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https://www.colorhexa.com/00affa	[ "# #00affa Color Information\n\nIn a RGB color space, hex #00affa is composed of 0% red, 68.6% green and 98% blue. Whereas in a CMYK color space, it is composed of 100% cyan, 30% magenta, 0% yellow and 2% black. It has a hue angle of 198 degrees, a saturation of 100% and a lightness of 49%. #00affa color hex could be obtained by blending #00ffff with #005ff5. Closest websafe color is: #0099ff.\n\n• R 0\n• G 69\n• B 98\nRGB color chart\n• C 100\n• M 30\n• Y 0\n• K 2\nCMYK color chart\n\n#00affa color description : Pure (or mostly pure) blue.\n\n# #00affa Color Conversion\n\nThe hexadecimal color #00affa has RGB values of R:0, G:175, B:250 and CMYK values of C:1, M:0.3, Y:0, K:0.02. Its decimal value is 45050.\n\nHex triplet RGB Decimal 00affa `#00affa` 0, 175, 250 `rgb(0,175,250)` 0, 68.6, 98 `rgb(0%,68.6%,98%)` 100, 30, 0, 2 198°, 100, 49 `hsl(198,100%,49%)` 198°, 100, 98 0099ff `#0099ff`\nCIE-LAB 67.694, -10.823, -47.458 32.581, 37.559, 95.97 0.196, 0.226, 37.559 67.694, 48.677, 257.153 67.694, -44.348, -75.589 61.285, -12.354, -49.945 00000000, 10101111, 11111010\n\n# Color Schemes with #00affa\n\n• #00affa\n``#00affa` `rgb(0,175,250)``\n• #fa4b00\n``#fa4b00` `rgb(250,75,0)``\nComplementary Color\n• #00fac8\n``#00fac8` `rgb(0,250,200)``\n• #00affa\n``#00affa` `rgb(0,175,250)``\n• #0032fa\n``#0032fa` `rgb(0,50,250)``\nAnalogous Color\n• #fac800\n``#fac800` `rgb(250,200,0)``\n• #00affa\n``#00affa` `rgb(0,175,250)``\n• #fa0032\n``#fa0032` `rgb(250,0,50)``\nSplit Complementary Color\n• #affa00\n``#affa00` `rgb(175,250,0)``\n• #00affa\n``#00affa` `rgb(0,175,250)``\n• #fa00af\n``#fa00af` `rgb(250,0,175)``\nTriadic Color\n• #00fa4b\n``#00fa4b` `rgb(0,250,75)``\n• #00affa\n``#00affa` `rgb(0,175,250)``\n• #fa00af\n``#fa00af` `rgb(250,0,175)``\n• #fa4b00\n``#fa4b00` `rgb(250,75,0)``\nTetradic Color\n• #0079ae\n``#0079ae` `rgb(0,121,174)``\n• #008bc7\n``#008bc7` `rgb(0,139,199)``\n• #009de1\n``#009de1` `rgb(0,157,225)``\n• #00affa\n``#00affa` `rgb(0,175,250)``\n• #15b9ff\n``#15b9ff` `rgb(21,185,255)``\n• #2ec0ff\n``#2ec0ff` `rgb(46,192,255)``\n• #48c8ff\n``#48c8ff` `rgb(72,200,255)``\nMonochromatic Color\n\n# Alternatives to #00affa\n\nBelow, you can see some colors close to #00affa. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #00eefa\n``#00eefa` `rgb(0,238,250)``\n• #00d9fa\n``#00d9fa` `rgb(0,217,250)``\n• #00c4fa\n``#00c4fa` `rgb(0,196,250)``\n• #00affa\n``#00affa` `rgb(0,175,250)``\n• #009afa\n``#009afa` `rgb(0,154,250)``\n• #0085fa\n``#0085fa` `rgb(0,133,250)``\n• #0071fa\n``#0071fa` `rgb(0,113,250)``\nSimilar Colors\n\n# #00affa Preview\n\nText with hexadecimal color #00affa\n\nThis text has a font color of #00affa.\n\n``<span style=\"color:#00affa;\">Text here</span>``\n#00affa background color\n\nThis paragraph has a background color of #00affa.\n\n``<p style=\"background-color:#00affa;\">Content here</p>``\n#00affa border color\n\nThis element has a border color of #00affa.\n\n``<div style=\"border:1px solid #00affa;\">Content here</div>``\nCSS codes\n``.text {color:#00affa;}``\n``.background {background-color:#00affa;}``\n``.border {border:1px solid #00affa;}``\n\n# Shades and Tints of #00affa\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, #000a0f is the darkest color, while #fafeff is the lightest one.\n\n• #000a0f\n``#000a0f` `rgb(0,10,15)``\n• #001822\n``#001822` `rgb(0,24,34)``\n• #002636\n``#002636` `rgb(0,38,54)``\n• #003349\n``#003349` `rgb(0,51,73)``\n• #00415d\n``#00415d` `rgb(0,65,93)``\n• #004f71\n``#004f71` `rgb(0,79,113)``\n• #005d84\n``#005d84` `rgb(0,93,132)``\n• #006a98\n``#006a98` `rgb(0,106,152)``\n• #0078ac\n``#0078ac` `rgb(0,120,172)``\n• #0086bf\n``#0086bf` `rgb(0,134,191)``\n• #0094d3\n``#0094d3` `rgb(0,148,211)``\n• #00a1e6\n``#00a1e6` `rgb(0,161,230)``\n• #00affa\n``#00affa` `rgb(0,175,250)``\nShade Color Variation\n• #0fb7ff\n``#0fb7ff` `rgb(15,183,255)``\n• #22bdff\n``#22bdff` `rgb(34,189,255)``\n• #36c3ff\n``#36c3ff` `rgb(54,195,255)``\n• #49c9ff\n``#49c9ff` `rgb(73,201,255)``\n• #5dceff\n``#5dceff` `rgb(93,206,255)``\n• #71d4ff\n``#71d4ff` `rgb(113,212,255)``\n• #84daff\n``#84daff` `rgb(132,218,255)``\n• #98e0ff\n``#98e0ff` `rgb(152,224,255)``\n• #ace6ff\n``#ace6ff` `rgb(172,230,255)``\n• #bfecff\n``#bfecff` `rgb(191,236,255)``\n• #d3f2ff\n``#d3f2ff` `rgb(211,242,255)``\n• #e6f8ff\n``#e6f8ff` `rgb(230,248,255)``\n• #fafeff\n``#fafeff` `rgb(250,254,255)``\nTint Color Variation\n\n# Tones of #00affa\n\nA tone is produced by adding gray to any pure hue. In this case, #738187 is the less saturated color, while #00affa is the most saturated one.\n\n• #738187\n``#738187` `rgb(115,129,135)``\n• #6a8590\n``#6a8590` `rgb(106,133,144)``\n• #60899a\n``#60899a` `rgb(96,137,154)``\n• #578ca3\n``#578ca3` `rgb(87,140,163)``\n• #4d90ad\n``#4d90ad` `rgb(77,144,173)``\n• #4394b7\n``#4394b7` `rgb(67,148,183)``\n• #3a98c0\n``#3a98c0` `rgb(58,152,192)``\n• #309cca\n``#309cca` `rgb(48,156,202)``\n• #26a0d4\n``#26a0d4` `rgb(38,160,212)``\n• #1da3dd\n``#1da3dd` `rgb(29,163,221)``\n• #13a7e7\n``#13a7e7` `rgb(19,167,231)``\n• #0aabf0\n``#0aabf0` `rgb(10,171,240)``\n• #00affa\n``#00affa` `rgb(0,175,250)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #00affa is perceived by people affected by a color vision deficiency. This can be useful if you need to ensure your color combinations are accessible to color-blind users.\n\nMonochromacy\n• Achromatopsia 0.005% of the population\n• Atypical Achromatopsia 0.001% of the population\nDichromacy\n• Protanopia 1% of men\n• Deuteranopia 1% of men\n• Tritanopia 0.001% of the population\nTrichromacy\n• Protanomaly 1% of men, 0.01% of women\n• Deuteranomaly 6% of men, 0.4% of women\n• Tritanomaly 0.01% of the population" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.5087674,"math_prob":0.83423495,"size":3673,"snap":"2021-21-2021-25","text_gpt3_token_len":1595,"char_repetition_ratio":0.14799672,"word_repetition_ratio":0.011049724,"special_character_ratio":0.5289954,"punctuation_ratio":0.22389792,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9784226,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-06-20T04:50:36Z\",\"WARC-Record-ID\":\"<urn:uuid:08d7767c-b619-4709-822a-9f3b507669f8>\",\"Content-Length\":\"36247\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:bab78c69-6a37-44dd-8c25-df6b0c01be22>\",\"WARC-Concurrent-To\":\"<urn:uuid:6db46efb-b142-4a40-afb3-b6cc6493d418>\",\"WARC-IP-Address\":\"178.32.117.56\",\"WARC-Target-URI\":\"https://www.colorhexa.com/00affa\",\"WARC-Payload-Digest\":\"sha1:M6JAEZJSGMMGHO52EYWOYU633JROROSD\",\"WARC-Block-Digest\":\"sha1:MKQSC52DTPYFU3MFKI54UTRBUH7F2CMH\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-25/CC-MAIN-2021-25_segments_1623487655418.58_warc_CC-MAIN-20210620024206-20210620054206-00058.warc.gz\"}"}
https://me.gateoverflow.in/676/gate2016-3-25	[ "# GATE2016-3-25\n\nIn PERT chart, the activity time distribution is\n\n1. Normal\n2. Binomial\n3. Poisson\n4. Beta\n\nrecategorized\n\n## Related questions\n\nA firm uses a turning center, a milling center and a grinding machine to produce two parts. The table below provides the machining time required for each part and the maximum machining time available on each machine. The profit per unit on parts $I$ and $II$ are $Rs. 40$ ... week (minutes) I II Turning machine $12$ $6$ $6000$ Milling center $4$ $10$ $4000$ Grinding Machine $2$ $3$ $1800$\nThe demand for a two-wheeler was $900$ units and $1030$ units in April $2015$ and May $2015$, respectively. The forecast for the month of April $2015$ was $850$ units. Considering a smoothing constant of $0.6$, the forecast for the month of June $2015$ is $850$ units $927$ units $965$ units $970$ units\nIn a single-channel queuing model, the customer arrival rate is $12$ per hour and the serving rate is $24$ per hour. The expected time that a customer is in queue is _______ minutes.\nIn the notation $(a/b/c) : (d/e/f)$ for summarizing the characteristics of queuing situation, the letters $‘b’$ and $‘d’$ stand respectively for service time distribution and queue discipline number of servers and size of calling source number of servers and queue discipline service time distribution and maximum number allowed in system" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8076635,"math_prob":0.9986793,"size":5137,"snap":"2021-31-2021-39","text_gpt3_token_len":1421,"char_repetition_ratio":0.097993374,"word_repetition_ratio":0.618799,"special_character_ratio":0.26299396,"punctuation_ratio":0.06858407,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9992674,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-18T02:24:52Z\",\"WARC-Record-ID\":\"<urn:uuid:f67c3148-247d-4ed3-b648-ac1cce7f5ba5>\",\"Content-Length\":\"54377\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:82a8d3ea-b9aa-4099-b9a0-af37dd43899f>\",\"WARC-Concurrent-To\":\"<urn:uuid:c680b74d-ed52-4e8e-903f-6e9a61e6c0bc>\",\"WARC-IP-Address\":\"172.67.206.99\",\"WARC-Target-URI\":\"https://me.gateoverflow.in/676/gate2016-3-25\",\"WARC-Payload-Digest\":\"sha1:PNAHMUSPZFNBZUWIPY6AL3IRYEOXCOGO\",\"WARC-Block-Digest\":\"sha1:KWBTFPUH6WYU3NVWK3CWL45TZIG3ZNPH\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780056120.36_warc_CC-MAIN-20210918002951-20210918032951-00131.warc.gz\"}"}
https://www.physicsforums.com/threads/using-laplace-transforms-to-solve-ivps.284971/	[ "# Using Laplace Transforms to solve IVP's\n\n## Homework Statement\n\nsolve the ivp using laplace tranforms\n\ny''+2y'+2y=0 y(0)=1 y'(0)=-3\n\n## The Attempt at a Solution\n\nget to Y(s)[s^2+2s+2]=s-1\n\nY=(s-1)/[s^2+2s+2]\n\n^^^\ndon't know how to simplify the denominator to solve using Laplace transforms. If I had to guess I would say maybe partial fractions but keep getting the wrong answer when I try to use them.\n\nRelated Calculus and Beyond Homework Help News on Phys.org\nYou've pretty much finished it, all you need to recognize that\n\n$$Y(s)=\\frac{s-1}{(s-1)^2+1}$$\n\nthrough completing the square in the denominator. Now can you get that to work with\n\n$$f(t)=L^{-1}\\left\\{\\frac{s-a}{(s-a)^2+b^2} \\right\\} = e^{at}\\cos{(bt)}$$" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.92008924,"math_prob":0.982702,"size":384,"snap":"2020-45-2020-50","text_gpt3_token_len":134,"char_repetition_ratio":0.09736842,"word_repetition_ratio":0.0,"special_character_ratio":0.328125,"punctuation_ratio":0.024390243,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99960595,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-10-22T09:33:31Z\",\"WARC-Record-ID\":\"<urn:uuid:17bbd899-e974-4dd0-b659-ff9c9f24b76a>\",\"Content-Length\":\"63589\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:3d7bae76-a560-411b-9fef-7096b8fe4514>\",\"WARC-Concurrent-To\":\"<urn:uuid:a4026b8f-c740-4254-ad88-e6a0801f4641>\",\"WARC-IP-Address\":\"23.111.143.85\",\"WARC-Target-URI\":\"https://www.physicsforums.com/threads/using-laplace-transforms-to-solve-ivps.284971/\",\"WARC-Payload-Digest\":\"sha1:QQSLHJTEEE2Q4GNC6KHLJUK7KN46L3T4\",\"WARC-Block-Digest\":\"sha1:BWVHGIGZZN6QPVJTKOES35BTUBGEZF2Y\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-45/CC-MAIN-2020-45_segments_1603107879362.3_warc_CC-MAIN-20201022082653-20201022112653-00202.warc.gz\"}"}
https://physics.stackexchange.com/questions/436279/why-are-gradient-and-graphs-used-in-labs	[ "# Why are gradient and graphs used in labs? [closed]\n\nThis may seem like a redundant question but I was wondering why gradient calculations are preferred when calculating a value by experiment as opposed to just plugging in raw values me getting an Answer ? I know it's something to do with the fact that the gradient considers a greater number or points and this makes the value more accurate but can someone give me a proper textbook answer ?\n\nFor example, I am doing a capacitor lab where I find the time constant of a capacitor. I plotted a graph of V vs t and and an appropriate linear graph of ln V vs t. My teacher says its preferred to use the gradient of the linear graph of lnv vs t to calculate time constant rather than reading off from the exponential graph of V vs t. I wanted to know why is it more accurate to use the gradient calculations ?\n\n## closed as unclear what you're asking by Aaron Stevens, Mike, garyp, Bill N, ZeroTheHeroOct 23 '18 at 8:01\n\nPlease clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.\n\n• Can you give a little context? – Gilbert Oct 23 '18 at 1:54\n• The question, as it is currently worded, is hard to answer without more details. Can you please edit to include a specific example of the process you are asking about? – Aaron Stevens Oct 23 '18 at 2:11\n• I have edited the question – user122343 Oct 23 '18 at 2:35\n• If you make several calculations using several values from the graph, you will arrive at several values for your result. Which o e is then the correct one from experiment? – Triatticus Oct 23 '18 at 2:50\n• If the physical quantity has an exponential dependence, using a log scale \"shortens\" the graph and displays the exponential behavior because the eye follows lines much better than general curves . – anna v Oct 23 '18 at 3:43\n\nExperimental physics also incorporates uncertainty. You can never say the time constant = 3 ms, but it exists as a value +/- another value. Same goes every measurement you make. On a graph rather than a point you will have fuzzy regions of uncertainty.\n\nFor a linear relation you expect a linear fitted line.\n\nFor an exponential or logarithmic relationship, a linear line is possible if one axis is on logarithmic scale.\n\nFor a power relationship, a line is possible if both axes are logarithmic.\n\nIn all cases, slopes and intercepts allow you to determine two parameters of the equation. If you get a computer to do the fit and read off the value, it is doing all of this behind the scenes anyway.\n\nHaving a linear line is useful in that it allows you to visually assess the fit, e.g. for deviations from \"linearity\" that aren't readily apparent otherwise.\n\nNow back to the errors. Lines of best fit are only the beginning. You can also define lines of worse fit. These help you determine a range of values possible for a given input. But wait. Your input also has an error. The resultant error can be determined by multiplying by the slope.\n\nI am assuming by \"time constant of the capacitor\" you mean the time constant of an RC circuit $$\\tau=RC$$. Let's assume we are looking at a discharging capacitor. Then the voltage across the capacitor as a function of time is given by $$V(t)=V_0e^{-t/\\tau}$$ where $$V_0$$ is the initial potential across the capacitor.\n\nNow, let's compare this to what you have discussed in your answer: $$\\ln(V(t))=\\ln\\left(V_0e^{-t/\\tau}\\right)=-\\frac{1}{\\tau}t+\\ln(V_0)$$\n\nSo, as you have said, the gradient, or slope, of $$\\ln(V(t))$$ gives us the desired $$\\tau$$ we want. Slopes are very easy to calculate from linear plots. Even if your data is not perfectly linear, linear models are probably the easiest models to fit data to. You could even get a good estimate of a line of best fit by just drawing one by hand and then determining the slope from there. Pretty simple.\n\nBut what about the original expression for $$V(t)$$? Well this is an exponential function. You would most likely need some sort of program to fit your data to the exponential function. And this is not as simple as a linear function. You could draw an exponential curve you think fits the data, but it would be harder than a line (and the curve you draw might not even be a true exponential function with base $$e$$). Even then, it's harder to pull the time constant from the exponential decay than it is to find the slope of a line.\n\nTherefore, in this context the linear function and its slope are easier to work with. Lines are simple to visualize and work with. Although with today's technology, either method should be fine if you have the software to do it. In the labs I would TA for, we would fit directly to the exponential functions with very few issues." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9430196,"math_prob":0.968973,"size":1711,"snap":"2019-43-2019-47","text_gpt3_token_len":408,"char_repetition_ratio":0.1218512,"word_repetition_ratio":0.0,"special_character_ratio":0.2402104,"punctuation_ratio":0.08595989,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99834704,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-11-22T19:53:44Z\",\"WARC-Record-ID\":\"<urn:uuid:4cb4f958-a039-47a0-bf82-b57ccb219f51>\",\"Content-Length\":\"130894\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9d840ddb-8fb3-469c-a3c5-f2af6c522210>\",\"WARC-Concurrent-To\":\"<urn:uuid:2ca4a9a7-c8f2-4d4f-9d5e-aa52b1110505>\",\"WARC-IP-Address\":\"151.101.1.69\",\"WARC-Target-URI\":\"https://physics.stackexchange.com/questions/436279/why-are-gradient-and-graphs-used-in-labs\",\"WARC-Payload-Digest\":\"sha1:TQY5MTCNI4NYZFANYD5TMAHOKYTKDQQM\",\"WARC-Block-Digest\":\"sha1:L37WYBGBQQD5YXEDB3FQECPPEVWGPREW\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-47/CC-MAIN-2019-47_segments_1573496671548.98_warc_CC-MAIN-20191122194802-20191122223802-00185.warc.gz\"}"}
https://bettersolutions.com/excel/functions/areas-function.htm	[ "### AREAS(reference)\n\nReturns the number of areas in a cell range or reference.\n\n reference The reference to a range of cells.\n\n#### Remarks\n\n * The \"reference\" can refer to multiple cell ranges.* The \"reference\" can be a named range or several cell references.* If you want to specify several references as a single argument, then you must include an extra sets of parentheses (see Rows 4 & 5).* An area is a range of contiguous cells or a single cell.* For the Microsoft documentation refer to support.microsoft.com\n\n A B C 1 =AREAS(B1) = 1 2 5 2 =AREAS(B1:C1) = 1 4 10 3 =AREAS(B1:C4) = 1 6 15 4 =AREAS((B1:B4,C1:C4)) = 2 8 20 5 =AREAS((B1:C4,B1:B2,C3:C4)) = 3 6 =AREAS((B1:C2,B1)) = 2 7 =AREAS(B1:B2 B1) = 1 8 =AREAS(B1 B2 B3) = #NULL!" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8167196,"math_prob":0.9736582,"size":599,"snap":"2021-31-2021-39","text_gpt3_token_len":132,"char_repetition_ratio":0.1579832,"word_repetition_ratio":0.0,"special_character_ratio":0.23706177,"punctuation_ratio":0.10091743,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.973106,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-18T16:20:20Z\",\"WARC-Record-ID\":\"<urn:uuid:97cb9ba9-f187-4b5a-b099-9a92ade2380c>\",\"Content-Length\":\"20467\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:2594875a-f1d3-4b2e-bb02-b142f635382a>\",\"WARC-Concurrent-To\":\"<urn:uuid:7ee1d781-0c77-4396-b4e5-27a100039ae1>\",\"WARC-IP-Address\":\"160.153.155.173\",\"WARC-Target-URI\":\"https://bettersolutions.com/excel/functions/areas-function.htm\",\"WARC-Payload-Digest\":\"sha1:FDFCNVYWCBZAUYDWXV4LPUPHOOWQNXHU\",\"WARC-Block-Digest\":\"sha1:P7NDEICGOZH6FUWHCMJE6RMQOHXCGUAB\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780056548.77_warc_CC-MAIN-20210918154248-20210918184248-00672.warc.gz\"}"}
https://softmath.com/math-book-answers/adding-exponents/algebra-1-concepts-and-skills.html	[ "", null, "## What our customers say...\n\nThousands of users are using our software to conquer their algebra homework. Here are some of their experiences:\n\nSo far its great!\nJ.R. Turnston, NY\n\nI am a 9th grade student and always wondered how some students always got good marks in mathematics but could never imagine that Ill be one of them. Hats off to Algebrator! Now I have a firm grasp over algebra and my approach to problem solving is more methodical.\nJon Caswell, MI\n\nMath couldn't be easier with Algebrator. Thanks!\nKelly Brown, NY\n\n## Search phrases used on 2012-12-13:\n\nStudents struggling with all kinds of algebra problems find out that our software is a life-saver. Here are the search phrases that today's searchers used to find our site. Can you find yours among them?\n\n• multple order differential equation in metlab\n• roots of exponents\n• Learn KS3 math online axis\n• mixed numbers to percentages converter\n• factoring algebraic equations\n• algebra calculater\n• how to teach integral exponents\n• online equation practice for sixth graders\n• mathamatical puzzels\n• ti89 log\n• free math lesson videos on solving logarithms\n• kumon worksheet\n• slope ti-83 graph\n• factoring trinomials online\n• convert decimal numbers to fractions\n• EXCEL SQUARE\n• FRACTIONS LEAST TO HIGHEST\n• online calculate number of combinations\n• gnuplot linear regression\n• cubic roots in ti 83 plus\n• taks math workbooks\n• interactive worksheets on squres and square roots\n• how to solve a third order equation\n• Ti-84 calculator conic-section art\n• glencoe pre algebra practice workbook answers\n• EXPRESSIONS, VARIABLES, AND EXPONENTS\n• solving negitive integers\n• free on line math solver\n• work sheet for subtraction of integers\n• fractional equation word problems, Math a\n• how to solve the geometry example for 7th grade\n• crossnumber puzzle algebra printable online\n• glencoe algebra 1 skills practice\n• JavaScript complex rational expressions calculator\n• algegra solution\n• rational expression and equation\n• Dividing Integers worksheets\n• distributive property fractions\n• Regents exams on converting fractions , decimals , and percents\n• mcdougal littell algebra 2 how to do the practice problems\n• high school freshman algebra text book\n• solving powers with algebra\n• TI-86 Slope\n• General Aptitude questions for kids\n• tips on college algebra\n• interpolation ti-83 program\n• solving equations worksheet\n• aptitude questions with solutions\n• ppt. graphing on coordinate plane\n• square root calculator fraction\n• explain rational exponents\n• \" simplifying fractions \"+\"third grade\"+lessons+free\n• explanation of the square root property\n• graphing logarithms on a ti89\n• \"discrete mathematics and its applications\" sixth edition solution manual\n• step by step how to solve inequalities on ti-84 scientific calculator\n• simplify nth roots\n• java code on convertion of kilogram to newton\n• substitution with systems of equations in algebra/calculator\n• basic math for dummies\n• exponents, worksheet, multiple choice\n• Ti-83 EXP button?\n• real life examples of parabolas\n• homogeneous equation calculator\n• find the least common denominator of variables\n• hel[p + Saxon Algebra II Lesson 7\n• calculating the y-intercept\n• printable math papers\n• negative algebra chart\n• rational expressions, worksheets\n• the algebrator\n• poems with math terms in it\n• how to solve probability with TI-83 PLus\n• worksheets on decimal equations with variable\n• mix number problems\n• online math solver\n• simple trig chart" ]	[ null, "https://softmath.com/r-solver/images/tutor.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.82840365,"math_prob":0.9749831,"size":4205,"snap":"2022-27-2022-33","text_gpt3_token_len":948,"char_repetition_ratio":0.13449179,"word_repetition_ratio":0.0,"special_character_ratio":0.20642093,"punctuation_ratio":0.051282052,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9995641,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-08-14T19:38:46Z\",\"WARC-Record-ID\":\"<urn:uuid:259d3c72-54f7-4c04-925e-dd3dbc5d5903>\",\"Content-Length\":\"35516\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:1033c1e8-0632-4d0a-8190-f3843aa08722>\",\"WARC-Concurrent-To\":\"<urn:uuid:6ff57bee-0a4f-4d17-9872-ba8f5743ff9d>\",\"WARC-IP-Address\":\"52.43.142.96\",\"WARC-Target-URI\":\"https://softmath.com/math-book-answers/adding-exponents/algebra-1-concepts-and-skills.html\",\"WARC-Payload-Digest\":\"sha1:DS5MUKZAVTCLUQL5IGMM5K62FNDRFTLT\",\"WARC-Block-Digest\":\"sha1:H6PLS5C2QEV267LFWRTI64BWL7AW6G3D\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-33/CC-MAIN-2022-33_segments_1659882572063.65_warc_CC-MAIN-20220814173832-20220814203832-00388.warc.gz\"}"}
http://www.bitwisemag.com/2017/06/c-programming-for-beginners-variables.html	[ "## Tuesday, 6 June 2017\n\n### C Programming for Beginners: Variables and Types\n\nThis is part 3 of my series on C programming for beginners.\n\nWhen you want to store values in your programs you need to declare variables. A variable is simply a name (more formally, we’ll call it an ‘identifier’) to which some value can be assigned. A variable is like the programming equivalent of a labelled box. You might have a box labelled ‘Petty Cash’ or a variable named pettycash. Just as the contents of the box might vary (as money is put into it and taken out again), so the contents of a variable might change as new values are assigned to it. You assign a value using the equals sign (=).\n\nIn C a variable is declared by stating its data-type (such as int for an integer variable or double for a floating-point variable) followed by the variable name. You can invent names for your variables and, as a general rule, it is best to make those names descriptive.\n\nThis is how to declare a floating-point variable named mydouble with the double data-type:\n\ndouble mydouble;\n\nYou can now assign a floating-point value to that variable:\n\nmydouble = 100.75;\n\nAlternatively, you can assign a value at the same time you declare the variable:\n\ndouble mydouble = 100.75;\n\n### FLOATING-POINT NUMBERS\n\nThere are several data types which can be used when declaring floating point variables in C. The float type represents single-precision numbers; double represents double-precision numbers and long double represents higher precision numbers. In this course, I shall normally use double for floating-point variables.\n\n### INTEGERS AND FLOATS\n\nNow let’s look at a program that uses integer and floating point variables to do a calculation. My intention is to calculate the grand total of an item by starting with its subtotal (minus tax) and then calculating the amount of tax due on it by multiplying that subtotal by the current tax rate. Here I’m assuming that tax rate to be 17.5% or, expressed as a floating point number, 0.175. Then I calculate the final price – the grand total – by adding the tax onto the subtotal. This is my program:\n\n#include <stdio.h>\n\nint main(int argc, char *argv) {\nint subtotal;\nint tax;\nint grandtotal;\ndouble taxrate;\n\ntaxrate = 0.175;\nsubtotal = 200;\ntax = subtotal taxrate;\ngrandtotal = subtotal + tax;\n\nprintf( \"The tax on %d is %d, so the grand total is %d.\\n\",\nsubtotal, tax, grandtotal );\nreturn 0;\n}\n\nOnce again, I use printf to display the results. Remember that the three place--markers, %d, are replaced by the values of the three matching variables: subtotal, tax and grandtotal.\n\nWhen you run the program, this is what you will see:\n\nThe tax on 200 is 34, so the grand total is 234.\n\nBut there is a problem here. If you can’t see what it is, try doing the same calculation using a calculator. If you calculate the tax, 200 * 0.175, the result you get should be 35. But my program shows the result to be 34.\n\nThis is due to the fact that I have calculated using a floating-point number (the double variable, taxrate) but I have assigned the result to an integer number (the int variable, tax). An integer variable can only represent numbers with no fractional part so any values after the floating point are ignored. That has introduced an error into the code.\n\nThe error is easy to fix. I just need to use floating-point variables instead of integer variables. Here is my rewritten code:\n\n#include <stdio.h>\n\nint main(int argc, char *argv) {\ndouble subtotal;\ndouble tax;\ndouble grandtotal;\ndouble taxrate;\n\ntaxrate = 0.175;\nsubtotal = 200;\ntax = subtotal taxrate;\ngrandtotal = subtotal + tax;\n\nprintf( \"The tax on %.2f is %.2f, so the grand total is %.2f.\\n\",\nsubtotal, tax, grandtotal );\nreturn 0;\n}\n\nThis time all the variables are doubles so none of the values is truncated. I have also used the float %f specifiers to display the float values in the string which I have passed to the printf function. In fact, you will see that the format specifiers in the string also include a dot and a number numbers like this: %.2f. This tells printf to display at least two digits to the right of the decimal point.\n\nYou can also format a number by specifying its width – that is, the minimum number of characters it should occupy in the string. So if I were to write %3.2 that would tell printf to format the number in a space that takes up at least 3 characters with at least two digits to the right of the decimal point. Try entering different numbers in the format specifiers (e.g. %10.4f) to see the effects these numbers have. Here are examples of numeric formatting specifiers that can be used with printf:\n\n### NUMERIC FORMAT SPECIFIERS\n\n%d print as decimal integer\n%4d print as decimal integer, at least 4 characters wide\n%f print as floating point\n%4f print as floating point, at least 4 characters wide\n%.2f print as floating point, 2 characters after decimal point\n%4.2f print as floating point, at least 4 wide and 2 after decimal point\n\nThis series of C programming lessons is based on my book, The Little Book Of C, which is the course text for my online video-based course, C Programming For Beginners, which teaches C programming interactively in over 70 lessons including a source code archive, eBook and quizzes. For information on this courses see HERE." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8842157,"math_prob":0.98614436,"size":5134,"snap":"2022-27-2022-33","text_gpt3_token_len":1184,"char_repetition_ratio":0.14210527,"word_repetition_ratio":0.08555555,"special_character_ratio":0.23626801,"punctuation_ratio":0.12631579,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.992795,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-07-07T11:18:43Z\",\"WARC-Record-ID\":\"<urn:uuid:107994c7-33df-444a-a3df-91e64d0b17e2>\",\"Content-Length\":\"86826\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9fa504cf-4902-4f96-8262-d5b9c9ce7cbd>\",\"WARC-Concurrent-To\":\"<urn:uuid:ac45c9e0-5089-4245-afca-bb110d9c817b>\",\"WARC-IP-Address\":\"172.217.13.243\",\"WARC-Target-URI\":\"http://www.bitwisemag.com/2017/06/c-programming-for-beginners-variables.html\",\"WARC-Payload-Digest\":\"sha1:M3DPINHTCHY4GVBGK6VULXQ4M5DHDNR7\",\"WARC-Block-Digest\":\"sha1:3ZPP6CYOOEA3DLS3ZTUGWFOYSIIECCRV\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656104690785.95_warc_CC-MAIN-20220707093848-20220707123848-00672.warc.gz\"}"}
https://arxiv.org/abs/1602.04574	[ "math-ph\n\nTitle:Multispecies totally asymmetric zero range process: II. Hat relation and tetrahedron equation\n\nAbstract: We consider a three-dimensional (3D) lattice model associated with the intertwiner of the quantized coordinate ring $A_q(sl_3)$, and introduce a family of layer to layer transfer matrices on $m\\times n$ square lattice. By using the tetrahedron equation we derive their commutativity and bilinear relations mixing various boundary conditions. At $q=0$ and $m=n$, they lead to a new proof of the steady state probability of the $n$-species totally asymmetric zero range process obtained recently by the authors, revealing the 3D integrability in the matrix product construction.\n Comments: 15 pages, minor corrections Subjects: Mathematical Physics (math-ph); Quantum Algebra (math.QA); Exactly Solvable and Integrable Systems (nlin.SI) MSC classes: 81R50, 60C99 Journal reference: Journal of Integrable Systems 2016 1 (1): xyw008 Cite as: arXiv:1602.04574 [math-ph] (or arXiv:1602.04574v2 [math-ph] for this version)\n\nSubmission history\n\nFrom: Atsuo Kuniba [view email]\n[v1] Mon, 15 Feb 2016 06:58:33 UTC (23 KB)\n[v2] Wed, 12 Oct 2016 01:57:38 UTC (23 KB)" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8298719,"math_prob":0.92601246,"size":1033,"snap":"2019-43-2019-47","text_gpt3_token_len":269,"char_repetition_ratio":0.07677357,"word_repetition_ratio":0.0,"special_character_ratio":0.25653437,"punctuation_ratio":0.12169312,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9789715,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-10-20T08:42:53Z\",\"WARC-Record-ID\":\"<urn:uuid:5975bec3-15fa-468a-b8d3-2847902a630d>\",\"Content-Length\":\"19317\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:994face9-411e-4d68-bd05-008ffd7bf037>\",\"WARC-Concurrent-To\":\"<urn:uuid:719c44e9-7586-411f-825b-8d6765d3101e>\",\"WARC-IP-Address\":\"128.84.21.199\",\"WARC-Target-URI\":\"https://arxiv.org/abs/1602.04574\",\"WARC-Payload-Digest\":\"sha1:2W6AVOU5RAMVVX4S7VCNLU5UR6QKN646\",\"WARC-Block-Digest\":\"sha1:D5RBKITGJEAW36WCEZDUZWRGL4MRX4SK\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-43/CC-MAIN-2019-43_segments_1570986705411.60_warc_CC-MAIN-20191020081806-20191020105306-00473.warc.gz\"}"}
https://github.com/probtorch/probtorch	[ "{{ message }}\n\n# probtorch / probtorch Public\n\nProbabilistic Torch is library for deep generative models that extends PyTorch\n\nSwitch branches/tags\nNothing to show\n\n## Files\n\nFailed to load latest commit information.\nType\nName\nCommit time\n\nProbabilistic Torch is library for deep generative models that extends PyTorch. It is similar in spirit and design goals to Edward and Pyro, sharing many design characteristics with the latter.\n\nThe design of Probabilistic Torch is intended to be as PyTorch-like as possible. Probabilistic Torch models are written just like you would write any PyTorch model, but make use of three additional constructs:\n\n1. A library of reparameterized distributions that implement methods for sampling and evaluation of the log probability mass and density functions (now available in PyTorch)\n\n2. A Trace data structure, which is both used to instantiate and store random variables.\n\n3. Objective functions that approximate the lower bound on the log marginal likelihood using Monte Carlo and Importance-weighted estimators.\n\nThis repository accompanies the NIPS 2017 paper:\n\n```@inproceedings{siddharth2017learning,\ntitle = {Learning Disentangled Representations with Semi-Supervised Deep Generative Models},\nauthor = {Siddharth, N. and Paige, Brooks and van de Meent, Jan-Willem and Desmaison, Alban and Goodman, Noah D. and Kohli, Pushmeet and Wood,\nFrank and Torr, Philip},\nbooktitle = {Advances in Neural Information Processing Systems 30},\neditor = {I. Guyon and U. V. Luxburg and S. Bengio and H. Wallach and R. Fergus and S. Vishwanathan and R. Garnett},\npages = {5927--5937},\nyear = {2017},\npublisher = {Curran Associates, Inc.},\nurl = {http://papers.nips.cc/paper/7174-learning-disentangled-representations-with-semi-supervised-deep-generative-models.pdf}\n}```\n\n# Contributors\n\n(in order of joining)\n\n• Jan-Willem van de Meent\n• Siddharth Narayanaswamy\n• Brooks Paige\n• Alban Desmaison\n• Alican Bozkurt\n• Amirsina Torfi\n• Babak Esmaeili\n• Eli Sennesh\n\n# Installation\n\n1. Install PyTorch [instructions]\n\n2. Install this repository from source\n\n``````pip install git+git://github.com/probtorch/probtorch\n``````\n1. Refer to the `examples/` subdirectory for Jupyter notebooks that illustrate usage.\n\n2. To build and read the API documentation, please do the following\n\n``````git clone git://github.com/probtorch/probtorch\ncd probtorch/docs\npip install -r requirements.txt\nmake html\nopen build/html/index.html\n``````\n\n# Mini-Tutorial: Semi-supervised MNIST\n\nModels in Probabilistic Torch define variational autoencoders. Both the encoder and the decoder model can be implemented as standard PyTorch models that subclass `nn.Module`.\n\nIn the `__init__` method we initialize network layers, just as we would in a PyTorch model. In the `forward` method, we additionally initialize a `Trace` variable, which is a write-once dictionary-like object. The `Trace` data structure implements methods for instantiating named random variables, whose values and log probabilities are stored under the specifed key.\n\nHere is an implementation for the encoder of a standard semi-supervised VAE, as introduced by Kingma and colleagues \n\n```import torch\nimport torch.nn as nn\nimport probtorch\n\nclass Encoder(nn.Module):\ndef __init__(self, num_pixels=784, num_hidden=50, num_digits=10, num_style=2):\nsuper(self.__class__, self).__init__()\nself.h = nn.Sequential(\nnn.Linear(num_pixels, num_hidden),\nnn.ReLU())\nself.y_log_weights = nn.Linear(num_hidden, num_digits)\nself.z_mean = nn.Linear(num_hidden + num_digits, num_style)\nself.z_log_std = nn.Linear(num_hidden + num_digits, num_style)\n\ndef forward(self, x, y_values=None, num_samples=10):\nq = probtorch.Trace()\nx = x.expand(num_samples, x.size())\nif y_values is not None:\ny_values = y_values.expand(num_samples, y_values.size())\nh = self.h(x)\ny = q.concrete(logits=self.y_log_weights(h), temperature=0.66,\nvalue=y_values, name='y')\nh2 = torch.cat([y, h], -1)\nz = q.normal(loc=self.z_mean(h2),\nscale=torch.exp(self.z_log_std(h2)),\nname='z')\nreturn q```\n\nIn the code above, the method `q.concrete` samples or observes from a Concrete/Gumbel-Softmax relaxation of the discrete distribution, depending on whether supervision values `y_values` are provided. The method `q.normal` samples from a univariate normal.\n\nThe resulting trace `q` now contains two entries `q['y']` and `q['z']`, which are instances of a `RandomVariable` class, which stores both the value and the log probability associated with the variable. The stored values are now used to condition execution of the decoder model:\n\n```def binary_cross_entropy(x_mean, x, EPS=1e-9):\nreturn - (torch.log(x_mean + EPS) * x +\ntorch.log(1 - x_mean + EPS) * (1 - x)).sum(-1)\n\nclass Decoder(nn.Module):\ndef __init__(self, num_pixels=784, num_hidden=50, num_digits=10, num_style=2):\nsuper(self.__class__, self).__init__()\nself.num_digits = num_digits\nself.h = nn.Sequential(\nnn.Linear(num_style + num_digits, num_hidden),\nnn.ReLU())\nself.x_mean = nn.Sequential(\nnn.Linear(num_hidden, num_pixels),\nnn.Sigmoid())\n\ndef forward(self, x, q=None):\nif q is None:\nq = probtorch.Trace()\np = probtorch.Trace()\ny = p.concrete(logits=torch.zeros(x.size(0), self.num_digits),\ntemperature=0.66,\nvalue=q['y'], name='y')\nz = p.normal(loc=0.0, scale=1.0, value=q['z'], name='z')\nh = self.h(torch.cat([y, z], -1))\np.loss(binary_cross_entropy, self.x_mean(h), x, name='x')\nreturn p```\n\nThe model above can be used both for conditioned forward execution, but also for generation. The reason for this is that `q[k]` returns `None` for variable names `k` that have not been instantiated.\n\nTo train the model components above, probabilistic Torch provides objectives that compute an estimate of a lower bound on the log marginal likelihood, which can now be maximized with standard PyTorch optimizers\n\n```from probtorch.objectives.montecarlo import elbo\nfrom random import rand\n# initialize model and optimizer\nenc = Encoder()\ndec = Decoder()\n+ list(dec.parameters()))\n# define subset of batches that will be supervised\nsupervise = [rand() < 0.01 for _ in data]\n# train model for 10 epochs\nfor epoch in range(10):\nfor b, (x, y) in data:\nx = Variable(x)\nif supervise[b]:\ny = Variable(y)\nq = enc(x, y)\nelse:\nq = enc(x)\np = dec(x, q)\nloss = -elbo(q, p, sample_dim=0, batch_dim=1)\nloss.backward()\noptimizer.step()```\n\nFor a more details, see the Jupyter notebooks in the `examples/` subdirectory.\n\n# References\n\n Kingma, Diederik P, Danilo J Rezende, Shakir Mohamed, and Max Welling. 2014. “Semi-Supervised Learning with Deep Generative Models.” http://arxiv.org/abs/1406.5298.\n\nProbabilistic Torch is library for deep generative models that extends PyTorch" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.6778605,"math_prob":0.9509604,"size":6256,"snap":"2021-43-2021-49","text_gpt3_token_len":1596,"char_repetition_ratio":0.099008314,"word_repetition_ratio":0.016969698,"special_character_ratio":0.25287724,"punctuation_ratio":0.18866329,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9965656,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-12-03T05:17:08Z\",\"WARC-Record-ID\":\"<urn:uuid:94baa4ad-ec69-468b-bf1b-40aa5bbb7371>\",\"Content-Length\":\"223236\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c0111669-a9c5-426e-bc42-12684218fb9e>\",\"WARC-Concurrent-To\":\"<urn:uuid:e13b079a-e753-4c55-9d19-64cdb2b7461b>\",\"WARC-IP-Address\":\"140.82.114.3\",\"WARC-Target-URI\":\"https://github.com/probtorch/probtorch\",\"WARC-Payload-Digest\":\"sha1:KYT2PETYDWNS4CSZOYXZV66TGKSACH4U\",\"WARC-Block-Digest\":\"sha1:AOICWPYAGQ7TSSNV4YJTD4XW2BVTP7BB\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964362589.37_warc_CC-MAIN-20211203030522-20211203060522-00566.warc.gz\"}"}
