URL
stringlengths 15
1.68k
| text_list
listlengths 1
199
| image_list
listlengths 1
199
| metadata
stringlengths 1.19k
3.08k
|
|---|---|---|---|
https://gamedev.stackexchange.com/questions/167628/how-to-efficiently-no-pointers-store-hex-grid-with-shared-edges-and-vertices/167630 | [
"# How to efficiently (no pointers) store hex grid with shared edges and vertices?\n\nI have a hex grid stored in a 2d array; now I want to add some properties to the edges of the grid (e.g. rivers between hexes) and also potentially to the vertices. How can I store edges and vertices\n\n1. Without duplication, i.e. I don't want to store a river between two hexes in both, or check neighboring hexes - I want to store it once in the edge.\n2. With a fast way to find mapping between things, i.e. from a given edge to two hexes it's dividing, or from a given vertex to its 3 edges, etc.\n3. Efficiently i.e. without tons of pointers in every object to every neighboring object?\n\nI was thinking there should be 3 2d arrays (hexes, vertices, edges) with arithmetic formulas to map things to things. However for the only schemes I can come up with to store vertices and edges (since they don't map naturally onto a 2d array), the mapping formulas between array coordinates are very convoluted. I wonder if there's some simple scheme that I'm missing?\n\n• Are your memory issues really bad enough to spend CPU cycles and complexity instead of RAM for the map? Also, if you're doing 3D rendering/texturing then shared vertices will cause UV mapping problems. – Patrick Hughes Jan 31 '19 at 0:38\n• @PatrickHughes a hex grid in a 2d array can still be stored in RAM. I also understand the desire to not have pointers everywhere: that introduces lots of potential bugs and can make random access more challenging. – Max Jan 31 '19 at 18:14\n\n## 1 Answer\n\nAssuming you have Hex objects stored in a 2d array, just make each Hex own some of its edges and vertices. For example, have each hex own the edges with arrows pointing to them as well as the two vertices between those edges:\n\n __\n/ \\ <3\n\\__/ <2\n^\n1\n\n\nWe only store three edges and two vertices, so where do the others come from? Well, continuing counter-clockwise, edge 4 would be edge 1 from the hex above this one, edge 5 would be edge 2 from the hex to the upper left, etc. Vertices would be accessed similarly. You could add some accessor methods to make this ownership more transparent by automatically going to the appropriate neighbor to find the needed edge/vertex.\n\nThe downside of this approach is that along some of the sides of your grid, there won't be edges and vertices. You can add a set of dummy hexes around these sides and give them a flag like isDummy to indicate that they only exist to provide edges and vertices to other hexes."
]
| [
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.9620404,"math_prob":0.776699,"size":953,"snap":"2021-04-2021-17","text_gpt3_token_len":229,"char_repetition_ratio":0.12012645,"word_repetition_ratio":0.0,"special_character_ratio":0.23084995,"punctuation_ratio":0.11940298,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95659375,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-04-18T00:15:52Z\",\"WARC-Record-ID\":\"<urn:uuid:d2f5550b-7234-4307-8f69-e6d55b57e6f6>\",\"Content-Length\":\"169816\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9d808087-02c8-45b2-9ae6-b5de2f3d6330>\",\"WARC-Concurrent-To\":\"<urn:uuid:353de4c1-2b41-4dbc-b60c-0f48d14fa8f5>\",\"WARC-IP-Address\":\"151.101.65.69\",\"WARC-Target-URI\":\"https://gamedev.stackexchange.com/questions/167628/how-to-efficiently-no-pointers-store-hex-grid-with-shared-edges-and-vertices/167630\",\"WARC-Payload-Digest\":\"sha1:TSIWQQEVT3KAUGSSVJX7RHPE7RVSN2FO\",\"WARC-Block-Digest\":\"sha1:RIJBD4GPTMMTCBR4V6OABAZZQN5FVSOX\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-17/CC-MAIN-2021-17_segments_1618038464065.57_warc_CC-MAIN-20210417222733-20210418012733-00104.warc.gz\"}"} |
https://www.teachoo.com/9003/2862/Question-27-(Or-2nd)/category/CBSE-Class-10-Sample-Paper-for-2019-Boards/ | [
"Question 27 (OR 2 nd question)\n\nThe angle of elevation of a cloud from a point 60 m above the surface of the water of a lake is 30° and the angle of depression of its shadow from the same point in water of lake is 60° . Find the height of the cloud from the surface of water.",
null,
"",
null,
"",
null,
"",
null,
"",
null,
"1. Class 10\n2. Solutions of Sample Papers for Class 10 Boards\n3. CBSE Class 10 Sample Paper for 2019 Boards\n\nTranscript\n\nQuestion 27 (OR 2nd question) The angle of elevation of a cloud from a point 60 m above the surface of the water of a lake is 30° and the angle of depression of its shadow from the same point in water of lake is 60° . Find the height of the cloud from the surface of water. Here, AB = 60 m C is the point of the cloud CF is the height of cloud from surface of water Let CF = h m FD is the shadow of cloud in water So, FD = height of cloud = h m And, Δ BEC & Δ BED are right angle triangles Also, EF = AB = 60 m Now, In right triangle Δ BEC tan B = (𝑆𝑖𝑑𝑒 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑡𝑜 𝑎𝑛𝑔𝑙𝑒\" \" 𝐵)/(𝑆𝑖𝑑𝑒 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑡𝑜 𝑎𝑛𝑔𝑙𝑒\" \" 𝐵) tan 30° = CE/BE 1/√3 = (ℎ − 60)/𝐵𝐸 BE = √3 (h – 60) m Now, In right triangle Δ BED tan B = (𝑆𝑖𝑑𝑒 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑡𝑜 𝑎𝑛𝑔𝑙𝑒\" \" 𝐵)/(𝑆𝑖𝑑𝑒 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑡𝑜 𝑎𝑛𝑔𝑙𝑒\" \" 𝐵) tan 60° = 𝐸𝐷/BE √3 = (ℎ + 60)/𝐵𝐸 √3 BE = h + 60 Putting BE = √3 (h – 60) from (1) √3 × √3 (h – 60) = h + 60 3 (h – 60) = h + 60 3h – 180 = h + 60 3h – h = 60 + 180 2h = 240 h = 240/2 h = 120 m So, height of Cloud = 120 m\n\nCBSE Class 10 Sample Paper for 2019 Boards\n\nClass 10\nSolutions of Sample Papers for Class 10 Boards",
null,
""
]
| [
null,
"https://d1avenlh0i1xmr.cloudfront.net/2e9085ae-1b87-47ec-9997-57075ec91836/slide108.jpg",
null,
"https://d1avenlh0i1xmr.cloudfront.net/d2c57ba6-cb6a-4e85-a510-33b4bcba2457/slide109.jpg",
null,
"https://d1avenlh0i1xmr.cloudfront.net/aa1e5a33-1e06-4707-92ac-45d71b21e544/slide110.jpg",
null,
"https://d1avenlh0i1xmr.cloudfront.net/26eb6360-271d-4e8c-b83d-904ebd3ff973/slide111.jpg",
null,
"https://d1avenlh0i1xmr.cloudfront.net/069987a4-b43e-4068-b81f-49c60412e3a3/slide112.jpg",
null,
"https://delan5sxrj8jj.cloudfront.net/misc/Davneet+Singh.jpg",
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.8349782,"math_prob":0.99798983,"size":1360,"snap":"2019-51-2020-05","text_gpt3_token_len":703,"char_repetition_ratio":0.13274336,"word_repetition_ratio":0.43209878,"special_character_ratio":0.3897059,"punctuation_ratio":0.04276316,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9983329,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12],"im_url_duplicate_count":[null,5,null,5,null,5,null,5,null,5,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-12-16T12:15:26Z\",\"WARC-Record-ID\":\"<urn:uuid:d85458b9-f398-4331-9141-b7a58dcd2ac1>\",\"Content-Length\":\"72620\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a515092c-3857-42b5-b9bc-e239fdea5a51>\",\"WARC-Concurrent-To\":\"<urn:uuid:c95c9fd4-633f-477f-9aac-b2acbd9fc4c0>\",\"WARC-IP-Address\":\"35.174.159.248\",\"WARC-Target-URI\":\"https://www.teachoo.com/9003/2862/Question-27-(Or-2nd)/category/CBSE-Class-10-Sample-Paper-for-2019-Boards/\",\"WARC-Payload-Digest\":\"sha1:NNGMELDQXOQKOIVKWFJWE52MXVBTROEP\",\"WARC-Block-Digest\":\"sha1:UKWB4SIDXRPNX6BSBP3LKQMEYJKDVZRI\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-51/CC-MAIN-2019-51_segments_1575540565544.86_warc_CC-MAIN-20191216121204-20191216145204-00314.warc.gz\"}"} |
http://jsdoc.inflectra.com/HelpReadingPane.ashx?href=js56jsmthgettime.htm | [
"JScript\n\n# getTime Method\n\nReturns the time value in a Date object.\n\n`dateObj.getTime() `\n\nThe required dateObj reference is a Date object.\n\n#### Remarks\n\nThe getTime method returns an integer value representing the number of milliseconds between midnight, January 1, 1970 and the time value in the Date object. The range of dates is approximately 285,616 years from either side of midnight, January 1, 1970. Negative numbers indicate dates prior to 1970.\n\nWhen doing multiple date and time calculations, it is frequently useful to define variables equal to the number of milliseconds in a day, hour, or minute. For example:\n\n```var MinMilli = 1000 * 60\nvar HrMilli = MinMilli * 60\nvar DyMilli = HrMilli * 24```\n\n#### Example\n\nThe following example illustrates the use of the getTime method.\n\n```function GetTimeTest(){\nvar d, s, t;\nvar MinMilli = 1000 * 60;\nvar HrMilli = MinMilli * 60;\nvar DyMilli = HrMilli * 24;\nd = new Date();\nt = d.`getTime`();\ns = \"It's been \"\ns += Math.round(t / DyMilli) + \" days since 1/1/70\";\nreturn(s);\n}```\n\nVersion 1"
]
| [
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.68326783,"math_prob":0.98078746,"size":1019,"snap":"2021-43-2021-49","text_gpt3_token_len":275,"char_repetition_ratio":0.13103448,"word_repetition_ratio":0.03409091,"special_character_ratio":0.28851816,"punctuation_ratio":0.15384616,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9851181,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-12-01T02:36:29Z\",\"WARC-Record-ID\":\"<urn:uuid:820b77bb-225b-4687-84cc-d424bb1c8f45>\",\"Content-Length\":\"3374\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:bc4a9719-8e1e-4e7f-8f44-775c7bff64f5>\",\"WARC-Concurrent-To\":\"<urn:uuid:81fc0348-0a11-4ed7-bfd9-9c22859b6eea>\",\"WARC-IP-Address\":\"3.218.117.200\",\"WARC-Target-URI\":\"http://jsdoc.inflectra.com/HelpReadingPane.ashx?href=js56jsmthgettime.htm\",\"WARC-Payload-Digest\":\"sha1:E7QTWAMGTW47QOBECGS3F6QILC22MPUI\",\"WARC-Block-Digest\":\"sha1:DKZFTWBSZE7NYBXJHFPDWE6WOYJKMBZJ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964359082.78_warc_CC-MAIN-20211201022332-20211201052332-00618.warc.gz\"}"} |
http://cran.revolutionanalytics.com/web/packages/credule/vignettes/credule.html | [
"We will first explain how credit curves are constructed (using a reduced-form model) before showing how the credule package can be used to build credit curves from the CDS quotes. We will look at 2 specific US Issuers as of 27 May 2014: Pfizer (Pfizer Inc - PFE) and Radioshak (RadioShack Corp - RSH).\n\n## Credit Curve Bootstrapping Methodology\n\n### Hazard rate and Survival Probability\n\nThe reduced-form model that we use here is based on the work of Jarrow and Turnbull (1995), who characterize a credit event as the first event of a Poisson counting process which occurs at some time $$t$$ with a probability defined as :\n\n$$\\text{Pr}\\left[\\tau the probability of a default occurring within the time interval [t,t+dt) conditional on surviving to time t, is proportional to some time dependent function \\(\\lambda(t)$$, known as the hazard rate, and the length of the time interval dt. We make the simplifying assumption that the hazard rate process is deterministic. By extension, this assumption also implies that the hazard rate is independent of interest rates and recovery rates. From this definition, we can calculate the continuous time survival probability to the time $$T$$ conditional on surviving to the valuation time $$t_{V}$$ by considering the limit $$\\rightarrow0$$. It can be shown that the survival probability is given by:\n\n$$Q(t_{V},T)=\\exp\\left(-\\int_{t_{V}}^{T}\\lambda(s)\\, ds\\right)$$\n\nFinally, we assume that the hazard rate function is a step-wise constant function. In the below example, the hazard rate between time 0 and 1Y is $$h_{0,1}=1\\%$$ and the hazard rate between between 1Y and 3Y is $$h_{1,3}=2.5\\%$$. Therefore we have the 1Y survival probability $$Q_{0,1}=exp(-h_{0,1}\\times1)=99\\%$$ and the 3Y survival probability $$Q_{1,3}=Q_{0,1}*exp(-h_{1,3}\\times2)=91.9\\%$$.\n\ntenor hazardrate survprob\n1 1.00% 99.00%\n3 2.50% 91.85%\n5 3.00% 79.06%\n7 4.00% 59.75%",
null,
"",
null,
"### Valuing the Premium Leg\n\nThe premium leg is the series of payments of the default swap spread made to maturity or to the time of the credit event, whichever occurs first. It also includes the payment of premium accrued from the previous premium payment date until the time of the credit event. Assume that there are $$n=1,…,N$$ contractual payment dates $$t_1,… ,t_N$$ where $$t_N$$ is the maturity date of the default swap.\n\nThe present value of the premium leg is given by:\n\n$$\\text{PL PV}(t_{V},t_{N})=S(t_{0},t_{N})\\sum_{n=1}^{N}\\Delta(t_{n-1},t_{n},B)Z(t_{V},t_{n})\\left[Q(t_{V},t_{n})+\\frac{1_{PA}}{2}(Q(t_{V},t_{n-1})-Q(t_{V},t_{n}))\\right]$$\n\nwhere:\n\n• N is the total number of premiums. $$n=1,\\ldots,N$$ is the index for these premiums; premium dates are denoted $$t_{1},\\ldots,t_{N}$$. $$t_{N}$$ is the maturity date of the swap\n• $$S(t_{0},t_{N})$$ is the spread of a Credit Default Swap that matures on $$t_{N}$$\n• $$\\Delta(t_{n-1},t_{n},B)$$ is the day count fraction between premium dates $$t_{n-1}$$ and $$t_{n}$$ in the appropriate basis convention denoted by B (ACT/360 in our model)\n• $$Q(t_{V},t_{n})$$ is the arbitrage-free survival probability of the reference entity from valuation time $$t_{V}$$ to premium time $$t_{n}$$\n• $$Z(t_{V},t_{n})$$ is the risk-free discount factor from valuation date to the premium date $$n$$\n• $1_{PA}$equals 1 if premium accrued (PA) are included in the premium leg calculation, 0 otherwise.\n\n### Valuing the Default Leg\n\nThe default leg (or protection leg) is the contingent payment of (100\\% - R) on the face value of the protection made following the credit event. R is the expected recovery rate. In pricing the default leg, it is important to take into account the timing of the credit event because this can have a significant effect on the present value of the protection leg especially for longer maturity default swaps. Within the hazard rate approach we can solve this timing problem by conditioning on each small time interval $$[s,s+ds]$$ between time $$t_V$$ and time $$t_N$$ at which the credit event can occur. We calculate the expected present value of the recovery payment as:\n\n$$\\text{DL PV}(t_{V},t_{N})=(1-R)\\int_{t_{V}}^{t_{N}}Z(t_{V},s)Q(t_{V},s)\\lambda(s)ds$$\n\nThe integral makes this expression tedious to evaluate. It is possible to show that we can, without any material loss of accuracy, simply assume that the credit event can only occur on a finite number M of discrete points per year. In this case, the default leg value can be expressed as:\n\n$$\\text{DL PV}(t_{V},t_{N})=(1-R)\\sum_{m=1}^{M\\times t_{N}}Z(t_{V},t_{m})\\left(Q(t_{V},t_{m-1})-Q(t_{V},t_{m})\\right)$$\n\nwhere:\n\n• M is the number of discrete points per year on which we assume a credit event can happen. $$t_{N}$$ is the maturity of the swap. $$m=1,\\ldots,M\\times t_{N}$$ is the index the discrete point on which a credit event can happen between the valuation date and the maturity of the swap. We choose $$M=12$$ (monthly discretization) to keep the computation lightweight. D. O\\'kane and S. Turnbull (2003) compare the results of a daily discretization ($$M=365$$) and a monthly discretization ($$M=12$$) and shows that monthly intervals has an absolute error which is well inside the bid-offer spread; hence monthly discretization can be used to keep the model fast and simple.\n• $$R$$ is the recovery rate\n• $$Q(t_{V},t_{m})$$ is the arbitrage-free survival probability of the reference entity from valuation time $$t_{V}$$ to premium time $$t_{m}$$\n• $$Z(t_{V},t_{m})$$ is the risk-free discount factor from valuation date to the date m\n\n### Calibration of the Credit Curve\n\nCalibration of the model imply finding an hazard rate (non-cumulative hazard rate) function that matches the market CDS spreads. We make the following assumptions:\n\n• Swap premium payments are made quarterly following a business day calendar\n• Hazard rate is a piece-wise constant function of time (i.e. hazard rates are independent from interest rates)\n• Recovery rate is constant\n\nThe construction of the hazard rate term structure is done by an iterative process called bootstrapping. Let's assume we have quotes for 1Y, 3Y, 5Y and 7Y for a given issuer. From the 1Y CDS spread $$s_{1Y}$$, we will find the hazard rate $$\\lambda_{0,1}$$ which equates the present value of the premium leg and of the protection leg. From the 3Y spread $$s_{3Y}$$, we will find the hazard rate $$\\lambda_{1,3}$$ which equates the present value of the premium leg to the present value of the protection leg. By iterating this process, we obtain the hazard rates: $$\\lambda_{0,1},\\lambda_{1,3},\\lambda_{3,5},\\lambda_{5,7}$$. We can also easily calculate the survival probabilities from this hazard rate term structure (as we have seen earlier).\n\n## Market data\n\nyieldcurveTenor = c(1.0,2.0,3.0,4.0,5.0,7.0,\n10.0,15.0,20.0,30.0)\nyieldcurveRate = c(0.002585,0.005034,0.008981,0.012954,0.016452,\n0.021811,0.027007,0.031718,0.033834,0.035056)\n\n1 2 3 4 5 7 10 15 20 30\n0.26% 0.50% 0.90% 1.30% 1.65% 2.18% 2.70% 3.17% 3.38% 3.51%\ncdsTenor = c(1,2,3,4,5,7,10,15,20,30)\ncdsSpread_PFE = c(0.0003,0.0009,0.0015,0.0021,0.0028,0.0043,0.0061,0.0063, 0.0068,0.0066)\n\n1 2 3 4 5 7 10 15 20 30\n3 9 15 21 28 43 61 63 68 66\ncdsTenor = c(1,2,3,4,5,7,10,15,20,30)\ncdsSpread_RSH = c(0.6405,0.5956,0.5511,0.5144,0.4894,0.4511,0.4156,0.3815,0.3657,0.3506)\n\n1 2 3 4 5 7 10 15 20 30\n6405 5956 5511 5144 4894 4511 4156 3815 3657 3506\n\n## Credit Curves\n\n### RadioShack Corp, 27 May 2014\n\ncreditcurve_RSH = bootstrapCDS(yieldcurveTenor, yieldcurveRate,\ncdsTenor, cdsSpread_RSH,\nRR, premiumFrequency, defaultFrequency, accruedPremium)\n\ntenor survprob hazrate\n1 34.18% 107.37%\n2 15.38% 79.84%\n3 10.22% 40.91%\n4 8.45% 18.95%\n5 6.95% 19.61%\n7 5.66% 10.22%\n10 4.39% 8.51%\n15 3.37% 5.29%\n20 2.13% 9.16%\n30 1.36% 4.48%",
null,
"### Pfizer Inc, 27 May 2014\n\ncreditcurve_PFE = bootstrapCDS(yieldcurveTenor, yieldcurveRate,\ncdsTenor, cdsSpread_PFE,\nRR, premiumFrequency, defaultFrequency, accruedPremium)\n\ntenor survprob hazrate\n1 99.95% 0.05%\n2 99.70% 0.25%\n3 99.25% 0.45%\n4 98.59% 0.66%\n5 97.65% 0.96%\n7 94.92% 1.42%\n10 89.69% 1.89%\n15 84.71% 1.14%\n20 78.37% 1.56%\n30 71.27% 0.95%",
null,
"## Conclusion\n\nIt is interesting to compare the credit curve of these 2 issuers. A quick look at the 5Y spreads (market benchmark) shows that one issuer is much more risky than the other. Pfizer 5Y spread is at 28bp (i.e. 0.28\\%), this is a pretty low CDS spreads (only 3bp more than the 5Y CDS on a sovereign like Finland, as of 3 August 2015). On the other hand, Radioshack 5Y spread is at 4894bp (i.e. 48.94\\%) which is a very high CDS spread. As a comparison, it is more than two times than the Greece 5Y CDS as of 3 August 2015 (2203.70bp).\n\nFuthermore, we can observe than the hazard rate does not have the same dynamic for both issuers. For Pfizer, the hazard rate curve is upward sloping (i.e hazard rate increase over time) whereas for Radioshack, the hazard rate curve is downward sloping. Downward sloping curve is commonly observed for stressed assets/speculative-grade firms (Radioshack rating is CCC as of 27 May 2014) and it translates the investors's expecation of a short term default. In other words, investors think that the issuer has room to improve with age (become less risky) or less potential to worsen considering that it is very risky today.\n\n## References\n\n• JP Morgan. Credit Derivatives: A Primer. JP Morgan Credit Derivatives and Quantitative Research (January 2005)\n• D. O\\'kane and S. Turnbull. Valuation of Credit Default Swaps. Lehman Brothers Quantitative Credit Research (Apr. 2003)\n• Standard CDS Examples. Supporting document for the Implementation of the ISDA CDS Standard Model (Oct. 2012)"
]
| [
null,
"data:image/png;base64,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",
null,
"data:image/png;base64,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",
null,
"data:image/png;base64,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",
null,
"data:image/png;base64,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",
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.90996283,"math_prob":0.9991422,"size":12157,"snap":"2019-35-2019-39","text_gpt3_token_len":3711,"char_repetition_ratio":0.13799062,"word_repetition_ratio":0.8423943,"special_character_ratio":0.32886404,"punctuation_ratio":0.14704804,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9991924,"pos_list":[0,1,2,3,4,5,6,7,8],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-08-17T21:02:41Z\",\"WARC-Record-ID\":\"<urn:uuid:eb7dfeb2-ca33-4993-ad37-a0365850b21f>\",\"Content-Length\":\"52041\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:55e54216-1fe1-4d97-a1f7-2115fdd53c19>\",\"WARC-Concurrent-To\":\"<urn:uuid:0af43ba7-4738-482f-8984-1eb7f78c0443>\",\"WARC-IP-Address\":\"13.66.202.75\",\"WARC-Target-URI\":\"http://cran.revolutionanalytics.com/web/packages/credule/vignettes/credule.html\",\"WARC-Payload-Digest\":\"sha1:OLYDFXLDN7W5XMM3NJU2HA5NV43YCFWE\",\"WARC-Block-Digest\":\"sha1:IQSY4PMMIBPONY62OCTZBKAXIY3RDZQB\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-35/CC-MAIN-2019-35_segments_1566027312025.20_warc_CC-MAIN-20190817203056-20190817225056-00341.warc.gz\"}"} |
http://www.gtmath.com/2016/02/ | [
"## AM-GM Inequality (Part 2)\n\nIn Part 1 of this post, I introduced the AM-GM Inequality and showed a few examples of how to apply it to prove certain inequalities and find the minimum value of certain functions. In Part 2, I will start off with the proof (which I skipped in Part 1) and then proceed to a few more miscellaneous examples of clever applications. If you're not interested in the proof, then you can skip the next section and go straight to the examples below.\n\n#### Proof of the AM-GM Inequality\n\nAM-GM Inequality: For $x_1, x_2, \\dotsc , x_n \\geq 0$, $$\\dfrac{x_1 + x_2 + \\dotsb + x_n}{n} \\geq \\sqrt[n]{x_1 x_2 \\dotsm x_n}$$ with equality if and only if $x_1 = x_2 = \\dotsb = x_n$.\n\nProof: The proof is by induction. The base case for induction is $n=1$, for which it's obvious that the inequality holds (in fact, it's an equality).\n\nNow suppose we have $n+1$ non-negative real numbers $x_1, x_2, \\dotsc , x_n, x_{n+1}$, and assume that for any $n$ non-negative real numbers, the AM-GM inequality holds. For notational simplicity, define $\\alpha$ to be the arithmetic mean of the $n+1$ numbers, $$\\alpha = \\dfrac{x_1 + x_2 + \\dotsb + x_n + x_{n+1}}{n+1}$$ which can be re-arranged as $$(n+1) \\alpha = x_1 + x_2 + \\dotsb + x_n + x_{n+1} \\tag{\\star}$$ If each of the $n+1$ numbers is equal to $\\alpha$, then the AM and GM are both $\\alpha$, so there is nothing to prove. Otherwise, there must be at least one of the $x_i$'s that is greater than $\\alpha$ and at least one that is less than $\\alpha$. Let's say $x_n > \\alpha$ and $x_{n+1} < \\alpha$ (if this isn't the case, rearrange the labels so that it is). So we know that $$(x_n - \\alpha)(\\alpha - x_{n+1}) > 0 \\tag{\\star \\star}$$ a fact which we will use later.\n\nNow define $y = x_n + x_{n+1} - \\alpha$. Note that $y$ is positive since it is greater than $x_n - \\alpha$. Furthermore, the AM of the $n$ numbers $x_1, x_2, \\dotsc , x_{n-1}, y$ is $\\alpha$: \\begin{align} \\dfrac{x_1 + x_2 + \\dotsb + x_{n-1} + y}{n} &= \\dfrac{x_1 + x_2 + \\dotsb + x_{n-1} + (x_n + x_{n+1} - \\alpha)}{n} \\\\[3mm] &= \\dfrac{(x_1 + x_2 + \\dotsb + x_{n-1} + x_n + x_{n+1}) - \\alpha}{n} \\\\[3mm] &= \\dfrac{(n+1)\\alpha - \\alpha}{n} \\ \\ \\ \\text{[by \\star]} \\\\[3mm] &= \\dfrac{n \\alpha}{n} \\\\[2mm] &= \\alpha \\end{align} By the induction hypothesis, the AM-GM inequality holds for the $n$ numbers $x_1, x_2, \\dotsc , x_{n-1}, y$, i.e. \\begin{align} &&\\sqrt[n]{x_1 x_2 \\dotsm x_{n-1} y} \\ &\\leq \\ \\alpha& \\\\ \\implies &&x_1 x_2 \\dotsm x_{n-1} y \\ &\\leq \\ \\alpha^n& \\\\ \\implies &&x_1 x_2 \\dotsm x_{n-1} y \\cdot \\alpha \\ &\\leq \\ \\alpha^{n+1}& \\tag{\\spadesuit} \\end{align} The last ingredient is the following: \\begin{align} (x_n - \\alpha)(\\alpha - x_{n+1}) = \\ &\\underbrace{(x_n + x_{n+1} - \\alpha)}_{y} \\alpha - x_n x_{n+1} \\\\[2mm] > \\ &0 \\ \\ \\ \\text{[by \\star \\star]} \\\\[2mm] \\implies y \\alpha > \\ &x_n x_{n+1} \\end{align} Plugging this into $( \\spadesuit )$ yields \\begin{align} \\alpha^{n+1} &\\geq x_1 x_2 \\dotsm x_{n-1} (y \\alpha) \\\\[2mm] &> x_1 x_2 \\dotsm x_{n-1} (x_n x_{n+1}) \\\\[2mm] \\implies \\alpha &> \\sqrt[n+1]{x_1 x_2 \\dotsm x_{n-1} x_n x_{n+1}} \\end{align} i.e. the arithmetic mean is greater than the geometric mean. This completes the proof.\n$\\square$\n\n#### Some more examples of AM-GM applications\n\nIn Part 1, I showed one easy example and two examples where we could apply the AM-GM inequality to find the minimum value of a function subject to a volume-like constraint. In this part, I'll show 3 more examples of clever applications, in order of increasing difficulty.\n\nExample 4: Let $a_1, a_2, \\dotsc , a_n$ be a sequence of positive numbers and $b_1, b_2, \\dotsc , b_n$ be a permutation of the $a_i$'s. Show that $$\\dfrac{a_1}{b_1} + \\dfrac{a_2}{b_2} + \\dotsb + \\dfrac{a_n}{b_n} \\geq n$$ Proof: The AM-GM inequality implies $$\\dfrac{1}{n} \\left[ \\dfrac{a_1}{b_1} + \\dfrac{a_2}{b_2} + \\dotsb + \\dfrac{a_n}{b_n} \\right] \\geq \\sqrt[n \\uproot4]{\\dfrac{a_1}{b_1} \\dfrac{a_2}{b_2} \\dotsm \\dfrac{a_n}{b_n}} = 1$$ where the last equality is true since the $a_i$'s and $b_i$'s are the same list of numbers, just rearranged. Multiplying both sides by $n$ yields the desired inequality.\n$\\square$\n\nIn Example 4, there was no volume-like constraint on the values of the $a_i$'s, but we were able to produce one on the fractions $\\frac{a_i}{b_i}$ by multiplying them all together. In Example 5, we can do something similar- this one is admittedly shamelessly set up for AM-GM application, but I still thought it was cool enough to show.\n\nExample 5: Find the positive solutions to the system of equations: $$\\begin{gather} x_1 + \\dfrac{1}{x_2} = 4 \\\\[3mm] x_2 + \\dfrac{1}{x_3} = 1 \\\\[3mm] \\vdots \\tag{\\diamondsuit}\\\\[3mm] x_{99} + \\dfrac{1}{x_{100}} = 4 \\\\[3mm] x_{100} + \\dfrac{1}{x_1} = 1 \\end{gather}$$ Solution: The sum $x_1 + \\frac{1}{x_2}$ is 2 times the arithmetic mean of the numbers $x_1$ and $\\frac{1}{x_2}$, so by the AM-GM inequality, we have (using the same logic on the other 98 variables as well): $$\\begin{gather} x_1 + \\dfrac{1}{x_2} \\geq 2 \\sqrt{\\dfrac{x_1}{x_2}} \\\\[3mm] x_2 + \\dfrac{1}{x_3} \\geq 2 \\sqrt{\\dfrac{x_2}{x_3}} \\\\[3mm] \\vdots \\tag{\\clubsuit} \\\\[3mm] x_{100} + \\dfrac{1}{x_1} \\geq 2 \\sqrt{\\dfrac{x_{100}}{x_1}} \\end{gather}$$ Note that putting the $x$'s under square roots is valid since we are only considering positive solutions. Now, both sides of $( \\clubsuit )$ are positive in each of these inequalities, so we can multiply them together without reversing the direction of the inequality: \\begin{align} \\left( x_1 + \\dfrac{1}{x_2} \\right) \\left( x_2 + \\dfrac{1}{x_3} \\right) \\dotsm \\left( x_{100} + \\dfrac{1}{x_1} \\right) &\\geq 2^{100} \\sqrt{\\dfrac{x_1}{x_2} \\dfrac{x_2}{x_3} \\dotsm \\dfrac{x_{100}}{x_1}} \\\\[2mm] &= 2^{100} \\cdot 1 \\\\[2mm] &= 2^{100} \\end{align} thus $$\\left( x_1 + \\dfrac{1}{x_2} \\right) \\left( x_2 + \\dfrac{1}{x_3} \\right) \\dotsm \\left( x_{100} + \\dfrac{1}{x_1} \\right) \\geq 2^{100} \\tag{\\heartsuit}$$ On the other hand, multiplying together the original system of equations $( \\diamondsuit )$ shows that the product on the left side of $( \\heartsuit )$ is equal to $4^{50} \\cdot 1^{50} = 2^{100}$. In other words, the inequality $( \\heartsuit )$ is an exact equality. This implies, in turn, that each of the inequalities in $( \\clubsuit )$ are exact equalities too, since otherwise $( \\heartsuit )$ would not be an exact equality.\n\nFinally, we can solve for the variables: \\begin{align} && x_1 + \\dfrac{1}{x_2} &= 2 \\sqrt{\\dfrac{x_1}{x_2}} \\\\[3mm] &\\implies& x_1 - 2 \\sqrt{\\dfrac{x_1}{x_2}} + \\dfrac{1}{x_2} &= 0 \\\\[3mm] &\\implies& \\left( \\sqrt{x_1} - \\sqrt{\\dfrac{1}{x_2}} \\right)^2 &= 0 \\\\[3mm] &\\implies& \\sqrt{x_1} &= \\sqrt{\\dfrac{1}{x_2}} \\\\[3mm] &\\implies& x_1 &= \\dfrac{1}{x_2} \\end{align} Similarly, $x_2 = \\frac{1}{x_3}, \\dotsc, x_{100} = \\frac{1}{x_1}$. These, combined with the original system of equations $( \\diamondsuit )$, show that the solution is $x_1 = 2, x_2 = \\frac{1}{2}, x_3 = 2, x_4 = \\frac{1}{2}, \\dotsc, x_{99} = 2, x_{100} = \\frac{1}{2}$.\n$\\square$\n\nThe sixth and final example will require proving a lemma, but I thought this example (and the lemma itself) was so neat that it would be worth the effort.\n\nLemma: Let $p(x) = a_n x^n + a_{n-1} x^{n-1} + \\dotsb + a_1 x + a_0$ be a polynomial in the variable $x$ (could be a real or complex variable and real or complex coefficients), and let $r_1, r_2, \\dotsc, r_n$ be the roots of $p(x)$ (in general, these are complex numbers, and there may be repeats). Then the sum of the squares of the roots is related to the coefficients by the following formula: $$r_1^2 + r_2^2 + \\dotsb + r_n^2 = \\left( \\dfrac{a_{n-1}}{a_n} \\right)^2 - 2 \\dfrac{a_{n-2}}{a_n}$$ Proof: The polynomial $p(x)$ can alternatively be written as $$p(x) = a_n (x-r_1)(x-r_2) \\dotsm (x-r_n) \\tag{\\dagger}$$ When expanding this product, we add terms consisting of $\\pm a_n$ times $n$ items, each either an $x$ or one of the $r_i$'s, then combine \"like terms,\" i.e. those with the same power of $x$. For $0 \\leq k \\leq n$, a term in the expansion contributes to the $x^k$ coefficient when, from the $n$ factors in the product $( \\dagger )$, we choose an $x$ from $k$ of them and the $-r_i$ from the other $n-k$. Therefore, the coefficient of $x^k$ in the expansion is $$a_n (-1)^{n-k} \\sum_{1 \\leq i_1 < i_2 < \\dotsb < i_{n-k} \\leq n}{r_{i_1} r_{i_2} \\dotsm r_{i_{n-k}}} \\tag{\\ddagger}$$ Since the product $( \\dagger )$ is an equivalent way of writing $p(x)$, the sum $( \\ddagger )$ must be equal to the coefficient $a_k$.\n\nOn the other hand, the sum of the squares of the roots can be written as \\begin{align} r_1^2 + r_2^2 + \\dotsb + r_n^2 &= (r_1 + r_2 + \\dotsb + r_n)^2 - 2 \\sum_{1 \\leq i_1 < i_2 \\leq n}{r_{i_1} r_{i_2}} \\\\[3mm] &= \\left( \\dfrac{1}{a_n (-1)^{n-(n-1)}}a_{n-1} \\right)^2 -2 \\dfrac{1}{a_n (-1)^{n-(n-2)}}a_{n-2} \\tag{\\maltese} \\\\[3mm] &= \\left( \\dfrac{a_{n-1}}{a_n} \\right)^2 - 2 \\dfrac{a_{n-2}}{a_n} \\end{align} where the equality $( \\maltese )$ was obtained by plugging in $(n-1)$ and $(n-2)$ in for $k$ in $( \\ddagger )$. This completes the proof.\n$\\square$",
null,
"Note: the result in the lemma is one of Newton's identities (or Newton sums) for polynomials, and the formulas $( \\ddagger )$ (one formula for each value of $k$) are called Vieta's formulas. There are similar Newton's identities for the sum of the third, fourth, fifth, etc. powers of the roots. The formulas are recursive, each depending on the lower-degree Newton sums.\n\nWith the lemma and a slick AM-GM application, the final example will be a piece of cake.\n\nExample 6: A polynomial with all coefficients equal to $\\pm 1$ and only real roots has degree at most 3.\n\nProof: Suppose we have a polynomial $p(x) = a_n x^n + a_{n-1} x^{n-1} + \\dotsb + a_1 x + a_0$ where each of the $a_i$'s is either $1$ or $-1$ and with roots $r_1, r_2, \\dotsc, r_n$. We can assume that the leading coefficient $a_n$ is $+1$ since the polynomials with $a_n = -1$ are just the negatives of polynomials with $a_n = +1$.\n\nBy the lemma (with $a_n = 1$), the sum of the squares of the roots is $a_{n-1}^2 - 2 a_{n-2}$. Also, plugging in $k=n$ in $( \\ddagger )$ tells us that the product of the squares of the roots is $a_0^2$.\n\nBy the AM-GM inequality, we have \\begin{align} &&\\dfrac{r_1^2 + r_2^2 + \\dotsb + r_n^2}{n} &\\geq \\sqrt[n \\uproot2]{r_1^2 r_2^2 \\dotsm r_n^2} \\\\[3mm] &\\implies& \\dfrac{a_{n-1}^2 - 2 a_{n-2}}{n} &\\geq \\sqrt[n]{a_0^2} \\\\[3mm] &\\implies& \\dfrac{1 \\pm 2}{n} &\\geq 1 \\ \\ \\ [\\text{since all coefficients are }\\pm 1 ] \\\\[3mm] &\\implies& 3 &\\geq n \\end{align} $\\square$\n\nThat will do it for this post. Please post any questions in the comments section. Thanks for reading, and stay tuned for part 3..."
