is mapped in between two subcarrier. Can somebody explain me this issue in details?\n\n## Edit:\n\nI was interested in how DFT and IDFT can be mapped onto 3GPP TS 36.211. In other words, which part of the standard implies usages of DFT and IDFT. Based on the answer given by AlexTP I understand that DFT usage is implied by chapter 5.3.3, whereas IDFT is described in chapter 6.12 (I am refering to version 13.4 of the standard). My first guess was that both DFT and IDFT are implied by chapter 6.12. Extra information that only PUSCH uses SC-FDMA was very useful.\n\nThere are some subtleties though. However, it is out of the scope of this question.\n\n• hm, not having the formula in front of us is really a show-stopper for answering this question, I'm afraid... Generally, SC-FDMA is usually not much different to OFDMA at all, as you noticed. – Marcus Müller May 1 '17 at 13:11\n\nThe section 5.6 SC-FDMA baseband signal generation and 6.12 OFDM baseband signal generation are IDFT. The difference is the values of $\\left\\lbrace a_{k,n} \\right\\rbrace$. Ignoring MIMO stuffs in DL, in OFDM case, $\\left\\lbrace a_{k,n} \\right\\rbrace$ are mapped directly to subcarrier. In SC-FDMA case, $\\left\\lbrace a_{k,n} \\right\\rbrace$ are outputs of a supplementary block Transform precoder as in the image below.", null, "" ]

[ null, "https://i.stack.imgur.com/QNMfb.png", null ]

[ "If you find any mistakes, please make a comment! Thank you.\n\n## Solution to Mathematics for Machine Learning Exercise 7.4\n\nConsider whether the following statements are true or false:\n\n(a) The sum of any two convex functions is convex.\n\nSolution: True. Let $f_1(\\mathbf x)$ and $f_2(\\mathbf x)$ be two convex functions. Suppose their domains are $\\mathcal C_1$ and $\\mathcal C_2$, respectively. Then $\\mathcal C_1$ and $\\mathcal C_2$ are convex sets, by Definition 7.3. Hence by Exercise 7.3 (a), the intersection $\\mathcal C_1\\cap \\mathcal C_2$ is also convex. Note that $\\mathcal C_1\\cap \\mathcal C_2$ is also the domain of $f_1+f_2$. Hence the domain of  $f_1+f_2$ is convex. Now we only need to check the condition (7.30) in the textbook.\n\nFor any $\\mathbf x,\\mathbf y$ in $\\mathcal C_1\\cap \\mathcal C_2$ and $0\\leq \\theta \\leq 1$, we have\\begin{align*}&\\ (f_1+f_2)(\\theta \\mathbf x+(1-\\theta)\\mathbf y) \\\\ = &\\ f_1((\\theta \\mathbf x+(1-\\theta)\\mathbf y))+f_2((\\theta \\mathbf x+(1-\\theta)\\mathbf y))\\\\ \\leq &\\ \\theta f_1(\\mathbf x)+(1-\\theta)f_1(\\mathbf y) +\\theta f_2(\\mathbf x)+(1-\\theta)f_2(\\mathbf y)\\\\ = &\\ \\theta f_1(\\mathbf x) +\\theta f_2(\\mathbf x)+(1-\\theta)f_1(\\mathbf y) +(1-\\theta)f_2(\\mathbf y)\\\\ = &\\ \\theta (f_1+f_2)(\\mathbf x)+(1-\\theta)(f_1+f_2)(\\mathbf y).\\end{align*}In the inequality, we used equation (7.30) for $f_1$ and $f_2$ as they are convex. Therefore, $f_1+f_2$ is also convex.\n\n(b) The difference of any two convex functions is convex.\n\nI will use the following well-known fact: If $f''(x)\\geq 0$ holds for all $x$ in the domain (assumed to be convex) of $f$, then $f(x)$ is convex.\n\nSolution: False. For example, let $f_1(x)=x^2$ and $f_2(x)=2x^2$. Then $f_1$ and $f_2(x)$ are convex, but $$(f_1-f_2)(x)=f_1(x)-f_2(x)=-x^2$$ is not convex. Please fill the details.\n\n(c) The product of any two convex functions is convex.\n\nSolution: False. Let $f_1(x)=x^2$ and $f_2(x)=x$. Then $f_1$ and $f_2(x)$ are convex, but $$(f_1f_2)(x)=f_1(x)f_2(x)=-x^3$$ is not convex. Please fill the details, e.g. use $\\theta=1/2$, $x=-1$ and $y=0$ to get a counterexample.\n\n(d) The maximum of any two convex functions is convex.\n\nSolution: True. Let $f_1(\\mathbf x)$ and $f_2(\\mathbf x)$ be two convex functions. Let $f(x):=\\max\\{f_1(x),f_2(x)\\}$ be the maximum of them. Suppose their domains are $\\mathcal C_1$ and $\\mathcal C_2$, respectively. Then $\\mathcal C_1$ and $\\mathcal C_2$ are convex sets, by Definition 7.3. Hence by Exercise 7.3 (a), the intersection $\\mathcal C_1\\cap \\mathcal C_2$ is also convex. Note that $\\mathcal C_1\\cap \\mathcal C_2$ is also the domain of $f(x)$. Hence the domain of  $f(x)$ is convex. Now we only need to check the condition (7.30) in the textbook.\n\nNote that for any $\\mathbf x$ in $\\mathcal C_1\\cap \\mathcal C_2$, we have $f_1(\\mathbf x)\\leq f(\\mathbf x)$ and $f_2(\\mathbf x)\\leq f(\\mathbf x)$.\n\nFor any $\\mathbf x,\\mathbf y$ in $\\mathcal C_1\\cap \\mathcal C_2$ and $0\\leq \\theta \\leq 1$, we have\\begin{align*}&\\ f_1(\\theta \\mathbf x+(1-\\theta)\\mathbf y) \\\\ \\leq &\\  f_1(\\mathbf x)+(1-\\theta)f_1(\\mathbf y) \\\\ \\leq &\\ \\theta f(\\mathbf x) +(1-\\theta)f(\\mathbf y).\\end{align*}In the inequality, we used equation (7.30) for $f_1$ as it is convex. Similarly, we have $$\\ f_2(\\theta \\mathbf x+(1-\\theta)\\mathbf y)\\leq \\theta f(\\mathbf x) +(1-\\theta)f(\\mathbf y).$$Therefore, we have $$f(\\theta \\mathbf x+(1-\\theta)\\mathbf y)\\leq \\theta f(\\mathbf x) +(1-\\theta)f(\\mathbf y).$$Hence $f$ is also convex.\n\nRemark: Here, we used the fact that if $x \\leq a$ and $y\\leq a$, then $\\max\\{x,y\\}\\leq a$." ]

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[ "# Chemical Process Calculation Questions and Answers – Material balances for a Combustion System-I\n\nThis set of Chemical Process Calculation Multiple Choice Questions & Answers (MCQs) focuses on “Material balances for a Combustion System-I”.\n\n1. 144 grams of C5H12 is burnt with 2 molesO2, and 1 mole of CO2 is produced.\nWhat is the percentage of excess O2?\na) 25%\nb) 50%\nc) 75%\nd) 100%\n\nExplanation: Percentage of excess oxygen = 100*(2 – 1.6)/1.6 = 25%.\n\n2. For a combustion system, incorrect statement is\na) Reaction of oxygen with materials containing hydrogen, carbon and sulfur\nb) Product gases are such as C3H8O, CO2, CO, SO2\nc) Most combustion processes use air as the source of oxygen.\nd) None of the mentioned\n\nExplanation: Properties of a combustion system.\n\n3. All the gases resulting from a combustion process including the water vapour is known as\na) Flue gas\nb) Dry basis\nc) Orsat analysis\nd) None of the mentioned\n\nExplanation: All the gases resulting from a combustion process including the water vapour is known as flue gas.\n\n4. Ethane is burnt with 150% of excess air then what is the percentage of O2 in the products?\na) 12.5%\nb) 33.3%\nc) 45.4%\nd) 52.9%\n\nExplanation: Percentage of O2 = 45/85*100 = 52.9%.\n\n5. All the gases resulting from a combustion process, not including the water vapour is known as\na) Flue gas\nb) Stack gas\nc) Orsat analysis\nd) None of the mentioned\n\nExplanation: All the gases resulting from a combustion process, not including the water vapour is known as Orsat analysis.\nNote: Join free Sanfoundry classes at Telegram or Youtube\n\n6. A complete combustion reaction is\na) Fuel producing CO2\nb) Fuel producing CO and C3H8O\nc) Fuel producing CO2 and C3H8O\nd) Fuel producing CO and C3H8\n\nExplanation: A complete combustion reaction is where Fuel is producing CO2 and C3H8O.\n\n7. Two statements are given as below\nA. A fuel is producing some CO from the carbon source is the partial combustion of that fuel.\nB. CO does not produce as much energy as would be in the case of CO2 formation in a combustion process.\na) Both statements are true\nb) Both statements are false\nc) A is correct and B is false\nd) B is correct and A is false\n\nExplanation: Statement I is correct for partial combustion and CO2 produces more energy than CO.\n\n8. The amount of air required to be brought in to the process for complete combustion is\na) Theoretical Air\nb) Excess Air\nc) Partial Air\nd) None of the mentioned\n\nExplanation: The amount of air required to be brought in to the process for complete combustion is called Theoretical Air.\n\n9. The amount of air in excess of the theoretical air required for complete combustion is\na) Theoretical Air\nb) Excess Air\nc) Partial Air\nd) None of the mentioned\n\nExplanation: The amount of air in excess of the theoretical air required for complete combustion is called Excess Air.\n\n10. Ethane is burnt with 50% of excess air, what is the percentage of CO2 in the products?\na) 11.11%\nb) 36.36%\nc) 66.66%\nd) 72.72%\n\nExplanation: Percentage of CO2 = 20/55*100 = 36.36%.\n\nSanfoundry Global Education & Learning Series – Chemical Process Calculation.\nTo practice all areas of Chemical Process Calculation, here is complete set of 1000+ Multiple Choice Questions and Answers.", null, "" ]

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[ "# A note on Cauchy-Lipschitz-Picard theorem\n\n## Abstract\n\nIn this note, we try to generalize the classical Cauchy-Lipschitz-Picard theorem on the global existence and uniqueness for the Cauchy initial value problem of the ordinary differential equation with global Lipschitz condition, and we try to weaken the global Lipschitz condition. We can also get the global existence and uniqueness.\n\n## 1 Introduction\n\nIn his famous book , Brezis gave a very sketchy and interesting proof on the classical Cauchy-Lipschitz-Picard theorem.\n\n### Theorem 1.1\n\nLet E be a Banach space and let $$F: E\\rightarrow E$$ be a Lipschitz map, i.e., there is a constant L such that\n\n$$\\bigl\\Vert F(u)-F(v)\\bigr\\Vert \\leq L\\Vert u-v\\Vert , \\quad \\forall u,v \\in E.$$\n(1.1)\n\nThen, for any given $$u_{0}\\in E$$, there exists a unique solution $$u\\in C^{1}([0,+\\infty),E)$$ of the problem:\n\n$$\\left \\{ \\textstyle\\begin{array}{l} \\frac{du(t)}{dt}=F(u(t)),\\quad \\forall t\\in(0,+\\infty), \\\\ u(0)=u_{0}. \\end{array}\\displaystyle \\right .$$\n(1.2)\n\nIt is well known that if we only assume the local Lipschitz condition, we can only get the local existence and uniqueness for Cauchy initial value problems.\n\nIn this paper, we try to weaken the global Lipschitz condition, but we also want to get the global existence and uniqueness; we have the following theorem.\n\n### Theorem 1.2\n\nLet E be a Banach space (with norm $$\\|\\cdot\\|$$) and let $$F: [0,+\\infty)\\times E\\rightarrow E$$ be a map satisfying\n\n$$\\bigl\\Vert F(t,u)-F(t,v)\\bigr\\Vert \\leq L(t)\\cdot p\\bigl(\\Vert u-v\\Vert \\bigr), \\quad \\forall t\\in[0,+\\infty), u,v\\in E,$$\n\nwhere $$L: [0,+\\infty)\\rightarrow[0,+\\infty)$$, $$p: (0,+\\infty )\\rightarrow[0,+\\infty)$$ are continuous and there are $$0\\leq a< +\\infty$$ such that\n\n\\begin{aligned}& L=L(t)\\leq a, \\\\& p=p(s)\\leq s, \\end{aligned}\n\nand $$p(s)$$ is an increasing function. Then, for any given $$u_{0}\\in E$$, there exists a unique solution $$u\\in C^{1}([0,+\\infty);E)$$ for the problem\n\n$$\\left \\{ \\textstyle\\begin{array}{l} \\frac{du(t)}{dt}=F(t,u(t)),\\quad \\forall t\\in(0,+\\infty), \\\\ u(0)=u_{0}. \\end{array}\\displaystyle \\right .$$\n(1.3)\n\n### Corollary 1.3\n\nIn Theorem  1.2, if we take $\\mathrm{Ł}\\left(s\\right)=a>0$, $$p(s)=s$$ or $$p(s)=\\ln s$$, then the conditions and the results of Theorem  1.2 hold.\n\n## 2 The proof of Theorem 1.2\n\n### Lemma 2.1\n\n, Banach contraction mapping principle\n\nLet X be a nonempty complete metric space and let $$T: X\\rightarrow X$$ be a strict contraction, i.e., there is $$0< k<1$$ such that $$d(T(x),T(y))\\leq kd(x,y)$$, $$\\forall x,y\\in X$$, then S has a unique fixed point $$u=T(u)$$.\n\n### Lemma 2.2\n\nGronwall \n\nLet $$x\\in C^{1}[a,b]$$. If $$R[a,b]$$ denotes the set of Riemann integrable functional on $$[a,b]$$; $$\\beta\\in R[a,b]$$, and\n\n$$x^{\\prime}(t)\\leq\\beta(t)\\cdot x(t), \\quad \\forall t\\in[a,b],$$\n\nthen\n\n$$x(t)\\leq x(a)\\cdot e^{\\int^{t}_{a}\\beta(s)\\,ds},\\quad \\forall t\\in[a,b].$$\n\nThe following lemma can be regarded as a natural generalization of the Gronwall inequality.\n\n### Lemma 2.3\n\nLet $$x\\in C^{1}[a,b]$$, $$\\alpha,\\beta\\in R[a,b]$$. If\n\n$$x^{\\prime}(t)\\leq\\alpha(t)+\\beta(t)x(t),\\quad \\forall t\\in[a,b],$$\n\nthen\n\n$$x(t)\\leq\\biggl[x(a)+ \\int^{t}_{a}e^{-\\int_{0}^{s}\\beta(\\tau)\\,d\\tau}\\alpha (s)\\,ds\\biggr] \\cdot e^{\\int^{t}_{a}\\beta(s)\\,ds}, \\quad \\forall t\\in[a,b].$$\n\n### Proof\n\nLet $$v=e^{\\int^{t}_{a}\\beta(s)\\,ds}$$, $$w=\\frac{x}{v}$$, then $$v^{\\prime }(t)=\\beta(t)v(t)$$,\n\n$$w^{\\prime}(t)=\\frac{x^{\\prime}v-xv^{\\prime}}{v^{2}} \\leq \\frac {(\\alpha+\\beta x)v-\\beta xv}{v^{2}}=\\frac{\\alpha v}{v^{2}} = \\frac{\\alpha}{v}.$$\n\nThen\n\n\\begin{aligned}& w(t)\\leq \\int^{t}_{a}e^{-\\int_{0}^{s}\\beta(\\tau)\\,d\\tau}\\alpha (s)\\,ds+w(a), \\\\& \\frac{x(t)}{v(t)}\\leq x(a)+ \\int^{t}_{a}e^{-\\int_{0}^{s}\\beta(\\tau)\\,d\\tau }\\alpha(s)\\,ds, \\\\& x(t)\\leq e^{\\int^{t}_{a}\\beta(s)\\,ds}\\cdot \\biggl[x (a)+ \\int^{t}_{a}e^{-\\int_{0}^{s}\\beta(\\tau)\\,d\\tau}\\alpha(s)\\,ds \\biggr]. \\end{aligned}\n\nNow to prove Theorem 1.2, we use some similar arguments to Brezis .\n\nLet $$k>0$$, which is to be determined, and assume\n\n$$X=\\Bigl\\{ u\\in C\\bigl( [0,+\\infty );E\\bigr)\\big|\\sup_{t\\geq0}e^{-kt} \\cdot\\bigl\\Vert u(t)\\bigr\\Vert < +\\infty \\Bigr\\} .$$\n\nThen it is easy to see that X is a Banach space for the norm\n\n$$\\Vert u\\Vert _{X}=\\sup_{t\\geq0}e^{-kt} \\cdot\\bigl\\Vert u(t)\\bigr\\Vert .$$\n\nFor $$\\forall u\\in X$$, we define\n\n$$\\Phi(u) (t)=u_{0}+ \\int^{t}_{0}F\\bigl(s,u(s)\\bigr)\\,ds.$$\n\nThen u is a solution of (1.3) if and only if $$\\Phi(u)=u$$, that is, u is a fixed point of Φ.\n\n(1) We now show that, for every $$u\\in X$$, $$\\Phi(u)$$ also belongs to X.\n\nIn fact,\n\n\\begin{aligned} \\Vert \\Phi u\\Vert _{X} =&\\sup_{t\\geq0}e^{-kt} \\bigl\\Vert (\\Phi u) (t)\\bigr\\Vert \\\\ \\leq&\\sup_{t\\geq0}e^{-kt}\\cdot \\Vert u_{0} \\Vert +\\sup_{t\\geq0}e^{-kt}\\cdot \\biggl\\Vert \\int^{t}_{0}F\\bigl(s,u(s)\\bigr)\\,ds\\biggr\\Vert . \\end{aligned}\n\nWe only need to prove\n\n$$\\sup_{t\\geq0}e^{-kt}\\cdot \\int^{t}_{0}\\bigl\\Vert F\\bigl(s,u(s)\\bigr)\\bigr\\Vert \\,ds< +\\infty.$$\n\nNotice that\n\n\\begin{aligned}& \\bigl\\Vert F\\bigl(s,u(s)\\bigr)-F(0,u_{0})\\bigr\\Vert \\leq L(s) \\cdot p\\bigl(\\bigl\\Vert u(s)-u_{0}\\bigr\\Vert \\bigr), \\\\& \\bigl\\Vert F\\bigl(s,u(s)\\bigr)\\bigr\\Vert \\leq L(s)\\cdot p\\bigl(\\bigl\\Vert u(s)-u_{0}\\bigr\\Vert \\bigr)+\\bigl\\Vert F(0,u_{0})\\bigr\\Vert . \\end{aligned}\n\nSince\n\n$$\\sup_{t\\geq0}e^{-kt}\\cdot \\int^{t}_{0}\\bigl\\Vert F(0,u_{0})\\bigr\\Vert \\,ds=\\bigl\\Vert F(0,u_{0})\\bigr\\Vert \\sup _{t\\geq0}e^{-kt}\\cdot t< +\\infty.$$\n\nHence we only need to prove\n\n$$\\sup_{t\\geq0}e^{-kt}\\cdot \\int^{t}_{0}L(s)p\\bigl(\\bigl\\Vert u(s)-u_{0}\\bigr\\Vert \\bigr)\\,ds< +\\infty.$$\n\nLet\n\n$$\\varphi(t)=e^{-kt} \\int^{t}_{0}L(s)p\\bigl(\\bigl\\Vert u(s)-u_{0}\\bigr\\Vert \\bigr)\\,ds,\\qquad \\varphi(0)=0,$$\n\nthen\n\n\\begin{aligned} \\varphi(t) \\leq& \\int_{0}^{t}\\bigl[L(s)e^{ks} \\bigr]e^{-ks}\\bigl(\\bigl\\Vert u(s)\\bigr\\Vert +\\Vert u_{0}\\Vert \\bigr)\\, dse^{-kt} \\\\ \\leq&\\Bigl[\\sup_{t\\geq0}e^{-ks}\\bigl\\Vert u(s)\\bigr\\Vert +\\sup_{t\\geq0}e^{-ks}\\Vert u_{0} \\Vert \\Bigr]\\sup_{t\\geq0}\\biggl[ \\int_{0}^{t}L(s)e^{ks}\\, dse^{-kt}\\biggr] \\\\ \\leq& C_{1}C_{2}=C_{3}< +\\infty. \\end{aligned}\n\nHence we have proved Φ is a self-mapping from X to X.\n\n(2) We prove the contraction property of Φ. We have\n\n\\begin{aligned}& \\Vert \\Phi u-\\Phi v\\Vert \\leq \\int^{t}_{0}\\bigl\\Vert F\\bigl(s,u(s)\\bigr)-F \\bigl(s,v(s)\\bigr)\\bigr\\Vert \\,ds \\leq \\int^{t}_{0}L(s)p\\bigl(\\bigl\\Vert u(s)-v(s)\\bigr\\Vert \\bigr)\\,ds, \\\\& \\Vert \\Phi u -\\Phi v\\Vert _{X} \\leq \\sup_{t\\geq0}e^{-kt} \\cdot \\int ^{t}_{0}L(s)p\\bigl(\\bigl\\Vert u(s)-v(s)\\bigr\\Vert \\bigr)\\,ds \\\\& \\hphantom{\\Vert \\Phi u -\\Phi v\\Vert _{X}}\\leq \\sup_{t\\geq0} \\int^{t}_{0}L(s)e^{ks} \\bigl[e^{-ks}\\bigl\\Vert u(s)-v(s)\\bigr\\Vert \\bigr]\\, dse^{-kt} \\\\& \\hphantom{\\Vert \\Phi u -\\Phi v\\Vert _{X}} \\leq \\sup_{t\\geq0}\\bigl[e^{-ks}\\bigl\\Vert u(s)-v(s)\\bigr\\Vert \\bigr]\\sup_{t\\geq0}e^{-kt} \\int ^{t}_{0}L(s)e^{ks}\\,ds \\\\& \\hphantom{\\Vert \\Phi u -\\Phi v\\Vert _{X}} \\leq \\sup_{t\\geq0}e^{-kt} \\int^{t}_{0}L(s)e^{ks}\\,ds \\Vert u -v \\Vert _{X}. \\end{aligned}\n\nHence we have\n\n$$\\|\\Phi u-\\Phi v\\|_{X} \\leq \\frac{L}{k}\\| u -v\\|_{X}, \\quad \\forall u,v\\in X.$$\n\nWe can choose $$k>L$$, then we use Banach contraction mapping principle to find that (1.3) has at least one solution on $$[0,+\\infty)$$.\n\nFurthermore, by Gronwall’s inequality, we can get the uniqueness.\n\nIn fact, let $$u_{1}$$, $$u_{2}$$ be two solutions of (1.3), then\n\n\\begin{aligned} \\bigl\\Vert u_{1}(t)-u_{2}(t)\\bigr\\Vert \\leq& \\int_{0}^{t}\\bigl\\Vert F\\bigl(s,u_{1}(s) \\bigr)-F\\bigl(s,u_{2}(s)\\bigr)\\bigr\\Vert \\\\ \\leq& \\int_{0}^{t}L(s)p\\bigl(\\bigl\\Vert u_{1}(s)-u_{2}(s)\\bigr\\Vert \\bigr)\\,ds \\\\ \\leq& \\int_{0}^{t}L(s) \\bigl(\\bigl\\Vert u_{1}(s)-u_{2}(s)\\bigr\\Vert \\bigr)\\,ds. \\end{aligned}\n\nBy Gronwall’s inequality, we have\n\n$$\\bigl\\Vert u_{1}(t)-u_{2}(t)\\bigr\\Vert \\leq0.$$\n\nHence for any $$t\\in[0,+\\infty)$$, we have $$u_{1}(t)=u_{2}(t)$$. □\n\n## 3 Examples\n\n### Example 3.1\n\n$$\\left \\{ \\textstyle\\begin{array}{l} \\frac{du(t)}{dt}=F(t,u(t))=\\frac{1}{1+t+|u|},\\quad \\forall t\\in (0,+\\infty), \\\\ u(0)=u_{0}. \\end{array}\\displaystyle \\right .$$\n(3.1)\n\nThen\n\n$$\\bigl\\vert F\\bigl(t,u(t)\\bigr)-F\\bigl(t,v(t)\\bigr)\\bigr\\vert = \\frac{\\|v|-|u\\|}{(1+t+|u|)(1+t+|v|)}\\leq\\| v|-|u\\|.$$\n\nBy the triangle inequality, we have\n\n$$\\bigl\\vert F\\bigl(t,u(t)\\bigr)-F\\bigl(t,v(t)\\bigr)\\bigr\\vert \\leq \\vert u-v \\vert .$$\n\nSo $$F(t,u(t))=\\frac{1}{1+t+|u|}$$ is Lipschitz with $$a=1$$ and $$p(s)=s$$.\n\n### Example 3.2\n\nLet $$p_{1}(t):[0,+\\infty)\\rightarrow[0,+\\infty)$$ continuous and there is $$0\\leq a<+\\infty$$ such that\n\n$$p_{1}(t)\\leq a.$$\n\nLet $$p_{2}(t):[0,+\\infty)\\rightarrow R$$ continuous.\n\nWe consider\n\n$$\\left \\{ \\textstyle\\begin{array}{l} \\frac{du(t)}{dt}=F(t,u(t))=p_{1}(t)|u-p_{2}(t)|,\\quad \\forall t\\in(0,+\\infty), \\\\ u(0)=u_{0}. \\end{array}\\displaystyle \\right .$$\n(3.2)\n\nThen\n\n$$\\bigl\\vert F\\bigl(t,u(t)\\bigr)-F\\bigl(t,v(t)\\bigr)\\bigr\\vert =p_{1}(t) \\bigl(\\bigl\\vert \\bigl\\vert u-p_{2}(t)\\bigr\\vert -\\bigl\\vert v-p_{2}(t)\\bigr\\vert \\bigr\\vert \\bigr).$$\n\nBy the triangle inequality, we have\n\n$$\\bigl\\vert F\\bigl(t,u(t)\\bigr)-F\\bigl(t,v(t)\\bigr)\\bigr\\vert \\leq p_{1}(t)\\vert u-v\\vert .$$\n\nSo $$F(t,u(t))=p_{1}(t)|u-p_{2}(t)|$$ is Lipschitz with $$L(t)=p_{1}(t)$$ and $$p(s)=s$$.\n\n## References\n\n1. Brezis, H: Functional Analysis, Sobolev Spaces and PDE, pp. 184-185. Springer, New York (2011)\n\n2. Thomas, HG: Note on the derivatives with respect to a parameter of the solutions of a system of differential equations. Ann. Math. 20, 292-296 (1919)\n\n## Acknowledgements\n\nThe authors sincerely thank referee for his/her valuable comments. This paper was partially supported by NSFC (No. 11671278 and No. 11426181) and the National Research Foundation for the Doctoral Program of Higher Education of China under Grant No. 20120181110060.\n\n## Author information\n\nAuthors\n\n### Corresponding author\n\nCorrespondence to Fengying Li.\n\n### Competing interests\n\nThe authors declare that they have no competing interests.\n\n### Authors’ contributions\n\nThe research and writing of this manuscript was a collaborative effort from all the authors. All authors read and approved the final manuscript.\n\n## Rights and permissions", null, "" ]

[ null, "https://journalofinequalitiesandapplications.springeropen.com/track/article/10.1186/s13660-016-1214-x", null ]

[ "", null, "", null, "", null, "", null, "", null, "", null, "", null, "A bug in SameQ\n\n• To: mathgroup at smc.vnet.net\n• Subject: [mg23356] A bug in SameQ\n• From: \"Ersek, Ted R\" <ErsekTR at navair.navy.mil>\n• Date: Thu, 4 May 2000 02:59:23 -0400 (EDT)\n• Sender: owner-wri-mathgroup at wolfram.com\n\n```I use Mathematica Version 4.0.1.0 under Windows 98.\nFor me SameQ gives contradicting results about x1 and x2 (two numbers close\nto zero). Certain other platforms probably have the same problem.\n-------------------\n\nIn:=\nx1= \\$MinMachineNumber;\nx2= x1(1+\\$MachineEpsilon);\n\nIn:=\n{x1===x2, -x1 === -x2}\n\nOut=\n{True, False}\n\nThis shows that even the most fundamental features are not completely free\nof problems. Also notice Equal uses a different test and doesn't give\n\nWRI Tech Support already confirmed that this is a bug, so don't bother\nforwarding it to them. I certainly hope they will make sure the next version\ndoes this correctly on all platforms!\n\n--------------------\nRegards,\nTed Ersek\n\nSee my Mathematica tips, tricks at\nhttp://www.verbeia.com/mathematica/tips/Tricks.html\n\n```\n\n• Prev by Date: The most efficient Fibonacci algorithim?\n• Next by Date: Re: collect derivatives\n• Previous by thread: Re: The most efficient Fibonacci algorithim?\n• Next by thread: Re: A bug in SameQ" ]

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[ "Definition:Strictly Precede\n\nDefinition\n\nDefinition 1\n\nLet $\\left({S, \\prec}\\right)$ be a strictly ordered set.\n\nLet $a, b \\in S$ and $a \\prec b$.\n\nThen $a$ strictly precedes $b$.\n\nDefinition 2\n\nLet $\\left({S, \\preceq}\\right)$ be an ordered set.\n\nLet $a \\preceq b$ such that $a \\ne b$.\n\nThen $a$ strictly precedes $b$.\n\nNotation\n\nWhen $a \\preceq b$ and $a \\ne b$, it is usual to denote this with the symbol:\n\n$a \\prec b$\n\nand similar derived notation for other ordering symbols.\n\nAlso known as\n\nThe statement $a$ strictly precedes $b$ can be expressed as $a$ is a strict predecessor of $b$.\n\nSome sources refer to a strict predecessor simply as a predecessor.\n\nWhen the underlying set $S$ of the ordered set $\\left({S, <}\\right)$ is one of the sets of numbers $\\N$, $\\Z$, $\\Q$, $\\R$ or a subset, the term is less than is usually used instead of (strictly) precedes." ]

[ null ]

[ "Sivakrishna\n☆\n\nIndia,\n2020-10-07 08:15\n(633 d 07:55 ago)\n\nPosting: # 21973\nViews: 1,891\n\n## Treatment effect justification [General Sta­tis­tics]\n\nDear Members,\n\nGood Morning!!\n\nIn the Bio equivalence study, the p value is < 0.05 for Treatment effect, i.e., Treatment effect was statistically significant at 5% level of significance and however the obtained 90% confidence intervals were with in 80.00% to 125.00% and power was >90%. Could you please provide your suggestions to justify the treatment effect.\n\nThanks and Regards\nG. Siva Krishna Teja.\n\nElMaestro\n★★★\n\nDenmark,\n2020-10-07 10:19\n(633 d 05:51 ago)\n\n@ Sivakrishna\nPosting: # 21974\nViews: 1,642\n\n## Treatment effect justification\n\nHi Siva Krishna,\n\nsomething like this:\n\n\"The test producing a significant treatment effect is that log(y)T=log(y)R where y is the dependent variable (AUC or Cmax). This is not the hypothesis evaluated for the conclusion of bioequivalence. It is entirely expected that Test and Reference could be associated with different levels of (logarithmic) Cmax or AUC as they are truly two different formulations. What the evaluation for bioequivalence shows is that the difference in rate and extent and absorption does not differ by a clinically relevant margin, where clinical relevance is determined by the regulatory convention. Therefore the significant formulation effect does not translate, in this case, into inability to conclude bioequivalence.\"\n\nGood luck.", null, "Pass or fail!\nElMaestro\nHelmut\n★★★", null, "", null, "Vienna, Austria,\n2020-10-07 11:07\n(633 d 05:03 ago)\n\n@ Sivakrishna\nPosting: # 21975\nViews: 1,625\n\n## statistically significant ≠ clinically relevant\n\nHi Siva Krishna,\n\nI guess 100% was not contained in the 90% CI, right?\nIf yes, you have a statistically significant difference which is clinically not relevant.1 We abandoned testing for a statistically significant difference (see ElMaestro’s post) 33 (‼) years ago with Schuirmann’s TOST.2 To quote Wasserstein et al.3\n\nDon’t Say “Statistically Significant”\n\nFor which power did you plan the study? It might well be that\n• the T/R-ratio was closer to 100% than assumed and/or\n• the CV was lower than assumed and/or\n• the dropout-rate was lower than anticipated.\nAny of those (and their combinations) will lead to higher power and increases the chance of a statistically significant treatment effect.\nSee also the second part of this post. If you are in the lower right quadrants, you have high power and a statistically significant treatment effect is likely.\n\n1. $$\\small{\\Delta}$$ = clinically relevant difference. Commonly 0.20 (20%). For NTIDs (EMA and other jurisdicions) $$\\small{\\Delta}$$ 0.10, for Cmax (Russian Federation, EEU, GCC) $$\\small{\\Delta}$$ 0.25. For HVD(P)s, where CVwR >30%, $$\\small{\\Delta}$$ >0.30 (scaled to the variability of the reference). The acceptance range for bioequivalence $$\\small{\\left \\{\\theta_1,\\theta_2\\right \\}}$$ is calculated by $$\\small{\\theta_1=1-\\Delta}$$, $$\\small{\\theta_2=(1-\\Delta)^{-1}}$$. If the 90% CI lies entirely within $$\\small{\\left \\{\\theta_1,\\theta_2\\right \\}}$$, the observed difference of the treatment effect is considered clinically not relevant – irrespective how wide the CI is or where the point estimate lies. A formulation with a PE of 100% (CI 80.00–125.00%) is as BE as another with a PE of 85% (CI 80.00–90.31%). In the former case you were extremely lucky and in the second you have a statistically significant difference (100% not contained in the CI).\n2. Schuirmann DJ. A comparison of the Two One-Sided Tests Procedure and the Power Approach for Assessing the Equivalence of Average Bioavailability. J Pharmacokin Biopharm. 1987; 15(6): 657–80. doi:10.1007/BF01068419.\n3. Wasserstein RL, Schirm AL, Lazar NA. Moving to a World Beyond “p < 0.05”. Am Stat. 2019; 73(sup1): 1–19. doi:10.1080/00031305.2019.1583913.", null, "Open access.\n\nDif-tor heh smusma 🖖", null, "Helmut Schütz", null, "The quality of responses received is directly proportional to the quality of the question asked. 🚮\nScience Quotes\nSivakrishna\n☆\n\nIndia,\n2020-10-09 10:24\n(631 d 05:45 ago)\n\n@ Helmut\nPosting: # 21985\nViews: 1,498\n\n## statistically significant ≠ clinically relevant\n\nI would like to say Thank you sir for your valuable information. This may be helpful to my question.\n\nThanks and Regards\nG. Siva Krishna Teja.\nHelmut\n★★★", null, "", null, "Vienna, Austria,\n2020-10-09 15:04\n(631 d 01:06 ago)\n\n@ Sivakrishna\nPosting: # 21986\nViews: 1,462\n\n## Problems with low variability\n\nHi Siva Krishna,\n\n» I would like to say Thank you sir for your valuable information. This may be helpful to my question.\n\nWelcome. Would you mind answering my previous questions:\n» » I guess 100% was not contained in the 90% CI, right?\n» » For which power did you plan the study?\n\nSometimes statistically significant differences are common, namely if the CV is low (say, <10%) and you plan for 80% power. Then you may end up with a sample size far below the regulatory minimum of twelve. Add more subjects to compensate for potential dropouts and…\nIn my protocols I state that extremely high power is expected and the CI might well contain not 100%.", null, "script:\n\nlibrary(PowerTOST) balance <- function(x, seqs) { # gives complete sequences   x <- ceiling(x) + ceiling(x) %% seqs   return(x) } CV       <- 0.10    # assumed (here 10%) theta0   <- 0.925   # assumed T/R-ratio target   <- 0.80    # target (desired) power (here at least 80%) do.rate  <- 0.10    # anticipated dropout rate (here 10%) design   <- \"2x2x2\" # can be any one given by known.designs() seqs     <- as.integer(substr(design, 3, 3)) # sequences n        <- sampleN.TOST(CV = CV, theta0 = theta0, targetpower = target,                          design = design, details = FALSE,                          print = FALSE)[[\"Sample size\"]] if (n < 12) n <- 12 # force to minimum acc. to GLs dosed    <- balance(n / (1 - do.rate), seqs) # adjust for dropout-rate & balance eligible <- dosed:n; dropouts <- rev(eligible - n) res      <- data.frame(dosed = dosed, dropouts = dropouts, eligible = eligible,                        power = NA, CL.lo = NA, CL.hi = NA,                        p.left = NA, p.right = NA) for (j in seq_along(eligible)) {   res\\$power[j] <- suppressMessages(                     signif(power.TOST(CV = CV, theta0 = theta0,                                       design = design, n = eligible[j]), 4))   res[j, 5:6]  <- round(100*CI.BE(pe = theta0, CV = CV,                                   design = design, n = eligible[j]), 2)   res[j, 7:8]  <- suppressMessages(                     signif(pvalues.TOST(pe = theta0, CV = CV,                                         design = design, n = eligible[j]), 4)) } print(res, row.names = FALSE)\n\nGives (if the assumptions about the CV and T/R-ratio are realized in the study):\n\n dosed dropouts eligible  power CL.lo CL.hi   p.left   p.right     14        0       14 0.9760 86.49 98.93 0.001154 1.913e-06     14        1       13 0.9652 86.22 99.23 0.001752 4.846e-06     14        2       12 0.9521 85.92 99.59 0.002569 1.166e-05\n\nDif-tor heh smusma 🖖", null, "Helmut Schütz", null, "The quality of responses received is directly proportional to the quality of the question asked. 🚮\nScience Quotes", null, "", null, "Ing. Helmut Schütz", null, "" ]