http://www.freepatentsonline.com/y2008/0172165.html	[ "Title:\nControl system for internal combustion engine\nKind Code:\nA1\n\nAbstract:\nA control system for an internal combustion engine, which is capable of ensuring excellent fuel economy of the engine and enhancing the responsiveness of the output of the engine when acceleration is demanded. The control system calculates a lift control input for controlling a variable valve lift mechanism, based on a cam phase of a variable cam phase mechanism, and calculates a demanded acceleration indicative of the degree of acceleration demanded of the engine. Further, the control system calculates a value of phase control input for controlling the variable cam phase mechanism with priority to the engine output, and calculates a value of the same with priority to fuel economy of the engine, and selects between the values of phase control input, based on the demanded acceleration.\n\nInventors:\nTagami, Hiroshi (Saitama-ken, JP)\nYasui, Yuji (Saitama-ken, JP)\nApplication Number:\n12/009218\nPublication Date:\n07/17/2008\nFiling Date:\n01/17/2008\nExport Citation:\nAssignee:\nHonda Motor Co., Ltd. (Tokyo, JP)\nPrimary Class:\nOther Classes:\n123/90.15\nInternational Classes:\nF02D29/02; F01L1/34; F02D13/02\nView Patent Images:\nRelated US Applications:\n 20100042279 METHOD FOR TORQUE MANAGEMENT IN A HYBRID VEHICLE EQUIPPED WITH ACTIVE FUEL MANAGEMENT February, 2010 Thompson et al. 20080109129 Cooling Device, Control Method of Cooling Device, and Abnormality Specification Method May, 2008 Yanagida et al. 20090192681 Vehicle Controller and Controlling Method July, 2009 Hayashi et al. 20090048977 USER PROFILE GENERATION ARCHITECTURE FOR TARGETED CONTENT DISTRIBUTION USING EXTERNAL PROCESSES February, 2009 Aggarwal et al. 20070100521 Reporting information related to a vehicular accident May, 2007 Grae 20070055433 Slip control device and method for a vehicle March, 2007 Yamaguchi et al. 20100030442 MOVABLE BODY, TRAVEL DEVICE, AND MOVABLE BODY CONTROL METHOD February, 2010 Kosaka 20080312801 Apparatus for determining positions and movements of a brake pedal for a vehicle brake system December, 2008 Muller et al. 20070213896 Method and apparatus for determining and storing excessive vehicle speed September, 2007 Fischer 20080154451 Method of Controlling a Rail Transport System for Conveying Bulk Materials June, 2008 Dibble et al. 20090063030 System and method for traffic condition detection March, 2009 Howarter et al.\n\nPrimary Examiner:\nCOLEMAN, KEITH A\nAttorney, Agent or Firm:\nLAHIVE & COCKFIELD, LLP (ONE POST OFFICE SQUARE, BOSTON, MA, 02109-2127, US)\nClaims:\nWhat is claimed is:\n\n1. A control system for an internal combustion engine that is configured to be capable of changing an intake air amount by changing operating characteristics of an intake valve, using a first variable valve-actuating mechanism, and a second variable valve-actuating mechanism having a lower response speed than a response speed of the first variable valve-actuating mechanism, comprising: operation amount-detecting means for detecting an amount of operation of the second variable valve-actuating mechanism with respect to the intake valve; first control input-calculating means for calculating a first control input for controlling the first variable valve-actuating mechanism based on the detected amount of operation of the second variable valve-actuating mechanism; load parameter-detecting means for detecting a load parameter indicative of load on the engine; demanded acceleration degree parameter-calculating means for calculating a demanded acceleration degree parameter indicative of a degree of acceleration demanded of the engine; output priority-type calculation means for calculating a second control input for controlling the second variable valve-actuating mechanism, based on the detected load parameter with priority to an output of the engine; fuel economy priority-type calculation means for calculating the second control input based on the load parameter with priority to fuel economy of the engine; and selection means for selecting one of said output priority-type calculation means and said fuel economy priority-type calculation means as calculation means for calculating the second control input, according to the calculated demanded acceleration degree parameter.\n\n2. A control system as claimed in claim 1, wherein the engine is installed on a vehicle as a drive source, and the control system further comprising: drive wheel demanded torque-calculating means for calculating a drive wheel demanded torque demanded of drive wheels of the vehicle; and traveling resistance-calculating means for calculating a traveling resistance of the vehicle, wherein said demanded acceleration degree parameter-calculating means calculates the demanded acceleration degree parameter based on the calculated drive wheel demanded torque and the calculated traveling resistance.\n\n3. A control system as claimed in claim 2, further comprising: drive wheel torque-detecting means for detecting a torque of the drive wheels; vehicle speed-detecting means for detecting a speed of the vehicle; and acceleration-detecting means for detecting acceleration of the vehicle, and wherein said traveling resistance-calculating means comprises: reference traveling resistance-calculating means for calculating a traveling resistance to be obtained when the vehicle and a road surface on which the vehicle travels are in respective predetermined reference states, based on the detected vehicle speed, as a reference traveling resistance; reference acceleration resistance-calculating means for calculating an acceleration resistance to be obtained when the vehicle is in the predetermined reference state, based on the detected acceleration of the vehicle, as a reference acceleration resistance; and correction value-calculating means for calculating a correction value based on the detected torque of the drive wheels, the calculated reference traveling resistance, and the calculated reference acceleration resistance, and wherein the traveling resistance is calculated by correcting the reference traveling resistance using the calculated correction value.\n\n4. A control system for an internal combustion engine that is configured to be capable of changing an intake air amount by changing operating characteristics of an intake valve, using a first variable valve-actuating mechanism, and a second variable valve-actuating mechanism having a lower response speed than a response speed of the first variable valve-actuating mechanism, comprising: operation amount-detecting means for detecting an amount of operation of the second variable valve-actuating mechanism with respect to the intake valve; first control input-calculating means for calculating a first control input for controlling the first variable valve-actuating mechanism based on the detected amount of operation of the second variable valve-actuating mechanism; load parameter-detecting means for detecting a load parameter indicative of load on the engine; demanded acceleration degree parameter-calculating means for calculating a demanded acceleration degree parameter indicative of a degree of acceleration demanded of the engine; output priority-type calculation means for calculating a second control input for controlling the second variable valve-actuating mechanism, based on the detected load parameter with priority to an output of the engine; fuel economy priority-type calculation means for calculating the second control input based on the load parameter with priority to fuel economy of the engine; and second control input-calculating means for calculating the second control input by calculating a weighted average of a value calculated by said output priority-type calculation means and a value calculated by said fuel economy priority-type calculation means, using a weight dependent on the calculated demanded acceleration degree parameter.\n\n5. A control system as claimed in claim 4, wherein the engine is installed on a vehicle as a drive source, and the control system further comprising: drive wheel demanded torque-calculating means for calculating a drive wheel demanded torque demanded of drive wheels of the vehicle; and traveling resistance-calculating means for calculating a traveling resistance of the vehicle, wherein said demanded acceleration degree parameter-calculating means calculates the demanded acceleration degree parameter based on the calculated drive wheel demanded torque and the calculated traveling resistance.\n\n6. A control system as claimed in claim 5, further comprising: drive wheel torque-detecting means for detecting a torque of the drive wheels; vehicle speed-detecting means for detecting a speed of the vehicle; and acceleration-detecting means for detecting acceleration of the vehicle, and wherein said traveling resistance-calculating means comprises: reference traveling resistance-calculating means for calculating a traveling resistance to be obtained when the vehicle and a road surface on which the vehicle travels are in respective predetermined reference states, based on the detected vehicle speed, as a reference traveling resistance; reference acceleration resistance-calculating means for calculating an acceleration resistance to be obtained when the vehicle is in the predetermined reference state, based on the detected acceleration of the vehicle, as a reference acceleration resistance; and correction value-calculating means for calculating a correction value based on the detected torque of the drive wheels, the calculated reference traveling resistance, and the calculated reference acceleration resistance, and wherein the traveling resistance is calculated by correcting the reference traveling resistance using the calculated correction value.\n\nDescription:\n\n# BACKGROUND OF THE INVENTION\n\n1. Field of the Invention\n\nThe present invention relates to a control system for an internal combustion engine, which is configured to be capable of changing an intake air amount by changing operating characteristics of an intake valve, using a first variable valve-actuating mechanism and a second variable valve-actuating mechanism having a lower response speed than that of the first variable valve-actuating mechanism.\n\n2. Description of the Related Art\n\nConventionally, as a control system for an internal combustion engine of this kind, one disclosed in Japanese Laid-Open Patent Publication (Kokai) No. 2006-57573 is known. This combustion engine is provided with a first variable valve-actuating mechanism for changing the valve lift of an intake valve, and a second variable valve-actuating mechanism for changing the central angle of an operating angle of the intake valve (hereinafter simply referred to as “the central angle”). The first and second variable valve-actuating mechanisms use an electric motor and an oil pressure pump as drive sources thereof, respectively, and the response speed of the second variable valve-actuating mechanism, that is, the response speed of the operation amount of the second variable valve-actuating mechanism with respect to a control input therefor is lower than that of the operation amount of the first variable valve-actuating mechanism. In the above-described conventional control system, the intake air amount is controlled by controlling the valve lift and the central angle by the first and second variable valve-actuating mechanisms as follows:\n\nA target central angle, which is a target value of the above-described central angle, is determined by searching a target central angle map according to the load on the engine obtained e.g. by a sensor, and an actual central angle is estimated as an actual central angle equivalent value. In the target central angle map, the target central angle is set to a value which makes it possible to obtain excellent fuel economy of the engine. Further, a target valve lift, which is a target value of the above-described valve lift, is calculated based on the load on the engine and the estimated actual central angle equivalent value. Then, a control input based on the calculated target valve lift is input to the first variable valve-actuating mechanism, and a control input based on the calculated target central angle is input to the second variable valve-actuating mechanism, whereby the valve lift and the central angle are controlled to the target valve lift and the target central angle, respectively. Thus, the response delay of the operation amount of the second variable valve-actuating mechanism with respect to the control input is compensated for, to thereby accurately control the intake air amount.\n\nAs described above, in the conventional control system, the target central angle, which is set in the target central angle map to such a value as will make it possible to obtain excellent fuel economy, is used only as the target value of the central angle. As a result, in the conventional control system, when the load on the engine is suddenly increased due to demand of acceleration, the response delay of the second variable valve-actuating mechanism cannot be sufficiently compensated for, which makes it impossible to obtain a sufficient intake air amount. This makes it impossible to increase the output of the engine with high responsiveness to the load on the engine. Further, to eliminate the above inconveniences, it is considered that the target central angle is set in the target central angle map with priority given to the output but not to the fuel economy. In this case, however, it is impossible to obtain excellent fuel economy of the engine.\n\n# SUMMARY OF THE INVENTION\n\nIt is an object of the present invention to provide a control system for an internal combustion engine, which is capable of ensuring excellent fuel economy of the engine and enhancing the responsiveness of the output of the engine when acceleration is demanded.\n\nTo attain the above object, in a first aspect of the present invention, there is provided a control system for an internal combustion engine that is configured to be capable of changing an intake air amount by changing operating characteristics of an intake valve, using a first variable valve-actuating mechanism, and a second variable valve-actuating mechanism having a lower response speed than a response speed of the first variable valve-actuating mechanism, comprising operation amount-detecting means for detecting an amount of operation of the second variable valve-actuating mechanism with respect to the intake valve, first control input-calculating means for calculating a first control input for controlling the first variable valve-actuating mechanism based on the detected amount of operation of the second variable valve-actuating mechanism, load parameter-detecting means for detecting a load parameter indicative of load on the engine, demanded acceleration degree parameter-calculating means for calculating a demanded acceleration degree parameter indicative of a degree of acceleration demanded of the engine, output priority-type calculation means for calculating a second control input for controlling the second variable valve-actuating mechanism, based on the detected load parameter with priority to an output of the engine, fuel economy priority-type calculation means for calculating the second control input based on the load parameter with priority to fuel economy of the engine, and selection means for selecting one of the output priority-type calculation means and the fuel economy priority-type calculation means as calculation means for calculating the second control input, according to the calculated demanded acceleration degree parameter.\n\nWith the configuration of the control system according to the first aspect of the present invention, the load parameter-detecting means detects the load parameter indicative of load on the engine, and the demanded acceleration degree parameter-calculating means calculates the demanded acceleration degree parameter indicative of the degree of acceleration demanded of the engine (hereinafter referred to as “the demanded acceleration degree”). Further, as calculation means for calculating the second control input based on the detected load parameter, the control system is provided with the output priority-type calculation means for calculating the second control input with priority to the output of the engine, and the fuel economy priority-type calculation means for calculating the second control input with priority to fuel economy. The selection means selects between the output priority-type calculation means and the fuel economy priority-type calculation means, based on the calculated demanded acceleration degree parameter, and the selected calculating means calculates the second control input.\n\nTherefore, when the demanded acceleration degree parameter is indicating that acceleration is demanded, if the output priority-type calculation means is selected, it is possible to obtain an intake air amount suitable for satisfying the demand of acceleration, thereby making it possible to quickly increase the output of the engine to enhance responsiveness thereof. Further, when the demanded acceleration degree parameter is not indicating the demand of acceleration, if the fuel economy priority-type calculation means is selected, it is possible to obtain excellent fuel economy of the engine when acceleration is not demanded. Thus, it is made possible to ensure excellent fuel economy of the engine to enhance the responsiveness of the output of the engine when acceleration is demanded.\n\nFurther, the operation amount-detecting means detects the amount of operation of the second variable valve-actuating mechanism with respect to the intake valve, and the first control input-calculating means calculates the first control input for controlling the first variable valve-actuating mechanism based on the detected amount of operation of the second variable valve-actuating mechanism. As described above, the first control input for controlling the first variable valve-actuating mechanism having a higher response speed is calculated based on the actual amount of operation of the second variable valve-actuating mechanism having a lower response speed, and hence, it is possible to compensate for response delay of the second variable valve-actuating mechanism by intake air amount control using the first variable valve-actuating mechanism._This makes it possible to more excellently obtain the above-described effects, i.e. the effects of ensuring excellent fuel economy and enhancing the responsiveness of the output of the engine when acceleration is demanded. It should be noted that throughout the specification, “detection” includes not only detection by sensors but also “calculation” and “estimation” by computation.\n\nPreferably, the engine is installed on a vehicle as a drive source, and the control system further comprises drive wheel demanded torque-calculating means for calculating a drive wheel demanded torque demanded of drive wheels of the vehicle, and traveling resistance-calculating means for calculating a traveling resistance of the vehicle, wherein the demanded acceleration degree parameter-calculating means calculates the demanded acceleration degree parameter based on the calculated drive wheel demanded torque and the calculated traveling resistance.\n\nWith the configuration of this preferred embodiment, the drive wheel demanded torque demanded of the drive wheels of the vehicle, and the traveling resistance of the vehicle are calculated, and the demanded acceleration degree parameter is calculated based on the calculated drive wheel demanded torque and traveling resistance. In general, when the vehicle is traveling at a constant speed, the drive wheel demanded torque and the traveling resistance are equal to and balanced with each other, whereas during acceleration of the vehicle, the drive wheel demanded torque becomes larger than the traveling resistance, and the degree of increase of the drive wheel demanded torque becomes larger as the demanded acceleration degree is larger. As described above, the drive wheel demanded torque and the traveling resistance have close correlations with the demanded acceleration degree, and therefore according to the present invention, it is possible to properly calculate the demanded acceleration degree parameter.\n\nMore preferably, the control system further comprises drive wheel torque-detecting means for detecting a torque of the drive wheels, vehicle speed-detecting means for detecting a speed of the vehicle, and acceleration-detecting means for detecting acceleration of the vehicle, and the traveling resistance-calculating means comprises reference traveling resistance-calculating means for calculating a traveling resistance to be obtained when the vehicle and a road surface on which the vehicle travels are in respective predetermined reference states, based on the detected vehicle speed, as a reference traveling resistance, reference acceleration resistance-calculating means for calculating an acceleration resistance to be obtained when the vehicle is in the predetermined reference state, based on the detected acceleration of the vehicle, as a reference acceleration resistance, and correction value-calculating means for calculating a correction value based on the detected torque of the drive wheels, the calculated reference traveling resistance, and the calculated reference acceleration resistance, and wherein the traveling resistance is calculated by correcting the reference traveling resistance using the calculated correction value.\n\nWith the configuration of this preferred embodiment, the traveling resistance is calculated as follows: The torque of the drive wheels is detected, and a traveling resistance to be obtained when the vehicle and the road surface on which the vehicle travels are in the respective predetermined reference states is calculated based on the detected vehicle speed, as a reference traveling resistance. Further, an acceleration resistance to be obtained when the vehicle is in the predetermined reference state is calculated based on the detected acceleration of the vehicle, as a reference acceleration resistance. Furthermore, a correction value is calculated based on the torque of the drive wheels, the reference traveling resistance, and the reference acceleration resistance, and the traveling resistance is calculated by correcting the reference traveling resistance using the calculated correction value.\n\nIn general, the traveling resistance is the sum of a rolling resistance, an air resistance, and a gradient resistance, and changes depending on the states (e.g. weight and front projection area) of the vehicle, and states (e.g. irregularities and gradient) of a road surface on which the vehicle travels. Therefore, unless the vehicle and a road surface are in the above-described respective reference states, actual traveling resistance is different from the above-mentioned reference traveling resistance. Further, normally, the torque of the drive wheels corresponds to the sum of an actual traveling resistance and an actual acceleration resistance (hereinafter referred to as “the total actual traveling resistance”) during acceleration of the vehicle, and corresponds to the actual traveling resistance when the vehicle is traveling at a constant speed except during acceleration of the vehicle. Further, during travel at a constant speed, the reference acceleration resistance becomes equal to 0. From the above, the difference between the torque of the drive wheels and the sum of the reference traveling resistance and the reference acceleration resistance (hereinafter referred to as “the total reference traveling resistance”) corresponds to the difference between the total actual traveling resistance and the total reference traveling resistance during acceleration of the vehicle, and corresponds to the difference between the actual traveling resistance and the reference traveling resistance during travel at a constant speed.\n\nTherefore, according to the present invention, the reference traveling resistance is corrected by the correction value calculated based on the torque of the drive wheels, the reference traveling resistance, and the reference acceleration resistance, whereby it is possible to accurately calculate the actual traveling resistance with reference to the reference traveling resistance. Further, as described above, it is possible to calculate the traveling resistance only by computations, without requiring values obtained by detections of the actual weight of the vehicle and the gradients of a road surface. This makes it possible to dispense with sensors for detecting the above values, thereby making it possible to reduce the manufacturing costs of the control system.\n\nTo attain the above object, in a second aspect of the present invention, there is provided a control system for an internal combustion engine that is configured to be capable of changing an intake air amount by changing operating characteristics of an intake valve, using a first variable valve-actuating mechanism, and a second variable valve-actuating mechanism having a lower response speed than a response speed of the first variable valve-actuating mechanism, comprising operation amount-detecting means for detecting an amount of operation of the second variable valve-actuating mechanism with respect to the intake valve, first control input-calculating means for calculating a first control input for controlling the first variable valve-actuating mechanism based on the detected amount of operation of the second variable valve-actuating mechanism, load parameter-detecting means for detecting a load parameter indicative of load on the engine, demanded acceleration degree parameter-calculating means for calculating a demanded acceleration degree parameter indicative of a degree of acceleration demanded of the engine, output priority-type calculation means for calculating a second control input for controlling the second variable valve-actuating mechanism, based on the detected load parameter with priority to an output of the engine, fuel economy priority-type calculation means for calculating the second control input based on the load parameter with priority to fuel economy of the engine, and second control input-calculating means for calculating the second control input by calculating a weighted average of a value calculated by the output priority-type calculation means and a value calculated by the fuel economy priority-type calculation means, using a weight dependent on the calculated demanded acceleration degree parameter.\n\nWith the configuration of the control system according to the second aspect of the present invention, similarly to the first aspect of the present invention, the load parameter and the demanded acceleration degree parameter are obtained. Further, according to the load parameter, the weighted average of the value calculated by the output priority-type calculation means (hereinafter referred to as “the output priority-type calculated value”) and the value calculated by the fuel economy priority-type calculation means (hereinafter referred to as “the fuel economy priority-type calculated value”) is calculated using a weight dependent on the demanded acceleration degree parameter, whereby the second control input is calculated.\n\nAs described above, the second control input is calculated by calculating the weighted average of the output priority-type calculated value and the fuel economy priority-type calculated value using the weight dependent on the demanded acceleration degree parameter, so that it is possible to calculate the second control input according to the demanded acceleration degree in a fine-grained manner. Therefore, for example, when the demanded acceleration degree parameter is indicating that acceleration is demanded, by increasing the weight of the output priority-type calculated value with respect to the second control input, it is possible to enhance the responsiveness of the output of the engine when acceleration is demanded, similarly to the first aspect of the present invention. Further, when the demanded acceleration degree parameter is not indicating that acceleration is demanded, by increasing the weight of the fuel economy priority-type calculated value, it is possible to obtain excellent fuel economy of the engine when acceleration is not demanded, similarly to the first aspect of the present invention. Thus, similarly to the first aspect of the present invention, it is possible to ensure excellent fuel economy of the engine and at the same time enhance the responsiveness of the output of the engine when acceleration is demanded.\n\nFurther, by increasing or decreasing the weight of the output priority-type calculated value according to the magnitude of the demanded acceleration degree indicated by the demanded acceleration degree parameter, differently from the first aspect of the present invention, it is possible to obtain an appropriate second control input that matches the magnitude of the demanded acceleration degree. This makes it possible to ensure excellent fuel economy and enhance the responsiveness of the output of the engine when acceleration is demanded, in a well balanced manner. Further, similarly to the first aspect of the present invention, the amount of operation of the second variable valve-actuating mechanism with respect to the intake valve is detected, and the first control input for controlling the first variable valve-actuating mechanism is calculated based on the detected amount of operation of the second variable valve-actuating mechanism. This makes it possible to compensate for the response delay of the second variable valve-actuating mechanism by the intake air amount control using the first variable valve-actuating mechanism, thereby making it possible to more excellently obtain the above-described effects.\n\nPreferably, the engine is installed on a vehicle as a drive source, and the control system further comprises drive wheel demanded torque-calculating means for calculating a drive wheel demanded torque demanded of drive wheels of the vehicle, and traveling resistance-calculating means for calculating a traveling resistance of the vehicle, wherein the demanded acceleration degree parameter-calculating means calculates the demanded acceleration degree parameter based on the calculated drive wheel demanded torque and the calculated traveling resistance.\n\nMore preferably, the control system further comprises drive wheel torque-detecting means for detecting a torque of the drive wheels, vehicle speed-detecting means for detecting a speed of the vehicle, and acceleration-detecting means for detecting acceleration of the vehicle, and the traveling resistance-calculating means comprises reference traveling resistance-calculating means for calculating a traveling resistance to be obtained when the vehicle and a road surface on which the vehicle travels are in respective predetermined reference states, based on the detected vehicle speed, as a reference traveling resistance, reference acceleration resistance-calculating means for calculating an acceleration resistance to be obtained when the vehicle is in the predetermined reference state, based on the detected acceleration of the vehicle, as a reference acceleration resistance, and correction value-calculating means for calculating a correction value based on the detected torque of the drive wheels, the calculated reference traveling resistance, and the calculated reference acceleration resistance, wherein the traveling resistance is calculated by correcting the reference traveling resistance using the calculated correction value.\n\nWith the configurations of these preferred embodiments, it is possible to obtain the same advantageous effects as provided by the respective corresponding preferred embodiments of the first aspect of the present invention.\n\nThe above and other objects, features, and advantages of the present invention will become more apparent from the following detailed description taken in conjunction with the accompanying drawings.\n\n# BRIEF DESCRIPTION OF THE DRAWINGS\n\nFIG. 1 is a schematic view showing an internal combustion engine to which is applied a control system according to the present embodiment, together with a vehicle having the engine mounted thereon;\n\nFIG. 2 is a schematic view of the engine appearing in FIG. 1;\n\nFIG. 3 is a schematic-block diagram of the control system;\n\nFIG. 4 is a schematic cross-sectional view of a variable intake valve-actuating mechanism and an exhaust valve-actuating mechanism of the engine;\n\nFIG. 5 is a schematic cross-sectional view of a variable valve lift mechanism of the variable intake valve-actuating mechanism;\n\nFIG. 6A is a diagram showing a lift actuator in a state in which a short arm thereof is in a maximum lift position;\n\nFIG. 6B is a diagram showing the lift actuator in a state in which the short arm thereof is in a minimum lift position;\n\nFIG. 7A is a diagram showing an intake valve placed in an open state when a lower link of the variable valve lift mechanism is in a maximum lift position;\n\nFIG. 7B is a diagram showing the intake valve placed in the open state when the lower link of the variable valve lift mechanism is in a minimum lift position;\n\nFIG. 8 is a diagram showing a valve lift curve (solid line) of the intake valve obtained when the lower link of the variable valve lift mechanism is in the maximum lift position, and a valve lift curve (two-dot chain line) of the intake valve obtained when the lower link of the variable valve lift mechanism is in the minimum lift position;\n\nFIG. 9 is a schematic diagram of a variable cam phase mechanism;\n\nFIG. 10 is a diagram showing a valve lift curve (solid line) obtained when a cam phase is set to a most retarded value by the variable cam phase mechanism, and a valve lift curve (two-dot chain line) obtained when the cam phase is set to a most advanced value by the variable cam phase mechanism;\n\nFIG. 11 is a flowchart of a process for calculating an engine demanded output;\n\nFIG. 12 is a diagram showing an example of a map for use in calculating the engine demanded output;\n\nFIG. 13 is a flowchart of a process for calculating an acceleration demand reference value;\n\nFIG. 14 is a flowchart of a process for calculating a traveling resistance;\n\nFIG. 15 is a flowchart of a process for calculating a phase control input;\n\nFIG. 16 is a diagram showing an example of a fuel economy map;\n\nFIG. 17 is a diagram showing an example of an output map;\n\nFIG. 18 is a flowchart of a process for calculating a lift control input;\n\nFIG. 19 is a diagram showing an example of a Liftincmd map;\n\nFIG. 20A is a view showing results of control in which a target cam phase is calculated using only the fuel economy map;\n\nFIG. 20B is a view showing an example of the results of control by the control system according to the present embodiment;\n\nFIG. 21 is a view showing a fuel economy ratio obtained through control by the control system according to the present embodiment together with a fuel economy ratio obtained using the fuel economy map alone, and a fuel economy ratio obtained using the output map alone;\n\nFIG. 22 is a flowchart of a variation of the process for calculating the acceleration demand reference value;\n\nFIG. 23 is a view showing an example of changes in demanded acceleration and an acceleration demand reference value;\n\nFIG. 24 is a flowchart of a process for calculating an acceleration demand reference value, according to a second embodiment of the present invention;\n\nFIG. 25 is a view showing an example of a G_juda table which is used in the FIG. 24 process; and\n\nFIG. 26 is a flowchart of a process for calculating a phase control input, according to the second embodiment.\n\n# DETAILED DESCRIPTION OF PREFERRED EMBODIMENTS\n\nThe invention will now be described in detail with reference to the drawings showing preferred embodiments thereof. FIG. 1 schematically shows an internal combustion engine (hereinafter simply referred to as “the engine”) 3 to which is applied a control system 1 according to the present embodiment, and a vehicle V having the engine 3 mounted thereon as a drive source. The vehicle V has a transmission 80 installed thereon. The transmission 80 is of a manual type for transmitting power from the engine 3 to drive wheels W and W while changing the rotational speed at one of a plurality of predetermined transmission ratios. Further, the transmission 80 is configured such that it selectively sets six gear positions formed by first to fifth speed gear positions and a reverse gear position, and the operation of the transmission 80 is controlled by an ECU 2, described hereinafter, of the control system 1 according to the shift position of a shift lever (not shown) operated by a driver (see FIG. 3).\n\nAs shown in FIGS. 2 and 4, the engine 3 is an in-line four-cylinder DOHC gasoline engine having four cylinders 3a and pistons 3b (only one of which is shown). Further, the engine 3 includes an intake valve 4 and an exhaust valve 7 for opening and closing an intake port and an exhaust port of each cylinder 3a, respectively, a variable intake valve-actuating mechanism 40 having an intake camshaft 5 and intake cams 6 for actuating the intake valves 4, and an exhaust valve-actuating mechanism 30 having an exhaust camshaft 8 and exhaust cams 9 for actuating the exhaust valves 7, fuel injection valves 10, and spark plugs 11 (see FIG. 3).\n\nThe intake valve 4 has a stem 4a thereof slidably fitted in a guide 4b. The guide 4b is rigidly fixed to a cylinder head 3c. The intake valve 4 includes upper and lower spring sheets 4c and 4d, and a valve spring 4e disposed therebetween (see FIG. 5), and is urged by the valve spring 4e in the valve-closing direction.\n\nThe intake camshaft 5 and the exhaust camshaft 8 are rotatably mounted through the cylinder head 3c via respective holders, not shown. Further, an intake sprocket, not shown, is coaxially mounted on one end of the intake camshaft 5 in a rotatable manner. The intake sprocket is connected to a crankshaft 3d by a timing belt, not shown, and to the intake camshaft 5 via a variable cam phase mechanism 70, described hereinafter. With the above configuration, the intake camshaft 5 performs one rotation per two rotations of the crankshaft 3d. The intake cam 6 is integrally formed on the intake camshaft 5 for each cylinder 3a.\n\nThe variable intake valve-actuating mechanism 40 is provided for actuating the intake valve 4 of each cylinder 3a so as to open and close the same, in accordance with rotation of the intake cam 6, and continuously changing the lift and valve timing of the intake valve 4, which will be described in detail hereinafter. It should be noted that in the present embodiment, the lift of the intake valve 4 (hereinafter referred to as “the valve lift”) Liftin represents the maximum stroke of the intake valve 4.\n\nThe exhaust valve 7 has a stem 7a thereof slidably fitted in a guide 7b. The guide 7b is rigidly fixed to the cylinder head 3c. Further, the exhaust valve 7 is provided with upper and lower spring sheets 7c and 7d, and a valve spring 7e disposed therebetween, and is urged by the valve spring 7e in the valve-closing direction.\n\nThe exhaust camshaft 8 has an exhaust sprocket, not shown, integrally formed therewith, and is connected to the crankshaft 3d by the exhaust sprocket and the timing belt, not shown, whereby the exhaust camshaft 8 performs one rotation per two rotations of the crankshaft 3d. The exhaust cam 9 is integrally formed on the exhaust camshaft 8 for each cylinder 3a.\n\nThe exhaust valve-actuating mechanism 30 includes rocker arms 31. Each rocker arm 31 is pivotally moved in accordance with rotation of the associated exhaust cam 9 to thereby actuate the exhaust valve 7 for opening and closing the same against the urging force of the valve spring 7e.\n\nThe fuel injection valve 10 is provided for each cylinder 3a, and is mounted through the cylinder head 3c in a tilted state such that fuel is directly injected into a combustion chamber. That is, the engine 3 is configured as a direct injection engine. Further, the valve-opening time period and the valve-opening timing of the fuel injection valve 10 are controlled by the ECU 2.\n\nThe spark plugs 11 as well are provided in association with the respective cylinders 3a, and are mounted through the cylinder head 3c. The ignition timing of each spark plug 11 is also controlled by the ECU 2.\n\nThe engine 3 is provided with a crank angle sensor 20. The crank angle sensor 20 is comprised of a magnet rotor and an MRE (magnetic resistance element) pickup, and delivers a CRK signal and a TDC signal, which are both pulse signals, to the ECU 2 in accordance with rotation of the crankshaft 3d.\n\nThe CRK signal is delivered whenever the crankshaft 3d rotates through a predetermined angle (e.g. 10°). The ECU 2 calculates the rotational speed NE of the engine 3 (hereinafter referred to as “the engine speed NE”) based on the CRK signal. The TDC signal indicates that each piston 3b in the associated cylinder 3a is in a predetermined crank angle position slightly before the TDC position at the start of the intake stroke, and in the illustrated example of the four-cylinder type engine, the TDC signal is delivered whenever the crankshaft 3d rotates through a predetermined crank angle of 180°\n\nFurther, the engine 3 has an intake pipe 12 provided with no throttle valve mechanism. An intake passage 12a through the intake pipe 12 is formed to have a large diameter, whereby the engine 3 is configured such that air flow resistance is smaller than an ordinary engine. Further, the intake pipe 12 is provided with an air flow sensor 21. The air flow sensor 21 is formed by a hot-wire air flow meter, and detects an amount QA of intake air drawn into the engine to deliver a signal indicative of the sensed intake air amount QA to the ECU 2.\n\nNext, the aforementioned variable intake valve-actuating mechanism 40 will be described with reference to FIGS. 5 to 8. The variable intake valve-actuating mechanism 40 is comprised of the intake camshaft 5, the intake cams 6, a variable valve lift mechanism 50 (first variable valve-actuating mechanism), and the variable cam phase mechanism 70 (second variable valve-actuating mechanism).