]
| [
null,
"https://4.bp.blogspot.com/-Xa5J9wcLsEI/VsKxd6JX2OI/AAAAAAAAAas/oo1z5dv0mEw/s200/Vieta-Newton.jpg",
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.7380454,"math_prob":1.000005,"size":20802,"snap":"2021-31-2021-39","text_gpt3_token_len":7973,"char_repetition_ratio":0.14804308,"word_repetition_ratio":0.96875906,"special_character_ratio":0.4057783,"punctuation_ratio":0.08237313,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.000004,"pos_list":[0,1,2],"im_url_duplicate_count":[null,9,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-26T02:46:40Z\",\"WARC-Record-ID\":\"<urn:uuid:ad598cf1-a577-4084-8d27-8440620e2eb0>\",\"Content-Length\":\"78750\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d6f29867-3f47-4b0b-8ebc-d32eb4f46bab>\",\"WARC-Concurrent-To\":\"<urn:uuid:9bccd54e-8c4f-4719-8b81-a94a5e62143d>\",\"WARC-IP-Address\":\"142.250.73.243\",\"WARC-Target-URI\":\"http://www.gtmath.com/2016/02/\",\"WARC-Payload-Digest\":\"sha1:WOTR3XMZTH2BYU74CPAADVJTOGQBDMLX\",\"WARC-Block-Digest\":\"sha1:RKVK5TNVH7ZGPDTTWFPXFXLDUTYNCC3S\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780057796.87_warc_CC-MAIN-20210926022920-20210926052920-00352.warc.gz\"}"} |
http://www.ovsa.njit.edu/wiki/index.php?title=Polarization_Calibration&diff=next&oldid=537 | [
"# Difference between revisions of \"Polarization Calibration\"\n\n## Linear to Circular Conversion\n\nAt EOVSA’s linear feeds, in the electric field the linear polarization, X and Y, relates to RCP and LCP (R and L) as:",
null,
"In terms of autocorrelation powers, we have the 4 polarization products XX*, YY*, XY* and YX*, where the * denotes complex conjugation. The quantities RR* and LL* are then",
null,
"One problem is that there is generally a non-zero delay in Y with respect to X. This creates phase slopes in XY* and YX* from which we can determine the delay very accurately. As a check,",
null,
"For completeness:",
null,
"When I plot the quantities I, V, R and L as measured (Figure 1) for geosynchronous satellite Ciel-2, the results look reasonable, except that there are parts of the band where R and L are mis-assigned, and others where they do not separate well.",
null,
"Figure 1: Amplitudes of Stokes I, Stokes V, RCP and LCP for Ciel-2 satellite. The red vertical lines show the center of RCP channels, while the green vertical lines show the center of LCP channels. There is pretty good separation of RCP and LCP at lower and higher frequencies in the range, but around 12.4 GHz the polarization has switched and is mis-assigned.\n\nThe problem is that residual phase slope of Y with respect to X, caused by a difference in delay between the two channels. This can be seen in the upper panel of Figure 2, which shows the uncorrected phases of XY* and YX*. To correct the phases, the RCP phase was fit by a linear least-squares routine, and then the phases were offset by π/2 for both XY* and YX* according to:",
null,
"where φ(v) is the phase fit by the linear function. The corrected phases are shown in the lower panel of Figure 2.",
null,
"Figure 2: Upper panel) Raw phases on XY* (blue) and YX* (green). Lower panel) Phases corrected by removing a delay slope of order 3 ns, and shifting the phase such that the XY* phases on RCP channels is -π/2. This adjustment automatically makes XY* phases on LCP channels become +π/2.\n\nWhen the corrected (primed) quantities are used in",
null,
"the resulting channel separation is shown in Figure 3, which shows pretty much perfect channel separation, with the correct polarity.",
null,
"Figure 3: Amplitudes of Stokes I, Stokes V, RCP and LCP for Ciel-2 satellite, after correcting for delay. The red vertical lines show the center of RCP channels, while the green vertical lines show the center of LCP channels. There is now PERFECT correspondence between RCP channels and the plot, and likewise for LCP channels.\n\nThis analysis can be performed on all antennas in a completely automated way, to produce the results of Fig. 3. The curves for each antenna are very similar, except that they show differences in amplitude (gain) as a function of frequency. When the curves are scaled to have the same Stokes I levels, the result is shown in Figure 4. Here the 13 antennas are shown in different colors. It is clear that matching Stokes I also causes an excellent match in the other polarizations. Remaining mismatch is likely due to not subtracting a no-signal background prior to scaling the Stokes I curves.",
null,
"Figure 4: Same as Fig. 3, but for all 13 atnennas, with the Stokes I curve for each antenna scaled so that they agree. The beacon frequency in marked in the lower panel.\n\nAlso marked in Figure 4 is the location of the beacon signal for Ciel-2, which is at 12.209 GHz. The beacon signal appears in precisely the correct bin (frequency subchannel), indicating that we have precise tuning (certainly not unexpected, but nice to see).\n\nJust to show that the approach is general, the results for a different communications satellite, Nimiq-5, is shown in Figure 5.\n\n## Polarization Mixing Correction\n\nDue to relative feed rotation between az-al mounted antennas and equatorial mounted antennas"
]
| [
null,
"http://www.ovsa.njit.edu/wiki/images/math/8/6/7/867e27ab704c33e2d376e475429d6ae8.png ",
null,
"http://www.ovsa.njit.edu/wiki/images/math/5/c/3/5c3716cb6a39ab9beff90b72c427e151.png ",
null,
"http://www.ovsa.njit.edu/wiki/images/math/3/4/e/34ed060e22f869860834b301c9d62136.png ",
null,
"http://www.ovsa.njit.edu/wiki/images/math/2/6/6/266d7ff2007d687f6e3280d5359cdddf.png ",
null,
"http://www.ovsa.njit.edu/wiki/images/8/8e/Figure1.jpg",
null,
"http://www.ovsa.njit.edu/wiki/images/math/9/8/0/980577f9102eec46e270245c8d73cc89.png ",
null,
"http://www.ovsa.njit.edu/wiki/images/thumb/1/1f/Linear_to_Circular_Conversion_Figure2.jpg/800px-Linear_to_Circular_Conversion_Figure2.jpg",
null,
"http://www.ovsa.njit.edu/wiki/images/math/6/e/b/6eb472f7011d3bfd7f807e77647c443b.png ",
null,
"http://www.ovsa.njit.edu/wiki/images/thumb/1/15/Linear_to_Circular_Conversion_Figure3.jpg/800px-Linear_to_Circular_Conversion_Figure3.jpg",
null,
"http://www.ovsa.njit.edu/wiki/images/thumb/1/16/Linear_to_Circular_Conversion_Figure4.jpg/800px-Linear_to_Circular_Conversion_Figure4.jpg",
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.89047134,"math_prob":0.9269942,"size":7954,"snap":"2019-26-2019-30","text_gpt3_token_len":2194,"char_repetition_ratio":0.114591196,"word_repetition_ratio":0.6360947,"special_character_ratio":0.28715113,"punctuation_ratio":0.10721511,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9581949,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-07-17T22:54:06Z\",\"WARC-Record-ID\":\"<urn:uuid:831eafab-678f-48cc-926d-55bda7304f4d>\",\"Content-Length\":\"43530\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f5c29d74-bf43-47e5-b127-763146cb8e94>\",\"WARC-Concurrent-To\":\"<urn:uuid:9e1e3cee-e9b7-4b19-8624-a2d864812f2e>\",\"WARC-IP-Address\":\"192.100.16.206\",\"WARC-Target-URI\":\"http://www.ovsa.njit.edu/wiki/index.php?title=Polarization_Calibration&diff=next&oldid=537\",\"WARC-Payload-Digest\":\"sha1:XFDGTYX4HN6FHHIYB6PSAJ4QAR4UDI6U\",\"WARC-Block-Digest\":\"sha1:CPULQL4L6RMBATGBXJ2PXYFMHLELPA54\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-30/CC-MAIN-2019-30_segments_1563195525414.52_warc_CC-MAIN-20190717221901-20190718003901-00454.warc.gz\"}"} |
https://www.agilequalitymadeeasy.com/post/what-is-equivalence-partitioning-technique | [
"top of page\nSearch\n\n# What is Equivalence Partitioning Technique?\n\nUpdated: Jan 19, 2022",
null,
"This Black-Box test technique can be used at any level of testing (Unit, Component, Integration, and system). It is used to determine the set of valid and invalid inputs specified for the tested object and divide it up into partitions of data inputs that, according to the application requirements, should behave the same when used as an input (the application behavior should be the same when tested with these inputs).\n\nThe main advantage of this method is that it allows the tester to validate the tested object with a narrow set of test inputs because he will use a single condition from each class instead of all available information (remember that we assume that all inputs in a single class will behave the same, therefore we can test only single value).\n\nThe assumptions are:\n\nIf a single value from a class is working, we can assume that all class values will work the same once used as input.\n\nIf a single value from a class is NOT working, we need to validate the object with further inputs of the same class.\n\n• The probability of uncovering defects with the selected test inputs is higher.\n\n• We will cover different domain inputs (Valid/Invalid).\n\n• We will reduce the number of test cases.\n\n• We will reduce the testing time.\n\nOne main disadvantage in ECP is that you must remember that testing is a big mistake! Therefore, you should use this method wisely and test more than a single value if you know that a specific area has more chances to trigger defects or when you know that there is a higher risk.\n\nSo How can we use this method?\n\nThe tested object is an input field where the user needs to specify his age; the requirements are:\n\n• The user can insert any number between 25 – 100.\n\n• The user can insert any number between 5 – 15.\n\n• The user cannot use negative numbers.\n\nAs you can see, we will not want to test all inputs that are available for testing; therefore, we will use the EP method to create the following partitions:\n\nClass no'1: 25 <= Input <= 100 (Positive)\n\nClass no'2: 5 <= Input <= 15 (Positive)\n\nClass no'3: 16 <= Input <= 24 (Invalid)\n\nClass no'4: Input > 100 (Invalid)\n\nClass no'5: Input < 0 (Invalid)\n\nClass no'6: 0 <= Input <= 4 (Invalid)\n\nNow, all you need to do to validate the tested input field is to use one set of dates that belongs to each partition (six values in total). You can see we covered the main scenarios in 6 tests instead of running a large number of tests that will not add any real benefit/value to the testing effort.\n\nTags:"
]
| [
null,
"https://static.wixstatic.com/media/bb3ad0_01b73315c3c448cf8bd986ffa2814c63~mv2.png/v1/fill/w_380,h_271,al_c,q_85,enc_auto/bb3ad0_01b73315c3c448cf8bd986ffa2814c63~mv2.png",
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.86880904,"math_prob":0.8187033,"size":2543,"snap":"2023-14-2023-23","text_gpt3_token_len":565,"char_repetition_ratio":0.128397,"word_repetition_ratio":0.030701755,"special_character_ratio":0.23279591,"punctuation_ratio":0.08704454,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9781612,"pos_list":[0,1,2],"im_url_duplicate_count":[null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-05-29T08:10:50Z\",\"WARC-Record-ID\":\"<urn:uuid:4ebf1f43-c92b-4357-ae48-f8262ae7fde1>\",\"Content-Length\":\"1050585\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b68632be-b087-42e7-bf5e-59f949fff3a6>\",\"WARC-Concurrent-To\":\"<urn:uuid:ddc84d71-27d0-46dc-bd71-69457c95f024>\",\"WARC-IP-Address\":\"146.75.33.84\",\"WARC-Target-URI\":\"https://www.agilequalitymadeeasy.com/post/what-is-equivalence-partitioning-technique\",\"WARC-Payload-Digest\":\"sha1:SV7CMIINKLJNVIGTFBXWZIEZUQDG5V63\",\"WARC-Block-Digest\":\"sha1:Q7LM3UUJOAPXQ23MMGKVX3RC5G6GKT2M\",\"WARC-Truncated\":\"length\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224644817.32_warc_CC-MAIN-20230529074001-20230529104001-00362.warc.gz\"}"} |
https://www.physicsforums.com/threads/are-x-y-and-x-y-independent.969049/ | [
"# Are (X+Y) and (X-Y) independent?\n\nHomework Statement:\njoint distribution p(x,y):\n\\begin{array}{|c|c|c|c|}\n\\hline & \\frac xy & 0 & 1 \\\\\n\\hline & 0 & .5 & .2 \\\\\n\\hline & 1 & .2 & .1 \\\\\n\\hline\n\\end{array}\nRelevant Equations:\ncompute the marginal pmf from joint pmf: for x sum all of the probablities in one column\nfor y: sum all probabilities in one row\n\ntest for independence: if x(n)y(n) = x,y(n) then for all n then it is independent\nI have x(0) = .7 x(1) = .3, y(0)=.7 y(1) = .3\n\nsince x,y(0) = .5 =/= x(0)y(0), x(0)y(0) = .49, x and y are dependent\n\nnow I need to determine whether x and y are independent or not\n\nfor x+y\n$\\begin{array}{|c|c|c|c|} \\hline x+y & 0 & 1 & 2 \\\\ \\hline p(x+y) & .49 & .42 & .09 \\\\ \\hline \\end{array}$\n\nbut how can I possibly determine x-y? since the domain will be 0 1 and -1 how can I determine -1?\n\nHomework Helper\nDearly Missed\nI have x(0) = .7 x(1) = .3, y(0)=.7 y(1) = .3\n\nsince x,y(0) = .5 =/= x(0)y(0), x(0)y(0) = .49, x and y are dependent\n\nnow I need to determine whether x and y are independent or not\n\nfor x+y\n$\\begin{array}{|c|c|c|c|} \\hline x+y & 0 & 1 & 2 \\\\ \\hline p(x+y) & .49 & .42 & .09 \\\\ \\hline \\end{array}$\n\nbut how can I possibly determine x-y? since the domain will be 0 1 and -1 how can I determine -1?\nWhat is stopping you from computing 0-1?\n\n•",
null,
"What is stopping you from computing 0-1?\n\nI'm guessing I just take the negative of 1? I'm not sure, it seems to me, like for example, if this was rolling dice, that you can't have a -1. I'm probably just not understanding the concept though..\n\nand how do I prove that X+Y and X-Y are independent anyways? I can't construct a joint distribution from marginal distributions (which X+Y and X-Y would be.)\n\nHomework Helper\nDearly Missed\nI'm guessing I just take the negative of 1? I'm not sure, it seems to me, like for example, if this was rolling dice, that you can't have a -1. I'm probably just not understanding the concept though..\n\nand how do I prove that X+Y and X-Y are independent anyways? I can't construct a joint distribution from marginal distributions (which X+Y and X-Y would be.)\n\nYou are over-thinking it. Just perform X-Y and let X and Y be whatever they want to be. If X > Y, X-Y will b e >0; if X < Y, X-Y will be < 0. That's all there is to it!\n\n•",
null,
"Hmm\n\nWhen computing X+Y, I have a domain of: 0 1 2\n\nI can compute 0 by the product of Px(0)Py(0) = .7*.7 = .49 = X+Y(0)\nI can compute 1 by Px(0)Py(1) + Px(1)Py(0) = 42/100 = .42 = X+Y(1)\nI can compute 2 by Px(1)Py(1) = .3*.3 = .09 = X+Y(2)\n\nNow X-Y has domain: -1, 0,1\n\nHow do I compute -1 given that my marginal probabilities are: Px(0) = .7, Px(1) = .3, Py(0) = .7, Py(1) = .3\n\nThere is no way to get -1 from 0+0, 0+1 or 1+1\n\nGold Member\n\n•",
null,
"Hmm I think what I was doing wrong is computing X+Y and X-Y from the marginal distribution, when I should be computing it from the joint distribution\n\nso for\n:x+y\n$\\begin{array}{|c|c|c|c|} \\hline x+y & 0 & 1 & 2 \\\\ \\hline p(x+y) & .5 & .4 & .1 \\\\ \\hline \\end{array}$\n\nnot\n\nfor x+y\n$\\begin{array}{|c|c|c|c|} \\hline x+y & 0 & 1 & 2 \\\\ \\hline p(x+y) & .49 & .42 & .09 \\\\ \\hline \\end{array}$\n\nHomework Helper\nDearly Missed\nHmm\n\nWhen computing X+Y, I have a domain of: 0 1 2\n\nI can compute 0 by the product of Px(0)Py(0) = .7*.7 = .49 = X+Y(0)\nI can compute 1 by Px(0)Py(1) + Px(1)Py(0) = 42/100 = .42 = X+Y(1)\nI can compute 2 by Px(1)Py(1) = .3*.3 = .09 = X+Y(2)\n\nNow X-Y has domain: -1, 0,1\n\nHow do I compute -1 given that my marginal probabilities are: Px(0) = .7, Px(1) = .3, Py(0) = .7, Py(1) = .3\n\nThere is no way to get -1 from 0+0, 0+1 or 1+1\nRight: for X+Y you can never get -1. However, that was not the issue: you want to compute X-Y, and for that you can certainly get -1 (but you can never get 2).\n\n•",
null,
"I see. So the table for X-Y will be:\n\nfor x+y\n$\\begin{array}{|c|c|c|c|} \\hline x-y & -1 & 0 & 1 \\\\ \\hline p(x-y) & .2 & .6 & .2 \\\\ \\hline \\end{array}$\n\nGahh I'm a moron. I think this is what it will be:\n\nx+y:\n$\\begin{array}{|c|c|c|c|} \\hline x+y & 0 & 1 & 2 \\\\ \\hline p(x+y) & .5 & .4 & .1 \\\\ \\hline \\end{array}$\n\nx-y:\n$\\begin{array}{|c|c|c|c|} \\hline x+y & -1 & 0 & 1 \\\\ \\hline p(x+y) & .6 & .4 & 0 \\\\ \\hline \\end{array}$\n\ndoes this seem right now\n\nGold Member\nI'm guessing I just take the negative of 1? I'm not sure, it seems to me, like for example, if this was rolling dice, that you can't have a -1. /QUOTE]\nNo, but you can have the difference in the values of the faces equal -1. Or -2. The sums, differences are not modeling throws of dice.\n\nLast edited:\nHomework Helper\nGold Member\nGahh I'm a moron. I think this is what it will be:\n\nx+y:\n$\\begin{array}{|c|c|c|c|} \\hline x+y & 0 & 1 & 2 \\\\ \\hline p(x+y) & .5 & .4 & .1 \\\\ \\hline \\end{array}$\n\nCorrect\n\nx-y:\n$\\begin{array}{|c|c|c|c|} \\hline x+y & -1 & 0 & 1 \\\\ \\hline p(x+y) & .6 & .4 & 0 \\\\ \\hline \\end{array}$\nCheck your second one. Should start like:\n$\\begin{array}{|c|c|c|c|} \\hline x\\color{red} - y & -1 & 0 & 1 \\\\ \\hline p(x\\color{red} -y) & ? & ? & ? \\\\ \\hline \\end{array}$\nand you need to check the probabilities.\n\nCorrect\n\nCheck your second one. Should start like:\n$\\begin{array}{|c|c|c|c|} \\hline x\\color{red} - y & -1 & 0 & 1 \\\\ \\hline p(x\\color{red} -y) & ? & ? & ? \\\\ \\hline \\end{array}$\nand you need to check the probabilities.\n\nAhh sorry, meant to write:\n\nx-y:\n$\\begin{array}{|c|c|c|c|} \\hline x-y & -1 & 0 & 1 \\\\ \\hline p(x-y) & .2 & .6 & .2 \\\\ \\hline \\end{array}$\n\nThe probabilities add up to 1 so I think its correct? Does this seem right to you?\n\nLast edited:\n•",
null,
""
]
| [
null,
"data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7",
null,
"data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7",
null,
"data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7",
null,
"data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7",
null,
"data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7",
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.7490451,"math_prob":0.9962177,"size":1194,"snap":"2023-14-2023-23","text_gpt3_token_len":471,"char_repetition_ratio":0.14285715,"word_repetition_ratio":0.5690377,"special_character_ratio":0.4313233,"punctuation_ratio":0.12671232,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9994161,"pos_list":[0,1,2,3,4,5,6,7,8,9,10],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-03-22T09:10:01Z\",\"WARC-Record-ID\":\"<urn:uuid:a5ae6e86-3ed6-44cc-868e-b565d305bb37>\",\"Content-Length\":\"114265\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:fccb4550-f9c3-4c94-93a6-d9c6cd9177b9>\",\"WARC-Concurrent-To\":\"<urn:uuid:2834f8ec-f5b5-4b76-b529-6f3cc765aba6>\",\"WARC-IP-Address\":\"104.26.15.132\",\"WARC-Target-URI\":\"https://www.physicsforums.com/threads/are-x-y-and-x-y-independent.969049/\",\"WARC-Payload-Digest\":\"sha1:KBTMUICEIQOUT2QB3S4N2X3I7JY4ESEV\",\"WARC-Block-Digest\":\"sha1:AUCKAINOKE4PCVHO66345GOY353FEJ5S\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-14/CC-MAIN-2023-14_segments_1679296943809.22_warc_CC-MAIN-20230322082826-20230322112826-00081.warc.gz\"}"} |
https://metanumbers.com/1718 | [
"# 1718 (number)\n\n1,718 (one thousand seven hundred eighteen) is an even four-digits composite number following 1717 and preceding 1719. In scientific notation, it is written as 1.718 × 103. The sum of its digits is 17. It has a total of 2 prime factors and 4 positive divisors. There are 858 positive integers (up to 1718) that are relatively prime to 1718.\n\n## Basic properties\n\n• Is Prime? No\n• Number parity Even\n• Number length 4\n• Sum of Digits 17\n• Digital Root 8\n\n## Name\n\nShort name 1 thousand 718 one thousand seven hundred eighteen\n\n## Notation\n\nScientific notation 1.718 × 103 1.718 × 103\n\n## Prime Factorization of 1718\n\nPrime Factorization 2 × 859\n\nComposite number\nDistinct Factors Total Factors Radical ω(n) 2 Total number of distinct prime factors Ω(n) 2 Total number of prime factors rad(n) 1718 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0\n\nThe prime factorization of 1,718 is 2 × 859. Since it has a total of 2 prime factors, 1,718 is a composite number.\n\n## Divisors of 1718\n\n1, 2, 859, 1718\n\n4 divisors\n\n Even divisors 2 2 1 1\nTotal Divisors Sum of Divisors Aliquot Sum τ(n) 4 Total number of the positive divisors of n σ(n) 2580 Sum of all the positive divisors of n s(n) 862 Sum of the proper positive divisors of n A(n) 645 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 41.4488 Returns the nth root of the product of n divisors H(n) 2.66357 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors\n\nThe number 1,718 can be divided by 4 positive divisors (out of which 2 are even, and 2 are odd). The sum of these divisors (counting 1,718) is 2,580, the average is 645.\n\n## Other Arithmetic Functions (n = 1718)\n\n1 φ(n) n\nEuler Totient Carmichael Lambda Prime Pi φ(n) 858 Total number of positive integers not greater than n that are coprime to n λ(n) 858 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 270 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares\n\nThere are 858 positive integers (less than 1,718) that are coprime with 1,718. And there are approximately 270 prime numbers less than or equal to 1,718.\n\n## Divisibility of 1718\n\n m n mod m 2 3 4 5 6 7 8 9 0 2 2 3 2 3 6 8\n\nThe number 1,718 is divisible by 2.\n\n## Classification of 1718\n\n• Arithmetic\n• Semiprime\n• Deficient\n\n### Expressible via specific sums\n\n• Polite\n• Non-hypotenuse\n\n• Square Free\n\n## Base conversion (1718)\n\nBase System Value\n2 Binary 11010110110\n3 Ternary 2100122\n4 Quaternary 122312\n5 Quinary 23333\n6 Senary 11542\n8 Octal 3266\n10 Decimal 1718\n12 Duodecimal bb2\n20 Vigesimal 45i\n36 Base36 1bq\n\n## Basic calculations (n = 1718)\n\n### Multiplication\n\nn×y\n n×2 3436 5154 6872 8590\n\n### Division\n\nn÷y\n n÷2 859 572.666 429.5 343.6\n\n### Exponentiation\n\nny\n n2 2951524 5070718232 8711493922576 14966346558985568\n\n### Nth Root\n\ny√n\n 2√n 41.4488 11.9768 6.43807 4.43613\n\n## 1718 as geometric shapes\n\n### Circle\n\n Diameter 3436 10794.5 9.27249e+06\n\n### Sphere\n\n Volume 2.12402e+10 3.70899e+07 10794.5\n\n### Square\n\nLength = n\n Perimeter 6872 2.95152e+06 2429.62\n\n### Cube\n\nLength = n\n Surface area 1.77091e+07 5.07072e+09 2975.66\n\n### Equilateral Triangle\n\nLength = n\n Perimeter 5154 1.27805e+06 1487.83\n\n### Triangular Pyramid\n\nLength = n\n Surface area 5.11219e+06 5.9759e+08 1402.74\n\n## Cryptographic Hash Functions\n\nmd5 fc6709bfdf0572f183c1a84ce5276e96 564fd4f695747a12f20fb2f28e3c8198ad3ba8de c0d2b0fe4d7671d1bb90703e6ab60ab14f01998648c45ba1da553283f8ab9dd6 779ff59161db3038f2859d4712fc84a3330c3ecc47fe499a8f91451ba621918f91713bb34e08999b3ef4fcf9d248f0f917cd1e3ac4c25cc883dc8a27cdc1a1b7 fbd0c522be0f657c58c91fd76932873ce1312d1b"
]
| [
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.6180332,"math_prob":0.97822726,"size":4472,"snap":"2023-14-2023-23","text_gpt3_token_len":1780,"char_repetition_ratio":0.12488809,"word_repetition_ratio":0.026354318,"special_character_ratio":0.4355993,"punctuation_ratio":0.07069409,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9956973,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-03-31T13:36:28Z\",\"WARC-Record-ID\":\"<urn:uuid:6e2f4195-490a-4322-9629-455171b82102>\",\"Content-Length\":\"40080\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:58838752-86ab-4245-aeba-49ed41d36395>\",\"WARC-Concurrent-To\":\"<urn:uuid:20b0f1f7-fec5-4d2b-8244-b0464044fd7c>\",\"WARC-IP-Address\":\"46.105.53.190\",\"WARC-Target-URI\":\"https://metanumbers.com/1718\",\"WARC-Payload-Digest\":\"sha1:SOWZV34GZEV7XDCBVUIWHNUK3SZ4GNVU\",\"WARC-Block-Digest\":\"sha1:27VWXFKKRLVP3B2OKG2DQST2YZY4HPTY\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-14/CC-MAIN-2023-14_segments_1679296949642.35_warc_CC-MAIN-20230331113819-20230331143819-00116.warc.gz\"}"} |
https://www.wavewalkerdsp.com/2022/07/06/single-pole-iir-filter-frequency-response/ | [
"# Wave Walker DSP\n\n## DSP Algorithms for RF Systems\n\nFoundations of Digital Signal Processing: Complex Numbers is available Amazon now! Great for young engineers looking for a simple explanation of complex numbers.\n\nSingle Pole IIR Filter Frequency Response\nJuly 6, 2022\n\n#### Introduction\n\nThis blog post demonstrates how to find the frequency response of the single pole IIR filter and then plots the magnitude and phase responses for multiple",
null,
"values.\n\nCheck out these other blogs on DSP:\n\n#### Single Pole IIR Impulse Response\n\nThe single pole infinite impulse response (IIR) filter is an incredibly efficient filter!\n\nThe impulse response of the single pole IIR is\n\n(1)",
null,
"The filter uses two weights,",
null,
"and",
null,
". The weight",
null,
"is applied to the input signal x[n] and the value is typically small (",
null,
"< 0.1 ). The weight",
null,
"is much larger than",
null,
"and is applied to the feedback y[n-1]. The smaller the",
null,
"the more impact the feedback has on the output and the smoother (and more narrowband) the filter.\n\nThe single pole IIR may also be referred to as an exponential moving average (EMA) filter.\n\n#### Single Pole IIR Frequency Response\n\nThe Z-transform is applied to (1),\n\n(2)",
null,
"Equation (2) is simplified by gathering like terms,\n\n(3)",
null,
"(4)",
null,
"The transfer function for (1) is written as the ratio of Y(z) to X(z),\n\n(5)",
null,
"The frequency response",
null,
"is created by substituting\n\n(6)",
null,
"into (5) resulting in\n\n(7)",
null,
"Simplify (7) by definining\n\n(8)",
null,
"such that the frequency response is now\n\n(9)",
null,
"#### Magnitude of the Frequency Response\n\nThe magnitude of the frequency response (9) is\n\n(10)",
null,
"Substituting (9) into (10),\n\n(11)",
null,
"#### Example Plots\n\nThe magnitude and phase of the frequency response (9) is given below. The plots show that as",
null,
"decreases the filter becomes more narrowband and has more phase linearity.\n\nThis blog demonstrated how to find the frequency response of the a single pole IIR filter and plotted the phase and magnitude responses for several",
null,
"values."