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[ "# [pushed] OpenMP parallel region scope tests\n\nMessage ID [email protected] New show\n\n## Commit Message\n\nSimon Marchi (Code Review) Dec. 10, 2019, 10:46 p.m. UTC\n```The original change was created by Kevin Buettner.\n\nChange URL: https://gnutoolchain-gerrit.osci.io/r/c/binutils-gdb/+/504\n......................................................................\n\nOpenMP parallel region scope tests\n\nAdd tests which check for accessibility of variables from within\nvarious OpenMP parallel regions.\n\nTested on Fedora 27, 28, 29, 30, and 31. I also tested with my OpenMP\nwork on Fedora 30. The test has been annotated with setup_xfail and\nsetup_kfail statements so that there are no unexpected failures on any\nof these platforms when using gcc. Better still, for my own testing\nanyway, is that there are also no XPASSes or KPASSes either. So,\nregardless of platform, when using gcc, and regardless of whether my\n(not yet public) OpenMP work is used, seeing a FAIL indicates a real\nproblem.\n\nFedora 27 results:\n\n# of expected passes 85\n# of expected failures 65\n\n(Note: I have not retested F27 since v1 of the patch; it's possible\nthat the numbers will be slightly different for v2.)\n\nFedora 28, 29, 30 results:\n\n# of expected passes 131\n# of expected failures 4\n# of known failures 16\n\nFedora 30, 31 results w/ my OpenMP work:\n\n# of expected passes 151\n\nThe above results all use gcc, either the system gcc or a development\ngcc (when testing against my OpenMP work in GDB). I've also tested\nwith clang 9.0.0 and icc 19.0.5.281 20190815 on Fedora 31.\n\nFedora 31, clang:\n\nFAIL: gdb.threads/omp-par-scope.exp: multi_scope: after parallel: print file_scope_var\n\nFedora 31, icc:\n\nFor both clang and icc, it turns out that there are some problems with\nthe DWARF that these compilers generate. Of the two, icc does at\nleast nest the subprogram of the outlined function representing the\nparallel region within the function that it's defined, but does not\nhandle inner scopes if they exist. clang places the subprogram for\nthe outlined function at the same level as the containing function, so\nvariables declared within the function aren't visible at all.\n\nI could call setup_xfail to avoid FAILs for clang and icc also, but I don't\nwant to further complicate the test.\n\ngdb/testsuite/ChangeLog:\n\nChange-Id: Icb9c991730d84ca7509380af817dfcc778e764ea\n---\nM gdb/testsuite/ChangeLog\n3 files changed, 461 insertions(+), 0 deletions(-)\n```\n\n## Patch\n\n```diff --git a/gdb/testsuite/ChangeLog b/gdb/testsuite/ChangeLog\nindex 4b2e132..8b846a1 100644\n--- a/gdb/testsuite/ChangeLog\n+++ b/gdb/testsuite/ChangeLog\n@@ -1,5 +1,10 @@\n2019-12-10 Kevin Buettner <[email protected]>\n\n+\n+2019-12-10 Kevin Buettner <[email protected]>\n+\n* lib/gdb.exp (support_nested_function_tests): New proc.\n\n2019-12-10 Kevin Buettner <[email protected]>\nnew file mode 100644\nindex 0000000..c60e376\n--- /dev/null\n@@ -0,0 +1,166 @@\n+/* This testcase is part of GDB, the GNU debugger.\n+\n+ Copyright 2017-2019 Free Software Foundation, Inc.\n+\n+ This program is free software; you can redistribute it and/or modify\n+ the Free Software Foundation; either version 3 of the License, or\n+ (at your option) any later version.\n+\n+ This program is distributed in the hope that it will be useful,\n+ but WITHOUT ANY WARRANTY; without even the implied warranty of\n+ MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the\n+ GNU General Public License for more details.\n+\n+ You should have received a copy of the GNU General Public License\n+ along with this program. If not, see <http://www.gnu.org/licenses/>. */\n+\n+#include <stdio.h>\n+#include <omp.h>\n+\n+/* Testcase for checking access to variables in a single / outer scope.\n+ Make sure that variables not referred to in the parallel section are\n+ accessible from the debugger. */\n+\n+void\n+single_scope (void)\n+{\n+ static int s1 = -41, s2 = -42, s3 = -43;\n+ int i1 = 11, i2 = 12, i3 = 13;\n+\n+#pragma omp parallel num_threads (2) shared (s1, i1) private (s2, i2)\n+ {\n+\n+ s2 = 100 * (thread_num + 1) + 2;\n+ i2 = s2 + 10;\n+\n+ #pragma omp critical\n+ printf (\"single_scope: thread_num=%d, s1=%d, i1=%d, s2=%d, i2=%d\\n\",\n+\t thread_num, s1, i1, s2, i2);\n+ }\n+\n+ printf (\"single_scope: s1=%d, s2=%d, s3=%d, i1=%d, i2=%d, i3=%d\\n\",\n+\t s1, s2, s3, i1, i2, i3);\n+}\n+\n+static int file_scope_var = 9876;\n+\n+ nested within more than one lexical scope. Of particular interest\n+ are variables which are not referenced in the parallel section. */\n+\n+void\n+multi_scope (void)\n+{\n+ int i01 = 1, i02 = 2;\n+\n+ {\n+ int i11 = 11, i12 = 12;\n+\n+ {\n+ int i21 = -21, i22 = 22;\n+\n+#pragma omp parallel num_threads (2) \\\n+\t\t firstprivate (i01) \\\n+\t\t shared (i11) \\\n+\t\t private (i21)\n+\t{\n+\t i21 = 100 * (thread_num + 1) + 21;\n+\n+\t #pragma omp critical\n+\t printf (\"multi_scope: thread_num=%d, i01=%d, i11=%d, i21=%d\\n\",\n+\t}\n+\n+\tprintf (\"multi_scope: i01=%d, i02=%d, i11=%d, \"\n+\t\t\"i12=%d, i21=%d, i22=%d\\n\",\n+\t\ti01, i02, i11, i12, i21, i22);\n+ }\n+ }\n+}\n+\n+/* Nested functions in C is a GNU extension. Some non-GNU compilers\n+ define __GNUC__, but they don't support nested functions. So,\n+ unfortunately, we can't use that for our test. */\n+#if HAVE_NESTED_FUNCTION_SUPPORT\n+\n+/* Testcase for checking access of variables from within parallel\n+ region in a lexically nested function. */\n+\n+void\n+nested_func (void)\n+{\n+ static int s1 = -42;\n+ int i = 1, j = 2, k = 3;\n+\n+ void\n+ foo (int p, int q, int r)\n+ {\n+ int x = 4;\n+\n+ {\n+ int y = 5, z = 6;\n+#pragma omp parallel num_threads (2) shared (i, p, x) private (j, q, y)\n+ {\n+\tint tn = omp_get_thread_num ();\n+\n+\tj = 1000 * (tn + 1);\n+\tq = j + 1;\n+\ty = q + 1;\n+\t#pragma omp critical\n+\tprintf (\"nested_func: tn=%d: i=%d, p=%d, x=%d, j=%d, q=%d, y=%d\\n\",\n+\t\t tn, i, p, x, j, q, y);\n+ }\n+ }\n+ }\n+\n+ foo (10, 11, 12);\n+\n+ i = 101; j = 102; k = 103;\n+ foo (20, 21, 22);\n+}\n+#endif\n+\n+/* Testcase for checking access to variables from within a nested parallel\n+ region. */\n+\n+void\n+nested_parallel (void)\n+{\n+ int i = 1, j = 2;\n+ int l = -1;\n+\n+ omp_set_nested (1);\n+ omp_set_dynamic (0);\n+#pragma omp parallel num_threads (2) private (l)\n+ {\n+ int num = omp_get_thread_num ();\n+ int nthr = omp_get_num_threads ();\n+ int off = num * nthr;\n+ int k = off + 101;\n+ l = off + 102;\n+#pragma omp parallel num_threads (2) shared (num)\n+ {\n+ int inner_num = omp_get_thread_num ();\n+ #pragma omp critical\n+ }\n+ #pragma omp critical\n+ printf (\"nested_parallel (outer threads) %d: k = %d, l = %d\\n\", num, k, l);\n+ }\n+}\n+\n+int\n+main (int argc, char **argv)\n+{\n+ single_scope ();\n+ multi_scope ();\n+#if HAVE_NESTED_FUNCTION_SUPPORT\n+ nested_func ();\n+#endif\n+ nested_parallel ();\n+ return 0;\n+}\n+\nnew file mode 100644\nindex 0000000..e237b91\n--- /dev/null\n@@ -0,0 +1,290 @@\n+# Copyright 2017-2019 Free Software Foundation, Inc.\n+\n+# This program is free software; you can redistribute it and/or modify\n+# the Free Software Foundation; either version 3 of the License, or\n+# (at your option) any later version.\n+#\n+# This program is distributed in the hope that it will be useful,\n+# but WITHOUT ANY WARRANTY; without even the implied warranty of\n+# MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the\n+# GNU General Public License for more details.\n+#\n+# You should have received a copy of the GNU General Public License\n+# along with this program. If not, see <http://www.gnu.org/licenses/>.\n+\n+# This file is part of the gdb testsuite.\n+\n+# Tests which verify (or not) that GDB can access in-scope variables\n+# when stopped within an OpenMP parallel region.\n+\n+standard_testfile\n+\n+set have_nested_function_support 0\n+set opts {openmp debug}\n+if [support_nested_function_tests] {\n+ set have_nested_function_support 1\n+}\n+\n+if {[prepare_for_testing \"failed to prepare\" \\$testfile \\$srcfile \\$opts]} {\n+ return -1\n+}\n+\n+# gdb_openmp_setup may be defined to set auto-load safe-path and possibly\n+# sysroot. These settings are required for gdb to be able to find\n+# the libgomp python plugin. (sysroot only needs to be defined for\n+# remote debugging.)\n+#\n+# This approach has both pros and cons. On the plus side, it's easy\n+# to automatically set a precise auto-load safe-path. (It's easy because\n+# the output of ldd on the binary may be examined to learn the location\n+# of libgomp.so.)\n+#\n+# However, making these settings is also a drawback due to potentially\n+# overriding settings made by a board file. Therefore, this proc\n+# is optional and will only be called if it's defined.\n+\n+if {[info procs gdb_openmp_setup] != \"\"} {\n+ if {[gdb_openmp_setup \\$binfile] != \"\"} {\n+\tuntested \"could not set up OpenMP environment\"\n+\treturn -1\n+ }\n+}\n+\n+if {![runto_main]} {\n+ untested \"could not run to main\"\n+ return -1\n+}\n+\n+# We want to invoke setup_kfail (and in some cases setup_xfail) when\n+# GDB does not yet have support for finding the values of variables in\n+# (non-master) threads. We'll check this by looking at the output of\n+# \"maint print thread-parent\". If this command is undefined, then GDB\n+# does not yet have thread parent support, and it makes sense to kfail\n+# tests which won't work. It's possible for GDB to have this support,\n+# but not work. E.g. it may be the case that the plugin doesn't\n+# constraints related to setting allow_kfail to 0. But, for the moment,\n+# this simple test should be sufficient.\n+\n+set allow_kfail 1\n+ -re \"Undefined maintenance print command.*\\$gdb_prompt\" {\n+\tpass \"maint print thread-parent (does not exist)\"\n+ }\n+ -re \"No parent found.*\" {\n+\tset allow_kfail 0\n+ }\n+}\n+\n+# Determine whether to xfail some of the tests based on GCC version.\n+#\n+# This may need to be tweaked somewhat. Testing shows that GCC 7.3.1\n+# needs the xfails. GCC 8.3.1 and 9.1.1 do not. The assumption made\n+# below is that all versions of gcc 8 and above won't require the\n+# XFAIL setup and that all versions of gcc 7 and below will, but it's\n+# possible that there are versions in between 7.3.1 and 8.3.1 for\n+# which this assumption is invalid.\n+\n+set have_older_gcc 0\n+if {[test_compiler_info {gcc-[0-7]-*}]} {\n+ set have_older_gcc 1\n+}\n+\n+# maybe_setup_kfail will set up a kfail for gdb/22214 when COND holds in\n+# addition to considering the values of \\$have_older_gcc and \\$allow_kfail.\n+#\n+# When \\$have_older_gcc evaluates to true, setup_xfail will invoked\n+\n+proc maybe_setup_kfail {cond} {\n+ global have_older_gcc allow_kfail\n+ if {\\$have_older_gcc} {\n+\tsetup_xfail *-*-*\n+ } elseif {[uplevel 1 [list expr \\$cond]] && \\$allow_kfail} {\n+\tsetup_kfail \"gdb/22214\" *-*-*\n+ }\n+}\n+\n+with_test_prefix \"single_scope\" {\n+\n+ gdb_breakpoint [gdb_get_line_number \"single_scope: s1=\"]\n+\n+\twith_test_prefix \\$pref {\n+\t gdb_continue_to_breakpoint \"at printf\"\n+\n+\t if {\\$have_older_gcc} { setup_xfail \"*-*-*\" }\n+\t if {\\$have_older_gcc} { setup_xfail \"*-*-*\" }\n+\t gdb_test \"print s1\" \"= -41\"\n+\t gdb_test \"print s2\" \"= \\[12\\]02\"\n+\t if {\\$have_older_gcc} { setup_xfail \"*-*-*\" }\n+\t gdb_test \"print s3\" \"= -43\"\n+\t gdb_test \"print i1\" \"= 11\"\n+\t gdb_test \"print i2\" \"= \\12\"\n+\t gdb_test \"print i3\" \"= 13\"\n+\t}\n+ }\n+\n+ with_test_prefix \"after parallel region\" {\n+\tgdb_continue_to_breakpoint \"at printf\"\n+\n+\tgdb_test \"print s1\" \"= -41\"\n+\tgdb_test \"print s2\" \"= -42\"\n+\tgdb_test \"print s3\" \"= -43\"\n+\tgdb_test \"print i1\" \"= 11\"\n+\tgdb_test \"print i2\" \"= 12\"\n+\tgdb_test \"print i3\" \"= 13\"\n+ }\n+\n+}\n+\n+with_test_prefix \"multi_scope\" {\n+ gdb_breakpoint [gdb_get_line_number \"multi_scope: i01=\"]\n+\n+\twith_test_prefix \\$pref {\n+\t gdb_continue_to_breakpoint \"at printf\"\n+\n+\t if {\\$have_older_gcc} { setup_xfail \"*-*-*\" }\n+\n+\t gdb_test \"print i01\" \"= 1\"\n+\t gdb_test \"print i02\" \"= 2\"\n+\t gdb_test \"print i11\" \"= 11\"\n+\t gdb_test \"print i12\" \"= 12\"\n+\t gdb_test \"print i21\" \"= \\[12\\]21\"\n+\t gdb_test \"print i22\" \"= 22\"\n+\t gdb_test \"print file_scope_var\" \"= 9876\"\n+\t}\n+ }\n+\n+ with_test_prefix \"after parallel\" {\n+\tgdb_continue_to_breakpoint \"at printf\"\n+\n+\tgdb_test \"print i01\" \"= 1\"\n+\tgdb_test \"print i02\" \"= 2\"\n+\tgdb_test \"print i11\" \"= 11\"\n+\tgdb_test \"print i12\" \"= 12\"\n+\tgdb_test \"print i21\" \"= -21\"\n+\tgdb_test \"print i22\" \"= 22\"\n+\tgdb_test \"print file_scope_var\" \"= 9876\"\n+ }\n+}\n+\n+# Nested functions in C are a GNU extension, so only do the nested function\n+# tests if compiling with -DHAVE_NESTED_FUNCTION_SUPPORT was successful.\n+\n+if \\$have_nested_function_support {\n+ with_test_prefix \"nested_func\" {\n+\tgdb_breakpoint [gdb_get_line_number \"nested_func: tn=\"]\n+\n+\tforeach call_prefix {\"1st call\" \"2nd call\"} {\n+\t with_test_prefix \\$call_prefix {\n+\t\t\tgdb_continue_to_breakpoint \"at printf\"\n+\n+\t\t\tif {\\$have_older_gcc} { setup_xfail \"*-*-*\" }\n+\t\t\tset thread_num [get_valueof \"\" \"tn\" \"unknown\"]\n+\n+\t\t\tgdb_test \"print file_scope_var\" \"= 9876\"\n+\t\t\tif {\\$have_older_gcc} { setup_xfail *-*-* }\n+\t\t\tgdb_test \"print s1\" \"= -42\"\n+\t\t\tif {\\$call_prefix eq \"1st call\"} {\n+\t\t\t gdb_test \"print i\" \"= 1\"\n+\t\t\t} else {\n+\t\t\t gdb_test \"print i\" \"= 101\"\n+\t\t\t}\n+\t\t\tgdb_test \"print j\" \"= \\[12\\]000\"\n+\t\t\tif {\\$call_prefix eq \"1st call\"} {\n+\t\t\t gdb_test \"print k\" \"= 3\"\n+\t\t\t} else {\n+\t\t\t gdb_test \"print k\" \"= 103\"\n+\t\t\t}\n+\t\t\tif {\\$call_prefix eq \"1st call\"} {\n+\t\t\t gdb_test \"print p\" \"= 10\"\n+\t\t\t} else {\n+\t\t\t gdb_test \"print p\" \"= 20\"\n+\t\t\t}\n+\t\t\tgdb_test \"print q\" \"= \\[12\\]001\"\n+\t\t\tif {\\$call_prefix eq \"1st call\"} {\n+\t\t\t gdb_test \"print r\" \"= 12\"\n+\t\t\t} else {\n+\t\t\t gdb_test \"print r\" \"= 22\"\n+\t\t\t}\n+\t\t\tgdb_test \"print x\" \"= 4\"\n+\t\t\tgdb_test \"print y\" \"= \\[12\\]002\"\n+\t\t\tgdb_test \"print z\" \"= 6\"\n+\t\t\tif {\\$have_older_gcc} { setup_xfail \"*-*-*\" }\n+\t\t\tgdb_test \"print tn\" \"= \\[01\\]\"\n+\t\t }\n+\t\t}\n+\t }\n+\t}\n+ }\n+}\n+\n+with_test_prefix \"nested_parallel\" {\n+ gdb_breakpoint [gdb_get_line_number \"nested_parallel (inner threads)\"]\n+\n+\tforeach pref {\"1st stop\" \"2nd stop\" \"3rd stop\" \"4th stop\"} {\n+\t with_test_prefix \\$pref {\n+\t\tgdb_continue_to_breakpoint \"at printf\"\n+\n+\t\t# Don't need setup_xfail here due to fact that num is made\n+\t\t# made known to the inner parallel region.\n+\t\tset thread_num [get_valueof \"\" \"num\" \"unknown\"]\n+\n+\t\tif {\\$have_older_gcc} { setup_xfail \"*-*-*\" }\n+\t\tset inner_thread_num [get_valueof \"\" \"inner_num\" \"unknown\"]\n+\n+\t\tgdb_test \"print file_scope_var\" \"= 9876\"\n+\t\tgdb_test \"print num\" \"= \\[01\\]\"\n+\t\tgdb_test \"print i\" \"= 1\"\n+\t\tgdb_test \"print j\" \"= 2\"\n+\t\tif {\\$have_older_gcc || (\\$inner_thread_num != 0 && \\$allow_kfail)} { setup_xfail *-*-* }\n+\t\tgdb_test \"print l\" \"= 10\\[24\\]\"\n+\t\tif {\\$have_older_gcc ||( \\$inner_thread_num != 0 && \\$allow_kfail)} { setup_xfail *-*-* }\n+\t\tgdb_test \"print k\" \"= 10\\[13\\]\"\n+\t }\n+\t}\n+ }\n+\n+\tgdb_breakpoint [gdb_get_line_number \"nested_parallel (outer threads)\"]\n+\n+\twith_test_prefix \"outer stop\" {\n+\t gdb_continue_to_breakpoint \"at printf\"\n+\n+\t if {\\$have_older_gcc} { setup_xfail \"*-*-*\" }\n+\t set thread_num [get_valueof \"\" \"num\" \"unknown\"]\n+\n+\t gdb_test \"print file_scope_var\" \"= 9876\"\n+\t if {\\$have_older_gcc} { setup_xfail \"*-*-*\" }\n+\t gdb_test \"print num\" \"= \\[01\\]\"" ]

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[ "# Convert 36 cm to inches\n\n## How many inches in a cm?\n\nWhen you wish to convert 36 centimeters into a number in inches, first, you must know how many inches 1 centimeter represents.\n\nThis is how I will tell you directly one centimeter is equal to 0.3937 inches.\n\n## Centimeter Definition\n\nThe centimeter unit is a base unit of length.\nIt equals to 10 millimeters.\nThis unit is used in CGS system, maps, home repaire and all areas in our life.\nA single centimeter is about the same as 39.37 inches.\n\n## Meaning of Inch\n\nThe inch is a unit of length in the UK and the US customary systems of measurement. An inch is equal to 1/12 of a foot or 1/36 yard.\n\n## How do you convert 1 cm into inches?\n\nFor 1cm to inches conversion, multiply 1cm using a conversion factor 0.3937.\n\nThis makes it much easier to convert 36 cm to inches.\n\nTherefore, 1 cm to inches = 1 x 0.3937 = 0.3937 inches.\n\nThis information will help you answer the following questions easily and clearly.\n\n• What is 1 cm equal to in inches?\n• What is cm to inches conversion formula?\n• How many inches are equal to 1 cm?\n• What is 1 cm in inches equal?\n\n### How do I convert 36 cm to inches?\n\nYou have fully understood cm to inches by the above.\n\nThis is it:\n\nValue in inches = value in cm × 0.3937\n\nSo, 36 cm to inches = 36 cm × 0.3937 = 14.1732 inches\n\nThese questions can be answered using this formula:\n\n• What’s the formula to convert inches from 36 cm?\n• How do I convert cm to inches?\n• How to change cm to inches?\n• How to calculate cm to inches?\n• What size are 36 cm into inches?\n\n cm inches 35.6 cm 14.01572 inches 35.65 cm 14.035405 inches 35.7 cm 14.05509 inches 35.75 cm 14.074775 inches 35.8 cm 14.09446 inches 35.85 cm 14.114145 inches 35.9 cm 14.13383 inches 35.95 cm 14.153515 inches 36 cm 14.1732 inches 36.05 cm 14.192885 inches 36.1 cm 14.21257 inches 36.15 cm 14.232255 inches 36.2 cm 14.25194 inches 36.25 cm 14.271625 inches 36.3 cm 14.29131 inches 36.35 cm 14.310995 inches 36.4 cm 14.33068 inches" ]

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[ "# Código fuente para qiskit.algorithms.time_evolvers.variational.variational_principles.imaginary_mc_lachlan_principle\n\n```# This code is part of Qiskit.\n#\n#\n# obtain a copy of this license in the LICENSE.txt file in the root directory\n#\n# Any modifications or derivative works of this code must retain this\n# copyright notice, and modified files need to carry a notice indicating\n# that they have been altered from the originals.\n\n\"\"\"Class for an Imaginary McLachlan's Variational Principle.\"\"\"\nfrom __future__ import annotations\n\nimport warnings\n\nfrom collections.abc import Sequence\n\nimport numpy as np\n\nfrom qiskit import QuantumCircuit\nfrom qiskit.circuit import Parameter\nfrom qiskit.primitives import Estimator\nfrom qiskit.quantum_info.operators.base_operator import BaseOperator\n\nfrom .imaginary_variational_principle import ImaginaryVariationalPrinciple\n\nfrom ....exceptions import AlgorithmError\nBaseQGT,\nDerivativeType,\nLinCombQGT,\n)\n\n[documentos]class ImaginaryMcLachlanPrinciple(ImaginaryVariationalPrinciple):\n\"\"\"Class for an Imaginary McLachlan's Variational Principle. It aims to minimize the distance\nbetween both sides of the Wick-rotated Schrödinger equation with a quantum state given as a\nparametrized trial state. The principle leads to a system of linear equations handled by a\nlinear solver. The imaginary variant means that we consider imaginary time dynamics.\n\"\"\"\n\ndef __init__(\nself,\nqgt: BaseQGT | None = None,\n) -> None:\n\"\"\"\nArgs:\nqgt: Instance of a the GQT class used to compute the QFI.\nIf ``None`` provided, ``LinCombQGT`` is used.\nIf ``None`` provided, ``LinCombEstimatorGradient`` is used.\n\nRaises:\nAlgorithmError: If the gradient instance does not contain an estimator.\n\"\"\"\n\ntry:\nexcept Exception as exc:\nraise AlgorithmError(\n\"The provided gradient instance does not contain an estimator primitive.\"\n) from exc\nelse:\nestimator = Estimator()\n\nif qgt is None:\nqgt = LinCombQGT(estimator)\n\nself,\nhamiltonian: BaseOperator,\nansatz: QuantumCircuit,\nparam_values: Sequence[float],\ngradient_params: Sequence[Parameter] | None = None,\n) -> np.ndarray:\n\"\"\"\nCalculates an evolution gradient according to the rules of this variational principle.\n\nArgs:\nhamiltonian: Operator used for Variational Quantum Time Evolution.\nansatz: Quantum state in the form of a parametrized quantum circuit.\nparam_values: Values of parameters to be bound.\ngradient_params: List of parameters with respect to which gradients should be computed.\nIf ``None`` given, gradients w.r.t. all parameters will be computed.\n\nReturns:\n\nRaises:\nAlgorithmError: If a gradient job fails.\n\"\"\"\n\ntry:\n.result()\n)\n\nexcept Exception as exc:\nraise AlgorithmError(\"The gradient primitive job failed!\") from exc\n\n@staticmethod" ]

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[ "# How to Do Hierarchical Clustering in Python ? 5 Easy Steps Only", null, "Hierarchical Clustering is a type of the Unsupervised Machine Learning algorithm that is used for labeling the dataset. When you hear the words labeling the dataset, it means you are clustering the data points that have the same characteristics. It allows you to predict the subgroups from the dataset. In this tutorial of “How to, ” you will learn How to Do Hierarchical Clustering in Python?\n\nBefore going to the coding part to learn Hierarchical Clustering in python more, you must know the some of the terms that give you more understanding. It’s just a brief summary.\n\n## What is Hierarchical Clustering?\n\nHierarchical Clustering uses the distance based approach between the neighbor datapoints for clustering. Each data point is linked to its nearest neighbors.  There are two ways you can do Hierarchical clustering Agglomerative that is bottom-up approach clustering and Divisive uses top-down approaches for clustering. In this tutorial, I will use the popular approach Agglomerative way.\n\nIn order to find the number of subgroups in the dataset, you use dendrogram. It allows you to see linkages, relatedness using the tree graph.\n\nYou will find many use cases for this type of clustering and some of them are DNA sequencing, Sentiment Analysis, Tracking Virus Diseases e.t.c. Popular Use Cases are Hospital Resource Management, Business Process Management, and Social Network Analysis.\n\n## Step 1: Import the necessary Libraries for the Hierarchical Clustering\n\n``````import numpy as np\nimport pandas as pd\n\nimport scipy\nfrom scipy.cluster.hierarchy import fcluster\nfrom scipy.cluster.hierarchy import cophenet\nfrom scipy.spatial.distance import pdist\n\nimport matplotlib.pyplot as plt\nfrom pylab import rcParams\nimport seaborn as sb\n\nimport sklearn\nfrom sklearn import datasets\nfrom sklearn.cluster import AgglomerativeClustering\nimport sklearn.metrics as sm\nfrom sklearn.preprocessing import scale``````\n\nHere we are importing dendrogram, linkage, cluster, and cophenet from the scipy.cluster.hierarchy packages.\n\n## Step 2: Import  the libraries for the Data Visualization\n\n``````#Configure the output\nnp.set_printoptions(precision=4,suppress=True)\n%matplotlib inline\nrcParams[\"figure.figsize\"] =20,10\nsb.set_style(\"whitegrid\")``````\n\nThe first line np.set_printoptions(precision=4,suppress=True ) method will tell the python interpreter to use float datapoints up to 4 digits after the decimal. I am not going about it in detail. If you want to read about it then here is the link for the numpy.set_printoptions.\n\n## Step 3: Load the Dataset\n\nYou can use your own dataset, but I am using only the default Iris dataset loaded from the Sklearn.\n\n``````iris = datasets.load_iris()\n#scale the data\ndata = scale(iris.data)\ntarget = pd.DataFrame(iris.target)\nvariable_names = iris.feature_names\ndata[0:10]``````", null, "Here data is the input variable(scaled data)  and the target is the output variable. In this case, data is sepal length, sepal length, petal length, and petal width. The target is Iris species type.\n\nPlease note that you should always scale the data for accurate prediction.\n\n## Step 4: Draw the Dendrogram of the dataset.\n\nIn order to estimate the number of centroids. You should verify the number of clusters visually. In this case, you will use the dendrogram.  Use the following code.\n\n``z = linkage(data,\"ward\")``\n``````#generate dendrogram\ndendrogram(z,truncate_mode= \"lastp\", p =12, leaf_rotation=45,leaf_font_size=15, show_contracted=True)\nplt.title(\"Truncated Hierachial Clustering Dendrogram\")\nplt.xlabel(\"Cluster Size\")\nplt.ylabel(\"Distance\")\n#divide the cluster\nplt.axhline(y=15)\nplt.axhline(5)\nplt.axhline(10)\nplt.show()``````\n\nThere are three ways you can link the data points Ward, Complete and Average. In my case, I am using ward linkage parameter. It’s just for the visualizing the dendrogram. But at the last, you will take the distance metrics and linkage parameters on the accuracy score of the model.", null, "Using the Iris dataset and its dendrogram, you can clearly see at distance approx y= 9 Line has divided into three clusters. And also the dataset has three types of species. It means you should choose k=3, that is the number of clusters.\n\n## Step 5: Generate the Hierarchical cluster.\n\nIn this step, you will generate a Hierarchical Cluster using the various affinity and linkage methods. Doing this you will generate different accuracy score. You will choose the method with the largest score.\n\n``````#based on the dendrogram we have two clusetes\nk =3\n#build the model\n#fit the model on the dataset\nHClustering.fit(data)\n#accuracy of the model\nsm.accuracy_score(target,HClustering.labels_)``````\n\nYou can see in the code I am using Agglomerative Clustering with 3 clusters, Euclidean distance parameters and ward as the linkage parameter.  Sklearn metrics sm gives the accuracy score of the model. You may Keep on changing the affinity (Euclidean, Manhatten, Cosine ) and linkage  (ward, complete, average) until you get the best accuracy scores.", null, "I get the highest accuracy score of 0.68 when used Euclidean as affinity and the average as linkage parameters.  Thus It’s obvious that I will choose the third one as Hierarchal Clustering model for the Iris Dataset.\n\n## Other Clustering Alternatives –\n\nApart from the above one technique for clustering you may choose K-mean clustering technique for large data also.\n\n## Conclusion\n\nHierarchical Clustering is a very good way to label the unlabeled dataset.  Hierarchical agglomerative clustering (HAC)  has a time complexity of O(n^3). Thus making it too slow. Therefore, the machine learning algorithm is good for the small dataset. Avoid it to apply it on the large dataset.\n\nWe hope now you now have fully understood the concepts of Hierarchical Clustering. If you have any questions regarding it then you can directly message on our Facebook Page. We are always ready to help you. And the last Don’t forget to subscribe us.\n\nThanks\n\nData Science Learner Team", null, "" ]

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[ "# Cost - Category¶\n\nWarning\n\nThese are proposed functions for next mayor release.\n\n• They are not officially in the current release.\n• They will likely officially be part of the next mayor release:\n• The functions make use of ANY-INTEGER and ANY-NUMERICAL\n• Name might not change. (But still can)\n• Signature might not change. (But still can)\n• Functionality might not change. (But still can)\n• pgTap tests have being done. But might need more.\n• Documentation might need refinement.\n\n## General Information¶\n\n### Characteristics¶\n\nThe main Characteristics are:\n\n• Each function works as part of the family it belongs to.\n• It does not return a path.\n• Returns the sum of the costs of the resulting path(s) for pair combination of nodes in the graph.\n• Process is done only on edges with positive costs.\n• Values are returned when there is a path.\n• The returned values are in the form of a set of (start_vid, end_vid, agg_cost).\n• When the starting vertex and ending vertex are the same, there is no path.\n• The agg_cost int the non included values (v, v) is 0.\n• When the starting vertex and ending vertex are the different and there is no path.\n• The agg_cost in the non included values (u, v) is $$\\infty$$.\n• Let be the case the values returned are stored in a table, so the unique index would be the pair: (start_vid, end_vid).\n• Depending on the function and its parameters, the results can be symmetric.\n• The agg_cost of (u, v) is the same as for (v, u).\n• Any duplicated value in the start_vids or in end_vids are ignored.\n• The returned values are ordered:\n• start_vid ascending\n• end_vid ascending\n\n### See Also¶\n\nIndices and tables" ]

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[ "# Is every real matrix conjugate to a semi antisymmetric matrix?\n\nIs it true to say that every matrix $A\\in M_n(\\mathbb{R})$ is similar (conjugate) to a matrix $B=(b_{ij})$ with $b_{ij}=-b_{ji}$ for all $i\\neq j$?(With some abuse of terminology,a matrix $B$ with this property is called \"Semi antisymmetric\").\n\n• Just to be clear: there is no restriction on the diagonal of B, so this is a somewhat unusual usage of the term \"antisymmetric matrix\", right? – Nathaniel Johnston Oct 5 '17 at 0:23\n• It seems that for an identity matrix all the conjugates are also identity matrices, so the statement can be true only if there is no restriction on the diagonal, e.g. the identity matrix is defined to be antisymmetric. – Sergey Dovgal Oct 5 '17 at 2:39\n• I also assume that only \"real\" basis changes are desired, because otherwise it can be proven without much effort that the statement is true. – Sergey Dovgal Oct 5 '17 at 2:50\n• @NathanielJohnston yes you are right. I called such matrix semi anti symmetric, sonce there is no restriction on the doagonal. – Ali Taghavi Oct 5 '17 at 6:42\n• @SergeyDovgal In my question there is no restriction on diagonal. The congugacy is assumed via real matrices not complex matrices. With such assumption, do you think the answer to my question is affirmative? – Ali Taghavi Oct 5 '17 at 6:46\n\nYes. Every matrix can be written as the sum of a symmetric plus an antisymmetric one: $A = \\frac{A+A^T}{2}+\\frac{A-A^T}{2}$. Now change basis such that the symmetric part is diagonal." ]

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[ "# ML\n\nML\n\nFew terms to begin with:-\n\nLoss Function: tells how good a prediction model is in terms of how close it is able to predict the expected outcome.\n\nObjective Function: algorithms rely on maximizing and minimizing a function called objective fn, and the group of functions that are minimized are known as loss function.\n\nGradient Descent: if a mountain with smooth slopes is loss function, then GD is sliding down that smooth mountain to reach the bottom-most point. It is the most common way to find the min. point of a function.\n\nCategorization of Loss functions:", null, "No need to remember all this at the moment. Just know that:\n\npredicts a label                                         |            predicts a quantity\n\nRegression Loss:\n\n-Most commonly used", null, "-sum of sq. distances of target variable and predicted variable\n\n— aka Mean Square Error (or Quadratic Loss, L2 Loss)\n\nFurther graphs are going to use the example, where:\n\n• target value is 100\n• predicted values range between -10,000 to 10,000\n• The MSE loss becomes minimum when actual value is 100 (i.e. predicted value^2-actual value^2 = 0)", null, "– Yet another loss function is where:\n\n• the sum of absolute differences between our target and predicted variables\n• it measures average magnitude of errors in a set of predictions\n— aka Mean Absolute Error", null, "", null, "" ]

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[ "src/Tools/eqsubst.ML\n author wenzelm Thu Jun 09 16:34:49 2011 +0200 (2011-06-09) changeset 43324 2b47822868e4 parent 41164 6854e9a40edc child 44095 3ea5fae095dc permissions -rw-r--r--\ndiscontinued Name.variant to emphasize that this is old-style / indirect;\n``` 1 (* Title: Tools/eqsubst.ML\n```\n``` 2 Author: Lucas Dixon, University of Edinburgh\n```\n``` 3\n```\n``` 4 A proof method to perform a substiution using an equation.\n```\n``` 5 *)\n```\n``` 6\n```\n``` 7 signature EQSUBST =\n```\n``` 8 sig\n```\n``` 9 (* a type abbreviation for match information *)\n```\n``` 10 type match =\n```\n``` 11 ((indexname * (sort * typ)) list (* type instantiations *)\n```\n``` 12 * (indexname * (typ * term)) list) (* term instantiations *)\n```\n``` 13 * (string * typ) list (* fake named type abs env *)\n```\n``` 14 * (string * typ) list (* type abs env *)\n```\n``` 15 * term (* outer term *)\n```\n``` 16\n```\n``` 17 type searchinfo =\n```\n``` 18 theory\n```\n``` 19 * int (* maxidx *)\n```\n``` 20 * Zipper.T (* focusterm to search under *)\n```\n``` 21\n```\n``` 22 exception eqsubst_occL_exp of\n```\n``` 23 string * int list * thm list * int * thm\n```\n``` 24\n```\n``` 25 (* low level substitution functions *)\n```\n``` 26 val apply_subst_in_asm :\n```\n``` 27 int ->\n```\n``` 28 thm ->\n```\n``` 29 thm ->\n```\n``` 30 (cterm list * int * 'a * thm) * match -> thm Seq.seq\n```\n``` 31 val apply_subst_in_concl :\n```\n``` 32 int ->\n```\n``` 33 thm ->\n```\n``` 34 cterm list * thm ->\n```\n``` 35 thm -> match -> thm Seq.seq\n```\n``` 36\n```\n``` 37 (* matching/unification within zippers *)\n```\n``` 38 val clean_match_z :\n```\n``` 39 theory -> term -> Zipper.T -> match option\n```\n``` 40 val clean_unify_z :\n```\n``` 41 theory -> int -> term -> Zipper.T -> match Seq.seq\n```\n``` 42\n```\n``` 43 (* skipping things in seq seq's *)\n```\n``` 44\n```\n``` 45 (* skipping non-empty sub-sequences but when we reach the end\n```\n``` 46 of the seq, remembering how much we have left to skip. *)\n```\n``` 47 datatype 'a skipseq = SkipMore of int\n```\n``` 48 | SkipSeq of 'a Seq.seq Seq.seq;\n```\n``` 49\n```\n``` 50 val skip_first_asm_occs_search :\n```\n``` 51 ('a -> 'b -> 'c Seq.seq Seq.seq) ->\n```\n``` 52 'a -> int -> 'b -> 'c skipseq\n```\n``` 53 val skip_first_occs_search :\n```\n``` 54 int -> ('a -> 'b -> 'c Seq.seq Seq.seq) -> 'a -> 'b -> 'c Seq.seq\n```\n``` 55 val skipto_skipseq : int -> 'a Seq.seq Seq.seq -> 'a skipseq\n```\n``` 56\n```\n``` 57 (* tactics *)\n```\n``` 58 val eqsubst_asm_tac :\n```\n``` 59 Proof.context ->\n```\n``` 60 int list -> thm list -> int -> tactic\n```\n``` 61 val eqsubst_asm_tac' :\n```\n``` 62 Proof.context ->\n```\n``` 63 (searchinfo -> int -> term -> match skipseq) ->\n```\n``` 64 int -> thm -> int -> tactic\n```\n``` 65 val eqsubst_tac :\n```\n``` 66 Proof.context ->\n```\n``` 67 int list -> (* list of occurences to rewrite, use for any *)\n```\n``` 68 thm list -> int -> tactic\n```\n``` 69 val eqsubst_tac' :\n```\n``` 70 Proof.context -> (* proof context *)\n```\n``` 71 (searchinfo -> term -> match Seq.seq) (* search function *)\n```\n``` 72 -> thm (* equation theorem to rewrite with *)\n```\n``` 73 -> int (* subgoal number in goal theorem *)\n```\n``` 74 -> thm (* goal theorem *)\n```\n``` 75 -> thm Seq.seq (* rewritten goal theorem *)\n```\n``` 76\n```\n``` 77\n```\n``` 78 val fakefree_badbounds :\n```\n``` 79 (string * typ) list ->\n```\n``` 80 term ->\n```\n``` 81 (string * typ) list * (string * typ) list * term\n```\n``` 82\n```\n``` 83 val mk_foo_match :\n```\n``` 84 (term -> term) ->\n```\n``` 85 ('a * typ) list -> term -> term\n```\n``` 86\n```\n``` 87 (* preparing substitution *)\n```\n``` 88 val prep_meta_eq : Proof.context -> thm -> thm list\n```\n``` 89 val prep_concl_subst :\n```\n``` 90 int -> thm -> (cterm list * thm) * searchinfo\n```\n``` 91 val prep_subst_in_asm :\n```\n``` 92 int -> thm -> int ->\n```\n``` 93 (cterm list * int * int * thm) * searchinfo\n```\n``` 94 val prep_subst_in_asms :\n```\n``` 95 int -> thm ->\n```\n``` 96 ((cterm list * int * int * thm) * searchinfo) list\n```\n``` 97 val prep_zipper_match :\n```\n``` 98 Zipper.T -> term * ((string * typ) list * (string * typ) list * term)\n```\n``` 99\n```\n``` 100 (* search for substitutions *)\n```\n``` 101 val valid_match_start : Zipper.T -> bool\n```\n``` 102 val search_lr_all : Zipper.T -> Zipper.T Seq.seq\n```\n``` 103 val search_lr_valid : (Zipper.T -> bool) -> Zipper.T -> Zipper.T Seq.seq\n```\n``` 104 val searchf_lr_unify_all :\n```\n``` 105 searchinfo -> term -> match Seq.seq Seq.seq\n```\n``` 106 val searchf_lr_unify_valid :\n```\n``` 107 searchinfo -> term -> match Seq.seq Seq.seq\n```\n``` 108 val searchf_bt_unify_valid :\n```\n``` 109 searchinfo -> term -> match Seq.seq Seq.seq\n```\n``` 110\n```\n``` 111 (* syntax tools *)\n```\n``` 112 val ith_syntax : int list parser\n```\n``` 113 val options_syntax : bool parser\n```\n``` 114\n```\n``` 115 (* Isar level hooks *)\n```\n``` 116 val eqsubst_asm_meth : Proof.context -> int list -> thm list -> Proof.method\n```\n``` 117 val eqsubst_meth : Proof.context -> int list -> thm list -> Proof.method\n```\n``` 118 val setup : theory -> theory\n```\n``` 119\n```\n``` 120 end;\n```\n``` 121\n```\n``` 122 structure EqSubst: EQSUBST =\n```\n``` 123 struct\n```\n``` 124\n```\n``` 125 (* changes object \"=\" to meta \"==\" which prepares a given rewrite rule *)\n```\n``` 126 fun prep_meta_eq ctxt =\n```\n``` 127 Simplifier.mksimps (simpset_of ctxt) #> map Drule.zero_var_indexes;\n```\n``` 128\n```\n``` 129\n```\n``` 130 (* a type abriviation for match information *)\n```\n``` 131 type match =\n```\n``` 132 ((indexname * (sort * typ)) list (* type instantiations *)\n```\n``` 133 * (indexname * (typ * term)) list) (* term instantiations *)\n```\n``` 134 * (string * typ) list (* fake named type abs env *)\n```\n``` 135 * (string * typ) list (* type abs env *)\n```\n``` 136 * term (* outer term *)\n```\n``` 137\n```\n``` 138 type searchinfo =\n```\n``` 139 theory\n```\n``` 140 * int (* maxidx *)\n```\n``` 141 * Zipper.T (* focusterm to search under *)\n```\n``` 142\n```\n``` 143\n```\n``` 144 (* skipping non-empty sub-sequences but when we reach the end\n```\n``` 145 of the seq, remembering how much we have left to skip. *)\n```\n``` 146 datatype 'a skipseq = SkipMore of int\n```\n``` 147 | SkipSeq of 'a Seq.seq Seq.seq;\n```\n``` 148 (* given a seqseq, skip the first m non-empty seq's, note deficit *)\n```\n``` 149 fun skipto_skipseq m s =\n```\n``` 150 let\n```\n``` 151 fun skip_occs n sq =\n```\n``` 152 case Seq.pull sq of\n```\n``` 153 NONE => SkipMore n\n```\n``` 154 | SOME (h,t) =>\n```\n``` 155 (case Seq.pull h of NONE => skip_occs n t\n```\n``` 156 | SOME _ => if n <= 1 then SkipSeq (Seq.cons h t)\n```\n``` 157 else skip_occs (n - 1) t)\n```\n``` 158 in (skip_occs m s) end;\n```\n``` 159\n```\n``` 160 (* note: outerterm is the taget with the match replaced by a bound\n```\n``` 161 variable : ie: \"P lhs\" beocmes \"%x. P x\"\n```\n``` 162 insts is the types of instantiations of vars in lhs\n```\n``` 163 and typinsts is the type instantiations of types in the lhs\n```\n``` 164 Note: Final rule is the rule lifted into the ontext of the\n```\n``` 165 taget thm. *)\n```\n``` 166 fun mk_foo_match mkuptermfunc Ts t =\n```\n``` 167 let\n```\n``` 168 val ty = Term.type_of t\n```\n``` 169 val bigtype = (rev (map snd Ts)) ---> ty\n```\n``` 170 fun mk_foo 0 t = t\n```\n``` 171 | mk_foo i t = mk_foo (i - 1) (t \\$ (Bound (i - 1)))\n```\n``` 172 val num_of_bnds = (length Ts)\n```\n``` 173 (* foo_term = \"fooabs y0 ... yn\" where y's are local bounds *)\n```\n``` 174 val foo_term = mk_foo num_of_bnds (Bound num_of_bnds)\n```\n``` 175 in Abs(\"fooabs\", bigtype, mkuptermfunc foo_term) end;\n```\n``` 176\n```\n``` 177 (* T is outer bound vars, n is number of locally bound vars *)\n```\n``` 178 (* THINK: is order of Ts correct...? or reversed? *)\n```\n``` 179 fun fakefree_badbounds Ts t =\n```\n``` 180 let val (FakeTs,Ts,newnames) =\n```\n``` 181 List.foldr (fn ((n,ty),(FakeTs,Ts,usednames)) =>\n```\n``` 182 let val newname = singleton (Name.variant_list usednames) n\n```\n``` 183 in ((RWTools.mk_fake_bound_name newname,ty)::FakeTs,\n```\n``` 184 (newname,ty)::Ts,\n```\n``` 185 newname::usednames) end)\n```\n``` 186 ([],[],[])\n```\n``` 187 Ts\n```\n``` 188 in (FakeTs, Ts, Term.subst_bounds (map Free FakeTs, t)) end;\n```\n``` 189\n```\n``` 190 (* before matching we need to fake the bound vars that are missing an\n```\n``` 191 abstraction. In this function we additionally construct the\n```\n``` 192 abstraction environment, and an outer context term (with the focus\n```\n``` 193 abstracted out) for use in rewriting with RWInst.rw *)\n```\n``` 194 fun prep_zipper_match z =\n```\n``` 195 let\n```\n``` 196 val t = Zipper.trm z\n```\n``` 197 val c = Zipper.ctxt z\n```\n``` 198 val Ts = Zipper.C.nty_ctxt c\n```\n``` 199 val (FakeTs', Ts', t') = fakefree_badbounds Ts t\n```\n``` 200 val absterm = mk_foo_match (Zipper.C.apply c) Ts' t'\n```\n``` 201 in\n```\n``` 202 (t', (FakeTs', Ts', absterm))\n```\n``` 203 end;\n```\n``` 204\n```\n``` 205 (* Matching and Unification with exception handled *)\n```\n``` 206 fun clean_match thy (a as (pat, t)) =\n```\n``` 207 let val (tyenv, tenv) = Pattern.match thy a (Vartab.empty, Vartab.empty)\n```\n``` 208 in SOME (Vartab.dest tyenv, Vartab.dest tenv)\n```\n``` 209 end handle Pattern.MATCH => NONE;\n```\n``` 210\n```\n``` 211 (* given theory, max var index, pat, tgt; returns Seq of instantiations *)\n```\n``` 212 fun clean_unify thry ix (a as (pat, tgt)) =\n```\n``` 213 let\n```\n``` 214 (* type info will be re-derived, maybe this can be cached\n```\n``` 215 for efficiency? *)\n```\n``` 216 val pat_ty = Term.type_of pat;\n```\n``` 217 val tgt_ty = Term.type_of tgt;\n```\n``` 218 (* is it OK to ignore the type instantiation info?\n```\n``` 219 or should I be using it? *)\n```\n``` 220 val typs_unify =\n```\n``` 221 SOME (Sign.typ_unify thry (pat_ty, tgt_ty) (Vartab.empty, ix))\n```\n``` 222 handle Type.TUNIFY => NONE;\n```\n``` 223 in\n```\n``` 224 case typs_unify of\n```\n``` 225 SOME (typinsttab, ix2) =>\n```\n``` 226 let\n```\n``` 227 (* is it right to throw away the flexes?\n```\n``` 228 or should I be using them somehow? *)\n```\n``` 229 fun mk_insts env =\n```\n``` 230 (Vartab.dest (Envir.type_env env),\n```\n``` 231 Vartab.dest (Envir.term_env env));\n```\n``` 232 val initenv =\n```\n``` 233 Envir.Envir {maxidx = ix2, tenv = Vartab.empty, tyenv = typinsttab};\n```\n``` 234 val useq = Unify.smash_unifiers thry [a] initenv\n```\n``` 235 handle ListPair.UnequalLengths => Seq.empty\n```\n``` 236 | Term.TERM _ => Seq.empty;\n```\n``` 237 fun clean_unify' useq () =\n```\n``` 238 (case (Seq.pull useq) of\n```\n``` 239 NONE => NONE\n```\n``` 240 | SOME (h,t) => SOME (mk_insts h, Seq.make (clean_unify' t)))\n```\n``` 241 handle ListPair.UnequalLengths => NONE\n```\n``` 242 | Term.TERM _ => NONE\n```\n``` 243 in\n```\n``` 244 (Seq.make (clean_unify' useq))\n```\n``` 245 end\n```\n``` 246 | NONE => Seq.empty\n```\n``` 247 end;\n```\n``` 248\n```\n``` 249 (* Matching and Unification for zippers *)\n```\n``` 250 (* Note: Ts is a modified version of the original names of the outer\n```\n``` 251 bound variables. New names have been introduced to make sure they are\n```\n``` 252 unique w.r.t all names in the term and each other. usednames' is\n```\n``` 253 oldnames + new names. *)\n```\n``` 254 fun clean_match_z thy pat z =\n```\n``` 255 let val (t, (FakeTs,Ts,absterm)) = prep_zipper_match z in\n```\n``` 256 case clean_match thy (pat, t) of\n```\n``` 257 NONE => NONE\n```\n``` 258 | SOME insts => SOME (insts, FakeTs, Ts, absterm) end;\n```\n``` 259 (* ix = max var index *)\n```\n``` 260 fun clean_unify_z sgn ix pat z =\n```\n``` 261 let val (t, (FakeTs, Ts,absterm)) = prep_zipper_match z in\n```\n``` 262 Seq.map (fn insts => (insts, FakeTs, Ts, absterm))\n```\n``` 263 (clean_unify sgn ix (t, pat)) end;\n```\n``` 264\n```\n``` 265\n```\n``` 266 (* FOR DEBUGGING...\n```\n``` 267 type trace_subst_errT = int (* subgoal *)\n```\n``` 268 * thm (* thm with all goals *)\n```\n``` 269 * (cterm list (* certified free var placeholders for vars *)\n```\n``` 270 * thm) (* trivial thm of goal concl *)\n```\n``` 271 (* possible matches/unifiers *)\n```\n``` 272 * thm (* rule *)\n```\n``` 273 * (((indexname * typ) list (* type instantiations *)\n```\n``` 274 * (indexname * term) list ) (* term instantiations *)\n```\n``` 275 * (string * typ) list (* Type abs env *)\n```\n``` 276 * term) (* outer term *);\n```\n``` 277\n```\n``` 278 val trace_subst_err = (Unsynchronized.ref NONE : trace_subst_errT option Unsynchronized.ref);\n```\n``` 279 val trace_subst_search = Unsynchronized.ref false;\n```\n``` 280 exception trace_subst_exp of trace_subst_errT;\n```\n``` 281 *)\n```\n``` 282\n```\n``` 283\n```\n``` 284 fun bot_left_leaf_of (l \\$ r) = bot_left_leaf_of l\n```\n``` 285 | bot_left_leaf_of (Abs(s,ty,t)) = bot_left_leaf_of t\n```\n``` 286 | bot_left_leaf_of x = x;\n```\n``` 287\n```\n``` 288 (* Avoid considering replacing terms which have a var at the head as\n```\n``` 289 they always succeed trivially, and uninterestingly. *)\n```\n``` 290 fun valid_match_start z =\n```\n``` 291 (case bot_left_leaf_of (Zipper.trm z) of\n```\n``` 292 Var _ => false\n```\n``` 293 | _ => true);\n```\n``` 294\n```\n``` 295 (* search from top, left to right, then down *)\n```\n``` 296 val search_lr_all = ZipperSearch.all_bl_ur;\n```\n``` 297\n```\n``` 298 (* search from top, left to right, then down *)\n```\n``` 299 fun search_lr_valid validf =\n```\n``` 300 let\n```\n``` 301 fun sf_valid_td_lr z =\n```\n``` 302 let val here = if validf z then [Zipper.Here z] else [] in\n```\n``` 303 case Zipper.trm z\n```\n``` 304 of _ \\$ _ => [Zipper.LookIn (Zipper.move_down_left z)]\n```\n``` 305 @ here\n```\n``` 306 @ [Zipper.LookIn (Zipper.move_down_right z)]\n```\n``` 307 | Abs _ => here @ [Zipper.LookIn (Zipper.move_down_abs z)]\n```\n``` 308 | _ => here\n```\n``` 309 end;\n```\n``` 310 in Zipper.lzy_search sf_valid_td_lr end;\n```\n``` 311\n```\n``` 312 (* search from bottom to top, left to right *)\n```\n``` 313\n```\n``` 314 fun search_bt_valid validf =\n```\n``` 315 let\n```\n``` 316 fun sf_valid_td_lr z =\n```\n``` 317 let val here = if validf z then [Zipper.Here z] else [] in\n```\n``` 318 case Zipper.trm z\n```\n``` 319 of _ \\$ _ => [Zipper.LookIn (Zipper.move_down_left z),\n```\n``` 320 Zipper.LookIn (Zipper.move_down_right z)] @ here\n```\n``` 321 | Abs _ => [Zipper.LookIn (Zipper.move_down_abs z)] @ here\n```\n``` 322 | _ => here\n```\n``` 323 end;\n```\n``` 324 in Zipper.lzy_search sf_valid_td_lr end;\n```\n``` 325\n```\n``` 326 fun searchf_unify_gen f (sgn, maxidx, z) lhs =\n```\n``` 327 Seq.map (clean_unify_z sgn maxidx lhs)\n```\n``` 328 (Zipper.limit_apply f z);\n```\n``` 329\n```\n``` 330 (* search all unifications *)\n```\n``` 331 val searchf_lr_unify_all =\n```\n``` 332 searchf_unify_gen search_lr_all;\n```\n``` 333\n```\n``` 334 (* search only for 'valid' unifiers (non abs subterms and non vars) *)\n```\n``` 335 val searchf_lr_unify_valid =\n```\n``` 336 searchf_unify_gen (search_lr_valid valid_match_start);\n```\n``` 337\n```\n``` 338 val searchf_bt_unify_valid =\n```\n``` 339 searchf_unify_gen (search_bt_valid valid_match_start);\n```\n``` 340\n```\n``` 341 (* apply a substitution in the conclusion of the theorem th *)\n```\n``` 342 (* cfvs are certified free var placeholders for goal params *)\n```\n``` 343 (* conclthm is a theorem of for just the conclusion *)\n```\n``` 344 (* m is instantiation/match information *)\n```\n``` 345 (* rule is the equation for substitution *)\n```\n``` 346 fun apply_subst_in_concl i th (cfvs, conclthm) rule m =\n```\n``` 347 (RWInst.rw m rule conclthm)\n```\n``` 348 |> IsaND.unfix_frees cfvs\n```\n``` 349 |> RWInst.beta_eta_contract\n```\n``` 350 |> (fn r => Tactic.rtac r i th);\n```\n``` 351\n```\n``` 352 (* substitute within the conclusion of goal i of gth, using a meta\n```\n``` 353 equation rule. Note that we assume rule has var indicies zero'd *)\n```\n``` 354 fun prep_concl_subst i gth =\n```\n``` 355 let\n```\n``` 356 val th = Thm.incr_indexes 1 gth;\n```\n``` 357 val tgt_term = Thm.prop_of th;\n```\n``` 358\n```\n``` 359 val sgn = Thm.theory_of_thm th;\n```\n``` 360 val ctermify = Thm.cterm_of sgn;\n```\n``` 361 val trivify = Thm.trivial o ctermify;\n```\n``` 362\n```\n``` 363 val (fixedbody, fvs) = IsaND.fix_alls_term i tgt_term;\n```\n``` 364 val cfvs = rev (map ctermify fvs);\n```\n``` 365\n```\n``` 366 val conclterm = Logic.strip_imp_concl fixedbody;\n```\n``` 367 val conclthm = trivify conclterm;\n```\n``` 368 val maxidx = Thm.maxidx_of th;\n```\n``` 369 val ft = ((Zipper.move_down_right (* ==> *)\n```\n``` 370 o Zipper.move_down_left (* Trueprop *)\n```\n``` 371 o Zipper.mktop\n```\n``` 372 o Thm.prop_of) conclthm)\n```\n``` 373 in\n```\n``` 374 ((cfvs, conclthm), (sgn, maxidx, ft))\n```\n``` 375 end;\n```\n``` 376\n```\n``` 377 (* substitute using an object or meta level equality *)\n```\n``` 378 fun eqsubst_tac' ctxt searchf instepthm i th =\n```\n``` 379 let\n```\n``` 380 val (cvfsconclthm, searchinfo) = prep_concl_subst i th;\n```\n``` 381 val stepthms = Seq.of_list (prep_meta_eq ctxt instepthm);\n```\n``` 382 fun rewrite_with_thm r =\n```\n``` 383 let val (lhs,_) = Logic.dest_equals (Thm.concl_of r);\n```\n``` 384 in searchf searchinfo lhs\n```\n``` 385 |> Seq.maps (apply_subst_in_concl i th cvfsconclthm r) end;\n```\n``` 386 in stepthms |> Seq.maps rewrite_with_thm end;\n```\n``` 387\n```\n``` 388\n```\n``` 389 (* distinct subgoals *)\n```\n``` 390 fun distinct_subgoals th =\n```\n``` 391 the_default th (SINGLE distinct_subgoals_tac th);\n```\n``` 392\n```\n``` 393 (* General substitution of multiple occurances using one of\n```\n``` 394 the given theorems*)\n```\n``` 395\n```\n``` 396\n```\n``` 397 exception eqsubst_occL_exp of\n```\n``` 398 string * (int list) * (thm list) * int * thm;\n```\n``` 399 fun skip_first_occs_search occ srchf sinfo lhs =\n```\n``` 400 case (skipto_skipseq occ (srchf sinfo lhs)) of\n```\n``` 401 SkipMore _ => Seq.empty\n```\n``` 402 | SkipSeq ss => Seq.flat ss;\n```\n``` 403\n```\n``` 404 (* The occL is a list of integers indicating which occurence\n```\n``` 405 w.r.t. the search order, to rewrite. Backtracking will also find later\n```\n``` 406 occurences, but all earlier ones are skipped. Thus you can use to\n```\n``` 407 just find all rewrites. *)\n```\n``` 408\n```\n``` 409 fun eqsubst_tac ctxt occL thms i th =\n```\n``` 410 let val nprems = Thm.nprems_of th in\n```\n``` 411 if nprems < i then Seq.empty else\n```\n``` 412 let val thmseq = (Seq.of_list thms)\n```\n``` 413 fun apply_occ occ th =\n```\n``` 414 thmseq |> Seq.maps\n```\n``` 415 (fn r => eqsubst_tac'\n```\n``` 416 ctxt\n```\n``` 417 (skip_first_occs_search\n```\n``` 418 occ searchf_lr_unify_valid) r\n```\n``` 419 (i + ((Thm.nprems_of th) - nprems))\n```\n``` 420 th);\n```\n``` 421 val sortedoccL =\n```\n``` 422 Library.sort (Library.rev_order o Library.int_ord) occL;\n```\n``` 423 in\n```\n``` 424 Seq.map distinct_subgoals (Seq.EVERY (map apply_occ sortedoccL) th)\n```\n``` 425 end\n```\n``` 426 end\n```\n``` 427 handle THM _ => raise eqsubst_occL_exp (\"THM\",occL,thms,i,th);\n```\n``` 428\n```\n``` 429\n```\n``` 430 (* inthms are the given arguments in Isar, and treated as eqstep with\n```\n``` 431 the first one, then the second etc *)\n```\n``` 432 fun eqsubst_meth ctxt occL inthms =\n```\n``` 433 SIMPLE_METHOD' (eqsubst_tac ctxt occL inthms);\n```\n``` 434\n```\n``` 435 (* apply a substitution inside assumption j, keeps asm in the same place *)\n```\n``` 436 fun apply_subst_in_asm i th rule ((cfvs, j, ngoalprems, pth),m) =\n```\n``` 437 let\n```\n``` 438 val th2 = Thm.rotate_rule (j - 1) i th; (* put premice first *)\n```\n``` 439 val preelimrule =\n```\n``` 440 (RWInst.rw m rule pth)\n```\n``` 441 |> (Seq.hd o prune_params_tac)\n```\n``` 442 |> Thm.permute_prems 0 ~1 (* put old asm first *)\n```\n``` 443 |> IsaND.unfix_frees cfvs (* unfix any global params *)\n```\n``` 444 |> RWInst.beta_eta_contract; (* normal form *)\n```\n``` 445 (* val elimrule =\n```\n``` 446 preelimrule\n```\n``` 447 |> Tactic.make_elim (* make into elim rule *)\n```\n``` 448 |> Thm.lift_rule (th2, i); (* lift into context *)\n```\n``` 449 *)\n```\n``` 450 in\n```\n``` 451 (* ~j because new asm starts at back, thus we subtract 1 *)\n```\n``` 452 Seq.map (Thm.rotate_rule (~j) ((Thm.nprems_of rule) + i))\n```\n``` 453 (Tactic.dtac preelimrule i th2)\n```\n``` 454\n```\n``` 455 (* (Thm.bicompose\n```\n``` 456 false (* use unification *)\n```\n``` 457 (true, (* elim resolution *)\n```\n``` 458 elimrule, (2 + (Thm.nprems_of rule)) - ngoalprems)\n```\n``` 459 i th2) *)\n```\n``` 460 end;\n```\n``` 461\n```\n``` 462\n```\n``` 463 (* prepare to substitute within the j'th premise of subgoal i of gth,\n```\n``` 464 using a meta-level equation. Note that we assume rule has var indicies\n```\n``` 465 zero'd. Note that we also assume that premt is the j'th premice of\n```\n``` 466 subgoal i of gth. Note the repetition of work done for each\n```\n``` 467 assumption, i.e. this can be made more efficient for search over\n```\n``` 468 multiple assumptions. *)\n```\n``` 469 fun prep_subst_in_asm i gth j =\n```\n``` 470 let\n```\n``` 471 val th = Thm.incr_indexes 1 gth;\n```\n``` 472 val tgt_term = Thm.prop_of th;\n```\n``` 473\n```\n``` 474 val sgn = Thm.theory_of_thm th;\n```\n``` 475 val ctermify = Thm.cterm_of sgn;\n```\n``` 476 val trivify = Thm.trivial o ctermify;\n```\n``` 477\n```\n``` 478 val (fixedbody, fvs) = IsaND.fix_alls_term i tgt_term;\n```\n``` 479 val cfvs = rev (map ctermify fvs);\n```\n``` 480\n```\n``` 481 val asmt = nth (Logic.strip_imp_prems fixedbody) (j - 1);\n```\n``` 482 val asm_nprems = length (Logic.strip_imp_prems asmt);\n```\n``` 483\n```\n``` 484 val pth = trivify asmt;\n```\n``` 485 val maxidx = Thm.maxidx_of th;\n```\n``` 486\n```\n``` 487 val ft = ((Zipper.move_down_right (* trueprop *)\n```\n``` 488 o Zipper.mktop\n```\n``` 489 o Thm.prop_of) pth)\n```\n``` 490 in ((cfvs, j, asm_nprems, pth), (sgn, maxidx, ft)) end;\n```\n``` 491\n```\n``` 492 (* prepare subst in every possible assumption *)\n```\n``` 493 fun prep_subst_in_asms i gth =\n```\n``` 494 map (prep_subst_in_asm i gth)\n```\n``` 495 ((fn l => Library.upto (1, length l))\n```\n``` 496 (Logic.prems_of_goal (Thm.prop_of gth) i));\n```\n``` 497\n```\n``` 498\n```\n``` 499 (* substitute in an assumption using an object or meta level equality *)\n```\n``` 500 fun eqsubst_asm_tac' ctxt searchf skipocc instepthm i th =\n```\n``` 501 let\n```\n``` 502 val asmpreps = prep_subst_in_asms i th;\n```\n``` 503 val stepthms = Seq.of_list (prep_meta_eq ctxt instepthm);\n```\n``` 504 fun rewrite_with_thm r =\n```\n``` 505 let val (lhs,_) = Logic.dest_equals (Thm.concl_of r)\n```\n``` 506 fun occ_search occ [] = Seq.empty\n```\n``` 507 | occ_search occ ((asminfo, searchinfo)::moreasms) =\n```\n``` 508 (case searchf searchinfo occ lhs of\n```\n``` 509 SkipMore i => occ_search i moreasms\n```\n``` 510 | SkipSeq ss =>\n```\n``` 511 Seq.append (Seq.map (Library.pair asminfo) (Seq.flat ss))\n```\n``` 512 (occ_search 1 moreasms))\n```\n``` 513 (* find later substs also *)\n```\n``` 514 in\n```\n``` 515 occ_search skipocc asmpreps |> Seq.maps (apply_subst_in_asm i th r)\n```\n``` 516 end;\n```\n``` 517 in stepthms |> Seq.maps rewrite_with_thm end;\n```\n``` 518\n```\n``` 519\n```\n``` 520 fun skip_first_asm_occs_search searchf sinfo occ lhs =\n```\n``` 521 skipto_skipseq occ (searchf sinfo lhs);\n```\n``` 522\n```\n``` 523 fun eqsubst_asm_tac ctxt occL thms i th =\n```\n``` 524 let val nprems = Thm.nprems_of th\n```\n``` 525 in\n```\n``` 526 if nprems < i then Seq.empty else\n```\n``` 527 let val thmseq = (Seq.of_list thms)\n```\n``` 528 fun apply_occ occK th =\n```\n``` 529 thmseq |> Seq.maps\n```\n``` 530 (fn r =>\n```\n``` 531 eqsubst_asm_tac' ctxt (skip_first_asm_occs_search\n```\n``` 532 searchf_lr_unify_valid) occK r\n```\n``` 533 (i + ((Thm.nprems_of th) - nprems))\n```\n``` 534 th);\n```\n``` 535 val sortedoccs =\n```\n``` 536 Library.sort (Library.rev_order o Library.int_ord) occL\n```\n``` 537 in\n```\n``` 538 Seq.map distinct_subgoals\n```\n``` 539 (Seq.EVERY (map apply_occ sortedoccs) th)\n```\n``` 540 end\n```\n``` 541 end\n```\n``` 542 handle THM _ => raise eqsubst_occL_exp (\"THM\",occL,thms,i,th);\n```\n``` 543\n```\n``` 544 (* inthms are the given arguments in Isar, and treated as eqstep with\n```\n``` 545 the first one, then the second etc *)\n```\n``` 546 fun eqsubst_asm_meth ctxt occL inthms =\n```\n``` 547 SIMPLE_METHOD' (eqsubst_asm_tac ctxt occL inthms);\n```\n``` 548\n```\n``` 549 (* syntax for options, given \"(asm)\" will give back true, without\n```\n``` 550 gives back false *)\n```\n``` 551 val options_syntax =\n```\n``` 552 (Args.parens (Args.\\$\\$\\$ \"asm\") >> (K true)) ||\n```\n``` 553 (Scan.succeed false);\n```\n``` 554\n```\n``` 555 val ith_syntax =\n```\n``` 556 Scan.optional (Args.parens (Scan.repeat Parse.nat)) ;\n```\n``` 557\n```\n``` 558 (* combination method that takes a flag (true indicates that subst\n```\n``` 559 should be done to an assumption, false = apply to the conclusion of\n```\n``` 560 the goal) as well as the theorems to use *)\n```\n``` 561 val setup =\n```\n``` 562 Method.setup @{binding subst}\n```\n``` 563 (Scan.lift (options_syntax -- ith_syntax) -- Attrib.thms >>\n```\n``` 564 (fn ((asmflag, occL), inthms) => fn ctxt =>\n```\n``` 565 (if asmflag then eqsubst_asm_meth else eqsubst_meth) ctxt occL inthms))\n```\n``` 566 \"single-step substitution\";\n```\n``` 567\n```\n``` 568 end;\n```" ]