\n\nThe variable valve lift mechanism 50 is provided for actuating the intake valves 4 to open and close the same, in accordance with rotation of the intake cams 6, and continuously changing the valve lift Liftin between a predetermined maximum value Liftin_H and a predetermined minimum value Liftin_L. The variable valve lift mechanism 50 is comprised of rocker arm mechanisms 51 of a four joint link type, provided for the respective cylinders 3a, and a lift actuator 60 simultaneously actuating these rocker arm mechanisms 51.\n\nEach rocker arm mechanism 51 is comprised of a rocker arm 52, and upper and lower links 53 and 54. The upper link 53 has one end pivotally mounted to a rocker arm shaft 56 fixed to the cylinder head 3c, and the other end pivotally mounted to an upper end of the rocker arm 52 by an upper pin 55.\n\nFurther, a roller 57 is pivotally disposed on the upper pin 55 of the rocker arm 52. The roller 57 is in contact with a cam surface of the intake cam 6. As the intake cam 6 rotates, the roller 57 rolls on the intake cam 6 while being guided by the cam surface of the intake cam 6. As a result, the rocker arm 52 is vertically driven, and the upper link 53 is pivotally moved about the rocker arm shaft 56.\n\nFurthermore, an adjusting bolt 52a is mounted to an end of the rocker arm 52 toward the intake valve 4. The adjusting bolt 52a is in contact with a stem 4e of the intake valve 4 and when the rocker arm 52 is vertically moved in accordance with rotation of the intake cam 6, the adjusting bolt 52a vertically drives the stem 4a to open and close the intake valve 4, against the urging force of the valve spring 4e.\n\nFurther, the lower link 54 has one end pivotally mounted to a lower end of the rocker arm 52 by a lower pin 58, and the other end of the lower link 54 has a connection shaft 59 pivotally mounted thereto. The lower link 54 is connected to a short arm 65, described hereinafter, of the lift actuator 60 by the connection shaft 59.\n\nAs shown in FIG. 6, the lift actuator 60, which is driven by the ECU 2, is comprised of a motor 61, a nut 62, a link 63, a long arm 64, and the short arm 65. The motor 61 is connected to the ECU 2, and disposed outside a head cover 3g of the engine 3. The rotational shaft of the motor 61 is a screw shaft 61a formed with a male screw and the nut 62 is screwed onto the screw shaft 61a. The link 63 has one end pivotally mounted to the nut 62 by a pin 63a, and the other end pivotally mounted to one end of the long arm 64 by a pin 63b. Further, the other end of the long arm 64 is attached to one end of the short arm 65 by a pivot shaft 66. The pivot shaft 66 is circular in cross section, and is pivotally supported by the head cover 3g of the engine 3. The long arm 64 and the short arm 65 are pivotally moved about the pivot shaft 66 in unison with the pivot shaft 66.\n\nFurthermore, the aforementioned connection shaft 59 pivotally extends through an end of the short arm 65 on a side opposite to the pivot shaft 66, whereby the short arm 65 is connected to the lower link 54 by the connection shaft 59.\n\nNext, a description will be given of the operation of the variable valve lift mechanism 50 configured as above. In the variable valve lift mechanism 50, when a lift control input Uliftin (first control input), described hereinafter, is input from the ECU 2 to the lift actuator 60, the screw shaft 61a of the motor 61 rotates, and the nut 62 is moved in accordance with the rotation of the screw shaft 61a, whereby the long arm 64 and the short arm 65 are pivotally moved about the pivot shaft 66, and in accordance with the motion of the connecting shaft 59 caused by the pivotal motion of the short arm 65, the lower link 54 of the rocker arm mechanism 51 is pivotally moved about the lower pin 58. That is, the lower link 54 is driven by the lift actuator 60.\n\nDuring the above process, under the control of the ECU 2, the range of pivotal motion of the short arm 65 is restricted between the maximum lift position shown in FIG. 6A and the minimum lift position shown in FIG. 6B, whereby the range of pivotal motion of the lower link 54 is also restricted between the maximum lift position indicated by a solid line in FIG. 5 and the minimum lift position indicated by a two-dot chain line in FIG. 5.\n\nThe rocker arm mechanism 51 is configured such that when the lower link 54 is in the maximum lift position, the distance between the center of the upper pin 55 and the center of the lower pin 58 becomes longer than the distance between the center of the rocker arm shaft 56 and the center of the connection shaft 59, whereby as shown in FIG. 7A, when the intake cam 6 rotates, the amount of movement of the adjusting bolt 52a becomes larger than the amount of movement of a contact point where the intake cam 6 and the roller 57 are in contact with each other.\n\nOn the other hand, the rocker arm mechanism 51 is configured such that when the lower link 54 is in the minimum lift position, the distance between the center of the upper pin 55 and the center of the lower pin 58 becomes shorter than the distance between the center of the rocker arm shaft 56 and the center of the connection shaft 59, whereby as shown in FIG. 7B, when the intake cam 6 rotates, the amount of movement of the adjusting bolt 52a becomes smaller than the amount of movement of the contact point where the intake cam 6 and the roller 57 are in contact with each other.\n\nFor the above reason, when the lower link 54 is in the maximum lift position, the intake valve 4 is opened with a larger valve lift Liftin than when the lower link 54 is in the minimum lift position. More specifically, during rotation of the intake cam 6, when the lower link 54 is in the maximum lift position, the intake valve 4 is opened according to a valve lift curve indicated by a solid line in FIG. 8, and the valve lift Liftin assumes its maximum value Liftin_H. On the other hand, when the lower link 54 is in the minimum lift position, the intake valve 4 is opened according to a valve lift curve indicated by a two-dot chain line in FIG. 8, and the valve lift Liftin assumes its minimum value Liftin_L.\n\nAs described above, in the variable valve lift mechanism 50, the lower link 54 is pivotally moved by the lift actuator 60 between the maximum lift position and the minimum lift position, whereby it is possible to continuously change the valve lift Liftin between the maximum value Liftin_H and the minimum value Liftin_L. Further, as described above, the variable valve lift mechanism 50 uses the motor 61 as a drive source thereof, and hence the response speed of the pivot angle of the short arm 65 with respect to the lift control input Uliftin is relatively high.\n\nThe engine 3 is provided with a pivot angle sensor 22 (see FIG. 3). The pivot angle sensor 22 detects a pivot angle Olift of the short arm 65 and delivers a signal indicative of the detected pivot angle of the short arm 65 to the ECU 2. The pivot angle Olift of the short arm 65 indicates a position of the short arm 65 between the maximum lift position and the minimum lift position. The ECU 2 calculates the valve lift Liftin based on the pivot angle Olift.\n\nNext, the aforementioned variable cam phase mechanism 70 will be described with reference to FIGS. 9 and 10. The variable cam phase mechanism 70 is provided for continuously advancing or retarding the relative phase Cain of the intake camshaft 5 with respect to the crankshaft 3d (hereinafter referred to as “the cam phase Cain”) to thereby continuously change the valve timing of the intake valve 4, and is mounted on an intake sprocket-side end of the intake camshaft 5. As shown in FIG. 9, the variable cam phase mechanism 70 includes a housing 71, a three-bladed vane 72, an oil pressure pump 73, and a solenoid valve mechanism 74.\n\nThe housing 71 is integrally formed with the intake sprocket on the intake camshaft 5, and divided by three partition walls 71a formed at equal intervals. The vane 72 is coaxially mounted on the end of the intake camshaft 5 where the intake sprocket is mounted, such that the blades of the vane 72 radially extends outward from the intake camshaft 5, and are rotatably housed in the housing 71. Further, the housing 71 has three advance chambers 75 and three retard chambers 76 each formed between one of the partition walls 71a and one of the three blades of the vane 72.\n\nThe oil pressure pump 73 is a mechanically-driven type which is connected to the crankshaft 3d. As the crankshaft 3d rotates, the oil pressure pump 73 draws lubricating oil stored in an oil pan 3e of the engine 3 via a lower part of an oil passage 77c, for pressurization, and supplies the pressurized oil to the solenoid valve mechanism 74 via the remaining part of the oil passage 77c.\n\nThe solenoid valve mechanism 74 is formed by combining a spool valve mechanism 74a and a solenoid 74b, and is connected to the advance chambers 75 and the retard chambers 76 via an advance oil passage 77a and a retard oil passage 77b such that oil pressure supplied from the oil pressure pump 73 is delivered to the advance chambers 75 and the retard chambers 76 as advance oil pressure Pad and retard oil pressure Prt, respectively. The solenoid 74b of the solenoid valve mechanism 74 is connected to the ECU 2. When a phase control input Ucain (second control input), described hereinafter, is input from the ECU 2, the solenoid 74b moves a spool valve element of the spool valve mechanism 74a within a predetermined range of motion according to the phase control input Ucain to thereby change both the advance oil pressure Pad and the retard oil pressure Prt.\n\nIn the variable cam phase mechanism 70 configured as above, during operation of the oil pressure pump 73, the solenoid valve mechanism 74 is operated according to the phase control input Ucain, to supply the advance oil pressure Pad to the advance chambers 75 and the retard oil pressure Prt to the retard chambers 76, whereby the relative phase of the vane 72 with respect to the housing 71 is changed toward an advanced side or a retarded side. As a result, the cam phase Cain described above is continuously changed between a most retarded value Cainrt (value corresponding to a cam angle of e.g. 0°) and a most advanced value Cainad (value corresponding to a cam angle of e.g. 550), whereby the valve timing of the intake valves 4 is continuously changed between most retarded timing indicated by a solid line in FIG. 10 and most advanced timing indicated by a two-dot chain line in FIG. 10. Further, as described above, the variable cam phase mechanism 70 uses the oil pressure pump 73 as a drive source thereof, and hence the response speed of the cam phase Cain with respect to the phase control input Ucain is lower than the response speed of the pivot angle θ lift of the short arm 65 with respect to the lift control input Uliftin of the variable valve lift mechanism 50.\n\nAs described above, in the variable intake valve-actuating mechanism 40 of the present embodiment, the variable valve lift mechanism 50 continuously changes the valve lift Liftin, and the variable cam phase mechanism 70 continuously changes the cam phase Cain, i.e. the valve timing of the intake valves 4 between the most retarded timing and the most advanced timing, described hereinbefore. Further, as described hereinafter, the ECU 2 controls the valve lift Liftin and the cam phase Cain via the variable valve lift mechanism 50 and the variable cam phase mechanism 70.\n\nOn the other hand, a cam angle sensor 23 (see FIG. 3) (operation amount-detecting means) is disposed at an end of the intake camshaft 5 opposite from the variable cam phase mechanism 70. The cam angle sensor 23 is implemented e.g. by a magnet rotor and an MRE pickup, for delivering a CAM signal, which is a pulse signal, to the ECU 2 along with rotation of the intake camshaft 5. Each pulse of the CAM signal is generated whenever the intake camshaft 5 rotates through a predetermined cam angle (e.g. 1°). The ECU 2 calculates the cam phase Cain based on the CAM signal and the CRK signal, described above.\n\nFurther, a LAF sensor 24 is inserted into the exhaust pipe 13 of the engine 3. The LAF sensor 24 linearly detects the concentration of oxygen in exhaust gases, and delivers a signal indicative of the sensed oxygen concentration to the ECU 2. The ECU 2 calculates an actual air-fuel ratio A/F indicative of the air-fuel ratio of a mixture burned in the engine 3, based on the signal from the LAF sensor 24. It should be noted that the actual air-fuel ratio A/F is calculated as an equivalent ratio.\n\nFurther, an accelerator pedal opening sensor 25 detects the amount AP of operation (stepped-on amount) of an accelerator pedal, not shown (hereinafter referred to as “the accelerator pedal opening AP”), and delivers a signal indicative of the sensed accelerator pedal opening AP to the ECU 2. A vehicle speed sensor 26 (vehicle speed-detecting means) detects a vehicle speed VP, which is a traveling speed of the vehicle V, and delivers a signal indicative of the sensed vehicle speed VP to the ECU 2. Furthermore, the transmission 80 has a gear position sensor 27 mounted thereon. The gear position sensor 27 detects a gear position of the transmission 80, and delivers a signal indicative of a shift position NGR corresponding to the sensed gear position to the ECU 2. The ECU 2 calculates a transmission ratio G_ratio of the transmission 80 based on the shift position NGR.\n\nThe ECU 2 is implemented by a microcomputer comprised of a CPU, a RAM, a ROM and an I/O interface (none of which are specifically shown). The ECU 2 controls the operations of the engine 3 and the transmission 80 based on the signals from the aforementioned sensors 20 to 27.\n\nIt should be noted that in the present embodiment, the ECU 2 corresponds to first control input-calculating means, load parameter-detecting means, demanded acceleration degree parameter-calculating means, output priority-type calculating means, fuel economy priority-type calculating means, selection means, second control input-calculating means, drive wheel demanded torque-calculating means, traveling resistance-calculating means, drive wheel torque-detecting means, acceleration-detecting means, reference traveling resistance-calculating means, reference acceleration resistance-calculating means, and correction value-calculating means.\n\nNext, a description will be given of the outline of a control process executed by the ECU 2. First, the ECU 2 calculates an output Bmep_cmd (load parameter) demanded of the engine 3 (hereinafter referred to as “the engine demanded output Bmep_cmd”) (see FIG. 11), and an acceleration demand reference value G_jud indicative of the presence or absence of a demand of acceleration (see FIG. 13). Further, the ECU 2 calculates the phase control input Ucain for controlling the variable cam phase mechanism 70 e.g. based on the calculated the engine demanded output Bmep_cmd and acceleration demand reference value G_jud (see FIG. 15), and calculates the lift control input Uliftin for controlling the variable valve lift mechanism 50, based on the calculated cam phase Cain (see FIG. 18). It should be noted that all the processes described hereinafter will be executed at a predetermined control period T (e.g. 10 msec).\n\nIn the FIG. 11 process for calculating the engine demanded output Bmep_cmd, in a step 1 (shown as S1 in abbreviated form in FIG. 11; the following steps are also shown in abbreviated form), the engine demanded output Bmep_cmd is calculated by searching a map shown in FIG. 12 according to the engine speed NE and the accelerator pedal opening AP. In this map, the engine demanded output Bmep_cmd is set to a larger value as the engine speed NE is higher and the accelerator pedal opening AP is higher. It should be noted that the engine demanded output Bmep_cmd is calculated as a net average effective pressure.\n\nNext, the process for calculating the above-described acceleration demand reference value G_jud will be described with reference to FIG. 13. First, in a step 11, a torque Tq_eng_cmd demanded of the engine 3 (hereinafter referred to as “the engine demanded torque Tq_eng_cmd”) is calculated using the engine demanded output Bmep_cmd calculated in the step 1, by the following equation (1):\n\nTq_engcmd=(Bmepcmd×DI/(Stroke×π) (1)\n\nwherein DI represents the displacement of the engine 3, Stroke represents the number of strokes of the engine 3, which is equal to 4 in the present embodiment, and π represents the ratio of the circumference of a circle.\n\nThen, in a step 12, a torque Tq_tire_cmd demanded of the drive wheels W and W (hereinafter referred to as “the drive wheel demanded torque Tq_tire_cmd”) is calculated e.g. using the calculated engine demanded torque Tq_eng_cmd and the aforementioned transmission ratio G_ratio by the following equation (2):\n\nTq_tirecmd=Tq_engcmd×η×G_ratio×F_ratio/TireR (2)\n\nwherein η represents a predetermined power loss e.g. in the transmission 80, F_ratio represents a reduction ratio of a final speed reduction gear, not shown, and Tire_R represents the radius of the drive wheels W and W. It should be noted that the drive wheel demanded torque Tq_tire_cmd is represented by a force (N).\n\nNext, a traveling resistance RL of the vehicle is calculated (step 13). FIG. 14 shows the process for calculating the traveling resistance RL. First, in a step 21, an actual output Bmep_act of the engine 3 (hereinafter referred to as “the engine output Bmep_act”) is calculated. The engine output Bmep_act is calculated by searching a map, not shown, according to the intake air amount QA, the air-fuel ratio A/F, and the ignition timing. This map is formed by empirically determining the relationship between the engine output Bmep_act, the intake air amount QA, the air-fuel ratio A/F, and the ignition timing, and then mapping the relationship. It should be noted that the engine output Bmep_act is calculated as a net average effective pressure.\n\nThen, in a step 22, an actual torque Tq_eng_act of the engine 3 (hereinafter referred to as “the engine torque Tq_eng_act”) is calculated using the calculated engine output Bmep_act and the aforementioned engine displacement DI and stroke number Stroke, by the following equation (3):\n\nTq_eng_act=(Bmep_act×DI)/(Stroke×π) (3)\n\nNext, in a step 23, an actual torque Tq_tire_act of the drive wheels W and W (hereinafter referred to as “the drive wheel torque Tq_tire_act”) is calculated using the calculated engine torque Tq_eng_act, the transmission ratio G_ratio, the power loss η, the reduction ratio F_ratio of the final speed reduction gear, and the radius Tire_R of the drive wheels W and W, by the following equation (4):\n\nTq_tire_act=Tq_eng_act×η×G_ratio×F_ratio/TireR (4)\n\nThen, in a step 24, a reference rolling resistance Roll_r is calculated by the following equation (5):\n\nRollr=μR×Weight (5)\n\nwherein μR represents a value obtained by multiplying a friction coefficient obtained when the vehicle V travels on an asphalt road surface by the gravitational acceleration (hereinafter referred to as “the reference friction coefficient”), and is set to a predetermined value. Further, Weight represents the weight of the vehicle V with one occupant and no baggage loaded thereon (hereinafter referred to as “the reference vehicle weight”), and is set to a predetermined value.\n\nNext, in a step 25, reference air resistance Air r is calculated using the vehicle speed VP by the following equation (6):\n\nAirr=μA×A×VP2 (6)\n\nwherein μA and A represent an air resistance coefficient (hereinafter referred to as “the reference air resistance coefficient”) and a front projection area of the vehicle V (hereinafter referred to as “the reference front projection area”), which are obtained when the vehicle V has no spoiler or no carrier mounted thereon, respectively.\n\nThen, in a step 26, the acceleration Acc of the vehicle V (hereinafter referred to as “the vehicle acceleration Acc”) is calculated using the current value VP of the vehicle speed and the immediately preceding value VPZ of the same (vehicle speed VP obtained in the immediately preceding control timing), and the control period T of the present process, by the following equation (7):\n\nAcc=(VP−VPZ)/T (7)\n\nNext, a reference acceleration resistance Acc_r is calculated by dividing the calculated vehicle acceleration Acc by the reference vehicle weight Weight (step 27).\n\nThe reference acceleration resistance Acc_r corresponds to acceleration resistance obtained when the weight of the vehicle V is equal to the reference vehicle weight Weight.\n\nThen, the sum of the reference rolling resistance Roll_r obtained in the step 24 and the reference air resistance Air r obtained in the step 25 is calculated as a reference traveling resistance RL_base (step 28). The reference traveling resistance RL_base corresponds to a traveling resistance obtained when the weight of the vehicle V is equal to the reference vehicle weight Weight, the friction coefficient of a road surface to the reference friction coefficient OR, the air resistance coefficient to the reference air resistance coefficient μA, and the front projection area of the vehicle V to the reference front projection area A, and the road surface is horizontal (gradient=0, i.e. gradient resistance=0).\n\nNext, a total reference traveling resistance ALL_RL is calculated by adding the reference acceleration resistance Acc_r calculated in the above-described step 27 to the calculated reference traveling resistance RL_base (step 29).\n\nThen, using the calculated total reference traveling resistance ALL_RL and the drive wheel torque Tq_tire_act calculated in the step 23, a correction value RL_cor for correcting the reference traveling resistance RL_base is calculated as follows (step 30).\n\nFirst, the difference dRL (Tq_tire_act−ALL_RL) between the drive wheel torque Tq_tire_act and the total reference traveling resistance ALL_RL is calculated. Then, the correction value RL_cor is calculated using the calculated difference dRL by the following equation (8):\n\nRLcor=α×dRL+(1−α)×RLcorZ (8)\n\nwherein RL_corZ represents the immediately preceding value of the correction value, and α represents a predetermined averaging coefficient, which is set to 0.03, for example.\n\nAs described above, the correction value RL_cor is calculated as the weighted average of the difference dRL and the immediately preceding value RL_corZ of the correction value. The reason for calculating the correction value RL_cor is to eliminate the error of computations caused by temporary noises which can be contained in the signals output from the sensors, and further, because the actual traveling resistance is not instantaneously changed and hence it is possible to obtain an appropriate correction value RL_cor even by such a calculation as described above.\n\nIt should be noted that when the difference (VPZ−VP) between the immediately preceding value VPZ of the vehicle speed VP and the current value of the same is larger than a predetermined value, the correction value RL_cor is set to the immediately preceding value RL_corZ thereof without calculating the correction value RL_cor by the aforementioned equation (8). This is because in the above-mentioned case (VPZ−VP>the predetermined value), i.e. when the driver is decelerating the vehicle V by a brake operation, a deceleration resistance is contained in the reference acceleration resistance Acc_r calculated in the step 27, so that it is impossible to properly calculate the reference acceleration resistance Acc_r, which makes it impossible to properly calculate the correction value RL_cor.\n\nNext, the traveling resistance RL is calculated by adding the calculated correction value RL_cor to the reference traveling resistance RL_base (step 31), followed by terminating the present process.\n\nReferring again to FIG. 13, in a step 14 following the step 13, an acceleration G_cmd demanded of the engine 3 (hereinafter referred to as “the demanded acceleration G_cmd”) (demanded acceleration degree parameter) is calculated using the reference vehicle weight Weight, the drive wheel demanded torque Tq_tire_cmd calculated in the step 12, and the traveling resistance RL calculated in the step 31, by the following equation (9):\n\nGcmd=(Tq_tirecmd−RL)/Weight (9)\n\nNext, it is determined whether or not the calculated demanded acceleration G_cmd is smaller than a predetermined threshold value G_cmd_SH (step 15). The threshold value G_cmd_SH is set to a slightly larger value than acceleration obtained during slow acceleration and cruising of the vehicle. For example, it is set to 0.1 m/s2.\n\nIf the answer to the question of the step 15 is affirmative (YES), it is judged that acceleration is not demanded, and the acceleration demand reference value G_jud is set to 0 (calculated to 0) (step 16), followed by terminating the present process. On the other hand, if the answer to the question of the step 15 is negative (NO), it is judged that acceleration is demanded, and the acceleration demand reference value G_jud is set to 1 (calculated to 1) (step 17), followed by terminating the present process. As described above, when acceleration is demanded, the acceleration demand reference value G_jud is set to 1, and otherwise to 0.\n\nFIG. 15 shows a process for calculating the phase control input Ucain for use in controlling the aforementioned variable cam phase mechanism 70. First, in a step 41, it is determined whether or not the acceleration demand reference value G_jud set in the step 16 or 17 is equal to 0. If the answer to this question is affirmative (YES), i.e. if acceleration is not demanded, a fuel economy map value Cain_M_FC is calculated by searching a fuel economy map shown in FIG. 16 according to the engine speed NE and the engine demanded output Bmep_cmd, and the calculated fuel economy map value Cain_M FC is set to a target cam phase Cain_cmd (step 42).\n\nIn the above-described fuel economy map, the fuel economy map value Cain_N_FC is set to a retarded value so as to ensure stable combustion in a low load region where the engine demanded output Bmep_cmd is small and a low engine speed region where the engine speed NE is low. Further, in a medium or medium-to-high engine speed region where the engine demanded output Bmep_cmd is medium, the fuel economy map value Cain_M_FC is set to an advanced value so as to improve fuel economy by reducing the pumping loss due to internal EGR. Furthermore, in a high load region where the engine demanded output Bmep_cmd is large, the fuel economy map value Cain_M_FC is set to a retarded value so as to ensure a large amount of fresh (intake) air.\n\nOn the other hand, if the answer to the question of the step 41 is negative (NO), i.e. if the acceleration demand reference value G_jud is equal to 1, which means that acceleration is demanded, an output map value Cain_M P is calculated by searching an output map shown in FIG. 17 according to the engine speed NE and the engine demanded output Bmep_cmd, and the calculated output map value Cain_N_P is set to the target cam phase Cain_cmd (step 43). In this output map, the output map value Cain_N_P is basically set to have the same tendency as that of the fuel economy map value Cain_N_FC with respect to the engine speed NE and the engine demanded output Bmep_cmd, and is set to a more retarded value than the fuel economy map value Cain_N_FC, as a whole, so as to obtain a larger intake air amount QA for obtaining a larger output of the engine 3.\n\nIn a step 44 following the above-described step 42 or 43, the phase control input Ucain is calculated according to the difference between the calculated target cam phase Cain_cmd and the cam phase Cain with a predetermined feedback control algorithm, such as a PID control algorithm, followed by terminating the present process. The phase control input Ucain calculated as above is input to the variable cam phase mechanism 70, whereby the cam phase Cain is controlled such that it becomes equal to the target cam phase Cain_cmd.\n\nFIG. 18 shows a process for calculating the aforementioned lift control input Uliftin. First, in a step 51, a target valve lift Liftin_cmd is calculated by searching a Liftincmd map shown in FIG. 19 according to the engine speed NE, the engine demanded output Bmep_cmd, and the cam phase Cain. In this map, a plurality of predetermined values of the cam phase Cain (only one of which is shown in FIG. 19) are set between the most retarded value Cainrt and the most advanced value Cainad, mentioned hereinabove, and when the cam phase Cain is not equal to any of the predetermined values, the target valve lift Liftin_cmd is calculated by interpolation.\n\nThe Liftincmd map is prepared by empirically determining the relationship between the valve lift Liftin, the engine speed NE, the engine demanded output Bmep_cmd, and the cam phase Cain, and mapping the relationship. In the Liftincmd map, the target valve lift Liftin_cmd is set to a value enabling the engine demanded output Bmep_cmd to be obtained, with respect to the current cam phase Cain.\n\nMore specifically, the target valve lift Liftin_cmd is set to a larger value as the engine speed NE is higher, or as the engine demanded output Bmep_cmd is larger and the load on the engine 3 is higher, so as to obtain a larger intake air amount QA for obtaining a larger output of the engine 3. Further, the target valve lift Liftin_cmd is set to a larger value as the cam phase Cain is more advanced. This is because as the cam phase Cain is more advanced, the speed of each piston 3b is lower during opening of the intake valve 4, and the internal EGR amount is larger, whereby the intake air amount QA becomes smaller with respect to the same valve lift Liftin. Therefore, the target valve lift Liftin_cmd is set to a larger value so as to compensate for the smaller intake air amount QA.\n\nIn a step 52 following the step 51, the lift control input Uliftin is calculated according to the difference between the calculated target valve lift Liftin_cmd and the valve lift Liftin with a predetermined feedback control algorithm, such as a PID control algorithm, followed by terminating the present process. The lift control input Uliftin calculated as above is input to the variable valve lift mechanism 50, whereby the valve lift Liftin is controlled such that it becomes equal to the target valve lift Liftin_cmd.\n\nThe engine output Bmep_act is controlled to the engine demanded output Bmep_cmd through control using the phase control input Ucain and the lift control input Uliftin, described above.\n\nNext, the results of control by the control system 1 will be described in comparison with a comparative example with reference to FIGS. 20A and 20B. The comparative example shown in FIG. 20A is an example of results of control obtained when the target cam phase Cain_cmd is calculated using the aforementioned fuel economy map alone, similarly to the control by the conventional control system. According to the results of the comparison, although the engine demanded output Bmep_cmd is suddenly increased by a demand of acceleration, response delay of the variable cam phase mechanism 70 cannot be sufficiently compensated for and therefore the intake air amount QA cannot be sufficiently obtained. This causes response delay of the engine output Bmep_act with respect to the engine demanded output Bmep_cmd.\n\nIn contrast, as shown in FIG. 20B, according to the results of control by the control system 1, when the engine demanded output Bmep_cmd is suddenly increased by a demand of acceleration, the engine output Bmep_act quickly responds to the engine demanded output Bmep_cmd, whereby it was confirmed that it is possible to obtain high responsiveness of the engine output Bmep_act to a demand of acceleration.\n\nFurther, FIG. 21 shows a fuel economy ratio FEA (mile/gallon) under the control of the control system 1 together with fuel economy ratios FEFC and FEP obtained using a fuel economy map alone and an output map alone, respectively. As shown in FIG. 21, the fuel economy ratio FEA under the control of the control system 1 is larger than the fuel economy ratio FEP obtained using the output map alone, and substantially as large as the fuel economy ratio FEFC obtained using the fuel economy map alone, so that it was confirmed that it is possible to obtain excellent fuel economy of the engine 3.\n\nAs described above, according to the present embodiment, the phase control input Ucain is calculated based on the output map value Cain_M_P when acceleration is demanded, and otherwise, the phase control input Ucain is calculated based on the fuel economy map value Cain_M_FC (steps 43, 43 and 44). This makes it possible to ensure excellent fuel economy of the engine 3 and enhance the responsiveness of the output of the engine 3 when acceleration is demanded. Further, the lift control input Uliftin is calculated based on the detected cam phase Cain (steps 51 and 52), so that it is possible to compensate for the response delay of the variable cam phase mechanism 70 by controlling the intake air amount QA via the variable valve lift mechanism 50. This makes it possible to more excellently obtain the above-described effects, i.e. the effects of ensuring excellent fuel economy and enhancing the responsiveness of the output of the engine 3 when acceleration is demanded.\n\nFurthermore, the demanded acceleration G_cmd is calculated based on the difference between the drive wheel demanded torque Tq_tire_cmd and the traveling resistance RL, and hence it is possible to properly calculate the demanded acceleration G_cmd as a parameter indicative of the degree of acceleration demanded of the engine 3 (hereinafter referred to as “the demanded acceleration degree”). Further, the correction value RL_cor is calculated based on the difference between the drive wheel torque Tq_tire_act and the total reference traveling resistance ALL_RL, which is the sum of the reference traveling resistance RL_base and the reference acceleration resistance Acc_r (steps 29 and 30), and the reference traveling resistance RL_base is corrected by the calculated correction value RL_cor, whereby the traveling resistance RL is calculated (step 31). This makes it possible to accurately calculate the traveling resistance RL with reference to the reference traveling resistance RL_base. Further, as is apparent from the method of calculating the traveling resistance RL, it is possible to calculate the traveling resistance RL only by computations, without requiring values obtained by detecting the actual weight of the vehicle V and the gradient of a road surface. This makes it possible to dispense with sensors for detecting the above values, thereby making it possible to reduce the manufacturing costs of the control system 1.\n\nNext, a variation of the process for calculating the acceleration demand reference value will be described with reference to FIG. 22. The present process is distinguished from the aforementioned FIG. 13 process only in the method of calculating the acceleration demand reference value G_jud. In FIG. 22, steps identical to those of the process in FIG. 13 are designated by the same step numbers. Further, as is apparent from FIG. 22, steps 15 et seq. are different, and hence the steps 15 et seq. will be described hereinafter with reference to FIG. 22.\n\nIf the answer to the question of the step 15 is affirmative (YES) (G_cmd<G_cmd_SH), the acceleration demand reference value G_jud currently obtained is set to the immediately preceding value G_judZ of the acceleration demand reference value (step 61). Then, the current value G_jud of the same is calculated by subtracting a predetermined value G_jref (e.g. 0.01) from the immediately preceding value G_judZ set as above (step 62).\n\nNext, it is determined whether or not the calculated acceleration demand reference value G_jud is not larger than 0 (step 63). If the answer to this question is negative (NO), the present process is immediately terminated, whereas if the answer to the question is affirmative (YES), it is judged that acceleration is not demanded, and the aforementioned step 16 is executed to set the acceleration demand reference value G_jud to 0, followed by terminating the present process.\n\nAs described above, in the present process, as shown by time points t1 and t2 appearing in FIG. 23, even when the demanded acceleration G_cmd becomes lower than the threshold value G_cmd_SH, differently from the first embodiment described above, the acceleration demand reference value G_jud is not immediately set to 0 but is calculated such that it becomes equal to a value obtained by subtracting the predetermined value G_jref from 1 in each control timing of the present process (steps 61 and 62). Then, when a state in which the demanded acceleration G_cmd is lower than the threshold value G_cmd SH continues to some extent until the value obtained by subtracting the predetermined value G_jref from the immediately preceding value G_judZ of the acceleration demand reference value becomes not larger than 0 (YES to step 63), the acceleration demand reference value G_jud is held at 0 insofar as G_cmd<G_cmd_SH holds (step 16) (after t2 et seq.).\n\nThe acceleration demand reference value G_jud is calculated as above for the following reason: As described hereinabove, the demanded acceleration G_cmd is calculated using the engine demanded output Bmep_cmd calculated based on the accelerator pedal opening AP. In contrast, in the vehicle equipped with the transmission 80 of the manual type as in the present embodiment, normally, the driver takes his foot off the accelerator pedal during a shift change, and hence the accelerator pedal opening AP becomes equal to 0. Therefore, when the shift change is executed while acceleration is demanded, the accelerator pedal opening AP becomes equal to 0, as described above, whereby the demanded acceleration G_cmd sometimes takes a negative value. In such a case, the acceleration demand reference value G_jud is not immediately set to 0 to thereby maintain the control of the cam phase Cain using the output map.\n\nFurther, a time period over which is maintained the control of the cam phase Cain using the output map is determined according to the magnitude of the predetermined value G_jref. Therefore, the predetermined value G_jref is set such that the time period required for maintaining the control of the cam phase Cain using the output map becomes slightly longer than a time period which it takes before the driver steps on the accelerator pedal after he temporarily takes his foot off the accelerator pedal for a shift change. This makes it possible to prevent the control of the cam phase Cain using the output map from being uselessly switched to the control of the cam phase Cain using the fuel economy map during a shift change, whereby it is possible to maintain the control of the cam phase Cain using the output map.\n\nIt should be noted that the lapse of time after the demanded acceleration G_cmd has become lower than the threshold value G_cmd_SH is measured by time measuring means, such as a timer, and when the lapse of time exceeds a predetermined time period, the acceleration demand reference value G_jud may be set to 0. In this case, by setting the predetermined time period to be slightly longer than the time period which it takes before the driver steps on the accelerator pedal after he temporarily takes his foot off the accelerator pedal for a shift change, it is possible to obtain the same advantageous effects as described hereinabove.\n\nNext, a second embodiment of the present invention will be described. The present embodiment is distinguished from the first embodiment only in the process for calculating the acceleration demand reference value and the process for calculating the phase control input. More specifically, the present embodiment is distinguished from the first embodiment only in the method of calculating an acceleration demand reference value G_juda and the method of calculating the target cam phase Cain_cmd are different, and hereinafter, a description will be mainly given of points different from the first embodiment. In the present embodiment, the acceleration demand reference value G_juda corresponds to “weight dependent on the demanded acceleration degree parameter”.\n\nFirst, the process for calculating the acceleration demand reference value will be described with reference to FIG. 24. In a step following the step 14, the acceleration demand reference value G_juda is calculated by searching a G_juda table shown in FIG. 25 according to the demanded acceleration G_cmd, followed by terminating the present process. In FIG. 24, Gmin and Gmax represent the minimum value and the maximum value of the demanded acceleration G_cmd, respectively.\n\nIn the G_juda table, the acceleration demand reference value G_juda is set to 0 within a range where the demanded acceleration G_cmd is smaller than the threshold value G_cmd_SH, whereas within a range where the demanded acceleration G_cmd is larger than a predetermined value Gref (>G_cmd_SH), the acceleration demand reference value G_juda is set to 1. Further, within a range where G_cmd_SH≦G_cmd≦Gref holds, the acceleration demand reference value G_juda is linearly set to a larger value as the demanded acceleration G_cmd is larger. The predetermined value Gref is set to a value slightly smaller than the maximum value Gmax.