]
| [
null,
"https://i0.wp.com/www.wavewalkerdsp.com/wp-content/ql-cache/quicklatex.com-5f44d9bbc8046069be4aa2989bff19aa_l3.png",
null,
"https://i0.wp.com/www.wavewalkerdsp.com/wp-content/ql-cache/quicklatex.com-043e2a5cf1f6f1af4a8587787c03de82_l3.png",
null,
"https://i0.wp.com/www.wavewalkerdsp.com/wp-content/ql-cache/quicklatex.com-5f44d9bbc8046069be4aa2989bff19aa_l3.png",
null,
"https://i0.wp.com/www.wavewalkerdsp.com/wp-content/ql-cache/quicklatex.com-5280f4c3615553fc7f5a96263d8db651_l3.png",
null,
"https://i0.wp.com/www.wavewalkerdsp.com/wp-content/ql-cache/quicklatex.com-5f44d9bbc8046069be4aa2989bff19aa_l3.png",
null,
"https://i0.wp.com/www.wavewalkerdsp.com/wp-content/ql-cache/quicklatex.com-5f44d9bbc8046069be4aa2989bff19aa_l3.png",
null,
"https://i0.wp.com/www.wavewalkerdsp.com/wp-content/ql-cache/quicklatex.com-5280f4c3615553fc7f5a96263d8db651_l3.png",
null,
"https://i0.wp.com/www.wavewalkerdsp.com/wp-content/ql-cache/quicklatex.com-5f44d9bbc8046069be4aa2989bff19aa_l3.png",
null,
"https://i0.wp.com/www.wavewalkerdsp.com/wp-content/ql-cache/quicklatex.com-5f44d9bbc8046069be4aa2989bff19aa_l3.png",
null,
"https://i0.wp.com/www.wavewalkerdsp.com/wp-content/ql-cache/quicklatex.com-55a9a56c17d868ba0f81d26d5888b211_l3.png",
null,
"https://i0.wp.com/www.wavewalkerdsp.com/wp-content/ql-cache/quicklatex.com-706d9fa93a08ff7ffba16d08bfc0efd2_l3.png",
null,
"https://i0.wp.com/www.wavewalkerdsp.com/wp-content/ql-cache/quicklatex.com-50ecdd411a3c35eefaee3e76574a31b1_l3.png",
null,
"https://i0.wp.com/www.wavewalkerdsp.com/wp-content/ql-cache/quicklatex.com-26ec7fe92d633c34fe92b66868c0fa58_l3.png",
null,
"https://i0.wp.com/www.wavewalkerdsp.com/wp-content/ql-cache/quicklatex.com-4d92bd499ded7c96734db9e94dbbde49_l3.png",
null,
"https://i0.wp.com/www.wavewalkerdsp.com/wp-content/ql-cache/quicklatex.com-1d42a8ae12820bbe1d3f5d2d7e70d080_l3.png",
null,
"https://i0.wp.com/www.wavewalkerdsp.com/wp-content/ql-cache/quicklatex.com-fd2aa254aa7beefacd3df6b6ac8eb117_l3.png",
null,
"https://i0.wp.com/www.wavewalkerdsp.com/wp-content/ql-cache/quicklatex.com-c7171d05ea4f1533d3ac50ac77730a74_l3.png",
null,
"https://i0.wp.com/www.wavewalkerdsp.com/wp-content/ql-cache/quicklatex.com-71afbe0f3fc702788d485ca43f926336_l3.png",
null,
"https://i0.wp.com/www.wavewalkerdsp.com/wp-content/ql-cache/quicklatex.com-25f3d51d8c6400ebebd3be8993f89416_l3.png",
null,
"https://i0.wp.com/www.wavewalkerdsp.com/wp-content/ql-cache/quicklatex.com-86d5a85d19fe582003a9c5a1828521d4_l3.png",
null,
"https://i0.wp.com/www.wavewalkerdsp.com/wp-content/ql-cache/quicklatex.com-5f44d9bbc8046069be4aa2989bff19aa_l3.png",
null,
"https://i0.wp.com/www.wavewalkerdsp.com/wp-content/ql-cache/quicklatex.com-5f44d9bbc8046069be4aa2989bff19aa_l3.png",
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.9077278,"math_prob":0.9306351,"size":1589,"snap":"2022-27-2022-33","text_gpt3_token_len":362,"char_repetition_ratio":0.1526814,"word_repetition_ratio":0.08178439,"special_character_ratio":0.22404027,"punctuation_ratio":0.058020476,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96639377,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44],"im_url_duplicate_count":[null,null,null,1,null,null,null,2,null,null,null,null,null,2,null,null,null,null,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-08-20T04:53:59Z\",\"WARC-Record-ID\":\"<urn:uuid:61fe6e14-4e39-481d-83b4-95837f24f582>\",\"Content-Length\":\"212178\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:88d8f53d-a2b7-4a50-a53a-bdb762968829>\",\"WARC-Concurrent-To\":\"<urn:uuid:7f68b8f3-b6e6-4199-ae1b-5475f768dc51>\",\"WARC-IP-Address\":\"107.180.236.188\",\"WARC-Target-URI\":\"https://www.wavewalkerdsp.com/2022/07/06/single-pole-iir-filter-frequency-response/\",\"WARC-Payload-Digest\":\"sha1:2XUS6VAMDZQTVG3H66DXTJYBQ7M4UFKT\",\"WARC-Block-Digest\":\"sha1:U3HSNQBPPKMBL5KT6YOPVKPNRBR25UAC\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-33/CC-MAIN-2022-33_segments_1659882573908.30_warc_CC-MAIN-20220820043108-20220820073108-00020.warc.gz\"}"} |
https://books.halfrost.com/leetcode/ChapterFour/0200~0299/0237.Delete-Node-in-a-Linked-List/ | [
"0237. Delete Node in a Linked List\n\n# 237. Delete Node in a Linked List#\n\n## 题目 #\n\nWrite a function to delete a node in a singly-linked list. You will not be given access to the `head` of the list, instead you will be given access to the node to be deleted directly.\n\nIt is guaranteed that the node to be deleted is not a tail node in the list.\n\nExample 1:",
null,
"``````Input: head = [4,5,1,9], node = 5\nOutput: [4,1,9]\nExplanation:You are given the second node with value 5, the linked list should become 4 -> 1 -> 9 after calling your function.\n\n``````\n\nExample 2:",
null,
"``````Input: head = [4,5,1,9], node = 1\nOutput: [4,5,9]\nExplanation:You are given the third node with value 1, the linked list should become 4 -> 5 -> 9 after calling your function.\n\n``````\n\nExample 3:\n\n``````Input: head = [1,2,3,4], node = 3\nOutput: [1,2,4]\n\n``````\n\nExample 4:\n\n``````Input: head = [0,1], node = 0\nOutput: \n\n``````\n\nExample 5:\n\n``````Input: head = [-3,5,-99], node = -3\nOutput: [5,-99]\n\n``````\n\nConstraints:\n\n• The number of the nodes in the given list is in the range `[2, 1000]`.\n• `1000 <= Node.val <= 1000`\n• The value of each node in the list is unique.\n• The `node` to be deleted is in the list and is not a tail node\n\n## 代码 #\n\n``````package leetcode\n\nimport (\n\"github.com/halfrost/leetcode-go/structures\"\n)\n\n// ListNode define\ntype ListNode = structures.ListNode\n\n/**\n* type ListNode struct {\n* Val int\n* Next *ListNode\n* }\n*/\nfunc deleteNode(node *ListNode) {\nnode.Val = node.Next.Val\nnode.Next = node.Next.Next\n}\n``````",
null,
"Apr 8, 2023",
null,
"Edit this page"
]
| [
null,
"https://assets.leetcode.com/uploads/2020/09/01/node1.jpg",
null,
"https://assets.leetcode.com/uploads/2020/09/01/node2.jpg",
null,
"https://books.halfrost.com/leetcode/svg/calendar.svg",
null,
"https://books.halfrost.com/leetcode/svg/edit.svg",
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.701709,"math_prob":0.9467436,"size":1505,"snap":"2023-40-2023-50","text_gpt3_token_len":526,"char_repetition_ratio":0.13524318,"word_repetition_ratio":0.08429119,"special_character_ratio":0.3282392,"punctuation_ratio":0.19335347,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97271955,"pos_list":[0,1,2,3,4,5,6,7,8],"im_url_duplicate_count":[null,7,null,5,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-10-01T00:00:57Z\",\"WARC-Record-ID\":\"<urn:uuid:6ec6028c-8c17-4092-95f7-11f39589b719>\",\"Content-Length\":\"192949\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:837590ea-e71a-43b9-b05c-d8f74ad55cdd>\",\"WARC-Concurrent-To\":\"<urn:uuid:04cf0c3f-d8e1-43a3-9a52-7399291bb851>\",\"WARC-IP-Address\":\"121.43.154.9\",\"WARC-Target-URI\":\"https://books.halfrost.com/leetcode/ChapterFour/0200~0299/0237.Delete-Node-in-a-Linked-List/\",\"WARC-Payload-Digest\":\"sha1:LYYSWHSPQYG3GQVEEC2GJPGOMX6WXH2Y\",\"WARC-Block-Digest\":\"sha1:SUGV3Q77KQY5Y3DGSQ4WEZVDIQZGZ5UT\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233510730.6_warc_CC-MAIN-20230930213821-20231001003821-00576.warc.gz\"}"} |
https://physics.stackexchange.com/questions/191451/what-will-happen-if-the-photon-has-non-zero-mass?noredirect=1 | [
"# What will happen if the photon has non-zero mass?\n\nI want to know the theoretical implication if photons have a non-zero mass.\n\n1. What happens to the Maxwell equations?\n\n2. What happens to QFT?\n\n3. If the photon have mass it can decade?\n\n• For one relativity would become meaningless. – Horus Jun 26 '15 at 15:01\n• Well the Proca Lagrangian would no longer be gauge invariant -- that's going to cause a lot of problems for the standard model as we know it. – sxwzd Jun 26 '15 at 15:29\n• @Horus: relativity would be just fine with massive particles. It doesn't need an actual implementation of a massless field that traces the outer limits of Lorentz transformations. – CuriousOne Jun 26 '15 at 18:00\n• Possible duplicates: physics.stackexchange.com/q/4700/2451 , physics.stackexchange.com/q/31994/2451 and links therein. – Qmechanic Jun 26 '15 at 19:17\n• The photon can acquire an effective mass due to a few different mechanisms, which respectively cause the following effects: 1. en.wikipedia.org/wiki/Meissner_effect 2. en.wikipedia.org/wiki/Debye_length – Siva Jun 26 '15 at 19:42\n\nI think that the most interesting part of this question is as follows:\n\n1. What happens to the Maxwell equations?\n\nThe speed of light $c$ is a fundamental constant used both in Maxwell equations and Special relativity .\n\nMaxwell equations allow for electromagnetic waves travelling at the speed $c$. Classical special relativity (SR) does not allow for particles with non-zero masses to travel at the speed $c$. From this we conclude that in classical physics photons should be massless.\n\nHowever, in quantum physics the situation is different. Massive particles are allowed to travel at speed even exceeding $c$. See, e.g., quote from Frank Wilczek's Nobel lecture:\n\n\"Imagine a particle moving on average at very nearly the speed of light, but with an uncertainty in position, as required by quantum theory. Evidently it there will be some probability for observing this particle to move a little faster than average, and therefore faster than light, which special relativity won’t permit. The only known way to resolve this tension involves introducing the idea of antiparticles.\"\n\nHence, in quantum theory we are not restricted by the SR's requirement that only massless particles can travel at the speed $c$.\n\nDo Maxwell equations allow for massive sources travelling at the speed $c$?\n\nMaxwell equations do not contain any mass terms. Particularly, in the case of electromagnetic waves in vacuum that are considered \"source-free\".\n\nIf we will assume that photons have non-zero mass, then in addition to electromagnetic fields $\\textbf{E}$, $\\textbf{B}$ we will also require matter fields for the source particle. One of the options is to assume that these matter fields are spinorial:\n\n\\begin{equation} \\psi{}(x)=\\left\\{\\xi{},\\ \\dot{\\eta{}}\\right\\}=\\ \\left\\{\\xi{},\\ 0\\right\\}+\\left\\{0,\\ \\dot{\\eta{}}\\right\\}=\\ \\left({\\begin{array}{ cc} {\\xi{}}^1(x) \\\\ {\\xi{}}^2(x)\\\\ {\\eta{}}_{\\dot{1}}(x) \\\\ {\\eta{}}_{\\dot{2}}(x) \\end{array}}\\right) \\end{equation}\n\nIn addition to Maxwell equations we will also need to introduce some new matter field equations that will be \"responsible\" for photon's kinematics.\n\nThe role of these matter field equations will be similar to the role of Dirac equations in case of electron motion.\n\n\\begin{equation} \\begin{array}{columns} {\\partial{}}^{\\mu{}\\dot{\\nu{}}} {\\eta{}}_{\\dot{\\nu{}}}\\ +\\ im \\ {\\xi{}}^{\\mu{}}=\\ 0 \\\\ \\\\ {\\partial{}}_{\\mu{}\\dot{\\nu{}}} {\\xi{}}^{\\mu{}}+\\ im \\ {\\eta{}}_{\\dot{\\nu{}}}=0 \\\\ \\\\ (Free \\ Dirac \\ equation \\ for \\ electron) \\end{array} \\end{equation}\n\nBelow I will show that there exist a Lorentz-invariant spinorial equation that:\n\n• makes electromagnetic waves \"massive\",\n• can be made not just consistent with, but equivalent to, Maxwell equations,\n• in the case of transverse plane electromagnetic waves allows only for massless solutions travelling at the speed $c$ (photons),\n• in the case of longitudinal plane waves allows for massive solutions travelling at the speed $c$ and satisfying Majorana condition (neutrino).\n\nI use here the spinor calculus developed by B. van der Waerden, G.E. Uhlenbeck and O. Laporte (detailed notations and all calculations can be found here). This is because many spinorial equations are much simpler than the corresponding tensorial equations. As one can see from expressions above and below, this applies equally to both Maxwell and Dirac equations.\n\n## Maxwell equations in spinorial form\n\nIn tensor algebra electromagnetic field strengths are expressed in the form of antisymmetric second rank electromagnetic field tensor $F_{\\mu \\nu}$.\n\nSimilarly, in spinor calculus electromagnetic field strengths are expressed in the form of two complex conjugated symmetric second rank spinors $f_{\\mu{}\\nu{}}$ and $\\dot{f}_{\\dot{\\mu{}}\\dot{\\nu{}}}$:\n\n\\begin{equation} f_{\\mu{}\\nu{}}=\\ f_{\\nu{}\\mu{}} \\end{equation}\n\n\\begin{equation} {\\dot{f}}_{\\dot{\\nu{}}}^{\\dot{\\mu{}}}=\\overline{f_{\\nu{}}^{\\mu{}}} \\end{equation}\n\nDue to symmetry of the spinors the field has only 3 complex components:\n\n\\begin{equation} f_{11}, \\hspace{10mm} f_{12}=\\ f_{21}, \\hspace{10mm} f_{22} \\end{equation}\n\nThis property enables us to introduce the structure of 3-dimentional complex space for electromagnetic field spinors\n\n\\begin{equation} f_{\\nu{}}^{\\mu{}}=\\ \\left[\\begin{array}{ cc} f_1^1 & f_2^1 \\\\ f_1^2 & f_2^2 \\end{array}\\right]=\\ \\left[\\begin{array}{ cc} F^3 & F^1-iF^2 \\\\ F^1+iF^2 & -F^3 \\end{array}\\right]=F^k{\\sigma{}}_k,\\hspace{10mm} k=1,2,3 \\end{equation}\n\nwhere \"coordinates\" $F^k$ can be decomposed into real and imaginary parts\n\n\\begin{equation} \\textbf{F}=\\textbf{E}-i\\textbf{B} \\end{equation}\n\nMaxwell equations in spinor form are very simple and elegant:\n\n\\begin{equation} \\begin{array}{columns} {\\partial{}}^{\\nu{}\\dot{\\rho{}}}f_{\\nu{}}^{\\mu{}}=\\ S^{\\mu{}\\dot{\\rho{}}} \\\\ \\\\ (Maxwell \\ equations \\ in \\ spinorial \\ from) \\end{array} \\end{equation}\n\nHere we use the spinorial form of the electromagnetic current density $S^{\\mu{}\\dot{\\rho{}}}$:\n\n\\begin{equation} S_{\\mu{}\\dot{\\nu{}}}=\\left[\\begin{array}{ cc} S_{1\\dot{1}} & S_{1\\dot{2}} \\\\ S_{2\\dot{1}} & S_{2\\dot{2}} \\end{array}\\right]=\\left[\\begin{array}{ cc} J^0+J^3 & J^1+iJ^2 \\\\ J^1-iJ^2 & J^0-J^3 \\end{array}\\right] \\\\ \\end{equation}\n\nComplex vector $J^k$ corresponding to spinor $S^{\\mu{}\\dot{\\rho{}}}$ can be decomposed into electric ($J_e$) and magnetic ($J_m$) current densities:\n\n\\begin{equation} J^k={J_e}^k-i{J_m}^k \\hspace{10mm} k=0, 1, 2, 3 \\end{equation}\n\nThe Lorentz force density spinor is as follows:\n\n\\begin{equation} {\\Lambda{}}_{\\mu{}\\dot{\\nu{}}}=-\\left[{\\dot{f}}_{\\dot{\\nu{}}}^{\\dot{\\rho{}}} \\ S_{\\mu{}\\dot{\\rho{}}}+f_{\\mu{}}^{\\delta{}} \\ {\\dot{S}}_{\\delta{}\\dot{\\nu{}}}\\right] \\end{equation}\n\nOf course, the force density spinor ${\\Lambda{}}_{\\mu{}\\dot{\\nu{}}}$ corresponds to the Lorentz force density 4-vector $\\mathcal{F}^{\\mu{}}$:\n\n\\begin{equation} {\\Lambda{}}_{\\mu{}\\dot{\\nu{}}}=\\left[\\begin{array}{ cc} {\\Lambda{}}_{1\\dot{1}} & {\\Lambda{}}_{1\\dot{2}} \\\\ {\\Lambda{}}_{2\\dot{1}} & {\\Lambda{}}_{2\\dot{2}} \\end{array}\\right]=\\left[\\begin{array}{ cc} \\mathcal{F}^0+\\mathcal{F}^3 & \\mathcal{F}^1+i\\mathcal{F}^2 \\\\ \\mathcal{F}^1-i\\mathcal{F}^2 & \\mathcal{F}^0-\\mathcal{F}^3 \\end{array}\\right] \\end{equation}\n\n## Matter field equations\n\nNow we need to introduce the matter field equations containing the mass terms of the source particle.\n\nWe will start with the well known Pauli coupling equation that looks very similar to free Dirac equation (see above):\n\n\\begin{equation} \\begin{array}{cols} {\\partial{}}^{\\mu{}\\dot{\\nu{}}}{\\eta{}}_{\\dot{\\nu{}}} + \\ if_{\\nu{}}^{\\mu{}} \\ {\\xi{}}^{\\nu{}} = 0\\\\ \\\\ {\\partial{}}_{\\mu{}\\dot{\\nu{}}}{\\xi{}}^{\\mu{}} + \\ i{\\dot{f}}_{\\dot{\\nu{}}}^{\\dot{\\mu{}}} \\ {\\eta{}}_{\\dot{\\mu{}}} = 0 \\\\ \\\\ (Pauli \\ coupling \\ equation) \\end{array} \\end{equation}\n\nIn this equation the spinor and co-spinor fields are coupled via second rank electromagnetic field spinors $f_{\\mu{}\\nu{}}$ and $\\dot{f}_{\\dot{\\mu{}}\\dot{\\nu{}}}$, while in a free Dirac equation the spinor and co-spinor fields are coupled via mass constant $m$.\n\nThe similarity between two equations can be made even stronger, if we demand that spinorial matter fields $\\xi{}$ and $\\dot{\\eta{}}$ are eigenvectors of the second rank electromagnetic field spinors $f_{\\nu{}}^{\\mu{}}$ and ${\\dot{f}}_{\\dot{\\nu{}}}^{\\dot{\\mu{}}}$ correspondingly:\n\n\\begin{equation} \\begin{array}{ccc} f_{\\nu{}}^{\\mu{}} \\ {\\xi{}}^{\\nu{}}=\\ \\lambda{}\\ {\\xi{}}^{\\mu{}} \\\\ \\\\ {\\dot{f}}_{\\dot{\\nu{}}}^{\\dot{\\mu{}}} \\ {\\eta{}}_{\\dot{\\mu{}}}=\\bar{\\lambda{}}\\ {\\eta{}}_{\\dot{\\nu{}}} \\end{array} \\end{equation}\n\nWith this condition our matter field equations become very simple\n\n\\begin{equation} \\begin{array}{cols} {\\partial{}}^{\\mu{}\\dot{\\nu{}}}{\\eta{}}_{\\dot{\\nu{}}}+\\ i\\lambda{}\\ {\\xi{}}^{\\mu{}}=0\\\\ \\\\ {\\partial{}}_{\\mu{}\\dot{\\nu{}}}{\\xi{}}^{\\mu{}}\\ +\\ i\\bar{\\lambda{}}\\ {\\eta{}}_{\\dot{\\nu{}}}=0 \\end{array} \\end{equation}\n\nand replicate the structure of the free Dirac equation where constant mass term $m$ is replaced by the variable \"mass density\" terms $\\lambda{}$ and $\\bar{\\lambda{}}$.\n\nTaking account the explicit form of electromagnetic field spinors $f_{\\nu{}}^{\\mu{}}$ and ${\\dot{f}}_{\\dot{\\nu{}}}^{\\dot{\\mu{}}}$ one can see that eigenvalues $\\lambda{}$ and $\\bar{\\lambda{}}$ are well known electromagnetic field invariants:\n\n\\begin{equation} \\begin{array}{ccc} {\\lambda{}}_{\\pm{}}=\\ \\pm{}\\sqrt{{\\left(F^1\\right)}^2+{\\left(F^2\\right)}^2+{\\left(F^3\\right)}^2} & & {{\\lambda{}}_{\\pm{}}}^2=\\ E^2-B^2-2i\\ \\textbf{E}\\cdot \\textbf{B} \\\\ \\\\ {\\overline{\\lambda{}}}_{\\pm{}}=\\ \\pm{}\\sqrt{{\\left(\\ \\bar{F^1}\\right)}^2+{\\left(\\ \\bar{F^2}\\right)}^2+{\\left(\\ \\bar{F^3}\\right)}^2} & & {{\\overline{\\lambda{}}}_{\\pm{}}}^2=\\ E^2-B^2+2i\\ \\textbf{E}\\cdot \\textbf{B} \\end{array} \\end{equation}\n\n## Eigenvectors\n\nLet us now derive the expressions for the eigenvectors of the electromagnetic field spinors $f_{\\nu{}}^{\\mu{}}$ and ${\\dot{f}}_{\\dot{\\nu{}}}^{\\dot{\\mu{}}}$.\n\nConsider arbitrary point $Q$ at the space-time. For the sake of convenience we can choose the reference frame (denoted as $M_{\\perp{}}$) in such a way that the fields $\\textbf{E}$, $\\textbf{B}$ at the point $Q$ will be orthogonal to the axis $\\textbf{e}_3$.",
null,
"There is, of course, infinite number of such frames, but all the considerations presented in below are valid for any of these frames.\n\nIn the reference frame $M_{\\perp{}}$ the expression for spinor $f_{\\nu{}}^{\\mu{}}$ at the point $Q$ will be:\n\n\\begin{equation} f_{\\nu{}}^{\\mu{}}=\\ \\left[\\begin{array}{ cc} 0 & F^1-iF^2 \\\\ F^1+iF^2 & 0 \\end{array}\\right] \\end{equation}\n\nbecause $F^3=0$ at the point $Q$.\n\nOne can easily check now that two spinors ${\\xi{}}_+$ and ${\\xi{}}_-$ defined as\n\n\\begin{equation} {\\xi{}}_{\\pm{}}=\\ \\left[\\begin{array}{ c} {\\xi{}}_{\\pm{}}^1 \\\\ \\\\ {\\xi{}}_{\\pm{}}^2 \\end{array}\\right]=\\ \\left[ \\begin{array}{ c} \\pm{}\\sqrt{F^1-iF^2} \\\\ \\\\ \\sqrt{F^1+iF^2} \\end{array}\\right] \\end{equation}\n\nwill be eigenvectors of the matrix $f_{\\nu{}}^{\\mu{}}$ at the point $Q$:\n\n\\begin{equation} f_{\\nu{}}^{\\mu{}} \\ {\\xi{}}_{\\pm{}}^{\\nu{}}=\\ {\\lambda{}}_{\\pm{}} \\ {\\xi{}}_{\\pm{}}^{\\mu{}} \\end{equation}\n\nHence, with the special choice of the reference frame we can write an explicit expression for the components of the spinorial field $\\xi{}$ satisfying eigenvector condition. The expressions for the field components in all other frames can be obtained by the appropriate Lorentz transformations.\n\nSimilarly one can show that at the reference frame $M_{\\perp{}}$ two co-spinors ${\\dot{\\eta{}}}_+$ and ${\\dot{\\eta{}}}_-$ defined as\n\n\\begin{equation} {\\dot{\\eta{}}}_{\\pm{}}=\\ \\left[ \\begin{array}{c} {{\\eta{}}_{\\pm{}}}_{\\dot{1}} \\\\ \\\\ {{\\eta{}}_{\\pm{}}}_{\\dot{2}} \\end{array}\\right]=\\ \\left[ \\begin{array}{c} \\pm{}\\sqrt{\\bar{F^1}-i\\bar{F^2}} \\\\ \\\\ \\sqrt{\\bar{F^1}+i\\bar{F^2}} \\end{array}\\right] \\end{equation}\n\nwill satisfy the condition\n\n\\begin{equation} {\\dot{f}}_{\\dot{\\nu{}}}^{\\dot{\\mu{}}} \\ {{\\eta{}}_{\\pm{}}}_{\\dot{\\mu{}}}={\\bar{\\lambda{}}}_{\\pm{}} \\ {{\\eta{}}_{\\pm{}}}_{\\dot{\\nu{}}} \\end{equation}\n\nat the point $Q$.\n\nFrom these explicit expressions for spinor components we can derive the \"mass square\" of the momentum density 4-vector $P_{\\mu{}}$\n\n\\begin{equation} \\left\\{P_{\\mu{}}\\right\\}\\rightarrow{}P^{\\mu{}\\dot{\\nu{}}}=\\ \\xi^{\\mu{}}\\xi^{\\dot{\\nu{}}}+\\eta^{\\mu{}}\\eta^{\\dot{\\nu{}}} \\end{equation}\n\nwhich is invariant under Lorentz transformations and hence has the same value in all reference frames:\n\n\\begin{equation} P^{\\mu{}}P_{\\mu{}}=4{\\left\\vert{}\\lambda{}\\right\\vert{}}^2 \\end{equation}\n\nIt is worth noting that the momentum density vector $P_{\\mu{}}$ is always time-like, and its time-like component $P_0$ is always positive, hence no solutions with negative energies are allowed.\n\n## Transverse plane electromagnetic waves\n\nNow we will consider the special case of Transverse plane electromagnetic waves in vacuum, assuming that both Maxwell equations and matter field equations are satisfied.\n\nThe major difficulty is in finding the balance between these two equations responsible for evolutions of the electromagnetic field and it's source: the Maxwell equations and the (spinorial) matter field equations correspondingly.\n\nThis balance can be achieved due to strong connection between spinorial matter fields and electromagnetic field established due to eigenvector condition. Using the approach developed by Belinfante and Ohanian (Am. J. Phys. 54 (6) (1986)), we will demonstrate that with this eigenvector condition the matter field equation can be reduced to Maxwell equations, so that evolutions of the electromagnetic field and it's source will be synchronized.\n\nConsider transverse plane waves propagating in the direction of the axis $\\textbf{e}_3$. In each point the electric and magnetic field vectors $\\textbf{E}$, $\\textbf{B}$ are orthogonal to the axis $\\textbf{e}_3$:\n\n\\begin{equation} \\begin{array}{c} \\textbf{E}, \\textbf{B} \\ \\perp{}\\ \\textbf{e}_3 \\\\ \\\\ F^3={\\bar{F}}^3\\equiv{}0 \\end{array} \\end{equation}\n\nThis enables us to use in our calculations the explicit expressions for spinorial filed components derived for the special choice of the reference frame $M_{\\perp{}}$, and rewrite the matter field equations as\n\n\\begin{equation} \\begin{array}{c} \\left({\\partial{}}_1-{i\\partial{}}_2\\right)\\left(F^1+iF^2\\right) = i \\ \\bar{\\lambda{}} \\ \\left(P^0+P^3\\right) \\\\ \\\\ \\left({\\partial{}}_0-{\\partial{}}_3\\right)\\left(F^1-iF^2\\right) = 0 \\\\ \\\\ \\left({\\partial{}}_0+{\\partial{}}_3\\right)\\left(F^1+iF^2\\right) = 0 \\\\ \\\\ \\left({\\partial{}}_1+i{\\partial{}}_2\\right)\\left(F^1-iF^2\\right) = i \\ \\bar{\\lambda{}} \\ \\left(P^0-P^3\\right) \\end{array} \\end{equation}\n\nAt the same time, the Maxwell equations for transverse plane waves are as follows:\n\n\\begin{equation} \\begin{array}{ c} \\left({\\partial{}}_1-{i\\partial{}}_2\\right)\\left(F^1+iF^2\\right)=J^0+J^3 \\\\ \\\\ \\left({\\partial{}}_0-{\\partial{}}_3\\right)\\left(F^1-iF^2\\right)=0 \\\\ \\\\ \\left({\\partial{}}_0+{\\partial{}}_3\\right)\\left(F^1+iF^2\\right)=0 \\\\ \\\\ \\left({\\partial{}}_1+i{\\partial{}}_2\\right)\\left(F^1-iF^2\\right)=J^0-J^3 \\end{array} \\end{equation}\n\nIf we demand the equivalence of the Maxwell and matter field equations, we conclude that in the case of the transverse plane waves there can be established the following relationship between the charge density current $J^{\\mu{}}$ and the momentum density $P^{\\mu{}}$:\n\n\\begin{equation} J^{\\mu{}}=i \\ \\bar{\\lambda{}}\\ P^{\\mu{}} \\end{equation}\n\nIn this expression the electromagnetic field invariant ($i\\bar{\\lambda{}}$) plays the role of the electromagnetic charge density scalar (It is clear that ($-i\\lambda{}$) plays the same role for anti-particles).\n\nGenerally $\\bar{\\lambda{}}$ is complex valued, hence allowing for both non-zero electric and magnetic charge densities.\n\nThe expression for the Lorentz force density acting on matter fields derived for transverse plane waves is as follows:\n\n\\begin{equation} {\\Lambda{}}_{\\mu{}\\dot{\\nu{}}}=-i\\left({\\lambda{}}^2-{\\bar{\\lambda{}}}^2\\right){P_A}_{\\mu{}\\dot{\\nu{}}} \\end{equation}\n\nIt is interesting that the Lorentz force $\\mathcal{F}^{\\mu{}}$ is proportional to the axial vector current ${P_A}^{\\mu{}}$.\n\nFrom expression above we can see that Lorentz force vanishes when $\\textbf{E}\\cdot \\textbf{B}=0$, i.e. when imaginary parts of the squared electromagnetic field invariants ${\\lambda{}}^2$ and ${\\bar{\\lambda{}}}^2$ are zero.\n\nWhen Lorentz force is zero, the momentum density of the matter field remains constant in the course of particle's motion, hence allowing for uniform motion of the particle.\n\nAs mentioned earlier, the value of the \"mass square\" of the momentum density 4-vector $P_{\\mu{}}$ is as follows:\n\n\\begin{equation} P^{\\mu{}}P_{\\mu{}}=4{\\left\\vert{}\\lambda{}\\right\\vert{}}^2 \\end{equation}\n\nThat means that even if the Lorentz force is zero, but $\\lambda$ is non-zero, the source particle will not be travelling at the speed $c$. This is the case of charged massive particles such as $W$ and $Z$ bosons.\n\nIn the case of $\\textbf{E} \\perp{} \\textbf{B}$, $E=B$ we have $\\lambda{}=\\bar{\\lambda{}}=0$, and matter field equations coincide with the \"source-free\" Maxwell equations. In this case the momentum density $P^{\\mu{}}$ of the matter field is non-zero, while the charge density current $J^{\\mu{}}$ is zero.\n\nIn this sense the electromagnetic wave in vacuum is not actually \"source-free\", i.e. despite of zero charge density there exist a spinorial matter field which is the source of electromagnetic field. However, only massless photons can travel at the speed $c$ (with $P^{\\mu{}}P_{\\mu{}}=0$).\n\nOn the contrary, as shown here, in the case of the longitudinal plane waves the same combination of Maxwell equations and matter field equations allows for solutions that:\n\n• satisfy Majorana condition,\n• have non-zero mass and charge densities, and\n• travel at the speed of light.\n\nHence, Maxwell equations allow for massless photons and massive neutrinos both travelling at the speed $c$.\n\n1. What happens to QFT?\n\nElectromagnetic field of a particle is determined by Maxwell equations, while the properties of the particle depending on its mass are determined by matter field equations, such as Dirac equation for electron etc.\n\nThe relationship between these two types of equations is usually established by an arbitrary constant called \"electric charge\", when the \"charge current\" used in Maxwell equations (as a source of electromagnetic field) is simply assumed proportional to mechanical momentum of a particle determined by the matter field equations:\n\n$J_\\mu = Q \\times P_\\mu$\n\n($J_\\mu$ - charge current, $P_\\mu$ - mechanical momentum, $Q$ - \"electric charge\" constant).\n\nThis approach works well in QFT, where all particles are considered point-like or structureless, and charge of a particle is always \"nailed\" to its mass.\n\nFundamentally, in QFT all processes are considered to be happening inside a \"black boxes\", and the theory is only aiming at predicting the outputs of the black box from given inputs. In QFT creation and annihilation operators (and their commutation relations) are used to get rid of input particles and to transform them into ouput particles. Black boxes are nothing but vertices of Feynman diagrams.",