[ null ]

[ "Popular Topics\n\n# SOLIDWORKS Simulation: Statics 101 Using Simulation\n\nRemember back when you took Engineering Statics and you had to find the forces by hand calculations? You may have said to yourself, why do I have to solve all this by hand when we now have simulation software to get the answers we need. Well, that is true that simulation software can be used to get the answers, but how would you use the software to find unknown forces? Hopefully after reading this blog, I will have cleared the smoke of how to solve the forces of a typical Statics problem by hand and how to use SOLIDWORKS Simulation software to prove what you solved mathematically.\n\nAn example to consider is shown in Figure 1 below which is a ring subjected to force F1 and F2. If it’s required that the resultant force have a magnitude of 1000 N, how would we determine the magnitude of F1 and F2 by hand and by using SOLIDWORKS Simulation software.", null, "## Fig. 1 Ring problem\n\nTo start to solve this problem by hand we need to remember vector addition according to the parallelogram law. From there we can use the tip to tail method to make one single vector triangle. See Figure 2 below.", null, "", null, "## Fig. 2 Vector addition setup\n\nNow that we have a single vector triangle, we can then use the Law of Sines to solve for F1 and F2. See Figure 3 below.", null, "", null, "## Fig. 3 Solved forces\n\nNow that we know what the force values are mathematically, how do we go about finding these forces in SOLIDWORKS Simulation software? Let’s first start with a design of the ring like the one shown in Figure 1 above. Taking advantage of configurations, I made one ring design using very small circular split faces to mimic point loads like they show in any Engineering Statics book and another ring design using larger split faces to mimic rope pulling at the shown angled values of Config two in Figure 4 below. The rectangular ring baseplate was left out of the design because it was not needed for this analysis.", null, "", null, "## Fig. 4 Ring dimensions\n\nFor configuration one the simulation setup is:\n\n• Plain carbon steel is the material type.\n• Reference geometry fixtures are applied to split circular faces A and B and using reference planes on the split circular faces. See Figure 5 below.\n• A force of 1000 Newtons is applied at the top face of the vertical part of the ring.\n• The mesher type used is the curvature based mesher.", null, "## Fig. 5 Config one sim setup\n\nFor configuration two the simulation setup is:\n\n• Plain carbon steel is the material type.\n• Reference geometry fixtures were applied to the split faces shown at A and B and using reference to the angles sketch lines. See Figure 6 below.\n• A force of 1000 Newtons is applied at the top face of the vertical part of the ring.\n• The mesher type used is the curvature based mesher.", null, "## Fig. 6 Config two sim setup\n\nAfter running the analysis on configuration one and two you can then view the forces in the X, Y, Z, and the resultant force by simply right clicking on the Results folder and choosing List Resultant Force. Notice the resultant force values equate to the force values we found earlier by hand calculations. See Figures 7 through 10 below.", null, "## Fig. 7 Config one force values", null, "## Fig. 8 Config one showing Displaying resultant force arrows only", null, "## Fig. 9 Config two force values", null, "## Fig. 10 Config two showing Display resultant force arrows only\n\nIn conclusion it turns out that no matter how you slice it, SOLIDWORKS Simulation gives the same values either way. I hope you have found this informative, and I have shed some light about how to correlate hand calculations to SOLIDWORKS Simulation.\n\nNick Pusateri\nSenior Application Engineer Specialist, Simulation\nComputer Aided Technology" ]

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[ "# Applications of Nonlinear Partial Differential Equations in by Edited by R. Finn", null, "By Edited by R. Finn\n\nRead or Download Applications of Nonlinear Partial Differential Equations in Mathematical Physics. Proceedings of Symposia in Applied Mathematics Volume XVII PDF\n\nSimilar applied books\n\nInteractions Between Electromagnetic Fields and Matter. Vieweg Tracts in Pure and Applied Physics\n\nInteractions among Electromagnetic Fields and subject offers with the rules and techniques which can magnify electromagnetic fields from very low degrees of indications. This ebook discusses how electromagnetic fields will be produced, amplified, modulated, or rectified from very low degrees to allow those for software in communique structures.\n\nKrylov Subspace Methods: Principles and Analysis\n\nThe mathematical conception of Krylov subspace tools with a spotlight on fixing structures of linear algebraic equations is given an in depth therapy during this principles-based e-book. ranging from the belief of projections, Krylov subspace equipment are characterized by way of their orthogonality and minimisation houses.\n\nSmart Structures and Materials: Selected Papers from the 7th ECCOMAS Thematic Conference on Smart Structures and Materials\n\nThis paintings used to be compiled with increased and reviewed contributions from the seventh ECCOMAS Thematic convention on clever buildings and fabrics, that used to be held from three to six June 2015 at Ponta Delgada, Azores, Portugal. The convention supplied a accomplished discussion board for discussing the present cutting-edge within the box in addition to producing idea for destiny principles particularly on a multidisciplinary point.\n\nExtra info for Applications of Nonlinear Partial Differential Equations in Mathematical Physics. Proceedings of Symposia in Applied Mathematics Volume XVII\n\nSample text\n\nV L m , —L v Ki v K2 v . . v K n Prom such a pair, the clause Li v L2 v . . v L m v Ki v K2 v . . K n can be inferred. This clause (called the resolvent) is then added to the set of clauses which have been accumulated from previous inferences. Furthermore, a substitution is considered only if it is the most general that could be made, thus maintaining the maximum degree of generality in the result. Another closely related method of inference is factoring: A substitution is sought such that (a) two or more of the literals of a clause will collapse into a single literal, and (b) no more general substitution would have the same effect.\n\nThe constraint imposed on 3 is: if the resolvent of C and D is a non-unit whose level is a specified bound ko, or a unit whose level exceeds ko, then it is not added to the corresponding list, and the pair is treated as if no resolvent were generated. To illustrate the difficulty thus avoided consider the set consisting of the clauses P(a), — P(x) v P(f(x)), which correspond to a subset of the Peano axioms, and some clause of length 3. , ad infinitum. We would be caught in an infinite loop which would continually present and execute the task of resolving a new unit with the same 2-clause instead of either passing to the proof recovery or to the non-unit section.\n\n6. 7. 8. 9. 10. 11. 12. 13. 14. P(x,e,x) P(e,x,x) - P ( x , y , u ) v -P(y,z,v) - P ( x , y , u ) v -P(y,z,v) P(x,x,e) P(a,b,c) -P(b,a,c) —P(y,z,v) v —P(e,z,w) - P ( y , w , v ) v P(y,v,w) —P(x,z,u) v —P(x,e,w) —P(w,z,u) v P(u,z,w) P(c,b,a) P(c,a,b) -P(c,a,b) v -P(u,z,w) v P(x,v,w) v -P(x,v,w) v P(u,z,w) v P(y,v,w) v P(u,z,w) (Al) (A2) (A3) (A4) (A5) (A6) (A7) (from (from (from (from (from (from (from 5 and 3) 2 and 8) 5 and 4) 1 and 10) 6 and 11) 12 and 9) 7 and 11) Since 13 and 14 are manifestly contradictory, the proof is complete." ]

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[ "", null, "Magicode", null, "# 03.05 Hierarchical Indexing", null, "Up to this point we've been focused primarily on one-dimensional and two-dimensional data, stored in Pandas Series and DataFrame objects, respectively. Often it is useful to go beyond this and store higher-dimensional data–that is, data indexed by more than one or two keys. While Pandas does provide Panel and Panel4D objects that natively handle three-dimensional and four-dimensional data (see Aside: Panel Data), a far more common pattern in practice is to make use of hierarchical indexing (also known as multi-indexing) to incorporate multiple index levels within a single index. In this way, higher-dimensional data can be compactly represented within the familiar one-dimensional Series and two-dimensional DataFrame objects.\nIn this section, we'll explore the direct creation of MultiIndex objects, considerations when indexing, slicing, and computing statistics across multiply indexed data, and useful routines for converting between simple and hierarchically indexed representations of your data.\nWe begin with the standard imports:", null, "import pandas as pd\nimport numpy as np\n\n### A Multiply Indexed Series\n\nLet's start by considering how we might represent two-dimensional data within a one-dimensional Series. For concreteness, we will consider a series of data where each point has a character and numerical key.\n\nSuppose you would like to track data about states from two different years. Using the Pandas tools we've already covered, you might be tempted to simply use Python tuples as keys:", null, "index = [('California', 2000), ('California', 2010),\n('New York', 2000), ('New York', 2010),\n('Texas', 2000), ('Texas', 2010)]\npopulations = [33871648, 37253956,\n18976457, 19378102,\n20851820, 25145561]\npop = pd.Series(populations, index=index)\npop\n(California, 2000) 33871648 (California, 2010) 37253956 (New York, 2000) 18976457 (New York, 2010) 19378102 (Texas, 2000) 20851820 (Texas, 2010) 25145561 dtype: int64\nWith this indexing scheme, you can straightforwardly index or slice the series based on this multiple index:", null, "pop[('California', 2010):('Texas', 2000)]\n(California, 2010) 37253956 (New York, 2000) 18976457 (New York, 2010) 19378102 (Texas, 2000) 20851820 dtype: int64\nBut the convenience ends there. For example, if you need to select all values from 2010, you'll need to do some messy (and potentially slow) munging to make it happen:", null, "pop[[i for i in pop.index if i == 2010]]\n(California, 2010) 37253956 (New York, 2010) 19378102 (Texas, 2010) 25145561 dtype: int64\nThis produces the desired result, but is not as clean (or as efficient for large datasets) as the slicing syntax we've grown to love in Pandas.\n\n#### The Better Way: Pandas MultiIndex\n\nFortunately, Pandas provides a better way. Our tuple-based indexing is essentially a rudimentary multi-index, and the Pandas MultiIndex type gives us the type of operations we wish to have. We can create a multi-index from the tuples as follows:", null, "index = pd.MultiIndex.from_tuples(index)\nindex\nMultiIndex(levels=[['California', 'New York', 'Texas'], [2000, 2010]], labels=[[0, 0, 1, 1, 2, 2], [0, 1, 0, 1, 0, 1]])\nNotice that the MultiIndex contains multiple levels of indexing–in this case, the state names and the years, as well as multiple labels for each data point which encode these levels.\nIf we re-index our series with this MultiIndex, we see the hierarchical representation of the data:", null, "pop = pop.reindex(index)\npop\nCalifornia 2000 33871648 2010 37253956 New York 2000 18976457 2010 19378102 Texas 2000 20851820 2010 25145561 dtype: int64\nHere the first two columns of the Series representation show the multiple index values, while the third column shows the data. Notice that some entries are missing in the first column: in this multi-index representation, any blank entry indicates the same value as the line above it.\nNow to access all data for which the second index is 2010, we can simply use the Pandas slicing notation:", null, "pop[:, 2010]\nCalifornia 37253956 New York 19378102 Texas 25145561 dtype: int64\nThe result is a singly indexed array with just the keys we're interested in. This syntax is much more convenient (and the operation is much more efficient!) than the home-spun tuple-based multi-indexing solution that we started with. We'll now further discuss this sort of indexing operation on hieararchically indexed data.\n\n#### MultiIndex as extra dimension\n\nYou might notice something else here: we could easily have stored the same data using a simple DataFrame with index and column labels. In fact, Pandas is built with this equivalence in mind. The unstack() method will quickly convert a multiply indexed Series into a conventionally indexed DataFrame:", null, "pop_df = pop.unstack()\npop_df\nNaturally, the stack() method provides the opposite operation:", null, "pop_df.stack()\nCalifornia 2000 33871648 2010 37253956 New York 2000 18976457 2010 19378102 Texas 2000 20851820 2010 25145561 dtype: int64\nSeeing this, you might wonder why would we would bother with hierarchical indexing at all. The reason is simple: just as we were able to use multi-indexing to represent two-dimensional data within a one-dimensional Series, we can also use it to represent data of three or more dimensions in a Series or DataFrame. Each extra level in a multi-index represents an extra dimension of data; taking advantage of this property gives us much more flexibility in the types of data we can represent. Concretely, we might want to add another column of demographic data for each state at each year (say, population under 18) ; with a MultiIndex this is as easy as adding another column to the DataFrame:", null, "pop_df = pd.DataFrame({'total': pop,\n'under18': [9267089, 9284094,\n4687374, 4318033,\n5906301, 6879014]})\npop_df\nIn addition, all the ufuncs and other functionality discussed in Operating on Data in Pandas work with hierarchical indices as well. Here we compute the fraction of people under 18 by year, given the above data:", null, "f_u18 = pop_df['under18'] / pop_df['total']\nf_u18.unstack()\nThis allows us to easily and quickly manipulate and explore even high-dimensional data.\n\n### Methods of MultiIndex Creation\n\nThe most straightforward way to construct a multiply indexed Series or DataFrame is to simply pass a list of two or more index arrays to the constructor. For example:", null, "df = pd.DataFrame(np.random.rand(4, 2),\nindex=[['a', 'a', 'b', 'b'], [1, 2, 1, 2]],\ncolumns=['data1', 'data2'])\ndf\nThe work of creating the MultiIndex is done in the background.\nSimilarly, if you pass a dictionary with appropriate tuples as keys, Pandas will automatically recognize this and use a MultiIndex by default:", null, "data = {('California', 2000): 33871648,\n('California', 2010): 37253956,\n('Texas', 2000): 20851820,\n('Texas', 2010): 25145561,\n('New York', 2000): 18976457,\n('New York', 2010): 19378102}\npd.Series(data)\nCalifornia 2000 33871648 2010 37253956 New York 2000 18976457 2010 19378102 Texas 2000 20851820 2010 25145561 dtype: int64\nNevertheless, it is sometimes useful to explicitly create a MultiIndex; we'll see a couple of these methods here.\n\n#### Explicit MultiIndex constructors\n\nFor more flexibility in how the index is constructed, you can instead use the class method constructors available in the pd.MultiIndex. For example, as we did before, you can construct the MultiIndex from a simple list of arrays giving the index values within each level:", null, "pd.MultiIndex.from_arrays([['a', 'a', 'b', 'b'], [1, 2, 1, 2]])\nMultiIndex(levels=[['a', 'b'], [1, 2]], labels=[[0, 0, 1, 1], [0, 1, 0, 1]])\nYou can construct it from a list of tuples giving the multiple index values of each point:", null, "pd.MultiIndex.from_tuples([('a', 1), ('a', 2), ('b', 1), ('b', 2)])\nMultiIndex(levels=[['a', 'b'], [1, 2]], labels=[[0, 0, 1, 1], [0, 1, 0, 1]])\nYou can even construct it from a Cartesian product of single indices:", null, "pd.MultiIndex.from_product([['a', 'b'], [1, 2]])\nMultiIndex(levels=[['a', 'b'], [1, 2]], labels=[[0, 0, 1, 1], [0, 1, 0, 1]])\nSimilarly, you can construct the MultiIndex directly using its internal encoding by passing levels (a list of lists containing available index values for each level) and labels (a list of lists that reference these labels):", null, "pd.MultiIndex(levels=[['a', 'b'], [1, 2]],\nlabels=[[0, 0, 1, 1], [0, 1, 0, 1]])\nMultiIndex(levels=[['a', 'b'], [1, 2]], labels=[[0, 0, 1, 1], [0, 1, 0, 1]])\nAny of these objects can be passed as the index argument when creating a Series or Dataframe, or be passed to the reindex method of an existing Series or DataFrame.\n\n#### MultiIndex level names\n\nSometimes it is convenient to name the levels of the MultiIndex. This can be accomplished by passing the names argument to any of the above MultiIndex constructors, or by setting the names attribute of the index after the fact:", null, "pop.index.names = ['state', 'year']\npop\nstate year California 2000 33871648 2010 37253956 New York 2000 18976457 2010 19378102 Texas 2000 20851820 2010 25145561 dtype: int64\nWith more involved datasets, this can be a useful way to keep track of the meaning of various index values.\n\n#### MultiIndex for columns\n\nIn a DataFrame, the rows and columns are completely symmetric, and just as the rows can have multiple levels of indices, the columns can have multiple levels as well. Consider the following, which is a mock-up of some (somewhat realistic) medical data:", null, "# hierarchical indices and columns\nindex = pd.MultiIndex.from_product([[2013, 2014], [1, 2]],\nnames=['year', 'visit'])\ncolumns = pd.MultiIndex.from_product([['Bob', 'Guido', 'Sue'], ['HR', 'Temp']],\nnames=['subject', 'type'])\n\n# mock some data\ndata = np.round(np.random.randn(4, 6), 1)\ndata[:, ::2] *= 10\ndata += 37\n\n# create the DataFrame\nhealth_data = pd.DataFrame(data, index=index, columns=columns)\nhealth_data\nHere we see where the multi-indexing for both rows and columns can come in very handy. This is fundamentally four-dimensional data, where the dimensions are the subject, the measurement type, the year, and the visit number. With this in place we can, for example, index the top-level column by the person's name and get a full DataFrame containing just that person's information:", null, "health_data['Guido']\nFor complicated records containing multiple labeled measurements across multiple times for many subjects (people, countries, cities, etc.) use of hierarchical rows and columns can be extremely convenient!\n\n### Indexing and Slicing a MultiIndex\n\nIndexing and slicing on a MultiIndex is designed to be intuitive, and it helps if you think about the indices as added dimensions. We'll first look at indexing multiply indexed Series, and then multiply-indexed DataFrames.\n\n#### Multiply indexed Series\n\nConsider the multiply indexed Series of state populations we saw earlier:", null, "pop\nstate year California 2000 33871648 2010 37253956 New York 2000 18976457 2010 19378102 Texas 2000 20851820 2010 25145561 dtype: int64\nWe can access single elements by indexing with multiple terms:", null, "pop['California', 2000]\n33871648\nThe MultiIndex also supports partial indexing, or indexing just one of the levels in the index. The result is another Series, with the lower-level indices maintained:", null, "pop['California']\nyear 2000 33871648 2010 37253956 dtype: int64\nPartial slicing is available as well, as long as the MultiIndex is sorted (see discussion in Sorted and Unsorted Indices):", null, "pop.loc['California':'New York']\nstate year California 2000 33871648 2010 37253956 New York 2000 18976457 2010 19378102 dtype: int64\nWith sorted indices, partial indexing can be performed on lower levels by passing an empty slice in the first index:", null, "pop[:, 2000]\nstate California 33871648 New York 18976457 Texas 20851820 dtype: int64\nOther types of indexing and selection (discussed in Data Indexing and Selection) work as well; for example, selection based on Boolean masks:", null, "pop[pop > 22000000]\nstate year California 2000 33871648 2010 37253956 Texas 2010 25145561 dtype: int64\nSelection based on fancy indexing also works:", null, "pop[['California', 'Texas']]\nstate year California 2000 33871648 2010 37253956 Texas 2000 20851820 2010 25145561 dtype: int64\n\n#### Multiply indexed DataFrames\n\nA multiply indexed DataFrame behaves in a similar manner. Consider our toy medical DataFrame from before:", null, "health_data\nRemember that columns are primary in a DataFrame, and the syntax used for multiply indexed Series applies to the columns. For example, we can recover Guido's heart rate data with a simple operation:", null, "health_data['Guido', 'HR']\nyear visit 2013 1 32.0 2 50.0 2014 1 39.0 2 48.0 Name: (Guido, HR), dtype: float64\nAlso, as with the single-index case, we can use the loc, iloc, and ix indexers introduced in Data Indexing and Selection. For example:", null, "health_data.iloc[:2, :2]\nThese indexers provide an array-like view of the underlying two-dimensional data, but each individual index in loc or iloc can be passed a tuple of multiple indices. For example:", null, "health_data.loc[:, ('Bob', 'HR')]\nyear visit 2013 1 31.0 2 44.0 2014 1 30.0 2 47.0 Name: (Bob, HR), dtype: float64\nWorking with slices within these index tuples is not especially convenient; trying to create a slice within a tuple will lead to a syntax error:", null, "health_data.loc[(:, 1), (:, 'HR')]\nFile \"\", line 1 health_data.loc[(:, 1), (:, 'HR')] ^ SyntaxError: invalid syntax\nYou could get around this by building the desired slice explicitly using Python's built-in slice() function, but a better way in this context is to use an IndexSlice object, which Pandas provides for precisely this situation. For example:", null, "idx = pd.IndexSlice\nhealth_data.loc[idx[:, 1], idx[:, 'HR']]\nThere are so many ways to interact with data in multiply indexed Series and DataFrames, and as with many tools in this book the best way to become familiar with them is to try them out!\n\n### Rearranging Multi-Indices\n\nOne of the keys to working with multiply indexed data is knowing how to effectively transform the data. There are a number of operations that will preserve all the information in the dataset, but rearrange it for the purposes of various computations. We saw a brief example of this in the stack() and unstack() methods, but there are many more ways to finely control the rearrangement of data between hierarchical indices and columns, and we'll explore them here.\n\n#### Sorted and unsorted indices\n\nEarlier, we briefly mentioned a caveat, but we should emphasize it more here. Many of the MultiIndex slicing operations will fail if the index is not sorted. Let's take a look at this here.\nWe'll start by creating some simple multiply indexed data where the indices are not lexographically sorted:", null, "index = pd.MultiIndex.from_product([['a', 'c', 'b'], [1, 2]])\ndata = pd.Series(np.random.rand(6), index=index)\ndata.index.names = ['char', 'int']\ndata\nchar int a 1 0.003001 2 0.164974 c 1 0.741650 2 0.569264 b 1 0.001693 2 0.526226 dtype: float64\nIf we try to take a partial slice of this index, it will result in an error:", null, "try:\ndata['a':'b']\nexcept KeyError as e:\nprint(type(e))\nprint(e)\n<class 'KeyError'> 'Key length (1) was greater than MultiIndex lexsort depth (0)'\nAlthough it is not entirely clear from the error message, this is the result of the MultiIndex not being sorted. For various reasons, partial slices and other similar operations require the levels in the MultiIndex to be in sorted (i.e., lexographical) order. Pandas provides a number of convenience routines to perform this type of sorting; examples are the sort_index() and sortlevel() methods of the DataFrame. We'll use the simplest, sort_index(), here:", null, "data = data.sort_index()\ndata\nchar int a 1 0.003001 2 0.164974 b 1 0.001693 2 0.526226 c 1 0.741650 2 0.569264 dtype: float64\nWith the index sorted in this way, partial slicing will work as expected:", null, "data['a':'b']\nchar int a 1 0.003001 2 0.164974 b 1 0.001693 2 0.526226 dtype: float64\n\n#### Stacking and unstacking indices\n\nAs we saw briefly before, it is possible to convert a dataset from a stacked multi-index to a simple two-dimensional representation, optionally specifying the level to use:", null, "pop.unstack(level=0)", null, "pop.unstack(level=1)\nThe opposite of unstack() is stack(), which here can be used to recover the original series:", null, "pop.unstack().stack()\nstate year California 2000 33871648 2010 37253956 New York 2000 18976457 2010 19378102 Texas 2000 20851820 2010 25145561 dtype: int64\n\n#### Index setting and resetting\n\nAnother way to rearrange hierarchical data is to turn the index labels into columns; this can be accomplished with the reset_index method. Calling this on the population dictionary will result in a DataFrame with a state and year column holding the information that was formerly in the index. For clarity, we can optionally specify the name of the data for the column representation:", null, "pop_flat = pop.reset_index(name='population')\npop_flat\nOften when working with data in the real world, the raw input data looks like this and it's useful to build a MultiIndex from the column values. This can be done with the set_index method of the DataFrame, which returns a multiply indexed DataFrame:", null, "pop_flat.set_index(['state', 'year'])\nIn practice, I find this type of reindexing to be one of the more useful patterns when encountering real-world datasets.\n\n### Data Aggregations on Multi-Indices\n\nWe've previously seen that Pandas has built-in data aggregation methods, such as mean(), sum(), and max(). For hierarchically indexed data, these can be passed a level parameter that controls which subset of the data the aggregate is computed on.", null, "health_data" ]

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[ "# Double and dual numbers\n\n(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)\nJump to: navigation, search\n\nHypercomplex numbers of the form", null, ", where", null, "and", null, "are real numbers, and where the double numbers satisfy the relation", null, ", while the dual numbers satisfy the relation", null, "(cf. Hypercomplex number). Addition of double and dual numbers is defined by", null, "Multiplication of double numbers is defined by", null, "and that of dual numbers by", null, "Complex numbers, double numbers and dual numbers are also called complex numbers of hyperbolic, elliptic and parabolic types, respectively. These numbers are sometimes used to represent motions in the three-dimensional spaces of Lobachevskii, Riemann and Euclid (see, for instance, Helical calculus).\n\nBoth double and dual numbers form two-dimensional (with base 1 and", null, ") associative-commutative algebras over the field of real numbers. As distinct from the field of complex numbers, these algebras comprise zero divisors, all these having the form", null, "in the algebra of double numbers. The algebra of double numbers may be split into a direct sum of two real number fields. Hence yet another name for double numbers — splitting complex numbers. Double numbers have yet another appellation — paracomplex numbers. The algebra of dual numbers is considered not only over the field", null, "of real numbers, but also over an arbitrary field or commutative ring. Let", null, "be a commutative ring and let", null, "be an", null, "-module. The direct sum of", null, "-modules", null, "equipped with the multiplication", null, "is a commutative", null, "-algebra and is denoted by", null, ". It is known as the algebra of dual numbers with respect to the module", null, ". The", null, "-module", null, "is identical with the ideal of the algebra", null, "which is the kernel of the augmentation homomorphism", null, "The square", null, "of this ideal is zero, while", null, ". If", null, "is a regular ring the converse is also true: If", null, "is an", null, "-algebra and", null, "is an ideal in", null, "such that", null, "and", null, ", then", null, ", where", null, "is regarded as an", null, "-module .\n\nIf", null, ", the algebra", null, "(then denoted by", null, ") is isomorphic to the quotient algebra of the algebra of polynomials", null, "by the ideal", null, ". Many properties of an", null, "-module may be formulated as properties of the algebra", null, "; as a result, many problems on", null, "-modules can be reduced to corresponding problems in the theory of rings .\n\nLet", null, "be an arbitrary", null, "-algebra, let", null, "be a homomorphism and let", null, "be a derivation (cf. Derivation in a ring) of", null, "with values in the", null, "-module", null, ", regarded as a", null, "-module with respect to the homomorphism", null, ". The mapping", null, "(", null, ") will then be a homomorphism of", null, "-algebras. Conversely, for any homomorphism of", null, "-algebras", null, "the composition", null, ", where", null, "is the projection of", null, "onto", null, ", is an", null, "-derivation of", null, "with values in", null, ", regarded as a", null, "-module with respect to the homomorphism", null, ". This property of double and dual numbers is utilized for the description of the tangent space to an arbitrary functor in the category of schemes , .\n\nHow to Cite This Entry:\nDouble and dual numbers. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Double_and_dual_numbers&oldid=46769\nThis article was adapted from an original article by I.V. Dolgachev (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. See original article" ]

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[ "", null, "###### Norm Prokup\n\nCornell University\nPhD. in Mathematics\n\nNorm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.\n\n##### Thank you for watching the video.\n\nTo unlock all 5,300 videos, start your free trial.\n\n# Graphing Polynomial Functions with Repeated Factors - Problem 1\n\nNorm Prokup", null, "###### Norm Prokup\n\nCornell University\nPhD. in Mathematics\n\nNorm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.\n\nShare\n\nI want to graph a polynomial function with repeated factors. Here is an example; graph y equals 5x² minus x³. At first we want to factor this is so that I can figure out what the x intercepts are. And there is a common factor of x² so let me pull that out.\n\nI pull that x² and I’m left with 5 minus x. That tells me that the x intercepts are (0, 0) and (5, 0). To get the end behavior I need to look at the leading term. The leading term is the term with the highest power of x and it's minus x³. Remember the leading term may not always be first. So the end behavior is determined by minus x³ and that means it’s going to be the reflection of a cubic. It’ll look something like this. So the left tail is going to go up and the right tail is going to go down.\n\nNow let me plot my x intercepts; (0, 0) (5, 0). And to get an idea of what the shape of this graph is, I want to plot a few points in between the intercepts. Here’s x and here’s x², 5 minus x. Let’s try, how about 2?\n\n2² is 4, 5 minus 2, 3 so we get 12. Let me make this 16 so each of these has an increment of 4. Se here’s 12, 2, 12 goes right here. And then let’s plot, just to get an idea, I have a feeling that it’s going to come in hit the x axis, come up and then go down again. Let me just check the value at 3. 3² is 9 and 5 minus 3 is 2, so goes up to 18, so 2 past 16, quite about there.\n\nNow let’s talk about the behavior near the intercepts. Remember when you have a repeated factor like x², the behavior near the intercept that corresponds to x² (0, 0) is going to look quadratic. It’s going to touch the x axis and just bounce off of it. Remember that the left tail is going to go up so that means that the graph is going to come in like this, bounce off the x axis, so let me draw that in, and then it will up again and it’s going to go up through this points and them come down through and then it will pass right through the x axis because this is just a simple factor, 5 minus x. Let’s see if I can do this, up through this guys and then straight down. Yeah that’s not bad. That’s our graph of y equals 5x² minus x³.\n\nRemember when you’ve got repeated factors the behavior near the intercept that goes with that factor is not going to be simple like this, there’ll be some kind of bouncing off or maybe a cubic behavior if you have an x³ but remember intercepts and behavior, plot some points and look for the behavior near the intercepts." ]

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[ "", null, "Competitions\n\n# Thylacine\n\nGiven the sequence of n integers. Find its nonempty subsequence of consecutive numbers, such that the sum of these numbers is maximal.\n\n#### Input\n\nFirst line contains number n (1n`106`). Second line contains the sequence elements, each of them is no more than 1000.\n\n#### Output\n\nPrint the maximum sum of numbers in non-empty sub-sequence of consecutive numbers.\n\nTime limit 1 second\nMemory limit 122.17 MiB\nInput example #1\n```6\n4 2 3 -7 8 1\n```\nOutput example #1\n```11\n```\nInput example #2\n```8\n2 -1 4 3 -8 6 -3 4\n```\nOutput example #2\n```8\n```\nInput example #3\n```3\n-12 -6 -1\n```\nOutput example #3\n```-1\n```" ]

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[ "# Lyapunov Exponents for independent-nonidentically distributed matrices?\n\nMy question is highlighted in bold at the end.\n\n$\\mathrm{\\underline{Background}}$\n\nConsider a product of i.i.d. $d\\times d$ random matrices $A_{i}$ (with $\\mathbb{E}\\log\\left\\Vert A_{i}\\right\\Vert <\\infty$) acting on a non-zero vecor $X$, i.e. $$A_{n}\\cdots A_{1}X.$$ The Lyapunov exponents are used to describe the exponential growth properties of $$\\left\\Vert A_{n}\\cdots A_{1}X\\right\\Vert .$$ Thus we define the the Lyapunov exponent as $$\\lambda\\left(X\\right):=\\lim_{n\\to\\infty}\\frac{1}{n}\\log\\left\\Vert A_{n}\\cdots A_{1}X\\right\\Vert \\;\\;(1).$$ According to the Muliplicative Ergodic Theorem (or Furstenberg-Kesten Theorem) the number of distinct values $p$ that ($1$) can take is at most $d$, i.e. we have $$\\lambda_{d}\\leq\\cdots\\leq\\lambda_{1}.$$ Now, let $\\sigma_{i,n}$ be the $i$th singular value of the matrix product $A_{n}\\cdots A_{1}$ such that $$\\sigma_{d,n}\\leq\\cdots\\leq\\sigma_{1,n}.$$ Furstenberg-Kesten Theorem states that $$\\lim_{n\\to\\infty}\\frac{1}{n}\\log\\sigma_{i,n}=\\lambda_{i}.\\;\\;(2)$$ The crux of my question (to follow) revolves around ($2$).\n\n$\\underline{\\mathrm{Question\\;Setup: Independent\\;nonidentical\\;matrices}}$\n\nConsider random scalars $a_{i},b_{i},c_{i}$ where $b_{i}s$ are i.i.d. with $\\mathbb{E}\\log\\left|b_{1}\\right|<\\infty$, $c_{i}$s are i.i.d. $\\mathbb{E}\\log\\left|c_{1}\\right|<\\infty$ and $a_{i}s$ are independent but for any $i$ there exists a finite non-zero not-necessarily unitary constant $\\alpha_{i}$ such that $a_{1}\\overset{d}{=}\\alpha_{i}a_{i}$ ($\\overset{d}{=}$ denotes equality in distribution) with $\\mathbb{E}\\log\\left|a_{1}\\right|<\\infty$ . It is easy to show that with $B_{i}$ defined as $$B_{i}:=\\left(\\begin{array}{cc} a_{i} & b_{i}\\\\ 0 & c_{i} \\end{array}\\right)$$ $$\\mathbb{E}\\log\\left\\Vert B_{i}\\right\\Vert <\\infty\\Longleftrightarrow\\mathbb{E}\\log\\left|a_{1}\\right|<\\infty,\\mathbb{E}\\log\\left|b_{1}\\right|<\\infty,\\mathbb{E}\\log\\left|c_{1}\\right|<\\infty.$$ The Lyapunov index of $a_{i}$ is given by $$\\lim_{n\\to\\infty}\\frac{1}{n}\\log\\left|a_{n}\\cdots a_{1}\\right|=\\lim_{n\\to\\infty}\\frac{1}{n}\\log\\left|\\alpha_{n}\\cdots\\alpha_{1}\\right|+\\mathbb{E}\\log\\left|a_{1}\\right|,$$ that of $b_{i}$ is given by $$\\mathbb{E}\\log\\left|b_{1}\\right|$$ and that of $c_{i}$ is given by $$\\mathbb{E}\\log\\left|c_{1}\\right|.$$\n\nWith $\\sigma_{i,n}$ $i=1,2$ defined as the singular values of $A_{n}\\cdots A_{1}$, it can be shown that $$\\frac{1}{n}\\log\\sigma_{1,n}\\to\\max\\left\\{ \\lim_{n\\to\\infty}\\frac{1}{n}\\log\\left|\\alpha_{n}\\cdots\\alpha_{1}\\right|+\\mathbb{E}\\log\\left|a_{1}\\right|,\\mathbb{E}\\log\\left|c_{1}\\right|\\right\\} \\;\\; (3)$$ and $$\\frac{1}{n}\\log\\sigma_{2,n}\\to\\min\\left\\{ \\lim_{n\\to\\infty}\\frac{1}{n}\\log\\left|\\alpha_{n}\\cdots\\alpha_{1}\\right|+\\mathbb{E}\\log\\left|a_{1}\\right|,\\mathbb{E}\\log\\left|c_{1}\\right|\\right\\} \\;\\; (4).$$ Here is my question: As stated previously, according to Furstenberg for i.i.d. matrices $A_{i}$, the limiting behavior of the singular values coincide with the Lyapunov exponents of $A_{i}$, see ($2$). Is this true if one considers my non-i.i.d. setup? In other words, is it right to say that ($3$) and ($4$) are the Lyapunov exponents of $B_i$?" ]