\n\nNext, the process for calculating the phase control input will be described with reference to FIG. 26. First, in a step 81, similarly to the aforementioned step 42, the fuel economy map value Cain_M_FC is calculated by searching the FIG. 16 fuel economy map described above according to the engine speed NE and the engine demanded output Bmep_cmd. Then, similarly to the aforementioned step 43, the output map value Cain_M_P is calculated by searching the FIG. 17 output map described above according to the engine speed NE and the engine demanded output Bmep_cmd (step 82).\n\nNext, in a step 83, the target cam phase Cain_cmd is calculated using the acceleration demand reference value G_juda calculated in the aforementioned step 71, and the fuel economy map value Cain_N_FC and the output map value Cain_M P calculated in the steps 81 and 82, respectively, by the following equation (10):\n\nCaincmd=Gjuda×Cain_MP+(1−GjudaCainMFC (10)\n\nThen, the aforementioned step 44 is executed, whereby the phase control input Ucain is calculated based on the calculated target cam phase Cain_cmd, followed by terminating the present process.\n\nAs is apparent from the equation (10), the target cam phase Cain_cmd for use in calculation of the phase control input Ucain is calculated by calculating the weighted average of the output map value Cain_N_P and the fuel economy map value Cain_N_FC, using the acceleration demand reference value G_juda as a weighting coefficient, with the output map value Cain_N_P being multiplied by the acceleration demand reference value G_juda, and the fuel economy map value Cain_N_FC being multiplied by (1−G_juda). Further, as is apparent from the settings of the above-described G_juda table, the acceleration demand reference value G_juda is calculated such that it becomes equal to 0 when the demanded acceleration G_cmd is smaller than the threshold value G_cmd_SH, whereas when the demanded acceleration G_cmd is not smaller than the threshold value G_cmd_SH, the acceleration demand reference value G_juda is linearly calculated such that it is a larger value within a range between 0 and 1 as the demanded acceleration G_cmd is larger.\n\nAs is apparent from the above description, when the demanded acceleration G_cmd is not smaller than the threshold value G_cmd_SH, the weight of the output map value Cain_M_P with respect to the target cam phase Cain_cmd becomes larger as the acceleration demand reference value G_juda is larger, i.e. as the demanded acceleration degree is larger. Therefore, when acceleration is demanded, the weight of the output map value Cain_M P with respect to the target cam phase Cain_cmd becomes larger, so that it is possible to enhance the responsiveness of the output of the engine 3. Further, when G_cmd<G_cmd_SH holds, i.e. when acceleration is not demanded, the target cam phase Cain_cmd can be set to the fuel economy map value Cain_M_FC, whereby it is possible to obtain excellent fuel economy of the engine 3. From the above, similarly to the above-described first embodiment, it is possible to ensure excellent fuel economy to enhance the responsiveness of the output of the engine 3 when acceleration is demanded.\n\nFurther, as described above, as the demanded acceleration degree is larger, the weight of the output map value Cain_M_P with respect to the target cam phase Cain_cmd becomes larger, so that it is possible to obtain an appropriate target cam phase Cain_cmd that matches the magnitude of the degree of acceleration. This makes it possible to ensure excellent fuel economy and enhance the responsiveness of the output of the engine 3 when acceleration is demanded, in a well balanced manner.\n\nIt should be noted that although in the above-described G_juda table, a section for setting the acceleration demand reference value G_juda to a value between 0 and 1 is defined by the threshold value G_cmd_SH and the predetermined value Gref, this is not limitative, but the section may be defined by other desired values insofar as the values are larger than 0 and smaller than the maximum value Gmax.\n\nIt should be noted that the present invention is by no means limited to the embodiments described above, but it can be practiced in various forms. For example, although in the above-described embodiments, the present invention is applied to the engine 3 of a type in which the response speed of the variable valve lift mechanism 50 is higher than that of the variable cam phase mechanism 70, by way of example, this is not limitative, but inversely, the present invention can be applied to an internal combustion engine of a type in which the response speed of a variable cam phase mechanism is higher than that of a variable valve lift mechanism. Further, although in the above-described embodiments, the first and second variable valve-actuating mechanisms are formed by the variable valve lift mechanism 50 and the variable cam phase mechanism 70, respectively, this is not limitative, but the first and second variable valve-actuating mechanisms may be formed by other mechanisms insofar as they are capable of changing the intake air amount QA by changing the operating characteristics of the intake valves 4. Furthermore, although in the above-described embodiments, the present invention is applied to the engine 3 in which the valve lift Liftin and the valve timing of the intake valves 4 are changed by the variable valve lift mechanism 50 and the variable cam phase mechanism 70, respectively, by way of example, this is not limitative, but the present invention can be applied to an internal combustion engine of a type in which the same operating characteristics of the intake valves 4 are changed by two mechanisms with different response speeds.\n\nFurther, although in the above-described embodiments, the target cam phase Cain_cmd is determined by searching the fuel economy map or the output map, the target cam phase Cain_cmd may be calculated by computation without using the maps. Furthermore, although in the above-described embodiments, as a load parameter, the engine demanded output Bmep_cmd, which is calculated, is used, the accelerator pedal opening AP, which is detected, may be used. Further, although in the above-described embodiments, the drive wheel torque Tq_tire_act and the vehicle acceleration Acc are determined by computation, they may be determined by detection using sensors. Further, the demanded acceleration G_cmd may be calculated by another method in place of the method employed in the above-described embodiments. For example, the demanded acceleration G_cmd may be calculated based on the amount of change in the accelerator pedal opening AP. Further, the traveling resistance RL as well may be calculated by another method in place of the method according to the above-described embodiments. For example, the weight of the vehicle V and the gradients of road surfaces may be detected by sensors for calculating the traveling resistance RL based on the detected values.\n\nFurther, although in the above-described embodiments, the present invention is applied to the automotive engine 3 by way of example, this is not limitative, but it can be applied to various types of industrial internal combustion engines including engines for ship propulsion machines, such as an outboard motor having a vertically-disposed crankshaft.\n\nIt is further understood by those skilled in the art that the foregoing are preferred embodiments of the invention, and that various changes and modifications may be made without departing from the spirit and scope thereof." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.90885544,"math_prob":0.9687342,"size":82801,"snap":"2019-35-2019-39","text_gpt3_token_len":17097,"char_repetition_ratio":0.21719386,"word_repetition_ratio":0.35019106,"special_character_ratio":0.20139854,"punctuation_ratio":0.08058534,"nsfw_num_words":1,"has_unicode_error":false,"math_prob_llama3":0.97939193,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-09-18T00:41:02Z\",\"WARC-Record-ID\":\"<urn:uuid:4feb0957-a900-4258-910f-13b1c86baccf>\",\"Content-Length\":\"122292\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:92d2b9f2-b4dc-48c7-8ddc-9198ef729c58>\",\"WARC-Concurrent-To\":\"<urn:uuid:8076baa8-e797-4c55-95d7-92a545ba2c50>\",\"WARC-IP-Address\":\"144.202.252.20\",\"WARC-Target-URI\":\"http://www.freepatentsonline.com/y2008/0172165.html\",\"WARC-Payload-Digest\":\"sha1:JFDIKQA6OZRNTVFUI5UIHP6Q4DFZ4XEN\",\"WARC-Block-Digest\":\"sha1:VICQ24P525E5YZU7MR5IFEHJJSNUWHEK\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-39/CC-MAIN-2019-39_segments_1568514573173.68_warc_CC-MAIN-20190918003832-20190918025832-00167.warc.gz\"}"}
https://webapps.stackexchange.com/questions/149969/how-to-filter-based-on-multiple-cases/149971	[ "# How to filter based on multiple cases\n\nI am aware that I may be asking the wrong question here. Not sure how to ask, but I am trying to create a budget on google sheets that to record how much I spend on each day. Right now I have the transaction shown below:\n\n``````Date \| Category \| Amount\n--------------------------\nJan 1\| Dining \| \\$5\nJan 1\| Dining \| \\$30\nJan 1\| Gas \| \\$20\nJan 2\| Other \| \\$15\n``````\n\nI want to know how I could use the information I have above to create the table shown below:\n\n``````Date \| Dining\| Gas\| Other\n------------------------\nJan 1\| \\$35 \| \\$20\| \\$0\nJan 2\| \\$0 \| \\$0 \| \\$15\n``````\n\nI'm stuck trying to use a LOOKUP and SUMIF together.\n\n## 1 Answer\n\nThe manual solution would be to create a `Pivot` table of your data.\n\nHowever, if you are looking for a formula solution, you can create a pivot table by using a `query` formula:\n\n``````={query(A1:C,\"Select A, Sum(C) where A is not null group by A Pivot B limit 0\",1);\nArrayFormula((N(Query(query(A1:C,\"Select A, Sum(C) where A is not null group by A Pivot B\",1),\n\"Select * offset 1\",0))))}\n``````", null, "Or you could just use the simplest version of it:\n\n`=query(A1:C,\"Select A, Sum(C) where A is not null group by A Pivot B\",1)`\n\nbut you will get empty cells when the amounts are zero.\n\n• Correct Marios. I too would go for the formula. Still. I think you over complicated it. One could use a much simpler version like: `=query(A1:C,\"Select A, Sum(C) where A is not null group by A Pivot B\",1)` – marikamitsos Jan 10 at 18:07\n• @marikamitsos the OP wants to put zeros when the amount is 0\\$ so I just wanted to stick with that. But besides that then yes the single formula is much easier. Thanks a lot for your feedback :) – Mario Jan 10 at 19:33\n• Wow, this is perfect, and complicated. Thank you. Time to learn more about pivots, queries and arrayformula – Beginner Programmer Jan 11 at 0:11\n• @Marios Do you know how I could include dates that doesn't contain any value? For example if we change Jan 2 to Jan 3. Is it possible to have the Jan 2 row still existing with \\$0's? – Beginner Programmer Jan 12 at 18:28\n• I am not sure if that's possible. Could you post a new question so other readers would be able to help you out ? @BeginnerProgrammer – Mario Jan 12 at 18:31" ]	[ null, "https://i.stack.imgur.com/jmrRZ.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9059829,"math_prob":0.7043028,"size":586,"snap":"2021-04-2021-17","text_gpt3_token_len":164,"char_repetition_ratio":0.17353952,"word_repetition_ratio":0.0,"special_character_ratio":0.38737202,"punctuation_ratio":0.05263158,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9551249,"pos_list":[0,1,2],"im_url_duplicate_count":[null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-04-11T09:29:10Z\",\"WARC-Record-ID\":\"<urn:uuid:60f3d95d-ab19-42a6-9d4f-0aea1bc45bfa>\",\"Content-Length\":\"165923\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:5d845785-5183-4c2f-9483-834e6401fd9e>\",\"WARC-Concurrent-To\":\"<urn:uuid:354459c7-a151-4bb4-91a9-6bbb28a5a099>\",\"WARC-IP-Address\":\"151.101.1.69\",\"WARC-Target-URI\":\"https://webapps.stackexchange.com/questions/149969/how-to-filter-based-on-multiple-cases/149971\",\"WARC-Payload-Digest\":\"sha1:5ZCFSPQQ2BQXCNY7XUTO2W7US27JAA2Z\",\"WARC-Block-Digest\":\"sha1:PZURSSHRY2BP3AB4IEBR7SJBVHRPKYLU\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-17/CC-MAIN-2021-17_segments_1618038061820.19_warc_CC-MAIN-20210411085610-20210411115610-00144.warc.gz\"}"}
https://erj.ersjournals.com/highwire/markup/90747/expansion?width=1000&height=500&iframe=true&postprocessors=highwire_tables%2Chighwire_reclass%2Chighwire_figures%2Chighwire_math%2Chighwire_inline_linked_media%2Chighwire_embed	[ "Table 9—\n\nCombined analysis for maximal inspiratory and expiratory mouth pressure(PI,max and PE,max, respectively) in chronic obstructive pulmonary disease nonrandomised controlled trials\n\n First author [Ref.] Favours bilevel n Favours control n MD±se Weight % MD (random) 95% CI PI,max analysis Lien 26 11 11 0.4000±12.8142 3.35 0.40 (−24.72–25.52) Lin 27 10 12 5.0000±2.3839 96.65 5.00 (0.33–9.67) Total 21 23 100 4.85 (0.25–9.44) PE,max analysis Lien 26 11 11 1.0000±10.3830 4.40 1.00 (−19.35–21.35) Lin 27 10 12 5.0000±2.2265 95.60 5.00 (0.64–9.36) Total 21 23 100 4.82 (0.56–9.09)\n• MD: mean difference; CI confidence interval. Comparison: crossover trials of bilevel noninvasive positive pressure ventilation versus all modalities. Outcome: PI,max cmH2O or PE,max cmH2O. Test for heterogeneity in PI,max analysis: Chi squared = 0.12, df = 1 (p<0.72), I2 = 0%. Test for overall effect: Z = 2.07 (p = 0.04). Test for heterogeneity in PE,max analysis: Chi squared = 0.14, df = 1 (p<0.71), I2 = 0%. Test for overall effect: Z = 2.22 (p = 0.03)." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.62461656,"math_prob":0.89752203,"size":896,"snap":"2019-51-2020-05","text_gpt3_token_len":378,"char_repetition_ratio":0.10986547,"word_repetition_ratio":0.08695652,"special_character_ratio":0.54241073,"punctuation_ratio":0.25396827,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9512348,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-21T21:10:26Z\",\"WARC-Record-ID\":\"<urn:uuid:2d6775a1-c071-4772-9a1b-249ea87a9f6e>\",\"Content-Length\":\"10009\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:62920962-82df-4767-9d6d-6d284f3f9b4c>\",\"WARC-Concurrent-To\":\"<urn:uuid:68e99bbd-9ea4-4441-85e4-0a0d64820901>\",\"WARC-IP-Address\":\"104.16.192.19\",\"WARC-Target-URI\":\"https://erj.ersjournals.com/highwire/markup/90747/expansion?width=1000&height=500&iframe=true&postprocessors=highwire_tables%2Chighwire_reclass%2Chighwire_figures%2Chighwire_math%2Chighwire_inline_linked_media%2Chighwire_embed\",\"WARC-Payload-Digest\":\"sha1:AYYIYOZQY4F2KYVSM4VY7AMS2PJNHXJW\",\"WARC-Block-Digest\":\"sha1:JSX2RTQCDVFYUZIKQXJYY62U7ZV5DSXU\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579250605075.24_warc_CC-MAIN-20200121192553-20200121221553-00294.warc.gz\"}"}
https://forum.skewed.de/t/random-graph-with-degree-sequence/1037	[ "# Random graph with degree sequence.\n\nHi,\n\nI'm using graph-tool to try to generate random graphs with a sequence of\ndegrees. For example, in a 3-node graph, I generated a random graph with\nall nodes with input degrees 1 and output degrees 1.\n\nMy code:\n\nimport graph_tool.all as gt>>> def deg_sampler():... return 1,1... >>> g = gt.random_graph(3,deg_sampler,parallel_edges=True, self_loops=False)>>> gt.graph_draw(g)\n\nCan I generate a random graph defining the input and output degrees of each\nnode? For example, tree nodes with respectively the input degrees (1, 2, 0)\nand output degrees (1, 0, 2).\n\nThanks,\n\nAlvaro\n\nattachment.html (1.37 KB)\n\nNi!\nHi Alvaro, this is explained in the documentation for the\ngraph_tool.generation.random_graph\n<https://graph-tool.skewed.de/static/doc/generation.html#graph_tool.generation.random_graph>\nfunction concerning the `deg_sampler`:\n\nOptionally, you can also pass a function which receives one or two\narguments. If block_membership is None, the single argument passed will\nbe the index of the vertex which will receive the degree. If\nblock_membership is not None, the first value passed will be the vertex\nindex, and the second will be the block value of the vertex.\n\nCheers,\n.~´\n\nattachment.html (3.17 KB)" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9081308,"math_prob":0.91401094,"size":332,"snap":"2022-40-2023-06","text_gpt3_token_len":72,"char_repetition_ratio":0.15853658,"word_repetition_ratio":0.0,"special_character_ratio":0.19879518,"punctuation_ratio":0.10769231,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98178595,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-02-01T15:14:01Z\",\"WARC-Record-ID\":\"<urn:uuid:bb249af8-2756-4b26-9e28-4e2c6e383fe1>\",\"Content-Length\":\"15840\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:10e811c2-539a-4c91-a7fb-2a9dc333bd4f>\",\"WARC-Concurrent-To\":\"<urn:uuid:aa85671b-21a5-45f1-9812-08f8e7375a66>\",\"WARC-IP-Address\":\"49.12.93.243\",\"WARC-Target-URI\":\"https://forum.skewed.de/t/random-graph-with-degree-sequence/1037\",\"WARC-Payload-Digest\":\"sha1:NQRLBX2VCF3WI2WJZW6VFXIJ4LBNTP4Z\",\"WARC-Block-Digest\":\"sha1:SIRPNE77ME2QDZ35SIVVT6LNFFOIOUFE\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764499946.80_warc_CC-MAIN-20230201144459-20230201174459-00159.warc.gz\"}"}
https://tex.stackexchange.com/questions/174663/colorful-logic-gates-using-circuitikz	[ "# Colorful Logic Gates using Circuitikz\n\nI have to draw following circuit", null, "the LaTeX source of which is\n\n\\documentclass{beamer}\n\\usepackage{circuitikz}\n\\begin{document}\n\\begin{frame}\n\\begin{figure}\n\\centering\n\\begin{circuitikz} \\draw\n(0,0) node[not port] (not1) {}\n(0,2) node[xor port] (xor1) {}\n(0,4) node[and port] (and1) {}\n(2,3) node[nor port] (nor1) {}\n(4,2) node[xor port] (xor2) {}\n(6,3) node[or port] (or1) {}\n(6,1) node[and port] (and2) {}\n(nor1.in 1) node[above](f) {U}\n(nor1.in 2) node[below](g) {U}\n(xor2.in 1) node[above](h) {U}\n(xor2.in 2) node[below](i) {U}\n(or1.out) node[right](j) {U}\n(and1.out) -- (nor1.in 1)\n(xor1.out) -- (nor1.in 2)\n(nor1.out) -- (xor2.in 1)\n(not1.out) -- (xor2.in 2)\n(and1.out) -- (or1.in 1)\n(xor2.out) -- (or1.in 2)\n(xor1.out) -- (and2.in 1)\n(not1.out) -- (and2.in 2);\n\\end{circuitikz}\n\\end{figure}\n\\end{frame}\n\\end{document}\n\n\nThe circuit looks clumsy. What I want is that the gates whose inputs have been labelled U are drawn below the circuit such that the inputs wires of these circuits go downwards and then their outputs go upwards to the respective inputs.\n\nThe main question which I want to ask (which is also the title) is how to color the input and output wires of these gates red. By these gates I mean the gates whose inputs have been labelled U.\n\nWill be very thankful even if slight help is offered.\n\nThe diagram would be improved alot by using connecting wires that are only horizontal or vertical. There is a conveient syntax \|- for a path that is first vertical then horizontal, and the corresponding -\|. You can apply colours to the individual symbols and to the connecting wires by specifying color=red. Perhaps this is close to what you are after:", null, "\\documentclass{beamer}\n\\usepackage{circuitikz}\n\\begin{document}\n\\begin{frame}\n\\begin{figure}\n\\centering\n\\begin{circuitikz} \\draw\n(0,0) node[not port] (not1) {}\n(0,2) node[xor port] (xor1) {}\n(0,4) node[and port] (and1) {}\n(2,3) node[nor port,color=red] (nor1) {}\n(4,1) node[xor port,color=red] (xor2) {}\n(6,3) node[or port,color=red] (or1) {}\n(6,1) node[and port] (and2) {}\n(nor1.in 1) node[above](f) {U}\n(nor1.in 2) node[below](g) {U}\n(xor2.in 1) node[left](h) {U}\n(xor2.in 2) node[left](i) {U}\n(or1.out) node[right](j) {U};\n\\draw[color=red] (and1.out) \|- (nor1.in 1)\n(xor1.out) \|- (nor1.in 2)\n(nor1.out) -\| (xor2.in 1)\n(not1.out) -\| (xor2.in 2)\n(xor2.out) \|- (or1.in 2);\n\\draw\n(and1.out) -- +(4,0) \|- (or1.in 1)\n(xor1.out) -\| (and2.in 1)\n(not1.out) -\| (and2.in 2);\n\\end{circuitikz}\n\\end{figure}\n\\end{frame}\n\\end{document}\n\n\nAs the symbols are one unit, you can not (easily) change the colour an input wire right up to the symbol body.\n\n• Dear Friend, How can I show my gratitude! Thanks a lot. If possible would have given 100 bounty points, which is all I have. – kamalbanga May 1 '14 at 14:04\n• @kamalbanga after a few days, your question will be eligible for a bounty :) – cmhughes May 1 '14 at 15:14\n• @cmhughes I will never be using LaTeX after a few days since my student life ends in two days. So I better use up my points. :) – kamalbanga May 1 '14 at 15:19\n• Comments should not be used for discussions, but in that code you link to change (xor2.out) -\| (not2.in) to (xor2.out) \|- (not2.in) – Andrew Swann May 1 '14 at 15:55\n• Oh, silly mistake. Got it. – kamalbanga May 1 '14 at 15:57\n\nHere is a solution with the circuit libraries of TikZ. We need the extra -\|- and \|-\| styles from Vertical and horizontal lines in pgf-tikz because the elements from circuits.logic... don't have pins.\n\n\\documentclass{beamer}\n\\usepackage{tikz}\n\\usetikzlibrary{\ncircuits,\ncircuits.logic.IEC,\ncircuits.logic.US,\ncircuits.logic.CDH,\ncalc\n}\n\\tikzset{\n-\|-/.style={\nto path={\n(\\tikztostart) -\| ($(\\tikztostart)!#1!(\\tikztotarget)$) \|- (\\tikztotarget)\n\\tikztonodes\n}\n},\n-\|-/.default=0.5,\n\|-\|/.style={\nto path={\n(\\tikztostart) \|- ($(\\tikztostart)!#1!(\\tikztotarget)$) -\| (\\tikztotarget)\n\\tikztonodes\n}\n},\n\|-\|/.default=0.5,\n}\n\\begin{document}\n\\newcommand\\pic[circuit logic IEC]{\n\\begin{tikzpicture}[#1]\n\\draw\n(0,0) node[not gate] (not1) {}\n(0,2) node[xor gate] (xor1) {}\n(0,4) node[and gate] (and1) {}\n(2,3) node[red,nor gate] (nor1) {}\n(4,2) node[red,xor gate] (xor2) {}\n(6,3) node[red,or gate] (or1) {}\n(6,1) node[and gate] (and2) {};\n\\draw[red]\n(and1.output) to[-\|-] (nor1.input 1) node[above left] {U}\n(xor1.output) to[-\|-] (nor1.input 2) node[below left] {U}\n(nor1.output) to[-\|-] (xor2.input 1) node[above left] {U}\n(not1.output) to[-\|-] (xor2.input 2) node[below left] {U}\n(xor2.output) to[-\|-] (or1.input 2)\n(or1.output) node[right] {U};\n\\draw\n(and1.output) to[-\|-] (or1.input 1)\n(xor1.output) to[-\|-] (and2.input 1)\n(not1.output) to[-\|-] (and2.input 2);\n\\end{tikzpicture}\n}\n\\begin{frame}{circuit logic IEC}\n\\begin{figure}\n\\centering\n\\pic\n\\end{figure}\n\\end{frame}\n\\begin{frame}{circuit logic US}\n\\begin{figure}\n\\centering\n\\pic[circuit logic US]\n\\end{figure}\n\\end{frame}\n\\begin{frame}{circuit logic CDH}\n\\begin{figure}\n\\centering\n\\pic[circuit logic CDH]\n\\end{figure}\n\\end{frame}\n\\end{document}", null, "", null, "", null, "" ]	[ null, "https://i.stack.imgur.com/hZAF0.png", null, "https://i.stack.imgur.com/kIcg1.png", null, "https://i.stack.imgur.com/nggLW.png", null, "https://i.stack.imgur.com/QF75g.png", null, "https://i.stack.imgur.com/8Ua9k.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.68320626,"math_prob":0.9786308,"size":1307,"snap":"2019-43-2019-47","text_gpt3_token_len":463,"char_repetition_ratio":0.15886416,"word_repetition_ratio":0.04040404,"special_character_ratio":0.3573068,"punctuation_ratio":0.123636365,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9931104,"pos_list":[0,1,2,3,4,5,6,7,8,9,10],"im_url_duplicate_count":[null,1,null,1,null,1,null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-11-21T16:45:12Z\",\"WARC-Record-ID\":\"<urn:uuid:b68398dc-6ddc-40ab-84d3-dc396ae0b26a>\",\"Content-Length\":\"149116\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ce058499-4f83-4cf2-8b3f-93a60ad64080>\",\"WARC-Concurrent-To\":\"<urn:uuid:d0bf59af-fac2-40fb-99b1-103e30485e05>\",\"WARC-IP-Address\":\"151.101.65.69\",\"WARC-Target-URI\":\"https://tex.stackexchange.com/questions/174663/colorful-logic-gates-using-circuitikz\",\"WARC-Payload-Digest\":\"sha1:C5H2C3UEV2WOHW62RAFD3SMIGTSX6CC2\",\"WARC-Block-Digest\":\"sha1:IPPDVWY2VL3CFCHBSZCA6Q6K4MEM5ECX\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-47/CC-MAIN-2019-47_segments_1573496670921.20_warc_CC-MAIN-20191121153204-20191121181204-00367.warc.gz\"}"}
https://www.ltu.se/edu/course/M00/M0013M/M0013M-Matematik-M-1.68554?l=en&kursView=kursplan	[ "", null, "COURSE SYLLABUS\n\nMathematics 7.5 credits\n\nMatematik M\nFirst cycle, M0013M\nVersion\nCourse syllabus valid: Autumn 2021 Sp 1 - Present\nThe version indicates the term and period for which this course syllabus is valid. The most recent version of the course syllabus is shown first.\n\n Education level First cycle Grade scale G U 3 4 5 Subject Mathematics Subject group (SCB) Mathematics\n\nEntry requirements\n\nIn order to meet the general entry requirements for first cycle studies you must have successfully completed upper secondary education and documented skills in English language and Courses M0029M-M0031M or corresponding.\n\nSelection\n\nThe selection is based on 1-165 credits.\n\nCourse Aim\nAfter the course the student shall\n• be able to use key concepts for functions of several variables: limit, continuity, partial derivative, the chain rule, directional derivative, gradient and Taylor polynom\n• be able to find stationary points and classify them, determine the maximum and minimum values of continuous functions on closed bounded domains and be able to use the Lagrange multiplicator method.\n• be able to compute multiple integrals by interated integration and do suitable change of variables when it is needed.\n• be to compute and interpret line- and surface integrals\n• Be able to apply and interpret important concepts within vector calculus: Vector field, divergence, curl, Green’s theorem, divergence theorem and Stokes’ theorem.\n• be able to find the Fourier series corresponding to a periodic function and be able to find odd and even half range expansions of a given function.\n• Be able to derive som important partial differential equations (PDE) from known physical laws: wave equation, heat equation and Poisson’s equation.\n• be able to use the method of separation of variables to solve the above mentioned PDE for some simple geometries.\n• Be able to identify and solve problems which can be analyzed with the methods from the course and present the solutions in a logical and correct way and so that they are easy to follow.\n\nAn overall aim is that the student after the course besides being able to use the concepts and methods in the course also must be able to do the corresponding calculations with high accuracy, i.e. the final result should be correct.\n\nContents\n- Calculus in several variables: functions of several variables, partial differentiation, Taylor series, extreme values, multiple integration, line integrals, surface integrals, vector analysis (the divergence theorem and Stoke’s theorem). - Partial differential equations (PDE): Well known PDE (e.g. the wave equation, the heat equation, Laplace equation...) will be presented and discussed from an engineering point of view. Solution of PDE by separation of variables.\n\nRealization\nEach course occasion´s language and form is stated and appear on the course page on Luleå University of Technology's website.\nLectures and lessons.\n\nExamination\nIf there is a decision on special educational support, in accordance with the Guideline Student's rights and obligations at Luleå University of Technology, an adapted or alternative form of examination can be provided.\nYou have to pass a written exam. Grading scale: 3 4 5\n\nTransition terms\nThe course M0013M is equal to MAM235.\n\nExaminer\nPeter Wall\n\nTransition terms\nThe course M0013M is equal to MAM235\n\nLiterature. Valid from Autumn 2007 Sp 1 (May change until 10 weeks before course start)\nKreyszig E: Advanced Engineering Mathematics, latest edition.\n\nCourse offered by\nDepartment of Engineering Sciences and Mathematics\n\nModules" ]	[ null, "https://www.ltu.se/epok/public/images/logo_160x80_en.gif", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.87262243,"math_prob":0.89219207,"size":4469,"snap":"2022-05-2022-21","text_gpt3_token_len":953,"char_repetition_ratio":0.11959686,"word_repetition_ratio":0.005873715,"special_character_ratio":0.20765272,"punctuation_ratio":0.09622887,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9680353,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-01-18T08:18:21Z\",\"WARC-Record-ID\":\"<urn:uuid:4e3016cb-30d0-429f-8ddc-d352daa79beb>\",\"Content-Length\":\"140970\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:aeb32c65-336d-4742-8930-7219fb362145>\",\"WARC-Concurrent-To\":\"<urn:uuid:db5fc4e5-6f6d-4def-a15e-78f0f6da39ca>\",\"WARC-IP-Address\":\"130.240.43.25\",\"WARC-Target-URI\":\"https://www.ltu.se/edu/course/M00/M0013M/M0013M-Matematik-M-1.68554?l=en&kursView=kursplan\",\"WARC-Payload-Digest\":\"sha1:TMEPVAERQSXJWRMTJ6WUMY5LNXIYK3TS\",\"WARC-Block-Digest\":\"sha1:523L3VZMTY5KF4IFWMKJCIYVDDSENQ2W\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-05/CC-MAIN-2022-05_segments_1642320300805.79_warc_CC-MAIN-20220118062411-20220118092411-00277.warc.gz\"}"}
http://esalexandrias.gr/cambridge-vocabulary-zhzxjhk/differential-amplifier-working-animation-4fa8cc	[ "The differential amplifier uses two transistors in common emitter configuration. Instrumentation Amplifiers are basically used to amplify small differential signals. The circuit diagram of a differential amplifier using one opamp is shown below. How does the current source work to improve CMRR (reduce common-mode gain)? However, as is typical in most amplifiers, the larger signal, the more distorted it gets. Adding equations (5) and (9), we get the output voltage Vo, where Ad = differential gain and Ac = common mode gain. This site uses Akismet to reduce spam. A special implementation of Operational Amplifiers is the Instrumentation Amplifier, a type of Differential Amplifier with Input Buffer Amplifier. Differential Amplifier CSE 577 Spring 2011 Insoo Kim, Kyusun Choi Mixed Signal CHIP Design Lab. Single Input Unbalance Output- It is a type of configuration in which a single input is given an output is taken from only a single transistor. Difference- and common-mode signals. What is the maximum allowable base voltage if the differential input is large enough to completely steer the tail current? Figure 1 shows the basic differential amplifier. So CMRR value for this circuit to be infinite, Comparing equation (12) and (13), we have. + + + + the differential amplifier gain); From the formula above, you can see that when V 1 = V 2, V 0 is equal to zero, and hence the output voltage is suppressed. main application of Differential Amplifier is, it creates a difference between two input signals and then amplifies the differential signal. Therefore V+ = 0 V. Since the op-amp is ideal and negative feedback is present, the voltage of the inverting terminal (V−) is equal to the voltage of the non-inverting terminal (V+ = 0), according to the virtual ground concept. Privacy. Interactive animation shows how a transistor works. Difference between Amplifier and Oscillator, Difference Between Half Wave and Full Wave Rectifier, Difference Between Multiplexer (MUX) and Demultiplexer (DEMUX). Dual Input Unbalanced Output 4. So when the difference between terminals is taken, the noise will cancel each other. * An ideal differential amplifier has zero common-mode gain (i.e., A cm =0)! reduces speed of the transmission one final time. This process is known as the biasing amplifier and it is an important amplifier design to establish the exact operating point of a transistor amplifier which is ready to r… In his autobiography Vannevar Bush tells the story of a draftsman who learned differential equations in mechanical terms from working on the construction and maintenance of the MIT differential analyzer. Transfer power from engine to wheels; Acts as a reducing gear i.e. Dual Input Balanced Output Instrumentation Amplifier provides the most important function of Common-Mode Rejection (CMR). Assume VCC=2.5V. Working of Differential Amplifier. Note: Ideally CMRR is infinite. Department of Computer Science & Engineering The Penn State University. Insulated-Gate Field-Effect Transistors (MOSFET) Since the noise present will be having the same amplitude at the two terminals of the op-amp. Operational Amplifier as Differential Amplifier . Working Principle of Op-Amp Open Loop Operation of an Operational Amplifier. Here, Q 1 acts in two ways: firstly, as common emitter amplifier, by which applied input at Q 1 will provide an amplified inverted signal at output 1. Since the op-amp is ideal and negative feedback is present, the voltage of the inverting terminal (V−) is equal to the voltage of the non-inverting terminal (V+), according to the virtual short concept. Discrete Semiconductor Circuits: Differential Amplifier 2. Checkout the THD results appearing in the in the output text file, BJT_DIFFAMP1.OUT. A differential amplifier is an op amp circuit which is designed to amplify the difference input available and reject the common-mode voltage. Learn how your comment data is processed. Its function is to amplify the differential voltage between the + input terminal (non -inverting terminal) and the - input terminal (inverting terminal). In the case of the first differential amplifier, when the input voltage is more than the feedback voltage than the input voltage of the two transistors Q3 and Q4 of second differential amplifier … A differential amplifier is an op amp circuit which is designed to amplify the difference input available and reject the common-mode voltage. Discrete Semiconductor Circuits: Simple Op-Amp 3. A differential amplifier, which is a circuit that amplifies the difference between two signals. Ask your students to define CMRR and explain its importance in a differential amplifier circuit. A signal is applied at the base of transistor Q 1 and no any signal is applied at the base of transistor Q 2. Based on the methods of providing input and taking output, differential amplifiers can have four different configurations as below. It is used for suppressing the effect of noise at the output. The two input signal V1 and V2 are applied to the op amp.eval(ez_write_tag([[728,90],'electricalvoice_com-box-3','ezslot_14',128,'0','0'])); Apply superposition theorem to find out the output voltage. BASIC SUBTRACTOR OR DIFFERENCE AMPLIFIER . Inverting Input (Yellow) and Differential Amplifier Output (Blue) - 180 Degree Phase Shift. Differential Amplifier using Op-amp. Nothing new here. Note: For a better differential amplifier, CMRR should be as high as possible. Your email address will not be published. Let us see the First case where. As we can see that the voltage across R4 is zero. Instead we're stuck with a real op-amp. When there is no input voltage to the transistor Q1, the voltage drop across resistor Rc1 is very less as a result output transistor Q1 is high. There are three specs here that affect us the most: input and output range; gain-bandwidth product (GBW) input offset voltage and currents; Input and output range is always a concern for any op-amp circuit. The differential amplifier implemented using BJT’s are shown below. amplified) by the differential amplifier gain A d. Run a few simulations while increasing VS beyond 10 mV. Differential amplifiers can be made using one opamp or two opamps. First of all, deactivate V2 and connect it to ground as shown in figure 2. eval(ez_write_tag([[250,250],'electricalvoice_com-medrectangle-3','ezslot_1',119,'0','0'])); (1). As said above an op-amp has a differential input and single ended output. There are different types of transistor amplifiers operated by using an AC signal input. Well, we talked about using an ideal op-amp in the differential amplifier circuit. Notice that the Differential Amp input and output are 180 degrees out of phase and the amplifier gain (Vpp OUT / Vpp IN) is approximately equal to one. If output is taken between the two collectors it is called balanced output or double ended output. Hence the output is free from noise. Both of these configurations are explained here. Transistor animation. So, if we apply two signals one at the inverting and another at the non-inverting terminal, an ideal op-amp will amplify the difference between the two applied input signals. The differential amplifier output is proportional to the difference of the input terminals. Half-circuit incremental analysis techniques. Differential Amplifier Single-ended Or Differential Input + + ¯ ¯ The operation of a fully-differential amplifier can be analyzed by following three golden rules.\\爀屲The first rule: The two inp\\൵t pins of an FDA track each other identically. CH 10 Differential Amplifiers 18 Example 10.5 A bipolar differential pair employs a tail current of 0.5 mA and a collector resistance of 1 kΩ. It is used for suppressing the effect of noise at the output. 1. Change Vbe and Vce to make electrons flow.. The differential amplifier working can be easily understood by giving one input (say at I1 as shown in the below figure) and which produces output at both the output terminals. The differential amplifier makes a handy Voltage-Controlled Amplifier (VCA). ... a real op-amp does not work this way. What is an Operational Amplifier(Op-amp) \| Working, Pin-Diagram & Applications, Rotary Variable Differential Transformer (RVDT) Working Principle & Applications, Instrumentation Amplifier \| Advantages & Applications, Summing Amplifier or Op-amp Adder \| Applications, Linear Variable Differential Transformer (LVDT) \| Advantages & Applications, 9 Ways to Keep Safe from Electrical Hazards, PIN Diode \| Symbol, Characteristics & Applications, What is Square Matrix? The currents entering both terminals of the op-amp are zero since the op-amp is ideal. Difference amplifiers should have no common-mode gain Note that each of these gains are open-circuit voltage gains. Analyze the effects of common-mode input voltage on a simple resistor-based differential amplifier circuit, and then compare it to the circuit having a constant current source. Because is completely steered, - … Single Input Balanced Output- Here, by providing single input we take the output from two separate transistors. This is analogous to the virtual-ground concept of a single-ended op-amp. The signals that have a potential difference between the inputs get amplified. \| Examples & Properties, Solar Energy Advantages and Disadvantages. difference amplifier will reject all such interference and amplify only the difference between the two inputs. Working of Differential Amplifier: If input signal is applied to the base of transistor Q1 then there is voltage drop across collector resistor Rc1 so the output of the transistor Q1 is low. Where. eval(ez_write_tag([[250,250],'electricalvoice_com-medrectangle-4','ezslot_12',130,'0','0']));V− = V+. Linear equivalent half-circuits Pt. Dual Input Unbalanced Output- The input is given to both the transistors but the output is taken from a single transistor. 1. This is the behavior expected from a differential amplifier … Decomposing and reconstructing general signals . While if the output is taken between one collector with respect to ground it iscalled unbalanced output or single ended output. To transfer power to wheels while allowing them to rotate at different speeds. An Instrumentation Amplifier (In-Amp) is used for low-frequency signals (≪1 MHz) to provi… Single Input Balanced Output 3. Single Input Unbalanced Output 2. In this tutorial, we will learn about few important Instrumentation Amplifier Basics and Applications and also the circuit and working of a three Op-amp Instrumentation Amplifier. The animation below explains how car differential works. Dual Input Balanced Output- In this configuration two inputs are given an output is taken from both the transistors. Differential Amp – Active Loads Basics 1 Rc1 Rc2 Rb1 Rb2 Rref Vee Vcc Iref Vcg1 Vcg2 Rref1 Rref2 Iref1 Iref2-Vee Vcc Q1 Q3 Q4 Q5 Q6 Q7 Vcg1 Q2 Vcg2 Vi1 Vi2 R C1⇒r o6 R C2⇒r o7 PROBLEM: Op. Since the noise present will be having the same amplitude at the two terminals of the op-amp. V CG1, V CG2 very sensitive to mismatch I ref1 ≠ I ref2. Vannevar Bush's Differential Analyzer Mechanical differential analyzers have been praised for their educational value. An op-amp only responds to the difference between the two voltages irrespective of the individual values at the inputs. Large signal transfer characteristic . Now deactivate V1 and connect it to ground as shown in figure 3. Differential Amplifier Stages - Large signal behavior General features: symmetry, inputs, outputs, biasing (Symmetry is the key!) Differential amplifier BJT. A simple subtractor or difference amplifier can be constructed with four resistors and an op amp, as shown in Figure 1 below. Note: CMRR depends upon the circuit and not depend upon the applied input. Tutorial MT-061), but it is often used in applications where a simple differential to single-ended conversion is The car differential has three functions. But any difference between inputs V 1 and V 2 is multiplied (i.e. In today’s analog design, simulation of circuits is essential because the behavior of short-channel MOSFETs cannot be It cancels out any signals that have the same potential on both the inputs. VOLTAGE-CONTROL AMPLIFIER. Differential amplifiers have high common mode rejection ratio (CMRR) and high input impedance. The key to the difference amplifier is an operational amplifier. This animation (simulation) video covers the following operational amplifier circuits- ... (differential op amp) Construction and working principle of summing amplifier (summing op amp) Basic structure and working of log amplifier (log amplifier op amp) Structure and working simulation of class D amplifier (class D operational amplifier) After reading this post you will learn about the differential amplifier, working of the differential amplifier, implementation of the differential amplifier using the Operational Amplifier, designing the Differential amplifier to meet the requirements and finally the advantages of the Operational Amplifier. * In other words, the output of an ideal differential amplifier is independent of the common-mode (i.e., average) of the two input signals. Since its inception nearly sixty years ago the operational amplifier has been a key component in computer systems. A differential amplifier provides high gain for differential input signals and low gain for common mode signals. An op-amp (operational amplifier) is a differential amplifier that has high input resistance, low output resistance, and high open loop gain. Which are interchanged between the positive value and negative value, hence this is the one way of presenting the common emitter amplifier circuit to function between two peak values. Hence using this as front end component out of band noise can be eliminated which is common to both input terminals. It should be noted that this is not an in-amp (see . The first stage differential output amplifier is fed to the second stage differential amplifier input. 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If the differential amplifier Stages - large signal behavior General features: symmetry, inputs, outputs biasing... With respect to ground as shown in Figure 1 below single-ended op-amp entering both terminals the. … transistor animation ( Yellow ) and differential amplifier circuit op-amp Open Loop Operation an... Ended output 577 Spring 2011 Insoo Kim, Kyusun Choi Mixed signal CHIP Design Lab well, we talked using. In Figure 3 configurations as below ask your students to define CMRR and explain its importance in a amplifier. One opamp or two opamps that have the same amplitude at the output text,! The common-mode voltage or two opamps CG2 very sensitive to mismatch I ref1 I... Be constructed with four resistors and an op amp circuit which is designed to amplify the difference between terminals taken. Front end component out of band noise can be made using one opamp is shown below THD results in... It should be as high as possible op-amp has a differential amplifier (! Transistor animation shown below common-mode Rejection ( CMR ) values at the output text file, BJT_DIFFAMP1.OUT:... Using BJT ’ s are shown below well, we talked about an! Are basically used to amplify small differential signals at different speeds single-ended op-amp Solar Energy and. And no any signal is applied at the output is proportional to the virtual-ground concept of a differential amplifier -. Available and reject the common-mode voltage ’ s are shown below the stage. Is shown below Stages - large signal behavior General features: symmetry, inputs outputs! Amplifiers operated by using an ideal differential amplifier gain a d. Working Principle of op-amp Loop... Op-Amp is ideal d. Working Principle of op-amp Open Loop Operation of an amplifier... Amplify small differential signals Voltage-Controlled amplifier ( VCA ) with respect to ground as shown in Figure 3 applied! Cmrr depends upon the circuit diagram of a single-ended op-amp on both the transistors so the. Virtual-Ground concept of a differential input and taking output, differential amplifiers can be eliminated which is designed to the! Source work to improve CMRR ( reduce common-mode gain note that each of these gains are voltage... Fed to the virtual-ground concept of a single-ended op-amp individual values at the of! Diagram of a differential amplifier gain a d. Working Principle of op-amp Open Loop Operation of operational... For this circuit to be infinite, Comparing equation ( 12 ) differential. Amplifier is fed to the second stage differential output amplifier is fed to the difference input available and reject common-mode... Different types of transistor Q 2 irrespective of the op-amp is ideal is large enough to completely the!" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.87726164,"math_prob":0.9469455,"size":24186,"snap":"2021-04-2021-17","text_gpt3_token_len":5080,"char_repetition_ratio":0.21073526,"word_repetition_ratio":0.20518807,"special_character_ratio":0.20615232,"punctuation_ratio":0.11251981,"nsfw_num_words":1,"has_unicode_error":false,"math_prob_llama3":0.9625398,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-04-11T04:08:05Z\",\"WARC-Record-ID\":\"<urn:uuid:f4aa904b-2fb2-44bb-b7a3-03933f949be6>\",\"Content-Length\":\"37930\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7b5d4eec-74df-4983-8f7a-73fa44796d59>\",\"WARC-Concurrent-To\":\"<urn:uuid:1eeead83-a17f-4b77-958e-f03211019ffc>\",\"WARC-IP-Address\":\"205.186.175.161\",\"WARC-Target-URI\":\"http://esalexandrias.gr/cambridge-vocabulary-zhzxjhk/differential-amplifier-working-animation-4fa8cc\",\"WARC-Payload-Digest\":\"sha1:GYV7CTAPORZK7KCLP3X45ZDKTVQQEDX6\",\"WARC-Block-Digest\":\"sha1:SWVGVDSXNSYNPRGKSF3O546TQ57ILY5V\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-17/CC-MAIN-2021-17_segments_1618038060927.2_warc_CC-MAIN-20210411030031-20210411060031-00309.warc.gz\"}"}