null,
"Due to this fundamental \"point-like\" approach to particles in QFT this theory cannot be used to address the issues related to particles' masses (the well known self-action difficulties in QFT, that are not restricted to the massive particles such as electron, but occur also for the photon). That is why all particles' masses in the Standard Model are just free parameters identified from experiment. For instance, the difference between electron and muon in the SM is only in the value of the coupling constant of the fermion field and Higgs field.\n\nQFT is an excellent theory, but it is not all-embracing, and it is not designed to (or suitable for) addressing the issues of particles' masses.\n\nRoughly speaking, QFT has nothing to do with masses of the particles, and hence nothing is going to happen to QFT if photons will be found to have non-zero masses (which I think is not going to happen). The Standard model is flexible enough to elaborate both massive and massless particles. For instance, the neutrinos in the SM can be made massive and massless, and even have oscillating masses.\n\nNevertheless, there is at least one benefit for the QFT arising from the model described above. In this model the vector current is preserved, while the divergence of the axial current is exactly as required by quantum perturbation theory. In QFT this property is not a consequence of the theory, but the result of an ambiguous choice (known as Axial anomaly).\n\nThere are very strict limits on the mass of the photon already, so it would only affect our understanding of physics on the largest scales.\n\nThe cosmologists would have some hard thinking to do, for instance.\n\nHowever, contrary to a comment, it would not affect relativity beyond requiring us to reconsider the usual name for $c$: not \"the speed of light\" but \"the ultimate speed\".\n\n• Or \"speed of gluons\" ;) – Ryan Unger Jun 26 '15 at 16:19\n• I would argue that technically a massive photon actually solves a lot of cosmological problems. It may completely eliminate heat/cold death scenarios and lead to an endless hierarchy of constantly expanding universes with self-similar laws (that don't have to be identical to ours). \"We\" would be basically someone else's inflationary era. That's pure speculation on my part, of course. – CuriousOne Jun 26 '15 at 18:05\n• @CuriousOne this is a very old question, but could you provide a reference for reading regarding an endless hierarchy of universes with self similar laws? – KF Gauss Jul 13 '17 at 4:44\n\nFrom experimental side, photons having a mass will lead to slight deviation from coulomb's law: $$F=\\frac{ke^2}{r^{2+x}}$$.\n\nFrom theory side, if photons were to have mass, the QED Lagrangian would contain a term $$A_\\mu A^\\mu$$ which will break gauge invariance.\n\nHowever, in condensed matter physics, there are situations where you call the photons gain effective mass. For example, you can say photons gain effective mass in Meissner effect. In fact, this is the historically the first thinking towards the Higgs mechanism. Photons acquire effective mass just as the W and Z bosons acquire mass in the Standard Model of Particle Physics.\n\n• That's not the correct deviation. The potential produced by exchange of massive particles is given by the Yukawa potential: $$U = \\frac{k e^2}{r} \\operatorname{e}^{-mcr / \\hbar}.$$ So the Yukawa force is: $$F = \\frac{ke^2}{r^2} \\operatorname{e}^{-mcr/\\hbar} \\left(\\frac{mcr}{\\hbar} + 1\\right).$$ – Sean E. Lake Sep 17 '16 at 3:50\n\n(1) It wouldn't be called a photon anymore. An example: neutrino. (2) If it is a particle with finite mass(and for simplicity has no spin), its behavior would be then described by Klein-Gordon (-Fock) equation, which is Lorentz-invariant. It will be still a part of well-known physics."
]
| [
null,
"https://i.stack.imgur.com/cSgqM.png",
null,
"https://i.stack.imgur.com/Cp1Eg.jpg",
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.76981574,"math_prob":0.9997218,"size":20640,"snap":"2019-35-2019-39","text_gpt3_token_len":6120,"char_repetition_ratio":0.18201202,"word_repetition_ratio":0.047040973,"special_character_ratio":0.30523255,"punctuation_ratio":0.06157229,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99996984,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,6,null,6,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-09-15T20:02:09Z\",\"WARC-Record-ID\":\"<urn:uuid:64808ed6-0b2a-4559-a7d2-d3ba015604e9>\",\"Content-Length\":\"188154\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7d1ffbfe-2b05-4380-953e-091ca932c51c>\",\"WARC-Concurrent-To\":\"<urn:uuid:223c340d-7a35-456f-b1a3-afd74675a392>\",\"WARC-IP-Address\":\"151.101.129.69\",\"WARC-Target-URI\":\"https://physics.stackexchange.com/questions/191451/what-will-happen-if-the-photon-has-non-zero-mass?noredirect=1\",\"WARC-Payload-Digest\":\"sha1:JPRDW2NZ3YO32JSY757ZJXZRAI7TVFZ4\",\"WARC-Block-Digest\":\"sha1:DUWYL2THBT22UK4AEAWOK4JOLVAQPAIL\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-39/CC-MAIN-2019-39_segments_1568514572289.5_warc_CC-MAIN-20190915195146-20190915221146-00394.warc.gz\"}"} |
https://usethinkscript.com/threads/remove-cloud-from-macd-custom-indicator.2714/ | [
"# Remove cloud from MACD custom indicator\n\nP\n\n#### Parker427\n\n##### New member\nCan someone remove the \"clouds\" that are behind this indicator?\n\nCode:\n``````declare lower;\n\ninput fastLength = 12;\ninput slowLength = 26;\ninput MACDLength = 9;\ninput showBreakoutSignals = yes;\ninput ma_length = 21; #Length(180-200 for floating S/R , 55 for swing entry)\n\n# Four Pole Filter\nscript g {\ninput length = 4;\ninput price = OHLC4;\ndef c;\ndef w;\ndef beta;\ndef alpha;\ndef G;\nc = price;\nw = (2 * Double.Pi / length);\nbeta = (1 - Cos(w)) / (Power(1.414, 2.0 / betaDev) - 1 );\nalpha = (-beta + Sqrt(beta * beta + 2 * beta));\nG = Power(alpha, 4) * c +\n4 * (1 – alpha) * G – 6 * Power( 1 - alpha, 2 ) * G +\n4 * Power( 1 - alpha, 3 ) * G - Power( 1 - alpha, 4 ) * G;\nplot Line = G;\n}\n# Modified MACD\nplot MACD_Value = g(length = fastLength) - g(length = slowLength);\nMACD_Value.Hide();\nplot MACD_Avg = g(price = MACD_Value, length = MACDLength);\nMACD_Avg.Hide();\n\nplot Diff = MACD_Value - MACD_Avg;\nplot ZeroLine = 0;\n\nMACD_Value.SetDefaultColor(GetColor(1));\nMACD_Avg.SetDefaultColor(GetColor(8));\nDiff.SetDefaultColor(GetColor(5));\nDiff.SetPaintingStrategy(PaintingStrategy.HISTOGRAM);\nDiff.SetLineWeight(3);\nDiff.DefineColor(\"Positive and Up\", Color.GREEN);\nDiff.DefineColor(\"Positive and Down\", Color.DARK_GREEN);\nDiff.DefineColor(\"Negative and Down\", Color.RED);\nDiff.DefineColor(\"Negative and Up\", Color.DARK_RED);\nDiff.AssignValueColor(if Diff >= 0 then if Diff > Diff then Diff.Color(\"Positive and Up\") else Diff.Color(\"Positive and Down\") else if Diff < Diff then Diff.Color(\"Negative and Down\") else Diff.Color(\"Negative and Up\"));\n\nZeroLine.SetDefaultColor(GetColor(4));\nZeroLine.HideTitle();\n\nplot UpSignal = if Diff crosses above ZeroLine then ZeroLine else Double.NaN;\nplot DownSignal = if Diff crosses below ZeroLine then ZeroLine else Double.NaN;\n\nUpSignal.SetHiding(!showBreakoutSignals);\nUpsignal.SetLineWeight(3);\nDownSignal.SetHiding(!showBreakoutSignals);\nDownsignal.SetLineWeight(3);\n\nUpSignal.SetDefaultColor(Color.UPTICK);\nUpSignal.SetPaintingStrategy(PaintingStrategy.ARROW_UP);\nUpSignal.HideTitle();\nDownSignal.SetDefaultColor(Color.DOWNTICK);\nDownSignal.SetPaintingStrategy(PaintingStrategy.ARROW_DOWN);\nDownSignal.HideTitle();\n\n# Forward / Reverse EMA\n# (c) 2017 John F. Ehlers\n# Ported to TOS 07.16.2017\n# Mobius\n\n# Inputs:\ninput AA = .1;\n\n# Vars:\ndef CC;\ndef RE1;\ndef RE2;\ndef RE3;\ndef RE4;\ndef RE5;\ndef RE6;\ndef RE7;\ndef RE8;\ndef EMA;\nplot EMA_Signal;\nplot plot0;\n\nCC = if CC == 0 then .9 else 1 – AA;\nEMA = AA * close + CC * EMA;\nRE1 = CC * EMA + EMA;\nRE2 = Power(CC, 2) * RE1 + RE1;\nRE3 = Power(CC, 4) * RE2 + RE2;\nRE4 = Power(CC, 8) * RE3 + RE3;\nRE5 = Power(CC, 16) * RE4 + RE4;\nRE6 = Power(CC, 32) * RE5 + RE5;\nRE7 = Power(CC, 64) * RE6 + RE6;\nRE8 = Power(CC, 128) * RE7 + RE7;\n\nEMA_Signal = EMA – AA * RE8;\nEMA_Signal.AssignValueColor(if EMA_Signal > EMA_Signal\nthen Color.GREEN\nelse Color.RED);\nEMA_Signal.SetLineWeight(3);\n\nplot0 = if IsNaN(close) then Double.NaN else 0;\nplot0.SetDefaultColor(Color.GRAY);\nplot0.HideTitle();\n\n# Multi-moving average Component\n\n#input length = 14; #hint Length: Number of periods to average the data.\ninput movingAverageType = {default \"Simple MA\", \"Exponential MA\", \"Wilders Smoothing\", \"Weighted MA\", \"Hull MA\", \"Adaptive MA\", \"Triangular MA\", \"Variable MA\", \"Dema MA\", \"Tema MA\", \"EHMA\", \"THMA\"};\ninput displace = 0;\n\ndef avg_MACD_Value = ExpAverage(2 * ExpAverage(MACD_Value, ma_length / 2) - ExpAverage(MACD_Value, ma_length), Round(Sqrt(ma_length)));\ndef avg_MACD_Avg = ExpAverage(2 * ExpAverage(MACD_Avg, ma_length / 2) - ExpAverage(MACD_Avg, ma_length), Round(Sqrt(ma_length)));\n\nplot X;\nX.SetDefaultColor(CreateColor(0, 102, 204));\nX.SetLineWeight(3);\nplot Y;\nY.SetDefaultColor(Color.WHITE);\nY.SetLineWeight(3);\n\nswitch (movingAverageType) {\ncase \"Simple MA\":\nX = Average(MACD_Value, ma_length);\nY = Average(MACD_Avg, ma_length);\n\ncase \"Exponential MA\":\nX = ExpAverage(MACD_Value, ma_length);\nY = ExpAverage(MACD_Avg, ma_length);\n\ncase \"Wilders Smoothing\":\nX = WildersAverage(MACD_Value, ma_length);\nY = WildersAverage(MACD_Avg, ma_length);\n\ncase \"Weighted MA\":\nX = wma(MACD_Value, ma_length);\nY = wma(MACD_Avg, ma_length);\n\ncase \"Hull MA\":\nX = HullMovingAvg(MACD_Value, ma_length);\nY = HullMovingAvg(MACD_Avg, ma_length);\n\ncase \"Triangular MA\":\nX = MovAvgTriangular(MACD_Value, ma_length);\nY = MovAvgTriangular(MACD_Avg, ma_length);\n\ncase \"Variable MA\":\nX = VariableMA(MACD_Value, ma_length);\nY = VariableMA(MACD_Avg, ma_length);\n\ncase \"Dema MA\":\nX = DEMA(MACD_Value, ma_length);\nY = DEMA(MACD_Avg, ma_length);\n\ncase \"Tema MA\":\nX = TEMA(MACD_Value, ma_length);\nY = TEMA(MACD_Avg, ma_length);\n\ncase EHMA:\nX = ExpAverage(2 * ExpAverage(MACD_Value, ma_length / 2) - ExpAverage(MACD_Value, ma_length), Round(Sqrt(ma_length)));\nY = ExpAverage(2 * ExpAverage(MACD_Avg, ma_length / 2) - ExpAverage(MACD_Avg, ma_length), Round(Sqrt(ma_length)));\n\ncase THMA:\nX = WMA(WMA(MACD_Value, (ma_length / 2) / 3) * 3 - WMA(MACD_Value, (ma_length / 2) / 2) - WMA(MACD_Value, (ma_length / 2)), (ma_length / 2));\nY = WMA(WMA(MACD_Avg, (ma_length / 2) / 3) * 3 - WMA(MACD_Avg, (ma_length / 2) / 2) - WMA(MACD_Avg, (ma_length / 2)), (ma_length / 2));\n}\n;``````\n\nLast edited by a moderator:",
null,
"#### BenTen\n\nStaff\nVIP\nGo to line 34 in the code and remove the following:\n\nCode:\n``addCloud(MACD_Value, MACD_Avg, Color.GREEN, Color.RED);``",
null,
"Remove Sub-Menu Bar from Charts Questions 2",
null,
"Moving average cloud? Questions 6",
null,
"Cloud above or below stock's opening price? Questions 2"
]
| [
null,
"https://usethinkscript.com/data/avatars/m/0/1.jpg",
null,
"https://usethinkscript.com/data/avatars/s/5/5929.jpg",
null,
"https://usethinkscript.com/data/avatars/s/5/5264.jpg",
null,
"https://usethinkscript.com/data/avatars/s/5/5929.jpg",
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.50313985,"math_prob":0.99513143,"size":5359,"snap":"2020-45-2020-50","text_gpt3_token_len":1789,"char_repetition_ratio":0.20784314,"word_repetition_ratio":0.0665742,"special_character_ratio":0.33289793,"punctuation_ratio":0.27137178,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9967929,"pos_list":[0,1,2,3,4,5,6,7,8],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-10-29T19:32:40Z\",\"WARC-Record-ID\":\"<urn:uuid:f13f6f26-4f0e-41f3-a277-8ba8cf64a10f>\",\"Content-Length\":\"56339\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:155b123d-6b25-41e4-b443-c6d09d5f84c2>\",\"WARC-Concurrent-To\":\"<urn:uuid:9afb23e5-103f-4d40-87e4-c571f3111017>\",\"WARC-IP-Address\":\"104.18.34.79\",\"WARC-Target-URI\":\"https://usethinkscript.com/threads/remove-cloud-from-macd-custom-indicator.2714/\",\"WARC-Payload-Digest\":\"sha1:LAWE4B7GJZQSDXWQE7MSH2UJBBKIUC5C\",\"WARC-Block-Digest\":\"sha1:2SQZHFHCWYLVSCYICU37CLLM45DNIH22\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-45/CC-MAIN-2020-45_segments_1603107905777.48_warc_CC-MAIN-20201029184716-20201029214716-00310.warc.gz\"}"} |
https://convertoctopus.com/477-ounces-to-pounds | [
"## Conversion formula\n\nThe conversion factor from ounces to pounds is 0.0625, which means that 1 ounce is equal to 0.0625 pounds:\n\n1 oz = 0.0625 lb\n\nTo convert 477 ounces into pounds we have to multiply 477 by the conversion factor in order to get the mass amount from ounces to pounds. We can also form a simple proportion to calculate the result:\n\n1 oz → 0.0625 lb\n\n477 oz → M(lb)\n\nSolve the above proportion to obtain the mass M in pounds:\n\nM(lb) = 477 oz × 0.0625 lb\n\nM(lb) = 29.8125 lb\n\nThe final result is:\n\n477 oz → 29.8125 lb\n\nWe conclude that 477 ounces is equivalent to 29.8125 pounds:\n\n477 ounces = 29.8125 pounds",
null,
"## Alternative conversion\n\nWe can also convert by utilizing the inverse value of the conversion factor. In this case 1 pound is equal to 0.033542976939203 × 477 ounces.\n\nAnother way is saying that 477 ounces is equal to 1 ÷ 0.033542976939203 pounds.\n\n## Approximate result\n\nFor practical purposes we can round our final result to an approximate numerical value. We can say that four hundred seventy-seven ounces is approximately twenty-nine point eight one three pounds:\n\n477 oz ≅ 29.813 lb\n\nAn alternative is also that one pound is approximately zero point zero three four times four hundred seventy-seven ounces.\n\n## Conversion table\n\n### ounces to pounds chart\n\nFor quick reference purposes, below is the conversion table you can use to convert from ounces to pounds\n\nounces (oz) pounds (lb)\n478 ounces 29.875 pounds\n479 ounces 29.938 pounds\n480 ounces 30 pounds\n481 ounces 30.063 pounds\n482 ounces 30.125 pounds\n483 ounces 30.188 pounds\n484 ounces 30.25 pounds\n485 ounces 30.313 pounds\n486 ounces 30.375 pounds\n487 ounces 30.438 pounds"
]
| [
null,
"https://convertoctopus.com/images/477-ounces-to-pounds",
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.81191856,"math_prob":0.99514925,"size":1663,"snap":"2020-34-2020-40","text_gpt3_token_len":436,"char_repetition_ratio":0.20072332,"word_repetition_ratio":0.0,"special_character_ratio":0.33734214,"punctuation_ratio":0.104166664,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9938862,"pos_list":[0,1,2],"im_url_duplicate_count":[null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-08-11T21:30:29Z\",\"WARC-Record-ID\":\"<urn:uuid:3b695bd5-2445-4666-8641-19f90cde68fa>\",\"Content-Length\":\"28295\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:1f348880-44d5-478a-bdfd-8e963fb1fe70>\",\"WARC-Concurrent-To\":\"<urn:uuid:ecfd3883-a607-497d-b06b-17840d8f690c>\",\"WARC-IP-Address\":\"172.67.208.237\",\"WARC-Target-URI\":\"https://convertoctopus.com/477-ounces-to-pounds\",\"WARC-Payload-Digest\":\"sha1:LIYHPBG3RJHCRRI52VQV6TSMEYK4ESPZ\",\"WARC-Block-Digest\":\"sha1:GWRTMJNDFYJGEA3FUHS62WE352IXWTIC\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-34/CC-MAIN-2020-34_segments_1596439738855.80_warc_CC-MAIN-20200811205740-20200811235740-00510.warc.gz\"}"} |
https://www.aepochadvisors.com/imcot/ | [
"",
null,
"# IMCOT\n\nIn this comprehensive guide, we will explore the IMCOT function in Excel, which is used to calculate the hyperbolic cotangent of a complex number. The IMCOT function is part of the Excel Engineering Functions and is particularly useful in various engineering, physics, and mathematics calculations. We will cover the syntax, examples, tips and tricks, common mistakes, troubleshooting, and related formulae for the IMCOT function.\n\n## IMCOT Syntax\n\nThe syntax for the IMCOT function in Excel is as follows:\n\nIMCOT(number)\n\nWhere the ‘number’ argument represents the complex number for which you want to calculate the hyperbolic cotangent. The complex number can be entered as a real number, a text string, or a cell reference containing the complex number.\n\n## IMCOT Examples\n\nLet’s look at some examples of using the IMCOT function in Excel:\n\nExample 1: Calculating the hyperbolic cotangent of a complex number\n\nSuppose you have a complex number “3+4i” and you want to calculate its hyperbolic cotangent. You can use the IMCOT function as follows:\n\n=IMCOT(“3+4i”)\n\nThis formula will return the hyperbolic cotangent of the complex number “3+4i”.\n\nExample 2: Using a cell reference as the argument\n\nIf you have the complex number stored in a cell, say A1, you can use the IMCOT function with a cell reference as follows:\n\n=IMCOT(A1)\n\nThis formula will return the hyperbolic cotangent of the complex number stored in cell A1.\n\n## IMCOT Tips & Tricks\n\nHere are some tips and tricks to help you use the IMCOT function more effectively:\n\nTip 1: The IMCOT function can handle complex numbers in both “a+bi” and “a+bj” formats. Excel will recognize both formats and return the correct result.\n\nTip 2: If you need to calculate the hyperbolic cotangent of multiple complex numbers, you can use the IMCOT function in an array formula. This will allow you to perform the calculation on an entire range of complex numbers at once.\n\n## Common Mistakes When Using IMCOT\n\nHere are some common mistakes that users make when using the IMCOT function:\n\nMistake 1: Entering the complex number in an incorrect format. Make sure to enter the complex number in either “a+bi” or “a+bj” format, where ‘a’ and ‘b’ are real numbers, and ‘i’ or ‘j’ represents the imaginary unit.\n\nMistake 2: Using the IMCOT function with a non-complex number. The IMCOT function is specifically designed for complex numbers, so using it with a non-complex number may produce unexpected results.\n\n## Why Isn’t My IMCOT Working?\n\nIf you’re having trouble with the IMCOT function, here are some possible reasons and solutions:\n\nProblem 1: The complex number is entered in an incorrect format. Make sure to enter the complex number in either “a+bi” or “a+bj” format.\n\nSolution: Double-check the format of your complex number and correct it if necessary.\n\nProblem 2: The IMCOT function is returning a #VALUE! error.\n\nSolution: This error usually occurs when the ‘number’ argument is not recognized as a valid complex number. Make sure the complex number is entered in the correct format and that the cell reference (if used) contains a valid complex number.\n\n## IMCOT: Related Formulae\n\nHere are some related formulae that you might find useful when working with the IMCOT function:\n\n1. IMCOSH: This function calculates the hyperbolic cosine of a complex number.\n\n=IMCOSH(number)\n\n2. IMSINH: This function calculates the hyperbolic sine of a complex number.\n\n=IMSINH(number)\n\n3. IMTAN: This function calculates the hyperbolic tangent of a complex number.\n\n=IMTAN(number)\n\n4. IMCSC: This function calculates the hyperbolic cosecant of a complex number.\n\n=IMCSC(number)\n\n5. IMSEC: This function calculates the hyperbolic secant of a complex number.\n\n=IMSEC(number)\n\nBy understanding the IMCOT function and its related formulae, you can perform complex calculations involving hyperbolic cotangent and other hyperbolic functions with ease. This comprehensive guide should provide you with all the information you need to use the IMCOT function effectively in Excel.\n\n## Related",
null,
"### Hard to find or retain a good accountant? Try cloud accounting solution\n\nForeign business owners or management team always take financial transparency as a pre-condition for good decision making and sustainable profitability. However, achieving the visualization of",
null,
"### Cloud Accounting Software Automates Compliance Service in China\n\nManaging accounting compliance in China can be a challenging task for businesses, as it involves dealing with complex regulations and paperwork. However, the advent of",
null,
""
]
| [
null,
"data:image/svg+xml;base64,PHN2ZyB4bWxucz0iaHR0cDovL3d3dy53My5vcmcvMjAwMC9zdmciIHdpZHRoPSIyMjQzIiBoZWlnaHQ9IjE0NDYiIHZpZXdCb3g9IjAgMCAyMjQzIDE0NDYiPjxyZWN0IHdpZHRoPSIxMDAlIiBoZWlnaHQ9IjEwMCUiIHN0eWxlPSJmaWxsOiNjZmQ0ZGI7ZmlsbC1vcGFjaXR5OiAwLjE7Ii8+PC9zdmc+",
null,
"data:image/svg+xml;base64,PHN2ZyB4bWxucz0iaHR0cDovL3d3dy53My5vcmcvMjAwMC9zdmciIHdpZHRoPSIzMDAiIGhlaWdodD0iMTY0IiB2aWV3Qm94PSIwIDAgMzAwIDE2NCI+PHJlY3Qgd2lkdGg9IjEwMCUiIGhlaWdodD0iMTAwJSIgc3R5bGU9ImZpbGw6I2NmZDRkYjtmaWxsLW9wYWNpdHk6IDAuMTsiLz48L3N2Zz4=",
null,
"data:image/svg+xml;base64,PHN2ZyB4bWxucz0iaHR0cDovL3d3dy53My5vcmcvMjAwMC9zdmciIHdpZHRoPSIzMDAiIGhlaWdodD0iMzAwIiB2aWV3Qm94PSIwIDAgMzAwIDMwMCI+PHJlY3Qgd2lkdGg9IjEwMCUiIGhlaWdodD0iMTAwJSIgc3R5bGU9ImZpbGw6I2NmZDRkYjtmaWxsLW9wYWNpdHk6IDAuMTsiLz48L3N2Zz4=",
null,
"data:image/svg+xml;base64,PHN2ZyB4bWxucz0iaHR0cDovL3d3dy53My5vcmcvMjAwMC9zdmciIHdpZHRoPSIxMDIwIiBoZWlnaHQ9IjEwMjAiIHZpZXdCb3g9IjAgMCAxMDIwIDEwMjAiPjxyZWN0IHdpZHRoPSIxMDAlIiBoZWlnaHQ9IjEwMCUiIHN0eWxlPSJmaWxsOiNjZmQ0ZGI7ZmlsbC1vcGFjaXR5OiAwLjE7Ii8+PC9zdmc+",
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.82285076,"math_prob":0.9580831,"size":3954,"snap":"2023-40-2023-50","text_gpt3_token_len":935,"char_repetition_ratio":0.22936709,"word_repetition_ratio":0.10223642,"special_character_ratio":0.20283258,"punctuation_ratio":0.10807113,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9941176,"pos_list":[0,1,2,3,4,5,6,7,8],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-23T22:07:51Z\",\"WARC-Record-ID\":\"<urn:uuid:63264d12-c40c-4c96-8873-6f5227cd33fa>\",\"Content-Length\":\"116202\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b5df0811-27ce-4868-8028-582e0f06a5be>\",\"WARC-Concurrent-To\":\"<urn:uuid:08f4b5bd-d726-424a-90fb-05c2c0e26fcd>\",\"WARC-IP-Address\":\"123.60.71.191\",\"WARC-Target-URI\":\"https://www.aepochadvisors.com/imcot/\",\"WARC-Payload-Digest\":\"sha1:OZR7I7MFVUQCYSSFJHYOVW7S3BKI5AT3\",\"WARC-Block-Digest\":\"sha1:XRM4DHUWDPEIB25UDDO3PFJSITQSCQH4\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233506528.3_warc_CC-MAIN-20230923194908-20230923224908-00465.warc.gz\"}"} |
http://blogs.geelongcollege.vic.edu.au/62821/2018/11/01/fays-nines-reflection/ | [
"# Fay’s Nines Reflection\n\nThis past few weeks we have been doing the Fay’s Nines problem. Here are some unique solutions:",
null,
"As you can see, there are 3 columns of numbers. The units column equals 19, the tens column equals 18 and the hundreds column equals 8. But for this to work, you have to carry (starting from the units column), then it all adds up to 999.\n\nWhen I started playing with this problem I was interested to see how many unique solutions their were. I tried to find different strategies to work out different solutions, but I was having trouble finding unique solutions fast.\n\nAt the end of the first session (during the class discussion) Sienna and I realised that the units column had to add up to 19, the tens column had to add up to 18 and the hundreds column had to add up to 8. Here is our working out for the first session.",
null,
"To break this problem into manageable parts, I got one unique solution and then turned it into another one by moving the numbers around. But with that strategy, I only got 2 solutions with similar numbers.\n\nI kept doing trail and error for two sessions. Then I realised that if I got one solution, I could get a lot more. I didn’t realise it at the time, but I could get 36 more solutions. So, to start off with I kept the same numbers in the units column, and shuffled the numbers around those ones. I found 20 solutions, but I had a feeling there was more. After Mr Henderson had told us that there were 36 solutions for each 3 numbers that were in the units column, I had a light-bulb moment. Because there are 36 solutions for each combinations for each 3 numbers in the units column, and there are 5 units combinations (that equal 19), you multiply 36 by 5 and that equals 180, which is the correct answer.\n\nThis problem was challenging because I didn’t understand the carrying. After Mr Hopper helped my understand, I kept trying to find the answer. But little did I know that I had to find the answer to a smaller problem. To find the number of solutions for each 3 numbers in the units columns. This problem was very challenging and fun, and I hope to do something similar in the future."