[ null ]

[ "Solutions by everydaycalculation.com\n\n## Subtract 60/70 from 1/10\n\n1/10 - 60/70 is -53/70.\n\n#### Steps for subtracting fractions\n\n1. Find the least common denominator or LCM of the two denominators:\nLCM of 10 and 70 is 70\n\nNext, find the equivalent fraction of both fractional numbers with denominator 70\n2. For the 1st fraction, since 10 × 7 = 70,\n1/10 = 1 × 7/10 × 7 = 7/70\n3. Likewise, for the 2nd fraction, since 70 × 1 = 70,\n60/70 = 60 × 1/70 × 1 = 60/70\n4. Subtract the two like fractions:\n7/70 - 60/70 = 7 - 60/70 = -53/70\n\nMathStep (Works offline)", null, "Download our mobile app and learn to work with fractions in your own time:" ]

[ null, "https://answers.everydaycalculation.com/mathstep-app-icon.png", null ]

[ "PROXY WHOIS RQUOTE TEXTS SOFT FOREX BBOARD\n Radio Music Philosophy Code Literature Russian\n\n= ROOT|Albert_Einstein|Relativity_the_Special_and_General_Theory-3332.txt =\n\npage 10 of 41\n\n system K then corresponds to the embankment, and a co-ordinate system K' to the train. An event, wherever it may have taken place, would be fixed in space with respect to K by the three perpendiculars x, y, z on the co-ordinate planes, and with regard to time by a time value t. Relative to K1, the same event would be fixed in respect of space and time by corresponding values x1, y1, z1, t1, which of course are not identical with x, y, z, t. It has already been set forth in detail how these magnitudes are to be regarded as results of physical measurements. Obviously our problem can be exactly formulated in the following manner. What are the values x1, y1, z1, t1, of an event with respect to K1, when the magnitudes x, y, z, t, of the same event with respect to K are given ? The relations must be so chosen that the law of the transmission of light in vacuo is satisfied for one and the same ray of light (and of course for every ray) with respect to K and K1. For the relative orientation in space of the co-ordinate systems indicated in the diagram (Fig. 2), this problem is solved by means of the equations : eq. 1: file eq01.gif y1 = y z1 = z eq. 2: file eq02.gif This system of equations is known as the \" Lorentz transformation.\" * If in place of the law of transmission of light we had taken as our basis the tacit assumptions of the older mechanics as to the absolute character of times and lengths, then instead of the above we should have obtained the following equations: x1 = x - vt y1 = y z1 = z t1 = t This system of equations is often termed the \" Galilei transformation.\" The Galilei transformation can be obtained from the Lorentz transformation by substituting an infinitely large value for the velocity of light c in the latter transformation. Aided by the following illustration, we can readily see that, in accordance with the Lorentz transformation, the law of the transmission of light in vacuo is satisfied both for the reference-body K and for the reference-body K1. A light-signal is sent along the positive x-axis, and this light-stimulus advances in accordance with the equation x = ct, i.e. with the velocity c. According to the equations of the Lorentz transformation, this simple relation between x and t involves a relation between x1 and t1. In point of fact, if we substitute for x the value ct in the first and fourth equations of the Lorentz transformation, we obtain: eq. 3: file eq03.gif eq. 4: file eq04.gif from which, by division, the expression x1 = ct1 immediately follows. If referred to the system K1, the propagation of light takes place according to this equation. We thus see that the velocity of transmission relative to the reference-body K1 is also equal to c. The same result is obtained for rays of light advancing in any other direction whatsoever. Of cause this is not surprising, since the equations of the Lorentz transformation were derived conformably to this point of view. Notes *) A simple derivation of the Lorentz transformation is given in Appendix I. THE BEHAVIOUR OF MEASURING-RODS AND CLOCKS IN MOTION Place a metre-rod in the x1-axis of K1 in such a manner that one end (the beginning) coincides with the point x1=0 whilst the other end (the end of the rod) coincides with the point x1=I. What is the length of the metre-rod relatively to the system K? In order to learn this, we need only ask where the beginning of the rod and the end of the rod lie with respect to K at a particular time t of the system K. By means of the first equation of the Lorentz transformation the values of these two points at the time t = 0 can be shown to be eq. 05a: file eq05a.gif eq. 05b: file eq05b.gif =10=\n\n 1.4|5|6|7|8|9| < PREV = PAGE 10 = NEXT > |11|12|13|14|15|16.41\n\nUP TO ROOT | UP TO DIR | TO FIRST PAGE", null, "Web soft.rosinstrument.com\n\n0.017297 wallclock secs ( 0.01 usr + 0.00 sys = 0.01 CPU)" ]

[ null, "http://www.google.com/logos/Logo_25gry.gif", null ]

[ "# Custom WebGL Operation in TensorFlow.js", null, "TensorFlow.js is a framework that enables us to run a deep learning model in the web browser easily and efficiently. As you may imagine, tensor manipulation requires a huge amount of computation power. The defacto standard way for that is currently using accelerator such as GPU. GPU provides us with a large amount of parallelism so that we can distribute the tensor calculation in the multiple threads. TensorFlow.js make it possible even in the web browser by using WebGL. WebGL is a standard API to use GPU from web browser mainly for graphical processing. Thanks to WebGL, we can see the rich web contents smoothly. Chrome experiments shows a bunch of fun experience realized by using WebGL.\n\nIn this article, I’m going to briefly explain how to create a custom operator that is leveraged by WebGL acceleration in TensorFlow.js. This explanation based on the official documentation so please refer “Creating custom WebGL operations” more detail.\n\n• Operators in TensorFlow.js\n• Using WebGL as GPGPU\n• Fragment Shader in TensorFlow.js\n• Configuration of GPGPU Program\n• Recap\n\n# Operators in TensorFlow.js\n\nFirst, it’s necessary to know the structure of the kernel in TensorFlow.js. Each operator is implemented as the collection of multiple basic kernels. Kernels are the minimal unit of tensor manipulation. For example, tf.oneHot operator runs oneHot kernel implemented for CPU and WebGL respectively. If WebGL is available, TensorFlow.js chooses the WebGL backend transparently so that the application can be leveraged by WebGL acceleration without rewriting application code.", null, "backend provides the interface to access the kernels. It has CPU and WebGL implementations. Once a kernel has WebGL implementation, it can be run on GPU without any modification to the above layer over backend.\n\nimport * as tf from '@tensorflow/tfjs';\n// oneHot operator is executed by the backend implementation.\ntf.oneHot(tf.tensor1d([0, 1], 'int32'), 3).print();\n\n\nCurrently, TensorFlow.js supports three types of backend officially.\n\n1. CPU\n2. WebGL\n3. Node.js\n\nCPU is just a JavaScript runtime running in the web browser so it’s the slowest implementation. TensorFlow.js should fallback to this implementation if any other backend implementation is not available. WebGL will be described later. You can use TensorFlow core implementation via Node.js. TensorFlow provides C API so that it can be bound by any other languages which support C extension. Node.js backend delegates the computation to TensorFlow core. It indicates that we can use any kind of functionality of TensorFlow technically by using Node.js backend. If you are interested in Node.js backend, the detail is described in the repository.\n\nSo how does WebGL look like?\n\n# Using WebGL as GPGPU\n\nSince WebGL was originally designed for acceleration of graphical processing, it requires some tricky technique to use WebGL for tensor computation. Input tensors are copied as texture to GPU memory. It can be regarded as the simple square and the computation is executed as fragment shader in WebGL pipeline. The following example shows how adding two tensors is done in WebGL. As you can see, tensor A and tensor B are represented as 2x2 images in GPU memory buffer. Hence, each element of these tensors are the pixels of the texture. In general, the texture has RGBA in each element since it’s an image. But it is redundant and memory consuming when it comes to GPGPU case. In the environment supporting WebGL 2.0, gl.R32F texture type is used to avoid allocating memory space for green, blue and alpha pixels. So basically WebGL backend only uses R channel in the tensor computation.", null, "# Fragment Shader in TensorFlow.js\n\nA fragment shader runs in parallel by each pixel, which is accelerated by underlying GPU threads. The function called in parallel in the fragment shader is named main. The main function in TensorFlow.js looks like for example.\n\nvoid main() {\n// This is the 2 element vector representing the coordination of output position.\nivec2 coords = getOutputCoords();\n\n// Get an element in tensor A whose position is exactly to the output.\nfloat a = getA(coords, coords);\nfloat b = getB(coords, coords);\n\nfloat result = a + b;\nsetOutput(result);\n}\n\n\nThis function is called in parallel so that the calculation of each tensor is much faster than single thread calculation by CPU. In order to mitigate the difficulty to write a shader program, TensorFlow.js shader compiler provides some utility functions such as getOutputCoords, getA etc. You can get an element from input tensor safely by using these functions. For example, getOutputCoords returns the position of the output element. By using that you can calculate the position of the input element. The previous code shows the example which is adding two tensors in element-wise. This is the core logic of fragment shader. But how can we decide the shape of the input and output? Can we specify the name of the input tensor?\n\nOf course, you can by using GPGPUProgram class defined in TensorFlow.js.\n\n# Configuration of GPGPU Program\n\nThe interface of GPGPUProgram is defined as follows.\n\ninterface GPGPUProgram {\nvariableNames: string[];\noutputShape: number[];\nuserCode: string;\n}\n\n\nvariableNames is the name list of input variables. It is case insensitive. If you define variableNames = ['A', 'B'], you can get the element of these tensors by calling getA or getB. GPGPUProgram automatically creates the function referred by this name. You can define the shape of the output tensor by defining outputShape. If the output tensor is a 2x2 tensor, the outputShape will be [2, 2]. userCode is the shader code compiled by TensorFlow.js shader compiler.\n\nLet’s say you want to implement a GPGPU program to calculate the squared value of each element. That program can look like this.\n\nclass SquaredProgram implements GPGPUProgram {\nvariableNames = ['A'];\noutputShape: number[];\nuserCode: string;\n\nconstructor(inputShape: number[]) {\n// Element-wise operator generates the tensor whose shape is exactly same with the input one.\nthis.outputShape = inputShape;\nthis.userCode = \nvoid main() {\nfloat a = getAAtOutCoords();\nfloat output = a * a;\nsetOutput(output);\n}\n;\n}\n}\n\n\nSince we defined variableNames as 'A', we can use getAAtOutCoords to get the element in A in the position same as the output position. If the function is called for the position [0, 2], getAAtOutCoords returns an element in the position [0, 2] in tensor A. The function is very helpful for us to write an error durable shader program. Other than that, there are several utility functions defined by TensorFlow.js side.\n\nname functionality\nsetOutput(float) Set the output value in the output position\ngetOutputCoords() Get the coordination of the output position\nisNaN(float) Check whether the value is NaN or not\nround(float) Round value to the nearest integer\n\nOf course, you can use the standard GLSL functions such as sin or cos in GPGPUProgram. Please visit the official OpenGL specification for more detail about the available built-in functions.\n\nOnce you can create your own GPGPUProgram, TensorFlow.js can compile and run the program on behalf of you. For instance, running a program to add two tensors looks like this. Since compileAndRun is defined in the backend implementation, all you need to do is just passing your own program, input, and output.\n\nconst program = new BinaryOpProgram(binaryop_gpu.ADD, a.shape, b.shape);\nconst output = this.makeOutputArray(program.outputShape, dtype) as Tensor;\nreturn this.compileAndRun<Tensor>(program, [a, b], output);\n\n\nThe code is available in src/kernels/backend_webgl.ts\n\n# Recap\n\nIn this article, I described how to write your own GPGPU program runnable in TensorFlow.js. It’s maybe topic diving into the internal of TensorFlow.js. If you are interested in the usage or application of TensorFlow.js, “Deep Learning in the Browser” is a good resource to learn that kind of thing. That covers the latest technologies to run a deep learning algorithm in modern web browsers.", null, "", null, "I hope this book would be helpful for you to start the journey to the world of high-performance deep learning in web browsers. Thanks!" ]

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[ "", null, "# HON Solo (Henry's Object Notation)... solo\n\nThis table serializer, with a name based on a notable Star Wars character, focuses on readability. And this is as readable as it gets… I hope.\nThis actually translates the table into something that lua can parse and run. There isn’t a decoder written, not yet, since all you need to do is RunString the encoded text.\n\nSpeed Test:\n\n``````\nStart HON Solo Deserializer\n===================================\nTest : 0.304469\nTest : 0.315747\nTest : 0.301665\nTest : 0.303268\nTest : 0.303499\nTest : 0.299851\nTest : 0.319738\nTest : 0.303351\nTest : 0.301949\nTest : 0.304352\nLength: 899\nAverage Time: 0.3057889\nAccuracy: pretty darn accurate\n===================================\n\nStart VON Deserializer\n===================================\nTest : 1.318248\nTest : 1.302694\nTest : 1.322732\nTest : 1.330334\nTest : 1.319991\nTest : 1.329704\nTest : 1.314146\nTest : 1.323041\nTest : 1.401673\nTest : 1.348156\nLength: 967\nAverage Time: 1.3310719\nAccuracy: pretty darn accurate\n===================================\n\n``````\n\nexample.lua:\n[lua]require(“honsolo”)\n\nlocal test = {\nrawr = “blah\\b”;\ntest = [=[a\\bsd\\afsdf()\ntest()]==]]]]\ntest “”’’]=];\nzombie = { test = “hi”; hello = “brah”; bloo = {}; ack = ‘aaasdasd’ };\n = “test”;\n[“hello there!”] = “zz”;\n}\n\nprint( honsolo.encode(test, true) );\n\n[/lua]\n\noutput:\n\n``````\n\n{\n = \"test\",\n[\"hello there!\"] = \"zz\",\ntest = [=[a\\bsd\\afsdf()\ntest()]==]]]]\ntest \"\"'']=],\nrawr = \"blah\\b\",\nzombie =\n{\ntest = \"hi\",\nhello = \"brah\",\nack = \"aaasdasd\",\nbloo =\n{\n}\n}\n}\n\n``````\n\nif you want to create a more compact one with less readability, don’t compile with the “pretty” argument on:\n[lua]\n– don’t add the second argument.\nhonsolo.encode(test);\n[/lua]\n\nthe output will have no indentation and other unnecessary stuff\n\nhonsolo.lua:\n[lua]-- “Beer-Ware License” (Revision: Pabst Blue Ribbon)\n– <[email protected]> wrote this file. As long as you retain this notice you\n– can do whatever with this stuff. Unlike Poul-Henning Kamp’s license, if we\n– meet some day, and you think this stuff is worth it, you can buy me a PBR\n– in return.\n\nlocal string, table, type, pairs, tonumber = string, table, type, pairs, tonumber;\nlocal print = print;\n\nmodule( “honsolo” );\n\nlocal regex_endquotes = “](=)]\";\nlocal regex_unescaped = \"^[%a_][%a%d_]\n\"; local regex_digits = \"^%d+”;\nlocal norm_dquotes = “”\";\nlocal norm_quotes = “’”;\nlocal norm_newline = \"\n\";\n\nextensions = {};\n\n– Name the child after the datatype. AKA whatever you get with type().\n– eg. string, number, Vector, Entity, Angle\n\n– The below extension is not guaranteed to work, this is due to me not having GMOD to test it out on.\nfunction extensions.Vector( data )\nreturn “Vector(” … data.x … “,” … data.y … “,” … data.z … “)”\nend\n\nfunction escape_key( str )\nif string.find(str, regex_unescaped) then\nreturn str;\nelse\nif tonumber(str) then\nreturn “[” … str … “]”;\nelse\nreturn “[” … escape_string(str) … “]”;\nend\nend\nend\n\nlocal controls = {\n[\"\"\"] = “\\”\";\n[\"\\a\"] = “\\a”;\n[\"\\b\"] = “\\b”;\n[\"\\f\"] = “\\”\";\n[\"\n“] = “\\””;\n[\"\\r\"] = “\\”\";\n[\" “] = “\\””;\n[\"\\v\"] = “\\”\";\n[\"\\\"] = “\\\\”;\n[\"’\"] = “\\’”;\n[\"\\0\"] = “\\0”;\n}\nlocal function control_escape( char )\nreturn controls[char];\nend\n\nfunction escape_string( str, out )\nlocal use_big_string_escape = false;\n\n``````local dquotes = string.find(str, norm_dquotes);\nlocal quotes = string.find(str, norm_quotes);\n\nif ( not quotes and dquotes ) then\nreturn norm_quotes .. str:gsub(\"%c\", control_escape) .. norm_quotes;\nend\n\nif ( not dquotes ) then\nreturn norm_dquotes .. str:gsub(\"%c\", control_escape) .. norm_dquotes;\nend\n\nlocal escape_count = grab_string_escape_count( str );\n\nreturn \"[\" .. (\"=\"):rep(escape_count) .. \"[\" .. str .. \"]\" .. (\"=\"):rep(escape_count) .. \"]\";\n``````\n\nend\n\nfunction grab_string_escape_count( str )\nlocal max_num = 0;\nlocal unescaped = {};\nlocal finding_unescapers = true\nlocal gend = 0;\nwhile ( finding_unescapers ) do\nlocal start, fend, match = string.find(str, regex_endquotes, gend );\ngend = fend;\nif not match then\nfinding_unescapers = false;\nelse\nunescaped[match:len()] = true;\nend\nwhile ( unescaped[max_num] ) do\nmax_num = max_num + 1;\nend\nend\nreturn max_num\nend\n\nfunction encode( data, pretty, indent, out, child )\n– first time calling initiate all variables\nindent = indent or 0;\nout = out or {};\n\n``````local t = type( data );\nif t == \"string\" then\nreturn escape_string(data);\nelseif t == \"number\" then\nreturn data;\nelseif t == \"boolean\" then\nif data then\nreturn \"true\";\nend\nreturn \"false\";\nelseif t == \"table\" then\nif ( pretty ) then table.insert( out, \"\n``````\n\n\" ); table.insert(out, (\" \"):rep(indent)) end\ntable.insert( out, “{” );\n\n`````` local first = true;\nfor key, value in pairs( data ) do\nindent = indent + 1;\nif not first then\ntable.insert( out, \",\" );\nend\nif ( pretty ) then table.insert( out, \"\n``````\n\n\" ); table.insert(out, (\" \"):rep(indent)) end\ntable.insert( out, escape_key(key) );\nif ( pretty ) then table.insert( out, \" \" ); end\ntable.insert( out, “=” );\nif ( pretty ) then table.insert( out, \" \" ); end\ntable.insert( out, encode(value, pretty, indent, out, true) );\nindent = indent - 1;\nfirst = false;\nend\n\n`````` if ( pretty ) then table.insert( out, \"\n``````\n\n\" ); table.insert(out, (\" \"):rep(indent)) end\ntable.insert( out, “}” );\n\n`````` if child then\nreturn \"\";\nend\nelseif extensions[t] then\nreturn extensions[t]( data, pretty, indent );\nend\n\nreturn table.concat(out);\n``````\n\nend\n\n[/lua]\n\nI quickly wrote this module on a macbook… in VIM. There’s no “decode” function, it converts the table into lua syntax. You have to RunString the text.\n\nHow does it compare to VoN?", null, "i dunno, i wrote it for fun.\nfor one, the string escaping isn’t broken.\n\nvON is better since it adds vector and angle support. i can probably do that but garry’s mod wont load on my mac", null, "Write it in plain lua then convert to gmod?\n\njust updated it with a quick extensions system for the honsolo.lua file.\nnow you can add different datatypes, like Vector or Angle.\nI added Vector as an example.\n\nUpdated it once again. It’s faster now.\n\nBenchmark test:\n\n``````\nStart HON Solo Serializer\n===================================\nTest : 1.126522\nTest : 1.152416\nTest : 1.106659\nTest : 1.101698\nTest : 1.116938\nTest : 1.094186\nTest : 1.149754\nTest : 1.099651\nTest : 1.105124\nTest : 1.124522\nLength: 861\nAverage Time: 1.117747\n===================================\n\nStart VON Serializer\n===================================\nTest : 1.188559\nTest : 1.121239\nTest : 1.117582\nTest : 1.124909\nTest : 1.148265\nTest : 1.146918\nTest : 1.120226\nTest : 1.118492\nTest : 1.126816\nTest : 1.166944\nLength: 930\nAverage Time: 1.137995\n===================================\n\n``````\n\nBenchmark source:\n[lua]require(“honsolo”)\nrequire(“von”)\n\nlocal test = {\nrawr = “blah\\b”;\ntest = [=[a\\bsd\\afsdf()\ntest()]==]]]]\ntest “”’’]=];\nzombie = { test = “hi”; hello = “brah”; bloo = {}; ack = ‘aaasdasd’; z = {}; };\n = “test”;\n[“hello there!”] = “zz”;\nagain = {\nrawr = “blah\\b”;\ntest = [=[a\\bsd\\afsdf()\ntest()]==]]]]\ntest “”’’]=];\nzombie = { test = “hi”; hello = “brah”; bloo = {}; ack = ‘aaasdasd’; z = {}; };\n = “test”;\n[“hello there!”] = “zz”;\nagain = {\nrawr = “blah\\b”;\ntest = [=[a\\bsd\\afsdf()\ntest()]==]]]]\ntest “”’’]=];\nzombie = { test = “hi”; hello = “brah”; bloo = {}; ack = ‘aaasdasd’; z = {}; };\n = “test”;\n[“hello there!”] = “zz”;\nbig_string = [====[\n\nlocal string, table, type, pairs, tonumber = string, table, type, pairs, tonumber;\nlocal print = print;\n\nmodule( “honsolo” );\n\nlocal regex_endquotes = “](=)]\";\nlocal regex_unescaped = \"^[%a_][%a%d_]\n\"; local regex_digits = \"^%d+”;\nlocal norm_dquotes = “”\";\nlocal norm_quotes = “’”;\nlocal norm_newline = \"\n\";\n\nextensions = {};\n\n]====];\n};\n};\n}\n\nprint( “” );\nprint( “Start HON Solo Serializer” );\nprint( “===================================” );\n–print( \" Output: \" … honsolo.encode(test) );\nlocal num = 0;\nfor i = 1, 10 do\nlocal x = os.clock();\nfor i = 1, 10000 do\nhonsolo.encode(test);\nend\nlocal y = os.clock();\nprint( \" Test [\"…i…\"]: \" … y - x );\nnum = num + ( y - x );\nend\nprint( \" Length: \" … honsolo.encode(test):len() );\nprint( \" Average Time: \" … num / 10 );\nprint( “===================================” );\nprint( “” );\nprint( “Start VON Serializer” );\nprint( “===================================” );\n–print( \" Output: \" … von.serialize(test) );\nlocal num = 0;\nfor i = 1, 10 do\nlocal x = os.clock();\nfor i = 1, 10000 do\nvon.serialize(test);\nend\nlocal y = os.clock();\nprint( \" Test [\"…i…\"]: \" … y - x );\nnum = num + ( y - x );\nend\nprint( \" Length: \" … von.serialize(test):len() );\nprint( \" Average Time: \" … num / 10 );\nprint( “===================================” );\nprint( “” );\n[/lua]\n\nPlease benchmark the deserialization too. I’m curious about the results. On vON, my main focus was deserialization.\n\nEdit: From the source, I see you didn’t localize some functions like table.insert. Doing so will make it even faster.\n\nThere is no deserialization\nyou just do RunString(“blah =” … encodedstuff)\n\nIt just translates the table into something readable by lua.\n\nWell, test that. I’m curious but too lazy to test for myself. xD\n\n``````\nStart HON Solo Deserializer\n===================================\nTest : 0.304469\nTest : 0.315747\nTest : 0.301665\nTest : 0.303268\nTest : 0.303499\nTest : 0.299851\nTest : 0.319738\nTest : 0.303351\nTest : 0.301949\nTest : 0.304352\nLength: 899\nAverage Time: 0.3057889\nAccuracy: pretty darn accurate\n===================================\n\nStart VON Deserializer\n===================================\nTest : 1.318248\nTest : 1.302694\nTest : 1.322732\nTest : 1.330334\nTest : 1.319991\nTest : 1.329704\nTest : 1.314146\nTest : 1.323041\nTest : 1.401673\nTest : 1.348156\nLength: 967\nAverage Time: 1.3310719\nAccuracy: pretty darn accurate\n===================================\n\n``````\n\nthis is using\nas the deserializer for HON, since that’s lua’s native way of evaluating strings.\n\nBenchmark Source:\n[lua]require(“honsolo”)\nrequire(“von”)\n\nlocal test = {\nrawr = “blah\\b”;\ntest = [=[a\\bsd\\afsdf()\ntest()]==]]]]\ntest “”’’]=];\nzombie = { test = “hi”; hello = “brah”; bloo = {}; ack = ‘aaasdasd’; z = {}; };\n = “test”;\n[“hello there!”] = “zz”;\nagain = {\nrawr = “blah\\b”;\ntest = [=[a\\bsd\\afsdf()\ntest()]==]]]]\ntest “”’’]=];\nzombie = { test = “hi”; hello = “brah”; bloo = {}; ack = ‘aaasdasd’; z = {}; };\n = “test”;\n[“hello there!”] = “zz”;\nagain = {\nrawr = “blah\\b”;\ntest = [=[a\\bsd\\afsdf()\ntest()]==]]]]\ntest “”’’]=];\nzombie = { test = “hi”; hello = “brah”; bloo = {}; ack = ‘aaasdasd’; z = {}; };\n = “test”;\n[“hello there!”] = “zz”;\n[“acur a see”] = “pretty darn accurate”;\nbig_string = [====[\n\nlocal string, table, type, pairs, tonumber = string, table, type, pairs, tonumber;\nlocal print = print;\n\nmodule( “honsolo” );\n\nlocal regex_endquotes = “](=)]\";\nlocal regex_unescaped = \"^[%a_][%a%d_]\n\"; local regex_digits = \"^%d+”;\nlocal norm_dquotes = “”\";\nlocal norm_quotes = “’”;\nlocal norm_newline = \"\n\";\n\nextensions = {};\n\n]====];\n};\n};\n}\n\nlocal hon_todecode = honsolo.encode(test);\nlocal von_todecode = von.serialize(test);\n\nprint( “” );\nprint( “Start HON Solo Deserializer” );\nprint( “===================================” );\nlocal num = 0;\nfor i = 1, 10 do\nlocal x = os.clock();\nfor i = 1, 10000 do\nend\nlocal y = os.clock();\nprint( \" Test [\"…i…\"]: \" … y - x );\nnum = num + ( y - x );\nend\nprint( \" Length: \" … honsolo.encode(test):len() );\nprint( \" Average Time: \" … num / 10 );\nprint( \" Accuracy: \" … z.again.again[“acur a see”] );\nprint( “===================================” );\nprint( “” );\nprint( “Start VON Deserializer” );\nprint( “===================================” );\nlocal num = 0;\nfor i = 1, 10 do\nlocal x = os.clock();\nfor i = 1, 10000 do\nz = von.deserialize(von_todecode);\nend\nlocal y = os.clock();\nprint( \" Test [\"…i…\"]: \" … y - x );\nnum = num + ( y - x );\nend\nprint( \" Length: \" … von.serialize(test):len() );\nprint( \" Average Time: \" … num / 10 );\nprint( \" Accuracy: \" … z.again.again[“acur a see”] );\nprint( “===================================” );\nprint( “” );\n[/lua]\n\nneeds a better way of detecting accuracy though\n\nI also gotta write something for booleans.\n\nAs I expected.", null, "Your deserializer is basically written in C. Mine is in Lua… :C\nNice work!", null, "waiting for an even faster serialization code to appear\n\nWhy not use JSon, it looks easier to read and does the same function?\n\nJSON is like 3x slower than this.\n\nAnd it’s bound to the JSON standard, which says stuff like all indices must be strings. Tables with numeric indices are turned to arrays and the rest of the keys are lost.\n\ninput:\n[lua]\nlocal test = {\nblah = {\nthis_is = “pretty readable to me.”;\nwhat_more = “can you want?”;\n[“if you know lua”] = “it’s pretty easy to figure it out.”;\n}\n}\n[/lua]\n\nout:\n\n``````\n{\nblah =\n{\nwhat_more = \"can you want?\",\n[\"if you know lua\"] = \"it's pretty easy to figure it out.\",\nthis_is = \"pretty readable to me.\"\n},\n}\n\n``````\n\norder gets messed up on the output though.\n\nLooks like i am switching to HON from VON.\n-.-\n\nExtremely useful, thanks. One minor optimization though. Localize only the functions you are using, not the entire libraries, that way there will be no table lookups." ]

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[ "# Linear Equations\n\n## Word Problems on simultaneous linear equations\n\nIn this chapter we will discuss some problems related to the concept of simultaneous linear equations. To solve the problems, you have to comprehend the statement carefully and then frame the linear equation. After getting two linear equations, you then have to get the value of variables to get the final solution. Simultaneous linear equations …\n\n## Solvability of linear equations\n\nIn this chapter we will learn about the solvability of linear equations with solved examples. Here for a given system of equation, we will check if the equations provides unique solution, multiple solution or no solution. Solvability of simultaneous equations Let two equations are given to us; Here, three conditions are possible; (a) The above …\n\n## Cross multiplication method for linear equations\n\nIn this chapter, we will discuss cross multiplication method of solving simultaneous linear equation. Solving linear equation using Cross Multiplication method Cross multiplication method works when two linear equations are given with two unknown variables. It basically provide you with formula that will help you find value of unknown variables. Let the two equations are; …\n\n## Solving systems of equations by elimination\n\nIn this chapter we will learn to solve linear equation with two variables using elimination method. Solving systems of equations using elimination The elimination method will work when two linear equations are given with two variables. To understand the process, let us imagine that we have been provided with linear equation with variables x and …\n\n## Substitution method of linear equation\n\nIn this chapter we will learn to solve linear equation with two variables using substitution method. To understand the chapter, you should have fair knowledge of linear and simultaneous equations. Substitution Method This method works when we have two linear equations with two variables. Using the method we can find common point that satisfies both …\n\n## Simultaneous Linear equation\n\nIn this post we will learn about the concept of simultaneous linear equation and some methods to solve them. At the end of the chapter, some solved examples are also provided for conceptual clarity. What are simultaneous linear equation ? The set of two linear equations containing two or more variables are called simultaneous equation. …\n\n## Word Problems on Linear equation in one variable\n\nIn this chapter we will solve some word problem questions related to linear equation in one variable. To solve the questions, you have to follow below steps; (a) Frame the equation after reading the question. (b) Solve the equation and find the unknown variable. Hence, apart from math skills, you should also posses reading comprehension …\n\n## Problems on Linear equation in one variable\n\nThis chapter is a collection of important problems related to linear equation in one variable. We have already discussed different method of solving linear equation. Click the below red link to learn about the same. What is Linear equation ? How to solve linear equation ? Solving linear equation using cross multiplication Problems on linear …\n\n## Cross multiplication method for linear equation\n\nIn this chapter we will learn cross multiplication method to solve linear equation with solved examples. What is cross multiplication method ? When two algebraic fractions are present on either side of the equation, then the equation can be solved by; ⟹ multiplying denominator on left to numerator on right. ⟹ multiplying denominator on right …\n\n## Solving linear equations\n\nIn this chapter we will learn to solve linear equations with one variable. All the questions are fully solved with step by step explanation. Method to solve linear equation with one variable Solving any linear equation means that we have to find the value of given variable which can satisfy the given equation. When solving …" ]

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[ "# 加拿大预测网-开奖|pc预测|加拿大28预测|专注比特币28预测_加拿大pc预测_官网数据！\n\n## 比特币28预测 永恒地址：28xd.com\n\n#### 第 11300397 期 距下期： 00 分 22 秒\n\n6 + 5 + 2 = 13 -\n\n##### 100期 正确：55 期 错误：45 期正确率：55%\n\n11300398 --- ---\n11300397 6+5+2=13\n11300396 3+4+4=11\n11300395 4+4+6=14\n11300394 1+4+2=07\n11300393 8+7+6=21\n11300392 8+3+5=16\n11300391 0+2+4=06\n11300390 0+6+0=06\n11300389 6+9+0=15\n11300388 8+0+5=13\n11300387 6+2+0=08\n11300386 4+8+4=16\n11300385 1+1+6=08\n11300384 9+1+8=18\n11300383 2+8+8=18\n11300382 7+6+3=16\n11300381 9+7+6=22\n11300380 1+6+1=08\n11300379 0+9+5=14\n11300378 4+9+9=22\n11300377 6+6+9=21\n11300376 5+3+5=13\n11300375 7+2+8=17\n11300374 7+8+9=24\n11300373 8+4+3=15\n11300372 9+5+6=20\n11300371 3+5+5=13\n11300370 1+1+7=09\n11300369 0+0+4=04\n11300368 8+3+7=18\n11300367 1+1+5=07\n11300366 2+1+1=04\n11300365 1+7+7=15\n11300364 5+2+5=12\n11300363 3+5+4=12\n11300362 1+7+1=09\n11300361 1+7+2=10\n11300360 0+1+2=03\n11300359 0+0+4=04\n11300358 2+1+6=09\n11300357 3+2+7=12\n11300356 1+7+4=12\n11300355 0+4+8=12\n11300354 1+1+1=03\n11300353 9+4+7=20\n11300352 9+5+4=18\n11300351 9+5+7=21\n11300350 5+2+4=11\n11300349 8+2+6=16\n11300348 5+1+2=08\n11300347 4+1+4=09\n11300346 5+2+1=08\n11300345 3+6+6=15\n11300344 0+2+9=11\n11300343 7+0+9=16\n11300342 4+3+7=14\n11300341 6+3+7=16\n11300340 8+1+7=16\n11300339 1+6+9=16\n11300338 0+0+6=06\n11300337 5+7+7=19\n11300336 3+2+3=08\n11300335 3+9+1=13\n11300334 0+8+3=11\n11300333 5+6+0=11\n11300332 2+8+5=15\n11300331 9+7+2=18\n11300330 4+5+8=17\n11300329 6+0+7=13\n11300328 6+7+3=16\n11300327 5+2+5=12\n11300326 9+8+4=21\n11300325 5+3+5=13\n11300324 6+5+2=13\n11300323 3+8+7=18\n11300322 6+6+0=12\n11300321 2+9+7=18\n11300320 4+4+6=14\n11300314 7+6+9=22\n11300313 9+4+8=21\n11300312 0+7+4=11\n11300311 5+7+0=12\n11300310 1+8+9=18\n11300309 6+2+0=08\n11300308 1+9+0=10\n11300307 5+0+4=09\n11300306 9+8+4=21\n11300305 4+2+9=15\n11300304 1+5+7=13\n11300303 9+8+3=20\n11300302 2+1+0=03\n11300301 3+6+7=16\n11300300 1+6+2=09\n11300299 1+2+3=06\n11300298 1+3+3=07\n11300297 2+9+6=17\n11300296 7+0+0=07\n11300295 3+6+4=13\n11300294 2+4+2=08\n11300293 6+7+3=16" ]

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[ "# ST_Union Method | Geospatial Data Types | Teradata Vantage - 17.10 - ST_Union Method - Advanced SQL Engine - Teradata Database\n\n## Teradata Vantage™ - Geospatial Data Types\n\nProduct\nRelease Number\n17.10\nRelease Date\nJuly 2021\nContent Type\nProgramming Reference\nPublication ID\nB035-1181-171K\nLanguage\nEnglish (United States)\n\nReturns an ST_Geometry value that represents the point set union of two ST_Geometry values.\n\n## Valid Data Types\n\nAll ST_Geometry types except geometry collections.\n\nThis method can be called on 3D geometries (those that include z coordinates). However, the z coordinate is ignored in method calculations. Consequently, any z coordinates returned by this method should be ignored. Teradata recommends using the Make_2D method to strip out the z coordinates of the return value.\n\n## Result Type\n\nThe type that the ST_Geometry return type represents is one from the possible set of types in the following table, depending on the parameter types.\n\na ∪ b Ø ST_Point ST_LineString, GeoSequence ST_Polygon ST_MultiPoint ST_MultiLineString ST_MultiPolygon\nØ Ø R01 R02 R03 R04 R05 R06\nST_Point R01 R20 R15 R22 R04 R17 R19\nST_LineString, GeoSequence R02 R15 R16 R21 R15 R16 R18\nST_Polygon R03 R22 R21 R23 R22 R21 R23\nST_MultiPoint R04 R04 R15 R22 R04 R17 R19\nST_MultiLineString R05 R17 R16 R21 R17 R16 R18\nST_MultiPolygon R06 R19 R18 R23 R19 R18 R23\nWhere:\n R09 = ST_Point R02 = ST_LineString R03 = ST_Polygon R04 = ST_MultiPoint R05 = ST_MultiLineString R06 = ST_MultiPolygon R15 = ST_LineString, ST_GeomCollection of ST_Point and ST_LineString values R16 = ST_LineString, ST_MultiLineString R17 = ST_MultiLineString, ST_GeomCollection of ST_Point and ST_LineString values R18 = ST_MultiPolygon, ST_GeomCollection of ST_LineString and ST_Polygon values R19 = ST_MultiPolygon, ST_GeomCollection of ST_Point and ST_Polygon values R20 = ST_Point, ST_MultiPoint R21 = ST_Polygon, ST_GeomCollection of ST_LineString and ST_Polygon values R22 = ST_Polygon, ST_GeomCollection of ST_Point ST_Polygon values R23 = ST_Polygon, ST_MultiPolygon\n\nVantage converts GeoSequence types that are involved in the ST_Union method to ST_LineString values. Therefore, ST_Union never returns a GeoSequence type." ]

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[ "# Properties\n\n Label 58800.m Number of curves 6 Conductor 58800 CM no Rank 1 Graph", null, "# Related objects\n\nShow commands for: SageMath\nsage: E = EllipticCurve(\"58800.m1\")\nsage: E.isogeny_class()\n\n## Elliptic curves in class 58800.m\n\nsage: E.isogeny_class().curves\nLMFDB label Cremona label Weierstrass coefficients Torsion order Modular degree Optimality\n58800.m1 58800gb6 [0, -1, 0, -15366808, -23180759888] 2 1572864\n58800.m2 58800gb4 [0, -1, 0, -960808, -361655888] 4 786432\n58800.m3 58800gb3 [0, -1, 0, -764808, 256136112] 2 786432\n58800.m4 58800gb5 [0, -1, 0, -666808, -587447888] 2 1572864\n58800.m5 58800gb2 [0, -1, 0, -78808, -1799888] 4 393216\n58800.m6 58800gb1 [0, -1, 0, 19192, -231888] 2 196608 $$\\Gamma_0(N)$$-optimal\n\n## Rank\n\nsage: E.rank()\n\nThe elliptic curves in class 58800.m have rank $$1$$.\n\n## Modular form 58800.2.a.m\n\nsage: E.q_eigenform(10)\n$$q - q^{3} + q^{9} - 4q^{11} - 2q^{13} - 6q^{17} + 4q^{19} + O(q^{20})$$\n\n## Isogeny matrix\n\nsage: E.isogeny_class().matrix()\n\nThe $$i,j$$ entry is the smallest degree of a cyclic isogeny between the $$i$$-th and $$j$$-th curve in the isogeny class, in the LMFDB numbering.\n\n$$\\left(\\begin{array}{rrrrrr} 1 & 2 & 8 & 4 & 4 & 8 \\\\ 2 & 1 & 4 & 2 & 2 & 4 \\\\ 8 & 4 & 1 & 8 & 2 & 4 \\\\ 4 & 2 & 8 & 1 & 4 & 8 \\\\ 4 & 2 & 2 & 4 & 1 & 2 \\\\ 8 & 4 & 4 & 8 & 2 & 1 \\end{array}\\right)$$\n\n## Isogeny graph\n\nsage: E.isogeny_graph().plot(edge_labels=True)\n\nThe vertices are labelled with LMFDB labels.", null, "" ]