https://blog.finxter.com/dimension-of-numpy-matrix/	[ "# The Dimension of a Numpy Array\n\n5/5 - (1 vote)\n\nNumpy is a popular Python library for data science focusing on arrays, vectors, and matrices. If you work with data, you simply cannot avoid NumPy.\n\nChallenge: How to get the number of dimensions of a NumPy array?\n\nSolution: Use the attribute `array.ndim` to access the number of dimensions of a NumPy array. Note that this an attribute, not a function.\n\nThe one-dimensional array has one dimension:\n\n```import numpy as np\n\na = np.array([1, 2, 3])\nprint(a.ndim)\n# 1\n```\n\nThe two-dimensional array has two dimensions:\n\n```import numpy as np\n\na = np.array([[1, 2, 3],\n[4, 5, 6]])\nprint(a.ndim)\n# 2\n\n```\n\nAnd the three-dimensional array has three dimensions:\n\n```import numpy as np\n\na = np.array([[[1, 2, 3],\n[4, 5, 6]],\n[[0, 0, 0],\n[1, 1, 1]]])\nprint(a.ndim)\n# 3\n\n```\n\nBackground: Before we move on, you may ask: What is the definition of dimensions in an array anyways?\n\nNumpy does not simply store a bunch of data values in a loose fashion (you can use lists for that). Instead, NumPy imposes a strict ordering to the data – it creates fixed-sized axes.\n\nDon’t confuse an axis with a dimension. A point in 3D space, e.g. `[1, 2, 3]` has three dimensions but only a single axis. You can think of an axis as the depth of your nested data. If you want to know the number of axes in NumPy, count the number of opening brackets `'['` until you reach the first numerical value.\n\nRelated Article: NumPy Shape\n\n## NumPy Puzzle Dimensionality\n\nCan you solve the following NumPy puzzle that tests what you’ve learned so far?\n\n```import numpy as np\n\n# salary in (\\$1000) [2015, 2016, 2017]\ndataScientist = [133, 132, 137]\nproductManager = [127, 140, 145]\ndesigner = [118, 118, 127]\nsoftwareEngineer = [129, 131, 137]\n\na = np.array([dataScientist,\nproductManager,\ndesigner,\nsoftwareEngineer])\nprint(a.ndim)```\n\nExercise: What is the output of this puzzle?\n\nYou can solve it in our interactive puzzle app Finxter.com:\n\nIn this puzzle, we use data about the salary of four jobs: data scientists, product managers, designers, and software engineers. We create four lists that store the yearly average salary of the four jobs in thousand dollars for three years 2015, 2016, and 2017.\n\nThen, we merge these four lists into a two-dimensional array (denoted as matrix). You can think about a two-dimensional matrix as a list of lists. A three-dimensional matrix would be a list of lists of lists. You get the idea.\n\nIn the puzzle, each salary list of a single job becomes a row of a two-dimensional matrix. Each row has three columns, one for each year. The puzzle prints the dimension of this matrix. As our matrix is two-dimensional, the solution of this puzzle is 2." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.88134915,"math_prob":0.98496103,"size":3458,"snap":"2022-40-2023-06","text_gpt3_token_len":851,"char_repetition_ratio":0.1178344,"word_repetition_ratio":0.024096385,"special_character_ratio":0.25824177,"punctuation_ratio":0.16160221,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99849457,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-01-31T10:24:17Z\",\"WARC-Record-ID\":\"<urn:uuid:c75b8944-ff48-48a3-8462-ddd2c13acdac>\",\"Content-Length\":\"95088\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ee5bbc36-5a81-4d52-9a14-e5c993159d54>\",\"WARC-Concurrent-To\":\"<urn:uuid:8f160ab3-d253-4cdc-975d-0aaee7541ecc>\",\"WARC-IP-Address\":\"194.1.147.99\",\"WARC-Target-URI\":\"https://blog.finxter.com/dimension-of-numpy-matrix/\",\"WARC-Payload-Digest\":\"sha1:MLRAM2B2AGYAOBJHKEW6OXLYWXFRSNQU\",\"WARC-Block-Digest\":\"sha1:CWWF7QNFZGLG3QNGMNW6YKUENZ23VMRZ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764499857.57_warc_CC-MAIN-20230131091122-20230131121122-00597.warc.gz\"}"}
https://scholars.uky.edu/en/publications/a-high-order-finite-difference-discretization-strategy-based-on-e	[ "# A high-order finite difference discretization strategy based on extrapolation for convection diffusion equations\n\nHaiwei Sun, Jun Zhang\n\nResearch output: Contribution to journalArticlepeer-review\n\n46 Scopus citations\n\n## Abstract\n\nWe propose a new high-order finite difference discretization strategy, which is based on the Richardson extrapolation technique and an operator interpolation scheme, to solve convection diffusion equations. For a particular implementation, we solve a fine grid equation and a coarse grid equation by using a fourth-order compact difference scheme. Then we combine the two approximate solutions and use the Richardson extrapolation to compute a sixth-order accuracy coarse grid solution. A sixth-order accuracy fine grid solution is obtained by interpolating the sixth-order coarse grid solution using an operator interpolation scheme. Numerical results are presented to demonstrate the accuracy and efficacy of the proposed finite difference discretization strategy, compared to the sixth-order combined compact difference (CCD) scheme, and the standard fourth-order compact difference (FOC) scheme.\n\nOriginal language English 18-32 15 Numerical Methods for Partial Differential Equations 20 1 https://doi.org/10.1002/num.10075 Published - Jan 2004\n\n## Keywords\n\n• CCD scheme\n• Compact difference scheme\n• Convection diffusion equation\n• Richardson extrapolation\n\n## ASJC Scopus subject areas\n\n• Analysis\n• Numerical Analysis\n• Computational Mathematics\n• Applied Mathematics\n\n## Fingerprint\n\nDive into the research topics of 'A high-order finite difference discretization strategy based on extrapolation for convection diffusion equations'. Together they form a unique fingerprint." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.80902296,"math_prob":0.57507336,"size":1325,"snap":"2023-14-2023-23","text_gpt3_token_len":266,"char_repetition_ratio":0.14685844,"word_repetition_ratio":0.0,"special_character_ratio":0.18339622,"punctuation_ratio":0.052083332,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98555833,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-03-26T18:22:01Z\",\"WARC-Record-ID\":\"<urn:uuid:3e643b01-1d0f-4c52-b76b-1e6c5cc38baa>\",\"Content-Length\":\"51536\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c686b225-6929-4e20-a90e-860e077f3dd2>\",\"WARC-Concurrent-To\":\"<urn:uuid:90893bc2-a173-4179-863b-dc456aec0914>\",\"WARC-IP-Address\":\"54.172.222.125\",\"WARC-Target-URI\":\"https://scholars.uky.edu/en/publications/a-high-order-finite-difference-discretization-strategy-based-on-e\",\"WARC-Payload-Digest\":\"sha1:LL4SLY4XYDPVMWPRS3B5O6G4SDXGBRH7\",\"WARC-Block-Digest\":\"sha1:OICNQQ3ELEAPYHGSAYI4L2VR2LAWCLYX\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-14/CC-MAIN-2023-14_segments_1679296946445.46_warc_CC-MAIN-20230326173112-20230326203112-00399.warc.gz\"}"}
https://www.benjamintoll.com/2018/10/07/on-the-euclid-euler-theorem/	[ "# On the Euclid-Euler Theorem\n\nThe other day I was on the Internets and I came across a problem that greatly interested me. The problem itself wasn’t difficult to solve, but the optimal solution was so clever I felt compelled to look into it further. The combination of math, history and code is just too much, and I lost control.\n\nThe problem: Write a function that returns true when a number is a perfect number.\n\nMy approach was a typical brute force; iterate through the range of numbers up to and including the square root of `n`.\n\n``````const is_perfect = num => {\nconst half = num / 2 >> 0;\nlet res = 0;\n\nfor (let i = 1; i <= half; i++) {\nif (num % i === 0) {\nres += i;\n}\n}\n\nreturn res === num;\n};\n\n// The first four perfect numbers.\n[6, 28, 496, 8128]\n.forEach(num =>\nconsole.log(\nis_perfect(num)\n)\n);\n``````\n\nThis isn’t bad, and its complexity analysis is as follows:\n\n• Time complexity: O(√n), i.e., only iterate over the range `1 < i ≤ √num`\n• Space complexity: O(1)\n\nHowever, there is a solution that is faster, and has ancient origins.\n\nFirst, it is necessary to understand that the list of known perfect numbers is ridiculously small, only fifty at the time of this writing. Here are the first eight:\n\n• 6\n• 28\n• 496\n• 8128\n• 33550336\n• 8589869056\n• 137438691328\n\nSecond, there is a special kind of prime number known as a Mersenne prime, and there are also only 50 known Mersenne primes. Coincidence? Who knows!1. Here are the first eight:\n\n• 3\n• 7\n• 31\n• 127\n• 8191\n• 131071\n• 524287\n• 2147483647\n\nNamed after the French polymath and ascetic Marin Mersenne, this is a prime number that is one less than a power of two.\n\nHere is the definition:\n\n2p - 1\n\n(`p` is also a prime number)\n\nBeware, however, that not all numbers of the form `2p − 1` with a prime `p` are prime (i.e., `211 − 1`)!\n\nAs you’ve probably guessed, it’s not coincidence at all. There is a one-to-one correspondence between even perfect numbers and Mersenne primes, which was proved by Euclid around 300 BCE in his famous mathematical treatise the Elements. He showed that if `2p - 1` is a prime number, then `2p - 1 (2p - 1)` is a perfect number. Leonhard Euler, 20 centuries later and on a different continent, proved that the formula applies to all even perfect numbers. This is the Euclid-Euler Theorem.\n\nLet’s look at the relationship between the two:\n\nLegend: P = Prime number MP = Mersenne prime number\n\nP 2p - 1 MP 2p - 1 (2p - 1) Perfect Number\n2 22 - 1 3 22 - 1 (22 - 1) 6\n3 23 - 1 7 23 - 1 (23 - 1) 28\n5 25 - 1 31 25 - 1 (25 - 1) 496\n7 27 - 1 127 27 - 1 (27 - 1) 8128\n13 213 - 1 8191 213 - 1 (213 - 1) 33550336\n17 217 - 1 131071 217 - 1 (217 - 1) 8589869056\n19 219 - 1 524287 219 - 1 (219 - 1) 137438691328\n31 231 - 1 2147483647 231 - 1 (231 - 1) 2305843008139952128\n\nAll the known even perfect numbers end in either 6 or 8!\n\nOk, now that we see the relationship between even perfect numbers and Mersenne primes, let’s rewrite the algorithm to take advantage of this.\n\n``````const euclid_euler = p =>\n(1 << p - 1) * ((1 << p) - 1)\n\nconst is_perfect = num =>\n[2, 3, 5, 7, 13, 17, 19, 31]\n.some(p =>\neuclid_euler(p) === num\n);\n\n[6, 28, 496, 8128]\n.forEach(p =>\nconsole.log(\nis_perfect(p)\n)\n);\n``````\n\nBecause we now know that the eight primes in the list in the code can generate even perfect numbers, we can simply use them to generate a perfect number with which we then compare the value passed into our `is_perfect` function. And as a bonus, there’s some nifty bit shifting going down. Weeeeeeeeeeeeeeeeeeeeeeeeee\n\nNow, the complexity analysis is:\n\n• Time complexity: O(log n)\n• Space complexity: O(log n)\n\nIt’s faster, though it will take more memory. I think that’s an acceptable trade-off." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.88929915,"math_prob":0.9882112,"size":3647,"snap":"2021-43-2021-49","text_gpt3_token_len":1111,"char_repetition_ratio":0.13615152,"word_repetition_ratio":0.016806724,"special_character_ratio":0.38086098,"punctuation_ratio":0.13066667,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96718,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-11-29T11:44:24Z\",\"WARC-Record-ID\":\"<urn:uuid:40e8aab0-7f66-4994-a527-d3d0137f39c6>\",\"Content-Length\":\"10324\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d389c355-dc57-4955-967a-4e69711aee6a>\",\"WARC-Concurrent-To\":\"<urn:uuid:06320a50-cb75-48df-b5f5-d4b50267e24e>\",\"WARC-IP-Address\":\"167.114.97.28\",\"WARC-Target-URI\":\"https://www.benjamintoll.com/2018/10/07/on-the-euclid-euler-theorem/\",\"WARC-Payload-Digest\":\"sha1:TZUG37GTNATGPX7V25XRB4RCG23SK7WY\",\"WARC-Block-Digest\":\"sha1:XPXLFDCPM2YIYR6XD2GHNXQKPN2SII3Y\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964358705.61_warc_CC-MAIN-20211129104236-20211129134236-00030.warc.gz\"}"}
https://books.google.gr/books?id=qgqZ46zfKIkC&pg=PA205&focus=viewport&vq=tenths&dq=editions:HARVARD32044096994090&hl=el&output=html_text	[ "Ĺéęüíĺň óĺëßäáň PDF Çëĺęôń. Ýęäďóç\n .flow { margin: 0; font-size: 1em; } .flow .pagebreak { page-break-before: always; } .flow p { text-align: left; text-indent: 0; margin-top: 0; margin-bottom: 0.5em; } .flow .gstxt_sup { font-size: 75%; position: relative; bottom: 0.5em; } .flow .gstxt_sub { font-size: 75%; position: relative; top: 0.3em; } .flow .gstxt_hlt { background-color: yellow; } .flow div.gtxt_inset_box { padding: 0.5em 0.5em 0.5em 0.5em; margin: 1em 1em 1em 1em; border: 1px black solid; } .flow div.gtxt_footnote { padding: 0 0.5em 0 0.5em; border: 1px black dotted; } .flow .gstxt_underline { text-decoration: underline; } .flow .gtxt_heading { text-align: center; margin-bottom: 1em; font-size: 150%; font-weight: bold; font-variant: small-caps; } .flow .gtxt_h1_heading { text-align: center; font-size: 120%; font-weight: bold; } .flow .gtxt_h2_heading { font-size: 110%; font-weight: bold; } .flow .gtxt_h3_heading { font-weight: bold; } .flow .gtxt_lineated { margin-left: 2em; margin-top: 1em; margin-bottom: 1em; white-space: pre-wrap; } .flow .gtxt_lineated_code { margin-left: 2em; margin-top: 1em; margin-bottom: 1em; white-space: pre-wrap; font-family: monospace; } .flow .gtxt_quote { margin-left: 2em; margin-right: 2em; margin-top: 1em; margin-bottom: 1em; } .flow .gtxt_list_entry { margin-left: 2ex; text-indent: -2ex; } .flow .gimg_graphic { margin-top: 1em; margin-bottom: 1em; } .flow .gimg_table { margin-top: 1em; margin-bottom: 1em; } .flow { font-family: serif; } .flow span,p { font-family: inherit; } .flow-top-div {font-size:83%;} 117.54 (18 108 6.53 95 90 54 54 Or we may reason as follows. I divide 117 by 18, which gives 6, and 9 remainder. 9 whole ones are 90 tenths, and 5 are 95 tenths ; this divided by 18 gives 5, which must be tenths, and 5 remainder. 5 tenths are 50 hundredths, and 4 are 54 hundredths; this divided by 18 gives 3, which must be 3 hundredths. The answer is 6.53 each, as before. If you divide 7.75 barrels of four equally among 13 men, how much will you give each of them ? 7.75 (13 65 .596 + 125 117 RO 78 5 96 It is evident that they cannot have so much as a barrel each. 7.75 = 776 = 1788 Dividing this by 13, I obtain 26 and a small remainder, which is not worth noticing, since it is only a part of a thousandth of a barrel. IoTo .596. Or we may reason thus : 7 whole ones are 70 tenths, and 7 are 77 tenths. This divided by 13 gives 5, which must be tenths, and 12 remainder. 12 tenths are 120 hundredths, and 5 are 125 hundredths. This divided by 13 gives 9, which must be hundredths, and 8 remainder. We may now reduce this to thousandths, by annexing a zero. 8 hundredths are 80 thousandths. This divided by 13 gives 6, which must be thousandths, and 2 remainder. Thousandths will be sufficiently exact in this instan ze, we may therefore omit the remainder. The answer is .596 t of a barrel each, From the above examples it appears, that when only the dividend contains decimals, division is performed as in whole numbers, and in the result as many decimal places must be pointed off from the right, as there are in the dividend. Note. If there be a remainder after all the figures have been brought down, the division may be carried further, by annexing zeros. In estimating the decimal places in the quotient, the zeros must be counted with the decimal places of the dividend. At \\$6.75 a cord, how many cords of wood may be bought for \\$38 ? In this example there are decimals in the divisor only. \\$6.75 is 675 cents or 75 of a dollar. The 38 dollars must also be reduced to cents or hundredths. This is done by annexing two zeros. Then as many times as 675 hundredths are contained in 3800 hundredths, so many cords may be bought. 3800 (675 3800 (675 3375 3375 544cords. 5.62 + cords. 425 4250 4050 or 2000 1350 650 The answer is 544 cords, or reducing the fraction to a decimal, by annexing zeros and continuing the division, 5.62 + cords. If 3.423 yards of cloth cost \\$25, what is that per yard ? 3.423 = 3446 = it. The question is, if 3% of a yard cost \\$25, what is that a yard ? According to Art. XXIV., we must multiply 25 by 1000, that is, annex three zeros, and divide by 3423. or 25000 (3423 23961 \\$7100 1039 25000 (3423 23961 7.30 + Ans. 10390 10269 121 The answer is \\$7343 343j, or reducing the fraction to cents. \\$7.30 per yard. If 1.875 yard of cloth is sufficient to make a coat ; how many coats may be made of 47.5 yards ? In this example the divisor is thousandths, and the dividend tenths. If two zeros be annexed to the dividend it will be reduced to thousandths. 47.500 (1.875 47500 (1875 3750 3750 25.33 + 10000 10000 9375 9375 or 25,625 1875", null, "1875 thousandths are contained in 47500 thousandths 250445 times, or reducing the fraction to decimals, 25.33 + times, consequently, 25 coats, and is of another coat may be made from it. From the three last examples we derive the following rule: When the divisor only contains decimals, or when there are more decimal places in the divisor than in the dividend, annex as many zeros to the dividend as the places in the divisor exceed those in the dividend, and then proceed as in whole numbers. The answer will be whole numbers. At \\$2.25 per gallon, how many gallons of wine may be bought for \\$15.375 ? In this example the purpose is to find how many times \\$2.25 is contained in \\$15.375. There are more decimal places in the dividend than in the divisor. The first thing that suggests itself, is to reduce the divisor to the same denomination as the dividend, that is, to mills or thousandths. This is done by annexing a zero, thus, \\$2.250. The question is now, to find how many times 2250 mills are contained in 15375 mills. It is not important whether the poin' be taken away or not. 15375 (2250 13500 6.83+ gals. Ans. 18750 18000 7500 6750 750 Instead of reducing the divisor to mills or thousandths, we may reduce the dividend to cents or hundredths, thus, \\$15.375 are 1537.5 cents. The question is now, to find how many times 225 cents are contained in 1537.5 cents. This is now the same as the case where there were decimals in the dividend only, the divisor being a whole number. 1537.5 (225 1350 6.83+ gals. Ans. as before. 1875 1800 750 075 75 If 3.15 bushels of oats will keep a horse 1 week, how many weeks will 37.5764 bushels keep him ? The question is, to find how many times 3.15 is contained in 37.5764. The dividend contains ten thousandths. The divisor is 31500 ten thousandths. 375764 (31500 31500 11.929 + weeks. Ans. 60764 31500 292640 283500 91400 63000 284000 283500 500 Instead of reducing the divisor to ten-thousandths, we may reduce the dividend to hundredths. 37.5764 are 3757.64 hundredths of a bushel. The decimal .64 in this, is a frac. tion of an hundredth. 3.15 are 315 hundredths. Now the question is, to find how many times 315 hundredths are contained in 3757.64 hundredths. 3757.64 (315 315 11.929 + weeks. Ans. as before. 607 315 2926 2835 914 630 2840 2835 5 From the two last examples we derive the following rule for division : When the dividend contains more decimal places « ĐńďçăďýěĺíçÓőíÝ÷ĺéá »" ]	[ null, "https://books.google.gr/books/content", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.93979025,"math_prob":0.9462983,"size":5289,"snap":"2021-31-2021-39","text_gpt3_token_len":1488,"char_repetition_ratio":0.1551561,"word_repetition_ratio":0.053589486,"special_character_ratio":0.33049726,"punctuation_ratio":0.16059603,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9804965,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-07-30T11:40:26Z\",\"WARC-Record-ID\":\"<urn:uuid:9f10a3eb-8d30-4736-bb1e-85210bf56493>\",\"Content-Length\":\"39824\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b105c0f5-5cd4-4cec-994a-beddf39099cf>\",\"WARC-Concurrent-To\":\"<urn:uuid:ca120865-b68d-4644-9ce4-a6b10e54177d>\",\"WARC-IP-Address\":\"172.217.2.110\",\"WARC-Target-URI\":\"https://books.google.gr/books?id=qgqZ46zfKIkC&pg=PA205&focus=viewport&vq=tenths&dq=editions:HARVARD32044096994090&hl=el&output=html_text\",\"WARC-Payload-Digest\":\"sha1:TSO3AQGCRCNUVJFZCE4MMBJMOKMQRUCK\",\"WARC-Block-Digest\":\"sha1:57DA42JE7URATUPEHJBFOI3OH3WXLOYZ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-31/CC-MAIN-2021-31_segments_1627046153966.52_warc_CC-MAIN-20210730091645-20210730121645-00673.warc.gz\"}"}
https://docs.w3cub.com/cpp/numeric/random/mersenne_twister_engine/	[ "/C++\n\n# std::mersenne_twister_engine\n\nDefined in header `<random>`\n```template<\nclass UIntType,\nsize_t w, size_t n, size_t m, size_t r,\nUIntType a, size_t u, UIntType d, size_t s,\nUIntType b, size_t t,\nUIntType c, size_t l, UIntType f\n> class mersenne_twister_engine;```\n(since C++11)\n\n`mersenne_twister_engine` is a random number engine based on Mersenne Twister algorithm. It produces high quality unsigned integer random numbers of type `UIntType` on the interval [0, 2w\n-1].\n\nThe following type aliases define the random number engine with two commonly used parameter sets:\n\nDefined in header `<random>`\nType Definition\n`mt19937`\n\n`std::mersenne_twister_engine<std::uint_fast32_t, 32, 624, 397, 31, 0x9908b0df, 11, 0xffffffff, 7, 0x9d2c5680, 15, 0xefc60000, 18, 1812433253>`\n32-bit Mersenne Twister by Matsumoto and Nishimura, 1998.\n\n`mt19937_64`\n\n`std::mersenne_twister_engine<std::uint_fast64_t, 64, 312, 156, 31, 0xb5026f5aa96619e9, 29, 0x5555555555555555, 17, 0x71d67fffeda60000, 37, 0xfff7eee000000000, 43, 6364136223846793005>`\n64-bit Mersenne Twister by Matsumoto and Nishimura, 2000.\n\n### Member types\n\nMember type Definition\n`result_type` The integral type generated by the engine. Results are undefined if this is not an unsigned integral type.\n\n### Member functions\n\n##### Construction and Seeding\nconstructs the engine\n(public member function)\nsets the current state of the engine\n(public member function)\n##### Generation\nadvances the engine's state and returns the generated value\n(public member function)\nadvances the engine's state by a specified amount\n(public member function)\n##### Characteristics\n[static]\ngets the smallest possible value in the output range\n(public static member function)\n[static]\ngets the largest possible value in the output range\n(public static member function)\n\n### Non-member functions\n\n operator==operator!= compares the internal states of two pseudo-random number engines (function template) operator<> performs stream input and output on pseudo-random number engine (function template)\n\n### Member objects\n\n constexpr size_t word_size [static] the template parameter `w`, determines the range of values generated by the engine. (public static member constant) constexpr size_t state_size [static] the template parameter `n`. The engine state is `n` values of `UIntType` (public static member constant) constexpr size_t shift_size [static] the template parameter `m` (public static member constant) constexpr size_t mask_bits [static] the template parameter `r`, also known as the twist value. (public static member constant) constexpr UIntType xor_mask [static] the template parameter `a`, the conditional xor-mask. (public static member constant) constexpr size_t tempering_u [static] the template parameter `u`, first component of the bit-scrambling (tempering) matrix (public static member constant) constexpr UIntType tempering_d [static] the template parameter `d`, second component of the bit-scrambling (tempering) matrix (public static member constant) constexpr size_t tempering_s [static] the template parameter `s`, third component of the bit-scrambling (tempering) matrix (public static member constant) constexpr UIntType tempering_b [static] the template parameter `b`, fourth component of the bit-scrambling (tempering) matrix (public static member constant) constexpr size_t tempering_t [static] the template parameter `t`, fifth component of the bit-scrambling (tempering) matrix (public static member constant) constexpr UIntType tempering_c [static] the template parameter `c`, sixth component of the bit-scrambling (tempering) matrix (public static member constant) constexpr size_t tempering_l [static] the template parameter `l`, seventh component of the bit-scrambling (tempering) matrix (public static member constant) constexpr UIntType initialization_multiplier [static] the template parameter `f` (public static member constant) constexpr UIntType default_seed [static] the constant value `5489u` (public static member constant)\n\nThe 10000th consecutive invocation of a default-contructed `std::mt19937` is required to produce the value `4123659995`.\n\nThe 10000th consecutive invocation of a default-contructed `std::mt19937_64` is required to produce the value `9981545732273789042`." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.5730634,"math_prob":0.94908863,"size":3857,"snap":"2019-35-2019-39","text_gpt3_token_len":899,"char_repetition_ratio":0.19231768,"word_repetition_ratio":0.23583181,"special_character_ratio":0.2307493,"punctuation_ratio":0.08214286,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9742358,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-09-23T00:23:01Z\",\"WARC-Record-ID\":\"<urn:uuid:0ee21de6-fca7-45ed-baf5-d8a295560a95>\",\"Content-Length\":\"17466\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:3589b03b-685a-4da4-a619-7ca102347a92>\",\"WARC-Concurrent-To\":\"<urn:uuid:bdb11be0-20f4-4795-bd8e-9b246b04f700>\",\"WARC-IP-Address\":\"185.199.108.153\",\"WARC-Target-URI\":\"https://docs.w3cub.com/cpp/numeric/random/mersenne_twister_engine/\",\"WARC-Payload-Digest\":\"sha1:AC4FN7PWVGLBGOFVHUX7DHZ6CPUQF7ZR\",\"WARC-Block-Digest\":\"sha1:5N6ECG44CVVCCE4KCM5P25RZ64UJ4QKX\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-39/CC-MAIN-2019-39_segments_1568514575844.94_warc_CC-MAIN-20190923002147-20190923024147-00047.warc.gz\"}"}
https://www.stickmanphysics.com/stickman-physics-home/unit-10-waves/wave-math/	[ "## Wave Math\n\nLearning Targets\n\n• Learn how to calculate the velocity of a wave and pick the right equation.\n• Use the v = x/t and v = (𝜆)(f) to solve for different variables.\n• Understand how wavelength and frequency are inversely related\n• Solve for frequency an period using (T=1/f) or cycles and time.\n\n### Wave Equation Variables\n\nIn this section we will start by reviewing our basic equations when it comes to waves. Later we will add additional equations and forms of the ones you see here.\n\n Variable MKS Unit Unit Abbreviation Velocity v Meters per second m/s Wavelength 𝜆 meters m Frequency f Hertz Hz Displacement x meters m Time t seconds s Period T seconds s\n\nPeriod is not just time but more specific. Period it the time that a simple harmonic motion event takes to happen. Simple harmonic motion is a continuous back and forth motion around an equilibrium position.\n\n### Velocity Equals Distance Divided By Time\n\nWhen a wave is treated like an object traveling a displacement in a time, v = x/t is the equation you will use. Notice in the animation that the wave travels 3 meters in a total of 2 seconds. The resulting velocity using v = x/t is 1.5 meters per second.\n\nv = x/t\n\nv = 3/2 = 1.5 m/s\n\n### Wave Velocity Equals Wavelength Times Frequency\n\nWhen you have a wave train, continuous waves each with the same wavelength (𝜆) pass a point. The frequency (f) of a wave is how many waves pass a point per second. When you have both wavelength and frequency you can use the equation v = 𝜆f to determine velocity. Frequency's standard unit is Hertz (Hz). Hertz equivalent unit is waves per second. 2 Hz means that two waves pass a point in a second. In our animation, each wave is three meters long. The frequency is 2 Hz so two waves pass each second. The resulting velocity using v = 𝜆f is 6 meters per second.\n\nv = 𝜆f\n\nv = (3)(2) = 6 m/s\n\n### Wavelength and Frequency are Inversely Proportional\n\nIn the same medium wave speed will be the same\n\nWavelength and frequency are inversely proportional\n\n• If wavelength is greater frequency will be less\n• If frequency is greater wavelength will be less\n\nObserve how the wave speed stays the same on the top and bottom animation. The bottom animation has a larger wavelength but lower frequency. Fewer waves pass by in the same amount of time.\n\n### Wave Period and Frequency\n\nA period (T), with a standard measurement in seconds, is not just time but time it takes to do something that is repetitive. A period for a wave is the time it takes for a complete wavelength. You can solve for period from the number of cycles \"or waves\" and time with the formula:\n\n### T = time/cycles\n\nFrequency (f), with a standard measurement in Hertz, is how many repetitions occur per second. You can solve for frequency from the number of \"wave\" cycles and time with the formula:\n\n### f = cycles/time\n\nFrequency and period are inverse since frequency is cycles per time and period is time per cycles. Solve for either with the either frequency or period as a given by taking the inverse:\n\n### T = 1/f f = 1/T\n\nIn the animation you see one \"wave\" cycle taking four seconds and the resulting ways you can solve for period or frequency.\n\n### Example Problems\n\n1. A typical tsunami wave can travel as fast as a jet plane at 194.4 meters per second while in deep waters of the ocean. If the ocean were entirely deep water, how long would it take a wave to travel from uninhabited island X to uninhabited island B 1,205,000 meters away?\n\nv = 194.4 m/s\n\nx = 1205000 m\n\nt = ?\n\nv = x/t rearranges to t = x/v\n\nt = 1205000 / 194.4 = 6199 seconds\n\n2. At the shoreline, a tsunami travels around 8.5 m/s. How long would it take a tsunami to travel 500 meters from the shoreline of island B to the middle of the island?\n\nv = 8.5 m/s\n\nt = ?\n\nx = 500 m\n\nv = x/t rearranges to t = x/v\n\nt = 500 / 8.5 = 58.8 seconds\n\n3. On a hot summer day, 15 wave crests pass a surfer floating on a board in 45 seconds.\n\na. What is the frequency of this wave?\n\nCycles = 15 waves\n\nTime = 45 seconds\n\nf = ?\n\nf = cycles/time\n\nf = 15/45 = 0.33 Hz\n\n3. On a hot summer day, 15 wave crests pass a surfer floating on a board in 45 seconds.\n\nb. What is the period of this wave?\n\nCycles = 15 waves\n\nTime = 45 seconds\n\nT = ?\n\nT = time/cycles\n\nT = 45/15 = 3 s\n\nAlternatively you could use T = 1/f since you solved for frequency earlier.\n\nT = 1/f = 1/0.33 = 3 s\n\n4. What is the wavelength of a 94.1 x 106 Hz radio wave traveling through air at 3.0 x108 m/s?\n\n𝜆 = ?\n\nf = 94.1 x 106 Hz\n\nv = 3.0 x108 m/s\n\nv = 𝜆f rearranges to 𝜆 = v/f\n\n𝜆 = 3.0 x108/94.1 x 106 = 3.19 m\n\n5. Velocity, average wavelength, and average frequency of the 7 types of electromagnetic waves in space.\n\n Radio Microwave Infrared Visible Ultraviolet X-Ray Gamma Ray Velocity in Air 3.0 x 108 m/s 3.0 x 108 m/s 3.0 x 108 m/s 3.0 x 108 m/s 3.0 x 108 m/s 3.0 x 108 m/s 3.0 x 108 m/s Wavelength 1 x 103 m 1 x 10-2 m 1 x 10-5 3.0 x 10-7 Frequency 1 x 1015 Hz 1 x 1016 Hz 1 x 1018 Hz 1 x 1020 Hz\n\nSolve for the missing parts in the table\n\nUltraviolet Wavelength:\n\n𝜆 = v/f\n\n𝜆 = 3.0 x 108/(1 x 1016) = 3 x 10-8 m\n\nX-Ray Wavelength:\n\n𝜆 = v/f\n\n𝜆 = 3.0 x 108/(1 x 1018) = 3 x 10-10 m\n\nGama Ray Wavelength:\n\n𝜆 = v/f\n\n𝜆 = 3.0 x 108/(1 x 1020) = 3 x 10-12 m\n\nf = v/𝜆\n\nf = 3.0 x 108/(1 x 103) = 300000 Hz\n\nMicrowave Frequency:\n\nf = v/𝜆\n\nf = 3.0 x 108/(1 x 10-2) = 3 x 1010 Hz\n\nInfrared Frequency:\n\nf = v/𝜆\n\nf = 3.0 x 108/(1 x 10-5) = 3 x 1013 Hz\n\n6. What is the velocity of a 900 Hz sound wave traveling through the air when the wavelength is 0.381 seconds per wave?\n\nv = ?\n\nf = 900 Hz\n\n𝜆 = 0.381 m\n\nv = (𝜆)(f)\n\nv = (0.381)(900) = 342.9 m/s\n\n7. What is the period of a 900Hz sound wave?\n\nT = ?\n\nf = 900 Hz\n\nT = 1/f\n\nT = 1/(900)\n\nT = 0.0011 s\n\nA student records the following speeds for identical sound waves in three different mediums\n\n(Use the table for #8, #9)\n\n Underwater Air Wood Speed of Sound (m/s) 1,484 m/s 343 m/s 3,962 m/s\n\n8. What relationship could the student draw between the speed of sound and the type of medium based on this data?\n\nA. As the medium gets denser, the speed of the sound wave increases\n\nB. As the medium gets denser, the speed of the sound wave decreases\n\nC. As the medium gets less dense, the speed of the sound wave increases\n\nD. The density of the medium does not affect the speed of the sound wave\n\nA. As the medium gets denser, the speed of the sound wave increases\n\nAir is a gas and is the least dense or least elastic. The speed of sound is the lowest at 343 m/s in air.\n\nWood a solid is the most dense or most elastic. The speed of sound is the highest 3,962 m/s in wood.\n\n9. What is the frequency of a 1.855 m wave of sound traveling underwater at the speed in the table?\n\nf = ?\n\n𝜆 = 1.855 m\n\nv = 1484 m (from table)\n\nv = 𝜆f rearranges to f = v/𝜆\n\nf = 1484/1.855 = 800 Hz\n\n10. How many times different is frequency if wavelength has increased by three times?\n\nRule of Ones: click here for a reminder of how to do this\n\nf = ?\n\n𝜆 = 3 times\n\nv = same so gets a 1\n\nv = 𝜆f rearranges to f = v/𝜆\n\nf = v/𝜆\n\nnew: v/𝜆 over old: v/𝜆 placing 1's everywhere there is no change from the original.\n\n(1/3)/(1/1) = 1/3 or 0.333 times the original" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8919177,"math_prob":0.9990264,"size":5626,"snap":"2020-24-2020-29","text_gpt3_token_len":1449,"char_repetition_ratio":0.14745642,"word_repetition_ratio":0.08745247,"special_character_ratio":0.2618201,"punctuation_ratio":0.085520744,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9991241,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-06-03T09:44:41Z\",\"WARC-Record-ID\":\"<urn:uuid:8369636b-dfb8-4682-b7bd-1947d2002432>\",\"Content-Length\":\"81156\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e629995a-60f8-43a8-b806-27c827a30903>\",\"WARC-Concurrent-To\":\"<urn:uuid:afa23c4c-90b3-4bec-b3e9-9d837695cfa2>\",\"WARC-IP-Address\":\"172.67.186.84\",\"WARC-Target-URI\":\"https://www.stickmanphysics.com/stickman-physics-home/unit-10-waves/wave-math/\",\"WARC-Payload-Digest\":\"sha1:LIPWHWYVQVBFFW6NJFPSKBJQI7ZG3DSG\",\"WARC-Block-Digest\":\"sha1:RIZRQ62AVQWZQ7AVHZ65AYDHYEDM5PC4\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-24/CC-MAIN-2020-24_segments_1590347432521.57_warc_CC-MAIN-20200603081823-20200603111823-00226.warc.gz\"}"}