]
| [
null,
"http://blogs.geelongcollege.vic.edu.au/62821/files/2018/11/WIN_20181102_10_29_31_Pro-2iy1hbt-300x169.jpg",
null,
"http://blogs.geelongcollege.vic.edu.au/62821/files/2018/11/Fays-lesson-1-data-1b2qmdm-300x225.jpg",
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.97647864,"math_prob":0.7655482,"size":2110,"snap":"2021-04-2021-17","text_gpt3_token_len":473,"char_repetition_ratio":0.15764482,"word_repetition_ratio":0.050890584,"special_character_ratio":0.22890995,"punctuation_ratio":0.094808124,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9723521,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,3,null,3,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-01-26T11:20:39Z\",\"WARC-Record-ID\":\"<urn:uuid:43e7a7d8-9055-4f8c-b0df-86fd2090ab79>\",\"Content-Length\":\"32583\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:5367bc8e-00d6-40aa-b71a-eecabab31217>\",\"WARC-Concurrent-To\":\"<urn:uuid:79ddc4cb-94d3-4df2-bc6e-4a67283d0f5b>\",\"WARC-IP-Address\":\"52.62.79.83\",\"WARC-Target-URI\":\"http://blogs.geelongcollege.vic.edu.au/62821/2018/11/01/fays-nines-reflection/\",\"WARC-Payload-Digest\":\"sha1:43OTQFIV4VCQPM5JXCIYH6NA3JNC7QEL\",\"WARC-Block-Digest\":\"sha1:I5IKEW3UE7HUH6O7KM36RQ5A43ZKIXSV\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-04/CC-MAIN-2021-04_segments_1610704799741.85_warc_CC-MAIN-20210126104721-20210126134721-00124.warc.gz\"}"} |
https://www.turito.com/learn/physics/force-between-charges-grade-8 | [
"#### Need Help?\n\nGet in touch with us",
null,
"",
null,
"# Force Between Charges\n\n## Key Concepts\n\n• The force between the charges\n• Force of attraction between the charges\n• Force of repulsion between the charges\n• Conductors, insulators and semiconductors\n\n## Introduction: Force Between Charges\n\n### The force between the charges\n\nConsider an electric field in which two charges, q1 and q2, are at the distance d from each other.\n\n### Coulomb inverse square law:\n\nCoulomb inverse square law states that charges which are present in the same electric field experience attraction or the repulsion force.\n\nThe force in the electric field is directly proportional to the charges present and inversely proportional to the square of the distance between them.\n\nConsider k=1\n\nThe units of the force are coulomb2/meter2\n\nWhich is also written as C2/m2\n\n### Points to be remembered in coulomb inverse square law:\n\n1. Both charges that experience force should have the same electric charge.\n1. There may be both positive or the both negative charges.\n1. The distance between the charges is always measured in the square.\n1. Force doesn’t have SI unit Newton in the inverse square law.\n\n### Coulomb’s inverse law if the charge has opposite charges:\n\nThe negative sign is added as the charges are opposite to each other, and the charges have the force of attraction.\n\n### Coulomb’s inverse law if the charge has same charges:\n\nThe positive sign indicates that the charges are the same and have a force of repulsion.\n\n### Conductors:\n\nSubstances or the objects which allow electricity to pass through them completely are called conductors.\n\n1. Iron\n1. Copper\n1. Aluminum\n\n### Semi-conductors:\n\nSubstances or objects that allow electricity to pass through them are partially called semi-conductors.\n\n1. Silicon\n1. Germanium\n\n### Insulators:\n\nSubstances or objects which doesn’t allow electricity to pass through them are called insulators\n\n1. Rubber\n2. Plastic\n\n### Summary:\n\nCoulomb inverse square law states that charges present in the same electric field experiences attraction or the repulsion force.\n\nThe force in the electric field is directly proportional to the charges present and inversely proportional to the square of the distance between them.\n\nThe units of force are coulomb²/meter\n\nWhich is also written as C²/m²\n\nPoints to be remembered in coulomb inverse square law:\n\n1. Both charges that experience force should have the same electric charge.\n2. There may be both positive or both negative charge.\n3. The distance between the charges is always measured in the square.\n4. Force doesn’t have SI unit Newton in the inverse square law.\n\nConductors:\n\nSubstances or objects which allow electricity to pass through them completely are called conductors.\n\nSemiconductors:\n\nSubstances or objects that allow electricity to pass through them are partially called semiconductors.\n\nInsulators:\n\nSubstances or objects which do not allow electricity to pass through them are called Insulators.\n\n#### Related topics",
null,
"#### Different Types of Waves and Their Examples\n\nIntroduction: We can’t directly observe many waves like light waves and sound waves. The mechanical waves on a rope, waves on the surface of the water, and a slinky are visible to us. So, these mechanical waves can serve as a model to understand the wave phenomenon. Explanation: Types of Waves: Fig:1 Types of waves […]",
null,
"#### Dispersion of Light and the Formation of Rainbow\n\nIntroduction: Visible Light: Visible light from the Sun comes to Earth as white light traveling through space in the form of waves. Visible light contains a mixture of wavelengths that the human eye can detect. Visible light has wavelengths between 0.7 and 0.4 millionths of a meter. The different colors you see are electromagnetic waves […]",
null,
"#### Force: Balanced and Unbalanced Forces\n\nIntroduction: In a tug of war, the one applying more force wins the game. In this session, we will calculate this force that makes one team win and one team lose. We will learn about it in terms of balanced force and unbalanced force. Explanation: Force Force is an external effort that may move a […]",
null,
"",
null,
"",
null,
""
]
| [
null,
"https://a.storyblok.com/f/128066/x/8fc40ab3b3/search18gray.svg",
null,
"https://a.storyblok.com/f/128066/x/907568df11/closegray12.svg",
null,
"data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7",
null,
"data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7",
null,
"data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7",
null,
"data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7",
null,
"data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7",
null,
"data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7",
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.93776345,"math_prob":0.95946795,"size":4409,"snap":"2023-40-2023-50","text_gpt3_token_len":929,"char_repetition_ratio":0.14120318,"word_repetition_ratio":0.29420084,"special_character_ratio":0.19573599,"punctuation_ratio":0.10519645,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9710133,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-02T09:10:37Z\",\"WARC-Record-ID\":\"<urn:uuid:665aa37f-b510-44e2-9070-5b7a0476f093>\",\"Content-Length\":\"130413\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:44875ce0-6d14-46ec-b0a7-689a19263d92>\",\"WARC-Concurrent-To\":\"<urn:uuid:ea5bb572-23fa-47ca-87f7-58899364d709>\",\"WARC-IP-Address\":\"3.111.119.132\",\"WARC-Target-URI\":\"https://www.turito.com/learn/physics/force-between-charges-grade-8\",\"WARC-Payload-Digest\":\"sha1:BHUH76WS3S6OLIEFIEACBJOSAEPM4HLK\",\"WARC-Block-Digest\":\"sha1:Q627WIYHNRCU43NVYO5S4G5FXIWKIO6H\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100381.14_warc_CC-MAIN-20231202073445-20231202103445-00792.warc.gz\"}"} |
https://www.themathdoctors.org/whats-the-point-of-limits/ | [
"# What’s the Point of Limits?\n\n#### (An archive question of the week)\n\nMany calculus courses start out with a chapter on limits; or they may be introduced in a “precalculus” course. But too often the concept is not sufficiently motivated. What good are limits? Why did they have to be invented? Are they as simple as they seem? Why is an epsilon-delta proof necessary?\n\n## Are limits useless?\n\nThe following question arrived in 2001:\n\nUnderstanding the Need for LimitsI searched all over your site and couldn't find anything about understanding limits.Basically I'm trying to find any extra help I can on understanding the concept of a limit. Most of the time I get the same vague description: \"The value that f(x) approaches as x approaches some number.\"I can see the need for a limit on things like: lim (x->0) [sinx/x]because you need to know what the value \"approaches\" not what it actually equals, because at 0 it basically doesn't exist. But why do they give us problems like: lim (x->3) [x^2]is this just for practice, or am I missing something?Basically what I'm asking:1. Should the concept of a limit be applicable to all functions, or is it really only useful for certain situations?2. Are there any other ways of describing a limit?3. Why is this concept so vague?One more thing. The definition of a derivative is: lim (h->0) [f(x+h) - f(x)] / hbut I don't see the need for a limit here because all you do is plug in the value of 0 for h once you've reduced the function into its simplest form. So why the limit? Why not just say, \"when h = 0\"?I really appreciate any time you spend on this. A service like this is very useful and informative.\n\nJosh basically understands what a limit is, but sees it as unnecessary or trivial in many cases. So, beyond the mechanics of finding or proving limits, what is the concept really about?\n\nHi, Josh. These are good questions, and too easily overlooked when this subject is taught too formally. I think it's important to have not only a good understanding of the formal definitions, but also a feel for how things work and what they are all about.\n\nBut why do they give us problems like:\n(x->3)lim [x^2]\nis this just for practice, or am I missing something?\n\nExactly: This is for practice with the concept of limits and the methods for proving them. But also, it introduces the concept of continuity. It is possible that such a function might NOT approach its value at 3, so you have to prove that it does. In case you haven't been introduced to continuity yet, a continuous function is one whose limit is in fact equal to its value; the fact that most familiar functions are continuous is the reason the concept seems a little silly to you.\n\nThis happens in many fields of math: to keep it simple, we have to give problems where the answer seems almost trivial. As Josh recognized, some limit questions, like $$\\displaystyle \\lim_{x\\rightarrow 0}\\frac{\\sin(x)}{x}$$, are important and non-obvious; but others seem hardly worth doing, because you can just plug in the number. I’m not sure all calculus students need to be able to prove limits (that’s for the math majors), but seeing and understanding such a proof is intended to help clarify the point of limits, which is, in part, that approaching a value is not the same as actually having that value.\n\n1. Should the concept of a limit be applicable to all functions, or is it really only useful for certain situations?\n\nAs I suggested in my mention of continuity, the limit concept is applicable to all functions, but is only \"interesting\" for those peculiar functions that either are not continuous, or are not defined at some points.\n\nFor a continuous function, evaluating a limit requires just evaluating the function (once you know it really is continuous). It just happens, as we’ll see, that the primary purpose of limits in calculus, the derivative, is a case where you can’t (just) plug in a number; that’s the interesting case.\n\n2. Are there any other ways of describing a limit?\n\nThe general idea of \"approaching,\" and the formal definition using delta and epsilon, are the two main ways to discuss limits. The latter is just a precise statement of the former. Actually, there's an even more formal and general (and therefore hard to follow) definition that takes it beyond functions of a single real variable, which generalizes the idea of a delta to \"open balls\" or \"neighborhoods.\" If you didn't find \"Formal Definition of a Limit\" in our archives, you may want to read it, because it attempts to demystify the formal definition. \n\nThat link is one that I gave last time, but didn’t quote.\n\n3. Why is this concept so vague?\n\nWhen calculus was first being developed, the concepts of differentiation and integration were far more vague, and needed careful definition before mathematicians could be comfortable working with them - there was no way to prove anything about such unformed concepts as \"sliding two points on a curve together.\" The delta-epsilon definition was introduced precisely so that these concepts did not have to be so vague. I'm not sure exactly what about limits seems vague to you. I think probably you don't mean \"vaguely defined,\" but something like \"vaguely related to anything else,\" or \"not clearly needed.\" I hope my answers to your other questions deal with that.\n\n## Limits and derivatives\n\nThe derivative, before the concept of limit, sounded like magic: the slope of the line formed by two points that have moved toward one another until the are really the same point and so can’t determine a line? That’s “vague”! This is what philosopher George Berkeley famously called “ghosts of departed quantities”, mocking the inadequate mathematical foundations of calculus as it was then. The limit concept is what rescued the sanity of mathematics.\n\nThe definition of a derivative is: lim (h->0) [f(x+h) - f(x)] / hI don't see the need for a limit here because all you do is plug in the value of 0 for h once you've reduced the function into its simplest form. So why the limit? Why not just say, \"when h = 0?\"Because \"reducing a function to its simplest form\" is not something we can do arbitrarily, or even define clearly in all cases. (You admitted that there is a need to use a limit for sin(x)/x, and that is exactly what you get if you work out the derivative of the sine at zero.) If we want a definition of the derivative, it has to apply to any differentiable function, not just to those we can work with easily. And in fact the limit concept clarifies exactly what you mean when you talk about the \"simplest form,\" in a way that is precisely defined.Think about the geometric model of the derivative as the slope of a curve. This makes it the limit of the slope of chords, as the endpoints move together. This can't be defined at all for h = 0; there is no chord in that case. So it has to be defined as a limit. The \"h\" form of the definition merely formalizes this definition; we can't drop the need for a limit just because it's now written in algebraic form.Feel free to continue this conversation - your questions are definitely worth asking, and I'd like to keep talking until you're confident that you understand the concepts.\n\nIn common introductory examples of finding derivatives from the definition, the limit is a simple one that can be found by simplifying the expression in a way that “fills the hole” of a removable discontinuity (see below). Such an example makes it look as if the limit is a trivial part of the definition. But this is not true of the derivative of the sine function, for what that previously mentioned limit of sin(x)/x is needed.\n\nJosh replied, to check his understanding:\n\nLet me start by thanking you for your response. I can't stress enough how great this service is!\n\nNow, on to more questions. I have read the simplified formal definition of the limit and it certainly makes things more clear.\n\nLet's see if I have this straight:\n\nA limit is when you take a value for an independent variable and show that the function produces a value for the dependent variable that is consistently close to (whatever value it produces) even if the independent variable's value is only close to it's initial value?\n\nSo if I put in a 3 and get out a 5, the limit exists if a 3.0001 produces a number close to 5, and 2.9999 produces a number close to 5? Which is basically like saying that the value was approached from both sides.\n\nThen you take this concept and use it to evaluate \"interesting\" functions where you need to know what value a function approaches but not what it equals at certain points. Or to show that a function does, or does not, remain continuous for all independent values.\n\nAm I close?\n\nI guess the hard part about limits is understanding what, or when, they are used. Our Calculus book covered them in the first couple of chapters, but then we haven't really seen them again since. Other than using the definition of a derivative, which now we just use the shortcut rules rather then the formal definition, when differentiating.\n\nThank you again. I hope I'm not coming off too thickheaded. I've always been the type to prefer understanding things thoroughly, not superficially.\n\n## What a limit is\n\nExcellent questions and comments! I responded:\n\nNever apologize for trying to understand something thoroughly! It's wonderful to see someone who isn't satisfied with the shallow stuff.I think you've basically got it. Of course, you just gave examples, so what you said is far from complete. It wouldn't be enough just to show that the value at 3.00001 is close to the limit; you have to show that this is true no matter how you define \"close.\" I forget just how the answer I pointed you to stated it, but here's how I explain the definition of a limit:You don't believe I'm right about the limit, and challenge me to prove that the value of the function, f(x), stays close to 5 whenever x is close to 3. Since \"close\" is not well defined, you're not satisfied to let me decide how close is close enough; in order to convince you, I let you decide how close it has to be. You say, okay, I want you to make f(x) within 0.00000000001 of 5. I do some calculating and say, as long as x is within 0.000000000000000001 of 3, it will work. Then you say, well, maybe I should have picked a smaller number, and then you couldn't do it! But I turn around and show you that I have a formula I can use to pick my distance, which will work no matter how small your definition of \"close\" is. At that point you give up. I've convinced you. We don't need to play the game any more. So the epsilon-delta proof says, no matter how close you consider close, there will be some range of x around the target value for which f(x) will be that close to the proposed limit.\n\nThis is similar to the explanations of the epsilon-delta formulation that I referred to last time.\n\n## Where a limit is needed\n\nYou're right that limits are used mostly at the definition level in elementary calculus; once you've been convinced that the shortcuts give the correct derivative or integral, you can ignore them. Whenever you come across a new type of function, of course, you'll have to go back to the definition; and there are some types of problems where the limit is an essential part of the problem (such as infinite series, which involve a related sort of limit). But the main use of the limit, as far as you are concerned right now, is just to make it possible to do calculus with confidence that everything is working right behind the scenes. Calculus was done before the limit concept was clearly defined; but mathematicians had a queasy feeling that their foundations weren't quite stable until that work was done. Now that the foundation has been stabilized, you can just shut the basement door and not worry what's going on down there (most of the time). It's still good to understand it, though, so you'll know what to think about when an earthquake hits.\n\nI can see where you're coming from on the definition you gave me. It makes perfect sense. All I need to do is get more comfortable with the mathematical proof side of the limit concept (the delta epsilon one.) I just need to let it sink in. Can I ask you a semi-personal question? At what point in your mathematics career did you first feel like you understood limits? Is this concept something that a first semester Calculus student should feel comfortable with? Or is it something that comes with time and more advanced knowledge? I really liked your last description of a limit. That makes a lot more sense.\n\nTo which I replied:\n\nI don't specifically recall being troubled by the limit concept when I first learned it (though it was a long time ago); but my understanding has certainly deepened with time and further courses. I suppose I've used it too much since then to be able to recall what that first exposure was like - just as you probably don't recall how amazing the concept of talking felt when you said your first word.\n\n## Why not just substitute?\n\nI’d like to add another answer, from half a year later, in which I referred to this one; it provides an example of the sort of removable discontinuity that Josh found uninspiring:\n\nGraphing LimitsI was hoping you could explain the concept of \"limits\" and how to read them from graphs, and why substituting the \"c\" number into the equation doesn't always work (i.e. lim (x^2-1)/(x-1) ) x-->1the answer is 2, but I didn't get 2, I got undefined. Help, please!\n\nHere, you can’t just substitute -1 into the formula, because it results in 0/0. Justin here is expecting all limits to be entirely trivial, the opposite of Josh’s concern! So here, I started by answering the broader question, about the meaning of limits, once again explaining the epsilon-delta “game”:\n\nHere's the basic idea of a limit: We have a function, which may or may not be defined at x = c. We want to know how it behaves near x = c, so we look at values of x closer and closer to c. Suppose the graph of y = f(x) looks like this: y | o : | o : | o : L+ - - - - - - - - - -?- - - - - - - | : o | : o | : +--------------------+-------------- x cThe closer x gets to c, the closer y gets to L. Then we say that f(x) approaches L as x approaches c; that is, L is the limit of f(x) as x approaches c. For the sort of functions we usually work with, L is in fact the value of f(c), so this seems like a waste of effort. But in some very important cases (those for which this concept was invented), f(c) is not even defined, so the limit is the only way we can talk about this idea of closeness. In a sense, the limit \"fills the gap\" in the definition of the function, allowing us to assign a value to f(c) that makes the function \"continuous.\"The trick is to define \"closer and closer\" precisely. We do this by seeing it as sort of a game, the sort of game you might have between two gamblers who don't trust one another. You say it's close - but how do I know that what you mean by \"close\" is really close enough?So I make a challenge: I'll believe that L is the limit, if for ANY definition of \"close\" I choose (that is, any distance I call \"close enough\"), you can give me a range around x = c within which the function is always that close to L. That's what the \"epsilon-delta definition\" of a limit means. If _you_ can keep the function as close to L as _I_ want it to be, I'll believe you. (It's sort of like having one kid cut the cake and the other choose the larger piece.)\n\n## Filling the hole\n\nThen I took a look specifically at the example:\n\nYour problem is lim (x^2-1)/(x-1) x-->1To find this limit, you have to simplify the expression so that it is defined for x = 1, and then substitute. Without simplifying, you get 0/0, which is indeterminate. (That is, it might have any value at all, and can't be determined without the additional information in the limit.) So this is an example where the function is not defined at x = 1, yet we can talk about its limit there. What we are doing is showing that the function is equal to a new function everywhere except at x = 1, so we can \"fill the gap\" by saying it is also equal to that function at x = 1. This value is the limit.\n\nNote that when we simplify such a rational function, we change its domain: we factor the numerator, making it $$\\displaystyle \\frac{(x+1)(x-1)}{x-1} = x+1$$, and find that the new function is defined for all values of x, but is equal to the original wherever it was defined. Therefore (since we know the new function is continuous, having previously proved this for all linear functions), we can find the limit by mere substitution. It looks almost trivial, as Josh saw it; but there is a lot going on behind the scenes to make it work.\n\nThis site uses Akismet to reduce spam. Learn how your comment data is processed."
]
| [
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.9658155,"math_prob":0.88290876,"size":16657,"snap":"2021-43-2021-49","text_gpt3_token_len":3794,"char_repetition_ratio":0.14327748,"word_repetition_ratio":0.06264424,"special_character_ratio":0.23305517,"punctuation_ratio":0.1014068,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9746701,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-19T22:01:50Z\",\"WARC-Record-ID\":\"<urn:uuid:d1c10801-ca5d-4fb0-a2de-f47b9eb6b24d>\",\"Content-Length\":\"95538\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:0297b4cc-31a6-4f49-97c6-12459877a5ad>\",\"WARC-Concurrent-To\":\"<urn:uuid:86e62c62-ef18-4cac-9f62-e0cee7ecf1a1>\",\"WARC-IP-Address\":\"66.228.34.45\",\"WARC-Target-URI\":\"https://www.themathdoctors.org/whats-the-point-of-limits/\",\"WARC-Payload-Digest\":\"sha1:FZ55CMZ3UA4F7WISJVUHGJSMHUCQD5EK\",\"WARC-Block-Digest\":\"sha1:6PYGFB2FGWQKJJS27PZVFLRSAR3X5RLT\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323585281.35_warc_CC-MAIN-20211019202148-20211019232148-00079.warc.gz\"}"} |
https://kr.mathworks.com/help/comm/ref/bchenc.html | [
"# bchenc\n\n## Syntax\n\n``code = bchenc(msg,N,K)``\n``code = bchenc(msg,N,K,paritypos)``\n\n## Description\n\nexample\n\n````code = bchenc(msg,N,K)` encodes the input message using an (`N`,`K`) BCH encoder that uses a narrow-sense generator polynomial. For a description of Bose–Chaudhuri–Hocquenghem (BCH) coding, see .```\n````code = bchenc(msg,N,K,paritypos)` appends or prepends the parity symbols to the encoded input message to form the output.```\n\n## Examples\n\ncollapse all\n\nSet the BCH parameters for a Galois array of GF(2).\n\n```M = 4; n = 2^M-1; % Codeword length k = 5; % Message length nwords = 10; % Number of words to encode```\n\nCreate a message.\n\n`msgTx = gf(randi([0 1],nwords,k));`\n\nFind the error-correction capability.\n\n`t = bchnumerr(n,k)`\n```t = 3 ```\n\nEncode the message.\n\n`enc = bchenc(msgTx,n,k);`\n\nCorrupt up to `t` bits in each codeword.\n\n`noisycode = enc + randerr(nwords,n,1:t);`\n\nDecode the noisy code.\n\n`msgRx = bchdec(noisycode,n,k);`\n\nValidate that the message was properly decoded.\n\n`isequal(msgTx,msgRx)`\n```ans = logical 1 ```\n\nIncrease the number of possible errors, and generate another noisy codeword.\n\n```t2 = t + 1; noisycode2 = enc + randerr(nwords,n,1:t2);```\n\n`[msgRx2,numerr] = bchdec(noisycode2,n,k);`\n\nDetermine if the message was properly decoded by examining the number of corrected errors, `numerr`. Entries of `-1` correspond to decoding failures, which occur when the codeword has more errors than can be corrected for the specified `[n,k]` pair.\n\n`numerr`\n```numerr = 10×1 1 2 -1 2 3 1 -1 4 2 3 ```\n\nTwo of the ten transmitted codewords were not correctly received.\n\n## Input Arguments\n\ncollapse all\n\nMessage to encode, specified as a Galois field array of symbols over GF(2). Each `K`-element row of `msg` represents a message word, where the leftmost symbol is the most significant symbol.\n\nExample: `msgTx = gf(randi([0 1],10,5))`, where `msgTx` is a 10-by-5 ```gf array```.\n\nCodeword length, specified as an integer of the form `N` = 2M–1, where M is an integer from 3 through 16. For more information, see Tips.\n\nExample: `15` for `M=4`\n\nMessage length, specified as an integer. `N` and `K` must produce a narrow-sense BCH code.\n\nExample: `5` specifies a Galois array with five elements\n\nParity position, specified as `'end'` or `'beginning'`. Parity symbols are at the end or beginning of each word in the output Galois array.\n\n## Output Arguments\n\ncollapse all\n\nEncoded message, returned as a Galois field array. Parity symbols are at the end or beginning of each word in the output Galois array. To specify the position of the parity symbols, use the `paritypos` argument.\n\n## Tips\n\n• To generate the list of valid (`N`,`K`) pairs along with the corresponding values of the error-correction capability, run `bchnumerr`(`N`).\n\n• Valid values for `N` = 2M–1, where M is an integer from 3 through 16. The maximum allowable value of `N` is 65,535.\n\n Clark, George C., Jr., and J. Bibb Cain. Error-Correction Coding for Digital Communications, New York: Plenum Press, 1981."