[ null, 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WrVund999V/Hx8ZQ/L8EEEGVyqdsCAAA804%2B3/QoMDNTGjRu5ctdLMAFEmfj4%2BMhut2v79u2aNWuW6TgAACeZOXOmduzYIbvdTvnzIkwAUS6PPfaYli1bpvz8fAUHB5uOAwCohGPHjiksLEz33nuv5syZYzoOqhATQJRLXFyciouLFRsbazoKAKCSYmNjVVJSori4ONNRUMUogCiXyy%2B/XC%2B%2B%2BKJefvll7du3z3QcAEAF7du3T6%2B88opefPFFNWnSxHQcVDFOAaPczp49q7Zt26pNmzZ6//33TccBAFRAr169tG/fPu3Zs0e1atUyHQdVjAkgyq1WrVpKSkrSsmXLtGzZMtNxAADltGzZMi1fvlzJycmUPy/FBBAVYlmWunfvrsOHD2vHjh3y8/MzHQkAUAYOh0Pt27dXs2bNtGrVKq789VJMAFEhNptNqampys/P1%2BTJk03HAQCU0SuvvKL9%2B/crNTWV8ufFmACiUqKjo5WWlqb8/HyFhoaajgMAuIijR48qLCxMAwcO5MW7l6MAolIKCwsVFham/v37a9q0aabjAAAuYvDgwVq4cKHy8/MVEhJiOg4M4hQwKiUkJEQxMTGaMWOGtm3bZjoOAOB/2LZtm1599VWNHj2a8gcmgKi8c%2BfOqUOHDmrcuLHWrl3Le0oAwM1YlqXbb79dBQUF2r59u2rWrGk6EgxjAohKq1mzplJTU7V%2B/Xq9%2B%2B67puMAAH7lnXfe0YcffqiUlBTKHyQxAYQT3Xfffdq5c6f27t2rgIAA03EAAJKKi4vVunVrXXvttVq8eLHpOHATTADhNMnJyTp8%2BLAmTpxoOgoA4AfJyck6cuSIkpOTTUeBG2ECCKf65z//qalTpyovL09NmzY1HQcAvNqXX36pVq1aKTo6WomJiabjwI1QAOFUJ06cUFhYmHr06KG5c%2BeajgMAXm3gwIH64IMPlJeXp3r16pmOAzfCKWA4Vb169TR%2B/HjNmzdPmzZtMh0HALzWxo0bNX/%2BfI0fP57yh99gAginu3DhgiIjI1WjRg1t3rxZPj68zgCAqlRaWqquXbuqtLRUn3zyiXx9fU1HgpthZ4bT%2Bfr6ym63KysrS/PmzTMdBwC8zty5c5Wdna3U1FTKH34XE0C4TP/%2B/bVhwwbl5uYqKCjIdBwA8ArfffedwsPDddttt%2Bmtt94yHQduigkgXCYhIUHHjh1TXFyc6SgA4DXGjx%2Bv48ePKyEhwXQUuDEKIFzmyiuv1L/%2B9S8lJyfrs88%2BMx0HAKq9Tz/9VBMnTtTzzz%2BvK664wnQcuDFOAcOlTp8%2BrdatWysyMlLvvfee6TgAUK3169dP2dnZys3NVWBgoOk4cGNMAOFStWvXVnx8vBYtWqQ1a9aYjgMA1dbq1auVnp6uhIQEyh8uiQkgXM6yLN1888367rvvtGXLFtWoUcN0JACoVs6fP6%2BOHTuqXr162rBhg2w2m%2BlIcHNMAOFyNptNdrtdO3fu1Kuvvmo6DgBUOzNmzNDu3btlt9spfygTJoCoMn/961%2B1ePFi5eXlqUGDBqbjAEC18O233yosLEx9%2B/bVrFmzTMeBh2ACiCozfvx4lZSUaPTo0aajAEC1ERMTo3PnzmncuHGmo8CDUABRZS677DK99NJLmjx5svbs2WM6DgB4vD179mjKlCl66aWXdNlll5mOAw/CKWBUqZKSErVr105XX321VqxYwXtVAKCCLMtSjx499Nlnn2n37t3y9/c3HQkehAkgqpS/v7%2BSk5P13//%2BV%2B%2B//77pOADgsZYuXaoPPvhAEydOpPyh3JgAospZlqW77rpLBw8e1O7du%2BXn52c6EgB4lJKSEl1zzTVq3ry5Vq5cydkUlBsTQFQ5m82mlJQUHThwQJMmTTIdBwA8zqRJk3TgwAGlpKRQ/lAhTABhzFNPPaW5c%2BcqLy9PjRs3Nh0HADzCN998o7CwMD322GN6%2BeWXTceBh6IAwpiioiKFhYXpj3/8IzeIBoAyeuKJJ7Ro0SLl5%2BdzT1VUGKeAYUzDhg0VGxurWbNmacuWLabjAIDby8nJ0WuvvabY2FjKHyqFCSCMOn/%2BvK677jrVr19fH374Ie9lAYD/wbIs3XLLLTp%2B/Li2bdvGz1VHpTABhFE1atRQamqqPvroIy1cuNB0HABwWwsWLNDHH38su91O%2BUOlMQGEW%2BjTp4%2B2bdumvXv3KjAw0HQcAHArZ86cUevWrdWpUyelp6ebjoNqgAkg3EJycrKOHDmipKQk01EAwO0kJibqm2%2B%2B4RgJp6EAwi20bNlSw4YN04QJE/TFF1%2BYjgMAbuPQoUOKj4/XsGHD1LJlS9NxUE1wChhu4%2BTJkwoPD1dUVJTS0tJMxwEAt/Dwww9r3bp1ysvLU1BQkOk4qCaYAMJt1K1bV3FxcXrzzTf18ccfm44DAMZ99NFHeuuttxQXF0f5g1MxAYRbKS0tVZcuXSRJn3zyiXx8eI0CwDuVlpYqMjJSPj4%2ByszM5HgIp%2BJfE9yKj4%2BPZs6cqdLSUi1dutR0HAAwZsmSJbIsSzNnzqT8wemYAAIAAHgZXlIAAAB4GQogAACAl6EAAgAAeBkKIAAAgJehAAIAAHgZCiAAAICXoQDCY8TFSZGRUlCQ1KiR1LevlJtrOhUAlB/HM5hGAYTHWL9eevJJafNm6YMPpPPnpbvukk6fNp0MAMqH4xlM40bQ8FgFBd%2B/cl6/Xrr1VtNpAKDiOJ6hqjEBhMc6ceL7Xxs0MJsDACqL4xmqGhNAeCTLkvr0kY4dkzZsMJ0GACqO4xlMqGE6AFARTz0l7dghffSR6SQAUDkcz2ACBRAe5%2BmnpcWLpQ8/lJo1M50GACqO4xlMoQDCY1jW9wfLRYukdeuk5s1NJwKAiuF4BtMogPAYTz4ppaVJGRnf3zvr66%2B//3i9elJAgNlsAFAeHM9gGheBwGPYbL//8dmzpccfr9IoAFApHM9gGgUQAADAy3AfQAAAAC9DAQQAAPAyFEAAAAAvQwEEAADwMhRAAAAAL0MBBAAA8DIUQAAAAC9DAQQAAPAyFEBUOy%2B99JK6d%2B%2Bu7777znQUAF7sxIkT6t69u/7zn/%2BYjgL8Bj8JBNXOV199pfDwcP3jH/9QUlKS6TgAvNRzzz2n6dOnKzc3V02bNjUdB/gFJoCodpo2baoRI0bIbrcrLy/PdBwAXig3N1eTJk3SiBEjKH9wS0wAUS0VFxerTZs2at%2B%2BvZYsWWI6DgAvc%2B%2B992r37t3au3evatWqZToO8BtMAFEtBQQEKCkpSUuXLtWKFStMxwHgRZYvX673339fSUlJlD%2B4LSaAqLYsy1K3bt30zTffaMeOHapZs6bpSACquXPnzql9%2B/Zq0qSJ1qxZI5vNZjoS8LuYAKLastlsSk1NVV5enqZMmWI6DgAvMHnyZOXn5ys1NZXyB7fGBBDV3pAhQ7RgwQLl5%2BcrJCTEdBwA1VRBQYHCwsL08MMPa%2BrUqabjABfFBBDV3pgxY2RZlkaOHGk6CoBq7MdjTGxsrOEkwKVRAFHthYaGatSoUZo%2Bfbp27NhhOg6Aamj79u2aMWOGYmJiFBoaajoOcEmcAoZXcDgc6tChgy6//HKtXr2a9%2BYAcBrLshQVFaWvv/6aC87gMZgAwiv4%2BfkpJSVFa9eu1aJFi0zHAVCNvPfee1q3bp1SUlIof/AYTADhVe655x7t2bOHm7MCcIqzZ8%2BqTZs2ateunZYuXWo6DlBmTADhVSZOnKgvv/xSKSkppqMAqAZ%2BPKYkJyebjgKUCxNAeJ1nn31WM2bMUF5eni6//HLTcQB4qK%2B%2B%2BkqtWrXS4MGDKYDwOBRAeJ3jx48rLCxMvXr10uuvv246DgAP9eijj2rFihXKy8tTcHCw6ThAuXAKGF4nODhY48aN0xtvvKHMzEzTcQB4oM2bN2vu3LkaN24c5Q8eiQkgvNKFCxcUERGhWrVqaePGjfLx4bUQgLIpLS3VDTfcIIfDoezsbPn6%2BpqOBJQbux68kq%2Bvr%2Bx2uzIzM5WWlmY6DgAPMn/%2BfH3yySey2%2B2UP3gsJoDwag888IA2btyo3Nxc1alTx3QcAG7u1KlTCg8P180336yFCxeajgNUGBNAeLXExEQVFRVpwoQJpqMA8ABxcXE6duyYEhISTEcBKoUCCK921VVXafjw4UpKStKBAwdMxwHgxj777DMlJydr%2BPDhuuqqq0zHASqFU8DweqdPn1arVq10/fXX65133jEdB4Cb%2BuMf/6jMzEzl5uaqdu3apuMAlcIEEF6vdu3amjBhgt59912tW7fOdBwAbmjt2rV67733FB8fT/lDtcAEEND3t3W46aabdObMGW3ZsoUr%2BwD85Pz584qIiFDt2rX18ccfy2azmY4EVBoTQECSj4%2BP7Ha7duzYoZkzZ5qOA8CNzJw5Uzt27JDdbqf8odpgAgj8zOOPP66lS5cqPz9f9evXNx0HgGHHjh1TWFiY7rvvPs2ePdt0HMBpmAACPxMXF6eSkhLFxsaajgLADYwePVolJSUaP3686SiAU1EAgZ9p0qSJRowYoVdeeUX79u0zHQeAQXv37tXkyZP14osvqkmTJqbjAE7FKWDgV86ePau2bduqdevWWrZsmek4AAywLEt333238vPztXv3btWqVct0JMCpmAACv1KrVi0lJydr%2BfLlFEDASy1btkwrV65UcnIy5Q/VEhNA4HdYlqU77rhDX375pXbu3Ck/Pz/TkQBUEYfDoWuuuUZXXHGFPvjgA678RbXEBBD4HTabTampqdq/f79eeeUV03EAVKGXX35Zn376qVJTUyl/qLaYAAIXER0drbS0NOXl5alRo0am4wBwsaNHjyosLEwDBw7U5MmTTccBXIYCCFxEYWGhwsLC9OCDD2r69Omm4wBwsUGDBumdd95Rfn6%2BGjZsaDoO4DKcAgYuIiQkRKNHj9arr76qbdu2mY4DwIW2bt2qmTNnavTo0ZQ/VHtMAIFLOHfunK699lqFhoZq3bp1vCcIqIYsy9Jtt92moqIibdu2TTVr1jQdCXApJoDAJdSsWVMpKSn68MMP9c4775iOA8AF3n77bW3YsEGpqamUP3gFJoBAGd13333auXOn9u7dq4CAANNxADhJcXGxWrdurWuvvVaLFy82HQeoEkwAgTJKTk7W4cOHlZycbDoKACdKSkrSkSNHWNvwKkwAgXIYPny4pkyZotzcXDVr1sx0HACV9MUXX6hVq1Z66qmnlJCQYDoOUGUogEA5nDhxQuHh4brzzjs1b94803EAVNKAAQO0atUq5efnq27duqbjAFWGU8BAOdSrV0/jx4/X/PnztWnTJtNxAFTCxo0blZaWpri4OMofvA4TQKCcLly4oMjISNWoUUObN2%2BWjw%2BvowBPU1paqq5du6q0tFRZWVmsY3gd/sUD5eTr6yu73a6srCzNnTvXdBwAFfDGG28oOztbdrud8gevxAQQqKCHHnpI69evV15enoKCgkzHAVBGJ0%2BeVHh4uLp166Y333zTdBzACF72ABWUkJCg48ePa/z48aajACiH8ePH6%2BTJk4qPjzcdBTCGAghU0BVXXKHnn39eEydO1Keffmo6DoAy%2BPTTT5WSkqLnn39eV1xxhek4gDGcAgYq4cyZM2rVqpUiIyP13nvvmY4D4BL69eunnJwc7du3T4GBgabjAMYwAQQqITAwUAkJCVq0aJFWr15tOg6Ai1i1apXS09OVkJBA%2BYPXYwIIVJJlWbrlllt04sQJbd26VTVq1DAdCcCvnD9/Xtddd52Cg4O1YcMG2Ww205EAo5gAApVks9lkt9u1e/duzZgxw3QcAL9j%2BvTp2rNnj%2Bx2O%2BUPEBNAwGn%2B9re/KT09Xfn5%2BWrQoMFFP9eyLBUUFOjMmTNyOBzy8/NTYGCgQkND2ZyA/6Gi6%2Bbbb79VWFiY%2BvXrp5kzZ1ZhYsB9UQABJ/n6668VHh6uv/zlL7Lb7b/4s8LCQq1Zs0Y5OTnKzs7Sli1bdPz4id88RnBwPXXq1EmdO0cqIiJCUVFRCgkJqapvAXArzlo3Q4cO1Zw5c5Sfn6/GjRtX5bcAuC0KIOBECQkJGjFihHbs2KE2bdpo8%2BbNmjJlihYuXCiHw6FmjRupc6uW6tS6pdo1v1JBgQHyq1lTjnPn9N2ZYu0%2B8Lm27Nuv7Nz9%2BvKbo/Lz81P//v0VHR2trl27Mh1EtWdZllPXTZ06dXTdddcpLi5Ow4cPN/3tAW6DAgg4UUlJidq1a6c6derIJmnb9u1q3rSJhvTtpYE9o9QkpGGZH%2BtIYZHmrVijaenLdOCrI%2Bp43XWKGT1avXv3dt03ABiUkZGhmFGjnLpugutZTGP8AAAgAElEQVTVU0BgoA4cOCB/f38Xpgc8CwUQcKKioiLdf//9%2BvDDD9U9sqOeffh%2B9egaUamfNVpaWqqVmTlKefM9rcraqkceeUSTJk1Sw4Zl3xQBd1ZUVKShQ4cqLS1Nd0R21DAnrpuk%2Be9obc521g3wKxRAwEnS09M1ZPBgOc4Wyz5siAb0iHLqKVvLsjRvxWo9kzJd/gEBmjZ9uvr27eu0xwdMYN0AZnAbGKCSLMvS%2BPHj1a9fP3UJb6Fd86dpYM/uTn%2B/ns1m05/vvkO706apS3gL9evXT3FxceI1HDwR6wYwyzcmJibGdAjAU1mWpREjRig2Nlajnhioqf96WkG1XfsTBoICA/XQnbdJkkYmpcrhcCgqyrlTE8CVWDeAeRRAoBLi4uIUGxurxKf/rhcee6jKNhObzabbO12rOoEBGplsl7%2B/v2655ZYqeW6gslg3gHkUQKCC0tPTNWjQII16YqBeeOwhIxlubN9WliyNTEpVx44d1bp1ayM5gLJi3QDugYtAgAooKipSu7Zt1SW8hdITRhk9jWRZlvoMj1FW/gHt3rOHqxzhtlg3gPvgIhCgAoYOHSrH2WJNe/5p4%2B8hstlsmvb8UJUUF%2BuZZ54xmgW4GNYN4D4ogEA5ZWRkKC0tTfZhQ8p1g1pXujy0oezDBmv%2B/PlavHix6TjAb7BuAPfCKWCgHCzLUqeOHRXi56OV9vHGpxg/Z1mWejwzQkXnLOVs2eJW2eDdWDeA%2B2ECCJTD5s2btW37dg17%2BH632yhsNpv%2B76F%2B2rptmzIzM03HAX7CugHcDwUQKIcpU6aoedMm6tE1wnSU39Wja4SaN22iKVOmmI4C/IR1A7gfCiBQRoWFhVq4cKGG9O1VqZ9R6kq%2Bvr4a3LeXFixYoMLCQtNxANYN4KbcczUCbmjNmjVyOBwa2DPKZc%2BxPf8zPTIyTlf0GajA23qr7UN/l31Berke4889o%2BRwOLR27VoXpaze4uKkyEgpKEhq1Ejq21fKzTWdynNVxbrxuaHnb/6b9t775XoM1g28TQ3TAQBPkZOTo2aNG7n0CsacffkKCa6nuaP%2BpT80DtXGnXs0eMIk%2Bfr46KkHepfpMZqENFTTRqHKycnRAw884LKs1dX69dKTT35fAs%2Bfl158UbrrLmnPHql2bdPpPE9VrBtJeu2lZ9Xz%2Bs4//b5eOf%2ByWDfwNhRAoIyys7LUuVXLcn1Nt%2Bjhuubq5vL18dEby1fJr0YNxQ56VAN6ROnp5Ml6Z%2B1HalQ/WC8/F627b4jUX%2B/r8Yuvb9G0iTbt3KtF6z8ucwGUpM6tWyo7O6tcWfG9FSt%2B%2BfvZs7%2BfBObkSLfeaiaTJyvvuinvmvlRcJ06uqxhg0plZd3Am3AKGCgDy7K0ZesWdWpdvgIoSW8sW6WQ4LrKnGXXUw/0VnTiK3rwxXG6oX1b5cx5RXd1jdCjoxN15uzZ3/36k6dPq0HdoHI9Z6dWLbUlZ4u4y1PlnTjx/a8NKtctvFJF101F1szTyVMU2vNBdfnr05r23vsqLS0td17WDbwJBRAog4KCAh0/fkLtml9Z7q%2B9Nqy5XvrLIwr7Q1O98Gh/Bfj7KaReXf29z90K%2B0NTjfzrIyo6cVI79h/4zddu2rlHC1dv0KC%2Bvcr1nNe0uErHjh9XQUFBufPi/7Ms6dlnpZtvlq65xnQaz1PRdVPeNRM76FEtHDdCH0yKU/87btc/X56h8a%2B/Ve68rBt4E04BA2Vw5swZSVJQYEC5v7b91c1/%2Bn9fX181rFdX11x91U8fa9ygviTp6LHjv/i63Z8dVN/nR%2Bs/f31Ed3bpVK7nrPNDzqysLDVp0qTcmauzgIAAtWnTpkyf%2B9RT0o4d0kcfXfpz9%2B7dq%2BLi4kqmq14OHz4sqfzrprxr5qW/PPLTn10XfrUkacxr83/x8bL4cd3w9whvQAEEysDhcEiS/GrWLPfX1qzxy2Vm%2B9XHfrwxbmnp/z/ttOfA5%2Br%2B1L/1RO%2B7y72JSZLfD49/7733lvtrq7uOHTtqy5Ytl/y8p5%2BWFi%2BWPvxQatbs0o87YMAAbd261QkJq5/yrpuKrJmfu75da508fUbffHvsp7JYppw/PEdJSUm58gKeiAIIlIGfn58kyXHunMufa/dnB9X9qX/r0V53aNyQxyv0GI7z5yVJS5cuZQL4KwEBF59GWdb35W/RImndOql584t%2B%2Bk/mz5/P5OhXDh8%2BrPvuu69K1s3Pbc37VLX8/BRcp3xXAv%2B4bvz9/V0RC3ArFECgDAIDAyVJ351x7Qa/%2B7ODinrqed3VpZOeffh%2BfV30rSTJ18dHofWDy/w4p37IGRkZqUaNGrkka3X15JNSWpqUkfH9vQC//vr7j9erJ12sO5b1tLI3afbD6NSV62bJhs36%2BttjuuGaNgrw99PaLdv10vQ5%2Bnvfu%2BX/wwu3svpx3VzqRQJQHVAAgTIIDQ1VcHA97T7wufrdfpPLnuftNRtUcOyE5q9cq/kr//8Naa%2B8rJEOLHqjzI%2Bz67ODqh8crNDQUFfErNamTv3%2B19tv/%2BXHZ8%2BWHn%2B8qtN4tqpYNzVr1NDUd5fquUkzVFpaqhaXN9Hovz%2BqJ/94X7kfi3UDb2KzuN4dKJPuUVGqe6FE78WPNB3lkvo9P1qnagZo1arVpqPAy7FuAPfEbWCAMuocGans3P2mY5RJ9r796tw58tKfCLgY6wZwTxRAoIwiIiL05TdHdaSwyHSUizpSWKSvjhYoIiLCdBSAdQO4KQogUEZRUVHy8/PTvBVrTEe5qLkr1sjPz0/dunUzHQVg3QBuigIIlFFISIgefPBBTUtfVqEfM1UVLly4oOnpy9S/f3%2BFhISYjgOwbgA3RQEEyiE6OloHvjqilZk5pqP8rpWZOTrw1RFFR0ebjgL8hHUDuB%2BuAgbKwbIsRXTqpIY1bVppH//TTyRwB5ZlqcczI/TteSk7J8etssG7sW4A98MEECgHm82mmNGjtSprq%2BavdK/3NM1bsVqrsrZqVEwMmxjcCusGcD9MAIEKGDBggJYvXaJd86epSUhD03F0uKBI1wwYont699a8efNMxwF%2BF%2BsGcB8UQKACioqK1K5tW3UJb6H0hFFGJweWZanP8Bhl5R/Q7j171LCh%2BY0V%2BD2sG8B9cAoYqICGDRtq2vTpWvLRZsW%2BNt9oltGz5mnpx5maPmMGmxjcGusGcB%2B%2BMTExMaZDAJ6odevW8vPz08ikVNUJDNCN7dtWeYbktHf10rQ5Gj9%2BvJ544okqf36gvFg3gHugAAKVcPPNN8vhcGhksl2WLN3WsUOVnNayLEujZ83TS9Pm6IUXXtCoUWZPpwHlwboBzKMAApVgs9kUFRUlf39/jUxK1dbcT3V7p/YKCgx02XMeLijSwFHxejVjueLi4tjE4HFYN4B5XAQCOElGRoYGDxqkkuJi2YcN1sCe3Z26wViWpXkrVuuZlOnyDwjQ9Bkz1KdPH6c9PmAC6wYwgwIIOFFRUZF69uyp7OxsdY/sqGEP9VPP6zvLx6fi11tduHBBKzNzlPrWIq3K2qoBAwZo0qRJatCggROTA%2BYUFRVp6NChSktL0x2RHfV/Tlw3yWnvam3OdtYN8CsUQMCJioqKFBYWps6dO6uwoEBbt21T86ZNNLhvL/25Z1S57n12pLBIc1es0fT0ZTrw1RF1vO46xYwerd69e7vwOwDMycjI0OiYGKeum9CQEBWfPavPP/%2Bc8gf8DAUQcKKnn35ab7zxhvLy8tSoUSNlZmZqypQpWrBggRwOh5o1bqSIVlerU6uWuqbFVaoTGCC/GjXkOH9ep84Ua9dnB7Uld7%2By9%2B3XV0cL5O/vr/79%2Bys6OlpdunThPUuo9izLcuq6ufLKKxUeHq7HH39ckyZNMv3tAW6DAgg4ya5du3TdddcpPj5ezz333C/%2BrLCwUGvXrlV2drZycrKVk5Oj48dP/OYxgoPrKSIiQp07RyoiIkLdunVTSEhIVX0LgFtx1rpJSkrSv//9b23fvl3t2rWrym8BcFsUQMAJLMvSnXfeqUOHDmnXrl3y8/O75OcXFBSouLhYJSUl8vf3V0BAgEJDQ5nyAf9DRdeNw%2BFQu3bt1Lx5c61cuZI1BogCCDhFRkaG%2BvbtqyVLlujee%2B81HQfAryxZskS9e/dWRkYG76MFRAEEKq2kpETt2rVTy5YttXz5cqYLgBuyLEs9e/bUZ599pl27dsnf3990JMAofhYwUEl2u10HDx5USkoK5Q9wUzabTSkpKTpw4AAXgwBiAghUytdff62wsDD97W9/U2pqquk4AC7hmWee0ezZs5WXl6fLLrvMdBzAGAogUAl//etftXjxYuXn56t%2B/fqm4wC4hG%2B//VZhYWHq27evZs2aZToOYAyngIEKys7O1uzZszVmzBjKH%2BAhGjRooDFjxmj27NnKyckxHQcwhgkgUAGWZenmm2/Wd999py1btqhGjRqmIwEoo/Pnz6tjx46qV6%2BeNmzYwHt34ZWYAAIV8NZbb2njxo1KTU2l/AEepkaNGkpNTdXHH3%2BsBQsWmI4DGMEEECin06dPq3Xr1urSpYveffdd03EAVND999%2BvrKws5ebmKjAw0HQcoEoxAQTKKSEhQUePHlViYqLpKAAqISkpSUePHlVCQoLpKECVowAC5fD5558rISFBzz33nFq0aGE6DoBKaNGihZ599lnFx8fr0KFDpuMAVYpTwEA5PPTQQ/rwww%2BVm5uroKAg03EAVNJ3332n8PBw3X777XrzzTdNxwGqDBNAoIw2bNigBQsWaMKECZQ/oJoICgrShAkT9NZbb2nDhg2m4wBVhgkgUAYXLlxQZGSkatasqU2bNsnHh9dOQHVRWlqq66%2B/XufPn1dWVpZ8fX1NRwJcjl0MKIPZs2dr69atstvtlD%2BgmvHx8ZHdbtfWrVs1Z84c03GAKsEEELiEEydOKCwsTD169NDcuXNNxwHgIgMHDtQHH3yg/Px81a1b13QcwKUYZQCXMHbsWJ0%2BfVoTJkwwHQWAC02YMEGnTp3S2LFjTUcBXI4CCFxEXl6e7Ha7RowYoaZNm5qOA8CFmjVrphdeeEGpqanKz883HQdwKU4BAxdx3333adeuXdqzZ48CAgJMxwHgYsXFxWrTpo06dOigxYsXm44DuAwTQOB/WLFihZYuXarExETKH%2BAlAgIClJiYqCVLlmjlypWm4wAuwwQQ%2BB3nzp1Thw4d1LhxY61du1Y2m810JABVxLIs3X777SooKND27dtVs2ZN05EAp2MCCPyOqVOnKi8vT6mpqZQ/wMvYbDbZ7Xbt27dP06ZNMx0HcAkmgMCvFBYWKiwsTP379%2BfgD3ixwYMHa%2BHChcrPz1dISIjpOIBTMQEEfmXkyJGyLEtjxowxHQWAQWPHjpVlWRo1apTpKIDTUQCBn9mxY4emT5%2BuUaNGKTQ01HQcAAaFhoZq5MiRmjZtmnbu3Gk6DuBUnAIGfmBZlrp3767Dhw9rx44d8vPzMx0JgGEOh0Pt27dXs2bNtGrVKt4TjGqDCSDwg/T0dK1du1YpKSmUPwCSJD8/P6WkpGjNmjXKyMgwHQdwGiaAgKSzZ8%2Bqbdu2atOmjd5//33TcQC4mV69eik3N1e7d%2B9WrVq1TMcBKo0JICApJSVFX3zxhSZOnGg6CgA3NHHiRB06dEipqammowBOwQQQXu/w4cMKDw/XoEGDKIAA/qdhw4bp1VdfVV5eni6//HLTcYBKoQDC6z322GNatmyZ8vPzFRwcbDoOADd17NgxhYeH65577tGcOXNMxwEqhVPA8GqffPKJ3njjDY0bN47yB%2BCi6tevr7Fjx%2Br1119XVlaW6ThApTABhNcqLS3VjTfeqLNnzyonJ0e%2Bvr6mIwFwcxcuXFCnTp0UGBiojRs3clsYeCwmgPBaaWlpyszMlN1up/wBKBNfX1/Z7XZt3rxZaWlppuMAFcYEEF7p1KlTatWqlW688Ua9/fbbpuMA8DB/%2BtOftGnTJuXm5qpOnTqm4wDlxgQQXmnChAkqKipSYmKi6SgAPFBiYqKKiooUHx9vOgpQIRRAeJ2DBw8qKSlJw4cP11VXXWU6DgAP1Lx5c/3zn/9UUlKSDh48aDoOUG6cAobXeeCBB346dVO7dm3TcQB4qB/fSnLTTTdp4cKFpuMA5cIEEF5l3bp1eueddxQfH0/5A1ApderUUXx8vN5%2B%2B22tX7/edBygXJgAwmtw%2BwYAzsbtpOCpmADCa8ycOVM7duyQ3W6n/AFwCh8fH9ntdm3fvl2zZs0yHQcoMyaA8ArHjx9XWFgYP8IJgEs89thjWr58ufLy8vipQvAITADhFWJjY3X27FnFxcWZjgKgGoqLi9OZM2c0ZswY01GAMqEAotrbt2%2BfXn75Zb344otq0qSJ6TgAqqHLL79cL774oiZNmqTc3FzTcYBL4hQwqr1evXpp37592rNnj2rVqmU6DoBq6uzZs2rbtq3atGmj999/33Qc4KKYAKJaW7ZsmZYvX67k5GTKHwCXqlWrlpKSkn467gDujAkgqi2Hw6EOHTqoadOmWrVqFVf%2BAnA5y7LUvXt3HT58WDt37lTNmjVNRwJ%2BFxNAVFuTJ09Wfn6%2BUlNTKX8AqoTNZlNqaqry8/M1efJk03GA/4kJIKqlo0ePKjw8XAMGDOAgDKDKRUdHKy0tTfn5%2BQoNDTUdB/gNCiCqpcGDB2vhwoXKz89XSEiI6TgAvExhYaHCwsLUv39/TZs2zXQc4Dc4BYxqZ9u2bXr11Vc1evRoyh8AI0JCQhQTE6NXX31V27dvNx0H%2BA0mgKhWLMtSt27ddPToUW3fvp03YAMw5ty5c%2BrQoYMuu%2BwyrVmzhvciw60wAUS1snr1ap08eVIzZ86k/AEwqmbNmpo1a5ZOnDihNWvWmI4D/AITQAAAAC/DBBAAAMDLUAABAAC8DAUQAADAy1AAAQAAvAwFEAAAwMtQAOEx4uKkyEgpKEhq1Ejq21fKzTWdCgDKj%2BMZTKMAwmOsXy89%2BaS0ebP0wQfS%2BfPSXXdJp0%2BbTgYA5cPxDKZxH0B4rIKC7185r18v3Xqr6TQAUHEcz1DVmADCY5048f2vDRqYzQEAlcXxDFWNCSA8kmVJffpIx45JGzaYTgMAFcfxDCbUMB0AqIinnpJ27JA%2B%2Bsh0EgCoHI5nMIECCI/z9NPS4sXShx9KzZqZTgMAFcfxDKZQAOExLOv7g%2BWiRdK6dVLz5qYTAUDFcDyDaRRAeIwnn5TS0qSMjO/vnfX1199/vF49KSDAbDYAKA%2BOZzCNi0DgMWy23//47NnS449XaRQAqBSOZzCNAggAAOBluA8gAACAl6EAAgAAeBkKIAAAgJehAAIAAHgZCiAAAICXoQACAAB4GQogAACAl6EAAgAAeBkKIAAAgJehAMKtFBcXq1evXnruuedMRwEA45599lndc889Ki4uNh0F1QwFEG4lISFBq1evVnR0tOkoAGBcdHS0PvjgAyUmJpqOgmqGAgi38cUXXyg%2BPl7Dhg3T1VdfbToOABjXsmVLDRs2TBMmTNAXX3xhOg6qEZtlWZbpEIAkPfLII1q7dq3y8vIUFBRkOg4AuIWTJ08qPDxc3bt31/z5803HQTXBBBBu4aOPPtKbb76puLg4yh8A/EzdunUVFxentLQ0ffzxx6bjoJpgAgjjSktLFRkZKR8fH2VmZsrHh9clAPBzpaWl6tKliyTpk08%2B4TiJSuNfEIybM2eOtmzZIrvdzkENAH6Hj4%2BP7Ha7cnJy9Prrr5uOg2qACSCMOnnypMLCwnTHHXfw3hYAuIRHHnlEa9asUV5enurWrWs6DjwY4xYYNW7cOJ06dUrx8fGmowCA24uPj9fJkyc1fvx401Hg4SiAMCY/P18pKSn697//rWbNmpmOAwBu7w9/%2BIP%2B/e9/KyUlRfv37zcdBx6MU8Awpk%2BfPtq2bZv27dungIAA03EAwCOcOXNGbdq0UceOHZWenm46DjwUE0AY8d///leLFy9WYmIi5Q8AyiEwMFAJCQnKyMjQBx98YDoOPBQTQFS58%2BfP69prr1XDhg21fv162Ww205EAwKNYlqVbb71Vx44d07Zt21SjRg3TkeBhmACiyk2bNk179%2B6V3W6n/AFABdhsNtntdu3Zs0fTp083HQceiAkgqlRRUZHCwsL0pz/9STNmzDAdBwA82t///ne9%2B%2B67ys/PV8OGDU3HgQdhAogqNWrUKF24cEFjx441HQUAPN7YsWN14cIFxcTEmI4CD0MBRJXZuXOnpk6dqpEjR6pRo0am4wCAx2vcuLH%2B85//aOrUqdq1a5fpOPAgnAJGlbAsS3feeacOHTqkXbt2yc/Pz3QkAKgWHA6HrrnmGl155ZX673//y3urUSZMAFElFi9erNWrVyslJYXyBwBO5Ofnp4kTJ2rVqlVasmSJ6TjwEEwA4XIlJSVq27atwsPDtWzZMl6dAoCTWZalu%2B%2B%2BW/v379fu3bvl7%2B9vOhLcHBNAuFxqaqoOHTqkiRMnUv4AwAVsNptSUlJ08OBB2e1203HgAZgAwqWOHDmi8PBw/e1vf1NqaqrpOABQrT3zzDN67bXXlJ%2Bfr8suu8x0HLgxCiBc6i9/%2BYuWLFmi/Px81a9f33QcAKjWjh07prCwMPXu3Vuvvfaa6ThwY5wChstkZWVpzpw5Gjt2LOUPAKpA/fr1NWbMGM2ZM0fZ2dmm48CNMQGES1iWpZtuukmnT5/Wli1b5OvrazoSAHiF8%2BfPq1OnTgoKCtJHH33Ee6/xu5gAwiXefPNNbdq0SampqZQ/AKhCNWrUUGpqqjZu3Ki33nrLdBy4KSaAcLrTp0%2BrVatW6tq1q959913TcQDAK91///3KysrSvn37VLt2bdNx4GaYAMLp4uPjVVhYqMTERNNRAMBrJSUl6ejRo0pISDAdBW6IAgin%2Bvzzz5WYmKjnnntOLVq0MB0HALxWixYt9NxzzykhIUGHDh0yHQduhlPAcKr%2B/ftrw4YNysvLU506dUzHAQCv9t1336lVq1a69dZbeT8gfoEJIJzmww8/1MKFCxUfH0/5AwA3EBQUpAkTJmjBggXasGGD6ThwI0wA4RQXLlxQ586d5efnp02bNsnHh9cWAOAOSktLdcMNN%2BjcuXPKysrizgyQxAQQTvLaa69p27ZtstvtlD8AcCM%2BPj5KTU3V1q1bNXv2bNNx4CaYAKLSTpw4obCwMPXs2VNvvPGG6TgAgN/x5z//Wf/973%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%2BFQhw4ddPnll2v16tW8p9uLMAFEmS1atEhr165VSkoK5Q8AqgE/Pz%2BlpKRo7dq1Sk9PNx0HVYgJIMrk7NmzatOmjdq1a6elS5eajgMAcKJ77rlHe/fu1Z49e7ivq5dgAogymThxor788kslJyebjgIAcLKJEyfqiy%2B%2BUEpKiukoqCJMAHFJhw8fVnh4uAYPHkwBBIBq6tlnn9WMGTOUl5enyy%2B/3HQcuBgFEJf02GOPafny5crLy1NwcLDpOAAAFzh%2B/LjCwsJ0zz33aM6cOabjwMU4BYyLyszM1BtvvKFx48ZR/gCgGgsODta4ceP0%2Buuv65NPPjEdBy7GBBD/U2lpqW688UaVlJQoOztbvr6%2BpiMBAFzowoULioiIUK1atbRx40b5%2BDAnqq74m8X/NH/%2BfGVmZsput1P%2BAMAL%2BPr6ym63KzMzU2lpaabjwIWYAFZzlmWpoKBAZ86ckcPhkJ%2BfnwIDAxUaGnrRG36eOnVK4eHhuvnmm7Vw4cIqTAwAMO2BBx7Qxo0blZubqzp16lz0cyu6z8AsCmA1U1hYqDVr1ignJ0fZ2VnasmWLjh8/8ZvPCw6up06dOqlz50hFREQoKirqFz8M/KWXXlJycrL27t2rq666qgq/AwCAaQcPHlTr1q01fPhwjRkz5hd/5qx9BmZRAKsBy7K0efNmTZkyRQsXLpTD4VCzxo3UuVVLdWrdUu2aX6mgwAD51awpx7lz%2Bu5MsXYf%2BFxb9u1Xdu5%2BffnNUfn5%2Bal///6Kjo5Wo0aN1LZtW/3rX/9SbGys6W8PAGDAf/7zHyUmJmrfvn268sornbrPdO3alemgYRRAD5eRkaGYUaO0bft2NW/aREP69tLAnlFqEtKwzI9xpLBI81as0bT0ZTrw1RE1qF9fstl06NAh1a5d24XpAQDu6vTp02rVqpWuvPJKnTl92qn7TMfrrlPM6NHq3bu3C78DXAwF0EMVFRVp6NChSktL0x2RHTXs4fvVo2tEpa7YKi0t1crMHCXNf0drc7brkUce0aRJk9SwYdkXOQCgeigqKlLv%2B%2B7Txk2b1D2yo5514j6T8uZ7WpW1lX3GIAqgB0pPT9eQwYPlOFss%2B7AhGtAjyqmjdMuyNG/Faj2TMl3%2BAQGaNn26%2Bvbt67THBwC4N/aZ6o/bwHgQy7I0fvx49evXT13CW2jX/Gka2LO7099HYbPZ9Oe779DutGnqEt5C/fr1U1xcnHitAADVG/uM9/CNiYmJMftTGvUAAAqFSURBVB0Cl2ZZlkaMGKHY2FiNemKgpv7raQXVDnTpcwYFBuqhO2%2BTJI1MSpXD4VBUlHNfBQIA3AP7jHehAHqIuLg4xcbGKvHpv%2BuFxx6qssVhs9l0e6drVScwQCOT7fL399ctt9xSJc8NAKg67DPehQLoAdLT0zVo0CCNemKgXnjsISMZbmzfVpYsjUxKVceOHdW6dWsjOQAAzsc%2B4324CMTNFRUVqV3btuoS3kLpCaOMjsUty1Kf4THKyj%2Bg3Xv2cNUWAFQD/6%2B9%2B4%2BqsrDjOP65gPcemCnKvTZCTdTSSCu5YLl11sCdpNqwds6kAttZs3ScAt3yOM0Q2hFXWUo/ENyptIBlqybW/LGRnLO1Mxlc0M1aLovaUk5yb4F6JK7As3%2BKrdUpLnjv88Dzfv0H3Od5vufAw%2Bdzn1%2BXnLEnbgKxuIKCAgU/7lLFqrtNvybC4XCoYlWBuru6VFhYaOosAIBzg5yxJwqghdXW1qqmpkZlK5aF9MDNcLrAk6CyFUtVXV2tXbt2mT0OAGAIyBn74hSwRRmGodQ5c%2BR2RmlfWanp78r%2Bl2EYWlC4RoGzhnzNzZaaDQAwMOSMvXEE0KIOHDigg4cOacUt37fcH77D4dDym29Sy8GDamhoMHscAMAgkDP2RgG0qPLyciUnJWrBlV6zR/lCC670KjkpUeXl5WaPAgAYBHLG3iiAFuT3%2B/X8889r2Y3XD%2BkzF8MpOjpaS2%2B8Xjt27JDf7zd7HABACMgZWPO3bnP79%2B9XMBhUXlZm2LcV6DypSdl5ipqXpY5Tp0NadnFWpoLBoOrr68M0HQAgHCKVM9t%2B93tdnrdMsdd8T4k33KK7Nj4R0vLkTPhQAC3I5/Np4vkTInJH1pLSTbpsevKglk10Jyhpgkc%2Bn%2B8cTwUACKdI5Mwjv35Rayu2a9XiHB2urlTdYxtCPt1MzoRPjNkD4POaGhuVNmP6gF%2Bfkb9Ss6YlKzoqSs/sqZMzJkb333mbchdk6u6Hn9AL9a9pwrh4PfazfF03L71/uS0vvaKOU6d13%2B252vOXxkHNmjZzupqaBrcsAMAc4c6Zj06e0n2Vz2jXQ8Wanz6nfz2XTp0S8qzkTHhwBNBiDMNQc0uzUmcOfMeUpGd218kdP0YNT5bprh9kK/%2Bhx7Xo3vWaNztFvm2P69orvbqt5CGd%2BfhjSdIbre/pF09Va3vRSkVFDf7ur9QZ09XsaxZPEwKA4SESOfOHv7aoz%2BjTsfaAUm6%2BQ5Oy85Rz73r9%2B4P2kOclZ8KDAmgx7e3t6ujo1KXJF4a03OUXJWvtj27VRZOStPq2HMW6nHKPHaM7Fl6niyYlqej2WxXoPKm/HW1VdzCoW4t%2BqQfvWqLJX58wpHlnTZ2ijzo61N4e%2Bk4NAIi8SOTMO8fb1NdnaMP257Rp%2BVL9pvRefXjylK4tWK3g2bMhbZecCQ8KoMWcOXNGknReXGxIy82e9t/r%2BKKjo5UwdoxmTZvS/73zx4%2BTJJ34qEOrtzytS6ZMVl7W/CHPO/qTObu6uoa8LgBA%2BEUiZ/r6DJ3t6VHZT3%2BiBVel6apZl6jm/p/rrfePq953KKTtkjPhwTWAFhMMBiVJzlGjQlpuVMxnf5WO//vepw/57OszVO87pL%2B//a5eqP%2BTJOnTo%2Bqe6xZpzQ9vUckdiwe8Xecn2%2Bju7g5pXgCAOSKRM4nu8ZKklOTJ/T/3jIuXe%2BwY/SvE08DkTHhQAC3G6XRKUsiHyEPxQuladXUH%2B79u/Mc/9eP1j%2BiPWzZqWtIFIa0r2NMjSXK5XOd0RgBAeEQiZ755WYok6ch772viBI8k6cPOU/J3ntSFIV56RM6EBwXQYuLi4iRJp86E71D3tImfLXn%2Bzk5J0iVTJiv%2BvNEhrev0J3PGxoZ2KgEAYI5I5MzFkydq4bfmafnmClWuKtSYr8VpzZanNfPCicrwXh7SusiZ8OAaQIvxeDyKjx%2Br11vfM3uUATn8zrsaFx8vj8dj9igAgAGIVM5sL7pHc1Nm6Lv3FOnb%2BSs1KiZaezat/9yp5K9CzoSHw%2BC%2BasuZn5mpMb3deumBIrNH%2BUo3rSrR6VGxqqt71exRAAADRM6AI4AWlJaerqYjR80eY0Ca3jyqtLT0r34hAMAyyBlQAC3I6/Xq/Q9OqM0fMHuUL9XmD%2BjYiXZ5vaF9tA8AwFzkDCiAFpSZmSmn06mqvfvNHuVLPbt3v5xOpzIyMsweBQAQAnIGFEALcrvdWrRokSp27lZfX5/Z43yh3t5eVe7crZycHLndbrPHAQCEgJwBBdCi8vPz1XqsTfsafGaP8oX2NfjUeqxN%2Bfn5Zo8CABgEcsbeuAvYogzDkDc1VQmjHNpXVtr/hHUrMAxDCwrX6MMeqcnns9RsAICBIWfsjSOAFuVwOFRcUqK6xhZV77PWNRpVe19VXWOL1hUXs1MCwDBFztgbRwAtLjc3V3teeVmHqyuU6E4wexwdbw9oVu4y3ZCdraqqKrPHAQAMETljTxRAiwsEAro0JUVzL56qnQ%2BuM/WdkGEYWriyWI1vter1N95QQoL5/ygAAENDztgTp4AtLiEhQRWVlXr5tQO6/6lqU2cpebJKr/y5QZVbt7JTAsAIQc7YU3RxcXGx2UPgy82cOVNOp1NFGzdrdFysvjE7JeIzPFzzotZWbFNpaamWLFkS8e0DAMKHnLEfCuAwcfXVVysYDKro4TIZMnTNnMsicpjeMAyVPFmltRXbtHr1aq1bZ%2B7pAQBAeJAz9kIBHCYcDocyMzPlcrlUtHGzWo68rW%2BnztZ5cXFh2%2Bbx9oDy1j2gX9Xu0YYNG9gpAWAEI2fshZtAhqHa2lotvfNOdXd1qWzFUuVlzT%2BnO4xhGKra%2B6oKN1XKFRuryq1btXDhwnO2fgCAtZEzIx8FcJgKBAIqKChQTU2NvpM%2BR8tvvklZV6UpKmrw9/X09vZqX4NPm5/7reoaW5Sbm6tHH31U48ePP4eTAwCGA3JmZKMADnO1tbUqKS5Wy8GDSk5K1NIbr9firMyQnuXU5g/o2b37Vblzt1qPtWnOFVeouKRE2dnZYZwcADAckDMjEwVwBDAMQw0NDSovL9eOHTsUDAY18fwJ8s6YptQZ0zVr6hSNjouVMyZGwZ4enT7TpcPvvKvmI0fV9OZRHTvRLpfLpZycHOXn52vu3LlcgwEA6EfOjDwUwBHG7/ervr5eTU1N8vma5PP51NHR%2BbnXxcePldfrVVpaurxerzIyMuR2u02YGAAwnJAzIwMFcIQzDEPt7e3q6upSd3e3XC6XYmNj5fF4ePcFABgycmZ4ogACAADYDB8FBwAAYDMUQAAAAJuhAAIAANgMBRAAAMBmKIAAAAA2QwEEAACwGQogAACAzVAAAQAAbIYCCAAAYDMUQAAAAJuhAAIAANgMBRAAAMBmKIAAAAA2QwEEAACwGQogAACAzVAAAQAAbIYCCAAAYDMUQAAAAJuhAAIAANgMBRAAAMBmKIAAAAA2QwEEAACwGQogAACAzVAAAQAAbIYCCAAAYDMUQAAAAJuhAAIAANgMBRAAAMBmKIAAAAA2QwEEAACwGQogAACAzVAAAQAAbIYCCAAAYDMUQAAAAJuhAAIAANjMfwD0VKwfNshR3wAAAABJRU5ErkJggg%3D%3D", null, 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WrVund999V/Hx8ZQ/L8EEEGVyqdsCAAA804%2B3/QoMDNTGjRu5ctdLMAFEmfj4%2BMhut2v79u2aNWuW6TgAACeZOXOmduzYIbvdTvnzIkwAUS6PPfaYli1bpvz8fAUHB5uOAwCohGPHjiksLEz33nuv5syZYzoOqhATQJRLXFyciouLFRsbazoKAKCSYmNjVVJSori4ONNRUMUogCiXyy%2B/XC%2B%2B%2BKJefvll7du3z3QcAEAF7du3T6%2B88opefPFFNWnSxHQcVDFOAaPczp49q7Zt26pNmzZ6//33TccBAFRAr169tG/fPu3Zs0e1atUyHQdVjAkgyq1WrVpKSkrSsmXLtGzZMtNxAADltGzZMi1fvlzJycmUPy/FBBAVYlmWunfvrsOHD2vHjh3y8/MzHQkAUAYOh0Pt27dXs2bNtGrVKq789VJMAFEhNptNqampys/P1%2BTJk03HAQCU0SuvvKL9%2B/crNTWV8ufFmACiUqKjo5WWlqb8/HyFhoaajgMAuIijR48qLCxMAwcO5MW7l6MAolIKCwsVFham/v37a9q0aabjAAAuYvDgwVq4cKHy8/MVEhJiOg4M4hQwKiUkJEQxMTGaMWOGtm3bZjoOAOB/2LZtm1599VWNHj2a8gcmgKi8c%2BfOqUOHDmrcuLHWrl3Le0oAwM1YlqXbb79dBQUF2r59u2rWrGk6EgxjAohKq1mzplJTU7V%2B/Xq9%2B%2B67puMAAH7lnXfe0YcffqiUlBTKHyQxAYQT3Xfffdq5c6f27t2rgIAA03EAAJKKi4vVunVrXXvttVq8eLHpOHATTADhNMnJyTp8%2BLAmTpxoOgoA4AfJyck6cuSIkpOTTUeBG2ECCKf65z//qalTpyovL09NmzY1HQcAvNqXX36pVq1aKTo6WomJiabjwI1QAOFUJ06cUFhYmHr06KG5c%2BeajgMAXm3gwIH64IMPlJeXp3r16pmOAzfCKWA4Vb169TR%2B/HjNmzdPmzZtMh0HALzWxo0bNX/%2BfI0fP57yh99gAginu3DhgiIjI1WjRg1t3rxZPj68zgCAqlRaWqquXbuqtLRUn3zyiXx9fU1HgpthZ4bT%2Bfr6ym63KysrS/PmzTMdBwC8zty5c5Wdna3U1FTKH34XE0C4TP/%2B/bVhwwbl5uYqKCjIdBwA8ArfffedwsPDddttt%2Bmtt94yHQduigkgXCYhIUHHjh1TXFyc6SgA4DXGjx%2Bv48ePKyEhwXQUuDEKIFzmyiuv1L/%2B9S8lJyfrs88%2BMx0HAKq9Tz/9VBMnTtTzzz%2BvK664wnQcuDFOAcOlTp8%2BrdatWysyMlLvvfee6TgAUK3169dP2dnZys3NVWBgoOk4cGNMAOFStWvXVnx8vBYtWqQ1a9aYjgMA1dbq1auVnp6uhIQEyh8uiQkgXM6yLN1888367rvvtGXLFtWoUcN0JACoVs6fP6%2BOHTuqXr162rBhg2w2m%2BlIcHNMAOFyNptNdrtdO3fu1Kuvvmo6DgBUOzNmzNDu3btlt9spfygTJoCoMn/961%2B1ePFi5eXlqUGDBqbjAEC18O233yosLEx9%2B/bVrFmzTMeBh2ACiCozfvx4lZSUaPTo0aajAEC1ERMTo3PnzmncuHGmo8CDUABRZS677DK99NJLmjx5svbs2WM6DgB4vD179mjKlCl66aWXdNlll5mOAw/CKWBUqZKSErVr105XX321VqxYwXtVAKCCLMtSjx499Nlnn2n37t3y9/c3HQkehAkgqpS/v7%2BSk5P13//%2BV%2B%2B//77pOADgsZYuXaoPPvhAEydOpPyh3JgAospZlqW77rpLBw8e1O7du%2BXn52c6EgB4lJKSEl1zzTVq3ry5Vq5cydkUlBsTQFQ5m82mlJQUHThwQJMmTTIdBwA8zqRJk3TgwAGlpKRQ/lAhTABhzFNPPaW5c%2BcqLy9PjRs3Nh0HADzCN998o7CwMD322GN6%2BeWXTceBh6IAwpiioiKFhYXpj3/8IzeIBoAyeuKJJ7Ro0SLl5%2BdzT1VUGKeAYUzDhg0VGxurWbNmacuWLabjAIDby8nJ0WuvvabY2FjKHyqFCSCMOn/%2BvK677jrVr19fH374Ie9lAYD/wbIs3XLLLTp%2B/Li2bdvGz1VHpTABhFE1atRQamqqPvroIy1cuNB0HABwWwsWLNDHH38su91O%2BUOlMQGEW%2BjTp4%2B2bdumvXv3KjAw0HQcAHArZ86cUevWrdWpUyelp6ebjoNqgAkg3EJycrKOHDmipKQk01EAwO0kJibqm2%2B%2B4RgJp6EAwi20bNlSw4YN04QJE/TFF1%2BYjgMAbuPQoUOKj4/XsGHD1LJlS9NxUE1wChhu4%2BTJkwoPD1dUVJTS0tJMxwEAt/Dwww9r3bp1ysvLU1BQkOk4qCaYAMJt1K1bV3FxcXrzzTf18ccfm44DAMZ99NFHeuuttxQXF0f5g1MxAYRbKS0tVZcuXSRJn3zyiXx8eI0CwDuVlpYqMjJSPj4%2ByszM5HgIp%2BJfE9yKj4%2BPZs6cqdLSUi1dutR0HAAwZsmSJbIsSzNnzqT8wemYAAIAAHgZXlIAAAB4GQogAACAl6EAAgAAeBkKIAAAgJehAAIAAHgZCiAAAICXoQDCY8TFSZGRUlCQ1KiR1LevlJtrOhUAlB/HM5hGAYTHWL9eevJJafNm6YMPpPPnpbvukk6fNp0MAMqH4xlM40bQ8FgFBd%2B/cl6/Xrr1VtNpAKDiOJ6hqjEBhMc6ceL7Xxs0MJsDACqL4xmqGhNAeCTLkvr0kY4dkzZsMJ0GACqO4xlMqGE6AFARTz0l7dghffSR6SQAUDkcz2ACBRAe5%2BmnpcWLpQ8/lJo1M50GACqO4xlMoQDCY1jW9wfLRYukdeuk5s1NJwKAiuF4BtMogPAYTz4ppaVJGRnf3zvr66%2B//3i9elJAgNlsAFAeHM9gGheBwGPYbL//8dmzpccfr9IoAFApHM9gGgUQAADAy3AfQAAAAC9DAQQAAPAyFEAAAAAvQwEEAADwMhRAAAAAL0MBBAAA8DIUQAAAAC9DAQQAAPAyFEBUOy%2B99JK6d%2B%2Bu7777znQUAF7sxIkT6t69u/7zn/%2BYjgL8Bj8JBNXOV199pfDwcP3jH/9QUlKS6TgAvNRzzz2n6dOnKzc3V02bNjUdB/gFJoCodpo2baoRI0bIbrcrLy/PdBwAXig3N1eTJk3SiBEjKH9wS0wAUS0VFxerTZs2at%2B%2BvZYsWWI6DgAvc%2B%2B992r37t3au3evatWqZToO8BtMAFEtBQQEKCkpSUuXLtWKFStMxwHgRZYvX673339fSUlJlD%2B4LSaAqLYsy1K3bt30zTffaMeOHapZs6bpSACquXPnzql9%2B/Zq0qSJ1qxZI5vNZjoS8LuYAKLastlsSk1NVV5enqZMmWI6DgAvMHnyZOXn5ys1NZXyB7fGBBDV3pAhQ7RgwQLl5%2BcrJCTEdBwA1VRBQYHCwsL08MMPa%2BrUqabjABfFBBDV3pgxY2RZlkaOHGk6CoBq7MdjTGxsrOEkwKVRAFHthYaGatSoUZo%2Bfbp27NhhOg6Aamj79u2aMWOGYmJiFBoaajoOcEmcAoZXcDgc6tChgy6//HKtXr2a9%2BYAcBrLshQVFaWvv/6aC87gMZgAwiv4%2BfkpJSVFa9eu1aJFi0zHAVCNvPfee1q3bp1SUlIof/AYTADhVe655x7t2bOHm7MCcIqzZ8%2BqTZs2ateunZYuXWo6DlBmTADhVSZOnKgvv/xSKSkppqMAqAZ%2BPKYkJyebjgKUCxNAeJ1nn31WM2bMUF5eni6//HLTcQB4qK%2B%2B%2BkqtWrXS4MGDKYDwOBRAeJ3jx48rLCxMvXr10uuvv246DgAP9eijj2rFihXKy8tTcHCw6ThAuXAKGF4nODhY48aN0xtvvKHMzEzTcQB4oM2bN2vu3LkaN24c5Q8eiQkgvNKFCxcUERGhWrVqaePGjfLx4bUQgLIpLS3VDTfcIIfDoezsbPn6%2BpqOBJQbux68kq%2Bvr%2Bx2uzIzM5WWlmY6DgAPMn/%2BfH3yySey2%2B2UP3gsJoDwag888IA2btyo3Nxc1alTx3QcAG7u1KlTCg8P180336yFCxeajgNUGBNAeLXExEQVFRVpwoQJpqMA8ABxcXE6duyYEhISTEcBKoUCCK921VVXafjw4UpKStKBAwdMxwHgxj777DMlJydr%2BPDhuuqqq0zHASqFU8DweqdPn1arVq10/fXX65133jEdB4Cb%2BuMf/6jMzEzl5uaqdu3apuMAlcIEEF6vdu3amjBhgt59912tW7fOdBwAbmjt2rV67733FB8fT/lDtcAEEND3t3W46aabdObMGW3ZsoUr%2BwD85Pz584qIiFDt2rX18ccfy2azmY4EVBoTQECSj4%2BP7Ha7duzYoZkzZ5qOA8CNzJw5Uzt27JDdbqf8odpgAgj8zOOPP66lS5cqPz9f9evXNx0HgGHHjh1TWFiY7rvvPs2ePdt0HMBpmAACPxMXF6eSkhLFxsaajgLADYwePVolJSUaP3686SiAU1EAgZ9p0qSJRowYoVdeeUX79u0zHQeAQXv37tXkyZP14osvqkmTJqbjAE7FKWDgV86ePau2bduqdevWWrZsmek4AAywLEt333238vPztXv3btWqVct0JMCpmAACv1KrVi0lJydr%2BfLlFEDASy1btkwrV65UcnIy5Q/VEhNA4HdYlqU77rhDX375pXbu3Ck/Pz/TkQBUEYfDoWuuuUZXXHGFPvjgA678RbXEBBD4HTabTampqdq/f79eeeUV03EAVKGXX35Zn376qVJTUyl/qLaYAAIXER0drbS0NOXl5alRo0am4wBwsaNHjyosLEwDBw7U5MmTTccBXIYCCFxEYWGhwsLC9OCDD2r69Omm4wBwsUGDBumdd95Rfn6%2BGjZsaDoO4DKcAgYuIiQkRKNHj9arr76qbdu2mY4DwIW2bt2qmTNnavTo0ZQ/VHtMAIFLOHfunK699lqFhoZq3bp1vCcIqIYsy9Jtt92moqIibdu2TTVr1jQdCXApJoDAJdSsWVMpKSn68MMP9c4775iOA8AF3n77bW3YsEGpqamUP3gFJoBAGd13333auXOn9u7dq4CAANNxADhJcXGxWrdurWuvvVaLFy82HQeoEkwAgTJKTk7W4cOHlZycbDoKACdKSkrSkSNHWNvwKkwAgXIYPny4pkyZotzcXDVr1sx0HACV9MUXX6hVq1Z66qmnlJCQYDoOUGUogEA5nDhxQuHh4brzzjs1b94803EAVNKAAQO0atUq5efnq27duqbjAFWGU8BAOdSrV0/jx4/X/PnztWnTJtNxAFTCxo0blZaWpri4OMofvA4TQKCcLly4oMjISNWoUUObN2%2BWjw%2BvowBPU1paqq5du6q0tFRZWVmsY3gd/sUD5eTr6yu73a6srCzNnTvXdBwAFfDGG28oOztbdrud8gevxAQQqKCHHnpI69evV15enoKCgkzHAVBGJ0%2BeVHh4uLp166Y333zTdBzACF72ABWUkJCg48ePa/z48aajACiH8ePH6%2BTJk4qPjzcdBTCGAghU0BVXXKHnn39eEydO1Keffmo6DoAy%2BPTTT5WSkqLnn39eV1xxhek4gDGcAgYq4cyZM2rVqpUiIyP13nvvmY4D4BL69eunnJwc7du3T4GBgabjAMYwAQQqITAwUAkJCVq0aJFWr15tOg6Ai1i1apXS09OVkJBA%2BYPXYwIIVJJlWbrlllt04sQJbd26VTVq1DAdCcCvnD9/Xtddd52Cg4O1YcMG2Ww205EAo5gAApVks9lkt9u1e/duzZgxw3QcAL9j%2BvTp2rNnj%2Bx2O%2BUPEBNAwGn%2B9re/KT09Xfn5%2BWrQoMFFP9eyLBUUFOjMmTNyOBzy8/NTYGCgQkND2ZyA/6Gi6%2Bbbb79VWFiY%2BvXrp5kzZ1ZhYsB9UQABJ/n6668VHh6uv/zlL7Lb7b/4s8LCQq1Zs0Y5OTnKzs7Sli1bdPz4id88RnBwPXXq1EmdO0cqIiJCUVFRCgkJqapvAXArzlo3Q4cO1Zw5c5Sfn6/GjRtX5bcAuC0KIOBECQkJGjFihHbs2KE2bdpo8%2BbNmjJlihYuXCiHw6FmjRupc6uW6tS6pdo1v1JBgQHyq1lTjnPn9N2ZYu0%2B8Lm27Nuv7Nz9%2BvKbo/Lz81P//v0VHR2trl27Mh1EtWdZllPXTZ06dXTdddcpLi5Ow4cPN/3tAW6DAgg4UUlJidq1a6c6derIJmnb9u1q3rSJhvTtpYE9o9QkpGGZH%2BtIYZHmrVijaenLdOCrI%2Bp43XWKGT1avXv3dt03ABiUkZGhmFGjnLpugutZTGP8AAAgAElEQVTVU0BgoA4cOCB/f38Xpgc8CwUQcKKioiLdf//9%2BvDDD9U9sqOeffh%2B9egaUamfNVpaWqqVmTlKefM9rcraqkceeUSTJk1Sw4Zl3xQBd1ZUVKShQ4cqLS1Nd0R21DAnrpuk%2Be9obc521g3wKxRAwEnS09M1ZPBgOc4Wyz5siAb0iHLqKVvLsjRvxWo9kzJd/gEBmjZ9uvr27eu0xwdMYN0AZnAbGKCSLMvS%2BPHj1a9fP3UJb6Fd86dpYM/uTn%2B/ns1m05/vvkO706apS3gL9evXT3FxceI1HDwR6wYwyzcmJibGdAjAU1mWpREjRig2Nlajnhioqf96WkG1XfsTBoICA/XQnbdJkkYmpcrhcCgqyrlTE8CVWDeAeRRAoBLi4uIUGxurxKf/rhcee6jKNhObzabbO12rOoEBGplsl7%2B/v2655ZYqeW6gslg3gHkUQKCC0tPTNWjQII16YqBeeOwhIxlubN9WliyNTEpVx44d1bp1ayM5gLJi3QDugYtAgAooKipSu7Zt1SW8hdITRhk9jWRZlvoMj1FW/gHt3rOHqxzhtlg3gPvgIhCgAoYOHSrH2WJNe/5p4%2B8hstlsmvb8UJUUF%2BuZZ54xmgW4GNYN4D4ogEA5ZWRkKC0tTfZhQ8p1g1pXujy0oezDBmv%2B/PlavHix6TjAb7BuAPfCKWCgHCzLUqeOHRXi56OV9vHGpxg/Z1mWejwzQkXnLOVs2eJW2eDdWDeA%2B2ECCJTD5s2btW37dg17%2BH632yhsNpv%2B76F%2B2rptmzIzM03HAX7CugHcDwUQKIcpU6aoedMm6tE1wnSU39Wja4SaN22iKVOmmI4C/IR1A7gfCiBQRoWFhVq4cKGG9O1VqZ9R6kq%2Bvr4a3LeXFixYoMLCQtNxANYN4KbcczUCbmjNmjVyOBwa2DPKZc%2BxPf8zPTIyTlf0GajA23qr7UN/l31Berke4889o%2BRwOLR27VoXpaze4uKkyEgpKEhq1Ejq21fKzTWdynNVxbrxuaHnb/6b9t775XoM1g28TQ3TAQBPkZOTo2aNG7n0CsacffkKCa6nuaP%2BpT80DtXGnXs0eMIk%2Bfr46KkHepfpMZqENFTTRqHKycnRAw884LKs1dX69dKTT35fAs%2Bfl158UbrrLmnPHql2bdPpPE9VrBtJeu2lZ9Xz%2Bs4//b5eOf%2ByWDfwNhRAoIyys7LUuVXLcn1Nt%2Bjhuubq5vL18dEby1fJr0YNxQ56VAN6ROnp5Ml6Z%2B1HalQ/WC8/F627b4jUX%2B/r8Yuvb9G0iTbt3KtF6z8ucwGUpM6tWyo7O6tcWfG9FSt%2B%2BfvZs7%2BfBObkSLfeaiaTJyvvuinvmvlRcJ06uqxhg0plZd3Am3AKGCgDy7K0ZesWdWpdvgIoSW8sW6WQ4LrKnGXXUw/0VnTiK3rwxXG6oX1b5cx5RXd1jdCjoxN15uzZ3/36k6dPq0HdoHI9Z6dWLbUlZ4u4y1PlnTjx/a8NKtctvFJF101F1szTyVMU2vNBdfnr05r23vsqLS0td17WDbwJBRAog4KCAh0/fkLtml9Z7q%2B9Nqy5XvrLIwr7Q1O98Gh/Bfj7KaReXf29z90K%2B0NTjfzrIyo6cVI79h/4zddu2rlHC1dv0KC%2Bvcr1nNe0uErHjh9XQUFBufPi/7Ms6dlnpZtvlq65xnQaz1PRdVPeNRM76FEtHDdCH0yKU/87btc/X56h8a%2B/Ve68rBt4E04BA2Vw5swZSVJQYEC5v7b91c1/%2Bn9fX181rFdX11x91U8fa9ygviTp6LHjv/i63Z8dVN/nR%2Bs/f31Ed3bpVK7nrPNDzqysLDVp0qTcmauzgIAAtWnTpkyf%2B9RT0o4d0kcfXfpz9%2B7dq%2BLi4kqmq14OHz4sqfzrprxr5qW/PPLTn10XfrUkacxr83/x8bL4cd3w9whvQAEEysDhcEiS/GrWLPfX1qzxy2Vm%2B9XHfrwxbmnp/z/ttOfA5%2Br%2B1L/1RO%2B7y72JSZLfD49/7733lvtrq7uOHTtqy5Ytl/y8p5%2BWFi%2BWPvxQatbs0o87YMAAbd261QkJq5/yrpuKrJmfu75da508fUbffHvsp7JYppw/PEdJSUm58gKeiAIIlIGfn58kyXHunMufa/dnB9X9qX/r0V53aNyQxyv0GI7z5yVJS5cuZQL4KwEBF59GWdb35W/RImndOql584t%2B%2Bk/mz5/P5OhXDh8%2BrPvuu69K1s3Pbc37VLX8/BRcp3xXAv%2B4bvz9/V0RC3ArFECgDAIDAyVJ351x7Qa/%2B7ODinrqed3VpZOeffh%2BfV30rSTJ18dHofWDy/w4p37IGRkZqUaNGrkka3X15JNSWpqUkfH9vQC//vr7j9erJ12sO5b1tLI3afbD6NSV62bJhs36%2BttjuuGaNgrw99PaLdv10vQ5%2Bnvfu%2BX/wwu3svpx3VzqRQJQHVAAgTIIDQ1VcHA97T7wufrdfpPLnuftNRtUcOyE5q9cq/kr//8Naa%2B8rJEOLHqjzI%2Bz67ODqh8crNDQUFfErNamTv3%2B19tv/%2BXHZ8%2BWHn%2B8qtN4tqpYNzVr1NDUd5fquUkzVFpaqhaXN9Hovz%2BqJ/94X7kfi3UDb2KzuN4dKJPuUVGqe6FE78WPNB3lkvo9P1qnagZo1arVpqPAy7FuAPfEbWCAMuocGans3P2mY5RJ9r796tw58tKfCLgY6wZwTxRAoIwiIiL05TdHdaSwyHSUizpSWKSvjhYoIiLCdBSAdQO4KQogUEZRUVHy8/PTvBVrTEe5qLkr1sjPz0/dunUzHQVg3QBuigIIlFFISIgefPBBTUtfVqEfM1UVLly4oOnpy9S/f3%2BFhISYjgOwbgA3RQEEyiE6OloHvjqilZk5pqP8rpWZOTrw1RFFR0ebjgL8hHUDuB%2BuAgbKwbIsRXTqpIY1bVppH//TTyRwB5ZlqcczI/TteSk7J8etssG7sW4A98MEECgHm82mmNGjtSprq%2BavdK/3NM1bsVqrsrZqVEwMmxjcCusGcD9MAIEKGDBggJYvXaJd86epSUhD03F0uKBI1wwYont699a8efNMxwF%2BF%2BsGcB8UQKACioqK1K5tW3UJb6H0hFFGJweWZanP8Bhl5R/Q7j171LCh%2BY0V%2BD2sG8B9cAoYqICGDRtq2vTpWvLRZsW%2BNt9oltGz5mnpx5maPmMGmxjcGusGcB%2B%2BMTExMaZDAJ6odevW8vPz08ikVNUJDNCN7dtWeYbktHf10rQ5Gj9%2BvJ544okqf36gvFg3gHugAAKVcPPNN8vhcGhksl2WLN3WsUOVnNayLEujZ83TS9Pm6IUXXtCoUWZPpwHlwboBzKMAApVgs9kUFRUlf39/jUxK1dbcT3V7p/YKCgx02XMeLijSwFHxejVjueLi4tjE4HFYN4B5XAQCOElGRoYGDxqkkuJi2YcN1sCe3Z26wViWpXkrVuuZlOnyDwjQ9Bkz1KdPH6c9PmAC6wYwgwIIOFFRUZF69uyp7OxsdY/sqGEP9VPP6zvLx6fi11tduHBBKzNzlPrWIq3K2qoBAwZo0qRJatCggROTA%2BYUFRVp6NChSktL0x2RHfV/Tlw3yWnvam3OdtYN8CsUQMCJioqKFBYWps6dO6uwoEBbt21T86ZNNLhvL/25Z1S57n12pLBIc1es0fT0ZTrw1RF1vO46xYwerd69e7vwOwDMycjI0OiYGKeum9CQEBWfPavPP/%2Bc8gf8DAUQcKKnn35ab7zxhvLy8tSoUSNlZmZqypQpWrBggRwOh5o1bqSIVlerU6uWuqbFVaoTGCC/GjXkOH9ep84Ua9dnB7Uld7%2By9%2B3XV0cL5O/vr/79%2Bys6OlpdunThPUuo9izLcuq6ufLKKxUeHq7HH39ckyZNMv3tAW6DAgg4ya5du3TdddcpPj5ezz333C/%2BrLCwUGvXrlV2drZycrKVk5Oj48dP/OYxgoPrKSIiQp07RyoiIkLdunVTSEhIVX0LgFtx1rpJSkrSv//9b23fvl3t2rWrym8BcFsUQMAJLMvSnXfeqUOHDmnXrl3y8/O75OcXFBSouLhYJSUl8vf3V0BAgEJDQ5nyAf9DRdeNw%2BFQu3bt1Lx5c61cuZI1BogCCDhFRkaG%2BvbtqyVLlujee%2B81HQfAryxZskS9e/dWRkYG76MFRAEEKq2kpETt2rVTy5YttXz5cqYLgBuyLEs9e/bUZ599pl27dsnf3990JMAofhYwUEl2u10HDx5USkoK5Q9wUzabTSkpKTpw4AAXgwBiAghUytdff62wsDD97W9/U2pqquk4AC7hmWee0ezZs5WXl6fLLrvMdBzAGAogUAl//etftXjxYuXn56t%2B/fqm4wC4hG%2B//VZhYWHq27evZs2aZToOYAyngIEKys7O1uzZszVmzBjKH%2BAhGjRooDFjxmj27NnKyckxHQcwhgkgUAGWZenmm2/Wd999py1btqhGjRqmIwEoo/Pnz6tjx46qV6%2BeNmzYwHt34ZWYAAIV8NZbb2njxo1KTU2l/AEepkaNGkpNTdXHH3%2BsBQsWmI4DGMEEECin06dPq3Xr1urSpYveffdd03EAVND999%2BvrKws5ebmKjAw0HQcoEoxAQTKKSEhQUePHlViYqLpKAAqISkpSUePHlVCQoLpKECVowAC5fD5558rISFBzz33nFq0aGE6DoBKaNGihZ599lnFx8fr0KFDpuMAVYpTwEA5PPTQQ/rwww%2BVm5uroKAg03EAVNJ3332n8PBw3X777XrzzTdNxwGqDBNAoIw2bNigBQsWaMKECZQ/oJoICgrShAkT9NZbb2nDhg2m4wBVhgkgUAYXLlxQZGSkatasqU2bNsnHh9dOQHVRWlqq66%2B/XufPn1dWVpZ8fX1NRwJcjl0MKIPZs2dr69atstvtlD%2BgmvHx8ZHdbtfWrVs1Z84c03GAKsEEELiEEydOKCwsTD169NDcuXNNxwHgIgMHDtQHH3yg/Px81a1b13QcwKUYZQCXMHbsWJ0%2BfVoTJkwwHQWAC02YMEGnTp3S2LFjTUcBXI4CCFxEXl6e7Ha7RowYoaZNm5qOA8CFmjVrphdeeEGpqanKz883HQdwKU4BAxdx3333adeuXdqzZ48CAgJMxwHgYsXFxWrTpo06dOigxYsXm44DuAwTQOB/WLFihZYuXarExETKH%2BAlAgIClJiYqCVLlmjlypWm4wAuwwQQ%2BB3nzp1Thw4d1LhxY61du1Y2m810JABVxLIs3X777SooKND27dtVs2ZN05EAp2MCCPyOqVOnKi8vT6mpqZQ/wMvYbDbZ7Xbt27dP06ZNMx0HcAkmgMCvFBYWKiwsTP379%2BfgD3ixwYMHa%2BHChcrPz1dISIjpOIBTMQEEfmXkyJGyLEtjxowxHQWAQWPHjpVlWRo1apTpKIDTUQCBn9mxY4emT5%2BuUaNGKTQ01HQcAAaFhoZq5MiRmjZtmnbu3Gk6DuBUnAIGfmBZlrp3767Dhw9rx44d8vPzMx0JgGEOh0Pt27dXs2bNtGrVKt4TjGqDCSDwg/T0dK1du1YpKSmUPwCSJD8/P6WkpGjNmjXKyMgwHQdwGiaAgKSzZ8%2Bqbdu2atOmjd5//33TcQC4mV69eik3N1e7d%2B9WrVq1TMcBKo0JICApJSVFX3zxhSZOnGg6CgA3NHHiRB06dEipqammowBOwQQQXu/w4cMKDw/XoEGDKIAA/qdhw4bp1VdfVV5eni6//HLTcYBKoQDC6z322GNatmyZ8vPzFRwcbDoOADd17NgxhYeH65577tGcOXNMxwEqhVPA8GqffPKJ3njjDY0bN47yB%2BCi6tevr7Fjx%2Br1119XVlaW6ThApTABhNcqLS3VjTfeqLNnzyonJ0e%2Bvr6mIwFwcxcuXFCnTp0UGBiojRs3clsYeCwmgPBaaWlpyszMlN1up/wBKBNfX1/Z7XZt3rxZaWlppuMAFcYEEF7p1KlTatWqlW688Ua9/fbbpuMA8DB/%2BtOftGnTJuXm5qpOnTqm4wDlxgQQXmnChAkqKipSYmKi6SgAPFBiYqKKiooUHx9vOgpQIRRAeJ2DBw8qKSlJw4cP11VXXWU6DgAP1Lx5c/3zn/9UUlKSDh48aDoOUG6cAobXeeCBB346dVO7dm3TcQB4qB/fSnLTTTdp4cKFpuMA5cIEEF5l3bp1eueddxQfH0/5A1ApderUUXx8vN5%2B%2B22tX7/edBygXJgAwmtw%2BwYAzsbtpOCpmADCa8ycOVM7duyQ3W6n/AFwCh8fH9ntdm3fvl2zZs0yHQcoMyaA8ArHjx9XWFgYP8IJgEs89thjWr58ufLy8vipQvAITADhFWJjY3X27FnFxcWZjgKgGoqLi9OZM2c0ZswY01GAMqEAotrbt2%2BfXn75Zb344otq0qSJ6TgAqqHLL79cL774oiZNmqTc3FzTcYBL4hQwqr1evXpp37592rNnj2rVqmU6DoBq6uzZs2rbtq3atGmj999/33Qc4KKYAKJaW7ZsmZYvX67k5GTKHwCXqlWrlpKSkn467gDujAkgqi2Hw6EOHTqoadOmWrVqFVf%2BAnA5y7LUvXt3HT58WDt37lTNmjVNRwJ%2BFxNAVFuTJ09Wfn6%2BUlNTKX8AqoTNZlNqaqry8/M1efJk03GA/4kJIKqlo0ePKjw8XAMGDOAgDKDKRUdHKy0tTfn5%2BQoNDTUdB/gNCiCqpcGDB2vhwoXKz89XSEiI6TgAvExhYaHCwsLUv39/TZs2zXQc4Dc4BYxqZ9u2bXr11Vc1evRoyh8AI0JCQhQTE6NXX31V27dvNx0H%2BA0mgKhWLMtSt27ddPToUW3fvp03YAMw5ty5c%2BrQoYMuu%2BwyrVmzhvciw60wAUS1snr1ap08eVIzZ86k/AEwqmbNmpo1a5ZOnDihNWvWmI4D/AITQAAAAC/DBBAAAMDLUAABAAC8DAUQAADAy1AAAQAAvAwFEAAAwMtQAOEx4uKkyEgpKEhq1Ejq21fKzTWdCgDKj%2BMZTKMAwmOsXy89%2BaS0ebP0wQfS%2BfPSXXdJp0%2BbTgYA5cPxDKZxH0B4rIKC7185r18v3Xqr6TQAUHEcz1DVmADCY5048f2vDRqYzQEAlcXxDFWNCSA8kmVJffpIx45JGzaYTgMAFcfxDCbUMB0AqIinnpJ27JA%2B%2Bsh0EgCoHI5nMIECCI/z9NPS4sXShx9KzZqZTgMAFcfxDKZQAOExLOv7g%2BWiRdK6dVLz5qYTAUDFcDyDaRRAeIwnn5TS0qSMjO/vnfX1199/vF49KSDAbDYAKA%2BOZzCNi0DgMWy23//47NnS449XaRQAqBSOZzCNAggAAOBluA8gAACAl6EAAgAAeBkKIAAAgJehAAIAAHgZCiAAAICXoQACAAB4GQogAACAl6EAAgAAeBkKIAAAgJehAMKtFBcXq1evXnruuedMRwEA45599lndc889Ki4uNh0F1QwFEG4lISFBq1evVnR0tOkoAGBcdHS0PvjgAyUmJpqOgmqGAgi38cUXXyg%2BPl7Dhg3T1VdfbToOABjXsmVLDRs2TBMmTNAXX3xhOg6qEZtlWZbpEIAkPfLII1q7dq3y8vIUFBRkOg4AuIWTJ08qPDxc3bt31/z5803HQTXBBBBu4aOPPtKbb76puLg4yh8A/EzdunUVFxentLQ0ffzxx6bjoJpgAgjjSktLFRkZKR8fH2VmZsrHh9clAPBzpaWl6tKliyTpk08%2B4TiJSuNfEIybM2eOtmzZIrvdzkENAH6Hj4%2BP7Ha7cnJy9Prrr5uOg2qACSCMOnnypMLCwnTHHXfw3hYAuIRHHnlEa9asUV5enurWrWs6DjwY4xYYNW7cOJ06dUrx8fGmowCA24uPj9fJkyc1fvx401Hg4SiAMCY/P18pKSn697//rWbNmpmOAwBu7w9/%2BIP%2B/e9/KyUlRfv37zcdBx6MU8Awpk%2BfPtq2bZv27dungIAA03EAwCOcOXNGbdq0UceOHZWenm46DjwUE0AY8d///leLFy9WYmIi5Q8AyiEwMFAJCQnKyMjQBx98YDoOPBQTQFS58%2BfP69prr1XDhg21fv162Ww205EAwKNYlqVbb71Vx44d07Zt21SjRg3TkeBhmACiyk2bNk179%2B6V3W6n/AFABdhsNtntdu3Zs0fTp083HQceiAkgqlRRUZHCwsL0pz/9STNmzDAdBwA82t///ne9%2B%2B67ys/PV8OGDU3HgQdhAogqNWrUKF24cEFjx441HQUAPN7YsWN14cIFxcTEmI4CD0MBRJXZuXOnpk6dqpEjR6pRo0am4wCAx2vcuLH%2B85//aOrUqdq1a5fpOPAgnAJGlbAsS3feeacOHTqkXbt2yc/Pz3QkAKgWHA6HrrnmGl155ZX673//y3urUSZMAFElFi9erNWrVyslJYXyBwBO5Ofnp4kTJ2rVqlVasmSJ6TjwEEwA4XIlJSVq27atwsPDtWzZMl6dAoCTWZalu%2B%2B%2BW/v379fu3bvl7%2B9vOhLcHBNAuFxqaqoOHTqkiRMnUv4AwAVsNptSUlJ08OBB2e1203HgAZgAwqWOHDmi8PBw/e1vf1NqaqrpOABQrT3zzDN67bXXlJ%2Bfr8suu8x0HLgxCiBc6i9/%2BYuWLFmi/Px81a9f33QcAKjWjh07prCwMPXu3Vuvvfaa6ThwY5wChstkZWVpzpw5Gjt2LOUPAKpA/fr1NWbMGM2ZM0fZ2dmm48CNMQGES1iWpZtuukmnT5/Wli1b5OvrazoSAHiF8%2BfPq1OnTgoKCtJHH33Ee6/xu5gAwiXefPNNbdq0SampqZQ/AKhCNWrUUGpqqjZu3Ki33nrLdBy4KSaAcLrTp0%2BrVatW6tq1q959913TcQDAK91///3KysrSvn37VLt2bdNx4GaYAMLp4uPjVVhYqMTERNNRAMBrJSUl6ejRo0pISDAdBW6IAgin%2Bvzzz5WYmKjnnntOLVq0MB0HALxWixYt9NxzzykhIUGHDh0yHQduhlPAcKr%2B/ftrw4YNysvLU506dUzHAQCv9t1336lVq1a69dZbeT8gfoEJIJzmww8/1MKFCxUfH0/5AwA3EBQUpAkTJmjBggXasGGD6ThwI0wA4RQXLlxQ586d5efnp02bNsnHh9cWAOAOSktLdcMNN%2BjcuXPKysrizgyQxAQQTvLaa69p27ZtstvtlD8AcCM%2BPj5KTU3V1q1bNXv2bNNx4CaYAKLSTpw4obCwMPXs2VNvvPGG6TgAgN/x5z//Wf/973%2BVl5enevXqmY4DwxjVoNLGjBmjM2fOKC4uznQUAMD/MGHCBJ06dUpjx441HQVugAKISsnNzZXdbteIESPUtGlT03EAAP9D06ZNNWLECNntduXl5ZmOA8M4BYxKuffee7V7927t3btXtWrVMh0HAHARxcXFatOmjdq3b68lS5aYjgODmACiwlasWKH3339fSUlJlD8A8AABAQFKSkrS0qVLtXLlStNxYBATQFTIuXPn1KFDB1122WVas2aNbDab6UgAgDKwLEvdunXT0aNHtX37dtWsWdN0JBjABBAVMmXKFOXl5Sk1NZXyBwAexGazKTU1Vbm5uZo6darpODCECSDKraCgQGFhYXr44Yc5eACAhxoyZIgWLFig/Px8hYSEmI6DKsYEEOU2cuRISVJsbKzhJACAihozZowsy/rpmA7vQgFEuezYsUMzZsxQTEyMQkNDTccBAFRQaGioRo0apenTp2vHjh2m46CKcQoYZWZZlrp3764jR45ox44dvHEYADycw%2BFQhw4ddPnll2v16tW8p9uLMAFEmS1atEhr165VSkoK5Q8AqgE/Pz%2BlpKRo7dq1Sk9PNx0HVYgJIMrk7NmzatOmjdq1a6elS5eajgMAcKJ77rlHe/fu1Z49e7ivq5dgAogymThxor788kslJyebjgIAcLKJEyfqiy%2B%2BUEpKiukoqCJMAHFJhw8fVnh4uAYPHkwBBIBq6tlnn9WMGTOUl5enyy%2B/3HQcuBgFEJf02GOPafny5crLy1NwcLDpOAAAFzh%2B/LjCwsJ0zz33aM6cOabjwMU4BYyLyszM1BtvvKFx48ZR/gCgGgsODta4ceP0%2Buuv65NPPjEdBy7GBBD/U2lpqW688UaVlJQoOztbvr6%2BpiMBAFzowoULioiIUK1atbRx40b5%2BDAnqq74m8X/NH/%2BfGVmZsput1P%2BAMAL%2BPr6ym63KzMzU2lpaabjwIWYAFZzlmWpoKBAZ86ckcPhkJ%2BfnwIDAxUaGnrRG36eOnVK4eHhuvnmm7Vw4cIqTAwAMO2BBx7Qxo0blZubqzp16lz0cyu6z8AsCmA1U1hYqDVr1ignJ0fZ2VnasmWLjh8/8ZvPCw6up06dOqlz50hFREQoKirqFz8M/KWXXlJycrL27t2rq666qgq/AwCAaQcPHlTr1q01fPhwjRkz5hd/5qx9BmZRAKsBy7K0efNmTZkyRQsXLpTD4VCzxo3UuVVLdWrdUu2aX6mgwAD51awpx7lz%2Bu5MsXYf%2BFxb9u1Xdu5%2BffnNUfn5%2Bal///6Kjo5Wo0aN1LZtW/3rX/9SbGys6W8PAGDAf/7zHyUmJmrfvn268sornbrPdO3alemgYRRAD5eRkaGYUaO0bft2NW/aREP69tLAnlFqEtKwzI9xpLBI81as0bT0ZTrw1RE1qF9fstl06NAh1a5d24XpAQDu6vTp02rVqpWuvPJKnTl92qn7TMfrrlPM6NHq3bu3C78DXAwF0EMVFRVp6NChSktL0x2RHTXs4fvVo2tEpa7YKi0t1crMHCXNf0drc7brkUce0aRJk9SwYdkXOQCgeigqKlLv%2B%2B7Txk2b1D2yo5514j6T8uZ7WpW1lX3GIAqgB0pPT9eQwYPlOFss%2B7AhGtAjyqmjdMuyNG/Faj2TMl3%2BAQGaNn26%2Bvbt67THBwC4N/aZ6o/bwHgQy7I0fvx49evXT13CW2jX/Gka2LO7099HYbPZ9Oe779DutGnqEt5C/fr1U1xcnHitAADVG/uM9/CNiYmJMftTGvUAAAqFSURBVB0Cl2ZZlkaMGKHY2FiNemKgpv7raQXVDnTpcwYFBuqhO2%2BTJI1MSpXD4VBUlHNfBQIA3AP7jHehAHqIuLg4xcbGKvHpv%2BuFxx6qssVhs9l0e6drVScwQCOT7fL399ctt9xSJc8NAKg67DPehQLoAdLT0zVo0CCNemKgXnjsISMZbmzfVpYsjUxKVceOHdW6dWsjOQAAzsc%2B4324CMTNFRUVqV3btuoS3kLpCaOMjsUty1Kf4THKyj%2Bg3Xv2cNUWAFQD/6%2B9%2B4%2BqsrDjOP65gPcemCnKvTZCTdTSSCu5YLl11sCdpNqwds6kAttZs3ScAt3yOM0Q2hFXWUo/ENyptIBlqybW/LGRnLO1Mxlc0M1aLovaUk5yb4F6JK7As3%2BKrdUpLnjv88Dzfv0H3Od5vufAw%2Bdzn1%2BXnLEnbgKxuIKCAgU/7lLFqrtNvybC4XCoYlWBuru6VFhYaOosAIBzg5yxJwqghdXW1qqmpkZlK5aF9MDNcLrAk6CyFUtVXV2tXbt2mT0OAGAIyBn74hSwRRmGodQ5c%2BR2RmlfWanp78r%2Bl2EYWlC4RoGzhnzNzZaaDQAwMOSMvXEE0KIOHDigg4cOacUt37fcH77D4dDym29Sy8GDamhoMHscAMAgkDP2RgG0qPLyciUnJWrBlV6zR/lCC670KjkpUeXl5WaPAgAYBHLG3iiAFuT3%2B/X8889r2Y3XD%2BkzF8MpOjpaS2%2B8Xjt27JDf7zd7HABACMgZWPO3bnP79%2B9XMBhUXlZm2LcV6DypSdl5ipqXpY5Tp0NadnFWpoLBoOrr68M0HQAgHCKVM9t%2B93tdnrdMsdd8T4k33KK7Nj4R0vLkTPhQAC3I5/Np4vkTInJH1pLSTbpsevKglk10Jyhpgkc%2Bn%2B8cTwUACKdI5Mwjv35Rayu2a9XiHB2urlTdYxtCPt1MzoRPjNkD4POaGhuVNmP6gF%2Bfkb9Ss6YlKzoqSs/sqZMzJkb333mbchdk6u6Hn9AL9a9pwrh4PfazfF03L71/uS0vvaKOU6d13%2B252vOXxkHNmjZzupqaBrcsAMAc4c6Zj06e0n2Vz2jXQ8Wanz6nfz2XTp0S8qzkTHhwBNBiDMNQc0uzUmcOfMeUpGd218kdP0YNT5bprh9kK/%2Bhx7Xo3vWaNztFvm2P69orvbqt5CGd%2BfhjSdIbre/pF09Va3vRSkVFDf7ur9QZ09XsaxZPEwKA4SESOfOHv7aoz%2BjTsfaAUm6%2BQ5Oy85Rz73r9%2B4P2kOclZ8KDAmgx7e3t6ujo1KXJF4a03OUXJWvtj27VRZOStPq2HMW6nHKPHaM7Fl6niyYlqej2WxXoPKm/HW1VdzCoW4t%2BqQfvWqLJX58wpHlnTZ2ijzo61N4e%2Bk4NAIi8SOTMO8fb1NdnaMP257Rp%2BVL9pvRefXjylK4tWK3g2bMhbZecCQ8KoMWcOXNGknReXGxIy82e9t/r%2BKKjo5UwdoxmTZvS/73zx4%2BTJJ34qEOrtzytS6ZMVl7W/CHPO/qTObu6uoa8LgBA%2BEUiZ/r6DJ3t6VHZT3%2BiBVel6apZl6jm/p/rrfePq953KKTtkjPhwTWAFhMMBiVJzlGjQlpuVMxnf5WO//vepw/57OszVO87pL%2B//a5eqP%2BTJOnTo%2Bqe6xZpzQ9vUckdiwe8Xecn2%2Bju7g5pXgCAOSKRM4nu8ZKklOTJ/T/3jIuXe%2BwY/SvE08DkTHhQAC3G6XRKUsiHyEPxQuladXUH%2B79u/Mc/9eP1j%2BiPWzZqWtIFIa0r2NMjSXK5XOd0RgBAeEQiZ755WYok6ch772viBI8k6cPOU/J3ntSFIV56RM6EBwXQYuLi4iRJp86E71D3tImfLXn%2Bzk5J0iVTJiv%2BvNEhrev0J3PGxoZ2KgEAYI5I5MzFkydq4bfmafnmClWuKtSYr8VpzZanNfPCicrwXh7SusiZ8OAaQIvxeDyKjx%2Br11vfM3uUATn8zrsaFx8vj8dj9igAgAGIVM5sL7pHc1Nm6Lv3FOnb%2BSs1KiZaezat/9yp5K9CzoSHw%2BC%2BasuZn5mpMb3deumBIrNH%2BUo3rSrR6VGxqqt71exRAAADRM6AI4AWlJaerqYjR80eY0Ca3jyqtLT0r34hAMAyyBlQAC3I6/Xq/Q9OqM0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null ]