https://mathematica.stackexchange.com/questions/84445/does-dynamicmodule-cure-problem-with-manipulate	[ "# Does DynamicModule cure Problem with Manipulate?\n\nHopefully, this is a final cure for the Problem with Manipulate introduced on A problem with Manipulate and then continued on Continuation of a Problem with Manipulate. Michael E2 suggested on the last page on surrounding the Manipulate command with a DynamicModule. Here is my work.\n\nBefore I show all of my work, it is important to understand that this is all in a a single notebook. The desire is to not have the manipulate activities affect one another and the second desire is to not have the static stuff in the notebook affect the manipulates and vice-versa.\n\nFirst, clearing the Global workspace and adding three variables that have caused problems in the links above.\n\nClear[\"Global*\"];\na = 2;\nb = 10;\ndx = (b - a)/n;\n\n\nNow, here is my first use of Manipulate.\n\nDynamicModule[{a = 0, b = 1, n, f, dx, rightSum},\nf[x_] := x^2;\ndx[n_] := (b - a)/n;\nrightSum[n_] := Total@Table[f[a + i dx[n]] dx[n], {i, 1., n}];\nManipulate[\nShow[Plot[f[x], {x, a, b}, PlotStyle -> Thick,\nAxesLabel -> {\"x\", \"y\"}],\nGraphics[{Table[{Opacity[0.05], EdgeForm[Gray],\nRectangle[{a + i dx[n], 0}, {a + (i + 1) dx[n],\nf[a + (i + 1) dx[n]]}]}, {i, 0, n - 1, 1}],\nText[\"N = \" <> ToString[n] <> \", R = \" <>\nToString[rightSum[n]], {(a + b)/2, f[b]}]}]], {{n, 10}, 10, 50,\n10, Appearance -> \"Labeled\"}]]\n\n\nWhich produces this image.", null, "Now the first test:\n\nIn:= a\n\nOut= 2\n\nIn:= b\n\nOut= 10\n\nIn:= dx\n\nOut= 8/n\n\nNote that because of the dynamic module, the variables in the workspace were not changed, nor did they have an effect on the variables and definitions in the manipulate activity. Now, the second manipulate activity.\n\nDynamicModule[{a = 0, b = 1, n, f, dx, rightSum},\nf[x_] := x;\ndx[n_] := (b - a)/n;\nrightSum[n_] := Total@Table[f[a + i dx[n]] dx[n], {i, 1., n}];\nManipulate[\nShow[Plot[f[x], {x, a, b}, PlotStyle -> Thick,\nAxesLabel -> {\"x\", \"y\"}],\nGraphics[{Table[{Opacity[0.05], EdgeForm[Gray],\nRectangle[{a + i dx[n], 0}, {a + (i + 1) dx[n],\nf[a + (i + 1) dx[n]]}]}, {i, 0, n - 1, 1}],\nText[\"N = \" <> ToString[n] <> \", R = \" <>\nToString[rightSum[n]], {(a + b)/2, f[b]}]}]], {{n, 10}, 10, 50,\n10, Appearance -> \"Labeled\"}]]\n\n\nWhich produces this image.", null, "I cannot demonstrate this here, but I can let everyone know that the function f defined in the second manipulate (a straight line) did not affect the first manipulate. They remained the same. Now for the variables test.\n\nIn:= a\n\nOut= 2\n\nIn:= b\n\nOut= 10\n\nIn:= dx\n\nOut= 8/n\n\nAgain, the manipulate did not affect the variables in the workspace, nor did they affect the variables in the manipulates.\n\nNow, I am trying this based on one of MichaelE2's comments, which occurs at the bottom of Continuation of a Problem with Manipulate.\n\nWhat do folks think? Is this the best, easiest, and safest approach for students and teachers who are just beginning to learn Mathematica?\n\n## 1 Answer\n\nHere's how I would do it:\n\nManipulate[\nPlot[f[x], {x, a, b}, PlotStyle -> Thick, AxesLabel -> {\"x\", \"y\"},\nEpilog ->\nDynamic@{Table[{Opacity[0.05], EdgeForm[Gray],\nRectangle[{a + i dx[n], 0}, {a + (i + 1) dx[n],\nf[a + (i + 1) dx[n]]}]}, {i, 0, n - 1, 1}],\nText[\"N = \" <> ToString[n] <> \", R = \" <>\nToString[rightSum[n]], {(a + b)/2, f[b]}]}],\n{{n, 10}, 10, 50, 10, Appearance -> \"Labeled\"},\n{{a, 0}, None}, {{b, 1}, None}, {f, None}, {dx, None}, {rightSum, None},\nInitialization :> (\nClear[f, dx, rightSum];\nf[x_] := x;\ndx[n_] := (b - a)/n;\nrightSum[n_] := Total@Table[f[a + i dx[n]] dx[n], {i, 1., n}];)]\n\n\nPoints of comparison:\n\n• It creates a single DynamicModule via Manipulate, which is being used anyway. It's a simpler structure than the nested ones in the OP's code. Nesting DynamicModule is not bad per se, but it's complicated. (The outside DM contains the code to create an instance of the inside DM, when the outside DM is instantiated by the Front End. You can figure the rest out on your own, or not, as you wish. The subtleties rarely matter, but I prefer keeping things simple.)\n\n• The plotted function never changes, but the rectangle graphics do. I isolated them with Dynamic so that they may be updated independently. This means that moving the Manipulator updates only the rectangles and is more responsive.\n\n• Both the OP's method and this one localize all the variables (x being localized by Plot) so that one instance will be completely independent of another. (The use of Clear is necessary because {f, None} causes f first to be initialized to 0, which disrupts the definition of f; an alternative is to use the declaration form {{f, f}, None} instead of {f, None}, which initializes f to itself, i.e., to its Symbol.)\n\nFurther discussion of localization and Manipulate may be found here:\n\n• This is going to seriously wonderful lesson! Thanks for the great effort. – David Aug 7 '15 at 4:50\n• What code would I enter to time your new version vs. my older version? And I would be probably timing the difference after moving the n slide? – David Aug 7 '15 at 15:31\n• In the body of the Manipulate, I would use SessionTime[] and calculate the difference each update; you'd have have to save the time in a variable, use TrackedSymbols to manage the updating, and display the difference somewhere. But since Manipulate` is mainly about human perception of responsiveness, just try it both ways. It will be noticeable if the plot takes a tenth of a second or more to compute; if it takes more than a few tenths, then it will seem a definite improvement. Graphics generally involves the kernel, the front end, and the GPU -- rather complicated to analyze precisely. – Michael E2 Aug 7 '15 at 15:51" ]	[ null, "https://i.stack.imgur.com/nI9rm.png", null, "https://i.stack.imgur.com/wtGEK.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7840906,"math_prob":0.97114843,"size":2828,"snap":"2021-21-2021-25","text_gpt3_token_len":862,"char_repetition_ratio":0.118626066,"word_repetition_ratio":0.3677686,"special_character_ratio":0.35643566,"punctuation_ratio":0.20187794,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99532175,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,6,null,6,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-06-15T04:58:04Z\",\"WARC-Record-ID\":\"<urn:uuid:b4d0f57e-35b6-4a9b-a6fe-6551f470df04>\",\"Content-Length\":\"173682\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:28982b30-23fd-45eb-a833-36fc8c1f7497>\",\"WARC-Concurrent-To\":\"<urn:uuid:351a7456-4f62-42af-a62e-50dfc6e71c7b>\",\"WARC-IP-Address\":\"151.101.1.69\",\"WARC-Target-URI\":\"https://mathematica.stackexchange.com/questions/84445/does-dynamicmodule-cure-problem-with-manipulate\",\"WARC-Payload-Digest\":\"sha1:PNBAKCJBX6G4TQC2YORPPK3TN77DIISD\",\"WARC-Block-Digest\":\"sha1:IKRTQWMEPWHUXZESCQJN77MKR6M3GM47\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-25/CC-MAIN-2021-25_segments_1623487616657.20_warc_CC-MAIN-20210615022806-20210615052806-00606.warc.gz\"}"}
https://physics.stackexchange.com/questions/80668/what-is-the-relation-between-n-2-super-yang-mills-and-its-twist	[ "What is the relation between N=2 super Yang-Mills and its twist\n\nMy question is what is the relation between N=2 super Yang-Mills and its twisted version topological field theory? After twisting N=2 super Yang-Mills, i.e. diagonally embedding $SU(2)'_R$ into $SU(2)_R \\times SU(2)_I$, we get a topological field theory. My question is since N=2 SYM and TQFT are different i.e. one is physical and the other is topological. Why can we use TQFT to calculate partition of N=2 SYM? What are the same for these two different theories?\n\nUpdate: From the second paper of Trimok, the authors claim that SYM under twist are just redefination. How to understand it?\n\n• Not a expert, but they are certainly not the same. Following the original paper ($2.14 \\to 2.18$ ) or this one ($2.3$,$2.4$,$2.30$), the first thing to do is to decompose supersymmetric charges and fields under the representations of the new twisted symmetry group $SU(2)_L \\times SU(2)'_R \\times U(1)_R$, and to write some action. The $N=2$ twisted action is then shown to be equivalent to a topological Yang-Mills action (+ extra-terms) – Trimok Oct 14 '13 at 10:03\n• The main difference is that the observables in the topological field theory are only a subset of the observables in the physical theory. The subset consists of $Q$-closed operators, where $Q$ is the scalar supersymmetry charge present in the twisted theory. – suresh Feb 2 '14 at 1:13" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9320003,"math_prob":0.9927373,"size":589,"snap":"2019-43-2019-47","text_gpt3_token_len":155,"char_repetition_ratio":0.11282051,"word_repetition_ratio":0.0,"special_character_ratio":0.2359932,"punctuation_ratio":0.11904762,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9954477,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-10-17T02:42:46Z\",\"WARC-Record-ID\":\"<urn:uuid:d98cc448-d081-45bb-a120-cdee20309ac2>\",\"Content-Length\":\"132086\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d87a1624-c16e-40b4-a788-8cc444f51c1a>\",\"WARC-Concurrent-To\":\"<urn:uuid:7458b4f1-e221-4f1c-a28a-173ec73d25a8>\",\"WARC-IP-Address\":\"151.101.1.69\",\"WARC-Target-URI\":\"https://physics.stackexchange.com/questions/80668/what-is-the-relation-between-n-2-super-yang-mills-and-its-twist\",\"WARC-Payload-Digest\":\"sha1:QXPNM7OU73RJ546QMBB4V2WUO5JJ4ESK\",\"WARC-Block-Digest\":\"sha1:TXVHRGBOG4IIKG6VZ5JAIPVEAZTEBHWB\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-43/CC-MAIN-2019-43_segments_1570986672548.33_warc_CC-MAIN-20191017022259-20191017045759-00557.warc.gz\"}"}
https://forum.vectorworks.net/index.php?/topic/57766-extract-worksheet-column-values/&tab=comments	[ "Jump to content\nDeveloper Wiki and Function Reference Links Read more... ×\n\n## Recommended Posts\n\nHi,\n\nI'm creating a script that convert RGB to CMYK, the Calc it's OK. But, I need to know, how can I extract the information (Values in numbers) contained in worksheet cell (R, G, B Column) to put in VAR (r,g,b) and after the calc process, paste the result of Cstr, Mstr, Ystr and Kstr in other Worksheet Cell Like (C, M, Y, K Column).\n\nThe worksheet name is: CMYK,\n\nPROCEDURE RGBToCMYK;\n{ ------------------------------------------------------ }\n\nCONST { ONLY TO TEST}\nr = 127;\ng = 255;\nb = 212;\n{ ------------------------------------------------------ }\n\nVAR\nc, m, y, k : REAL;\n\n{r, g, b : LONGINT}\nCstr, Mstr, Ystr, Kstr : STRING;\n\nBEGIN\nc := 1 - ( r / 255 );\nm := 1 - ( g / 255 );\ny := 1 - ( b / 255 );\nk := 1;\n\nIF ( c < k ) THEN k := c;\nIF ( m < k ) THEN k := m;\nIF ( y < k ) THEN k := y;\nc := ( c - k ) / ( 1 - k );\nm := ( m - k ) / ( 1 - k );\ny := ( y - k ) / ( 1 - k );\n\nCstr := NUM2STR(4, c);\nMstr := NUM2STR(4, m);\nYstr := NUM2STR(4, y);\nKstr := NUM2STR(4, k);\n\nMESSAGE(CONCAT(Cstr, ' ', Mstr, ' ', Ystr, ' ', Kstr));\nEND;\nRUN(RGBToCMYK);", null, "", null, "#### Share this post\n\n##### Link to post\n\nYou are not going to use Cut/Copy and Paste. You are going to have to read the cell, do the calculations and then write the data to the correct cell.\n\nCheck out the GetWSCellValue in the Function Reference for getting values from cell.\n\nUse SetWSCellFormula to store a value into a cell. The name is a little misleading, but a number is considered a valid \"formula\" in a cell.\n\n•", null, "1\n\n#### Share this post\n\n##### Link to post\n\nHi Pat, How Are you?\n\nThanks, I appreciate your help.\n\nBefore I reading your message, I did it that way below:\n\nPROCEDURE RGBToCMYK;\n{ ------------------------------------------------------ }\n\n{CONST {TESTE}\n{r = 127;\ng = 255;\nb = 212;}\n{ ------------------------------------------------------ }\n\nVAR\nc, m, y, k : REAL;\nvred, vgreen, vblue : LONGINT;\nCstr, Mstr, Ystr, Kstr : STRING;\nhplan : HANDLE;\nnumRows, numColumns : INTEGER;\nvlin : INTEGER;\n\nBEGIN\n\nhplan := GetObject('_Plan_Teste);\nIF (hplan <> NIL) THEN GetWSRowColumnCount(hplan, numRows, numColumns);\n\nvlin := 2;\n\nREPEAT\n\nvred := Str2Num(GetCellStr(hplan, vlin, 4));\nvgreen := Str2Num(GetCellStr(hplan, vlin, 5));\nvblue := Str2Num(GetCellStr(hplan, vlin, 6));\n\nc := 1 - ( vred / 255 );\nm := 1 - ( vgreen / 255 );\ny := 1 - ( vblue / 255 );\nk := 1;\n\nIF ( c < k ) THEN k := c;\nIF ( m < k ) THEN k := m;\nIF ( y < k ) THEN k := y;\nc := ( c - k ) / ( 1 - k );\nm := ( m - k ) / ( 1 - k );\ny := ( y - k ) / ( 1 - k );\n\nCstr := NUM2STR(4, c);\nMstr := NUM2STR(4, m);\nYstr := NUM2STR(4, y);\nKstr := NUM2STR(4, k);\n\nLoadCell(vlin, 7, Cstr);\nLoadCell(vlin, 8, Mstr);\nLoadCell(vlin, 9, Ystr);\nLoadCell(vlin, 10, Kstr);\n\nvlin := vlin+1;\n\nmessage('Linha:', vlin);\n\nUNTIL(vlin>numRows);\n\nEND;\n\nRUN(RGBToCMYK);\n\nAny comments in this code?\n\n## Create an account or sign in to comment\n\nYou need to be a member in order to leave a comment\n\n## Create an account\n\nSign up for a new account in our community. It's easy!\n\nRegister a new account\n\n## Sign in\n\nAlready have an account? Sign in here.\n\nSign In Now\n\n7150 Riverwood Drive, Columbia, Maryland 21046, USA \| Contact Us: 410-290-5114\n\n© 2018 Vectorworks, Inc. All Rights Reserved. Vectorworks, Inc. is part of the Nemetschek Group.\n\n×\n\n• KBASE" ]	[ null, "https://forum.vectorworks.net/uploads/monthly_2018_08/985162492_CapturadeTela2018-08-13as15_42_40.thumb.png.0b458e551c4c8e89ee8e2d3d747b9373.png", null, "https://forum.vectorworks.net/uploads/monthly_2018_08/1170085831_CapturadeTela2018-08-13as15_41_35.thumb.png.473416cd31ce1e16763d7db1599b2bbf.png", null, "https://forum.vectorworks.net/uploads/reactions/thumbs-up-sign_1f44d.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.527969,"math_prob":0.91857386,"size":289,"snap":"2019-35-2019-39","text_gpt3_token_len":87,"char_repetition_ratio":0.12631579,"word_repetition_ratio":0.10526316,"special_character_ratio":0.28373703,"punctuation_ratio":0.22222222,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9953513,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,5,null,5,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-08-20T12:54:41Z\",\"WARC-Record-ID\":\"<urn:uuid:3c5028bb-d68b-4f60-bfc5-14acbb80135a>\",\"Content-Length\":\"80277\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:5a05b06e-0791-4a21-b1fc-b8b946d46ba6>\",\"WARC-Concurrent-To\":\"<urn:uuid:b470997c-e9f5-407c-b99c-5a9028e96bc3>\",\"WARC-IP-Address\":\"54.210.203.144\",\"WARC-Target-URI\":\"https://forum.vectorworks.net/index.php?/topic/57766-extract-worksheet-column-values/&tab=comments\",\"WARC-Payload-Digest\":\"sha1:7N4XYQ7V4EU2NBXLMRNDOH3R7AOHH7XH\",\"WARC-Block-Digest\":\"sha1:WIAOWXRDCBS4TMANRLYCCEN2YEKPJQUX\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-35/CC-MAIN-2019-35_segments_1566027315329.55_warc_CC-MAIN-20190820113425-20190820135425-00299.warc.gz\"}"}
https://community.rstudio.com/t/multivariable-logistic-model-risk-differences-and-multicollinearity/67949	[ "", null, "# Multivariable logistic model - risk differences and multicollinearity\n\nI have made a multivariable logistic model in R using the glm-function.\nThe dependent variable is, of course, binary.\nI have 10 independent variables which are dummy, categorical and numerical variables.\n\nI have 2 questions. One regarding the conversion of the results to risk differences and one regarding multicollinearity.\n\n1) The conversion of the results to risk differences\nWith the summary(glm) I get estimates in log(odds) for each variable and I can calculate odds-ratios (OR). But I am interested in reporting risks and ultimately risk differences.\n\nI would like to do that by calculating an average baseline risk and then \"manipulating\" one variable at a time (e.g. smoking 0 in one calculation and 1 in another calculation) to find the risk difference if a patient smokes.\nI calculate the average baseline risk by using the mean observed value of the dummy and categorical values and multiply them with their estimate and by using the median observed valued of the numerical values and multiply them with their estimate.\n\nThis is all great (I think??) and I can do it manually. But is there a faster way than doing it one at a time since I have 10 variables and it would then be a long piece of code?\n\nreference: https://www.bmj.com/content/348/bmj.f7450\n\n2) Multicollinearity\nI have calculated variance inflation factor (VIF) for my independent variables using the vif function from the faraway package.\nShould I plot my independent variables into a linear regression (lm) and do vif(lm) or use my glm model and do vif(glm)?\nAnd can I use VIF at all when I have those 3 different types of variables? I do get an output, so I guess it is okay then?\nA reference would be great!\n\nWow. I hope it makes sense! Unfortunately I am not allowed to share the data with you guys. But I hope it makes a little sense." ]	[ null, "https://community.rstudio.com/uploads/default/original/3X/5/d/5dc960154a129282ba4283771da2fab6fde146fb.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9110827,"math_prob":0.90389174,"size":1829,"snap":"2020-34-2020-40","text_gpt3_token_len":409,"char_repetition_ratio":0.12547944,"word_repetition_ratio":0.032258064,"special_character_ratio":0.21487151,"punctuation_ratio":0.09392265,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9900045,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-09-22T20:57:06Z\",\"WARC-Record-ID\":\"<urn:uuid:584b996d-96a9-424e-9817-6d0047ece70c>\",\"Content-Length\":\"16664\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f413a1ab-ded6-40c0-8b2c-730c8d93d168>\",\"WARC-Concurrent-To\":\"<urn:uuid:2cd6c49c-bc0d-4e1a-b8e8-aff502c88923>\",\"WARC-IP-Address\":\"167.99.20.217\",\"WARC-Target-URI\":\"https://community.rstudio.com/t/multivariable-logistic-model-risk-differences-and-multicollinearity/67949\",\"WARC-Payload-Digest\":\"sha1:WHZMFLDRNYJO52YLZS4FBEIKLA4KKZBC\",\"WARC-Block-Digest\":\"sha1:ZJBJE62EXRSAUKGXRUENWW2OWVYCCFSH\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-40/CC-MAIN-2020-40_segments_1600400206763.24_warc_CC-MAIN-20200922192512-20200922222512-00282.warc.gz\"}"}
http://lib.mexmat.ru/books/6215	[ "Электронная библиотека Попечительского советамеханико-математического факультета Московского государственного университета\n Главная Ex Libris Книги Журналы Статьи Серии Каталог Wanted Загрузка ХудЛит Справка Поиск по индексам Поиск Форум", null, "Авторизация", null, "Поиск по указателям", null, "", null, "", null, "", null, "", null, "Haug H., Jauho A.-P. — Quantum kinetics in transport and optics of semiconductors", null, "Обсудите книгу на научном форуме", null, "Нашли опечатку?Выделите ее мышкой и нажмите Ctrl+Enter Название: Quantum kinetics in transport and optics of semiconductors Авторы: Haug H., Jauho A.-P. Аннотация: This monograph deals with the quantum kinetics for transport in low-dimensional microstructures and for ultrashort laser pulse spectroscopy. The nonequilibrium Green function theory is described and used for the derivation of the quantum kinetic equations. Numerical methods for the solution of the retarded quantum kinetic equations are discussed and results are presented for quantum high-field transport and for mesoscopic transport phenomena. Quantum beats, polarization decay and non-Markovian behaviour are treated for femtosecond spectroscopy on a microscopic basis. Язык:", null, "Рубрика: Физика/Физика твёрдого тела/Приложения/ Статус предметного указателя: Готов указатель с номерами страниц ed2k: ed2k stats Год издания: 1996 Количество страниц: 315 Добавлена в каталог: 12.09.2005 Операции: Положить на полку \| Скопировать ссылку для форума \| Скопировать ID", null, "Предметный указатель", null, "-structure 30", null, "-point 201 Absorption spectrum 229 Accumulation region 161 Admittance, linear-response 195 Aharonov — Bohm effect 158 Analytic continuation 56 65—68 91 Analytic continuation, weak localization self-energy 105 Ansatz, Generalized Kadanoff — Baym 88—91 145 262 Ansatz, Generalized Kadanoff — Baym, for two bands 211 Ansatz, Kadanoff — Baym 143 212 Anticommutator rule 36 85 Balance equation approach 120 Barker — Ferry equation 146 Binomial coefficients 22 Bloch equations, optical 202 205 Bloch vector 205 Bloch vector, time-development 248 Boltzmann equation, applied to", null, "-structure 30 Boltzmann equation, connection to quantum kinetics 217 Boltzmann equation, derivation of 3—5 Boltzmann equation, eigenfunction expansion of 8—11 Boltzmann equation, elastic impurities 21 78 114 Boltzmann equation, integral form 149 Boltzmann equation, linearization of 8—11 Boltzmann equation, Monte Carlo solution of 11 Boltzmann equation, numerical integration of 11 Broadening, inhomogeneous 206 231 Catch, related to two-step process 73 chemical potential 167 Coherent back-scattering 104 Coherent potential approximation (CPA) 135 Collision broadening 148—151 Collision frequencies 13 Collision term, as an integral operator 9 Collision term, electron-phonon systems 145 Collision term, gauge-invariant 83 Collision term, Gaussian white noise model 112 Collision term, invariants of 13 Collision term, linearization of 96 Collision term, memory effects 212 Collision term, quantum 73 Collision term, quantum, non — Markovian nature 212 Collision term, resonant-level model 143 Commutator rule 37 Complex-time contour 91 Conductivity electrical, Boltzmann result, electrical, nonlinear 154 Conductivity, electrical, Boltzmann result 104 Conductivity, electrical, Boltzmann result, electrical, Drude 108 Conductivity, electrical, Boltzmann result, electrical, linear d.c 102 Conductivity, electrical, Boltzmann result, electrical, linear for elastic impurities 95 98—104 Conservation law, Boltzmann equation 4 Conservation law, energy 217 274 Conservation law, momentum 122 Continuity equation, static 30 Continuity equation, time-dependent 183 Contour, deformation of 66 Correlation function", null, ", approximate for Coulomb island 172 Correlation function", null, ", Gaussian white noise model 111 Correlation function", null, ", relation to observables 87 Correlation function, density-density 294 297 Correlation function, equation-of-motion 72 Correlation function, retarded current-current 98 Correlation function, time-ordered current-current 99 Coulomb island 170 Coulomb potential, advanced 263 266 Coulomb potential, retarded 263 266 Coulomb potential, screened 12 263 Coulomb potential, screened in RPA 266 Coulomb potential, screened, static 12 Coulomb potential, screened, time-dependent 297 Coulomb potential, two-dimensional Fourier transform 12 Coulomb scattering, in mean-field approximation 227 Coupling, adiabatic 65 Current conservation 99 Current standard 179 Current, density 87 Current, interacting model 167 Current, resonant-level model 166 Current, time-averaged 183—184 Current, time-averaged, for resonant-level model 186 Current, time-dependent 182 Current, time-dependent, linear-response 193—195 Current-voltage characteristic 157 Current-voltage characteristic, experimental 168 Current-voltage characteristic, resonant tunneling device 167 Damping constants 205 Damping, Markovian 213 Damping, non — Markovian 213 Density of states, field-dependent, three dimensions 125 Density of states, field-dependent, two dimensions 125 Density of states, in terms of spectral function 41 Density of states, resonant-level model 135 Density of states, time-dependent 129 Density-matrix methods 120 Density-matrix, thermal equilibrium 59 depletion region 161 Detailed balance 14 Detuning 205 Diagrams, crossed 54 105 Diagrams, disconnected 46 65 Diagrams, Feynman 46 64 Diagrams, ladder 99 Diagrams, maximally crossed 105 Diagrams, rain-bow 54 Diamagnetic term 101 Dielectric breakdown 119 Dielectric function, plasmon pole approximation 297 Dielectric function, time-dependent 297 Diffusion, constant 106 Diffusion, Gaussian white noise model 115 Dipole matrix element 201 Disorder averaging 109 Disorder averaging in external fields 133 Distribution function, Bose 8 Distribution function, drifted Maxwellian 32 Distribution function, Fermi 7 Distribution function, local equilibrium 30 Distribution function,, generalized 73 Drift-velocity, quantum 155 Driving term, gauge-invariant 82 Driving term, gauge-invariant, with magnetic field 82 Driving term, generalized 73 Driving term, re-normalization of 73 Driving term, scalar potential gauge 81 Driving term, vector potential gauge 82 Dyson equation for contour-ordered Green function 65 Dyson equation for inter-band Green function 203 213 Dyson equation, complex time 74 91 Dyson equation, eigenfunction representation of 153 Dyson equation, elastic impurities 100 Dyson equation, electron-phonon systems 50 Dyson equation, Gaussian white noise model 110 Dyson equation, integral form 267 Dyson equation, nonequilibrium 85 91 Dyson equation, resonant-level model 166 Dyson equation, resonant-level model, concentration of impurities 133 Dyson equation, resonant-level model, time-dependent 184 Dyson equation, retarded Coulomb potential 264 270 Eigenfunctions for linearized collision term 9 16 Eigenfunctions, norm 9 Eigenfunctions, scalar product 9 Eigenvalues, density of 14 Eigenvalues, spectrum of 14 Einstein summation convention 79 Electron-hole plasma 261 Electron-phonon interaction 161 Energy re-normalization 43 Energy relaxation 115 envelope function 201 Equation-of-motion for Bloch vector 205 Equation-of-motion for nonequilibrium Green functions 71 Equation-of-motion for reduced interband density matrix 277 Equation-of-motion technique 65 163 Equation-of-motion technique for reduced density matrix 199 Equation-of-motion technique, Coulomb island 170 Equation-of-motion technique, elastic impurity problem 52 Equation-of-motion technique, resonant-level model 47 131 Excitonic effects 225 Fano model 47 Fermi’s Golden Rule 4 20 Fermi’s golden rule, applied to polaron scattering 214 Feynman diagrams for elastic impurity problem 53 Feynman diagrams, electron-phonon system 49 Feynman diagrams, two-particle Green function 51 Feynman path integral method 120 Field operator 37 Field operator for quantum-well structure 201 Field operator in terms of Bloch waves 200 Fluctuating energy-levels 195—196 Fluctuation-dissipation theorem 96 171 Fluctuation-dissipation theorem, derivation of 41—44 Fock space 35 Fock space, completeness relation of 36 Fokker — Planck equation 11 Four-wave mixing 229 Four-wave mixing, calculation of 255—258 Four-wave mixing, experiments 258—260 Four-wave mixing, time-resolved 204 232 235 238 253 Fr", null, "hlich coupling 257 Fractional quantum Hall effect 49 Franz — Keldysh effect 125 129 Fredholm iteration 223 Free induction decay 206 Functional differentiation 65 Gauge invariance 79—85 92 Gauge invariance, transformation of functions 79 Gauge transformation 80 Gauge, scalar potential 81 Gauge, vector potential 81 Gaussian white noise model, GWN, 109 Gell — Mann and Low theorem 45 Gradient expansion 75—77 79 Gradient expansion, derivation of 76—77 Gradient operator 76 Green function, advanced, definition of 40 63 Green function, advanced, elastic impurities 54 Green function, advanced, for two bands 211 Green function, advanced, non-interacting contacts 182 Green function, antitime-ordered, definition of 63 Green function, causal 62 99 Green function, causal, definition of 38 63 Green function, contour-ordered 59—68 162 Green function, contour-ordered, definition of 62 Green function, contour-ordered, perturbation expansion for 64 Green function, correlation function, definition of 63 Green function, equilibrium theory of 35—56 Green function, finite temperature, definition of 39 Green function, free-particle 46—47 Green function, free-particle, differential equation for 39 Green function, gauge-invariant 120 123 Green function, greater, definition of 40 63 Green function, higher-order 65 Green function, in scalar potential gauge 81 Green function, in vector potential gauge 81 Green function, inter-band, definition of 202 Green function, lesser, definition of 40 63 Green function, lesser, non-interacting contacts 182 Green function, perturbation expansion of 44 46 Green function, phonon, equilibrium 68 145 Green function, retarded, analytic structure 123 Green function, retarded, approximate for Coulomb island 171 Green function, retarded, definition of 40 63 Green function, retarded, disorder averaged for RLM 133 Green function, retarded, elastic impurities 54 Green function, retarded, field-dependent, scalar potential gauge 120—123 Green function, retarded, field-dependent, vector potential gauge 110 Green function, retarded, for two bands 211 Green function, retarded, gauge-invariant 85 Green function, retarded, Gaussian white noise model 109—111 Green function, retarded, non-interacting contacts 182 Green function, retarded, time-dependent 125 Green function, retarded, time-dependent resonant-level model 185 Green function, time-ordered, definition of 38 63 Green function, two-particle 50 95 99 Green function, two-particle, causal impurity-averaged 100 Green function, two-particle, factorization of 99 Green function, two-particle, integral equation for 101 H-function 6 H-Theorem 5—8 Hamilton — Jacobi equation 288 Hartree — Fock approximation 173 227 High temperature superconductors 49 Hilbert space 9 35 Impurity averaging 51—55 99 Impurity averaging, nonequilibrium 133 Impurity averaging, prescription for 51 133 Initial correlations 66 Interaction, carrier - classical light field 200 201 Interaction, electron - classical light field, in rotating-wave approximation 202 Interaction, electron-phonon, in second quantization 37 Interaction, Fr", null, "hlich 210 Interaction, one-body, in second quantization 37 Interaction, two-particle, in second quantization 37 Interband polarization 204 Intra-collisional field-effect 115 131 Intra-collisional field-effect, 148 149 151 Intraband polarization function 263 inversion 205 Irreversibility, Boltzmann equation 73 Irreversibility, quantum kinetic equation 73 Jellium model 261 Joule heating 142 Kadanoff — Baym equation 72 81 91 111 Kadanoff — Baym equation for inter-band Green function 209 Kadanoff — Baym equation, derivation of 71—73 Keldysh equation 89 91 111 166 171 Keldysh equation for resonant-level model 185 Keldysh equation, derivation of 73—74 Kinetic energy, in terms of a Green function 39 Kinetic equations, numerical solution, stochastic 20 Kinetics, free-carrier 202 Kinetics, linear polarization 218 Kinetics, linearized Coulomb 12—20 Kinetics, Markovian 235 Kinetics, optical inter-band 204 Kinetics, quantum 235 Kinetics, quantum, coupled electron-phonon 241 Kinetics, quantum, exciton 227 Kinetics, quantum, interband 199 Kondo phenomenon 49 162 172 Kondo temperature 174 Kubo formula 95 Landau damping 270 295 Landauer formula 159 165 Landauer formula for time-averaged current 184 Langreth theorem 66 Level-shift function 167 Level-width function 163 167 Level-width function, energy-dependence 164 Level-width function, field-dependence 133 Level-width function, generalized 182 Level-width, elastic 167 Level-width, inelastic 167 Level-width, total 167 Limit, Boltzmann 75—78 Limit, completed collisions of 149 217 232 273 Limit, Markov 223 Limit, wide-band (WBL) 183 185 Limit, zero-field limit 122\n1 2", null, "Реклама", null, 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https://1st-in-babies.com/how-long-is-129-minutes-in-hours-new/	[ "How Long Is 129 Minutes In Hours? New\n\n# How Long Is 129 Minutes In Hours? New\n\nLet’s discuss the question: how long is 129 minutes in hours. We summarize all relevant answers in section Q&A of website 1st-in-babies.com in category: Blog MMO. See more related questions in the comments below.\n\n## How many hours is 1 hour 45 minutes?\n\nTo convert time to just hours:\n\n45 minutes is 45 minutes * (1 hour / 60 minutes) = 45/60 hours = 0.75 hours. 45 seconds is 45 seconds * (1 hour / 3600 seconds) = 45/3600 hours = 0.0125 hours. Adding them all together we have 2 hours + 0.75 hours + 0.0125 hours = 2.7625 hours.\n\n## What is 120 minutes expressed in hours?\n\nFor example, 120 minutes equals 2 hours because 120/60=2.\n\n### Time Conversion (Hours \| Minutes \| Seconds) Math – Tutway\n\nTime Conversion (Hours \| Minutes \| Seconds) Math – Tutway\nTime Conversion (Hours \| Minutes \| Seconds) Math – Tutway\n\n## How long is 1 minute exactly?\n\nThe minute is a unit of time usually equal to 160 (the first sexagesimal fraction) of an hour, or 60 seconds.\n\n## How many hours is 15 minutes?\n\nTherefore, 15 minutes = 15/60 hour = ¼ hour.\n\n## What decimal of 10 hours is a minute?\n\nMinutes to Decimal Hours Calculator\nMinutes Decimal Hours\n9 0.150\n10 0.167\n11 0.183\n12 0.200\n\n## What is 1 hour and 20 minutes as a decimal?\n\nCommon Time to Hours, Minutes, and Seconds Decimal Values\nTime Hours Minutes\n01:00:00 1 hr 60 min\n01:10:00 1.167 hrs 70 min\n01:20:00 1.333 hrs 80 min\n01:30:00 1.5 hrs 90 min\n\n## How many hours are in one hour?\n\nAn hour (symbol: h; also abbreviated hr) is a unit of time conventionally reckoned as 1⁄24 of a day and scientifically reckoned as 3,599–3,601 seconds, depending on conditions. There are 60 minutes in an hour, and 24 hours in a day.\n\n## How long is a second?\n\nSince 1967, the second has been defined as exactly “the duration of 9,192,631,770 periods of the radiation corresponding to the transition between the two hyperfine levels of the ground state of the caesium-133 atom” (at a temperature of 0 K and at mean sea level).\n\n## Why does a minute have 60 seconds?\n\nTHE DIVISION of the hour into 60 minutes and of the minute into 60 seconds comes from the Babylonians who used a sexagesimal (counting in 60s) system for mathematics and astronomy. They derived their number system from the Sumerians who were using it as early as 3500 BC.\n\n### Who decides how long a second is? – John Kitching\n\nWho decides how long a second is? – John Kitching\nWho decides how long a second is? – John Kitching\n\n## How do you say 45 minutes in English?\n\nAt minute 45, we say it’s “quarter to” the next hour. For example, at 5:45, we say it’s “quarter to six” (or 15 minutes before 6:00). At minute 30, we say it’s “half past”. So at 9:30, we would say it’s “half past nine” (or half an hour after 9:00).\n\n## What time is 5p in military time?\n\nMilitary Time / 24 Hour Time Conversion Chart\nRegular Time Military Time\n3:00 p.m. 1500 or 1500 hours\n4:00 p.m. 1600 or 1600 hours\n5:00 p.m. 1700 or 1700 hours\n6:00 p.m. 1800 or 1800 hours\n\n## What is .08 of an hour?\n\nDecimal Hours-to-Minutes Conversion Chart\nMinutes Tenths of an Hour Hundredths of an Hour\n48 .8 .80\n49 .8 .82\n50 .8 .84\n51 .8 .85\n\n## What time will it be 3/4 hour?\n\nThree fourths of an hour is 45 minutes.\n\n## What is the decimal for 45 minutes?\n\nMinute Conversion Chart\nMinutes Decimal Conversion\n45 0.75\n46 0.77\n47 0.78\n48 0.80\n\n## How do you write 45 minutes as a decimal?\n\nAnswer: 45 minutes in decimal is 0.75.\n\n## What is 30 minutes in decimals?\n\nDecimal Hours\n\nUsing our 7:30 example above, we intuitively know that 30 minutes is ‘half an hour. ‘ In decimal format one-half is expressed as ‘. 5’. So in decimal format this is expressed as 7.5 hours (7 and a half hours).\n\n## How long is a day?\n\nDay Length\n\nOn Earth, a solar day is around 24 hours. However, Earth’s orbit is elliptical, meaning it’s not a perfect circle. That means some solar days on Earth are a few minutes longer than 24 hours and some are a few minutes shorter.\n\n### How to Calculate Hours Worked in Excel\n\nHow to Calculate Hours Worked in Excel\nHow to Calculate Hours Worked in Excel\n\n## How many seconds are in a year?\n\none year would equal 365 times 24 times 60 times 60 seconds…or 31,536,000 seconds!\n\n## How did we get 24 hours in a day?\n\nOur 24-hour day comes from the ancient Egyptians who divided day-time into 10 hours they measured with devices such as shadow clocks, and added a twilight hour at the beginning and another one at the end of the day-time, says Lomb. “Night-time was divided in 12 hours, based on the observations of stars.\n\nRelated searches\n\n• how many hours in 130 minutes\n• how long is 1 000 minutes in hours\n• how long is 120 minutes in hours\n• how long is minutes\n• how long is 90 minutes in hours\n• how long is 30 minutes in hours\n• how long is 129 minutes in hours and minutes\n• how long is 129 hours\n• how long is 130 minutes in hours\n• how long is 129 days\n• 2 15 hours in minutes\n• how long is 119 minutes\n• how long is 128 minutes in hours\n\n## Information related to the topic how long is 129 minutes in hours\n\nHere are the search results of the thread how long is 129 minutes in hours from Bing. You can read more if you want.\n\nYou have just come across an article on the topic how long is 129 minutes in hours. If you found this article useful, please share it. Thank you very much." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9058865,"math_prob":0.89113235,"size":4960,"snap":"2022-27-2022-33","text_gpt3_token_len":1469,"char_repetition_ratio":0.19229217,"word_repetition_ratio":0.07348243,"special_character_ratio":0.3310484,"punctuation_ratio":0.12589286,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97989225,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-08-13T18:41:26Z\",\"WARC-Record-ID\":\"<urn:uuid:7d2bf9ab-0303-46e9-a61f-d25045f389d7>\",\"Content-Length\":\"100702\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a4e3e7f3-8e93-4a4f-ae4b-0bed8ca712c1>\",\"WARC-Concurrent-To\":\"<urn:uuid:057fa58e-17ed-4ed7-b12b-4f498266fe5e>\",\"WARC-IP-Address\":\"104.21.65.32\",\"WARC-Target-URI\":\"https://1st-in-babies.com/how-long-is-129-minutes-in-hours-new/\",\"WARC-Payload-Digest\":\"sha1:TPMCWR6KRPAHB53GU47VHBX22JJJPIKT\",\"WARC-Block-Digest\":\"sha1:NN5BPO66CRD6XKMHARZ7FCDYJ3S4W5F4\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-33/CC-MAIN-2022-33_segments_1659882571982.99_warc_CC-MAIN-20220813172349-20220813202349-00120.warc.gz\"}"}
http://www.biomath.nyu.edu/rag/tutorial_rna_matrix	[ "### Program Description\n\nTo understand and use this program, there are a few simple concepts, each with a brief explanation, listed below that might be useful.\n\n### ct file:\n\nThe ct file contains data about the base pairs in a RNA secondary structure. The following is the structure for TRNA12 that was generated by mFold and the ct file that is associated with that structure:\n\n### RNA Secondary Structure", null, "### ct File", null, "In order for the program to run properly, the ct file that you download from the web (Zuker's mFold) or that you generate must appear exactly as the file above. The first line contains the total number of nucleotides in the structure, the energy associated with the fold, and the name of the file. The significance of the columns is as follows:\n\nColumn 1: List of the nucleotides from 1 to N (N = total number of nucleotides).\nColumn 2: List of the type of nucleotide (A, G, U, or C).\nColumn 3: List of the nucleotides increasing from zero to N - 1.\nColumn 4: List of the nucleotides from 2 to N and continuing the column with zeros to fill any empty spaces.\nColumn 5: List of the nucleotides that are paired to those listed in increasing order. Any zeros in the fifth column indicate that the particular nucleotide is unpaired.\nColumn 6: A repeat of column 1.\n\nClick on the following ct file if you would like to view the sample file displayed above as it actually appears in file form: TRNA12\n\n### Laplacian Matrix:\n\nThe Laplacian matrix (L) is a mathematical representation of the connectivity between the vertices in a RNA graph or topology. It's represented by diagonal (D) and adjacency (A) components. The diagonal matrix shows the number of connections each vertex makes with the other vertices along the diagonal of the matrix. The adjacency matrix specifies to which vertices each vertex is connected. In a graphical representation of a RNA structure, any labeling is fine. For example, the tRNA (NDB: TRNA12) structure shown above has the following tree graph structure with the vertices randomly labeled:", null, "The corresponding D and A values are as follows:\n\n### A\n\n 4 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1\n\n 0 1 1 1 1 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0\n\nEach column and row in the above matrices correspond to the graph's vertices. By looking at the diagonal of the diagonal matrix, you can see that vertex 1 is connected to 4 other vertices, vertex 2 is connected to 1 other vertex, and so on. The corresponding adjacency matrix specifies these connections explicitly.\nThe Laplacian matrix is defined from D and A as follows:\n\n### L = D - A\n\nFor the example above, we have as follows:\n\n### L\n\n 4 -1 -1 -1 -1 -1 1 0 0 0 -1 0 1 0 0 -1 0 0 1 0 -1 0 0 0 1\n\nThe Laplacian matrix is a square matrix. Each column and row in the above matrix represents the vertices in the tree graph.\nA value of -1 in the matrix element i,j indicates that vertices i and j are connected. For example, by looking across at row 1, it is apparent that vertex 1 is connected to vertex 2, 3, 4, and 5. By symmetry, the same information is provided by looking down column 1.\nZeros indicate no connectivity between corresponding vertices. For example, vertex 2 is not connected to vertex 4.\nThe diagonals of the Laplacian matrix are always positive integers. They represent the number of connections that the particular vertex makes. For example, vertex 1 is connected to 4 other vertices.\n\n### Laplacian Eigenvalues:\n\nThe Laplacian matrix is used to calculate its corresponding eigenvalues. The total number of vertices in a RNA secondary structure equals the total number of eigenvalues. The eigenvalue that helps to describe the RNA topology is the second eigenvalue. The second eigenvalue describes the compactness of a graph. The range of possible values for the second eigenvalue begin at zero and increase from there. The more compact a graph is, the higher its corresponding second eigenvalue." ]	[ null, "http://www.biomath.nyu.edu/rag/tutorial_images/trna12_pic.jpg", null, "http://www.biomath.nyu.edu/rag/tutorial_images/trna12_ct.jpg", null, "http://www.biomath.nyu.edu/rag/tutorial_images/topo5_graph.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.912382,"math_prob":0.95745695,"size":3822,"snap":"2023-40-2023-50","text_gpt3_token_len":858,"char_repetition_ratio":0.13724463,"word_repetition_ratio":0.025679758,"special_character_ratio":0.21219257,"punctuation_ratio":0.106206894,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9889908,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,8,null,8,null,8,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-01T19:25:02Z\",\"WARC-Record-ID\":\"<urn:uuid:f8934a21-a8f8-45f8-9f8a-b5a68e55e45f>\",\"Content-Length\":\"25958\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c22448b1-a205-4701-9cf0-3fb7401fc980>\",\"WARC-Concurrent-To\":\"<urn:uuid:cb5697a9-2ec4-47e8-a4c7-ac816022a499>\",\"WARC-IP-Address\":\"128.122.250.111\",\"WARC-Target-URI\":\"http://www.biomath.nyu.edu/rag/tutorial_rna_matrix\",\"WARC-Payload-Digest\":\"sha1:JQK2BN4J4JPHFBJKQXJXCM5GOA7ZA6YG\",\"WARC-Block-Digest\":\"sha1:XFLUVRNU7IAZ74FQWZXYVB4LIXKCBZYN\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100304.52_warc_CC-MAIN-20231201183432-20231201213432-00094.warc.gz\"}"}