]
| [
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.6781083,"math_prob":0.97578275,"size":1690,"snap":"2021-43-2021-49","text_gpt3_token_len":493,"char_repetition_ratio":0.11150652,"word_repetition_ratio":0.0,"special_character_ratio":0.27100593,"punctuation_ratio":0.18457301,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99246055,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-12-09T12:47:04Z\",\"WARC-Record-ID\":\"<urn:uuid:0cddc3a2-122b-4da9-b38a-519a2100c771>\",\"Content-Length\":\"90027\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:fdfaae6a-405b-4caf-baf3-61a6db698a29>\",\"WARC-Concurrent-To\":\"<urn:uuid:a5b4c019-4130-4ad4-b876-dc3eeeb4d695>\",\"WARC-IP-Address\":\"104.110.193.39\",\"WARC-Target-URI\":\"https://kr.mathworks.com/help/comm/ref/bchenc.html\",\"WARC-Payload-Digest\":\"sha1:CVCIQL4ZG4J45GIVLWWZEQN3WCNKUR3W\",\"WARC-Block-Digest\":\"sha1:VI5QGWS7ZRA34KA7OX6U2FN67YCBI2HV\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964364169.99_warc_CC-MAIN-20211209122503-20211209152503-00348.warc.gz\"}"} |
http://blog.geomblog.org/2004/03/paths-of-planets.html | [
"## Wednesday, March 17, 2004\n\n### The Paths of the Planets\n\nCalculus is a heavy hammer, and can be used to solve many problems. Sometimes, a much more elegant solution to a problem can be found using basic geometric techniques.\n\nOne such example is Kepler's observations of the nature of planetary motion. Although Kepler's observations were accurate, it was not apparent why the planets should behave the way they did. Newton proved using calculus and his laws of motion that the orbits of planets are elliptical and that they sweep out equal areas in fixed time periods. However, he also showed this using nothing but the laws of motion and plane geometry, presumably so as to convince an audience that might be skeptical of calculus.\n\nRichard Feynman rederived Newton's method and presented it a lecture that is now part of a collection called Feynman's Lost Lectures. Nigel Higson and Rachel Hall at Penn State included this as part of coursework they developed for an honors calculus class.\n\nThe technique itself can be found here. It is a beautiful use of plane geometry, one among many such examples of the power of elementary geometric reasoning (let us not forget that the first theorems were geometric proofs)"
]
| [
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.96343297,"math_prob":0.8986206,"size":1184,"snap":"2019-51-2020-05","text_gpt3_token_len":233,"char_repetition_ratio":0.10254237,"word_repetition_ratio":0.0,"special_character_ratio":0.1858108,"punctuation_ratio":0.074418604,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95639753,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-12-12T13:02:12Z\",\"WARC-Record-ID\":\"<urn:uuid:5de3e685-7d94-4582-b328-7411a9a945dc>\",\"Content-Length\":\"133732\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:63454bf7-2182-47e6-9ed7-17a0990f5d0c>\",\"WARC-Concurrent-To\":\"<urn:uuid:533467e4-d552-4f17-b19c-45160f9aaf89>\",\"WARC-IP-Address\":\"172.217.5.243\",\"WARC-Target-URI\":\"http://blog.geomblog.org/2004/03/paths-of-planets.html\",\"WARC-Payload-Digest\":\"sha1:MOEQXTLG2UKADUFPMOV6SY24SX6L67AP\",\"WARC-Block-Digest\":\"sha1:NDK6GLDNF3ASZOOZRTCCY5QGRVKV3U6M\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-51/CC-MAIN-2019-51_segments_1575540543850.90_warc_CC-MAIN-20191212130009-20191212154009-00310.warc.gz\"}"} |
https://community.khronos.org/t/display-list/16903 | [
"# Display List\n\nI want to show my vertices with GLUspheres, to make them better visible. Therefore i created a display list.\nvoid MyGLWidget::drawCircle (GLfloat x,GLfloat y,GLfloat z) {\nglPushMatrix();\nglTranslatef(x,y,z);\nglCallList(circleList);\nglPopMatrix();\n}\nThe display list is created during the init:\nvoid MyGLWidget::createCircleList (void) {\nfloat radius = 15.0f;\nint slices = 10;\nint stacks = 10;\ncircleList = glGenLists(1);\nglNewList (circleList, GL_COMPILE);\nglEndList();\n}\nFor each node i call the following code:\nfor(int i=0;i<numVertices;i++){\nglBegin (GL_POINTS);\nglVertex3f(vertices[i*3],vertices[i*3+1],vertices[i*3+2]);\nif (bigVertices) {\ndrawCircle (vertices[i*3],vertices[i*3+1],vertices[i*3+2]);\n}\nglEnd();\n}\nBut the speres are not drawen on the nodes?? What is wrong defined, or is there a better way to define visble nodes!\n\nThanks for your help\nJuergen\n\nSlight misunderstanding on how this all works! I’ll try and help here:\n\nvoid MyGLWidget::drawCircle (GLfloat x,GLfloat y,GLfloat z) {\nglPushMatrix();\nglTranslatef(x,y,z);\nglCallList(circleList);\nglPopMatrix();\n}\nvoid MyGLWidget::createCircleList (void) {\n\nfloat radius = 15.0f;\nint slices = 10;\nint stacks = 10;\ncircleList = glGenLists(1);\nglNewList (circleList, GL_COMPILE);\nglEndList();\n}\n\nFor each node:\n\nif ( bigVertices)\n{\nfor(int i=0;i<numVertices;i++)\n{\nDrawCircle(vertices[i*3],vertices [i*3+1],vertices[i*3+2]);\n}\n}\nelse\n{\nglBegin ( GL_POINTS);\nfor(int i=0;i<numVertices;i++)\n{\nglVertex3f(vertices[i*3],vertices[i*3+1],vertices[i*3+2]);\n}\nglEnd ();\n}\n\nAlso, I didn’t see you delete that quadric object after use? (memory leak?).\n\nHi Robbo!\nThanks for your reply! I made the changes like you sayed, but i can not see any spheres? Nothing is displayed at the vertex positions!\nJuergen\n\nJust out of curiosity, how exactly are your vertices being stored?\n\nI note:\n\nVertices [i*3]\nVertices [i*3+1]\nVertices [i*3+2]\n\nI am assuming your vertices are something like:\n\nfloat Vertices [SOME LARGE NUMBER];\n\nTry printing out or debugging say the first 5 or so vertex values (ie. floats 0 to 14) and paste them in here so we can have a look. They might be incorrect - .\n\nOf course, you can always change the point size of the GL_POINTS in OpenGL.\n\nI am sorry but the error was the too big radius !!\nsorry\njuergen"
]
| [
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.5849406,"math_prob":0.93791026,"size":920,"snap":"2022-40-2023-06","text_gpt3_token_len":264,"char_repetition_ratio":0.1430131,"word_repetition_ratio":0.0,"special_character_ratio":0.27065217,"punctuation_ratio":0.23913044,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9826538,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-10-06T11:04:33Z\",\"WARC-Record-ID\":\"<urn:uuid:e8f87d93-d8af-4c8a-937c-755c82738491>\",\"Content-Length\":\"26392\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:eb44189c-7171-462a-a1f2-960171c48af7>\",\"WARC-Concurrent-To\":\"<urn:uuid:2829512d-8c1b-46bc-94ac-c899d69f2b8e>\",\"WARC-IP-Address\":\"68.183.55.73\",\"WARC-Target-URI\":\"https://community.khronos.org/t/display-list/16903\",\"WARC-Payload-Digest\":\"sha1:C7QDQBGS6MUXYYO7YFJHXVQ2Z72WP6EI\",\"WARC-Block-Digest\":\"sha1:C4XN3HOT4A2FENGF5BCKZ2C4CNKT5BWO\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030337803.86_warc_CC-MAIN-20221006092601-20221006122601-00002.warc.gz\"}"} |
https://www.idealancy.pk/easy-t-shirt-organizer | [
"",
null,
"###### SHOPPING CART",
null,
"NO PRODUCTS IN THE CART.\n\n# Easy T-Shirt Organizer\n\nWarranty\n\n14 days Return/Exchange\n\nHome Delivery\n\n2 to 6 working days\n\nAvailability\n\nIn Stock\n\nRs. 4,399\n\nRs. 7,699\nRs. 8,699 -11%\nRs. 249\nRs. 399 -38%\nRs. 199\n\nRs. 3,499\nRs. 3,899 -10%\nRs. 1,099\n\nRs. 999\nRs. 1,199 -17%\nRs. 2,599\nRs. 2,999 -13%\nRs. 2,799\nRs. 2,999 -7%\nRs. 499\n\nRs. 899\n\nRs. 899\nRs. 999 -10%\nRs. 2,999\nRs. 3,399 -12%\nRs. 3,199\nRs. 3,399 -6%\nRs. 2,199\n\nRs. 149\nRs. 199 -25%\nRs. 249\n\nRs. 149\nRs. 200 -26%\nRs. 299\nRs. 349 -14%\nRs. 99\nRs. 199 -50%\nRs. 299\nRs. 399 -25%\nRs. 249\n\nRs. 249\nRs. 299 -17%\nRs. 799\n\nRs. 699\nRs. 799 -13%\nRs. 499\nRs. 799 -38%\nRs. 299\nRs. 349 -14%\nRs. 299\nRs. 399 -25%\nRs. 199\nRs. 299 -33%\nRs. 149\nRs. 199 -25%\nRs. 299\nRs. 349 -14%\nRs. 249\nRs. 299 -17%\nRs. 249\n\nRs. 499\nRs. 749 -33%\nRs. 549\nRs. 649 -15%\nRs. 249\nRs. 299 -17%\nRs. 299\nRs. 349 -14%\nRs. 149\nRs. 199 -25%\nRs. 499\nRs. 599 -17%\nRs. 149\nRs. 199 -25%\nRs. 299\nRs. 399 -25%\nRs. 399\nRs. 449 -11%\nRs. 249\nRs. 499 -50%\nRs. 349\nRs. 399 -13%\nRs. 199\nRs. 299 -33%\nRs. 249\nRs. 399 -38%\nRs. 249\nRs. 299 -17%"
]
| [
null,
"https://www.facebook.com/tr",
null,
"https://www.idealancy.pk/images/cart-cut-icon.svg",
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.9082461,"math_prob":0.9203953,"size":878,"snap":"2022-05-2022-21","text_gpt3_token_len":193,"char_repetition_ratio":0.12242563,"word_repetition_ratio":0.0,"special_character_ratio":0.20273349,"punctuation_ratio":0.09638554,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98773384,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-29T08:58:42Z\",\"WARC-Record-ID\":\"<urn:uuid:aac9596f-0fa0-4314-b04c-80e4d443d701>\",\"Content-Length\":\"471614\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a1eaffc5-7bd3-42c7-b80c-7cdcb1925778>\",\"WARC-Concurrent-To\":\"<urn:uuid:51c34c9e-c8cc-48a3-8ad0-1cca7dc94640>\",\"WARC-IP-Address\":\"172.67.138.214\",\"WARC-Target-URI\":\"https://www.idealancy.pk/easy-t-shirt-organizer\",\"WARC-Payload-Digest\":\"sha1:LKX5JMGBOLUM22GL2V5MKYSEUWNKB4GW\",\"WARC-Block-Digest\":\"sha1:RQAPXBJV7ZURNJGKRN57LDOYZ3VOY7GR\",\"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-20220529072915-20220529102915-00299.warc.gz\"}"} |
https://www.media4math.com/StandardsFinder?page=5 | [
"# Standards Finder\n\n## To help you with the dot notation, refer to the table on the right. For example, if you are a grade 3 teacher looking for algebra resources, simply input in the field below \"Content.3.OA\" and you'll see a drop-down list of possible CCSS for that grade and topic. Then select the appropriate one and press the Apply button. Alternatively, you can copy and paste the CCSS in dot notation to the field and then select it from the drop-down menu.\n\n### Topic\n\nCC Counting and Cardinality K\nOA Operations and Algebraic Thinking K-8\nNBT Numbers and Operations in Base Ten K-5\nMD Measurement and Data K-8\nG Geometry K-8\nNFA Numbers and Operations--Fractions 4-5\nRP Ratios and Proportional Relationships 6-7\nNS Number System 6-8\nEE Expressions and Equations 6-8\nSP Statistics and Probability 6-8\nF Functions 8\nRN Real Number System HS\nHAS High School Algebra HS\nHSF High School Functions HS\nHSG High School Geometry HS\nHSS High School Statistics and Probability HS\nTotal Number of Resources: 44389\nCommon Core Standard Thumbnail Title Description Curriculum Topic\n\n## CCSS.MATH.CONTENT.6.RP.A.1",
null,
"## Video Tutorial: Ratios, Video 17\n\nVideo Tutorial: Ratios: Visual Models for Ratios and Percents\n\nThis is part of a collection of video tutorial\n\nRatios and Rates\n\n## CCSS.MATH.CONTENT.6.RP.A.1",
null,
"## Video Transcript: Ratios and Rates: Rates from Data\n\nVideo Transcript: Ratios and Rates: Rates from Data\n\nThis is the transcript that goes with the video segment\n\nRatios and Rates\n\n## CCSS.MATH.CONTENT.6.RP.A.1",
null,
"## Video Transcript: Ratios: Visual Models for Ratios and Percents\n\nVideo Transcript: Ratios: Visual Models for Ratios and Percents Ratios and Rates\n\n## CCSS.MATH.CONTENT.6.RP.A.1",
null,
"## Closed Captioned Video: Ratios and Proportions: Scale Drawings\n\nClosed Captioned Video: Ratios and Proportions: Scale Drawings\n\nVideo Tutorial: Ratios and Proportions: Scale\n\nRatios and Rates\n\n## CCSS.MATH.CONTENT.6.RP.A.1",
null,
"## Definition--Ratios, Proportions, and Percents Concepts--Ratios in Simplest Form\n\nDefinition--Ratios, Proportions, and Percents Concepts--Ratios in Simplest Form\n\nThis is part of a collection\n\nApplications of Ratios, Proportions, and Percents\n\n## CCSS.MATH.CONTENT.6.RP.A.1",
null,
"## Quizlet Flash Cards: Ratios from Shapes, Set 03\n\nIn this set of interactive flash cards, find the ratios for a given set of shapes.\n\nRatios and Rates\n\n## CCSS.MATH.CONTENT.6.RP.A.1",
null,
"## Video Tutorial: Rational Numbers: Video 3\n\nVideo Tutorial: Rational Numbers: Rational Numbers on a Number Line\n\nThis is part of a collection of video tu\n\nRational Expressions\n\n## CCSS.MATH.CONTENT.6.RP.A.1",
null,
"## Math Example--Ratios and Rates--Example 11\n\nMath Example--Ratios and Rates--Example 11\n\nThis is part of a collection of math examples that focus on ratio\n\nRatios and Rates\n\n## CCSS.MATH.CONTENT.6.RP.A.1",
null,
"## Video Tutorial: Ratios, Video 2\n\nVideo Tutorial: Ratios: Ratios with Three Items\n\nThis is part of a collection of video tutorials on the topic\n\nRatios and Rates\n\n## CCSS.MATH.CONTENT.6.RP.A.1",
null,
"## Math Clip Art--Ratios, Proportions, Percents--Ratios 11\n\nMath Clip Art--Ratios, Proportions, Percents--Ratios 11\n\nThis is part of a collection of math clip art images\n\nRatios and Rates\n\n## CCSS.MATH.CONTENT.6.RP.A.1",
null,
"## Math Example--Ratios and Rates--Ratios with Fractions--Example 10\n\nMath Example--Ratios and Rates--Ratios with Fractions--Example 10\n\nThis is part of a collection of math examp\n\nRatios and Rates\n\n## CCSS.MATH.CONTENT.6.RP.A.1",
null,
"## Math Example--Ratios and Rates--Example 27\n\nMath Example--Ratios and Rates--Example 27\n\nThis is part of a collection of math examples that focus on ratio\n\nRatios and Rates\n\n## CCSS.MATH.CONTENT.6.RP.A.1",
null,
"## Video Transcript: Ratios: Application of Ratios: Roofs and Ramps\n\nVideo Transcript: Ratios: Application of Ratios: Roofs and Ramps Applications of Ratios, Proportions, and Percents\n\n## CCSS.MATH.CONTENT.6.RP.A.1",
null,
"## Math Example--Ratios and Rates--Example 2\n\nMath Example--Ratios and Rates--Example 2\n\nThis is part of a collection of math examples that focus on ratios\n\nRatios and Rates\n\n## CCSS.MATH.CONTENT.6.RP.A.1",
null,
"## Definition--Ratios, Proportions, and Percents Concepts--Ratios with Decimals\n\nDefinition--Ratios, Proportions, and Percents Concepts--Ratios with Decimals\n\nThis is part of a collection of\n\nRatios and Rates\n\n## CCSS.MATH.CONTENT.6.RP.A.1",
null,
"## INSTRUCTIONAL RESOURCE: Algebra Application: Why Are Wildfires So Dangerous?\n\nINSTRUCTIONAL RESOURCE: Algebra Application: Why Are Wildfires So Dangerous?\n\nIn this Algebra Application, st\n\nLaws of Exponents and Applications of Ratios, Proportions, and Percents\n\n## CCSS.MATH.CONTENT.6.RP.A.1",
null,
"## Definition--Ratios, Proportions, and Percents Concepts--Percent of an Unknown\n\nDefinition--Ratios, Proportions, and Percents Concepts--Percent of an Unknown\n\nThis is part of a collection o\n\nPercents\n\n## CCSS.MATH.CONTENT.6.RP.A.1",
null,
"## Video Transcript: Rational Numbers: What Are Rational Numbers?\n\nVideo Transcript: Rational Numbers: What Are Rational Numbers? Rational Expressions\n\n## CCSS.MATH.CONTENT.6.RP.A.1",
null,
"## Video Transcript: Rational Numbers: Multiplying Rational Numbers\n\nVideo Transcript: Rational Numbers: Multiplying Rational Numbers\n\nThis is the transcript that goes with the v\n\nRational Expressions\n\n## CCSS.MATH.CONTENT.6.RP.A.1",
null,
"## Video Tutorial: Ratios, Video 3\n\nVideo Tutorial: Ratios: Fractional Ratios\n\nThis is part of a collection of video tutorials on the topic of Ra\n\nRatios and Rates\n\n## CCSS.MATH.CONTENT.6.RP.A.1",
null,
"## Math Example--Ratios and Rates--Example 18\n\nMath Example--Ratios and Rates--Example 18\n\nThis is part of a collection of math examples that focus on ratio\n\nRatios and Rates\n\n## CCSS.MATH.CONTENT.6.RP.A.1",
null,
"## Definition--Ratios, Proportions, and Percents Concepts--Proportional\n\nDefinition--Ratios, Proportions, and Percents Concepts--Proportional\n\nThis is part of a collection of definit\n\nProportions\n\n## CCSS.MATH.CONTENT.6.RP.A.1",
null,
"## Video Transcript: Ratios, Proportions, and Percents: Calculating Percents\n\nVideo Transcript: Ratios, Proportions, and Percents: Calculating Percents\n\nThis is the transcript that goes w\n\nRatios and Rates\n\n## CCSS.MATH.CONTENT.6.RP.A.1",
null,
"## Video Transcript: Ratios: What Are Ratios?\n\nVideo Transcript: Ratios: What Are Ratios? Ratios and Rates\n\n## CCSS.MATH.CONTENT.6.RP.A.1",
null,
"## Closed Captioned Video: Ratios and Rates: Ratios as Decimals\n\nClosed Captioned Video: Ratios and Rates: Ratios as Decimals\n\nVideo Tutorial: Ratios and Rates: Ratios as Dec\n\nRatios and Rates\n\n## CCSS.MATH.CONTENT.6.RP.A.1",
null,
"## Video Transcript: Rational Numbers: Comparing and Ordering Rational Numbers\n\nVideo Transcript: Rational Numbers: Comparing and Ordering Rational Numbers\n\nThis is the transcript that goes\n\nRational Expressions\n\n## CCSS.MATH.CONTENT.6.RP.A.1",
null,
"## Closed Captioned Video: Ratios and Rates: Rates and Slopes of Lines\n\nClosed Captioned Video: Ratios and Rates: Rates and Slopes of Lines\n\nVideo Tutorial: Ratios and Rates: Rates\n\nRatios and Rates\n\n## CCSS.MATH.CONTENT.6.RP.A.1",
null,
"## Definition--Ratios, Proportions, and Percents Concepts--Equivalent Ratios\n\nDefinition--Ratios, Proportions, and Percents Concepts--Equivalent Ratios\n\nThis is part of a collection of de\n\nRatios and Rates\n\n## CCSS.MATH.CONTENT.6.RP.A.1",
null,
"## Definition--Ratios, Proportions, and Percents Concepts--Scale Factor\n\nDefinition--Ratios, Proportions, and Percents Concepts--Scale Factor\n\nThis is part of a collection of definit\n\nProportions\n\n## CCSS.MATH.CONTENT.6.RP.A.1",
null,
"## Quizlet Flash Cards: Ratios from Shapes, Set 06\n\nIn this set of interactive flash cards, find the ratios for a given set of shapes.\n\nRatios and Rates\n\n## CCSS.MATH.CONTENT.6.RP.A.1",
null,
"## Definition--Ratios, Proportions, and Percents Concepts--Solving Proportions\n\nDefinition--Ratios, Proportions, and Percents Concepts--Solving Proportions\n\nThis is part of a collection of\n\nProportions\n\n## CCSS.MATH.CONTENT.6.RP.A.1",
null,
"## Math Example--Ratios and Rates--Example 9\n\nMath Example--Ratios and Rates--Example 9\n\nThis is part of a collection of math examples that focus on ratios\n\nRatios and Rates\n\n## CCSS.MATH.CONTENT.6.RP.A.1",
null,
"## Video Tutorial: Rational Numbers: Video 5\n\nVideo Tutorial: Rational Numbers: Comparing and Ordering Rational Numbers\n\nThis is part of a collection of vi\n\nRational Expressions\n\n## CCSS.MATH.CONTENT.6.RP.A.1",
null,
"## Video Tutorial: Ratios, Video 4\n\nVideo Tutorial: Ratios: Ratios and Fractions\n\nThis is part of a collection of video tutorials on the topic of\n\nRatios and Rates\n\n## CCSS.MATH.CONTENT.6.RP.A.1",
null,
"## Math Clip Art--Ratios, Proportions, Percents--Ratios 01\n\nMath Clip Art--Ratios, Proportions, Percents--Ratios 01\n\nThis is part of a collection of math clip art images\n\nRatios and Rates\n\n## CCSS.MATH.CONTENT.6.RP.A.1",
null,
"## Video Tutorial: Ratios, Video 9\n\nVideo Tutorial: Ratios: Unit Rates\n\nThis is part of a collection of video tutorials on the topic of Ratios an\n\nRatios and Rates\n\n## CCSS.MATH.CONTENT.6.RP.A.1",
null,
"## Video Tutorial: Ratios, Video 12\n\nVideo Tutorial: Ratios and Proportions: Scale Drawings\n\nThis is part of a collection of video tutorials on th\n\nRatios and Rates\n\n## CCSS.MATH.CONTENT.6.RP.A.1",
null,
"## Math Example--Ratios and Rates--Example 25\n\nMath Example--Ratios and Rates--Example 25\n\nThis is part of a collection of math examples that focus on ratio\n\nRatios and Rates\n\n## CCSS.MATH.CONTENT.6.RP.A.1",
null,
"## Video Transcript: Ratios and Proportions: Scale Drawings\n\nVideo Transcript: Ratios and Proportions: Scale Drawings\n\nThis is the transcript that goes with the video seg\n\nRatios and Rates\n\n## CCSS.MATH.CONTENT.6.RP.A.1",
null,
"## Video Transcript: Ratios: Ratios with Three Items\n\nVideo Transcript: Ratios: Ratios with Three Items Ratios and Rates\n\n## CCSS.MATH.CONTENT.6.RP.A.1",
null,
"## Closed Captioned Video: Ratios: Visual Models for Ratios and Percents\n\nClosed Captioned Video: Ratios: Visual Models for Ratios and Percents Ratios and Rates\n\n## CCSS.MATH.CONTENT.6.RP.A.1",
null,
"## Closed Captioned Video: Ratios and Proportions: What Are Proportions?\n\nClosed Captioned Video: Ratios and Proportions: What Are Proportions?\n\nVideo Tutorial: Ratios and Proportions\n\nRatios and Rates\n\n## CCSS.MATH.CONTENT.6.RP.A.1",
null,
"## Definition--Ratios, Proportions, and Percents Concepts--Dimensional Analysis\n\nDefinition--Ratios, Proportions, and Percents Concepts--Dimensional Analysis\n\nThis is part of a collection of\n\nApplications of Ratios, Proportions, and Percents\n\n## CCSS.MATH.CONTENT.6.RP.A.1",
null,
"## INSTRUCTIONAL RESOURCE: Algebra Application: Why Are Receipts So Long?\n\nINSTRUCTIONAL RESOURCE: Algebra Application: Why Are Receipts So Long?\n\nIn this Algebra Application, students\n\nLaws of Exponents and Data Analysis\n\n## CCSS.MATH.CONTENT.6.RP.A.1",
null,
"## Definition--Ratios, Proportions, and Percents Concepts--Percent of a Number\n\nDefinition--Ratios, Proportions, and Percents Concepts--Percent of a Number\n\nThis is part of a collection of\n\nPercents\n\n## CCSS.MATH.CONTENT.6.RP.A.1",
null,
"## Math Example--Ratios and Rates--Example 16\n\nMath Example--Ratios and Rates--Example 16\n\nThis is part of a collection of math examples that focus on ratio\n\nRatios and Rates\n\n## CCSS.MATH.CONTENT.6.RP.A.1",
null,
"## Video Tutorial: Ratios, Video 5\n\nVideo Tutorial: Ratios: Visual Models for Ratios\n\nThis is part of a collection of video tutorials on the topi\n\nRatios and Rates\n\n## CCSS.MATH.CONTENT.6.RP.A.1",
null,
"## Math Clip Art--Ratios, Proportions, Percents--Ratios 04\n\nMath Clip Art--Ratios, Proportions, Percents--Ratios 04\n\nThis is part of a collection of math clip art images\n\nRatios and Rates\n\n## CCSS.MATH.CONTENT.6.RP.A.1",
null,
"## Math Example--Ratios and Rates--Ratios with Fractions--Example 3\n\nMath Example--Ratios and Rates--Ratios with Fractions--Example 3\n\nThis is part of a collection of math exampl\n\nRatios and Rates\n\n## CCSS.MATH.CONTENT.6.RP.A.1",
null,
"## Video Tutorial: Ratios, Video 16\n\nVideo Tutorial: Ratios, Proportions, and Percents: Calculating Percents\n\nThis is part of a collection of vide\n\nRatios and Rates"
]
| [
null,
"https://www.media4math.com/sites/default/files/styles/max_325x325/public/library_asset/images/VideoTutorial--Ratios17Thumbnail.jpg",
null,
"https://www.media4math.com/sites/default/files/styles/max_325x325/public/library_asset/images/VideoTranscripts.jpg",
null,
"https://www.media4math.com/sites/default/files/styles/max_325x325/public/library_asset/images/VideoTranscripts.jpg",
null,
"https://www.media4math.com/sites/default/files/styles/max_325x325/public/library_asset/images/VideoTutorial--Ratios12Thumbnail.jpg",
null,
"https://www.media4math.com/sites/default/files/styles/max_325x325/public/library_asset/images/Definition--RatiosProportionsPercents--RatiosInSimplestForm.png",
null,
"https://www.media4math.com/sites/default/files/styles/max_325x325/public/library_asset/images/QuizletFlashCards.jpg",
null,
"https://www.media4math.com/sites/default/files/styles/max_325x325/public/library_asset/images/VideoTutorial--RationalNumbersVideoThumbnail03.jpg",
null,
"https://www.media4math.com/sites/default/files/styles/max_325x325/public/library_asset/images/RatiosAndRates11.jpg",
null,
"https://www.media4math.com/sites/default/files/styles/max_325x325/public/library_asset/images/VideoTutorial--Ratios2Thumbnail.jpg",
null,
"https://www.media4math.com/sites/default/files/styles/max_325x325/public/library_asset/images/MathClipArt--Ratios--11.png",
null,
"https://www.media4math.com/sites/default/files/styles/max_325x325/public/library_asset/images/RatiosWithFractions--Example10.png",
null,
"https://www.media4math.com/sites/default/files/styles/max_325x325/public/library_asset/images/RatiosAndRates27.jpg",
null,
"https://www.media4math.com/sites/default/files/styles/max_325x325/public/library_asset/images/VideoTranscripts.jpg",
null,
"https://www.media4math.com/sites/default/files/styles/max_325x325/public/library_asset/images/RatiosAndRates2.jpg",
null,
"https://www.media4math.com/sites/default/files/styles/max_325x325/public/library_asset/images/Definition--RatiosProportionsPercents--RatiosWithDecimals.png",
null,
"https://www.media4math.com/sites/default/files/styles/max_325x325/public/WildfiresGrafx--Title.png",
null,
"https://www.media4math.com/sites/default/files/styles/max_325x325/public/Definition--RatiosProportionsPercents--PercentOfAnUnknown.png",
null,
"https://www.media4math.com/sites/default/files/styles/max_325x325/public/library_asset/images/VideoTranscripts.jpg",
null,
"https://www.media4math.com/sites/default/files/styles/max_325x325/public/library_asset/images/VideoTranscripts.jpg",
null,
"https://www.media4math.com/sites/default/files/styles/max_325x325/public/library_asset/images/VideoTutorial--Ratios3Thumbnail.jpg",
null,
"https://www.media4math.com/sites/default/files/styles/max_325x325/public/library_asset/images/RatiosAndRates18.jpg",
null,
"https://www.media4math.com/sites/default/files/styles/max_325x325/public/library_asset/images/Definition--RatiosProportionsPercents--Proportional.png",
null,
"https://www.media4math.com/sites/default/files/styles/max_325x325/public/library_asset/images/VideoTranscripts.jpg",
null,
"https://www.media4math.com/sites/default/files/styles/max_325x325/public/library_asset/images/VideoTranscripts.jpg",
null,
"https://www.media4math.com/sites/default/files/styles/max_325x325/public/library_asset/images/VideoTutorial--Ratios18Thumbnail.jpg",
null,
"https://www.media4math.com/sites/default/files/styles/max_325x325/public/library_asset/images/VideoTranscripts.jpg",
null,
"https://www.media4math.com/sites/default/files/styles/max_325x325/public/library_asset/images/VideoTutorial--Ratios14Thumbnail.jpg",
null,
"https://www.media4math.com/sites/default/files/styles/max_325x325/public/library_asset/images/Definition--RatiosProportionsPercents--EquivalentRatios.png",
null,
"https://www.media4math.com/sites/default/files/styles/max_325x325/public/library_asset/images/Definition--RatiosProportionsPercents--ScaleFactor.png",
null,
"https://www.media4math.com/sites/default/files/styles/max_325x325/public/library_asset/images/QuizletFlashCards.jpg",
null,
"https://www.media4math.com/sites/default/files/styles/max_325x325/public/library_asset/images/Definition--RatiosProportionsPercents--SolvingProportions.png",
null,
"https://www.media4math.com/sites/default/files/styles/max_325x325/public/library_asset/images/RatiosAndRates9.jpg",
null,
"https://www.media4math.com/sites/default/files/styles/max_325x325/public/library_asset/images/VideoTutorial--RationalNumbersVideoThumbnail05.jpg",
null,
"https://www.media4math.com/sites/default/files/styles/max_325x325/public/library_asset/images/VideoTutorial--Ratios4Thumbnail.jpg",
null,
"https://www.media4math.com/sites/default/files/styles/max_325x325/public/library_asset/images/MathClipArt--Ratios--01.png",
null,
"https://www.media4math.com/sites/default/files/styles/max_325x325/public/library_asset/images/VideoTutorial--Ratios9Thumbnail.jpg",
null,
"https://www.media4math.com/sites/default/files/styles/max_325x325/public/library_asset/images/VideoTutorial--Ratios12Thumbnail.jpg",
null,
"https://www.media4math.com/sites/default/files/styles/max_325x325/public/library_asset/images/RatiosAndRates25.jpg",
null,
"https://www.media4math.com/sites/default/files/styles/max_325x325/public/library_asset/images/VideoTranscripts.jpg",
null,
"https://www.media4math.com/sites/default/files/styles/max_325x325/public/library_asset/images/VideoTranscripts.jpg",
null,
"https://www.media4math.com/sites/default/files/styles/max_325x325/public/library_asset/images/VideoTutorial--Ratios17Thumbnail.jpg",
null,
"https://www.media4math.com/sites/default/files/styles/max_325x325/public/library_asset/images/VideoTutorial--Ratios10Thumbnail.jpg",
null,
"https://www.media4math.com/sites/default/files/styles/max_325x325/public/library_asset/images/Definition--RatiosProportionsPercents--DimensionalAnalysis.png",
null,
"https://www.media4math.com/sites/default/files/styles/max_325x325/public/TheReceiptProblem--Title.png",
null,
"https://www.media4math.com/sites/default/files/styles/max_325x325/public/Definition--RatiosProportionsPercents--PercentOfNumber.png",
null,
"https://www.media4math.com/sites/default/files/styles/max_325x325/public/library_asset/images/RatiosAndRates16.jpg",
null,
"https://www.media4math.com/sites/default/files/styles/max_325x325/public/library_asset/images/VideoTutorial--Ratios5Thumbnail.jpg",
null,
"https://www.media4math.com/sites/default/files/styles/max_325x325/public/library_asset/images/MathClipArt--Ratios--04.png",
null,
"https://www.media4math.com/sites/default/files/styles/max_325x325/public/library_asset/images/RatiosWithFractions--Example03.png",
null,
"https://www.media4math.com/sites/default/files/styles/max_325x325/public/library_asset/images/VideoTutorial--Ratios16Thumbnail.jpg",
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.6108227,"math_prob":0.48568696,"size":11346,"snap":"2023-14-2023-23","text_gpt3_token_len":3309,"char_repetition_ratio":0.2453712,"word_repetition_ratio":0.5223301,"special_character_ratio":0.2304777,"punctuation_ratio":0.21886952,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9939334,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50,51,52,53,54,55,56,57,58,59,60,61,62,63,64,65,66,67,68,69,70,71,72,73,74,75,76,77,78,79,80,81,82,83,84,85,86,87,88,89,90,91,92,93,94,95,96,97,98,99,100],"im_url_duplicate_count":[null,2,null,null,null,null,null,3,null,3,null,10,null,2,null,1,null,4,null,2,null,2,null,1,null,null,null,1,null,3,null,3,null,3,null,null,null,null,null,3,null,1,null,2,null,null,null,null,null,2,null,null,null,1,null,3,null,3,null,10,null,2,null,1,null,2,null,4,null,2,null,2,null,3,null,1,null,null,null,null,null,2,null,1,null,1,null,1,null,1,null,2,null,2,null,2,null,2,null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-05-30T08:05:16Z\",\"WARC-Record-ID\":\"<urn:uuid:c232339f-d6cb-47a4-81dc-03892876a979>\",\"Content-Length\":\"136769\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:1f34bbcd-3fac-4974-9746-c254286f640a>\",\"WARC-Concurrent-To\":\"<urn:uuid:6a694c0e-7a4c-47cc-85b9-8a1e0eec6ab2>\",\"WARC-IP-Address\":\"54.221.183.199\",\"WARC-Target-URI\":\"https://www.media4math.com/StandardsFinder?page=5\",\"WARC-Payload-Digest\":\"sha1:YG4QYJRH2TTAWU3DT7G2XZT75W2E4LUN\",\"WARC-Block-Digest\":\"sha1:VNPCOQYFTGF4SJBYCZ6KED4L4F774RUH\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224645417.33_warc_CC-MAIN-20230530063958-20230530093958-00615.warc.gz\"}"} |
http://www.interviewquestionspdf.com/2015/10/sql-server-storagesizecapacity-related.html | [
"# SQL SERVER STORAGE/SIZE/CAPACITY RELATED INTERVIEW QUESTIONS ANSWERS\n\n1). What is the fundamental unit of storage in SQL Server data files and it's size?\nAns: A page with a size of 8k\n\n2). How many (maximum) no. of columns can be created in a MS SQL Table?\nAns: Max Columns per 'nonwide' table: 1,024\nMax Columns per 'wide' table: 30,000\n\n3). What is the difference between Wide and Nonwide tables in SQL Server?\nAns: 1) Wide table can contain 30,000 columns, Non-wide table(basic table) can contain only 1024 columns.\n2) Wide Tables are considered to be denormalized tables, Non-wide tables are considered to be Normalized tables.\n3) Wide tables are used in OLAP systems, Narrow tables are used in OLTP system.\n4) Wide table is new feature in SQL Server 2008. To over come the problem of having only 1024 columns in Narrow tables.\n5) Wide tables don't work with transactional or merge replication, but Non-wide can work.\n\n4). Maximum how many rows can be in the SQL Server tables?\nAns: According to Microsoft specification:\nRows per table: Limited by available storage\n\nBut there are some cases where SQL Server will prevent you from adding more rows\n• If you have an IDENTITY column and you hit the end of the range for the data type, e.g. 255 for TINYINT, 2,147,483,647 for INT, some ungodly number that starts with a 9 - possibly the number of inches to the sun and back - for BIGINT, etc. When you try to insert the next row, you'll get error message 815 about overflowing the type.\n• If you have a heap with a non-unique index, or a clustered index that is not unique, you won't be able to store more than 2 * 2,147,483,647 unique index key combinations. When you try to insert (2 * 2,147,483,647) + 1 rows with a value of 1 in an INT column that is the only column in a clustered index, you will get error message 666 about exhausting the uniqueifier. This is because the uniqueifier (which helps SQL Server identify a row when there is no true key) is only 4 bytes, which means it can't exceed the capacity of an INT (it does use both positive and negative, unlike IDENTITY unless configure it as such, which is why you get double). Now why you would ever do this, <shrug>... but you could.\n• In the VLDB space, a database can only be 524,272 terabytes. Again a very edge case, but if you have a humongous data warehouse then obviously at some point the number of rows - depending on row size - will put you near this limit.\n5). What is the maximum size of a varchar(max) variable?\nAns: Maximum size for a varchar(max) is 2GB, or looked up a more exact figure (2^31-1, or 2147483647).\n\n6). What are the difference Between varchar(8000) and varchar(max)?\n• Varchar(8000) stores a maximum of 8000 characters. Varchar(max) stores a maximum of 2 147 483 647 characters.\n• VARCHAR(MAX) uses the normal datapages until the content actually fills 8k of data as varchar(8000). When overflow happens, data is stored as old TEXT, IMAGE and a pointer is replacing the old content.\n• Columns that are of the large object (LOB) data types ntext, text, varchar(max), nvarchar(max), varbinary(max), xml, or image cannot be specified as key columns for an index\n• VARCHAR(MAX) has some ambiguity, if the size of the cell is < 8000 chars, it will be treated as Row data. If it's greater, it will be treated as a LOB for storage purposes. You can know this by querying RBAR.\n7). How can i query my sql server to only get the size of database?\nAns: Use \"YourDatabaseName\"\nexec sp_spaceused\n\n8). What would be the LEN and DATALENGTH of NULL value in SQL Server?\nAns: Both above function will return NULL as the length of NULL.\n\n9). How much size “Null” value takes in SQL Server?\nAns:\n• If the field is fixed width storing NULL takes the same space as any other value - the width of the field.\n• If the field is variable width the NULL value takes up no space.\n10). What would be the output of the following script?\nSelect LEN('A value') --Without space at end\nSelect LEN('A value ') --With 2 space at end\nAns: Both will return 7 because LEN function not including trailing spaces in SQL Server.\n\n11). How you will find the LEN in above case?\nAns: We can use following tick\nSelect LEN('A value ' + 'x') - 1\n\n12). Difference between Len() and DataLength()?\nAns: DATALENGTH()- returns the length of the string in bytes, including trailing spaces.\nLEN()- returns the length in characters, excluding trailing spaces.\n\nFor example\nSELECT LEN('string'), LEN('string '), DATALENGTH('string'), DATALENGTH('string '),\nLEN(N'string'), LEN(N'string '), DATALENGTH(N'string'), DATALENGTH(N'string ')\n\nwill return 6, 6, 6, 9, 6, 6, 12, 18"
]
| [
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.76057696,"math_prob":0.9164867,"size":6488,"snap":"2019-43-2019-47","text_gpt3_token_len":1670,"char_repetition_ratio":0.12492289,"word_repetition_ratio":0.59839714,"special_character_ratio":0.27820593,"punctuation_ratio":0.14337966,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9669548,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-11-15T15:31:56Z\",\"WARC-Record-ID\":\"<urn:uuid:9bdf8097-0a90-4591-8764-e490b71ab4e4>\",\"Content-Length\":\"71851\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f3b4b61a-e1c8-4b22-9cda-e55448b3c567>\",\"WARC-Concurrent-To\":\"<urn:uuid:a5971fba-ce84-48f8-bac5-58ad3a31c809>\",\"WARC-IP-Address\":\"172.217.15.115\",\"WARC-Target-URI\":\"http://www.interviewquestionspdf.com/2015/10/sql-server-storagesizecapacity-related.html\",\"WARC-Payload-Digest\":\"sha1:IZ33OSATJUVQMGUQGNZRMUQOSBO7BBCE\",\"WARC-Block-Digest\":\"sha1:N5YM3FESVMXATECEEUQTI24PFRUEHLU3\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-47/CC-MAIN-2019-47_segments_1573496668682.16_warc_CC-MAIN-20191115144109-20191115172109-00543.warc.gz\"}"} |
https://wmaproperty.com/blog/8-property-investment-calculations-every-property-investors-should-learn/ | [
"• Home /\n• Investment\n• / 8 Property Investment Calculations Every Property Investors Should Learn\n\n# 8 Property Investment Calculations Every Property Investors Should Learn",
null,
"By Kim Wong / 3 years ago",
null,
"A successful property investor is not only know how to invest wisely, but good at calculating the numbers. Property investment calculations are another knowledge which every property investors must learn how to calculate the numbers. Here we share you these 8 basic property investment calculations which can aid you in making confident property investments.\n\n### 1. Per Square Foot Price (PSF)\n\nThis is the most basic calculation that most people should be aware of! Here’s how you calculate:",
null,
"### 2. Gross Potential Income / Rental Yield\n\nThis is the second most common calculation known by most property buyers. If you’re keen to learn the calculation of a rental yield or how much one can rent it out, here’s how you should answer!",
null,
"### 3. Estimated Leverage Ability\n\nThis formula calculates the total amount from your income which can be used for loan purposes.",
null,
"### 4. Debt Service Ratio (DSR)\n\nThis formula calculates your current debt. A bank usually determines your loan approval after calculating your DSR, which differs among the banks.",
null,
"### 5. Return On Investment (ROI) / Capital Investment\n\nThis refers to the returns of invested capital. It is also what most real estate flipping investors look into since this is their main goal.",
null,
"### 6. Compound Annual Growth Rate (CAGR)\n\nThis is the annual growth rate of the investment property purchased at an earlier time.",
null,
"### 7. Return On Equity / Cash-On-Cash Return (COCR)\n\nThis concept is almost the same as the ROI formula, but you can use it to gauge your rental property. The real profit is here for keeper investors.",
null,
"### 8. Gross Rent Multiplier (GRM)\n\nThis is one of the tools to decide the rent prices when purchasing a property in a nearby area.",
null,
"Via PropertyGuru\n\nEvery property investors must learn these calculations and grasp of the details which enables you to make quick, effective decisions."