[ "# 10.26: Hypothesis Test for a Population Mean (5 of 5)\n\n$$\\newcommand{\\vecs}{\\overset { \\rightharpoonup} {\\mathbf{#1}} }$$ $$\\newcommand{\\vecd}{\\overset{-\\!-\\!\\rightharpoonup}{\\vphantom{a}\\smash {#1}}}$$$$\\newcommand{\\id}{\\mathrm{id}}$$ $$\\newcommand{\\Span}{\\mathrm{span}}$$ $$\\newcommand{\\kernel}{\\mathrm{null}\\,}$$ $$\\newcommand{\\range}{\\mathrm{range}\\,}$$ $$\\newcommand{\\RealPart}{\\mathrm{Re}}$$ $$\\newcommand{\\ImaginaryPart}{\\mathrm{Im}}$$ $$\\newcommand{\\Argument}{\\mathrm{Arg}}$$ $$\\newcommand{\\norm}{\\| #1 \\|}$$ $$\\newcommand{\\inner}{\\langle #1, #2 \\rangle}$$ $$\\newcommand{\\Span}{\\mathrm{span}}$$ $$\\newcommand{\\id}{\\mathrm{id}}$$ $$\\newcommand{\\Span}{\\mathrm{span}}$$ $$\\newcommand{\\kernel}{\\mathrm{null}\\,}$$ $$\\newcommand{\\range}{\\mathrm{range}\\,}$$ $$\\newcommand{\\RealPart}{\\mathrm{Re}}$$ $$\\newcommand{\\ImaginaryPart}{\\mathrm{Im}}$$ $$\\newcommand{\\Argument}{\\mathrm{Arg}}$$ $$\\newcommand{\\norm}{\\| #1 \\|}$$ $$\\newcommand{\\inner}{\\langle #1, #2 \\rangle}$$ $$\\newcommand{\\Span}{\\mathrm{span}}$$$$\\newcommand{\\AA}{\\unicode[.8,0]{x212B}}$$\n\nLearning Objectives\n\n• Interpret the P-value as a conditional probability.\n\nWe finish our discussion of the hypothesis test for a population mean with a review of the meaning of the P-value, along with a review of type I and type II errors.\n\n## Review of the Meaning of the P-value\n\nAt this point, we assume you know how to use a P-value to make a decision in a hypothesis test. The logic is always the same. If we pick a level of significance (α), then we compare the P-value to α.\n\n• If the P-value ≤ α, reject the null hypothesis. The data supports the alternative hypothesis.\n• If the P-value > α, do not reject the null hypothesis. The data is not strong enough to support the alternative hypothesis.\n\nIn fact, we find that we treat these as “rules” and apply them without thinking about what the P-value means. So let’s pause here and review the meaning of the P-value, since it is the connection between probability and decision-making in inference.\n\n## Birth Weights in a Town\n\nLet’s return to the familiar context of birth weights for babies in a town. Suppose that babies in the town had a mean birth weight of 3,500 grams in 2010. This year, a random sample of 50 babies has a mean weight of about 3,400 grams with a standard deviation of about 500 grams. Here is the distribution of birth weights in the sample.", null, "Obviously, this sample weighs less on average than the population of babies in the town in 2010. A decrease in the town’s mean birth weight could indicate a decline in overall health of the town. But does this sample give strong evidence that the town’s mean birth weight is less than 3,500 grams this year?\n\nWe now know how to answer this question with a hypothesis test. Let’s use a significance level of 5%.\n\nLet μ = mean birth weight in the town this year. The null hypothesis says there is “no change from 2010.”\n\n• H0: μ < 3,500\n• Ha: μ = 3,500\n\nSince the sample is large, we can conduct the T-test (without worrying about the shape of the distribution of birth weights for individual babies.)", null, "$T\\text{}=\\text{}\\frac{\\mathrm{3,400}-\\mathrm{3,500}}{\\frac{500}{\\sqrt{50}}}\\text{}\\approx \\text{}-1.41$\n\nStatistical software tells us the P-value is 0.082 = 8.2%. Since the P-value is greater than 0.05, we fail to reject the null hypothesis.\n\nOur conclusion: This sample does not suggest that the mean birth weight this year is less than 3,500 grams (P-value = 0.082). The sample from this year has a mean of 3,400 grams, which is 100 grams lower than the mean in 2010. But this difference is not statistically significant. It can be explained by the chance fluctuation we expect to see in random sampling.\n\n## What Does the P-Value of 0.082 Tell Us?\n\nA simulation can help us understand the P-value. In a simulation, we assume that the population mean is 3,500 grams. This is the null hypothesis. We assume the null hypothesis is true and select 1,000 random samples from a population with a mean of 3,500 grams. The mean of the sampling distribution is at 3,500 (as predicted by the null hypothesis.) We see this in the simulated sampling distribution.", null, "In the simulation, we can see that about 8.6% of the samples have a mean less than 3,400. Since probability is the relative frequency of an event in the long run, we say there is an 8.6% chance that a random sample of 500 babies has a mean less than 3,400 if the population mean is 3,500. We can see that the corresponding area to the left of T = −1.41 in the T-model (with df = 49) also gives us a good estimate of the probability. This area is the P-value, about 8.2%.\n\nIf we generalize this statement, we say the P-value is the probability that random samples have results more extreme than the data if the null hypothesis is true. (By more extreme, we mean further from value of the parameter, in the direction of the alternative hypothesis.) We can also describe the P-value in terms of T-scores. The P-value is the probability that the test statistic from a random sample has a value more extreme than that associated with the data if the null hypothesis is true.\n\n## What Does a P-Value Mean?\n\nDo women who smoke run the risk of shorter pregnancy and premature birth? The mean pregnancy length is 266 days. We test the following hypotheses.\n\n• H0: μ = 266\n• Ha: μ < 266\n\nSuppose a random sample of 40 women who smoke during their pregnancy have a mean pregnancy length of 260 days with a standard deviation of 21 days. The P-value is 0.04.\n\nWhat probability does the P-value of 0.04 describe? Label each of the following interpretations as valid or invalid.\n\nhttps://assessments.lumenlearning.co...sessments/3654\n\nhttps://assessments.lumenlearning.co...sessments/3655\n\nhttps://assessments.lumenlearning.co...sessments/3656\n\n## Review of Type I and Type II Errors\n\nWe know that statistical inference is based on probability, so there is always some chance of making a wrong decision. Recall that there are two types of wrong decisions that can be made in hypothesis testing. When we reject a null hypothesis that is true, we commit a type I error. When we fail to reject a null hypothesis that is false, we commit a type II error.\n\nThe following table summarizes the logic behind type I and type II errors.", null, "It is possible to have some influence over the likelihoods of committing these errors. But decreasing the chance of a type I error increases the chance of a type II error. We have to decide which error is more serious for a given situation. Sometimes a type I error is more serious. Other times a type II error is more serious. Sometimes neither is serious.\n\nRecall that if the null hypothesis is true, the probability of committing a type I error is α. Why is this? Well, when we choose a level of significance (α), we are choosing a benchmark for rejecting the null hypothesis. If the null hypothesis is true, then the probability that we will reject a true null hypothesis is α. So the smaller α is, the smaller the probability of a type I error.\n\nIt is more complicated to calculate the probability of a type II error. The best way to reduce the probability of a type II error is to increase the sample size. But once the sample size is set, larger values of α will decrease the probability of a type II error (while increasing the probability of a type I error).\n\nGeneral Guidelines for Choosing a Level of Significance\n\n• If the consequences of a type I error are more serious, choose a small level of significance (α).\n• If the consequences of a type II error are more serious, choose a larger level of significance (α). But remember that the level of significance is the probability of committing a type I error.\n• In general, we pick the largest level of significance that we can tolerate as the chance of a type I error.\n\n### Try It\n\nLet’s return to the investigation of the impact of smoking on pregnancy length.\n\nRecap of the hypothesis test: The mean human pregnancy length is 266 days. We test the following hypotheses.\n\n• H0: μ = 266\n• Ha: μ < 266\n\nhttps://assessments.lumenlearning.co...sessments/3778\n\nhttps://assessments.lumenlearning.co...sessments/3779\n\nhttps://assessments.lumenlearning.co...sessments/3780\n\n## Let’s Summarize\n\nIn this “Hypothesis Test for a Population Mean,” we looked at the four steps of a hypothesis test as they relate to a claim about a population mean.\n\n### Step 1: Determine the hypotheses.\n\n• The hypotheses are claims about the population mean, µ.\n• The null hypothesis is a hypothesis that the mean equals a specific value, µ0.\n• The alternative hypothesis is the competing claim that µ is less than, greater than, or not equal to the", null, "${\\mathrm{μ}}_{0}$ .\n• When", null, "${H}_{a}$ is", null, "$μ$ <", null, "${μ}_{0}$ or", null, "$μ$ >", null, "${μ}_{0}$ , the test is a one-tailed test.\n• When", null, "${H}_{a}$ is", null, "$μ$", null, "${μ}_{0}$ , the test is a two-tailed test.\n\n### Step 2: Collect the data.\n\nSince the hypothesis test is based on probability, random selection or assignment is essential in data production. Additionally, we need to check whether the t-model is a good fit for the sampling distribution of sample means. To use the t-model, the variable must be normally distributed in the population or the sample size must be more than 30. In practice, it is often impossible to verify that the variable is normally distributed in the population. If this is the case and the sample size is not more than 30, researchers often use the t-model if the sample is not strongly skewed and does not have outliers.\n\n### Step 3: Assess the evidence.\n\n• If a t-model is appropriate, determine the t-test statistic for the data’s sample mean.", null, "$\\frac{\\mathrm{sample}\\text{}\\mathrm{mean}-\\mathrm{population}\\text{}\\mathrm{mean}}{\\mathrm{estimated}\\text{}\\mathrm{standard}\\text{}\\mathrm{error}}\\text{}=\\text{}\\frac{\\stackrel{¯}{x}-μ}{s/\\sqrt{n}}$\n\n• Use the test statistic, together with the alternative hypothesis, to determine the P-value.\n• The P-value is the probability of finding a random sample with a mean at least as extreme as our sample mean, assuming that the null hypothesis is true.\n• As in all hypothesis tests, if the alternative hypothesis is greater than, the P-value is the area to the right of the test statistic. If the alternative hypothesis is less than, the P-value is the area to the left of the test statistic. If the alternative hypothesis is not equal to, the P-value is equal to double the tail area beyond the test statistic.\n\n### Step 4: Give the conclusion.\n\nThe logic of the hypothesis test is always the same. To state a conclusion about H0, we compare the P-value to the significance level, α.\n\n• If P ≤ α, we reject H0. We conclude there is significant evidence in favor of Ha.\n• If P > α, we fail to reject H0. We conclude the sample does not provide significant evidence in favor of Ha.\n• We write the conclusion in the context of the research question. Our conclusion is usually a statement about the alternative hypothesis (we accept Ha or fail to acceptHa) and should include the P-value.\n\n## Other Hypothesis Testing Notes\n\n• Remember that the P-value is the probability of seeing a sample mean at least as extreme as the one from the data if the null hypothesis is true. The probability is about the random sample; it is not a “chance” statement about the null or alternative hypothesis.\n• Hypothesis tests are based on probability, so there is always a chance that the data has led us to make an error.\n• If our test results in rejecting a null hypothesis that is actually true, then it is called a type I error.\n• If our test results in failing to reject a null hypothesis that is actually false, then it is called a type II error.\n• If rejecting a null hypothesis would be very expensive, controversial, or dangerous, then we really want to avoid a type I error. In this case, we would set a strict significance level (a small value of α, such as 0.01).\n• Finally, remember the phrase “garbage in, garbage out.” If the data collection methods are poor, then the results of a hypothesis test are meaningless." ]