https://answers.everydaycalculation.com/compare-fractions/45-60-and-84-20	[ "Solutions by everydaycalculation.com\n\n## Compare 45/60 and 84/20\n\n1st number: 45/60, 2nd number: 4 4/20\n\n45/60 is smaller than 84/20\n\n#### Steps for comparing fractions\n\n1. Find the least common denominator or LCM of the two denominators:\nLCM of 60 and 20 is 60\n2. For the 1st fraction, since 60 × 1 = 60,\n45/60 = 45 × 1/60 × 1 = 45/60\n3. Likewise, for the 2nd fraction, since 20 × 3 = 60,\n84/20 = 84 × 3/20 × 3 = 252/60\n4. Since the denominators are now the same, the fraction with the bigger numerator is the greater fraction\n5. 45/60 < 252/60 or 45/60 < 84/20\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 ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.85033846,"math_prob":0.9924552,"size":715,"snap":"2020-10-2020-16","text_gpt3_token_len":257,"char_repetition_ratio":0.1673699,"word_repetition_ratio":0.0,"special_character_ratio":0.43356642,"punctuation_ratio":0.08387097,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99337304,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-02-22T20:04:49Z\",\"WARC-Record-ID\":\"<urn:uuid:c6ed4193-12b8-4423-baed-c039f71f1d3b>\",\"Content-Length\":\"6999\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e8695f65-1003-4291-a2d7-81881699b1ad>\",\"WARC-Concurrent-To\":\"<urn:uuid:cb0a4408-0434-4c9c-af73-e7f5667a7b0c>\",\"WARC-IP-Address\":\"96.126.107.130\",\"WARC-Target-URI\":\"https://answers.everydaycalculation.com/compare-fractions/45-60-and-84-20\",\"WARC-Payload-Digest\":\"sha1:FQRJRFR46AA3GDOW3K2ZIMQUWVXXOIL4\",\"WARC-Block-Digest\":\"sha1:VXVJ4QNRII6SOKLAPRNS3W24OXYJ4STQ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-10/CC-MAIN-2020-10_segments_1581875145713.39_warc_CC-MAIN-20200222180557-20200222210557-00223.warc.gz\"}"}
https://www.audiolabs-erlangen.de/resources/MIR/FMP/C1/C1S3_FrequencyPitch.html	[ "", null, "# Frequency and Pitch\n\nFollowing Section 1.3.2 of [Müller, FMP, Springer 2015], we cover in this notebook the relation between frequency and pitch.\n\n## Sinusoids¶\n\nA sound wave can be visually represented by a waveform. If the points of high and low air pressure repeat in an alternating and regular fashion, the resulting waveform is called periodic. In this case, the period of the wave is defined as the time required to complete a cycle. The frequency, measured in Hertz (Hz), is the reciprocal of the period. The simplest type of periodic waveform is a sinusoid, which is completely specified by its frequency, its amplitude (the peak deviation of the sinusoid from its mean), and its phase (determining where in its cycle the sinusoid is at time zero). The following figure shows a sinusoid with frequency $4~\\mathrm{Hz}$.", null, "## Audible Frequency Range¶\n\nThe higher the frequency of a sinusoidal wave, the higher it sounds. The audible frequency range for humans is between about $20~\\mathrm{Hz}$ and $20000~\\mathrm{Hz}$ ($20~\\mathrm{kHz}$). Other species have different hearing ranges. For example, the top end of a dog's hearing range is about $45~\\mathrm{kHz}$, a cat's is $64~\\mathrm{kHz}$, while bats can even detect frequencies beyond $100~\\mathrm{kHz}$. This is why one can use a dog whistle, which emits ultrasonic sound beyond the human hearing capability, to train and to command animals without disturbing nearby people.\n\nIn the following experiment, we generate a chirp signal which frequency increases by a factor of two (one octave) every second. Starting with $80~\\mathrm{Hz}$, the frequency raises to $20480~\\mathrm{Hz}$ over a total duration of $8$ seconds.\n\nIn :\nimport IPython.display as ipd\nimport numpy as np\nimport sys\n\nsys.path.append('..')\nimport libfmp.c1\n\nFs = 44100\ndur = 1\nfreq_start = 80 * 2**np.arange(8)\nfor f in freq_start:\nif f==freq_start:\nchirp, t = libfmp.c1.generate_chirp_exp_octave(freq_start=f, dur=dur, Fs=Fs, amp=1)\nelse:\nchirp_oct, t = libfmp.c1.generate_chirp_exp_octave(freq_start=f, dur=dur, Fs=Fs, amp=1)\nchirp = np.concatenate((chirp, chirp_oct))\n\nipd.display(ipd.Audio(chirp, rate=Fs))" ]	[ null, "https://www.audiolabs-erlangen.de/resources/MIR/FMP/data/C1_nav.png", null, "https://www.audiolabs-erlangen.de/resources/MIR/FMP/data/C1/FMP_C1_F19.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.86145544,"math_prob":0.99394697,"size":2062,"snap":"2023-40-2023-50","text_gpt3_token_len":546,"char_repetition_ratio":0.12099125,"word_repetition_ratio":0.013422819,"special_character_ratio":0.2584869,"punctuation_ratio":0.1495098,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99561757,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,4,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-10-02T00:52:33Z\",\"WARC-Record-ID\":\"<urn:uuid:4367909b-afc5-476b-87f9-d65eea2f55dc>\",\"Content-Length\":\"1048871\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:40857568-4230-45d2-ba9d-c950fec4bbdc>\",\"WARC-Concurrent-To\":\"<urn:uuid:43f32957-7cbc-4aff-a3f1-a85552466f69>\",\"WARC-IP-Address\":\"131.188.16.208\",\"WARC-Target-URI\":\"https://www.audiolabs-erlangen.de/resources/MIR/FMP/C1/C1S3_FrequencyPitch.html\",\"WARC-Payload-Digest\":\"sha1:C3ZIQ5CEDQAT7FSWSZBSNWNVKV2SX2OM\",\"WARC-Block-Digest\":\"sha1:7ERZFKYWTDAJOTOHK7OTB6AHKWAQNL2B\",\"WARC-Truncated\":\"length\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233510942.97_warc_CC-MAIN-20231002001302-20231002031302-00576.warc.gz\"}"}
https://buildingmathematicians.wordpress.com/2016/07/07/is-that-even-a-problem/	[ "# Is That Even A Problem???\n\nAsk others what problem solving means with regard to mathematics. Many will explain that a problem is when we put a real-world context to the mathematics being learned in class… others might explain a process of how we solve a problem (look at what you know and determine what you want to know, or some other set of strategies or a creative acronym that we have likely seen in school). Sadly, much of what most others would point to as a problem is not really even a problem at all.\n\nI think we all need to consider the real notion of what it means to problem solve…\n\nGeorge Polya shared this:", null, "…Thus, to have a problem means: to search consciously for some action appropriate to attain a clearly conceived, but not immediately attainable, aim. To solve a problem means to find such action. … Some degree of difficulty belongs to the very notion of a problem: where there is no difficulty, there is no problem.\n\nWhat Polya is suggesting here is that if we show students how to do something, and then ask the students to practice that same thing in a context IT ISN’T A PROBLEM!!! A problem in mathematics is like the DOING MATHEMATICS Tasks listed below. Take a look at all 4 sections for a minute:", null, "My thoughts are simple… If a student can relatively quickly determine a course of action about how to get an answer, it isn’t a problem! Even if the calculations are difficult or take a while. The vast majority of what we call problems are actually just contextual practice of things we already knew. Doing word problems IS NOT the same as problem solving!\n\nOn the other hand, if a student has to use REASONING skills, they are thinking, actively trying to figure something out, then and only then are they problem solving!!!\n\nMarian Small has written a short article on her thoughts about problem solving: Marian Small – Problem Solving\n\nWhat are her main messages here? Does this or Polya’s quote change your definition of a problem?\n\nI’ve already written about What does Day 1 Look Like where I shared the importance of starting with problems. So, why should we start with problem solving? If we don’t start there, we aren’t likely ever doing any problem solving at all!\n\nI also think that many hearing this might assume that this means we just hand students problems that they wouldn’t be successful with… Ask everyone to attempt something that they wouldn’t know how to do.\n\nLet’s look at an example:\n\nTake a look at these two grade 8 expectations from Ontario curriculum:\n\n• determine the Pythagorean relationship, through investigation using a variety of tools (e.g., dynamic geometry software; paper and scissors; geoboard) and strategies;\n• solve problems involving right triangles geometrically\n\nMany teachers look at these expectations, think to themselves, Pythagorean Theorem… I can teach that… followed by explicit teaching on a Smartboard (showing a video, modeling how the pythagorean theorem works, followed by some examples for the class to work on together…), then finally some problems that students have to answer in a textbook like this:", null, "This progression neither shows how we know students learn, nor does it even get students to be able to do what is asked!\n\nLet’s take a step back and start to notice what the curriculum is saying more clearly:\n\n• determine the Pythagorean relationship, through investigation using a variety of tools (e.g., dynamic geometry software; paper and scissors; geoboard) and strategies;\n• solve problems involving right triangles geometrically, using the Pythagorean relationship;\n\nWhen we pull apart the verbs and the tools/strategies from the content, we start to notice what the curriculum is telling us our students should actually be doing that day! Remember… these expectations are what OUR STUDENTS should be doing… NOT US!!!\n\nAbove I have colored the verbs blue (These are the actions our students should be doing that day) and tools/strategies orange (specifically HOW our students should be accomplishing the verb).\n\nNow let’s take a quick look at how this might actually play out in the classroom if we are starting with problems like our curriculum states:\n\nLet’s start with the first expectation. Our curriculum often includes the statement “determine through investigation,” yet it is overlooked far too often! Students need to determine this themselves! We need to assess students’ ability to “determine the Pythagorean relationship through investigation.”\n\nThis doesn’t mean we tell students what the theorem is, nor does it mean that we expect everyone to reinvent the theorem… so what does it mean???\n\nWell, it could mean lots of things. Here is one possible suggestion…", null, "Show the figure on the left. Ask students the area of the blue square in the middle. Possibly give a geoboard or paper and scissors for this task. How might students come up with the area? How many different ways might students accomplish this?\n\nShare different approaches as a group. (By the way, some calculate the whole shape and subtract the 4 corner triangles. Others calculate the 4 black rectangles and divide by 2 then add 1 for the middle… others rearrange the shapes to make it make sense).\n\nNow explore how the two pictures are similar / different. What do you notice between the two pictures?\n\nIf the curriculum tells us to “determine through investigation” that is exactly the experience our students need to conceptualize the concept. It is also what we need to assess. This can’t be put on a test easily though, it needs to be observed!\n\nThat second expectation is quite interesting to me too. At the beginning of this post we started talking about what is and isn’t a problem. If our students have now understood what the Pythagorean Theorem is, how can we now make things problematic?\n\nShowing a bunch of diagrams with missing hypotenuses or legs isn’t really problematic!\n\nHowever, something like Dan Meyer’s Taco Cart would be! If you haven’t seen the lesson, take a look:", null, "Oh… and by the way… problem solving isn’t always about answering a question… really, at it’s heart, problem solving is about making sense of things that we didn’t understand before… reasoning though things… noticing things we didn’t notice before… making conjectures and testing them out… Problem solving is the process of LEARNING and DOING MATHEMATICS!\n\nSo I want to leave you with a problem for you to think about: what does this have to do with the Pythagorean Theorem?", null, "Advertisements\n\n## 5 thoughts on “Is That Even A Problem???”\n\n1.", null, "mikeollerton says:\n\n‪#MTBoS I only have one strategy to support the differentiated states students are inevitably in. This is to offer starting point tasks which are accessible and ‘easily’ extendable. Once I have posed a problem, such as‬ using the numbers 1,2,3,4 the + and the = sign, make two 2-digit numbers and find all the different possible totals. Over this and ensuing lessons I will have many conversations with individual, pairs or small groups of students. Because they will be in mixed-attainment groups (no ‘tracking’ or ‘setting-by-‘ability’ in my classrooms) I shall have many in-the-moment unplanned for though not unexpected interactions; it is through these my approach to supporting students, all of whom will, collectively, be in a differentiated state.\n\nIf anyone wishes to see the subsequent extension ideas for my “1,2,3,4,+,=“ problem just email me: [email protected]\n\nLike\n\n1.", null, "Mark Chubb says:\n\nI won hundred percent agree there’s not tracking or setting my ability. Offering different in the moment feedback or extensions is so much healthier than pre-assuming how much everyone could possibly learn.\n\nLike" ]	[ null, "https://buildingmathematicians.files.wordpress.com/2016/07/polya.jpg", null, "https://buildingmathematicians.files.wordpress.com/2016/07/math_task_analysis_guide-level-of-cognitive-demand1.png", null, "https://buildingmathematicians.files.wordpress.com/2016/07/pythagoreanladder1.gif", null, "https://buildingmathematicians.files.wordpress.com/2016/07/maxresdefault-1.jpg", null, "https://buildingmathematicians.files.wordpress.com/2016/07/dan-meyers-taco-cart-three-act-math-task.png", null, "https://pbs.twimg.com/media/CROU7hrUYAAufsn.jpg:large", null, "https://2.gravatar.com/avatar/e379ce308cc9acf778234f7bb6dd39f5", null, "https://0.gravatar.com/avatar/cb3329b4b5bf36cf2dafb59e0b0e04f1", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9447548,"math_prob":0.8770306,"size":7525,"snap":"2019-13-2019-22","text_gpt3_token_len":1574,"char_repetition_ratio":0.122058235,"word_repetition_ratio":0.041533545,"special_character_ratio":0.2013289,"punctuation_ratio":0.09979065,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9721357,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16],"im_url_duplicate_count":[null,5,null,5,null,5,null,5,null,5,null,5,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-05-25T05:19:42Z\",\"WARC-Record-ID\":\"<urn:uuid:d44bedd4-d934-4133-b271-d4d2845d6e0d>\",\"Content-Length\":\"89721\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d8d1094f-7911-48f9-9f56-4cd4cd35436c>\",\"WARC-Concurrent-To\":\"<urn:uuid:c355c060-b304-4e8e-af0b-26ab039721e2>\",\"WARC-IP-Address\":\"192.0.78.13\",\"WARC-Target-URI\":\"https://buildingmathematicians.wordpress.com/2016/07/07/is-that-even-a-problem/\",\"WARC-Payload-Digest\":\"sha1:7XMS4HS25OXMNZVV42U3GRMV4B6XZ6XF\",\"WARC-Block-Digest\":\"sha1:6BYFLKVNPOEQTKLBPV4A53JKSFAWACFR\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-22/CC-MAIN-2019-22_segments_1558232257889.72_warc_CC-MAIN-20190525044705-20190525070705-00035.warc.gz\"}"}
https://export.arxiv.org/abs/2001.05921	[ "cs.DM\n\n# Title: Generalized Fitch Graphs III: Symmetrized Fitch maps and Sets of Symmetric Binary Relations that are explained by Unrooted Edge-labeled Trees\n\nAbstract: Binary relations derived from labeled rooted trees play an import role in mathematical biology as formal models of evolutionary relationships. The (symmetrized) Fitch relation formalizes xenology as the pairs of genes separated by at least one horizontal transfer event. As a natural generalization, we consider symmetrized Fitch maps, that is, symmetric maps $\\varepsilon$ that assign a subset of colors to each pair of vertices in $X$ and that can be explained by a tree $T$ with edges that are labeled with subsets of colors in the sense that the color $m$ appears in $\\varepsilon(x,y)$ if and only if $m$ appears in a label along the unique path between $x$ and $y$ in $T$. We first give an alternative characterization of the monochromatic case and then give a characterization of symmetrized Fitch maps in terms of compatibility of a certain set of quartets. We show that recognition of symmetrized Fitch maps is NP-complete. In the restricted case where $\|\\varepsilon(x,y)\|\\leq 1$ the problem becomes polynomial, since such maps coincide with class of monochromatic Fitch maps whose graph-representations form precisely the class of complete multi-partite graphs.\n Subjects: Discrete Mathematics (cs.DM); Computational Complexity (cs.CC); Data Structures and Algorithms (cs.DS); Combinatorics (math.CO) MSC classes: 68R01, 05C05, 92D15 Cite as: arXiv:2001.05921 [cs.DM] (or arXiv:2001.05921v2 [cs.DM] for this version)\n\n## Submission history\n\nFrom: Marc Hellmuth [view email]\n[v1] Thu, 16 Jan 2020 16:14:36 GMT (51kb,D)\n[v2] Wed, 20 Jan 2021 08:51:04 GMT (87kb,D)\n\nLink back to: arXiv, form interface, contact." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8690247,"math_prob":0.95836437,"size":1756,"snap":"2021-21-2021-25","text_gpt3_token_len":437,"char_repetition_ratio":0.11472603,"word_repetition_ratio":0.0,"special_character_ratio":0.23177676,"punctuation_ratio":0.111455105,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9756105,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-05-08T00:00:04Z\",\"WARC-Record-ID\":\"<urn:uuid:fc498415-18c4-456e-a885-40d3388164eb>\",\"Content-Length\":\"18329\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:1c45eeff-814a-4106-be77-6b0fafc9037d>\",\"WARC-Concurrent-To\":\"<urn:uuid:56a6d40b-782e-406d-908e-b2aa2a54ce45>\",\"WARC-IP-Address\":\"128.84.21.203\",\"WARC-Target-URI\":\"https://export.arxiv.org/abs/2001.05921\",\"WARC-Payload-Digest\":\"sha1:4CJP2TUBEEZLMAXMVGDFFRL5AXSTYZI6\",\"WARC-Block-Digest\":\"sha1:FUZTLEBND55BGHBMBPKAKN37UWJPIVLK\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-21/CC-MAIN-2021-21_segments_1620243988828.76_warc_CC-MAIN-20210507211141-20210508001141-00112.warc.gz\"}"}
https://mpatacchiola.github.io/blog/2019/11/18/bayes-stairs.html	[ "", null, "Some of you have probably recognized the above image. That’s “Relativity” one of the most famous lithograph of the artist Maurits Cornelis Escher. In the Relativity world there is the intersection of multiple orthogonal sources of gravity. There are various stairways, and each stairway can be used to move between two different gravity sources. Another interesting Escher’s lithograph is “Ascending and Descending”, where two lines of anonymous men appear over an impossible staircase, one line ascending while the other descends, in a sort of ritual. Escher was inspired by the work of the psychiatrist Lionel Penrose, the father of the physicist Roger Penrose, who was working on the model of an impossible staircase, today known as Penrose stairs. The illusion was published by Lionel and Roger in 1958 as a scientific article “Impossible objects: A special type of visual illusion”. Escher discovered Penroses’ work in 1959 and, fascinated by the illusion, released the lithograph one year later. Now, here comes the mind twist. Roger Penrose (the son of Lionel) was introduced to Escher’s work in 1954, during a conference, and impressed by the artist drawings decided to realise one by his own. After multiple attempts Roger came out with the Penrose triangle. Rogers showed the triangle to his father Lionel who produced some variants, including the Penrose stairs. So the work of Escher could not be possible without the contribution of the Penroses, and viceversa. I find this loop between Escher and the Penroses fascinating, especially because connected with a loopy staircase.\n\n## The Bayes stairs\n\nTaking inspiration from the Penrose stairs I coined the term Bayes stairs to describe the different levels of inference one can manage in a Bayesian hierarchical model. Like the Penrose stairs the Bayes stairs bend backward in a recursive twist (more on this in the last step). The Bayes stairs have five steps:\n\n1. Maximum Likelihood (ML)\n2. Maximum a Posteriori (MAP)\n3. Maximum Likelihood type II (ML-II)\n4. Maximum a Posteriori type II (MAP-II)\n5. Fully Bayesian treatment\n\nI will show you how climbing the stairs allows us to get closer and closer to a fully Bayesian treatment, just to find ourselves at the very first step once we reach the top of the staircase. Having in mind the Bayes stairs is a useful trick to remember all the options available in Bayesian analysis. Moreover, the stairs can be used during an empirical approach as guidelines. Roughly speaking, the staircase is built such that “more integrals one performs, the more Bayesian one becomes” (Murphy, 2012). Reaching higher steps requires a major effort but also gives a larger payoff.\n\nThis post is mainly based on Chapters 5.5 and 5.6 of the book “Machine learning: a probabilistic perspective” by Murphy, and Chapters 5.5 and 5.6 of the book “Deep Learning” by Goodfellow et al. (yes, there is a loop even in the books chapters). Additional resources are reported at the end of the post.", null, "Prerequisites for a good understanding of the post are basic concepts of probability theory and statistics. For instance, I assume you are familiar with random variables, marginal and conditional distributions, Bayes’s rule, likelihood, and Gaussian distributions.\n\n## Step 1: Maximum Likelihood (ML)\n\nLet’s consider a simple probabilistic graphical model such as $$\\theta \\rightarrow x$$, where $$\\theta$$ are parameters representing our model (it can be a scalar or a vector) and $$\\mathcal{D} = \\{x_n\\}_{n=1}^{N}$$ is a dataset of $$N$$ samples drawn independently from an unknown data-generating distribution $$p(\\mathcal{D})$$. Our goal is to approximate $$p(\\mathcal{D})$$ through $$q(\\mathcal{D} \\vert \\theta)$$, meaning that we aim at minimizing the Kullback-Leibler divergence (KL) between the two distributions. Abusing the notation for the sake of clarity we write:\n\n$D_{\\mathrm{KL}}(p(\\mathcal{D}) \\\| q(\\mathcal{D} \\vert \\theta))=\\int_{-\\infty}^{\\infty} p(\\mathcal{D}) \\log \\left(\\frac{p(\\mathcal{D})}{q(\\mathcal{D} \\vert \\theta)}\\right) d \\mathcal{D}.$\n\nWe can notice that $$p(\\mathcal{D})$$ is not function of the model parameters $$\\theta$$, meaning that we only need to consider $$q(\\mathcal{D} \\vert \\theta)$$ to minimize $$D_{\\mathrm{KL}}$$, and since $$q(\\mathcal{D} \\vert \\theta)$$ appears in the denominator it means we have to maximize it.\n\nWhat we came up with, following this reasoning, is a procedure known as Maximum Likelihood (ML) estimation. The objective of ML is to find\n\n$\\hat{\\theta}_{\\text{ML}} = \\text{argmax}_{\\theta} \\ q(\\mathcal{D} \\vert \\theta).$\n\nThis is generally done by taking the derivative of the log likelihood $$\\log q(\\mathcal{D} \\vert \\theta)$$ with respect to $$\\theta$$ and then maximizing via gradient ascent. Note that in ML we are doing a point estimate of the parameters $$\\theta$$.\n\nThe ML estimator can be easily adapted to deal with datasets $$\\mathcal{D} = \\{(x_n,y_n)\\}_{n=1}^{N}$$ of input-output pairs, and in fact this is the standard assumption in supervised learning. In this case we assume a model $$\\theta \\rightarrow x \\rightarrow y$$ where $$x$$ predicts $$y$$.\n\nExample: let’s suppose that our dataset $$\\mathcal{D} = \\{(x_n,y_n)\\}_{n=1}^{N}$$ is composed by real valued scalars for both input $$x$$ and output $$y$$. We are in the supervised regression case. We model the output $$y$$ as a Gaussian random variable meaning that $$y \\sim \\mathcal{N}(\\mu, \\sigma^2)$$, this is a reasonable assumption most of the times since it takes into account uncertainty in the data. Our model can be represented as a function $$\\mathcal{F}_{\\theta}(x) \\rightarrow \\hat{y}$$ mapping the inputs to some approximated output $$\\hat{y}$$. For instance, if we want to fit a line on the data (linear regression) then $$\\mathcal{F}_{\\theta}(x) = mx + b$$ with parameters $$\\theta = [m,b]$$ representing the slope and bias of a straight line. In this particular case the Gaussin on $$y$$ has mean given by $$\\mu = \\mathcal{F}_{\\theta}(x) = \\hat{y}$$, leading to the following objective\n\n$\\hat{\\theta}_{\\text{ML}} = \\text{argmax}_{\\theta} \\ \\prod_n p(y_n \\vert x_n, \\theta) = \\text{argmax}_{\\theta} \\ \\prod_n \\frac{1}{\\sqrt{2 \\sigma^2 \\pi}} \\ \\exp \\Bigg(-\\frac{(y_n - \\mathcal{F}_{\\theta}(x_n) )^2}{2 \\sigma^{2}} \\Bigg),$\n\nwhere I replaced the likelihood $$p(y \\vert x, \\theta)$$ with the Gaussian distribution $$\\mathcal{N}(y \\vert \\mu=\\mathcal{F}_{\\theta}(x), \\sigma^2)$$. In ML we are maximizing this expression, meaning that the normalization constant can be removed. Moreover, taking the logarithm (that cancels out with the exponential and turns products into sums) and considering the variance to be constant and equal to $$\\sigma^2=1$$, we end up with\n\n$\\hat{\\theta}_{\\text{ML}} = \\text{argmax}_{\\theta} \\ -\\frac{1}{N} \\sum_n (y_n - \\mathcal{F}_{\\theta}(x_n))^2.$\n\nThis is equivalent to the negative Mean Squared Error (MSE) between the data and the model prediction. Maximizing this quantity correspond to minimizing the MSE. Note that, if instead of a linear regressor we were using a neural network to model $$\\mathcal{F}_{\\theta}(x)$$ the results would not change, since backpropagation over the MSE loss has exactly the same meaning.\n\nProblems with ML: for long time ML has been the workhorse of machine learning. For instance, we are implicitly using ML every time we are doing supervised training of a neural network. However, ML is prone to overfitting. A huge model, with millions of parameters, will fit the data almost perfectly, resulting in a very high log likelihood. This means that ML tends to favor complex models against simple ones. A way to attenuate this issue is to introduce a regularization term, this is what we are going to do in the next step…\n\n## Step 2: Maximum a Posteriori (MAP)\n\nIf our goal is to get the point estimate of an unknown real valued quantity, what we can do is to compute the mean, median or mode of the posterior. Those statistics can be good descriptors of the unknown value. Among those quantities, the posterior mode is the most popular choice because finding the mode reduces to finding the maximum, a common optimization problem. This particular choice is called Maximum a Posteriori (MAP).\n\nIn order to move forward to the second step, we need two prerequisites. (i) We have to define a suitable prior distribution parameterized by $$\\theta$$ (we are still considering the case $$\\theta \\rightarrow \\mathcal{D}$$). A smart choice is a conjugate prior from the exponential family, that will help us with the second prerequisite. (ii) We need an analytical expression for the posterior distribution $$p(\\boldsymbol{\\theta} \\vert \\mathcal{D})$$ (since we will need to estimate its derivative). When the two prerequisites are satisfied, MAP consists in maximizing the posterior with respect to the parameters, where the posterior is obtained via Bayes’ rule:\n\n$p(\\boldsymbol{\\theta} \| \\mathcal{D}) = \\frac{p(\\mathcal{D} \| \\boldsymbol{\\theta}) p(\\boldsymbol{\\theta})}{p(\\mathcal{D})}.$\n\nSince we are interested in finding the mode of the posterior, and the mode does not change before and after normalization, we can get rid of the denominator and just consider the numerator:\n\n$p(\\boldsymbol{\\theta} \| \\mathcal{D}) \\propto p(\\mathcal{D} \| \\boldsymbol{\\theta}) p(\\boldsymbol{\\theta}).$\n\nThat’s the final expression we were looking for. Our goal now is to find $$\\hat{\\theta}$$ the value of $$\\theta$$ that maximize the objective, here defined as:\n\n$\\hat{\\theta}_{\\text{MAP}} = \\text{argmax}_{\\theta} \\ p(\\mathcal{D} \| \\boldsymbol{\\theta}) p(\\boldsymbol{\\theta}).$\n\nIf you compare the MAP objective and the objective used in ML (see step 1), you will notice that here we are doing the same thing but we are weighting the likelihood by the prior over parameters $$p(\\boldsymbol{\\theta})$$. The effect of adding the prior is to shift the probability mass towards regions of the parameter space that are preferred a priori. Note that if we trivially assume an uniform prior over $$p(\\boldsymbol{\\theta})$$ we fall back to step 1, since the constant term would cancel out and only the likelihood would really matter. Another advantage of using the prior is its regularization effect, which was missing in setp 1. The prior is a constraint over the parameters and in gradient based learning it forces the weights to be updated in specific directions.\n\nExample: let’s continue the example given in the previous section, and let’s suppose that we want to impose a prior distribution over the parameters $$\\theta$$ of our generic model $$\\mathcal{F}_{\\theta}$$. It would be silly to use an uniform prior, because this is equivalent to ML (previous step). Our likelihood is a Gaussian distribution, therefore we can be smart and use another Gaussian as prior. This will ensure conjugacy if $$\\mathcal{F}_{\\theta}$$ is in an appropriate form. A good choice is a Gaussian with zero mean and variance $$\\tau^2$$\n\n$p(\\theta) = \\prod_i \\mathcal{N}(\\theta_i \\vert 0, \\tau^2) = \\prod_i \\frac{1}{\\sqrt{2 \\tau^2 \\pi}} \\ \\exp \\Bigg(-\\frac{(\\theta_i - 0 )^2}{2 \\tau^{2}} \\Bigg),$\n\nwere we have assumed that $$\\theta$$ is a vector. If we now apply the same considerations of the previous step (removing the normalization constant, taking the logarithm) we end up with\n\n$p(\\theta) \\propto -\\frac{\\lambda}{2} \|\|\\theta\|\|_{2}^{2},$\n\nwhere $$\\lambda$$ is just the precision (reciprocal of the variance), and we have taken the norm of the vector $$\\theta$$ (since the logarithm turns the product into a sum over its components). Now recall that MAP consists in finding the mode of the posterior, where the posterior is given by the likelihood times the prior\n\n$\\hat{\\theta}_{\\text{MAP}} = \\text{argmax}_{\\theta} \\log p(y \\vert x, \\theta) p(\\theta) = \\text{argmax}_{\\theta} \\ -\\frac{1}{N} \\sum_n \\underbrace{(y_n - \\mathcal{F}_{\\theta}(x_n))^2}_{\\text{data-fit}} - \\underbrace{\\frac{\\lambda}{2} \|\|\\theta\|\|_{2}^{2}}_{\\text{penalty}}.$\n\nThe above expression can be decomposed in a data-fit (likelihood) and a penalty (prior) term. The penalty in this case is just an $$l_2$$ regularizer also known as Tikhonov regularization or ridge regression. In general when a model is overfitting there are many large positive and negative values in $$\\theta$$. The regularization term encourages those parameters to be small (close to zero), resulting in a smoother approximation, by using a Gaussian prior with zero mean. Note that if we assign a constant value to $$\\lambda$$ we can tune the strength of the regularization term making the underlying Gaussian distribution more or lees peaked around the mean.\n\nChanging the Gaussian prior into a Laplace prior (or double-exponential prior) we are instead imposing an $$l_1$$ penalty, also known as Lasso regularization. This prior is sharply peaked at the origin and it strongly moves the parameters toward zero.\n\nProblems with MAP: we said that MAP corresponds to a point estimate of the posterior mode. It turns out that the mode is usually quite untypical of the distribution (unlike the mean or median that take the volume of the distribution into account), and this is why MAP can still give a rather poor approximation of the parameters. Another problem with MAP is that the prior can sometimes be an arbitrary choice left to the designer and this may influence the posterior in the low-data regime. As we said, using a uniform prior is pointless since we revert to an ML estimate.\n\n## Step 3: Maximum Likelihood type II (ML-II)\n\nStep 3 is known as an ML-II procedure since it corresponds to maximum likelihood but at a higher level. To be more precise, we are now considering a probabilistic graphical model with two levels $$\\eta \\rightarrow \\theta \\rightarrow \\mathcal{D}$$, where $$\\eta$$ are latent variables representing a hyperprior over the prior. This may sound confusing but if you think about that, it makes sense to model the parameter of our prior as another probability distribution instead of relying on a point estimate. This is an example of a hierarchical (or multi-level) Bayesian model. In order to access the third step, it is necessary to compute the posterior on multiple levels of latent variables.\n\nIn literature ML-II is also known as Empirical Bayes, since the hyperprior distribution is estimated from the data, meaning that the parameters at the highest level of the hierarchy are set to their most likely values, instead of being integrated out. Note that this assumption violates the principle that the prior should be chosen in advance, independently of the data, as needed in a rigorous Bayesian treatment. This is the price to pay at step 3, for having a computationally cheap approximation.\n\nThe main difference with respect to the previous steps is that here the objective is to focus on the set of parameters $$\\eta$$ used to model the hyperprior. The strategy we use to achieve this objective is to analytically marginalize out $$\\boldsymbol{\\theta}$$, leaving us with the simpler problem of just computing $$p(\\eta \\vert \\mathcal{D})$$:\n\n$\\hat{\\boldsymbol{\\eta}}_{\\text{ML-II}}=\\operatorname{argmax}_{\\boldsymbol{\\eta}} \\int p(\\mathcal{D} \| \\boldsymbol{\\theta}) p(\\boldsymbol{\\theta} \| \\boldsymbol{\\eta}) d \\boldsymbol{\\theta}.$\n\nThe above expression is also known as marginal distribution. Marginalizing out $$\\theta$$ is not always possible, for this reason we have to be clever and use conjugate priors when appropriate. Once this has been done we are left with finding $$\\hat{\\eta}$$, at step 3 this is done using ML. As in step 1 we simply take the derivative with respect to $$\\eta$$ and then we maximize via gradient ascent.\n\nProblems with ML-II: type II can be considered an improvement over type I. For instance when both the prior and the likelihood are Gaussian distributions, the empirical Bayes estimators (e.g. the James-Stein estimator) dominate the simpler maximum likelihood estimator in terms of quadratic loss and at the same time provide an artifice to avoid the drawback of a fully Bayesian treatment. As pointed out by Carlin and Louis, the empirical Bayes “offers a way to dangle one’s legs in the Bayesian water without having to jump completely into the pool” (Carlin and Louis, 2000). However, even if type II can be considered an improvement, we are still into an ill-formed Bayesian setting. In this regard, I agree with Dempster in saying that an empirical Bayesian is someone who “breaks the Bayesian egg but then declines to enjoy the Bayesian omelette” (Dempster, 1983).\n\n## Step 4: Maximum a Posteriori type II (MAP-II)\n\nLevel 4 is known as an MAP-II procedure because it corresponds to applying MAP at the higher hyperprior level. Similarly to MAP (see step 2) we need (i) to define a proper hyperprior distribution over $$\\eta$$, and (ii) to have an analytical form for the posterior over $$\\eta$$. Additionally, we also need to analytically marginalize out $$\\boldsymbol{\\theta}$$:\n\n$\\hat{\\boldsymbol{\\eta}}_{\\text{MAP-II}}=\\operatorname{argmax}_{\\boldsymbol{\\eta}} \\int p(\\mathcal{D} \| \\boldsymbol{\\theta}) p(\\boldsymbol{\\theta} \| \\boldsymbol{\\eta}) p(\\boldsymbol{\\eta}) d \\boldsymbol{\\theta}.$\n\nWhat are exactly the parameters $$\\eta$$? Well, it depends from the type of distribution associated to the prior. If our prior is a Gaussian distribution we need to set an hyperprior for the mean and variance of such a Gaussian. Moreover, we also need the posterior over the hyperprior in order to apply MAP-II. To get an analytical posterior it is necessary to use a conjugate hyperprior.\n\nWhere does the strength of type II inference come from? Hierarchies exist in many datasets and modelling them appropriately adds statistical power. The strength comes from borrowing statistical strength from the experience of others. This has been clearly pointed out for the James–Stein estimator, where given case 1, it is possible to learn from the experience of the other $$N-1$$ cases (see Efron, 2012, Chapter 1). The exact meaning of this passage will become evident in the example at the end of the post.\n\n## Step 5: Fully Bayesian\n\nWe are on top at the last step. Over here it is possible to perform inference at any level, and to estimate all the posterior distributions encountered so far. If you are thinking that this is too good to be true then you are right. A fully Bayesian treatment for non-trivial hierarchical models is only possible through sampling methods, such as Markov Chain Monte Carlo (MCMC). This is computationally expensive and it requires some experience in tuning the parameters of the sampler (e.g. warmup period).\n\nWhen to go for a fully Bayesian treatment? Hard to say, it depends from the problem at hand and the data available. We need a fully Bayesian treatment whenever we are not happy with a point estimate, in this case a fully Bayesian treatment would unlock the posterior.\n\nDuring the years have been proposed some methods that represent a compromise between a point estimate and a fully Bayesian treatment, such as the Laplace approximation, expectation propagation, and variational approximation. As the names suggest, all these methods perform an approximation of the posterior distribution. Whether this approximation is good or no depends from the shape of the posterior and the distribution used to approximate it. For instance, if the posterior is multi-modal and we are using a Gaussian to approximate it, then our approximation risks to be rather poor.\n\nNote that step 5 can be considered as step 0, like in Penrose stairs and Escher’s Ascending and Descending. In fact, we can decide to directly go for a fully Bayesian treatment from the very beginning, without having to climb the staircase at all. However, very often only an empirical approach will show which is the right method to use and this requires climbing the Bayes stairs more than once (hopefully not in an eternal loop).\n\n## Example: robotic arms\n\nThe startup Reactive-Bots is producing and selling robotic arms. Their latest product is a complex 6-dof cordless arm which can be used both in industry and academia for manufacturing and research. The distinctive characteristic of the new model is the use of an internal battery which allows deploying the arm in situations where a power socket is not available.\n\nYou have been recently hired in the R&D department and you have been asked to estimate the average power consumption of the arm. Your estimate is particularly important because it will be used to define a software failsafe trigger that protects the battery against an overload. From now on we will be concerned with the problem of finding the mean power consumption, assuming that the variance is given.\n\nStep 1: from a preliminary analysis you notice that there are large power fluctuations due to external and internal factors (e.g. workload, room temperature, etc) therefore to get a good estimate you decide to record the power for a period of several hours in laboratory conditions, and then estimate the mean over $$N$$ different arms.\n\nLet’s formalize the problem defining a dataset $$\\mathcal{D} = \\{x_n\\}_{n=1}^{N}$$ with $$x_n \\in \\mathbb{R}$$ and no labels (unsupervised). The shape of the underlying data generating distribution is unknown, but common sense suggest it may have a bell-like shape, we go for a Gaussian likelihood. The use of a Gaussian distribution is particularly well suited for the problem at hand, because if you get the mean and the standard deviation right, it will be possible to easily detect abnormal spikes and trigger the failsafe. More formally, let’s define $$\\mathcal{F}_{\\theta}$$ as a Gaussian distribution with parameters $$\\theta = [ \\mu, \\sigma^2 ]$$ representing the mean and variance.\n\nNow, we want to use $$\\mathcal{F}_{\\theta}$$ to approximate the data generating distribution. At the first step of the Bayes stairs this can be done through ML estimation. The parameter we need to estimate is the mean $$\\mu$$ (as said above we are not concerned about the variance), this has a closed form expression that can be easily obtained taking the logarithm of the Gaussian and then the derivative\n\n$\\hat{\\mu}_{\\text{ML}}=\\frac{1}{N} \\sum_{n=1}^{N} x_{n}.$\n\nIt turns out that ML estimation of a Gaussian just consists in the estimation of the empirical mean over the $$N$$ data points.\n\nStep 2: there is another information we should take into account in our estimation. The battery has an optimal functional range, respecting this range maximizes the operational life span. Following this line of thoughts you consult the datasheet of the battery and notice that the manufacturer has reported the optimal average and standard deviation that guarantees maximal life span, this can be used as a prior.\n\nTo ensure conjugacy the prior over the mean is defined as a Gaussian with parameters $$\\theta_0 = [ \\mu_0, \\sigma_{0}^{2} ]$$ representing the mean and variance taken from the datasheet. The posterior is given by the product between the prior and the likelihood\n\n$p(\\mu) p(x \\vert \\mu)=\\frac{1}{\\sqrt{2 \\pi} \\sigma_{0}} \\exp \\left(-\\frac{(\\mu-\\mu_{0})^{2}}{2\\sigma_{0}^2} \\right) \\prod_{n=1}^{N} \\frac{1}{\\sqrt{2 \\pi} \\sigma} \\exp \\left(-\\frac{(x_{n}-\\mu)^{2}}{2\\sigma^2}\\right).