]
| [
null,
"https://secure.gravatar.com/avatar/856b65e177e6f2676208365a66ebb602",
null,
"https://wmaproperty.com/blog/wp-content/uploads/2017/01/8-Property-Investment-Calculations-Every-Property-Investors-Should-Learn.jpg",
null,
"http://wmaproperty.com/blog/wp-content/uploads/2017/01/Per-Square-Foot-Price-PSF-1.png",
null,
"http://wmaproperty.com/blog/wp-content/uploads/2017/01/Gross-Potential-Income-Rental-Yield.png",
null,
"http://wmaproperty.com/blog/wp-content/uploads/2017/01/Estimated-Leverage-Ability.png",
null,
"http://wmaproperty.com/blog/wp-content/uploads/2017/01/4.-Debt-Service-Ratio-DSR.png",
null,
"http://wmaproperty.com/blog/wp-content/uploads/2017/01/5.-Return-On-Investment-ROI-Capital-Investment.png",
null,
"http://wmaproperty.com/blog/wp-content/uploads/2017/01/6.-Compound-Annual-Growth-Rate-CAGR.png",
null,
"http://wmaproperty.com/blog/wp-content/uploads/2017/01/7.-Return-On-Equity-Cash-On-Cash-Return-COCR.png",
null,
"http://wmaproperty.com/blog/wp-content/uploads/2017/01/8.-Gross-Rent-Multiplier-GRM.png",
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.91431767,"math_prob":0.9406532,"size":2414,"snap":"2019-51-2020-05","text_gpt3_token_len":501,"char_repetition_ratio":0.14232366,"word_repetition_ratio":0.03,"special_character_ratio":0.2013256,"punctuation_ratio":0.0864745,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98236877,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20],"im_url_duplicate_count":[null,null,null,7,null,10,null,10,null,10,null,10,null,10,null,10,null,10,null,10,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-19T01:51:29Z\",\"WARC-Record-ID\":\"<urn:uuid:9008ae28-ab1e-44d0-8156-088b5adbb50d>\",\"Content-Length\":\"107806\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:aa293b8f-53da-4ddb-a21c-bfd56b337083>\",\"WARC-Concurrent-To\":\"<urn:uuid:d6c53909-87a7-4c61-b36c-dc197a1ce925>\",\"WARC-IP-Address\":\"52.77.53.120\",\"WARC-Target-URI\":\"https://wmaproperty.com/blog/8-property-investment-calculations-every-property-investors-should-learn/\",\"WARC-Payload-Digest\":\"sha1:S7GPFTENI4BPNVB2QRPCRVJB2E3ABFLN\",\"WARC-Block-Digest\":\"sha1:PRZUWIRGB2YVUR4FIOIRM227JGPELCAX\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579250594101.10_warc_CC-MAIN-20200119010920-20200119034920-00536.warc.gz\"}"} |
http://skogenbrinner.com/1st-grade-math-worksheets-eog-addition-and-subtraction-two-numbers/ | [
"",
null,
"1st Grade Math Worksheets Eog Addition And Subtraction Two Numbers With 4th Place Value 0",
null,
"1st Grade Math Worksheets Eog Addition And Subtraction Two Numbers With 4th Place Value 1",
null,
"1st Grade Math Worksheets Eog Addition And Subtraction Two Numbers With 4th Place Value 2",
null,
"1st Grade Math Worksheets Eog Addition And Subtraction Two Numbers With UNIT 1 SOLVE PROBLEMS WITH ADDITION SUBTRACTION Ppt Video 3",
null,
"1st Grade Math Worksheets Eog Addition And Subtraction Two Numbers With UNIT 1 SOLVE PROBLEMS WITH ADDITION SUBTRACTION Ppt Video 4",
null,
"1st Grade Math Worksheets Eog Addition And Subtraction Two Numbers With Pearson Education 5 For All",
null,
"1st Grade Math Worksheets Eog Addition And Subtraction Two Numbers With Free For Ratio Word Problems 6",
null,
"",
null,
"",
null,
"",
null,
"",
null,
"",
null,
""
]
| [
null,
"http://skogenbrinner.com/wp-content/uploads/2019/01/1st-grade-math-worksheets-eog-addition-and-subtraction-two-numbers-with-4th-place-value-0.gif",
null,
"http://skogenbrinner.com/wp-content/uploads/2019/01/1st-grade-math-worksheets-eog-addition-and-subtraction-two-numbers-with-4th-place-value-1.gif",
null,
"http://skogenbrinner.com/wp-content/uploads/2019/01/1st-grade-math-worksheets-eog-addition-and-subtraction-two-numbers-with-4th-place-value-2.gif",
null,
"http://skogenbrinner.com/wp-content/uploads/2019/01/1st-grade-math-worksheets-eog-addition-and-subtraction-two-numbers-with-unit-1-solve-problems-with-addition-subtraction-ppt-video-3.jpg",
null,
"http://skogenbrinner.com/wp-content/uploads/2019/01/1st-grade-math-worksheets-eog-addition-and-subtraction-two-numbers-with-unit-1-solve-problems-with-addition-subtraction-ppt-video-4.jpg",
null,
"http://skogenbrinner.com/wp-content/uploads/2019/01/1st-grade-math-worksheets-eog-addition-and-subtraction-two-numbers-with-pearson-education-5-for-all.jpg",
null,
"http://skogenbrinner.com/wp-content/uploads/2019/01/1st-grade-math-worksheets-eog-addition-and-subtraction-two-numbers-with-free-for-ratio-word-problems-6.gif",
null,
"http://skogenbrinner.com/wp-content/uploads/2019/01/1st-grade-math-worksheets-eog-addition-and-subtraction-two-numbers-with-fractions-5th-khan-academy-7.png",
null,
"http://skogenbrinner.com/wp-content/uploads/2019/01/1st-grade-math-worksheets-eog-addition-and-subtraction-two-numbers-with-fractions-4th-khan-academy-8.png",
null,
"http://skogenbrinner.com/wp-content/uploads/2019/01/1st-grade-math-worksheets-eog-addition-and-subtraction-two-numbers-with-fractions-5th-khan-academy-9.png",
null,
"http://skogenbrinner.com/wp-content/uploads/2019/01/1st-grade-math-worksheets-eog-addition-and-subtraction-two-numbers-with-multiplication-word-problems-4th-10.gif",
null,
"http://skogenbrinner.com/wp-content/uploads/2019/01/1st-grade-math-worksheets-eog-addition-and-subtraction-two-numbers-with-nwea-practice-1545704-science-for-all-11.jpg",
null,
"http://0.gravatar.com/avatar/cb9c8b2e673c34a8b01dcdd0559609e4",
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.76253974,"math_prob":0.6395887,"size":1421,"snap":"2019-26-2019-30","text_gpt3_token_len":345,"char_repetition_ratio":0.21453775,"word_repetition_ratio":0.5701358,"special_character_ratio":0.1963406,"punctuation_ratio":0.0044247787,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99426347,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26],"im_url_duplicate_count":[null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-06-25T14:26:55Z\",\"WARC-Record-ID\":\"<urn:uuid:432429d6-dcfc-4759-8a15-6c56760b8f75>\",\"Content-Length\":\"55265\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:bcbb363a-47f3-439c-959c-b1325add090e>\",\"WARC-Concurrent-To\":\"<urn:uuid:d7ff75b4-d3f6-4b7b-a20e-f2fc50d499d1>\",\"WARC-IP-Address\":\"104.28.20.130\",\"WARC-Target-URI\":\"http://skogenbrinner.com/1st-grade-math-worksheets-eog-addition-and-subtraction-two-numbers/\",\"WARC-Payload-Digest\":\"sha1:66BQFLERIBSOKD7SNSOEZFKVVA6BO457\",\"WARC-Block-Digest\":\"sha1:AR7T65NSEZ2QQXBJXLYFHP5G7M4T24YP\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-26/CC-MAIN-2019-26_segments_1560627999838.27_warc_CC-MAIN-20190625132645-20190625154645-00114.warc.gz\"}"} |
http://en.verysource.com/topic/nn/ | [
"",
null,
"切换至中文 Over 1 million code package, 10 million code file free download\n• [Visual C++ (VC++)] NN_graduate.rar an expert in VC NN procedures. Very good. First running Matrixbase.dsw, again arithmetic.dsw. next nnbp.dsw, final neuralNetWork.dsw.\n• [Visual C++ (VC++)] nnsagafuzzy.rar nn sa ga fuzzy algorithms such as a summary of things, these algorithms have an initial understanding and helpful\nCategory: Windows Develop Upload User：cao6303 Size：267K\n• [C/C++] useGAtodealwithNN.rar genetic algorithm to solve arbitrary function approximation NN\n• [Matlab] nnliu1seperate0.rar Bp realize a return nn procedures, neural networks can learn from the reunification of the application"
]
| [
null,
"http://en.verysource.com/images/cn.gif",
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.58623075,"math_prob":0.8314505,"size":1535,"snap":"2022-05-2022-21","text_gpt3_token_len":436,"char_repetition_ratio":0.13520575,"word_repetition_ratio":0.01746725,"special_character_ratio":0.257329,"punctuation_ratio":0.17105263,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9591675,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-29T08:51:52Z\",\"WARC-Record-ID\":\"<urn:uuid:0fe82977-51bf-4d35-b18b-e5616a52b164>\",\"Content-Length\":\"11280\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f0b595b3-1d70-4a84-a7eb-75cf2091fa05>\",\"WARC-Concurrent-To\":\"<urn:uuid:92d2cf62-ef1d-4e28-9706-c7994738aa9e>\",\"WARC-IP-Address\":\"123.206.51.233\",\"WARC-Target-URI\":\"http://en.verysource.com/topic/nn/\",\"WARC-Payload-Digest\":\"sha1:4SXYIUM6FWH62LJX5GHDH7KL6WUMEJ7X\",\"WARC-Block-Digest\":\"sha1:7JJOVDOAUSYMPGFZKI6Q5HHK6WOIOC74\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652663048462.97_warc_CC-MAIN-20220529072915-20220529102915-00664.warc.gz\"}"} |
https://hackage.haskell.org/package/compdata-0.12/docs/Data-Comp-Multi-Generic.html | [
"compdata-0.12: Compositional Data Types\n\nCopyright (c) 2011 Patrick Bahr BSD3 Patrick Bahr experimental non-portable (GHC Extensions) None Haskell98\n\nData.Comp.Multi.Generic\n\nDescription\n\nThis module defines type generic functions and recursive schemes along the lines of the Uniplate library. All definitions are generalised versions of those in Data.Comp.Generic.\n\nSynopsis\n\n# Documentation\n\nsubterms :: forall f. HFoldable f => Term f :=> [E (Term f)] Source #\n\nThis function returns a list of all subterms of the given term. This function is similar to Uniplate's universe function.\n\nsubterms' :: forall f g. (HFoldable f, g :<: f) => Term f :=> [E (g (Term f))] Source #\n\nThis function returns a list of all subterms of the given term that are constructed from a particular functor.\n\ntransform :: forall f. HFunctor f => (Term f :-> Term f) -> Term f :-> Term f Source #\n\nThis function transforms every subterm according to the given function in a bottom-up manner. This function is similar to Uniplate's transform function.\n\ntransformM :: forall f m. (HTraversable f, Monad m) => NatM m (Term f) (Term f) -> NatM m (Term f) (Term f) Source #\n\nMonadic version of transform.\n\nquery :: HFoldable f => (Term f :=> r) -> (r -> r -> r) -> Term f :=> r Source #\n\nsubs :: HFoldable f => Term f :=> [E (Term f)] Source #\n\nsubs' :: (HFoldable f, g :<: f) => Term f :=> [E (g (Term f))] Source #\n\nsize :: HFoldable f => Cxt h f a :=> Int Source #\n\nThis function computes the generic size of the given term, i.e. the its number of subterm occurrences.\n\ndepth :: HFoldable f => Cxt h f a :=> Int Source #\n\nThis function computes the generic depth of the given term."
]
| [
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.61667025,"math_prob":0.93538195,"size":1941,"snap":"2020-45-2020-50","text_gpt3_token_len":580,"char_repetition_ratio":0.1874032,"word_repetition_ratio":0.47027028,"special_character_ratio":0.32972693,"punctuation_ratio":0.19794345,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.980727,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-11-29T19:34:51Z\",\"WARC-Record-ID\":\"<urn:uuid:9f6f218e-b921-4ebb-8a3c-fa35f7cc943b>\",\"Content-Length\":\"15588\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d49fbfa4-391c-49fe-bec6-a175c7da84c4>\",\"WARC-Concurrent-To\":\"<urn:uuid:b14720a4-de6d-491b-97fa-bd3a8e285646>\",\"WARC-IP-Address\":\"199.232.64.68\",\"WARC-Target-URI\":\"https://hackage.haskell.org/package/compdata-0.12/docs/Data-Comp-Multi-Generic.html\",\"WARC-Payload-Digest\":\"sha1:G2NSZWVFAISZ2ZNW6J4Z4HUTAEVJQ2CT\",\"WARC-Block-Digest\":\"sha1:SDEC2VJ2LRZ5V2RHDJQKZPISZU32W5IS\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141202590.44_warc_CC-MAIN-20201129184455-20201129214455-00027.warc.gz\"}"} |
https://www.aimsciences.org/article/doi/10.3934/proc.2011.2011.117 | [
"",
null,
"",
null,
"",
null,
"",
null,
"2011, 2011(Special): 117-125. doi: 10.3934/proc.2011.2011.117\n\n## Existence results to a quasilinear and singular parabolic equation\n\n 1 Laboratoire LMA, UMR CNRS 5142, Université de Pau et des Pays de l’Adour, 64013 Pau Cedex 2 LMAP (UMR 5142), Bat. IPRA, Université de Pau et des Pays de l'Adour, Avenue de l'Université, 64013 cedex Pau, France, France\n\nReceived July 2010 Revised January 2011 Published October 2011\n\nWe investigate the following quasilinear parabolic and singular equation, $u_t-\\Delta_p u &=\\frac{1}{u^\\delta}+f(t,x)\\;\\text{ in }\\,Q_T=(0,T]\\times\\Omega\\\\ u>0 \\text{ in }\\, Q_T\\; , u &=0\\,\\text{ on} \\;\\Gamma=[0,T]\\times\\partial\\Omega,\\\\ u(0,x) &=u_0(x)\\;\\text{ in }\\Omega$ where $\\Omega$ is an open bounded domain with smooth boundary in ${\\rm R}^N$, $1 < p< \\infty$ and $0<\\delta$, $T>0$, $f\\in L^\\infty(Q_T)$ and $u_0\\in L^\\infty(\\Omega)\\cap W^{1,p}_0(\\Omega)$. For any $\\delta\\in (0,2+\\frac{1}{p-1})$, $u_0$ satisfying a cone condition defined below and any $T>0$, In this paper we prove the existence and the uniqueness of a weak solution $u$ to $({\\rm P_t})$.\nCitation: Mehdi Badra, Kaushik Bal, Jacques Giacomoni. Existence results to a quasilinear and singular parabolic equation. Conference Publications, 2011, 2011 (Special) : 117-125. doi: 10.3934/proc.2011.2011.117\n Kei Matsuura, Mitsuharu Otani. Exponential attractors for a quasilinear parabolic equation. Conference Publications, 2007, 2007 (Special) : 713-720. doi: 10.3934/proc.2007.2007.713 Zongming Guo, Long Wei. A fourth order elliptic equation with a singular nonlinearity. Communications on Pure & Applied Analysis, 2014, 13 (6) : 2493-2508. doi: 10.3934/cpaa.2014.13.2493 Dung Le. Strong positivity of continuous supersolutions to parabolic equations with rough boundary data. Discrete & Continuous Dynamical Systems - A, 2015, 35 (4) : 1521-1530. doi: 10.3934/dcds.2015.35.1521 Fouad Hadj Selem, Hiroaki Kikuchi, Juncheng Wei. Existence and uniqueness of singular solution to stationary Schrödinger equation with supercritical nonlinearity. Discrete & Continuous Dynamical Systems - A, 2013, 33 (10) : 4613-4626. doi: 10.3934/dcds.2013.33.4613 Galina V. Grishina. On positive solution to a second order elliptic equation with a singular nonlinearity. Communications on Pure & Applied Analysis, 2010, 9 (5) : 1335-1343. doi: 10.3934/cpaa.2010.9.1335 Zongming Guo, Yunting Yu. Boundary value problems for a semilinear elliptic equation with singular nonlinearity. Communications on Pure & Applied Analysis, 2016, 15 (2) : 399-412. doi: 10.3934/cpaa.2016.15.399 Tomasz Cieślak, Kentarou Fujie. Global existence in the 1D quasilinear parabolic-elliptic chemotaxis system with critical nonlinearity. Discrete & Continuous Dynamical Systems - S, 2020, 13 (2) : 165-176. doi: 10.3934/dcdss.2020009 Laurence Cherfils, Stefania Gatti, Alain Miranville. A doubly nonlinear parabolic equation with a singular potential. Discrete & Continuous Dynamical Systems - S, 2011, 4 (1) : 51-66. doi: 10.3934/dcdss.2011.4.51 Shota Sato, Eiji Yanagida. Asymptotic behavior of singular solutions for a semilinear parabolic equation. Discrete & Continuous Dynamical Systems - A, 2012, 32 (11) : 4027-4043. doi: 10.3934/dcds.2012.32.4027 Haixia Li. Lifespan of solutions to a parabolic type Kirchhoff equation with time-dependent nonlinearity. Evolution Equations & Control Theory, 2020 doi: 10.3934/eect.2020088 Jong-Shenq Guo. Blow-up behavior for a quasilinear parabolic equation with nonlinear boundary condition. Discrete & Continuous Dynamical Systems - A, 2007, 18 (1) : 71-84. doi: 10.3934/dcds.2007.18.71 M. Sango. Weak solutions for a doubly degenerate quasilinear parabolic equation with random forcing. Discrete & Continuous Dynamical Systems - B, 2007, 7 (4) : 885-905. doi: 10.3934/dcdsb.2007.7.885 Bendong Lou. Self-similar solutions in a sector for a quasilinear parabolic equation. Networks & Heterogeneous Media, 2012, 7 (4) : 857-879. doi: 10.3934/nhm.2012.7.857 Yinbin Deng, Yi Li, Xiujuan Yan. Nodal solutions for a quasilinear Schrödinger equation with critical nonlinearity and non-square diffusion. Communications on Pure & Applied Analysis, 2015, 14 (6) : 2487-2508. doi: 10.3934/cpaa.2015.14.2487 Jacques Giacomoni, Tuhina Mukherjee, Konijeti Sreenadh. Existence and stabilization results for a singular parabolic equation involving the fractional Laplacian. Discrete & Continuous Dynamical Systems - S, 2019, 12 (2) : 311-337. doi: 10.3934/dcdss.2019022 Amal Attouchi, Eero Ruosteenoja. Gradient regularity for a singular parabolic equation in non-divergence form. Discrete & Continuous Dynamical Systems - A, 2020, 40 (10) : 5955-5972. doi: 10.3934/dcds.2020254 Aníbal Rodríguez-Bernal, Enrique Zuazua. Parabolic singular limit of a wave equation with localized boundary damping. Discrete & Continuous Dynamical Systems - A, 1995, 1 (3) : 303-346. doi: 10.3934/dcds.1995.1.303 Shota Sato, Eiji Yanagida. Singular backward self-similar solutions of a semilinear parabolic equation. Discrete & Continuous Dynamical Systems - S, 2011, 4 (4) : 897-906. doi: 10.3934/dcdss.2011.4.897 Xiumei Deng, Jun Zhou. Global existence and blow-up of solutions to a semilinear heat equation with singular potential and logarithmic nonlinearity. Communications on Pure & Applied Analysis, 2020, 19 (2) : 923-939. doi: 10.3934/cpaa.2020042 Liping Wang. Arbitrarily many solutions for an elliptic Neumann problem with sub- or supercritical nonlinearity. Communications on Pure & Applied Analysis, 2010, 9 (3) : 761-778. doi: 10.3934/cpaa.2010.9.761\n\nImpact Factor:"
]
| [
null,
"https://www.aimsciences.org:443/style/web/images/white_google.png",
null,
"https://www.aimsciences.org:443/style/web/images/white_facebook.png",
null,
"https://www.aimsciences.org:443/style/web/images/white_twitter.png",
null,
"https://www.aimsciences.org:443/style/web/images/white_linkedin.png",
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.57430744,"math_prob":0.8726858,"size":4974,"snap":"2020-45-2020-50","text_gpt3_token_len":1655,"char_repetition_ratio":0.16780683,"word_repetition_ratio":0.12912482,"special_character_ratio":0.3522316,"punctuation_ratio":0.24239351,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97432256,"pos_list":[0,1,2,3,4,5,6,7,8],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-10-30T04:29:01Z\",\"WARC-Record-ID\":\"<urn:uuid:8c4c4bc2-f655-4964-97a3-69e99f7ae6db>\",\"Content-Length\":\"59661\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9079cc58-7fa3-45e8-a389-d17a31684866>\",\"WARC-Concurrent-To\":\"<urn:uuid:293fffba-1082-45c7-89e6-8cf4acd3b4bd>\",\"WARC-IP-Address\":\"107.161.80.18\",\"WARC-Target-URI\":\"https://www.aimsciences.org/article/doi/10.3934/proc.2011.2011.117\",\"WARC-Payload-Digest\":\"sha1:WMJNYTH5CFXUU77EZL5RCAFSBXOUEMR4\",\"WARC-Block-Digest\":\"sha1:CYLL47ZBVNDEA2TUSCALZOCJEIM5GLYV\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-45/CC-MAIN-2020-45_segments_1603107907213.64_warc_CC-MAIN-20201030033658-20201030063658-00126.warc.gz\"}"} |
https://goprep.co/how-many-grams-of-common-salt-are-dissolved-in-this-500-ml-i-1nklr7 | [
"Q. 5 A5.0( 1 Vote )\n\n# 500 mL of a 1M solution of common salt is taken.How many grams of common salt are dissolved in this?\n\nGiven: Volume of solution of NaCl (common salt) = 500mL\n\n= 0.5L\n\nMolarity of solution = 1M",
null,
"We can write,\n\nNumber of moles = Molarity × volume in litres\n\nNumber of moles = 1M × 0.5L\n\nNumber of moles = 0.5\n\nNow, to find the mass in grams, apply the formula given:",
null,
"Gram molecular mass of NaCl = GMM of Na + GMM of Cl\n\n= 23g + 35.5g\n\n= 58.5g\n\nTherefore,",
null,
"Mass of NaCl in grams = 0.5 × 58.5g\n\nMass of NaCl in grams = 29.25g\n\nThus, 29.25 grams of common salt are dissolved.\n\nNote: If one litre of the solution contains 1mol of solute, it is called 1M (molar) solution.\n\nRate this question :\n\nHow useful is this solution?\nWe strive to provide quality solutions. Please rate us to serve you better.\nRelated Videos",
null,
"",
null,
"Basic Concept of Trigonometry43 mins",
null,
"",
null,
"Concept of Quadratic equation47 mins",
null,
"",
null,
"Master The Concept of Voice39 mins",
null,
"",
null,
"The Concept of Reflex Action46 mins",
null,
"",
null,
"Foundation | Concept of Lenz's Law44 mins",
null,
"",
null,
"Master The Concept of Voice39 mins",
null,
"",
null,
"Understand The Concept of Quadratic Equation45 mins",
null,
"",
null,
"Strengthen The Concept of Magnetic Field42 mins",
null,
"",
null,
"Learn the Concept of Past Tense121 mins",
null,
"",
null,
"NCERT | Concept Based Questions on Triangles31 mins\nTry our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts\nDedicated counsellor for each student\n24X7 Doubt Resolution\nDaily Report Card\nDetailed Performance Evaluation",
null,
"view all courses",
null,
""
]
| [
null,
"https://gradeup-question-images.grdp.co/liveData/PROJ18640/1530854870990939.png",
null,
"https://gradeup-question-images.grdp.co/liveData/PROJ18640/1530854871761595.png",
null,
"https://gradeup-question-images.grdp.co/liveData/PROJ18640/1530854872470884.png",
null,
"https://grdp.co/cdn-cgi/image/quality=90,fit=contain,width=350,f=auto/https://gs-post-images.grdp.co/2020/10/5eff50265f090602a5dd654ddf593a3cd18934a11b34ea264fdb2ed6eae9e5c0poster-high-webp.png",
null,
"https://gs-post-images.grdp.co/2020/8/group-2x-img1598864374422-78.png-rs-high-webp.png",
null,
"https://grdp.co/cdn-cgi/image/quality=90,fit=contain,width=350,f=auto/https://gs-post-images.grdp.co/2020/10/baf6f5fed7a3270a1862e4e987cbd8fe8ebed1783094d79cf66e3a4fc36217d9poster-high-webp.png",
null,
"https://gs-post-images.grdp.co/2020/8/group-2x-img1598864374422-78.png-rs-high-webp.png",
null,
"https://grdp.co/cdn-cgi/image/quality=90,fit=contain,width=350,f=auto/https://gs-post-images.grdp.co/2020/10/1ed4e086f6d3efdaa663524bf37408463ee130197a71406db6943861bc6ae510poster-high-webp.png",
null,
"https://gs-post-images.grdp.co/2020/8/group-2x-img1598864374422-78.png-rs-high-webp.png",
null,
"https://grdp.co/cdn-cgi/image/quality=90,fit=contain,width=350,f=auto/https://gs-post-images.grdp.co/2020/10/71a150da6b7f150ea4a8ea7021de26ec8753cc7587d9358d499abb5d2747df1eposter-high-webp.png",
null,
"https://gs-post-images.grdp.co/2020/8/group-2x-img1598864374422-78.png-rs-high-webp.png",
null,
"https://grdp.co/cdn-cgi/image/quality=90,fit=contain,width=350,f=auto/https://gs-post-images.grdp.co/2020/10/bd9bd1dd4e4b4dee0b4300c8524f97f8559a530cac2c7e4270795b9b7af1f32bposter-high-webp.png",
null,
"https://gs-post-images.grdp.co/2020/8/group-2x-img1598864374422-78.png-rs-high-webp.png",
null,
"https://grdp.co/cdn-cgi/image/quality=90,fit=contain,width=350,f=auto/https://gs-post-images.grdp.co/2020/11/f340288c4a9443c51c9be911bc10d0ab2e2158d61c3bb5d48c4e2ae00f866b88poster-high-webp.png",
null,
"https://gs-post-images.grdp.co/2020/8/group-2x-img1598864374422-78.png-rs-high-webp.png",
null,
"https://grdp.co/cdn-cgi/image/quality=90,fit=contain,width=350,f=auto/https://gs-post-images.grdp.co/2020/10/0bc2245e9f0129df6c70396c77ed5483955f6a77be1b1c4446095990445562f1poster-high-webp.png",
null,
"https://gs-post-images.grdp.co/2020/8/group-2x-img1598864374422-78.png-rs-high-webp.png",
null,
"https://grdp.co/cdn-cgi/image/quality=90,fit=contain,width=350,f=auto/https://gs-post-images.grdp.co/2020/10/8ae6773be1f73f0e0fc95f0d96eed5eb179afc1ab8f23d0322e4c3e3759ee88fposter-high-webp.png",
null,
"https://gs-post-images.grdp.co/2020/8/group-2x-img1598864374422-78.png-rs-high-webp.png",
null,
"https://grdp.co/cdn-cgi/image/quality=90,fit=contain,width=350,f=auto/https://gs-post-images.grdp.co/2020/10/9a0fa84420d662a826a109fc5c9753f6ee94253bc8a0999238b576d6036500f6poster-high-webp.png",
null,
"https://gs-post-images.grdp.co/2020/8/group-2x-img1598864374422-78.png-rs-high-webp.png",