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[ "", null, "## Solving 3 Variable Systems of Equations by Elimination\n\nWarning: Missing argument 2 for ted_filter_oembed_amp_iframe() in /home/customer/www/torontotutorteam.ca/public_html/wp-content/plugins/jetpack/modules/shortcodes/ted.php on line 108\n\nIn this tutorial we’ll explain solving 3 variable systems of equations by elimination. In other tutorials we’ve covered solving 2 variable systems of equations by elimination, and solving systems of equations by substitution.\n\nFor example, what is the value of x, y and z for 3x + 2y – z = 8, x – 3y + 2z = 7 and x – 2y + z = 4\n\nLet’s number our three equations:\n\n1) 3x + 2y – z = 8\n2) x – 3y + 2z = 7\n3) x – 2y + z = 4\n\nThe first step is to get rid of one of the variables by adding or subtracting two equations.\nSince x has the same coefficient in equations 2 and 3, let’s subtract them:\n\n(x – 3y + 2z = 7) – (x – 2y + z = 4)\nx – 3y + 2z – x + 2y – z = 7 – 4\n-3y + 2y + 2z – z = 3\n-y + z = 3\n\nWe’ll call this equation 4:\n\n4) -y + z = 3\n\nThis equation does not involve variable x, but we still can’t solve it because it has 2 unknowns, so we need 2 unique equations. The way to do this is to find another equation that does not involve variable x through a different combination of equations. We already used equations 2 and 3 to find equation 4, now we’ll use equations 1 and 3.\n\nWe can multiply both sides of equation 3 by 3 to get the same coefficients for x, resulting in 3x – 6y + 3z = 12\nNow we subtract this from equation 1:\n\n(3x + 2y – z = 8) – (3x – 6y + 3z = 12)\n3x + 2y – z – 3x + 6y – 3z = 8 – 12\n2y + 6y – z – 3z = -4\n8y – 4z = -4\nWe can actually simplify this by dividing everything by 4 to get\n2y – z = -1\n\nWe’ll call this equation 5:\n\n5) 2y – z = -1\n\nNow we have 2 equations that only involve y and z, so we can use the elimination method to solve. Since equation 4 has a +z term and equation 5 has a -z term, let’s add them together:\n\n(-y + z = 3) + (2y – z = -1)\n-y + z + 2y – z = 3 – 1\n-y + 2y = 2\ny = 2\n\nNow we can plug the value of y into equation 4 or 5 to solve for z. Let’s plug it into equation 4:\n\n-y + z = 3\n-2 + z = 3\nz = 5\n\nIf you were to plug y into equation 5 you also get z = 5. Now that we have the values of y and z, we can simply plug them into equation 1 or 2 to find x. Let’s use equation 2:\n\nx – 3y + 2z = 7\nx – 3(2) + 2(5) = 7\nx – 6 + 10 = 7\nx + 4 = 7\nx = 3\n\nIf you want to check that this is correct we can plug our answers into the original equations:\n\n1) 3x + 2y – z = 8\n2) x – 3y + 2z = 7\n3) x – 2y + z = 4\n\n1) 3(3) + 2(2) – (5) = 8\n2) (3) – 3(2) + 2(5) = 7\n3) (3) – 2(2) + (5) = 4\n\nHere’s a practice problem to try on your own:\nWhat is the value of x, y and z for 5x – 3y – 2z = -42, 2x + 3y – 7z = -37, and 4x – y + z = -3\nThe answer is x = -1, y = 7, z = 8\n\nAnd that’s it for solving 3 variable systems of equations using elimination!\n\nIf you prefer to watch a video version of this tutorial:\n\nYou can also check out more great math tutorials.\n\n#### Recent Comments", null, "### Contact Info\n\nReady to start learning? Give us a call or send us an email to book a tutoring session today.\n\n+1 855-775-4473\[email protected]\n\nUse the message icon below to live chat with a member of the Team right now!\n\nCopyright 2016 © All Rights Reserved" ]

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[ "The ” Nigam Scholar Workbooks Mathematics Standard Std – I “ Workbooks are used for solving extra problems and concepts which students have already studied from a textbook. The term workbook is also used to describe other compilations of questions that require the reader to complete scratch-work when dealing with higher-level mathematics. The mathematics in question is basic, arithmetic, computation, algebra, geometry, calculus, combinatorics, probability, and logic, all initially at an introductory level.\n\nProduct Sample Pages" ]

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[ "Note: _a1w_ denotes that first order adjoints are computed in working precision; this has the corresponding argument type nagad_a1w_w_rtype. Further implementations, for example for higher order differentiation or using the tangent linear approach, may become available at later marks of the NAG AD Library. The method of codifying AD implementations in the routine name and corresponding argument types is described in the NAG AD Library Introduction.\n\n## 1Purpose\n\ne01aa_a1w_f is the adjoint version of the primal routine e01aaf.\n\n## 2Specification\n\nFortran Interface\n Subroutine e01aa_a1w_f ( ad_handle, a, b, c, n1, n2, n, x, ifail)\n Integer, Intent (In) :: n1, n2, n Integer, Intent (Inout) :: ifail Type (nagad_a1w_w_rtype), Intent (In) :: x Type (nagad_a1w_w_rtype), Intent (Inout) :: a(n+1), b(n+1) Type (nagad_a1w_w_rtype), Intent (Out) :: c(n*(n+1)/2) Type (c_ptr), Intent (In) :: ad_handle\nThe routine may be called by the names e01aa_a1w_f or nagf_interp_dim1_aitken_a1w.\n\n## 3Description\n\ne01aa_a1w_f is the adjoint version of the primal routine e01aaf.\ne01aaf interpolates a function of one variable at a given point $x$ from a table of function values ${y}_{i}$ evaluated at equidistant or non-equidistant points ${x}_{i}$, for $\\mathit{i}=1,2,\\dots ,n+1$, using Aitken's technique of successive linear interpolations. For further information see Section 3 in the documentation for e01aaf.\nNone.\n\n## 5Arguments\n\nIn addition to the arguments present in the interface of the primal routine, e01aa_a1w_f includes some arguments specific to AD.\nA brief summary of the AD specific arguments is given below. For the remainder, links are provided to the corresponding argument from the primal routine. A tooltip popup for all arguments can be found by hovering over the argument name in Section 2 and in this section.\n1: ad_handle – Type (c_ptr) Input\nOn entry: a handle to the AD configuration data object, as created by x10aa_a1w_f.\n2: a($\\mathbf{n}+1$) – Type (nagad_a1w_w_rtype) array Input/Output\n3: b($\\mathbf{n}+1$) – Type (nagad_a1w_w_rtype) array Input/Output\n4: c($\\mathbf{n}×\\left(\\mathbf{n}+1\\right)/2$) – Type (nagad_a1w_w_rtype) array Output\n5: n1 – Integer Input\n6: n2 – Integer Input\n7: n – Integer Input\n8: Input\n9: ifail – Integer Input/Output\nOn entry: must be set to $\\mathrm{0}$, $-\\mathrm{1}\\text{ or }\\mathrm{1}$. On exit: any errors are indicated as described in Section 6.\n\n## 6Error Indicators and Warnings\n\nThere are no specific error codes from e01aaf, however e01aa_a1w_f can return:\n$\\mathbf{ifail}=-89$\nSee Section 4.5.2 in the NAG AD Library Introduction for further information.\n$\\mathbf{ifail}=-899$\nDynamic memory allocation failed for AD.\nSee Section 4.5.1 in the NAG AD Library Introduction for further information.\n\nNot applicable.\n\n## 8Parallelism and Performance\n\ne01aa_a1w_f is not threaded in any implementation." ]

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[ "# SQL CEILING Function (Transact SQL)\n\nBy: Kris Wenzel   |   Updated: March 6, 2022\nWorks With:\n\nThe SQL CEILING function returns the smallest integer value greater than or equal to the input value.\n\n## Description\n\nThe CEILING function evaluates the right side of the decimal value and returns an integer value that is least greater than or equal to the input value. CEILING is another SQL function for approximating numerical values like FLOOR and ROUND.\n\nThe CEILING is useful when you have a float or decimal value and you need to find the next lowest or highest integer.\n\nThe CEILING is the next highest integer value; whereas, floor is the next lowest.\n\nCEILING and FLOOR have a practical use when you’re working with discrete quantities and averages.\n\nFor instance, let’s assume the sales manager is having a sales convention for all the sales people.  They want to rent cars to be able to get around town.  Assuming each car holds 4 people how many cars do he need to rent for the business trip?\n\nTo figure this out, we can take the number of sales people and divide by four\n\n```SELECT COUNT(*) / 4.0 as NumberCars\nFROM   Sales.SalesPerson```\n\nThis result is 4.25 cars.  As you know you cannot rent nor drive a quarter of a car!\n\nTo get around this we need to round up to the nearest whole car.  We can use CEILING to do so.  Here is the query to use\n\n```SELECT CEILING(COUNT(*) / 4.0) as NumberCars\nFROM   Sales.SalesPerson```\n\nAlso, did you notice I used 4.0 rather than 4 in the calculation?  This is to ensure the result is a float; otherwise, the result is returned as an integer, and would have been implicitly converted  to an integer value of 4.\n\n## SQL CEILING Usage Notes\n\nThe only argument is a numeric value to be approximated to the smallest greater value.\n\nThe data type of input expression is from either exact numeric or approximate numeric categories. The bit data type is invalid.\n\nThe return type is the same as the input value’s data type.\n\n## Syntax\n\nCEILING (numeric_value)\n\n## SQL CEILING Examples\n\nThe following examples explain the use of the CEILING function for positive, negative, and zero input values.\n\nselect ceiling(22.3) Example1, ceiling(-22.75) Example2, ceiling(0.0) Example3\n```/* Answer */\nselect ceiling(22.3) Example1, ceiling(-22.75) Example2, ceiling(0.0) Example3```\n\nIn each case an integer is returned.\n\n• The result for Example1 is intuitive. CEILING(22.3) is 23.\n• However, notice Example2 returns -22 and not -23. Keep in mind -22 is greater than -23, and ceiling return the next greatest integer.\n\nThe output data type is similar to the input data type, money.\n\nThe following example uses the SQL CEILING function on the AdvetureWorks2019 database column. It returns the smallest value greater than or equal to the standard cost of each product.\n\nselect CEILING(StandardCost) StandardCostCeiling from Production.ProductCostHistory\n```/* Answer */\nselect CEILING(StandardCost) StandardCostCeiling\nfrom Production.ProductCostHistory```" ]

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[ "• CBSE sample papers\n• CBSE Class 8 Sample Papers\n• CBSE Class 8 Maths Sample Papers\n• CBSE Sample Paper Class 8 Maths Set 3", null, "## CBSE Class 8 Maths Sample Papers Set 3", null, "## Access Other Sets of CBSE Class 8 Maths Sample Papers\n\nHere are the other sets of CBSE Class 8 Sample Papers Maths. Click on the link below to access them.\n\nWe hope this information on ‘CBSE Sample Paper Class 8 Maths Set 3’ helped students in their exam preparation. Keep learning and stay tuned for further updates on CBSE and competitive exams. Download BYJU’S App and subscribe to the YouTube Channel to access interactive Maths and Science videos.\n\nRequest OTP on Voice Call\n\nPost My Comment", null, "• Share Share\n\nRegister with byju's & watch live videos.", null, "Talk to our experts\n\n1800-120-456-456\n\n• CBSE Sample Papers for Class 8 Maths with Solution 2023-24\n• Sample Papers", null, "## CBSE Sample Questions Paper for Class 8 Maths with Solutions - Free PDF Download\n\nMathematics is not like any other theory subject, which means mere reading or having clarity of concept is not sufficient for one to excel in the exam. To gain command over Maths, a  student needs to practice regularly. To simplify the process of practicing and help the students grow, Vedantu provides you with a complete set of CBSE Class 8 Maths question papers with solutions.\n\nAlso, check CBSE Class 8 Sample Papers for other Subjects:\n\n## CBSE Sample Questions Paper for Class 8 Maths with Solutions", null, "## Sample Paper Class 8 Syllabus Overview for Maths\n\nFollowing are the 11 chapters in Maths Class 8 CBSE book -\n\nSquares and Squares Roots, Cubes and Cube Roots\n\nLinear Equations in one variable\n\nMensuration\n\nAlgebraic Expressions, Identities, and Factorization\n\nRatio, proportion, and percentage\n\nDirect and Inverse Proportion\n\nRepresenting 3D in 2D\n\nPlaying with Numbers\n\nData Handling\n\nExponents and Powers\n\n## Sample Paper Class 8 Maths\n\nWe have segregated the weightage given to each chapter. therefore signifying which chapters demand more practice..\n\nLinear Equations in One Variable - 12 marks\n\nRational Numbers - 12 marks\n\nPlaying with numbers - 6 marks\n\nSquare and square roots, cube and cube roots - 22 marks\n\nComparing quantity - 14 marks\n\n## Benefits of solving Class 8 Maths CBSE Papers -\n\nStrengths and Weaknesses - by practicing more questions, students can identify the topics they are strong at and can divert their time and energy to work upon areas they are uncomfortable with\n\nTime Management - a regular statement students make after the exam - “I knew the paper, but I was short of time.” To avoid such a situation, rigorous practice is needed.\n\nSelf-Improvement - practice papers with solutions provide an opportunity to correct your mistakes before you appear for the exam.\n\nTargeted Study - if students have mock papers to solve, they will automatically finish the syllabus on time.\n\nEffective Revision Tool - mock papers will help students revise the topics. And multiple revisions will also add to their retention capability.\n\nFight Fear and Anxiety - prepare the student for the exam-like condition. So, the actual fear of exams is no more.\n\nBetter Output - practice will help one recollect and reproduce the content effectively.\n\nBoost Your Score and Confidence - as a well-prepared mind has less probability of committing silly mistakes.\n\n## Why should you choose Vedantu for downloading the CBSE Solved Sample Papers for Class 8 Maths?\n\nStudents of Class 8 can easily score the top score in Maths by practicing Sample Paper by Vedantu. All the Sample Papers are prepared according to the guidelines given by the CBSE . Students can easily download the Sample Papers in PDF format. We also have our app where all the NCERT solutions and study materials can be downloaded for free. So wait no further, download the Vedantu app from the play store/app store now and get access to free live master classes as well. At Vedantu, all educational material can be accessed 24x7 by students. This is a great feature because it offers the kind of versatility to students that would allow them to study whenever they want.\n\n## Important Related Links for CBSE Class 8", null, "## FAQs on CBSE Sample Papers for Class 8 Maths with Solution 2023-24\n\n1. Why Sample Paper Class 8 Maths are Important for students?\n\nSample papers provide a lot of practice on the concepts of Class 8 Maths before the exams. So, solving these Sample Paper of Class 8 Maths will help students to brush up on their basic concepts and also make them confident about the subject before the exams.\n\n2. What are the important chapters covered in the CBSE Sample Papers for Class 8 Maths?\n\nThe important chapters along with the breakage of the marks of each chapter asked in Sample papers for Class 8 Maths is given below:\n\nLinear Equations in One Variable - 12 marks\n\nRational Numbers - 12  marks\n\nPlaying with numbers - 6 marks\n\n3. Where can I find the CBSE Class 8 Maths Sample Papers?\n\nStudents can download the free PDF of Sample Paper of Class 8 Maths available on Vedantu Platform. These sample papers are provided with solutions in a more descriptive way to make students understand the concepts easily.\n\n4. Why should you refer to the CBSE Sample Paper Class 8 Maths from Vedantu?\n\nOn Vedantu, students can find multiple sample sets of CBSE Class 8 Maths papers with solutions at one stop. Apart from your regular coursebook, these Sample Papers can add value to your existing knowledge. It proves to be a win-win situation, as it strengthens your core knowledge along with boosting your grades.\n\n5. What is the level of difficulty of the CBSE Sample Paper Class 8 Maths?\n\nThe question-level range from easy scoring to moderate and some hard to attempt questions. Like, say, the questions from the first part of the basic concept and the theories are asked in the simple form that mostly has a solution of one or two liners. However, at times such a simple-looking question could be very tricky.\n\nThe questions would not be difficult for someone who has sincerely practiced the Sample Paper, explained with all the concepts by Vedantu. Understanding and practicing is the only mantra!\n\n6. Why should a student work hard and solve CBSE Sample Paper Class 8 Maths? Isn’t this needed for only the board's exams?\n\nHard work is not confined to a particular exam. The teachers at Vedantu do not merely focus on improving your grades. Rather they aim to build a stronger foundation for each student of all the classes. Practicing through Sample Papers keeps your study schedule in discipline, prepares you for higher studies and competition, and builds a strong knowledge base. Thus, it is one of the best ways to learn and evaluate one’s learning.\n\n7. Isn't the Class 8 NCERT Maths Book sufficient to score good marks? Why do we need to solve CBSE Sample Paper Class 8 Maths?\n\nOf course, the NCERT Class 8 Maths book is sufficient for you to score good marks. However, at times we miss covering certain important topics from the NCERT, and when such topics appear on the paper we fail to answer them,\n\nThus, Vedantu has compiled the set of CBSE Class 8 Maths Sample Papers so that you don't have to miss any important topics.\n\n8. What should be the way to solve the CBSE Sample Paper Class 8 Maths?\n\nStudents first need to finish understanding the topics and concepts from the NCERT books. Secondly, you should finish solving the practice questions given after every chapter in the textbook. Having done this, you should take one set of sample question papers at a time and try to solve it like a real exam paper. Evaluate your answers from the solution provided by Vedantu.\n\nIdentify your weak areas and rework on clearing those topics. Reach out to our experts at Vedantu if needed. But focus on clarifying your doubts and strengthening your knowledge base.\n\n## CBSE Sample Papers class 8 Maths\n\n• Active page\n\nThe following Sample papers are prepared chapter wise. Each chapter consist of questions based on CBSE Syllabus . Our academic team uploaded quality questions with detail solutions of each and every question. CBSE Sample Papers class 8 Maths must be solve after reading the theory of recommended text book.", null, "## List of Sample Papers for Class 8 Maths\n\n• Maths Class 8 Sample Paper - 1\n• Maths Class 8 Sample Paper - 2\n• Maths Class 8 Sample Paper - 3 Objective Type Questions\n• Maths Class 8 Sample Paper - 4 Objective Type Questions\n• Maths Class - 8 Sample Paper - 5 Objective Type Questions\n\n## How CBSE Sample Papers Class 8 Maths can help you to build concepts\n\nMaths is subject of practice. The best way to practice class 8 maths is solving questions. To help you in this Academic team of Physics Wallah uploaded CBSE Sample Papers class 8 Maths.Each Sample paper consist of solved and unsolved questions with solutions.\n\nFor additional information related to the subject you can check the Maths Formula and Maths Doubts section.\n\n## Why one must solve CBSE Sample Papers Class 8 Maths\n\nTo become expert in any Maths students must focus on solving questions. To do effective practice you need quality of questions in Maths and solve all questions by yourself for more effectiveness try to solve each sheet in the given period of time.\n\n## How to use Class 8 Maths CBSE Sample Papers\n\nCBSE Sample Papers class 8 Maths must be use after proper revision of your lesion. Start form first chapter read it thoroughly from your text book try to revise your notes. Once you are confident that you know and understand the chapter start solving the CBSE Sample Papers class 8 Maths . Do solve questions from NCERT exercise with the help of  NCERT Solutions for Class 8 Maths  prepared by Physics Wallah.\n\nPhysics Wallah uploaded lot of resource for class 8 Maths apart from worksheet. Like for class 8 Maths you can download Important questions for class 8 Maths, reference book solution for class 8 Maths , Maths formulas sheet, online test for class 8 Maths, detail and in-depth theory for JEE and NEET aspirants for class 8 Maths.\n\nStudents can also access the RS Aggarwal Class 8 Maths Solutions  from here.\n\nQ1. Why CBSE class 8 maths sample papers are important?\n\nAns.  The mathematical class 8 CBSE sample papers are a perfect test for students to verify the exam preparation. They could see where they are making mistakes and so they can work on those areas. This will improve their overall performance. Practicing sample papers is an effective way to get rid of exam anxiety. It gives students a sense of confidence and frees them from all worries on the day of the exam.\n\nQ2. What are the benefits of solving CBSE class 8 maths sample papers?\n\nAns.  When students solve CBSE sample papers, they learn how well prepared they are for the math exam and self-evaluate themselves by solving the latest CBSE sample papers.\n\n• They will speed up their question-solving speed so they can solve all the questions during the exam.\n• Solving questions over and over again puts all the important formulas at your fingertips. This will save time on the exam and resolve the paper quickly.\n• By solving the class 8 sample papers students can revise all the major sections in a quick sequence of time.\n\nQ3. What are CBSE class 8 maths sample papers?\n\nAns.  CBSE class 8 sample papers are the craved material for exam purposes which makes students neighborly with actual exam paper. In class 8 weightage of the question, time management, exam pattern, and preparation strategy play a key role. Repeated practice of sample papers will boost your confidence and reduce your stress level in exams.\n\nQ4. Does Physics Wallah provide solutions for the CBSE sample papers for class 8 maths?\n\nAns.  We at Physics Wallah offer you class 8 maths sample mock papers. You can access it through our website or mobile app in PDF format for easy download. Another advantage is that they are free. All of these solutions have been prepared by our academic excellent subject matter experts, so you can easily introduce to them without a doubt.\n\nQ5. How many sample papers should be practiced for CBSE class 8 maths?\n\nAns.  It depends on whether your preparation is good enough and you have completed the entire class 8 maths syllabus and if you are confident enough in almost the topics that you don't have to solve many sample papers. But if some topics are good and you are unsure of the others, you need to solve 7-9 articles so that you can build confidence for the exam.\n\nQ6. Which is the best website for class 8 maths sample papers?\n\nAns.  We will only say that quality speaks, Students can find CBSE class 8 maths sample papers in PDF format. At Physics Wallah, we provide sample documents for class 8 maths with solutions from our experienced academic teachers. Class 8 maths sample papers students can download from our website.\n\n• NCERT Solutions\n• Maths Formula\n• Maths Doubts\n• Important Question Maths\n• NCERT solutions Maths\n• Maths Notes\n• RS Aggarwal Solutions\n• Science Sample Papers\n\n## Talk to Our counsellor", null, "", null, "• Chhattisgarh\n• West Bengal\n• Maharashtra\n• Jammu & Kashmir\n• NCERT Books 2022-23\n• NCERT Solutions\n• NCERT Notes\n• NCERT Exemplar Books\n• NCERT Exemplar Solution\n• States UT Book\n• School Kits & Lab Manual\n• NCERT Books 2021-22\n• NCERT Books 2020-21\n• NCERT Book 2019-2020\n• NCERT Book 2015-2016\n• RD Sharma Solution\n• TS Grewal Solution\n• DK Goel Solution\n• TR Jain Solution\n• Selina Solution\n• Frank Solution\n• ML Aggarwal Solution\n• Lakhmir Singh and Manjit Kaur Solution\n• I.E.Irodov solutions\n• ICSE - Goyal Brothers Park\n• ICSE - Dorothy M. Noronhe\n• Sandeep Garg Textbook Solution\n• Micheal Vaz Solution\n• S.S. 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Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.\n\nMany students don’t achieve or score high marks in their annual examination because they don’t prepare for the exam properly. Therefore, subject matter experts at Selfstudys have prepared the CBSE Class 8 Maths Sample Papers to help students in their exam preparation.\n\nThese class 8 CBSE sample paper Maths are prepared by referring to the prescribed CBSE Class 8 syllabus. Those students who are preparing for their Class 8 annual examination will have to cover their whole syllabus as well as practice tons of questions before the board exam starts.\n\n## CBSE Class 8 Maths Sample Papers with Solutions\n\nSince the Maths questions vary in difficulty levels, students may face trouble in answering all the given problems. Therefore, our subject matter experts have solved the entire set of CBSE Class 8 Maths Sample Papers in an easy to understand manner.\n\nHere on this page, we provide the CBSE Class 8 Maths sample papers with solutions in PDF file format. Referring to these CBSE class 8 Maths sample papers with solutions PDF, students can easily enhance their performance in the Maths examination, because the questions are solved by our subject matter experts keeping in mind the understanding level of a student. They have also kept in mind the CBSE Class 8 Syllabus and exam pattern.\n\n## Set Wise CBSE Class 8 Maths Sample Papers\n\nThe more you practice the better you become, our experts believe in this quote and that is why they have prepared the CBSE Class 8 Maths Sample papers in set wise format. On Selfstudys, students are free to download the sample question paper for class 8 Maths CBSE in a set wise format.\n\nCBSE Class 8 Maths Sample Papers (Term-1) 2021-22 Set - 1\n\nCBSE Class 8 Maths Sample Papers  (Term-1) 2021-22 Set - 2\n\nCBSE Class 8 Maths Sample Papers  (Term-1) 2021-22 Set - 3\n\nCBSE Class 8 Maths Sample Papers  (Term-1) 2021-22 Set - 4\n\nCBSE Class 8 Maths Sample Papers  (Term-1) 2021-22 Set - 5\n\nCBSE Class 8 Maths Sample Papers  Term-I 2021 Set - 1\n\nCBSE Class 8 Maths Sample Papers  Term-I 2021 Set - 2\n\nCBSE Class 8 Maths Sample Papers  Term-I 2021 Set - 3\n\nCBSE Class 8 Maths Sample Papers  Term-II 2021 Set - 1\n\nCBSE Class 8 Maths Sample Papers  Term-II 2021 Set - 2\n\nThe great advantage of referring to the Set Wise CBSE Class 8 Maths Sample Papers is that it provides numerous questions that follow the actual annual exam pattern and the syllabus. Apart from this, the PDF that we provide here contains CBSE class 8 Maths sample papers with solutions.\n\nIt is absolutely free of cost and easy to Download CBSE Class 8 Maths Sample Papers with solutions here at Selfstudys. Follow the below given steps to know how to download\n\n• Open Selfstudys website in your browser", null, "• A popup will appear where you need to click on Sample papers\n• Just after clicking on sample papers, a new page will open where you to click on Class “ 8th ”; like this", null, "", null, "## Why Someone Needs CBSE Class 8 Maths Sample Papers\n\nThose who want to learn about the questions, their types, marking weightage, etc. need to refer to the CBSE Class 8 Maths Sample Papers. It not only helps in this, but Maths Sample Question Papers are very helpful in practicing tons of unique questions.\n\nAlso, if someone has fear of exams and its environments, they can refer to the sample paper of class 8 Maths CBSE SA2, SA1 or any types of class 8 Maths papers that can help them to be thorough with the whole Maths syllabus.\n\n## Who Needs CBSE Class 8 Maths Sample Papers\n\n• Those Class 8 students who are studying at a CBSE affiliated school need CBSE Class 8 Maths Sample papers.\n• Students who want to be thorough with plenty of questions\n• Students looking for additional Maths questions to practice apart from their CBSE Class 8 Maths Book.\n• Those who have fear of classroom tests or annual examination\n\n## Is it Worth to Use CBSE Class 8 Maths Sample Papers\n\nSolving the CBSE Class 8 Maths Sample Papers is worthy as it helps students to understand the questions' difficulty. It also allows them to solve the Maths questions that are given in their CBSE Class 8 Maths Syllabus.\n\nBeing thorough with the CBSE sample paper of class 8 Maths enables students to revise the topics and subtopics they covered earlier in their classrooms. Apart from all these worthy points, the class 8 CBSE sample paper Maths is helpful in doing the last minute exam preparation.\n\n## When is The Right Time to Solve Class 8 CBSE Sample Paper Maths\n\nA CBSE class 8 student can solve class 8 CBSE sample paper Maths anytime they want, but ideally it should be referred about 3 months before the CBSE class 8 Maths half yearly exam or annual exam begins.\n\nHowever, there is no hard and fast rule for this tool to use. The sample papers are beneficial for solving various types of questions and so a student can refer to it whenever they want.\n\n## Use Class 8 Maths Sample Papers to Know Everything About Annual Exam Before Appearing in Them\n\nOur subject matter experts and the teachers suggest the students: use a sample question paper for class 8 Maths CBSE before appearing in the annual examination. Doing so helps understand what types of Maths questions are asked. Whether the questions are from the syllabus or outside the syllabi. If the questions are from the syllabus, then what Maths topics are necessary to cover to score good marks? And how much time is required to solve the whole question paper of Class 8 Maths?\n\nThis information is helpful and assists in preparing for the CBSE Class 8 Maths annual examination. Therefore, students should use class 8 Maths sample papers to know everything about annual exams before appearing in them.\n\n## Benefits of Practicing CBSE Class 8 Maths Sample Papers\n\nA set of CBSE class 8 Maths sample papers means a collection of the questions that follow the final exam pattern. Solving such questions before the final examination provides students with support in developing a deeper understanding on how the questions should be answered in the Maths examination. Apart from this, there are a few more benefits of practicing CBSE class 8 Maths sample papers:\n\n• It helps in revision\n• Assist students to be thorough with the Maths questions so that answering them in the exam could be easier.\n• CBSE sample paper of class 8 Maths also enables the students to do the self-analysis. Means they can analyse their level of understanding very easily.\n\n## Study Tips for Annual Exam Using CBSE Class 8 Maths Sample Papers with Solution\n\nMaths subjects require an immense attention of a learner. It is an easy subject to score good marks, but students will have to pay very close attention to types of questions that are asked in the examination such as MCQs, Short answer types of questions, Long answer types of questions, fill in the blanks, etc. However, here are some of the study tips for the annual exam preparation using CBSE Class 8 Maths Sample Papers with Solution.\n\n• Before using the CBSE sample paper of class 8 Maths to prepare for the exam, firstly cover the whole CBSE Class 8 Syllabus for Maths.\n• Once done with the Maths syllabus go for a quick revision of the entire topics you have studied.\n• Then refer to the sample question paper for class 8 Maths CBSE to check where you lack in the preparation. The CBSE class 8 Maths annual exam sample papers is an ideal study tool that helps students discover the advanced level of questions that may be asked in the actual annual examination.\n• NCERT Solutions for Class 12 Maths\n• NCERT Solutions for Class 10 Maths\n• CBSE Syllabus 2023-24\n• Social Media Channels", null, "• Second click on the toggle icon", null, "## CBSE Class 8 Math Solution,Notes,MCQ,Sample Questions\n\nAlgebraic expressions and identities.\n\n• Important Questions\n• NCERT Solutions\n• Sample Questions\n• Sample Papers\n\n## Comparing Quantities\n\nCubes and cube roots, data handling, direct and inverse proportions, exponents and powers, factorization, introduction to graphs, mensuration, playing with numbers, practical geometry, squares and square roots, visualizing solid shapes, linear equations in one variable, rational numbers, understanding quadrilaterals.\n\n• Monday, November 06, 2023 11:04:55 IST", null, "## Search form", null, "## Kendriya Vidyalaya Sangathan Regional Office, Chandigarh An Autonomous Body Under Ministry of Education, Government of India", null, "## Class-VIII Sample Paper Maths Question Paper Set-1\n\n• Sample Paper\n• Question Paper\n• NCERT Solutions\n• NCERT Books\n• NCERT Audio Books\n• NCERT Exempler\n• Solved Sample Papers\n• Writing Skill Format\n• RD Sharma Solutions\n• HC Verma Solutions\n• CG Board Solutions\n• UP Board Solutions\n• Careers Opportunities\n• Courses & Career\n• Courses after 12th\n\nHome » Extras » Class 8 Maths Sample Paper Half Yearly 2023-24 (PDF) – 8th Maths Half Yearly Model Question Paper\n\n## Class 8 Maths Sample Paper Half Yearly 2023-24 (PDF) – 8th Maths Half Yearly Model Question Paper", null, "Class 8 Maths Sample Paper Half Yearly 2023-24 is now available. You can now download the 8th Maths Half Yearly Model Question Paper PDF here at aglasem. This class 8 half yearly sample paper for Maths contains specimen questions from latest class 8 Maths syllabus and gives you a good idea of what to expect in Maths paper in half yearly exams. Therefore by solving the Maths sample paper, you can aim for better marks in class 8 half yearly test.\n\n## Class 8 Maths Sample Paper Half Yearly 2023-24\n\nWhat is Class 8 Maths Sample Paper Half Yearly 2023-24?\n\nThe class 8 Maths half yearly exam sample question paper is the specimen paper for Maths subject. It has model questions from class 8 half yearly syllabus for Maths subject. It helps you become familiar with 8th std half yearly assessment for Maths.\n\n## Class 8 Maths Sample Paper Half Yearly 2023-24 PDF\n\nThe direct link to download 8th Maths model question paper half yearly is given above. While you can easily download the Maths sample paper from aglasem. In case you want to read the complete half yearly specimen paper for Maths, then that provision is also available on this page of aglasem. The complete class 8 Maths half yearly sample paper pdf is given below.", null, "## Class 8 Sample Paper Half Yearly\n\nBesides the Maths sample paper, there are model test papers for other subjects too. Here are all the sample question papers for half yearly exams of 8th standard.\n\n• Computer Science\n• Physical Education\n• Social Science\n\n## Half Yearly Sample Paper\n\nAll schools hold half yearly tests for various classes. The model question papers for different classes are as follows.\n\n## 8th Maths Half Yearly Model Question Paper – An Overview\n\nHighlights of the sample paper are as below.\n\n## Class 7 Physical Education Sample Paper Half Yearly 2023-24 (PDF) – 7th Physical Education Half Yearly Model Question Paper\n\nClass 9 information technology sample paper half yearly 2023-24 (pdf) – 9th information technology half yearly model question paper, related posts.", null, "", null, "", null, "## Karnataka 2nd PUC Chemistry Mid Term Question Paper 2023 (PDF)\n\n• CBSE Date Sheet\n• CBSE Result\n• CBSE Syllabus\n• CBSE Sample Papers\n• CBSE Question Papers\n• CBSE Practice Papers\n\n• CISCE Time Table\n• CISCE Results\n• CISCE Specimen Papers\n• CISCE Syllabus\n• CISCE Question Papers\n\n## Class Wise Study Material\n\nBoard exams 2023.\n\n• Revision Notes\n• State Board\n\n## Study Material\n\n• Class Notes\n• Courses After Class 12th\n• JEE Main 2023\n• Fashion & Design", null, "#### IMAGES\n\n1. CBSE Class 8 Maths Sample Papers Set 2 -Downlaod Now", null, "2. CBSE Class 8 Maths Sample Papers Set 1 Solution", null, "3. CBSE Class 8 Mathematics Question Paper Set D", null, "4. NCERT Solutions for Class 8 Maths Chapter 1 Rational Numbers", null, "5. CBSE Class 8 Maths Sample Papers Set 3 -Downlaod Now", null, "6. CBSE Class 8 Maths Sample Papers Set 4 -Downlaod Now", null, "#### VIDEO\n\n1. Class 8 Maths Sample Paper with solutions\n\n2. Class 8 Maths sample paper 22-23 with solutions कक्षा 8 गणित part-1 important questions for mid term\n\n3. Q: 1 & 2\n\n4. Class 8 Maths sample paper 23-24 with solutions कक्षा 8 गणित important questions for mid term\n\n5. Q: 3 & 4\n\n6. Class 8 Maths sample paper 23-24 with solutions कक्षा 8 गणित important questions for mid term\n\n1. What Is a Sample Methodology in a Research Paper?\n\nThe sample methodology in a research paper provides the information to show that the research is valid. It must tell what was done to answer the research question and how the research was done.\n\n2. Improve Your IELTS Score with the Right Sample Papers\n\nAre you preparing for the IELTS exam? If so, you know that practice makes perfect. One of the best ways to prepare for the IELTS is to use sample papers. Sample papers can help you get familiar with the format of the exam, practice your ski...\n\n3. The Ultimate Resource for IELTS Sample Papers\n\nAre you preparing for the International English Language Testing System (IELTS) exam? If so, you’re likely looking for the best resources to help you ace the test. One of the most important resources you can use to prepare for the IELTS is ...\n\n4. CBSE Sample Papers for Class 8 Maths\n\nCBSE Sample Papers for Class 8 Maths. CBSE Sample Papers for Class 8 Maths are a perfect trial run for the students through which they can check their exam\n\n5. CBSE Sample Paper for Class 8 Maths with Solutions\n\nSample Paper of Class 8 Math enables the students to understand the exam pattern,Chapter-wise weightage and also assists them to make their strong foundation in\n\n6. CBSE Class 8 Maths Sample Papers Set 3 -Downlaod Now\n\nClick here to download CBSE Class 8 Maths Sample Papers Set 3 in PDF format prepared by experts. Solve it and know how ready you are for the exam.\n\n7. CBSE Sample Papers for Class 8 Maths with Solution 2023-24\n\nSample Paper Class 8 Syllabus Overview for Maths · Squares and Squares Roots, Cubes and Cube Roots · Linear Equations in one variable · Understanding\n\n8. CBSE Class 8 Sample Paper 2023-24 with Solutions PDF\n\nCBSE Class 8 Maths Sample Papers. Maths is a subject that requires an immense level of practice to score better marks in the upcoming Class 8 final exam. For\n\n9. CBSE Sample Papers for Class 8 Maths\n\nAns. The mathematical class 8 CBSE sample papers are a perfect test for students to verify the exam preparation. They could see where they are making mistakes\n\n11. CBSE Class 8 Math Solution,Notes,MCQ,Sample Questions\n\nCL provides CBSE Math prep material for class 8 students. Get free Key Notes, MCQs, Tests, Sample Papers, NCERT Solutions, NCERT Solutions\n\n12. Class-VIII Sample Paper Maths Question Paper Set-1\n\nHome » Class-VIII Sample Paper Maths Question Paper Set-1. Class-VIII Sample Paper Maths Question Paper Set-1. Class-VIII Sample Paper Maths Question Paper\n\n13. Class 8 Maths Sample Paper Half Yearly 2023-24\n\nClass 8 Maths Sample Paper Half Yearly 2023-24 is now available. You can now download the 8th Maths Half Yearly Model Question Paper PDF\n\n14. CBSE Sample Paper for Class 8 Mathematics 2023-24" ]

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[ "# Is this a nested logit when my regression equations are similar to that of a multinomial regression?\n\nLet the mode of transport be a categorical random variable with three categories, representing Car, Bus and Train respectively. The goal is to predict for the $n$-th person, what mode of transport they use to commute to work.\n\nSuppose that I think that when making the choice of transport to take, a person first decides whether they will catch public transport or not. Then, if they catch public transport, they have the choice of going by bus or train. This hierarchical/nested relationship can be understood using a modification of Figure 4.1 (page 90) from Chapter 4: GEV of Train:", null, "This leads to a model that produces a vector of probabilities $(p_{0}, p_{1}, p_{2}) = (p_{n0}, p_{n1}, p_{n2})$ for the modes Car, Bus, and Train respectively via the regression equations (I am going to abuse notation and drop subscripts involving $n$ from now on):\n\n$$\\begin{split} \\operatorname{ln}(\\frac{p_0}{1 - p_0}) &= y_0 \\\\ \\operatorname{ln}(\\frac{p_1}{p_2}) &= y_1\\\\ p_0 + p_1 + p_2 &= 1 \\end{split}$$ where $y_0 = \\alpha_0 + \\alpha_1 x_{01} + \\alpha_2 x_{02} + \\dots$ and $y_1 = \\beta_0 + \\beta_1 x_{11} + \\beta_2 x_{12} + \\dots$ are linear combination of covariates, possibly shared. The expression on the LHS of the first equation is the usual logit and the LHS of equation two is the generalised logit as seen in the Wikipedia article on multinomial logistic regression.\n\nMy regression equations describe two logistic regressions of sorts, one nested within the other. The first equation mimics a standard logistic regression problem for predicting whether a person catches public transport or not, and the second equation is a logistic regression (adapter so that probabilities sum to one) for whether a person catches a bus or a train, conditional on them catching public transport.\n\nMy question is whether this model is a useful discrete choice model for the question at hand, as it doesn't seem like it is a nested logit (GEV) model, nor does it seem like a multinomial regression model. It doesn't seem like a GEV model for the following reasons. Following Chapter 4.2 of Train, linking my first equation for the \"upper\" model to the \"lower\" model seems to require $\\lambda$, a parameter for the cumulative GEV distribution (Equation 4.1, reproduced below), to be set to zero, which does not seem to make sense:\n\n$$\\exp\\left(-\\sum_{k=1}^K\\left(\\sum_{j \\in B_k} e^{-\\epsilon_{nj}/\\lambda_k}\\right)^{\\lambda_k}\\right)$$\n\nIt also is not a multinomial logit as the relationship between the covariates and the predicted probabilities are different, when covariates are shared, i.e. $x_{0j} = x_{1j}, \\forall j$. My first equation is subtly different to the one given by multinomial logistic regression, when assuming that Train is the reference class:\n\n$$\\operatorname{ln}(\\frac{p_0}{p_2}) = y_0$$\n\nTo simplify notation, let $Y_0 = \\exp(y_0)$ and $Y_1 = \\exp(y_1)$. Then, solving for the probabilities from my equations:\n\n$$\\begin{split} p_0 &= \\frac{Y_0}{1+Y_0}\\\\ p_1 &= \\frac{(1 - p_0) Y_1}{1+Y_1}\\\\ p_2 &= 1 - \\frac{p_0}{1+Y_1} - \\frac{Y_1}{1 + Y_1} \\end{split}$$\n\nIn contrast, with multinomial logit one obtains:\n\n$$\\begin{split} p_0 &= \\frac{Y_0}{1+Y_0 + Y_1}\\\\ p_1 &= \\frac{Y_1}{1+Y_0 + Y_1}\\\\ p_2 &= \\frac{1}{1+Y_0 + Y_1} \\end{split}$$" ]