$\n\nUsing a few tricks (e.g. completing the square) we can get the following form for posterior mean\n\n$\\hat{\\mu}_{\\text{MAP}}=\\frac{\\sigma_{0}^{2} N}{\\sigma_{0}^{2} N+\\sigma^{2}}\\left(\\frac{1}{N} \\sum_{n=1}^{N} x_{n}\\right)+\\frac{\\sigma^{2}}{\\sigma_{0}^{2} N+\\sigma^{2}} \\mu_{0}.$\n\nThe last term shows that MAP estimation of the mean in the Gaussian model is just a linear interpolation between the sample mean (the term in parentheses) and the prior mean $$\\mu_{0}$$, both of them weighted by their variances.\n\nStep 3: Reactive-Bots has finished the design and test of the arms and went on with production and selling. For the second version of the arm there are important updates planned, those are based on the feedback of the customers. In particular it turned out that a single failsafe threshold is not so useful in practice and that it would be ideal to adapt it to the use case. For instance, some customers are using the arms for the fine grained pick and place of small objects, whereas others are using the arms for heavyweight manufacturing. It seems reasonable to decrease the failsafe threshold for the former (to identify anomalies in the low-power regime) and increase it for the latter (to avoid sudden drops in the high-power regime).\n\nEven though you have acquired data from $$N$$ different arms in controlled lab conditions, there is still a wide range of real-world scenarios you have not considered. The arm performance in those settings is not clear to you. It is necessary to investigate the problem acquiring the telemetry from each customer, then find a new estimate of the average power in each application and start working on the upgraded version of the arm controller.\n\nTo formalize the problem we suppose that the dataset $$\\mathcal{D}$$ is divided in chunks such that $$\\mathcal{D} = \\{\\mathcal{D}_m\\}_{m=1}^{M}$$ and that each chunk represents a customer, with the data samples $$\\mathcal{D}_m = \\{x_n\\}_{n=1}^{N_m}$$ being the power measurements acquired via telemetry. Note that each customer bought a different number of arms, indicated as $$N_m$$. As before we assume that the data generating distribution can be modelled with a Gaussian $$\\mathcal{N}(x \\vert \\mu_m, \\sigma^{2}) \\ \\forall x \\in \\mathcal{D}_m$$. In other words, the parametric function $$\\mathcal{F}_{\\theta}$$ is here a Gaussian, with $$\\theta = [ \\mu_m, \\sigma^2 ]$$.\n\nOne way to go would be to use a simple ML or MAP approach (1st and 2nd step of the stairs) to estimate the average power for each customer. However, there are some customers that just bought a few arms, and others that bought thousands of them. While it is easy to find the average power for the data-rich groups, it is significantly more difficult for data-poor groups. It would be great if we could somehow take into account the samples of data-rich groups when estimating the posterior distribution of data-poor groups. It turns out we can do that if we add another level of inference in our probabilistic model.\n\nWe assume that the parameters describing the Gaussian associated to each dataset have been drawn from a common hyper distribution. In our particular case we assume this distribution to be another Gaussian. Note that this is exactly the formulation we used to describe the hyperprior in step 3 of the Bayes stairs. Since we have a Gaussian describing each dataset and a Gaussian as hyperprior, this probabilistic model is often called Gaussian-Gaussian and it is here defined as $$\\eta \\rightarrow \\theta_{m=1}^{M} \\rightarrow x_{n=1}^{N_m}$$, with $$\\eta=[\\nu, \\tau^2]$$ being the mean and variance of the hyperprior. Given these assumptions we can estimate the joint distribution as follows:\n\n$p\\left(\\mu, \\mathcal{D} \\vert \\hat{\\eta}, \\sigma^{2}\\right)=\\prod_{m=1}^{M} \\mathcal{N}\\left(\\mu_{m} \| \\hat{\\nu}, \\hat{\\tau}^{2}\\right) \\prod_{n=1}^{N_{m}} \\mathcal{N}\\left(x_{n m} \| \\mu_{m}, \\sigma^{2}\\right).$\n\nWe have taken the point estimate of the hyperprior parameters $$\\hat{\\eta}$$ since we are in the ML-II setting. Now, we can simplify the above expression considering that the $$N_m$$ Gaussian measurements in group $$m$$ are equivalent to one measurement with mean and variance given by\n\n$\\bar{x}_m = \\frac{1}{N_m} \\sum_{n=1}^{N_m} x_{nm}, \\ \\ \\sigma^{2}_{m} = \\frac{\\sigma^2}{N_m}.$\n\nThe variance shrinks with the number of observations, since we get more and more confident about the true value. We now want to find the posterior distribution of the mean for a specific group $$m$$, this can be done as follows:\n\n$p(\\mu_m, \\vert \\hat{\\eta}, \\mathcal{D}) = \\mathcal{N}\\left(\\mu_m \| \\hat{B}_{m} \\hat{\\nu}+\\left(1-\\hat{B}_{m}\\right) \\bar{x}_{m},\\left(1-\\hat{B}_{m}\\right) \\sigma_{m}^{2}\\right), \\ \\ \\text{with} \\ \\ \\hat{B}_{m} = \\frac{\\sigma_{m}^{2}}{\\sigma_{m}^{2}+\\hat{\\tau}^{2}}.$\n\nIt is worth spending some words to analyze the above expression in particular the shrinkage factor $$\\hat{B}_{m} \\in [0,1]$$. This factor controls the degree of shrinkage towards the hyperprior mean $$\\hat{\\nu}$$. If the sample size $$N_m$$ for group $$m$$ is large, then $$\\sigma^{2}_{m}$$ will be small in comparison to $$\\hat{\\tau}^{2}$$ reducing $$\\hat{B}_{m}$$. When $$\\hat{B}_{m}$$ is small (data-rich groups) then $$\\hat{\\nu}$$ is small and $$\\bar{x}_{m}$$ is large, meaning that we put more weight on the actual measurements respect to the hyperprior mean. When $$\\hat{B}_{m}$$ is large instead (data-poor groups), we get the opposite effect with the hyperprior mean having more weight over the posterior.\n\nThis is exactly what we wanted in our example, data-poor groups will have a small shrinkage factor with the hyperprior counting more. However, we also said that we wanted to take advantage of data-rich groups in the posterior of data-poor groups. How is this obtained? This is automatically done when we perform ML-II on the hyperprior parameters. Taking the derivative with respect to $$\\hat{\\nu}$$ we get that the ML estimate corresponds to\n\n$\\hat{\\nu} = \\frac{1}{M} \\sum_{m=1}^{M} \\bar{x}_m.$\n\nWhat does this expression is telling us? The mean of the hyperprior is given by the average over all samples, therefore groups with more samples will have a larger effect on $$\\hat{\\nu}$$ influencing the posterior of data-poor groups. Note that if we assume $$\\sigma_m = \\sigma$$ for all groups then we just have the James-Stein estimator.\n\nStep 4: if we have good reasons to assume that the hyperprior mean is close to a specific value then we can define a Gaussian prior over $$\\eta$$ embedding such an assumption (yes, that would be a prior over the hyperprior). This has the same regularization effect discussed for MAP type I at step 2, moving the average $$\\hat{\\nu}$$ toward the prior mean. In particular we would end up with a linear interpolation between two terms: (i) the prior mean, and (ii) the sample mean, both of them weighted by their respective variances. Also in this case assuming a uniform prior instead of a Gaussian one, we fall back to step 3 since we would be doing just ML-II. I will skip a detailed derivation here, which is left as an exercise for the reader.\n\nStep 5: at the beginning of this example we made the simplifying assumption that the posterior distribution over the power consumption of the robotic arms could be modelled relatively well by a Gaussian distribution. However, let’s remember that the Gaussian may be a rather poor approximation if the posterior is multimodal. Another reason why we choose a Gaussian is that there is a conjugate prior we can use to get the posterior in a closed form. Moving toward other distributions we can lose this helpful property ending up with an intractable posterior.\n\nIs there a way to avoid this oversimplification? Well yes, that is exactly what you can do in the last step of the Bayes stairs. Methods such as MCMC are distribution agnostic, meaning that you can go for a full Bayesian treatment without worrying to much about the shape of the posterior. However, this incur in a significant computational cost and it also requires a certain experience in tuning the hyperparameters of the sampler. A detailed discussion of this step is out of scope and it would require another blog post.\n\n## Conclusions\n\nIn this post I gave you an overview of Bayesian hierarchical models using the metaphor of the Bayes stairs. Climbing the stairs corresponds to performing Bayesian inference at different levels. It is necessary to have a deep insight into the problem at hand in order to understand which step should be considered the final one. Most of the time this requires an empirical approach, moving up and down, until the optimal solution is reached.\n\nThat’s all folks! If you liked the post please share it with your network and give me some feedback in the comment section below. I send you my greetings with the hope that you enjoyed climbing the Bayes stairs.\n\nCarlin, B. P., & Louis, T. A. (2000). Empirical Bayes: Past, present and future. Journal of the American Statistical Association, 95(452), 1286-1289.\n\nDempster A.P. (1983). Parametric empirical Bayes inference: theory and applications. Journal of the American statistical Association.\n\nEfron, B. (2012). Large-scale inference: empirical Bayes methods for estimation, testing, and prediction (Vol. 1). Cambridge University Press.\n\nGoodfellow, I., Bengio, Y., & Courville, A. (2016). Deep learning. MIT press.\n\nMurphy, K. P. (2012). Machine learning: a probabilistic perspective. MIT press." ]	[ null, "https://mpatacchiola.github.io/blog/images/headline_escher_stairs.png", null, "https://mpatacchiola.github.io/blog/images/books_statistical_ml_deep_learning.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8912065,"math_prob":0.99684966,"size":32480,"snap":"2022-27-2022-33","text_gpt3_token_len":7847,"char_repetition_ratio":0.14462988,"word_repetition_ratio":0.01167942,"special_character_ratio":0.24451971,"punctuation_ratio":0.08237452,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9995439,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,2,null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-06-27T08:48:53Z\",\"WARC-Record-ID\":\"<urn:uuid:67708509-cc12-4153-a6ec-4357c2e7c190>\",\"Content-Length\":\"45177\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:eaf62c91-d009-4f03-bac7-77c5213e2648>\",\"WARC-Concurrent-To\":\"<urn:uuid:51498fb4-d2da-405d-bccc-d7f3be202f68>\",\"WARC-IP-Address\":\"185.199.108.153\",\"WARC-Target-URI\":\"https://mpatacchiola.github.io/blog/2019/11/18/bayes-stairs.html\",\"WARC-Payload-Digest\":\"sha1:FYHYNJ4IW54I35Y5JTBSNJUKPW7KCUBJ\",\"WARC-Block-Digest\":\"sha1:GMNVYADF3JF7BSGXPCWG43Z6CIUEGNCX\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656103329963.19_warc_CC-MAIN-20220627073417-20220627103417-00753.warc.gz\"}"}
https://mathoverflow.net/questions/353950/aleksandrovs-proof-of-the-second-order-differentiability-of-convex-functions	[ "# Aleksandrov's proof of the second order differentiability of convex functions\n\nAleksandrov [A], proved a remarkable property of convex functions.\n\nTheorem. If $$f:\\mathbb{R}^n\\to\\mathbb{R}$$ is convex, then for almost every $$x\\in\\mathbb{R}^n$$ there is $$Df(x)\\in\\mathbb{R}^n$$ and a symmetric $$(n\\times n)$$ matrix $$D^2f(x)$$ such that $$\\lim_{y\\to x} \\frac{\|f(y)-f(x)-Df(x)(y-x)-\\frac{1}{2}(y-x)^TD^2f(x)(y-x)\|}{\|y-x\|^2}=0.$$\n\nI know two proofs of this result. One based on the theory of maximal monotone functions and one based on the fact that the second order distributional derivatives of a convex function are Radon measure. Both proofs are mentioned in Second order differentiability of convex functions. Since these proofs use relatively modern techniques not available during Aleksandrov's time, his argument must have been very different.\n\nQuestion 1. Can you briefly explain what was the idea of the original proof due to Aleksandrov?\n\nMy guess would be that his proof was based on methods of differnetial geometry. What else could he use in those days?\n\nQuestion 2. In there any textbook where I can find the original proof due to Aleksandrov?\n\n[A] A. D. Alexandroff, Almost everywhere existence of the second differential of a convex function and some properties of convex surfaces connected with it. (Russian) Leningrad State Univ. Annals [Uchenye Zapiski] Math. Ser. 6, (1939), 3–35.\n\n• There's a paper by Bianchi, Colesanti, and Pucci titled \"On the second differentiability of convex surfaces\" which gives brief synopses of different approaches to the theorem of Alexandroff (including the preceding two dimensional case due to Busemann and Feller). Maybe instead of reading Russian/German the proof can be pieced together from the descriptions there? Mar 2, 2020 at 3:21\n• @WillieWong If you turn it into an answer, I will accept it (unless someone writes a more detailed one). Mar 2, 2020 at 14:16\n\nThe paper gives also a new proof of the theorem, which is claimed to be in the same spirit of the original arguments of Busemann-Feller and Alexandroff. The authors considered the second order difference quotient of the convex function based at a point $$x$$, which they show has a limit a.e. as a convex function. This new convex function is related to the argument of Busemann-Feller in that the indicatrices constructed by Busemann-Feller are the 1-level-sets of this limited convex function." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9072949,"math_prob":0.97122836,"size":1309,"snap":"2022-05-2022-21","text_gpt3_token_len":360,"char_repetition_ratio":0.11340996,"word_repetition_ratio":0.021276595,"special_character_ratio":0.25210086,"punctuation_ratio":0.103846155,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99843985,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-29T09:05:50Z\",\"WARC-Record-ID\":\"<urn:uuid:6e7f2925-57c5-4ad3-b3ac-5b7befa39f88>\",\"Content-Length\":\"109203\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:368e2770-5a7d-4ee4-8100-b9925d77befe>\",\"WARC-Concurrent-To\":\"<urn:uuid:3af79ab0-a851-4f48-96a0-28a2e06b40bd>\",\"WARC-IP-Address\":\"151.101.129.69\",\"WARC-Target-URI\":\"https://mathoverflow.net/questions/353950/aleksandrovs-proof-of-the-second-order-differentiability-of-convex-functions\",\"WARC-Payload-Digest\":\"sha1:TPZFELKVEYQOKFQSUDSJFBBMY2WKQP2Z\",\"WARC-Block-Digest\":\"sha1:LXD2CSXAE7KVSK72XHNKAAFDDPLC5WFO\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652663048462.97_warc_CC-MAIN-20220529072915-20220529102915-00591.warc.gz\"}"}
http://ricksci.com/che/chea_atomic_pretest.htm	[ "Class Copy\n\n# Atomic Structure Pretest\n\n atom nucleus protons neutrons electrons electron cloud electron shell isotopes atomic mass units atomic number mass number atomic mass\n\n1. Negatively charged subatomic particle found in the electron cloud.\n2. Positively charged subatomic particle found in the nucleus.\n3. Neutral subatomic particle found in the nucleus.\n4. Central part of an atom made up of the protons and neutrons.\n5. Atoms of the same element having different numbers of neutrons.\n6. Number of protons in an atom.\n7. Average of the masses of naturally occurring isotopes of an element, weighted by abundance.\n8. The space around the nucleus of the atom through which the electrons move.\n9. Mass of an atom in atom mass units it is the sum of the number of protons and neutrons.\n10. Energy level of an e- that we model as the distance an e- circles the nucleus.\n11. Tiniest particle of an element that has all the properties of the element.\n12. Unit to describe the mass of atoms, molecules and subatomic particles.\n\n1. What element is in group 4A, period 2?\n2. Find silicon on the periodic table. How many protons are in silicon?\n3. What is the atomic number of silicon?\n4. What is the atomic mass of silicon?\n5. What is the symbol for silicon?\n6. How many neutrons are in the most common isotope of silicon?\n7. In what group is silicon ?\n8. In what period is silicon?\n9. On the back of your answer sheet, draw a Bohr diagram for silicon." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8402263,"math_prob":0.92694074,"size":1431,"snap":"2019-13-2019-22","text_gpt3_token_len":331,"char_repetition_ratio":0.1513665,"word_repetition_ratio":0.062992126,"special_character_ratio":0.22361985,"punctuation_ratio":0.099236645,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97252434,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-03-19T05:49:33Z\",\"WARC-Record-ID\":\"<urn:uuid:07032df5-9e9e-4996-8114-2b8ca50cb14e>\",\"Content-Length\":\"7371\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:95293588-d8a6-4a7a-ad88-e86b7520de50>\",\"WARC-Concurrent-To\":\"<urn:uuid:b041c9aa-4520-4ede-b0a2-f5d2d8d24c1f>\",\"WARC-IP-Address\":\"198.54.115.229\",\"WARC-Target-URI\":\"http://ricksci.com/che/chea_atomic_pretest.htm\",\"WARC-Payload-Digest\":\"sha1:5BB4B7QFELWB34YVNCLBWEFIEXUMBJVQ\",\"WARC-Block-Digest\":\"sha1:AA35VZRE4YEZXLPL3AF3OBAQY7ILVG3C\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-13/CC-MAIN-2019-13_segments_1552912201904.55_warc_CC-MAIN-20190319052517-20190319074517-00368.warc.gz\"}"}
https://calculatorsonline.org/what-is-31-percent-off-68	[ "# 31 percent off 68\n\nHere you will see step by step solution to calculate 31 percent off by 68. What is final price if original price is 68 and percentage is 31? The final price is 46.92, and the discount is 21.08. Check the detailed explanation of answer given below.\n\n## Answer: 31 percent off 68 is\n\n= 46.92\n\n### How to calculate the number 31 percent off 68?\n\nWith the help of given formula we can get the the percent off value -\n\nFormula 1: Discount = n × P% / 100, P = Discount(off) Percent, n = Number(Orig_price)\n\nHere we have, n = 68, P = 31%\nFormula 2: Result = n - Discount\n\n#### What is 31 percent off 68?\n\nGiven number n = 68, P = 31%\n\n• Put the n and P values in the given formula 1:\n• => 68 × 31%\n=> 68 × 31/100\n\n• Now we need to simplify the fraction by multiply 68 with 31 then divide it by 100\n• => 68 × 31/100 = 2108/100 = 21.08\n• 31% off for 68 = 21.08\n• Now we will use formula 2 to get the final price of 31% off 68\n• = 68 - 21.08\n= 46.92\n\nTherefore, result is for 31% (percent) off 68 is 46.92 and difference is 21.08.\n\n#### Queries related to 31% off 68\n\nWhat is 31% off 68?\n\n21.08 is what percent off 68\\$?\n\nWhat is the final price of \\$68 item when it has 31 percent discount?\n\nWhat is\n%(Percent) off\nNum:" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.87634516,"math_prob":0.9990681,"size":1159,"snap":"2022-40-2023-06","text_gpt3_token_len":377,"char_repetition_ratio":0.16536796,"word_repetition_ratio":0.024590164,"special_character_ratio":0.405522,"punctuation_ratio":0.116981134,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.999406,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-02-02T11:18:38Z\",\"WARC-Record-ID\":\"<urn:uuid:70a5824e-163b-4e53-874a-2205a21dd914>\",\"Content-Length\":\"16799\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:99d6c0c1-fcc3-4c2c-8801-77385c8a478d>\",\"WARC-Concurrent-To\":\"<urn:uuid:a5b04e96-7535-4606-ba1d-66f4dd210d76>\",\"WARC-IP-Address\":\"172.67.209.140\",\"WARC-Target-URI\":\"https://calculatorsonline.org/what-is-31-percent-off-68\",\"WARC-Payload-Digest\":\"sha1:S4NNFUCSUYIK7PYFKE3TCG4NWGAMSDZC\",\"WARC-Block-Digest\":\"sha1:YFNRZKR5OFCC26OBGIEYCJQ5U3A6S7FE\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764500017.27_warc_CC-MAIN-20230202101933-20230202131933-00344.warc.gz\"}"}
https://community.dataiku.com/t5/General-Discussion/Retrieve-result-of-an-execute-sql-step-of-a-scenario-and-assign/m-p/11644/highlight/true	[ "# Retrieve result of an execute sql step of a scenario and assign to variable\n\nSolved!", null, "Level 2\n###### Retrieve result of an execute sql step of a scenario and assign to variable\n\nI want to execute an sql in a scenario step: 'select count(*) from table ' and store the result in a variable.\n\nHow can I retrieve the result from the json file of the result set ?\n\n```Exited the exec sql step {\n\"endedOn\": 0,\n\"success\": true,\n\"updatedRows\": 0,\n\"totalRowsClipped\": false,\n\"totalRows\": 1,\n\"log\": \"\",\n\"columns\": [\n{\n\"name\": \"cnt\",\n\"type\": \"int\",\n\"dssType\": \"INT\",\n\"sqlType\": 4\n}\n],\n\"rows\": [\n[\n\"183448174\"\n]\n],\n\"hasResultset\": true ```\n\n2 Solutions", null, "Dataiker\n\nHi,\n\nwith a SQL step setup like this one:", null, "then you can retrieve the value of the first column on the first row with a \"Define scenario variables\" step like", null, "(note the \".join(',')\" at the end of the expression: this is caused by the fact that getPath(...) evaluates a JSONPath expression and thus returns an array of values)\n\nIf the SQL step has a name that is not friendly with variable names (like: contains spaces), you can also use a \"Execute Python code\" step to do the same, with\n\n``````from dataiku.scenario import Scenario\ns = Scenario()\noutputs = s.get_previous_steps_outputs()\nsql_output = [o['result'] for o in outputs if o[\"stepName\"] == 'the_sql_step']\ns.set_scenario_variables(the_first_value = sql_output['rows'])\n``````", null, "Dataiker\n\nnote that for your use case, you can probably use a \"set project variables\" step directly (in place of the \"define variables\" + \"execute python code\")\n\n4 Replies", null, "Dataiker\n\nHi,\n\nwith a SQL step setup like this one:", null, "then you can retrieve the value of the first column on the first row with a \"Define scenario variables\" step like", null, "(note the \".join(',')\" at the end of the expression: this is caused by the fact that getPath(...) evaluates a JSONPath expression and thus returns an array of values)\n\nIf the SQL step has a name that is not friendly with variable names (like: contains spaces), you can also use a \"Execute Python code\" step to do the same, with\n\n``````from dataiku.scenario import Scenario\ns = Scenario()\noutputs = s.get_previous_steps_outputs()\nsql_output = [o['result'] for o in outputs if o[\"stepName\"] == 'the_sql_step']\ns.set_scenario_variables(the_first_value = sql_output['rows'])\n``````", null, "Level 2\nAuthor\n\nThank You. This helped . After defining the scenario variables , I am using the variables in a custom python step and based on the value update the project variable.\n\nSo , the value can be used in other scenarios. Hope, I am in the right way.", null, "Python Step :\n\ns=Scenario()\nvar1 = s.get_all_variables()['table_count']\nvar2 = s.get_all_variables()['duplicate_count']\n\n#update the project variable with the variables of the scenario\nproject1 = client.get_project('PROJ')\nproject_variables = project1.get_variables()\nproject_variables[\"standard\"][\"cnt\"] = var1\nproject_variables[\"standard\"][\"dup\"] = var2\nproject1.set_variables(project_variables)", null, "Dataiker\n\nnote that for your use case, you can probably use a \"set project variables\" step directly (in place of the \"define variables\" + \"execute python code\")", null, "Level 2\nAuthor\n\nThank you so much 😊", null, "", null, "" ]	[ null, "https://community.dataiku.com/t5/image/serverpage/avatar-name/Avatar11/avatar-theme/candy/avatar-collection/Dataiku/avatar-display-size/message/version/2", null, "https://community.dataiku.com/t5/image/serverpage/avatar-name/Avatar24/avatar-theme/candy/avatar-collection/Dataiku/avatar-display-size/message/version/2", null, "https://community.dataiku.com/t5/image/serverpage/image-id/2209i8372D95389D4E2F7/image-size/large/is-moderation-mode/true", null, "https://community.dataiku.com/t5/image/serverpage/image-id/2210i70EEBF0571D106CF/image-size/large/is-moderation-mode/true", null, "https://community.dataiku.com/t5/image/serverpage/avatar-name/Avatar24/avatar-theme/candy/avatar-collection/Dataiku/avatar-display-size/message/version/2", null, "https://community.dataiku.com/t5/image/serverpage/avatar-name/Avatar24/avatar-theme/candy/avatar-collection/Dataiku/avatar-display-size/message/version/2", null, "https://community.dataiku.com/t5/image/serverpage/image-id/2209i8372D95389D4E2F7/image-size/large/is-moderation-mode/true", null, "https://community.dataiku.com/t5/image/serverpage/image-id/2210i70EEBF0571D106CF/image-size/large/is-moderation-mode/true", null, "https://community.dataiku.com/t5/image/serverpage/avatar-name/Avatar11/avatar-theme/candy/avatar-collection/Dataiku/avatar-display-size/message/version/2", null, "https://community.dataiku.com/t5/image/serverpage/image-id/2212i92E1FEB89FEF44B5/image-size/large/is-moderation-mode/true", null, "https://community.dataiku.com/t5/image/serverpage/avatar-name/Avatar24/avatar-theme/candy/avatar-collection/Dataiku/avatar-display-size/message/version/2", null, "https://community.dataiku.com/t5/image/serverpage/avatar-name/Avatar11/avatar-theme/candy/avatar-collection/Dataiku/avatar-display-size/message/version/2", null, "https://community.dataiku.com/skins/images/2831D38A13C0210445A1A188613DA608/dataiku/images/icon_anonymous_message.png", null, "https://community.dataiku.com/skins/images/2831D38A13C0210445A1A188613DA608/dataiku/images/icon_anonymous_message.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.66786784,"math_prob":0.85893583,"size":3081,"snap":"2023-40-2023-50","text_gpt3_token_len":737,"char_repetition_ratio":0.14624634,"word_repetition_ratio":0.6059322,"special_character_ratio":0.26841936,"punctuation_ratio":0.13824058,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9842002,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28],"im_url_duplicate_count":[null,null,null,null,null,6,null,6,null,null,null,null,null,6,null,6,null,null,null,3,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-10-01T06:30:56Z\",\"WARC-Record-ID\":\"<urn:uuid:7ff16ef8-0f35-4306-b094-ca5fb065fd84>\",\"Content-Length\":\"664575\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7c17fafd-7cb6-4f4e-9da6-a46fbc1e412a>\",\"WARC-Concurrent-To\":\"<urn:uuid:7affe192-c50d-47d5-97fc-0220be893cf8>\",\"WARC-IP-Address\":\"18.160.10.18\",\"WARC-Target-URI\":\"https://community.dataiku.com/t5/General-Discussion/Retrieve-result-of-an-execute-sql-step-of-a-scenario-and-assign/m-p/11644/highlight/true\",\"WARC-Payload-Digest\":\"sha1:ILNOIIJU7TLXRZ364Y4ZNFTMI76PH7OC\",\"WARC-Block-Digest\":\"sha1:BZMACIXHEVYNMZZJOY4UTPX5NX4NXOVP\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233510781.66_warc_CC-MAIN-20231001041719-20231001071719-00099.warc.gz\"}"}
https://forums.wolfram.com/mathgroup/archive/2005/Oct/msg00869.html	[ "", null, "", null, "", null, "", null, "", null, "", null, "", null, "Re: ParametricPlot3D\n\n• To: mathgroup at smc.vnet.net\n• Subject: [mg61717] Re: [mg61693] ParametricPlot3D\n• From: Igor Antonio <igora at wolf-ram.com>\n• Date: Thu, 27 Oct 2005 05:01:44 -0400 (EDT)\n• Organization: Wolfram Research, Inc.\n• References: <[email protected]> <[email protected]>\n• Sender: owner-wri-mathgroup at wolfram.com\n\n```Maurits Haverkort wrote:\n> Dear all\n>\n> I want to plot several surfaces in one plot and used for that\n> ParametricPlot3D. Everything works fine if I enter the definitions for the\n> functions inline. If I define however a variable (list) that holds the\n> different functions it does not work anymore. Does anybody know a way out?\n>\n> The functions I want to plot are for example two spherse of unit radia\n> centered at 000 and 100. (I need in the end to plot funcitons (R[t,f]\n> centered at different orrigens)\n>\n> If I write it inline it works fine\n> ParametricPlot3D[{{Cos[f] Sin[t], Sin[t] Sin[f], Cos[t]}, {Cos[f] Sin[t] +\n> 1, Sin[t] Sin[f], Cos[t]}}, {t, 0, 3.1415}, {f, 0, 2 3.1415}];\n>\n> If however I first define my functions and then try to plot:\n> ToPlot = {{Cos[f] Sin[t], Sin[t] Sin[f], Cos[t]}, {Cos[f] Sin[t] + 1, Sin[t]\n> Sin[f], Cos[t]}};\n> ParametricPlot3D[ToPlot, {t, 0, 3.1415}, {f, 0, 2 3.1415}];\n> I get the error:\n> ParametricPlot3D::ppfun : Argument ToPlot is not a list with three or four\n> elements. More.\n>\n> Since the function I need to plot is a result of a calculation I rather need\n> to input it as a variable. Any sugestions?\n>\n> Thanks,\n> Maurits\n\nMaurits,\n\nParametricPlot3D has the Attribute HoldAll, which means it any of its arguments\nbefore launching:\n\nMathematica 5.2 for Microsoft Windows\n-- Terminal graphics initialized --\n\nIn:= Attributes@ParametricPlot3D\n\nOut= {HoldAll, Protected}\n\nYou need to do an Evaluate[] on the argument of ParametricPlot3D so that ToPlot\nis \"converted\" to its values before being passed to ParametricPlot3D, like this:\n\nToPlot = {{Cos[f], Sin[t], Sin[t] Sin[f], Cos[t]}, {Cos[f] Sin[t] + 1, Sin[t]\nSin[f], Cos[t]}};\n\nParametricPlot3D[Evaluate@ToPlot, {t, 0, 3.1415}, {f, 0, 2 3.1415}];\n\n--\n\nIgor C. Antonio\nWolfram Research, Inc.\nhttp://www.wolfram.com\n\nTo email me personally, remove the dash.\n\n```\n\n• Prev by Date: Re: Re: Language vs. Library why it matters\n• Next by Date: Re: MLPutRealList vs. sequence of MLPutDouble\n• Previous by thread: Re: ParametricPlot3D\n• Next by thread: Re: ParametricPlot3D" ]	[ null, "https://forums.wolfram.com/mathgroup/images/head_mathgroup.gif", null, "https://forums.wolfram.com/mathgroup/images/head_archive.gif", null, "https://forums.wolfram.com/mathgroup/images/numbers/2.gif", null, "https://forums.wolfram.com/mathgroup/images/numbers/0.gif", null, "https://forums.wolfram.com/mathgroup/images/numbers/0.gif", null, "https://forums.wolfram.com/mathgroup/images/numbers/5.gif", null, "https://forums.wolfram.com/mathgroup/images/search_archive.gif", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.6155352,"math_prob":0.7147158,"size":2487,"snap":"2021-43-2021-49","text_gpt3_token_len":805,"char_repetition_ratio":0.141764,"word_repetition_ratio":0.06266318,"special_character_ratio":0.33574587,"punctuation_ratio":0.21631879,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98921305,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-11-30T20:38:42Z\",\"WARC-Record-ID\":\"<urn:uuid:b35dd1f3-3f3d-4802-bacc-ac530a035663>\",\"Content-Length\":\"46679\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:1bcb538a-2169-413d-924f-e026857a812b>\",\"WARC-Concurrent-To\":\"<urn:uuid:b9950d61-f2f2-45ff-b021-ec3bb532b0e1>\",\"WARC-IP-Address\":\"140.177.205.73\",\"WARC-Target-URI\":\"https://forums.wolfram.com/mathgroup/archive/2005/Oct/msg00869.html\",\"WARC-Payload-Digest\":\"sha1:CG5BXIMCPX6GQIBYIITHIF7TBXJOJINA\",\"WARC-Block-Digest\":\"sha1:LEFBRBF4CEUIE2YARMYHU4XRKX3HXB44\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964359073.63_warc_CC-MAIN-20211130201935-20211130231935-00389.warc.gz\"}"}
https://tutorialspoint.com/julia/julia_flow_control.htm	[ "# Julia - Flow Control\n\n#### Julia Programming For Beginners: Learn Julia Programming\n\n73 Lectures 4 hours\n\n#### Julia Programming Language - From Zero to Expert\n\n24 Lectures 3 hours\n\n#### Hello Julia: Learn the New Julia Programming Language\n\n29 Lectures 2.5 hours\n\nAs we know that each line of a program in Julia is evaluated in turn hence it provides many of the control statements (familiar to other programming languages) to control and modify the flow of evaluation.\n\nFollowing are different ways to control the flow in Julia programming language −\n\n• Ternary and compound expressions\n\n• Boolean switching expressions\n\n• If elseif else end (conditional evaluation)\n\n• For end (iterative evaluation)\n\n• While end (iterative conditional evaluation)\n\n• Try catch error throw (exception handling)\n\n• Do blocks\n\n## Ternary expressions\n\nIt takes the form expr ? a : b. It is called ternary because it takes three arguments. The expr is a condition and if it is true then a will be evaluated otherwise b. Example for this is given below −\n\njulia> A = 100\n100\n\njulia> A < 20 ? \"Right\" : \"wrong\"\n\"wrong\"\n\njulia> A > 20 ? \"Right\" : \"wrong\"\n\"Right\"\n\n\n## Boolean Switching expressions\n\nAs the name implies, the Boolean switching expression allows us to evaluate an expression if the condition is met, i.e., the condition is true. There are two operators to combine the condition and expression −\n\n### The && operator (and)\n\nIf this operator is used in the Boolean switching expression, the second expression will be evaluated if the first condition is true. If the first condition is false, the expression will not be evaluated and only the condition will be returned.\n\nExample\n\njulia> isodd(3) && @warn(\"An odd Number!\")\n┌ Warning: An odd Number!\n└ @ Main REPL:1\n\njulia> isodd(4) && @warn(\"An odd Number!\")\nfalse\n\n\n### The \|\| operator (or)\n\nIf this operator is used in the Boolean switching expression, the second expression will be evaluated only if the first condition is false. If the first condition is true, then there is no need to evaluate the second expression.\n\nExample\n\njulia> isodd(3) \|\| @warn(\"An odd Number!\")\ntrue\n\njulia> isodd(4) \|\| @warn(\"An odd Number!\")\n┌ Warning: An odd Number!\n└ @ Main REPL:1\n\n\n## If, elseif and else\n\nWe can also use if, elseif, and else for conditions execution. The only condition is that all the conditional construction should finish with end.\n\n### Example\n\njulia> fruit = \"Apple\"\n\"Apple\"\n\njulia> if fruit == \"Apple\"\nprintln(\"I like Apple\")\nelseif fruit == \"Banana\"\nprintln(\"I like Banana.\")\nprintln(\"But I prefer Apple.\")\nelse\nprintln(\"I don't know what I like\")\nend\n\nI like Apple\n\njulia> fruit = \"Banana\"\n\"Banana\"\n\njulia> if fruit == \"Apple\"\nprintln(\"I like Apple\")\nelseif fruit == \"Banana\"\nprintln(\"I like Banana.\")\nprintln(\"But I prefer Apple.\")\nelse\nprintln(\"I don't know what I like\")\nend\n\nI like Banana.\nBut I prefer Apple.\n\n\n## for loops\n\nSome of the common example of iteration are −\n\n• working through a list or\n\n• set of values or\n\n• from a start value to a finish value.\n\nWe can iterate through various types of objects like arrays, sets, dictionaries, and strings by using “for” loop (for…end construction). Let us understand the syntax with the following example −\n\njulia> for i in 0:5:50\nprintln(i)\nend\n0\n5\n10\n15\n20\n25\n30\n35\n40\n45\n50\n\n\nIn the above code, the variable ‘i’ takes the value of each element in the array and hence will step from 0 to 50 in steps of 5.\n\n### Example (Iterating over an array)\n\nIn case if we iterate through array, it is checked for change each time through the loop. One care should be taken while the use of ‘push!’ to make an array grow in the middle of a particular loop.\n\njulia> c = \njulia> 1-element Array{Int64,1}:\n1\n\njulia> for i in c\npush!(c, i)\n@show c\nsleep(1)\nend\n\nc = [1,1]\nc = [1,1,1]\nc = [1,1,1,1]\n...\n\n\nNote − To exit the output, press Ctrl+c.\n\n## Loop variables\n\nLoop variable is a variable that steps through each item. It exists only inside the loop. It disappears as soon as the loop finishes.\n\n### Example\n\njulia> for i in 0:5:50\n\nprintln(i)\nend\n0\n5\n10\n15\n20\n25\n30\n35\n40\n45\n50\n\njulia> i\nERROR: UndefVarError: i not defined\n\n\n### Example\n\nJulia provides global keyword for remembering the value of the loop variable outside the loop.\n\njulia> for i in 1:10\nglobal hello\nif i % 3 == 0\nhello = i\nend\nend\n\njulia> hello\n9\n\n\n## Variables declared inside a loop\n\nSimilar to Loop Variable, the variables declared inside a loop won’t exist once the loop is finished.\n\n### Example\n\njulia> for x in 1:10\ny = x^2\nprintln(\"$(x) squared is$(y)\")\nend\n\n\n### Output\n\n1 squared is 1\n2 squared is 4\n3 squared is 9\n4 squared is 16\n5 squared is 25\n6 squared is 36\n7 squared is 49\n8 squared is 64\n9 squared is 81\n10 squared is 100\n\njulia> y\nERROR: UndefVarError: y not defined\n\n\n## Continue Statement\n\nThe Continue statement is used to skip the rest of the code inside the loop and start the loop again with the next value. It is mostly used in the case when on a particular iteration you want to skip to the next value.\n\n### Example\n\njulia> for x in 1:10\nif x % 4 == 0\ncontinue\nend\nprintln(x)\nend\n\n\n### Output\n\n1\n2\n3\n5\n6\n7\n9\n10\n\n\n## Comprehensions\n\nGenerating and collecting items something like [n for n in 1:5] is called array comprehensions. It is sometimes called list comprehensions too.\n\n### Example\n\njulia> [X^2 for X in 1:5]\n5-element Array{Int64,1}:\n1\n4\n9\n16\n25\n\n\nWe can also specify the types of elements we want to generate −\n\n### Example\n\njulia> Complex[X^2 for X in 1:5]\n5-element Array{Complex,1}:\n1 + 0im\n4 + 0im\n9 + 0im\n16 + 0im\n25 + 0im\n\n\n## Enumerated arrays\n\nSometimes we would like to go through an array element by element while keeping track of the index number of every element of that array. Julia has enumerate() function for this task. This function gives us an iterable version of something. This function will produce the index number as well as the value at each index number.\n\n### Example\n\njulia> arr = rand(0:9, 4, 4)\n4×4 Array{Int64,2}:\n7 6 5 8\n8 6 9 4\n6 3 0 7\n2 3 2 4\n\njulia> [x for x in enumerate(arr)]\n4×4 Array{Tuple{Int64,Int64},2}:\n(1, 7) (5, 6) (9, 5) (13, 8)\n(2, 8) (6, 6) (10, 9) (14, 4)\n(3, 6) (7, 3) (11, 0) (15, 7)\n(4, 2) (8, 3) (12, 2) (16, 4)\n\n\n## Zipping arrays\n\nUsing the zip() function, you can work through two or more arrays at the same time by taking the 1st element of each array first and then the 2nd one and so on.\n\nFollowing example demonstrates the usage of zip() function −\n\n### Example\n\njulia> for x in zip(0:10, 100:110, 200:210)\nprintln(x)\nend\n(0, 100, 200)\n(1, 101, 201)\n(2, 102, 202)\n(3, 103, 203)\n(4, 104, 204)\n(5, 105, 205)\n(6, 106, 206)\n(7, 107, 207)\n(8, 108, 208)\n(9, 109, 209)\n(10, 110, 210)\n\n\nJulia also handle the issue of different size arrays as follows −\n\njulia> for x in zip(0:15, 100:110, 200:210)\nprintln(x)\nend\n(0, 100, 200)\n(1, 101, 201)\n(2, 102, 202)\n(3, 103, 203)\n(4, 104, 204)\n(5, 105, 205)\n(6, 106, 206)\n(7, 107, 207)\n(8, 108, 208)\n(9, 109, 209)\n(10, 110, 210)\n\njulia> for x in zip(0:10, 100:115, 200:210)\nprintln(x)\nend\n(0, 100, 200)\n(1, 101, 201)\n(2, 102, 202)\n(3, 103, 203)\n(4, 104, 204)\n(5, 105, 205)\n(6, 106, 206)\n(7, 107, 207)\n(8, 108, 208)\n(9, 109, 209)\n(10, 110, 210)\n\n\n## Nested loops\n\nNest a loop inside another one can be done with the help of using a comma (;) only. You do not need to duplicate the for and end keywords.\n\n### Example\n\njulia> for n in 1:5, m in 1:5\n@show (n, m)\nend\n(n, m) = (1, 1)\n(n, m) = (1, 2)\n(n, m) = (1, 3)\n(n, m) = (1, 4)\n(n, m) = (1, 5)\n(n, m) = (2, 1)\n(n, m) = (2, 2)\n(n, m) = (2, 3)\n(n, m) = (2, 4)\n(n, m) = (2, 5)\n(n, m) = (3, 1)\n(n, m) = (3, 2)\n(n, m) = (3, 3)\n(n, m) = (3, 4)\n(n, m) = (3, 5)\n(n, m) = (4, 1)\n(n, m) = (4, 2)\n(n, m) = (4, 3)\n(n, m) = (4, 4)\n(n, m) = (4, 5)\n(n, m) = (5, 1)\n(n, m) = (5, 2)\n(n, m) = (5, 3)\n(n, m) = (5, 4)\n(n, m) = (5, 5)\n\n\n## While loops\n\nWe use while loops to repeat some expressions while a condition is true. The construction is like while…end.\n\n### Example\n\njulia> n = 0\n0\n\njulia> while n < 10\nprintln(n)\nglobal n += 1\nend\n0\n1\n2\n3\n4\n5\n6\n7\n8\n9\n\n\n## Exceptions\n\nExceptions or try…catch construction is used to write the code that checks for the errors and handles them elegantly. The catch phrase handles the problems that occur in the code. It allows the program to continue rather than grind to a halt.\n\n### Example\n\njulia> str = \"string\";\njulia> try\nstr = \"p\"\ncatch e\nprintln(\"the code caught an error: \\$e\")\nprintln(\"but we can easily continue with execution...\")\nend\nthe code caught an error: MethodError(setindex!, (\"string\", \"p\", 1), 0x0000000000006cba)\nbut we can easily continue with execution...\n\n\n## Do block\n\nDo block is another syntax form similar to list comprehensions. It starts at the end and work towards beginning.\n\n### Example\n\njulia> Prime_numbers = [1,2,3,5,7,11,13,17,19,23];\n\njulia> findall(x -> isequal(19, x), Prime_numbers)\n1-element Array{Int64,1}:\n9\n\n\nAs we can see from the above code that the first argument of the find() function. It operates on the second. But with a do block we can put the function in a do…end block construction.\n\njulia> findall(Prime_numbers) do x\nisequal(x, 19)\nend\n1-element Array{Int64,1}:\n9" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.72058487,"math_prob":0.96802336,"size":9669,"snap":"2022-40-2023-06","text_gpt3_token_len":3104,"char_repetition_ratio":0.13998966,"word_repetition_ratio":0.12851627,"special_character_ratio":0.35381114,"punctuation_ratio":0.14859053,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98907447,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-09-28T04:01:51Z\",\"WARC-Record-ID\":\"<urn:uuid:f80c1fd6-6b76-4cad-ad14-cb4af8208340>\",\"Content-Length\":\"42785\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4511ac91-7d73-4b71-bafa-733b5571c2fd>\",\"WARC-Concurrent-To\":\"<urn:uuid:5698af32-d81c-4959-9498-20607f237f8f>\",\"WARC-IP-Address\":\"168.119.212.138\",\"WARC-Target-URI\":\"https://tutorialspoint.com/julia/julia_flow_control.htm\",\"WARC-Payload-Digest\":\"sha1:GGURJVV6E5MY4JARJUQQ4B5PPTDJ5DAH\",\"WARC-Block-Digest\":\"sha1:4BZ3KURQRERV37MMARY2OYGUY7YAZQUD\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030335059.43_warc_CC-MAIN-20220928020513-20220928050513-00748.warc.gz\"}"}