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[ "molecular equation of calcium metal and in indonesia\n\n7.1 Writing and Balancing Chemical Equations – General\n\nThis equation represents the reaction that takes place when sodium metal is placed in water. The solid sodium reacts with liquid water to produce molecular hydrogen gas and the ionic compound sodium hydroxide (a solid in pure form, but readily dissolved in water).\n\nCalcium | Ca - PubChem\n\nCalcium plays a vital role in the anatomy, physiology and biochemistry of organisms and of the cell, particularly in signal transduction pathways. More than 500 human proteins are known to bind or transport calcium.The skeleton acts as a major mineral storage site …\n\nWhat is the chemical equation for calcium and water? - Quora\n\nCalcium is a metal and a typical reaction of a metal with water will be as follows: Metal + Water → Metal Hydroxide + Hydrogen Gas Thus, the chemical reaction for calcium and water will be: [Word Equation] Calcium + Water → Calcium hydroxide + Hyd\n\nWriting and Balancing Chemical Equations | CHEM 1305\n\nBalancing Equations. A balanced chemical is equation has equal nuers of atoms for each element involved in the reaction are represented on the reactant and product sides.This is a requirement the equation must satisfy to be consistent with the law of conservation of matter. It may be confirmed by simply summing the nuers of atoms on either side of the arrow and comparing these sums to\n\nCHEMICAL PRECIPITATION: WATER SOFTENING\n\nlevel resulted as for each magnesium ion removed calcium ion is added. Thus to complete the hardness removal process, sodium carbonate required to be added to precipitate the calcium as presented in the equation below. + → + + + − − + − − + Cl SO Na CO CaCO s Na Cl SO Ca 2 ( ) 2 2 2 4 2 3 3 2 2 4 (6)\n\nNet Ionic Equations answers - Mrs. Norquist''s Chemistry\n\nMolecular: NH 4 I (aq) + Pb(C 2 H 3 O 2) 2 (aq) à NH 4 C 2 H 3 O 2 (aq) + PbI 2 (aq) NO rxn 12. magnesium metal + hydrochloric acid. Molecular: Mg (s) + 2 HCl (aq) → MgCl 2­(aq) + H 2(g) Net ionic: Mg (s) + 2 H + (aq) → Mg 2+ ­(aq) + H 2(g) 13. calcium metal + water. Molecular: Ca(s) + 2H 2 O(l) ——> Ca(OH) 2 (aq) + H 2 (g)\n\nMolecular, Ionic and Net Ionic Equations\n\nWrite molecular, ionic, and net ionic equations for this reaction. Milk of magnesia is a suspension of solid magnesium hydroxide, Mg(OH)2, in water. This solid can be made by adding a solution of sodium hydroxide, NaOH, to a solution of magnesium chloride, MgCl2, which causes Mg(OH)2 to precipitate and leaves sodium chloride in solution.\n\ngcse Reactivity series of metals, metallic activity order\n\nThe Reactivity Series of Metals (reactivity of calcium and compared with the non-metals carbon and hydrogen) Calcium burns quite fast with a brick red flame when strongly heated in air/oxygen to form the white powder calcium oxide. calcium + oxygen ==> calcium oxide; 2Ca(s) + O 2(g) ==> 2CaO(s) Calcium is oxidised, oxygen gain, oxidation reaction.\n\nHow to Balance: Ca + HCl = CaCl2 + H2| Breslyn\n\nWord equation: Calcium + Hydrochloric acid → Calcium chloride + Hydrogen gas. Type of Chemical Reaction: For this reaction we have a single displacement reaction. Balancing Strategies: In this reaction Calcium (Ca) metal is reacting with Hydrochloric acid (HCl).the Ca replaces the H in HCl and we end up with CaCl2 and H2. Overall it is a fairly straightforward equation to balance.\n\nAcid and Metal Word Equations Chemistry Tutorial\n\nWrite a word equation for the chemical reaction between calcium metal and hydrochloric aicd. PAUSE: PAUSE to Prepare a Game Plan (1) What information (data) have you been given in the question? (a) Name of the reactants: calcium metal and hydrochloric acid (b) Name of the products: hydrogen gas\n\nQuestion #a5e9d + Example - Socratic\n\nJul 02, 2016· Calcium metal burns in the presence of bromine gas to form calcium bromide \"Ca\"_ ((s)) + \"Br\"_ (2(g)) -> \"CaBr\"_ (2(s)) As a final note, the aqueous state suggests that the compound is dissolved in water. Therefore, you cannot use the aqueous state unless water is present.\n\ngcse Reactivity series of metals, metallic activity order\n\nThe Reactivity Series of Metals (reactivity of calcium and compared with the non-metals carbon and hydrogen) Calcium burns quite fast with a brick red flame when strongly heated in air/oxygen to form the white powder calcium oxide. calcium + oxygen ==> calcium oxide; 2Ca(s) + O 2(g) ==> 2CaO(s) Calcium is oxidised, oxygen gain, oxidation reaction.\n\nWebElements Periodic Table » Calcium » reactions of elements\n\nCalcium is a silvery white metal. The surface of calcium metal is covered with a thin layer of oxide that helps protect the metal from attack by air, but to a lesser extent than the corresponding layer in magnesium. Once ignited, calcium metal burns in air to give a mixture of white calcium oxide, CaO, and calcium nitride, Ca 3 N 2. Calcium\n\nUnit 5 Academic HW Key - Mrs. Horne''s Science Site\n\nl. Write/balance chemical equations (using syols) for each of the following synthesis reactions: a. Boron + Fluorine b. Germanium Sulfur c. Zirconium + Nitrogen d. Tetraphosphorus decaoxide + Water Phosphoric acid Write/balance chemical equations for the coustion of each of the iòllowing substances. a. Liquid acetone (C3H60) C b.\n\nAssessment Chapter Test A - Ed W. Clark High School\n\nMar 29, 2016· a. the amount of product formed by a chemical reaction. b. whether or not a specific chemical reaction is possible. c. the coefficients needed to balance a chemical equation. d. the amount of energy needed to start a chemical reaction. _____24. If calcium is listed above magnesium in the activity series, and\n\nMolecular, Ionic and Net Ionic Equations\n\nWrite molecular, ionic, and net ionic equations for this reaction. Milk of magnesia is a suspension of solid magnesium hydroxide, Mg(OH)2, in water. This solid can be made by adding a solution of sodium hydroxide, NaOH, to a solution of magnesium chloride, MgCl2, which causes Mg(OH)2 to precipitate and leaves sodium chloride in solution.\n\nMolecular and Ionic Equations | Chemistry for Non-Majors\n\nThis equation is called an ionic equation , an equation in which dissolved ionic compounds are shown as free ions.. Some other double-replacement reactions do not produce a precipitate as one of the products. The production of a gas and/or a molecular compound such as …\n\nWhat is the reaction of calcium with water? - Quora\n\nCLICK HERE for Cal­ci­um car­bide – re­ac­tion with wa­ter video When cal­ci­um car­bide re­acts with wa­ter, acety­lene is re­leased: 2H₂O + CaC₂ → C₂H₂↑ + Ca(OH)₂ Acety­lene is an in­dus­tri­al sub­stance with an un­pleas­ant smell, which is bro\n\nWrite chemical equation for the reaction a piece of\n\nJan 19, 2018· Calcium reacts slowly with water. The reaction forms calcium hydroxide, Ca(OH)2and hydrogen gas (H2). The calcium metal sinks in water and after an hour or so bubbles of hydrogen are evident, stuck to the surface of the metal.\n\n(e) Chemical formulae, equations and calculations Archives\n\n2:16 understand how metals can be arranged in a reactivity series based on their displacement reactions between: metals and metal oxides, metals and aqueous solutions of metal salts 2:17 know the order of reactivity of these metals: potassium, sodium, lithium, calcium, magnesium, aluminium, zinc, iron, copper, silver, gold\n\nMolecular, Ionic and Net Ionic Equations ~ Ionic Compounds\n\nJul 28, 2011· Write molecular, ionic, and net ionic equations for this reaction. Write molecular, ionic, and net ionic equations for any reactions that occur between the following pairs of compounds. If no reaction occurs, write ''N.R.'' CuCl 2(aq) and (NH 4) 2 CO 3(aq) HCl (aq) and MgCO 3(aq) ZnCl 2(aq) and AgC 2 H 3 O 2(aq)\n\nWriting and Balancing Chemical Equations – Chemistry\n\nBalancing Equations. The chemical equation described in section 4.1 is balanced, meaning that equal nuers of atoms for each element involved in the reaction are represented on the reactant and product sides.This is a requirement the equation must satisfy to be consistent with the law of …\n\nHow to Balance: Ca + HCl = CaCl2 + H2| Breslyn\n\nWord equation: Calcium + Hydrochloric acid → Calcium chloride + Hydrogen gas. Type of Chemical Reaction: For this reaction we have a single displacement reaction. Balancing Strategies: In this reaction Calcium (Ca) metal is reacting with Hydrochloric acid (HCl).the Ca replaces the H in HCl and we end up with CaCl2 and H2. Overall it is a fairly straightforward equation to balance.\n\nA piece of calcium metal is dropped in water write a\n\nJan 02, 2018· Find an answer to your question A piece of calcium metal is dropped in water write a chemical equation beenafaisalhb7485 beenafaisalhb7485 02.01.2018 Chemistry Secondary School A piece of calcium metal is dropped in water write a chemical equation 1 See answer beenafaisalhb7485 is waiting for your help. Add your answer and earn points.\n\nChelation - Wikipedia\n\nEquilibrium. log β. Δ G ⊖ {\\displaystyle \\Delta G^ {\\ominus }} Δ H ⊖ / k J m o l − 1 {\\displaystyle \\Delta H^ {\\ominus }\\mathrm {/kJ\\ mol^ {-1}} } − T Δ S ⊖ / k J m o l − 1 {\\displaystyle -T\\Delta S^ {\\ominus }\\mathrm {/kJ\\ mol^ {-1}} } Cu 2+ + 2 MeNH 2 ⇌ Cu (MeNH 2) 22+.\n\nOxalate - Wikipedia\n\nOxalate (IUPAC: ethanedioate) is the dianion with the formula C 2 O 2− 4, also written (COO) 2− 2.Either name is often used for derivatives, such as salts of oxalic acid, for example sodium oxalate Na 2 C 2 O 4, or dimethyl oxalate ((CH 3) 2 C 2 O 4).Oxalate also forms coordination compounds where it is sometimes abbreviated as ox.. Many metal ions form insoluble precipitates with oxalate\n\nChemical Reactions - Chemistry Encyclopedia - water\n\nA chemical reaction is a process in which one set of chemical substances (reactants) is converted into another (products). It involves making and breaking chemical bonds and the rearrangement of atoms. Chemical reactions are represented by balanced chemical equations, with chemical formulas syolizing reactants and products." ]

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[ "# JavaScript Functions\n\nA function is a set of statements, which perform a specific task. These functions are reusable and can be called anywhere in your program.\n\nFunctions allow a programmer to divide a big program into a number of small and manageable functions.\n\nFor example to design a program of calculator in spite of designing whole thing from starting , it is good to use small module like addition , subtraction , square root etc.\n\nThere are two type of functions in JavaScript:\n\n• Built-in Functions\n• User Define Functions\n\nBuilt-in function are that functions which are predefined and supplied along with the compiler.\n\nProgrammer can also define functions to do a specific task relevant to their programs. Such functions are called user-defined functions. In this topic we will discuss user-defined functions.\n\nFunction Definition: Before using any function , we have to define it. We need to tell compiler about the name , return type and argument list of function.\n\nSyntax: Following is syntax of function:\n\n```<script type = \"text/javascript\">\n<\nfunction functionname(parameter-list) {\nstatements\n}\n>\n</script>```\n\nExplanation:\n\n• ‘function’ is a keyword, which tell the compiler that one function is defined.\n• ‘functionname’ will be the name given by programmer.\n• Parameter-list is list of arguments which will pass during calling.\n\nExample:\n\n```<script type = \"text/javascript\">\n<\nfunction first()\n{\n}\n>\n</script```\n\nExplanation: Here we define a function named first, and it has single print statement.\n\nFunction Calling: We can call or invoke the function anywhere in program by simply  to write the name of that function.\n\n```<html>\n\n<body>\n<script type = \"text/javascript\">\nfunction first1() //function define\n{\ndocument.write (\"Hello javaScript\");\n}\nfirst1(); //function calling\n</script>\n</body>\n</html>```\n\nExplanation:\n\n• Here we define a function named first1().\n• This function has one write statement.\n• We simply write the name of function and put round braces, this is the syntax to call or invoke any function.\n\nFunction Parameter: We can pass arguments into function. When we define a function we pass the list of parameters. When we call the function we pass the real values that are known as arguments. Syntax is as following:\n\n```function functionName(parameter1, parameter2, parameter3)\n{\n// code to be executed\n}```\n\nRules:\n\n• In JavaScript during function definition we need not to specify data-type parameter.\n• In JavaScript functions do not perform type checking on the passed arguments.\n• In JavaScript functions do not check the number of arguments received.\n\nExample:\n\n```<html>\n<body>\n<script type = \"text/javascript\">\nfunction info(name, age)\n{\ndocument.write (name + \" is \" + age + \" years old.\");\n}\ninfo(\"easecod\",2);\n</script>\n</body>\n</html>```\n\nExplanation:\n\n• We define a function named info and pass two parameters(name,age).\n• There is single print statement.\n• We call info function and pass two arguments (string,integer).\n• name=”easecod” , age=2.\n\nOutput:\n\n```easecod is 2 years old.\n\n```" ]

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[ "", null, "", null, "# Facets of geometry: manifolds, dynamics, and curvature\n\n## Topics and goals\n\nThis research area aims for deep insights about manifolds, their automorphism groups and moduli spaces, and generalizations to metric measure spaces. This requires tools from various fields such as surgery theory, algebraic K-theory, measured and geometric group theory, harmonic analysis, and L2-invariants. We will combine forces to transfer techniques from one area to another. Among our main goals are the extension of the Farrell–Jones conjecture to reductive p-adic groups and new classification results for four-dimensional manifolds.\n\n## State of the art, our expertise\n\nThe Farrell–Jones conjecture. This conjecture has been proved by Lück and his coauthors for the algebraic K- and L-theory of group rings for a large class of discrete groups, for instance hyperbolic groups, CAT(0)-groups, lattices in locally compact second countable Hausdorff groups, and fundamental groups of three-manifolds. They have given striking applications to the classification of manifolds and their automorphism groups as well as to geometric group theory, e.g., to hyperbolic groups with spheres as boundary and to fibering manifolds [BFL14, BL12, BLRR14].\n\nL2-invariants. The Lück approximation theorem relates L2-Betti numbers to ordinary Q-Betti numbers. Partial results in characteristic p have been obtained by Lück and his coauthors. A systematic study of twisting L2-invariants by not necessarily unitary finite-dimensional representations has been initiated in [Lüc17]. Joint work with Friedl has led in dimension three to interesting connections between L2-torsion twisted by a one-parameter family of such representations and the Thurston polytope using the proof of Thurston’s virtual fibering conjecture [FL17].\n\nLow-dimensional manifolds. The failure of the s-cobordism theorem and surgery theory in dimensions < 5 implies that techniques such as the ones arising from the work on the Farrell–Jones conjecture have to be supplemented in low dimensions. For example, the usual intersection form is enriched by the theory of Whitney towers which was developed by Teichner and his coauthors over the last years. Whitney towers are understood inductively by higher-order intersection invariants _ , see the figure, with values in a certain group generated by trivalent trees [CST14]. In the four-ball, Whitney towers with boundary give geometric filtrations of classical links. It is proven in [CST12] that these filtrations can be computed by Milnor invariants, Sato–Levine invariants and higher-order Arf invariants. The theory has been applied to the classification of string links and gives a geometric understanding of Cochran’s link invariants [CST17]. Schneiderman–Teichner [ST14] showed that for an arbitrary four-manifold M the vanishing of the non-repeating part of the higher-order intersection invariants is equivalent to representing homotopy classes in _2(M) by disjoint two-spheres.\n\nModuli spaces and mapping spaces. The classical moduli space of curves and of abelian differentials has been studied with tools from algebraic and differential geometry and from dynamical systems. Geometric group theory and low-dimensional topology are linked; for example, mapping class groups and three-manifolds are related by the mapping torus of a surface automorphism. Hamenstädt’s contribution to this very fast developing area mainly concentrates on the dynamical aspects of the theory [Ham13] and on the link between metrics on Teichmüller spaces and the cohomology of mapping class groups [Ham14]. Dynamical as well as arithmetic aspects also play a central role in Huybrechts’ study of moduli spaces of hyperkähler manifolds and K3 surfaces [Huy12]. Related from the perspective of topological string theory is Klemm’s work on topological A- and B-models on Calabi–Yau manifolds [KKP16].\n\nThe study of harmonic maps with values in metric spaces of nonpositive curvature, so-called CAT(0)-spaces, was initiated by Korevaar–Schoen and Jost; recently it was enhanced by Zhang– Zhu. The most natural setting for domain spaces are CD spaces, i.e., metric measure spaces with synthetic lower bounds on the Ricci curvature. Much progress about harmonic functions and heat flows on such spaces was made in the last decade, e.g., for Finslerian spaces by Ohta– Sturm [OS14].\n\n## Research program\n\nThe Farrell–Jones conjecture. We expect many new applications of the Farrell–Jones conjecture to geometric and topological questions, for instance to automorphism groups of manifolds and to the classification of certain classes of manifolds such as torus bundles over lens spaces. We want to identify a group, or a property of groups, which has the potential to give a counterexample. A new long-term project is to formulate the algebraic K-theoretic analogue for a reductive p-adic group G. This would be the first instance where the Farrell–Jones conjecture is considered for a topological group. A potential proof will exploit the CAT(0)-structure and the resulting flow space on the associated Bruhat–Tits building. A consequence will be that the canonical map from colimK_G K0(H(K)) to K0(H(G)) is a bijection, where K runs through the compact open subgroups of G and H denotes the global Hecke algebra. This is closely related to the theory of smooth representations of reductive p-adic groups, as such representations correspond to modules over the global Hecke algebra, and thus links to RA A2. Another topic is to formulate a version of the Baum–Connes conjecture for Fréchet group algebras and prove it for classes of groups, e.g., SLn(Z), for which the Baum–Connes conjecture is not known. Computations of the K- and L-groups for specific groups, which will have consequences to questions in geometry, topology and operator algebras, will be carried out using methods from equivariant homotopy theory; see also RA A2.\n\nL2-invariants. A concrete goal concerning L2-invariants is to attack the Lück approximation theorem in characteristic p and the conjecture that the first L2-Betti number of a group is equal to its cost and to the rank gradient. The Bergeron–Venkatesh conjecture relates the L2-torsion to the growth of the torsion of the homology. Lück [Lüc16] proposed both algebraic and analytic methods for its proof. The conjecture is especially interesting in dimension three, where it relates the torsion growth of the first homology to the volume. Hamenstädt intends to use methods from low-dimensional topology by investigating manifolds fibering over the circle and to develop an effective Cˇ ech cohomology model on graphs for the thick part of a fibered hyperbolic threemanifold. Twisted L2-torsion shall be investigated for three-manifolds and knots as a function on the representation space of finite-dimensional linear representations of their fundamental groups. Its impact on high-dimensional manifolds and geometric groups will be investigated.\n\nLow-dimensional manifolds. Ray and Teichner will study knot and link concordance and its relation to the classification of four-manifolds. The focus will be on fundamental groups for which the four-dimensional topological s-cobordism theorem holds, on smooth versus topological disks in the four-ball and on finding counterexamples to the smooth four-dimensional Poincaré conjecture. Using recent progress in a Bousfield–Kan spectral sequence that describes configuration space approximations to links, Teichner wants to understand the possible values for his new higher-order Arf invariants and solve the remaining problems concerning Whitney towers in the four-ball. He will try to generalize the spectral sequence to concordance spaces to find new obstructions for representing classes in _2(M4) by embeddings of two-spheres. This is intimately related to the classification of smooth four-manifolds, the main open problem in manifold topology. A joint goal of Lück and Teichner is a general classification in the stable case, where one allows connected sums with S2 _ S2. Teichner conjectures that in the spin case, algebraic invariants like the (stable, equivariant) intersection form and a cubic form lead to a complete stable classification. This has been achieved for abelian and two-dimensional fundamental groups by Teichner and coauthors, the four-torus being a particularly interesting case. For other aspherical four-manifolds, the Farrell–Jones conjecture will be an important tool in this classification.\n\nModuli spaces and mapping spaces. The differential geometry of moduli space of curves will be further developed by Hamenstädt. A main goal is a better geometric understanding of the Schottky locus via its intersection with specific Hilbert modular varieties determined by pseudo- Anosov maps with prescribed trace field. The intriguing similarities between the moduli space of curves and moduli spaces of K3 surfaces, with relations to RA A1 and IRU D2, and of hyperkähler manifolds will be a focus of Huybrechts’ work. He also aims for new topological boundedness and finiteness results for hyperkähler manifolds. Many of these aspects also permeate Klemm’s further attempts to calculate topological string amplitudes.\n\nThe study of mappings from CD-spaces into CAT(0)-spaces will be pushed forward by Sturm. Existence and regularity results, especially Lipschitz continuity, of energy minimizers shall be derived as well as gradient estimates and equilibration rates for the associated time evolution. Moreover, in collaboration with RA B1 also suitable random evolutions on mapping spaces, in particular on loop spaces, will be studied.\n\n## Summary\n\nRA A3 will advance the understanding of manifolds, their automorphism groups and moduli spaces, and of metric measure spaces. One important innovative aspect is to combine the PIs’ expertise and to transfer tools between separate fields such as surgery theory, algebraic K- theory, measured and geometric group theory, harmonic analysis, and L2-invariants. Examples are the desired proofs of the Farrell–Jones conjecture using input from CAT(0)-spaces, controlled topology, homotopy theory, and dynamical systems, and proofs of new classification results in low-dimensional topology using algebraic and geometric methods as well as L2-invariants.\n\n## Bibliography\n\n[BFL14] A. Bartels, F. T. Farrell, and W. Lück. The Farrell–Jones Conjecture for cocompact lattices in virtually connected Lie groups. J. Amer. Math. Soc., 27(2):339–388, 2014.\n\n[BL12] A. Bartels and W. Lück. The Borel conjecture for hyperbolic and CAT(0)-groups. Ann. of Math. (2), 175:631–689, 2012.\n\n[BLRR14] A. Bartels, W. Lück, H. Reich, and H. Rüping. K- and L-theory of group rings over GLn(Z). Inst. Hautes Études Sci. Publ. Math., 119:97–125, 2014.\n\n[CST12] J. Conant, R. Schneiderman, and P. Teichner. Whitney tower concordance of classical links. Geom. Topol., 16(3):1419–1479, 2012.\n\n[CST14] J. Conant, R. Schneiderman, and P. Teichner. Milnor invariants and twisted Whitney towers. J. Topol., 7(1):187–224, 2014.\n\n[CST17] J. Conant, R. Schneiderman, and P. Teichner. Cochran’s i-invariants via twisted Whitney towers. J. Knot Theory Ramifications, 26(2):1740012, 28, 2017.\n\n[FL17] S. Friedl and W. Lück. Universal L2-torsion, polytopes and applications to 3-manifolds. Proc. Lond. Math. Soc. (3), 114(6):1114–1151, 2017.\n\n[Ham13] U. Hamenstädt. Bowen’s construction for the Teichmüller flow. J. Mod. Dyn., 7(4):489–526, 2013.\n\n[Ham14] U. Hamenstädt. Lines of minima in outer space. Duke Math. J., 163(4):733–776, 2014.\n\n[Huy12] D. Huybrechts. A global Torelli theorem for hyperkähler manifolds [after M. Verbitsky]. Astérisque, 348:375–403, 2012. Séminaire Bourbaki: Vol. 2010/2011, Exp. No. 1040.\n\n[KKP16] S. Katz, A. Klemm, and R. Pandharipande. On the motivic stable pairs invariants of K3 surfaces. In K3 surfaces and their moduli, volume 315 of Progr. Math., pages 111–146. Birkhäuser/Springer, Cham, 2016. With an appendix by R. P. Thomas.\n\n[Lüc16] W. Lück. Approximating L2-invariants by their classical counterparts. EMS Surv. Math. Sci., 3(2):269–344, 2016.\n\n[Lüc17] W. Lück. Twisting L2-invariants with finite-dimensional representations. J. Topol. Anal., 2017. DOI:10.1142/S1793525318500279.\n\n[OS14] S.-i. Ohta and K.-T. Sturm. Bochner–Weitzenböck formula and Li–Yau estimates on Finsler manifolds. Adv. Math., 252:429–448, 2014.\n\n[ST14] R. Schneiderman and P. Teichner. Pulling apart 2-spheres in 4-manifolds. Doc. Math., 19:941–992, 2014." ]

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[ "# Javascript Variables & Constants Tutorial\n\nIn this Javascript tutorial we learn about variables and constants. Single element temporary data containers that can be either be changed or must stay constant.\n\nWe cover how to declare and initialize, mutate and use variables and constants.\n\n## What is a variable\n\nA variable is a data container that stores behind-the-scenes data temporarily in a system’s memory (RAM) while the program runs. We give these variables meaningful names to allow us to easily work with the data inside.\n\nA variable can store a single type of data, such as a string of characters that form a sentence or a number.\n\nIn many other programming languages such as C or C#, we must explicitly declare what type of data the variable will hold. In Javascript however, the translator will automatically infer the data type from the value stored inside the variable.\n\n## How to declare a variable\n\nTo declare a variable, we write the keyword var , followed by the name we want to give the variable and we terminate the statement with a semicolon ( ; ).\n\nSyntax:\n``var unique_name;``\n\nThis will create an empty variable with whatever unique name we give it. This variable is now ready to accept data to store.\n\nExample:\n``var message;``\n\n## How to assign data to a variable\n\nNow that we have a variable, we can add some data inside of it. To assign data to a variable, we write its name, followed by the = operator and then any data we want to give it. Lastly, we end the statement with a semicolon again.\n\nSyntax:\n``unique_name = value;``\n\nNow the variable will have data stored inside it and we can use it.\n\nExample:\n``````// declaration\nvar message;\n\n// assign data\nmessage = \"Hello World\";``````\n\n## How to initialize a variable with a value\n\nIf we know the data we want to store beforehand, we can directly assign data to the variable when we declare it. This is called variable initialization.\n\nTo initialize a variable, we write the var keyword and the name we want to give it, then the assignment operator ( = ) and the data we want to assign to the variable. This statement needs to be ended with a semicolon as well.\n\nSyntax:\n``var unique_name = data;``\nExample:\n``````// initialization\nvar message = \"Hello World\";``````\n\n## How to use a variable\n\nTo use a variable, we simply refer to it by its name. The translator will pull the data from the variable and substitute it wherever we use it.\n\nWe can modify the code from our first application.\n\nExample:\n``````<script>\n\nvar message = \"Hello World\";\n\ndocument.write(message);\n// document.write(\"Hello World\");\n\n</script>``````\n\nIn the example above, we create a variable called message with the words “Hello World”. Where we previously wrote the words directly into the document.write() function, we now write the variable name.\n\nWhen we save the document and open it in the browser, we can see that the words inside the variable are displayed on the page.\n\nAs another example, let’s do some simple math with variables.\n\nExample:\n``````<script>\n\nvar x = 5;\nvar y = 3;\nvar result = x + y;\n\ndocument.write(\"5 + 3 = \", result);\n\n</script>``````\n\nIn the example above, we add the value of y to the value of x and store the result in a variable called result. Then we print the result to the webpage.\n\nSo the translator substitutes the variable name with its value. The example above would essentially look like this.\n\nExample:\n``````8 = 5 + 3\ndocument.write(\"5 + 3 = \", 8);``````\n\n## How to change data in a variable\n\nA variable is mutable, which means that we can change the data inside the variable at any time we need. We don’t have to set up a new variable each time we need data.\n\nTo change, or mutate, the data inside a variable, we simply assign the new data to it with the assignment operator.\n\nExample:\n``````<script>\n\nvar message = \"<p>Hello World</p>\";\ndocument.write(message);\n\n// mutate data\nmessage = \"<p>Greetings</p>\";\ndocument.write(message);\n\n</script>``````\n\nIn the example above, we mutate the message variable to hold a new value.\n\n## What is a constant\n\nA constant is the same as a variable except that it’s value cannot change during runtime.\n\nThis is useful so that the application doesn’t accidentally change a value that shouldn’t be changed, such as the value of pi or gravity.\n\n## How to declare/initialize a constant\n\nWe cannot declare a constant without a value. If we want to create a constant, we must initialize it with a value.\n\nTo initialize a constant, we use the keyword const instead of the keyword var.\n\nSyntax:\n``const UNIQUE_NAME = value;``\n\nWe write the name in all capital letters as a convention. It allows us to easily distinguish between regular variables and constants.\n\nExample:\n``const PI = 3.14;``\n\n## How to use a constant\n\nUsing a constant is the same as using a variable. We simply refer to the constant’s name where we need to.\n\nExample:\n``````<script>\n\nconst PI = 3.14;\n\n// calculate the area of a circle\n\ndocument.write(area);\n\n</script>``````\n\nIn the example above, we use the constant PI to calculate the area of a circle.\n\n## Variable & Constant naming rules and conventions\n\nThere are a few rules we should consider when naming our variables and constants.\n\n1. A name must start with a letter (a to z or A to Z), an underscore ( _ ), or a dollar sign ( \\$ ).\n2. A name cannot start with a number, but may contain numbers.\n3. A name cannot be the same as a Javascript keyword.\n4. Names are case sensitive. For example var a and var A are different variables.\nExample:\n``````// correct\nvar message;\nvar _message;\nvar \\$message;\n// incorrect\nvar #LostSock#TheStruggleIsReal;\nvar (^_^);\n\n// correct\nvar MamboNumber5;\n// incorrect\nvar 21JumpStreet;\n\n// correct\nvar a_unique_name;\n// incorrect\nvar const;\n\n// case sensitive\nvar learning;\nvar Learning;\nvar LEARNING;``````\n\nIt’s also a good idea to always follow the naming conventions, especially when working as part of a team.\n\n1. Variable names should use snake_case. All letters of the name is lowercase and multiple words are separated by underscores.\n2. Constant names should be in ALL_CAPS and multiple should also be separated by underscores.\n3. Names should not be abbreviated.\n4. Names should be singular, not plural.\nExample:\n``````// correct\nvar this_is_a_variable;\nconst THIS_IS_A_CONSTANT;\n// incorrect\nvar ThisIsPascalCaseNaming;\nvar thisIsCamelCaseNaming;\n\n// correct\nvar delete_me;\n// incorrect\nvar del_me;\n\n//correct\nvar employee_list;\n// incorrect\nvar employees;``````\n\nIt should be noted though that you may be required to follow different conventions by your employer. It doesn’t really matter which convention you follow, the key is consistency.\n\nIn this tutorial course we will be following the conventions above, as well as some others that we’ll get to when needed.\n\n## Summary: Points to remember\n\n• Variables and constants are temporary data containers, when the application stops (for whatever reason) the data inside is lost.\n• Variables can be mutated at runtime, whereas constants cannot.\n• Variables use the var keyword at declaration and initialization, and constants use the const keyword.\n• Constants cannot be empty, they must be initialized with a value." ]

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[ "# Area & Preimeter\n\n## Exercise 11.4\n\nQuestion 1: A garden is 90 m long and 75 m broad. A path 5 m wide is to be built outside and around it. Find the area of the path. Also find the area of the garden in hectare.\n\nSolution:Given; Length of the garden = 90m, Width of the garden = 75m, Width of the path around the garden = 5m\n\nArea of rectangle = Length xx Width\n\n= 90 xx75= 6750 m2\n\nWe know that, 10000 m2 = 1 hectare\n\nHence; 6750 m2 = 6750 ÷ 10000 = 0.675 hectare\n\nLength of the garden with path = 90 m + 5m + 5m = 100 m\n\nWidth of the garden with path = 75 m + 5m + 5m = 95 m\n\nArea of the garden with path = Length xx Width\n\n= 100xx95= 9500 m\n\nNow, Area of the path = Area of the garden with path — Area of the garden\n\n= 9500 m2 — 6750 m2 = 2750 m2\n\nHence, Area of the path = 2750 m2, and area of the garden = 0.675 hectare\n\nQuestion 2: A 3 m wide path runs outside and around a rectangular park of length 125 m and breadth 65 m. Find the area of the path.", null, "Solution: Given, Length of the park = 125m, Width of the park = 65m, Width of the path around the park = 3m\n\nTherefore, Length of the park with path = 125m + 3m + 3m = 131m\n\nWidth of the park with park = 65m + 3m + 3m = 71m\n\nArea of a rectangle = Length xx Width\n\nArea of the park = 125 xx 65 = 8125 m2\n\nArea of the park with path = Length xx Width\n\n= 131xx71 = 9301 m2\n\nNow, Area of the path = Area of the park with path — Area of the park\n\n= 9301m2 — 8125m2 = 1176 m2\n\nQuestion 3: A picture is painted on a cardboard 8 cm long and 5 cm wide such that there is a margin of 1.5 cm along each of its sides. Find the total area of the margin.", null, "Solution: Given, Length of the cardboard = 8cm, Width of the cardboard = 5cm, Margin left along each side of cardboard = 1.5cm\n\nHence, Length of the cardboard without margin = 8cm — 1.5cm — 1.5cm = 5cm\n\nWidth of the cardboard without margin = 5cm — 1.5cm — 1.5cm = 2cm\n\nArea of the cardboard = Length xx Width\n\n= 8xx5= 40 cm2\n\nArea of the cardboard without margin = Length × Width\n\n= 5xx2= 10 cm2\n\nTherefore, Area of margin = Area of the cardboard — Area of the cardboard without margin\n\n= 40cm2 — 10cm2= 30cm2\n\nQuestion 4: A verandah of width 2.25m is constructed all along outside a room which is 5.5m long and 4m wide. Find\n(i) The area of the verandah.\n(ii) The cost of cementing the floor of the verandah at the rate of Rs 200 per m2.", null, "Solution: Given, Length of room = 5.5m, Width of room = 4m, Width of verandah which is constructed all along outside of room = 2.25m, Cost of cementing of verandha = Rs 200 per m2\n\nTherefore, Length of room with verandah = 5.5m + 2.25m + 2.25m = 10m\n\nWidth of room with verandah = 4m + 2.25m + 2.25m = 8.5m\n\nArea of room = Length xx width\n\n= 5.5 xx 4= 22.0 m2\n\nArea of room with verandah = Length xx Width\n\n=10xx8.5 = 85 m2\n\nArea of verandah = Area of room with verandah — Area of room\n\n= 85m2 —22m2 = 63 m2\n\nCost of cementing of verandah = Area × Rate\n\nRs. 200 xx 63 = Rs. 12600" ]

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[ "Cody\n\n# Problem 14. Find the numeric mean of the prime numbers in a matrix.\n\nSolution 193044\n\nSubmitted on 18 Jan 2013\nThis solution is locked. To view this solution, you need to provide a solution of the same size or smaller.\n\n### Test Suite\n\nTest Status Code Input and Output\n1   Pass\n%% x = 3; y_correct = 3; assert(isequal(meanOfPrimes(x),y_correct))\n\n2   Pass\n%% x = [1 2 3]; y_correct = 2.5; assert(isequal(meanOfPrimes(x),y_correct))\n\n3   Fail\n%% x = [3 3; 3 3]; y_correct = 3; assert(isequal(meanOfPrimes(x),y_correct))\n\nError: Assertion failed.\n\n4   Pass\n%% x = [7 3 8 8]'; y_correct = 5; assert(isequal(meanOfPrimes(x),y_correct))\n\n### Community Treasure Hunt\n\nFind the treasures in MATLAB Central and discover how the community can help you!\n\nStart Hunting!" ]

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[ "# Modeling and analysis of cash-flow bullwhip in supply chain\n\nRattachut Tangsucheeva, Vittaldas V. Prabhu\n\nResearch output: Contribution to journalArticle\n\n20 Citations (Scopus)\n\n### Abstract\n\nThe bullwhip effect that occurs in supply chain inventory can distort demand forecasts and lead to inefficiencies such as excessive inventory, stock-outs, and backorders. In this paper we theorize that inventory bullwhip also leads to cash-flow bullwhip (CFB). Specifically, this paper focuses on studying CFB by developing mathematical and simulation models to analyze the relationship between inventory and cash-flow bullwhip by using Cash Conversion Cycle (CCC) as a metric. The mathematical models for inventory bullwhip are developed for two-stage and generic multi-stage supply chains, and then by extending these inventory models, the CFB models are developed for two-stage. CFB predicted by the proposed mathematical models approximately differ 14% from detailed simulation models. We find that increasing variability increases inventory and cash-flow bullwhip along with lead time, whereas increasing the demand observation period has the opposite effect. The average marginal impact of the bullwhip effect on the CFB is approximately 20%. Additionally, the CFB is also an increasing function of an expected value of inventory and a decreasing function of an expected value of demand.\n\nOriginal language English (US) 431-447 17 International Journal of Production Economics 145 1 https://doi.org/10.1016/j.ijpe.2013.04.054 Published - Sep 1 2013\n\n### Fingerprint\n\nSupply chains\nMathematical models\nBullwhip\nSupply chain\nModeling\nCash flow\n\n### All Science Journal Classification (ASJC) codes\n\n• Economics and Econometrics\n• Management Science and Operations Research\n• Industrial and Manufacturing Engineering\n\n### Cite this\n\n@article{0d5b638e3b2047df8bde7899a2c4e65d,\ntitle = \"Modeling and analysis of cash-flow bullwhip in supply chain\",\nabstract = \"The bullwhip effect that occurs in supply chain inventory can distort demand forecasts and lead to inefficiencies such as excessive inventory, stock-outs, and backorders. In this paper we theorize that inventory bullwhip also leads to cash-flow bullwhip (CFB). Specifically, this paper focuses on studying CFB by developing mathematical and simulation models to analyze the relationship between inventory and cash-flow bullwhip by using Cash Conversion Cycle (CCC) as a metric. The mathematical models for inventory bullwhip are developed for two-stage and generic multi-stage supply chains, and then by extending these inventory models, the CFB models are developed for two-stage. CFB predicted by the proposed mathematical models approximately differ 14{\\%} from detailed simulation models. We find that increasing variability increases inventory and cash-flow bullwhip along with lead time, whereas increasing the demand observation period has the opposite effect. The average marginal impact of the bullwhip effect on the CFB is approximately 20{\\%}. Additionally, the CFB is also an increasing function of an expected value of inventory and a decreasing function of an expected value of demand.\",\nauthor = \"Rattachut Tangsucheeva and Prabhu, {Vittaldas V.}\",\nyear = \"2013\",\nmonth = \"9\",\nday = \"1\",\ndoi = \"10.1016/j.ijpe.2013.04.054\",\nlanguage = \"English (US)\",\nvolume = \"145\",\npages = \"431--447\",\njournal = \"International Journal of Production Economics\",\nissn = \"0925-5273\",\npublisher = \"Elsevier\",\nnumber = \"1\",\n\n}\n\nModeling and analysis of cash-flow bullwhip in supply chain. / Tangsucheeva, Rattachut; Prabhu, Vittaldas V.\n\nIn: International Journal of Production Economics, Vol. 145, No. 1, 01.09.2013, p. 431-447.\n\nResearch output: Contribution to journalArticle\n\nTY - JOUR\n\nT1 - Modeling and analysis of cash-flow bullwhip in supply chain\n\nAU - Tangsucheeva, Rattachut\n\nAU - Prabhu, Vittaldas V.\n\nPY - 2013/9/1\n\nY1 - 2013/9/1\n\nN2 - The bullwhip effect that occurs in supply chain inventory can distort demand forecasts and lead to inefficiencies such as excessive inventory, stock-outs, and backorders. In this paper we theorize that inventory bullwhip also leads to cash-flow bullwhip (CFB). Specifically, this paper focuses on studying CFB by developing mathematical and simulation models to analyze the relationship between inventory and cash-flow bullwhip by using Cash Conversion Cycle (CCC) as a metric. The mathematical models for inventory bullwhip are developed for two-stage and generic multi-stage supply chains, and then by extending these inventory models, the CFB models are developed for two-stage. CFB predicted by the proposed mathematical models approximately differ 14% from detailed simulation models. We find that increasing variability increases inventory and cash-flow bullwhip along with lead time, whereas increasing the demand observation period has the opposite effect. The average marginal impact of the bullwhip effect on the CFB is approximately 20%. Additionally, the CFB is also an increasing function of an expected value of inventory and a decreasing function of an expected value of demand.\n\nAB - The bullwhip effect that occurs in supply chain inventory can distort demand forecasts and lead to inefficiencies such as excessive inventory, stock-outs, and backorders. In this paper we theorize that inventory bullwhip also leads to cash-flow bullwhip (CFB). Specifically, this paper focuses on studying CFB by developing mathematical and simulation models to analyze the relationship between inventory and cash-flow bullwhip by using Cash Conversion Cycle (CCC) as a metric. The mathematical models for inventory bullwhip are developed for two-stage and generic multi-stage supply chains, and then by extending these inventory models, the CFB models are developed for two-stage. CFB predicted by the proposed mathematical models approximately differ 14% from detailed simulation models. We find that increasing variability increases inventory and cash-flow bullwhip along with lead time, whereas increasing the demand observation period has the opposite effect. The average marginal impact of the bullwhip effect on the CFB is approximately 20%. Additionally, the CFB is also an increasing function of an expected value of inventory and a decreasing function of an expected value of demand.\n\nUR - http://www.scopus.com/inward/record.url?scp=84880918216&partnerID=8YFLogxK\n\nUR - http://www.scopus.com/inward/citedby.url?scp=84880918216&partnerID=8YFLogxK\n\nU2 - 10.1016/j.ijpe.2013.04.054\n\nDO - 10.1016/j.ijpe.2013.04.054\n\nM3 - Article\n\nAN - SCOPUS:84880918216\n\nVL - 145\n\nSP - 431\n\nEP - 447\n\nJO - International Journal of Production Economics\n\nJF - International Journal of Production Economics\n\nSN - 0925-5273\n\nIS - 1\n\nER -" ]

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[ "", null, "", null, "", null, "", null, "# Re: st: ivlogit Function Question\n\n From \"Brian P. Poi\" To [email protected] Subject Re: st: ivlogit Function Question Date Fri, 23 Jun 2006 09:00:58 -0500 (CDT)\n\n```On Fri, 23 Jun 2006 [email protected] wrote:\n\n```\n```Hi,\n\nI have an inquiry as to why there is no ivlogit command in STATA. Since\nivprobit and ivtobit are available, is there some theoretical reason for\nthe absence of ivlogit? I have looked a bit, but is there a user-defined\nivlogit function? All help is appreciated.\n\nNicola\n```\nNicola,\n\nThe maximum likelihood estimators used by -ivprobit- and -ivtobit- are derived by assuming the error term in the structural equation and the error term in the reduced-form equation for the endogenous regressor are jointly normally distributed. The derivation of the two-step estimators also assumes bivariate normal errors.\n\nIf you specify the structural equation as logit instead of probit, then you would need to figure out the appropriate bivariate distribution to use for the error terms, and my guess is that it would be much more complicated to work with than the bivariate normal distribution.\n\n-- Brian Poi\n-- [email protected]\n\n*\n* For searches and help try:\n* http://www.stata.com/support/faqs/res/findit.html\n* http://www.stata.com/support/statalist/faq\n* http://www.ats.ucla.edu/stat/stata/" ]

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[ "# Singular Value Decomposition (SVD)\n\nTo actively decompose a given matrix, Singular Value Decomposition (SVD) utilizes three matrices. The SVD technique is widely used in machine learning for dimensionality reduction. By utilizing the decomposed matrices, we can actively approximate the original matrix with a lower-rank representation. The process involves passively decomposing the original matrix into three matrices using SVD. Then actively using them to obtain the lower-rank approximation.\n\nGiven an m x n matrix A, SVD factorizes A into three matrices:\n\nA = U * Σ * V^T\n\nwhere U is an m x r matrix whose columns are the left singular vectors of A, and Σ is an r x r diagonal matrix containing the singular values of A. V is an n x r matrix whose columns are the right singular vectors of A.\n\nTo obtain a lower-rank approximation of matrix A, we can use the singular values in Σ. It is arranged in decreasing order. By selecting the first k singular values and their corresponding left and right singular vectors. We can obtain a set of matrices: the first k columns of U, the first k rows of Σ, and the first k rows of V^T. These matrices are multiplying together using matrix multiplication to obtain a new matrix B with dimensions k x n.\n\nB = Uk * Σk * Vk^T\n\nIn Singular Value Decomposition (SVD), we can use it to reduce dimensionality by selecting a subset of the largest singular values and their corresponding singular vectors. This will allow us to approximate the original matrix with a lower-rank representation. The process involves actively selecting the subset of singular values and singular vectors. Then using the obtain the lower-rank approximation of the original matrix. This can be useful for reducing the computational complexity of a dataset and identifying important features.\n\nPosted\n\nin\n\nby" ]

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[ "# Quizbowl Math\n\nSome quizbowl tournaments these days have stopped using computational math in their games. Their theory is– the rest of the game is quick recall of information, and computation uses a different skill set which makes it different from the rest of the game. Math questions where the answer is “integers” or “point of inflection” are still OK; just don’t make anyone do any calculations.\n\nWhatever your personal opinion on this, some states (including Illinois, Missouri, Georgia and Kansas) require some calculation in each game, so if no one on your team can calculate well, you’re at a disadvantage.\n\n1) During practice, go over the math questions from past tournament sets. Give your students the same amount of time they would get in a game for each problem.\n\nIf you have used our tournaments in the past, they usually have the solution written under the answer. This allows students to learn how to do them in practice. If the set you’re using has no solutions provided, ask a math teacher to come in for a few minutes and demonstrate the basic techniques. Concentrate on shortcuts that help problems get done more quickly.\n\nLet the students see each problem on the page instead of just reading it aloud. You may discover that they understand the problems fine but that simple arithmetic (like dividing 42 by 6) really slows them down. In that case, flash cards can really help.\n\n2) The first 10-20 problems in each section of a math book are easy and intended to be done quickly so students gain confidence. Many books have answers to odd-numbered problems in the back. So give the math expert 30 seconds (timed with a stopwatch) for each of the first 10 odd-numbered problems. If he gets 7-8 of them right in the time limit, move on to the next section. Continue for up to 15 minutes while other team members are studying their area of expertise. You may be surprised at the progress a student can make in the month before your next tournament.\n\n3) Use our math quizbowl set that has 200 questions, divided by general math subjects. Each question is designed to be answered in 30 seconds or less.\n\n4) Our weekly subscription sets each have several math questions in them to simulate a real game. Ensure your math experts can do all of the problems in each set, using the solution information after each answer.\n\n5) Use our Common Core math sets that map to the Common Core standards but are written in typical quizbowl style. You can use them for practice with both your math class and your quizbowl team. Create more questions by changing the numbers in the Word file we send you and just keeping the wording the same. Each Common Core set can be used by all the teachers at one school. We have sets for grades 4-8, but the 6th-grade set can also be used with 7th and 8th graders because questions at the 6th-grade level do appear even in 8th-grade quizbowl tournaments. How embarrassing would it be for your 8th-grade math expert to miss a 6th-grade math question ?" ]

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