null,
"https://grdp.co/cdn-cgi/image/quality=90,fit=contain,width=350,f=auto/https://gs-post-images.grdp.co/2020/11/dbf5b9cafc2712d40b1800c10581f5adb5478888caa623c10e4c386ec71943a2poster-high-webp.png",
null,
"https://gs-post-images.grdp.co/2020/8/group-2x-img1598864374422-78.png-rs-high-webp.png",
null,
"https://grdp.co/cdn-cgi/image/height=128,quality=80,f=auto/https://gs-post-images.grdp.co/2020/8/group-7-3x-img1597928525711-15.png-rs-high-webp.png",
null,
"https://gs-post-images.grdp.co/2020/8/group-img1597139979159-33.png-rs-high-webp.png",
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.8451849,"math_prob":0.95847094,"size":1520,"snap":"2020-45-2020-50","text_gpt3_token_len":513,"char_repetition_ratio":0.1292876,"word_repetition_ratio":0.2672956,"special_character_ratio":0.33552632,"punctuation_ratio":0.14640884,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9912546,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50],"im_url_duplicate_count":[null,1,null,1,null,1,null,null,null,null,null,6,null,null,null,2,null,null,null,2,null,null,null,3,null,null,null,2,null,null,null,null,null,null,null,1,null,null,null,2,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-11-28T08:08:57Z\",\"WARC-Record-ID\":\"<urn:uuid:bbd02133-ad06-4edc-bce0-6cebc65e9804>\",\"Content-Length\":\"184378\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:90b96919-0ce1-4169-8f74-3dfb6ec4cb50>\",\"WARC-Concurrent-To\":\"<urn:uuid:81d534f3-1151-463c-a60f-0999db21936c>\",\"WARC-IP-Address\":\"104.18.24.35\",\"WARC-Target-URI\":\"https://goprep.co/how-many-grams-of-common-salt-are-dissolved-in-this-500-ml-i-1nklr7\",\"WARC-Payload-Digest\":\"sha1:DXMQKZXW75WJERSNK52SM2ESLLNKFZSE\",\"WARC-Block-Digest\":\"sha1:NXRRZABQYOCVEC2EGQ54VPNTNPBLLVWH\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141195198.31_warc_CC-MAIN-20201128070431-20201128100431-00704.warc.gz\"}"} |
https://socratic.org/questions/how-do-you-solve-the-quadratic-equation-3x-9-2-12-by-the-square-root-property | [
"# How do you solve the quadratic equation (3x - 9)^2 = 12 by the square root property?\n\nAug 31, 2015\n\n$x = 3 \\pm \\frac{2 \\sqrt{3}}{3}$\n\n#### Explanation:\n\nThe square root property tells you that if ${x}^{2}$ is equal to a positive number $n$, then you have\n\n$\\textcolor{b l u e}{x = \\pm \\sqrt{n}}$\n\nYou can use $3$ as a common factor to rewrite the expression that's being squared like this\n\n${\\left[3 \\left(x - 3\\right)\\right]}^{2} = {3}^{2} \\cdot {\\left(x - 3\\right)}^{2} = 9 \\cdot {\\left(x - 3\\right)}^{2}$\n\nThe equation can thus be written as\n\n$\\frac{\\textcolor{red}{\\cancel{\\textcolor{b l a c k}{9}}} \\cdot {\\left(x - 3\\right)}^{2}}{\\textcolor{red}{\\cancel{\\textcolor{b l a c k}{9}}}} = \\frac{12}{9}$\n\n${\\left(x - 3\\right)}^{2} = \\frac{4}{3}$\n\nThe square root property tells you that\n\n$x - 3 = \\pm \\sqrt{\\frac{4}{3}}$\n\n$x - 3 = \\pm \\frac{2}{\\sqrt{3}} = \\pm \\frac{2 \\sqrt{3}}{3}$\n\nThis means that you get\n\n$x = 3 \\pm \\frac{2 \\sqrt{3}}{3}$\n\nThe two solutions to the equation will be\n\n${x}_{1} = 3 + \\frac{2 \\sqrt{3}}{3} \\text{ }$ and $\\text{ } {x}_{2} = 3 - \\frac{2 \\sqrt{3}}{3}$"
]
| [
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.7096236,"math_prob":1.0000095,"size":782,"snap":"2019-43-2019-47","text_gpt3_token_len":288,"char_repetition_ratio":0.12596402,"word_repetition_ratio":0.061068702,"special_character_ratio":0.42966753,"punctuation_ratio":0.025974026,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.0000015,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-11-15T04:28:17Z\",\"WARC-Record-ID\":\"<urn:uuid:2fa8ca4c-c39e-4bd8-853b-74869b26453a>\",\"Content-Length\":\"34741\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4e7b3518-cbac-4d6c-baa5-82fcca27a252>\",\"WARC-Concurrent-To\":\"<urn:uuid:c980ffad-6ef4-4c73-b8d6-3133cf6a66ea>\",\"WARC-IP-Address\":\"54.221.217.175\",\"WARC-Target-URI\":\"https://socratic.org/questions/how-do-you-solve-the-quadratic-equation-3x-9-2-12-by-the-square-root-property\",\"WARC-Payload-Digest\":\"sha1:RW3XI3W4GYU7BSYAGQJLJEMJ2H3RNE7T\",\"WARC-Block-Digest\":\"sha1:QSN3FYDPOBTFYKZ7THUKXVY6BTM4KOVM\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-47/CC-MAIN-2019-47_segments_1573496668585.12_warc_CC-MAIN-20191115042541-20191115070541-00539.warc.gz\"}"} |
https://www.askdrcallahan.com/category/1-algebra/chapter-4/ | [
"Posted on\n\n## Algebra 1 Ch4.4 #6\n\nQuestion from Kai:\n\nJacobs Algebra 1, Chapter 4, Lesson 4 (pg 167), problem 6 (c,d,e)\nI don’t understand why the answers are negative.\n\nThank You\n\nBecause in each case you have a minus sign in front of the fraction.\n\nso – 1/2 is the same as -0.5.\n\nIf you put this in the calculator, you need to include the minus sign on the numerator side. So I would put in these keystrokes for -1/2\n\n1\n\n/\n\n2\n\n=\n\nHope this helps,\n\ndwc"
]
| [
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.8938121,"math_prob":0.7163618,"size":436,"snap":"2021-43-2021-49","text_gpt3_token_len":128,"char_repetition_ratio":0.106481485,"word_repetition_ratio":0.0,"special_character_ratio":0.29587156,"punctuation_ratio":0.14423077,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9563089,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-12-03T11:10:49Z\",\"WARC-Record-ID\":\"<urn:uuid:9f9bfd42-035d-4152-8dfa-1d3804164442>\",\"Content-Length\":\"77852\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:6e125b09-dfcc-49b5-804e-3e8f69349d52>\",\"WARC-Concurrent-To\":\"<urn:uuid:b09e9929-3259-442f-8e15-75f6df6eb326>\",\"WARC-IP-Address\":\"104.21.91.194\",\"WARC-Target-URI\":\"https://www.askdrcallahan.com/category/1-algebra/chapter-4/\",\"WARC-Payload-Digest\":\"sha1:RWIKPVAUCNVKFZWNNH2IEVW5Q2WALR4W\",\"WARC-Block-Digest\":\"sha1:36Y5LRVYNI4PNXFCDLZFDG6APY7VYEEO\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964362619.23_warc_CC-MAIN-20211203091120-20211203121120-00037.warc.gz\"}"} |
https://www.bsraya.com/blog/storing-data | [
"Series\n\n# Storing Data\n\nPutting some turbo boost into our gradient descent code\n\n• Bijon Setyawan Raya\n\n• July 18, 2022\n\n7 mins",
null,
"`gd_with_df()` is the function that I originally ran to show my experimentations. However, the time required to run the code is too long whenever the number of iteration increases.\n\nIn this post, I will be presenting three means of storing data into dataframes.\n\n``````def gd_with_df(x, y, epochs, df, alpha = 0.01):\nintercept, coefficient = 2.0, -7.5\npredictions = predict(intercept, coefficient, x)\nsum_error = np.sum((predictions - y) ** 2) / (2 * len(x))\ndf.loc = [intercept, coefficient, sum_error]\nfor epoch in range(1, epochs + 1):\npredictions = predict(intercept, coefficient, x)\nb0_error = (1/len(x)) * np.sum(predictions - y)\nb1_error = (1/len(x)) * np.sum((predictions - y) * x)\nintercept = intercept - alpha * b0_error\ncoefficient = coefficient - alpha * b1_error\nsum_error = np.sum((predictions - y) ** 2) / (2 * len(x))\ndf.loc[epoch] = [intercept, coefficient, sum_error]\nsum_error = 0\nreturn df\n``````\n\n## List#\n\n``````def gd_with_list(x, y, epochs, df, alpha = 0.01):\nintercepts = list()\ncoefficients = list()\nsum_errors = list()\n\nintercept, coefficient = 2.0, -7.5\npredictions = predict(intercept, coefficient, x)\nsum_error = np.sum((predictions - y) ** 2) / (2 * len(x))\n\nintercepts.append(intercept)\ncoefficients.append(coefficient)\nsum_errors.append(sum_error)\n\nfor epoch in range(1, epochs+1):\npredictions = predict(intercept, coefficient, x)\nb0_error = (1/len(x)) * np.sum(predictions - y)\nb1_error = (1/len(x)) * np.sum((predictions - y) * x)\nintercept = intercept - alpha * b0_error\ncoefficient = coefficient - alpha * b1_error\nsum_error = np.sum((predictions - y) ** 2) / (2 * len(x))\n\nintercepts.append(intercept)\ncoefficients.append(coefficient)\nsum_errors.append(sum_error)\n\nsum_error = 0\n\ndf['intercept'] = intercepts\ndf['coefficient'] = coefficients\ndf['sum_error'] = sum_errors\n\nreturn df\n``````\n\n## Dictionary#\n\n``````def gd_with_dict(x, y, epochs, df, alpha = 0.01):\nintercept, coefficient = 2.0, -7.5\npredictions = predict(intercept, coefficient, x)\nsum_error = np.sum((predictions - y) ** 2) / (2 * len(x))\n\nresult = {\n'intercept': [intercept],\n'coefficient': [coefficient],\n'sum_error': [sum_error]\n}\n\nfor epoch in range(1, epochs+1):\npredictions = predict(intercept, coefficient, x)\nb0_error = (1/len(x)) * np.sum(predictions - y)\nb1_error = (1/len(x)) * np.sum((predictions - y) * x)\nintercept = intercept - alpha * b0_error\ncoefficient = coefficient - alpha * b1_error\nsum_error = np.sum((predictions - y) ** 2) / (2 * len(x))\n\nresult['intercept'].append(intercept)\nresult['coefficient'].append(coefficient)\nresult['sum_error'].append(sum_error)\n\nsum_error = 0\n\n# convert dict to dataframe\ndf = pd.DataFrame(result)\n\nreturn df\n``````\n\n## Observation#\n\nIterationsDataframe (mins)List (mins)Dictionary (mins)\n10000.03030.00140.0013\n100000.31230.01140.0112\n1000005.76110.10450.1064\n1000000I can take a nap1.10751.1141\n10000000\\$&!^@&#@(10.98710.521\n\nSumming the numbers up into a single picture, we can see that working with both lists and dictionaries is much faster than working with dataframes."
]
| [
null,
"https://images.unsplash.com/photo-1526374965328-7f61d4dc18c5",
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.5986814,"math_prob":0.9937609,"size":3200,"snap":"2023-40-2023-50","text_gpt3_token_len":974,"char_repetition_ratio":0.24155194,"word_repetition_ratio":0.49095023,"special_character_ratio":0.34125,"punctuation_ratio":0.1845878,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9999931,"pos_list":[0,1,2],"im_url_duplicate_count":[null,6,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-11T18:35:17Z\",\"WARC-Record-ID\":\"<urn:uuid:2576c327-97fc-4716-a0b6-bba3783cacee>\",\"Content-Length\":\"170710\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:75502373-f3ee-417c-a190-6fe9ca9f6539>\",\"WARC-Concurrent-To\":\"<urn:uuid:24a6cb31-d50f-46de-96c7-5359888b115f>\",\"WARC-IP-Address\":\"76.76.21.22\",\"WARC-Target-URI\":\"https://www.bsraya.com/blog/storing-data\",\"WARC-Payload-Digest\":\"sha1:ASSM7UY5MROGKNLIHHWROW4ZLCNOXBHQ\",\"WARC-Block-Digest\":\"sha1:CN6OJFVTNIUKZIAI55GLTO25S2M7OPXV\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679516047.98_warc_CC-MAIN-20231211174901-20231211204901-00265.warc.gz\"}"} |
https://studyres.com/doc/24454466/thermal-noise-of-mos-transistors | [
"Survey\n\n* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project\n\nDocument related concepts\n\nImmunity-aware programming wikipedia , lookup\n\nSwitched-mode power supply wikipedia , lookup\n\nCurrent source wikipedia , lookup\n\nOhm's law wikipedia , lookup\n\nStray voltage wikipedia , lookup\n\nSound level meter wikipedia , lookup\n\nAlternating current wikipedia , lookup\n\nBuck converter wikipedia , lookup\n\nVoltage optimisation wikipedia , lookup\n\nSemiconductor device wikipedia , lookup\n\nOpto-isolator wikipedia , lookup\n\nResistive opto-isolator wikipedia , lookup\n\nThermal copper pillar bump wikipedia , lookup\n\nMains electricity wikipedia , lookup\n\nDither wikipedia , lookup\n\nHistory of the transistor wikipedia , lookup\n\nThermal runaway wikipedia , lookup\n\nTransistor wikipedia , lookup\n\nWhite noise wikipedia , lookup\n\nTranscript\n```R 642\nPhilips Res. f-epts 22, 505-514, 1967\nTHERMAL NOISE OF MOS TRANSISTORS\nby F. M. KLAASSEN and J. PRINS\nAbstract\nA theory is given for the thermal noise of MOS transistors which takes\ninto account the influence of substrate doping. It is shown that the noise\nresistance is a simple function of the transconductance, the effective\ngate voltage and the pinch-off voltage. In practical transistors the value\nof the noise resistance may be a factor of 10 higher than the value\npredicted by the simple theory (zero substrate dope). Measurements are\ndiscussed that support the theory.\n1. Introduetion\nThe influence of the substrate doping upon the DC characteristics of silicon\nMOS transistors has recently been the subject of several theoretical and experimental investigations. Within the scope of this paper we report the results for\nthe transconductance obtained by Sah and Pao 1) 'and independently by Van\n, Nielen and Memelink 2). They have shown that only in the saturation range\ndoes the substrate doping have a marked effect upon the value of the transconductance gmO' In this case gmO is no longer proportional to the gate-source\nvoltage Vg, as it is for a MOS transistor with zero substrate doping, but is\nproportional to a certain voltage Vp ~ Vg, called pinch-off voltage, which is\nthe channel potential for which the channel charge-at the drain becomes zero.\nSo\nwhere the proportion constant f3 has been determined by the gate capacitance\nand the channel geometry. The value of Vp is decreased by increasing the doping\nor the oxide thickness (cf. ref. 2).\nBesides on the DC characteristic the substrate doping also has some influence\nupon the thermal noise of the channel. Sah et al. 4) have analysed the noise\nfor small values of doping, but the mathematics was rather involved and\nrestricted to very small or very large drain voltages. However, in the case of\nheavy doping it is not so difficult to 'understand the influence upon the noise.\nTo begin with we expect that, analogous to a normal resistor, the noise will\nbe proportional to the total integrated current charge in the channel. When the\nsubstrate doping is large the potential in the channel everywhere is much smaller\nthan the gate-source voltage, because of the space charge in the substrate\ndepletion layer; so we expect that this integral will be nearly proportional to\nthe voltage Vg. Thus we obtain the simple result\n506\nF. M. KLÀASSEN and J. PRINS\nwhere Vp and gmO are dependent on the substrate doping and (3 is a proportion\nconstant for gmO' This expression differs by a factor of Vg/Vp from the result\nvalid for a MOS transistor without substrate 3). For moderate values of doping\nthe integral is not so simple and a careful analysis is necessary.\nIn this paper a closed-form expression for the noise is given without restrictions for the drain voltage, and the results of measurements are' reported for\nMOS transistors with substrate doping in the range from 1014 to 1017 cmr \".\n2. Theoretical analysis\nWhen we assume a constant surface-charge density CoxVT, which is present\nirrespective of the potential difference Vg __:_Vex), the mobile-charge density in\nthe n-type channel of a MOS transistor will be given by 2) (see fig. 1)\nam(x) = -Cox [Vg -\nVT\n-\n+ a[V(x) + Va]1/2;\nVex)]\n.\ncapacitance of the oxide layer per unit area,\napplied gate potential,\nVg\noffset gate potential,\nVT\npotential\nin the channel at location x,\nVeX)\ntotal\ninduced\nsurface charge,\nCox[Vg - VT- V(x)]:\ndepletion\ncharge\nof the p-type substrate,\na[V(x) + VaJl/2\nCox\nOl:\n8\n(2 eqNA)1/2,\ndielectric constant of silicon,\nelectron charge,\nacceptor density of the substrate,\ndiffusion potential between channel and bulk.\n+\\iI\n+Var\n-.x\n_ Subsfraie'\nL = contact\nFig. 1. Schematic representation\n,\nI\n\"I\nof an n-channel MOS transistor.\n(1)\nTHERMAL\nNOISE OF MOS TRANSISTORS\n507\nThe channel current I at position x and time t will be\nI(x,t)\ndV\n-p,WGm(x) .\ndx\n=\n+ h(x,t),\n(2)\nwhere W is the width of the channel, p, 'the mobility of electrons and h(x,t) a\nrandom noise-current source at location x.\nThe preceding equation is used as a starting point for the derivation of the\nnoise spectrum. Assuming that the fluctuations of all quantities appearing in\n(2) are small, we may split up all time-dependent quantities into a large steadystate value and a superimposed small time-dependent signal\nI(x,t)\n= 10\nV(x,t)\n=\n+ i(x,t),\nVo(x)\n+ v(x,t),\n(3)\ndG\nGm(x,t) = Gm(Vo)\n+ --\ndVo\nv(x,t).\nBy making use of the relation (2) we obtain for the steady-state part 10:\n(4)\nAssuming that Vo(x) is a monotonically increasing function of x! relation (4)\nis a transformation formula between the variables x and Voo Moreover, this\nrelation yields an expression for the current 10 if one integrates from the source\nto the drain:\n(5)\nwhere L is the source-drain distance and the subscript d refers to the drain.\nSubstituting (3) into (2) we obtain for the fluctuation\nd\ni(x,t) = -p,W\n.\n- [G,nCVo)v(x,t)]\ndx\n.\n+ h(x,t),\n(6)\nif we neglect second-order terms.\nUsually the noise current of the device is measured when the output is shortcircuited for a.c. voltages. Thus for the short-circuit current noise of the channel\nwe have\nL\nL\nJ i(x,t)dx = J h(x,t)d~.\no'\n.\n0\n.\n(7)\n, 508\nF. M. KLAASSEN\nand J. PRINS\nSince we are more interested in the spectral density of the noise current we write\ni(x,t) ~ ~\n+ jOOl,(x) + ...l exp (joot),)\n(8)\n= hex) exp (jwt),\nh(x,t)\nin which we assume that the low-frequency spectrum io is independent of x,\nif the spectrum of the source hex) is white in the frequency range to be considered. Proof of this assumption is beyond the scope of this paper and will\nbe given elsewhere 5).\nConsequently by equating terms with equal power in w we obtain from (7)\nfor the spectral density of the short-circuit noise current at the drain\nL\n<fa. fa*)\nL\n= L-2 J dx J < h(x).\no\nh*(x'»\ndx'.\n(9)\n0\nIf we restrict ourselves to the frequency range in which thermal noise is the\ndominating noise' source we have\n<hex) . h*(x'»\n= 4kTf-t WO'III(x)5(x -\nx'),\n(IQ)\nwhere 5 is the Dirac delta function.\nSubstituting this expression into (9) and using expression (4) in order to\neliminate the variable x we obtain\nVOd\nf 0'1II2(Vo)dVo\n_ _\n4kTp,W 0\n<t«. io*) =\n-----L\nVOd\nf O'III(Vo)dVo\no\nThe integrals appearing\nis a simple function of\nUsually the channel\nresistance at the input,\n(11)\nin (11) can be computed immediately, because O'm(VO)\nthe stationary voltage Vo.\nnoise is transferred. to the thermal noise of a certain\ncalled the noise-equivalent resistance Rn:\n.(12)\nwhere\n(13)\nElaborating the integrals in (11) and using (1), (1.2)and (13) we finally obtain\n(14)\nTHERMAL\nNOISE OF MOS TRANSISTORS\nin which the dimensionless factor\na\n509\nis given by\na=\nVg'2 -!(a/Cox) Vg' VOd'l/2 + t,(a2/Cox2 -2V9')Vod'\nwhere\na\n= (2 eqNA)1/2,\nVg'\nVo/\n=\n=\n+ ~(a/CoJVo/3/2 + tVOd'2\n(15)\nVg- VT + Va),\nVOd + Va.\nAfterwards we shall see that his factor varies between tand 10 in practical\nsituations. Because the expression for a is not very clear some special cases\nwill be considered:\n(a) Transistor operating far below saturation. In this case we have\nand\na\nV'9 ././\n\" -C\nV.Od '3/2 ,\nox\nfrom which assumption we obtain\n.,\n(16)\nI\n(0) Transistor\nwith low substrate\ndoping. In this case we may neglect all terms\nwith the factor a, which results in\nV'2\n9\nV.Od 'V'9\n+ ~v. '2\n\"3\"\nOd\n(17)\n•If this transistor also operates in saturation we have VOd' ~ Vg' and we\nobtain the well-known result 3)\n(18)\n(c) Of special interest is the general case when the transistor operates in satu-\niation. Following Sah 1) and Van Nielen 2) we then have\n.......,.•, and\nI.\n510\nF •.M. KLAASSEN\nand J PRINS\nConsequently R; may be approximated by\nR\nR:!\nn\n1__ V.'\n0\n( 2 Vp'\n+_1)\n6\ng-l\n(19)\nm\nin which\nand\nThis is indeed a closed-form expression for a MOS transistor operating\nin saturation and it is valid for all possible values of substrate doping and\noxide thickness. As a general result we may conclude that the noise-equiv- _\nalent resistance is a monotonically increasing function of the substrate dope.\nand the oxide thickness, because the pinch-off voltage is a monotonically\ndecreasing function of the same parameters 2) (cf. also figs 4a and b).\nBesides the noise of the channel current it is also possible to compute the f\nnoise that is induced in the gate circuit at higher frequencies. However, since'\nwe have found that the substrate has little influence upon the gate noise, this ~\nsubject will be left out of consideration here.\nIn the next section the expressions for the noise resistance are compared with.\nthe results of experiments. .\n~\n3. Comparison with experimental results\nAlthough silicon MOS transistors also produce flicker noise, it is still possible, .\napplying a proper technology and choosing suitable dimensions, to make tran- ;\nsistors in which thermal noise predominates in a wide frequency range :\n(f> 10 kc/s). In fig. 2 we have illustrated this by plotting the power spectrum .~\nof the channel noise as a function of frequency for various drain currents. At. ;\nall times the transistor operated in saturation .\n. In order to test our equation (19) we first consider the thermal noise of several\nMOS transistors with different substrate doping (O·08-30Qcm material) and\nthe same oxide layer. In fig. 3 we have plotted the experimental values of ot:\nas a function of the acceptor density for an effective gate voltage between 5\nand 10 V. Moreover we compare the experimental values with the theoretical\nTHERMAL\n511\nMOSlr E3 III\nesuIJstraW =30.a cm-:-\n5\n<,\n1'oci>l1>\n<,\n<,\nr-,\n2 I' \"~\n\"f'..i'\n10*\nr-.\n--\n,\nId-:,f5pA\n1\n0\"\"\"\"'\" r~~\n5\n2\n1iT\n1\n.\nNOISE OF MOS TRANSISTORS\n5\n2\n10\n2\n-\n100pA •\nK I\n0·5mA\n~\n3·0mA\n5\n102\n5 1r1\n-f(kc/s)\n2\n.\n-\"-\no\n5\n2\nFig. 2. Noise spectrum of an n-channel MOS transistor.\n14\n~gm~\n12\nT\nMOST E3-8\nhox= 0·2p\nl{fo>Vp\noV =5V\n.V =fOV\n10\nvg\",,5~V·\ng\ng\n8\n/\n6\ny/\n4\n2\no ,Iq.\n10\n~\n2\n5\n,7~\n10\n2\n5\n0\n~1-fOV-\n~\np--\n~16\n10\n2\n5\n,11\n-NA\nFig. 3. Thermal noise vs substrate doping. Drawn curve represents theory.\nvalues (drawn curves). These curves were obtained by substituting the appropriate values of Vp in eq. (19). The pinch-off voltage as a function of the substrate doping and the oxide thickness is given by 2)\nF. M. KLAASSEN\n512\n(a.\n~ \"\n1\nVp - Vg - - 2\nCox\nand J. PRINS\n)2{(\nCox)'1/2 - 1}.,\n4Vg'\n1+\na\nwhere\nand\nIn figs 4a and b we have plotted the values of Vp, which we have used. Inspecting\n..\n18\n16\n\\f:>CII.\n'4\nI\nÇ)l/\nh=0.2e\nI~\nV ~\"\n.:V /\n12\nlr \\\" /'\nV\n10\n/\n8\n/\n./\n[7/ V\n6\n~\n~\nV/V\n/'\nVV\nl~ V »: r~\n~\n~\n6\n2\nI--- I--8\n10\nv\n-\nI--~\n12\n~\n14\nV\n\\1>\n/V Y\nV\nV\nV\nV\nI~/\n16 =007\n14\n./\na\nV\n,óV\n~\n_.,. ~\nVV\nI-- I-16\n18\n__..\n20\n--\nid'(_.,.-\np0-\nV\n22 24\nIY -Vr(V)\nNr1018\n26 4'J\n~:~=+==X=~==4=~==~~Mll-~\n~~\nNA=3.1d%m3\n__ ~~~=~0~.7~V-+\n__\n-+__~ __~~~~\nVp\nr\n5\nb\nro'~--+---+-~~--+-~+---~--~~\n20\n25\n---~\n30\n-Vr\n35\n4()\nFig. 4. The pinch-off voltage vs the 'effective gate voltage;\ndoping; (b) for different oxide-layer thicknesses.\n(a): for different substrate\nTHERMAL\n5.13\nNOISE OF MOS TRANSISTORS\nfig. 3 we may conclude that the agreement between theory and experiments is\nquite good. Moreover we observe that ais only a very weak function ofthe gate\nvoltage.\n.\nBesides the above experiments we also studied the thermal noise of several\nMOS transistors with oxide layers of different thickness. In fig. 5 we have plotted\nRn gmo\nt\n10\n/\n8\n~/\n/\n6\nV\n42\nL\nMOST E12-16\n= 3.1015cm-3 Vs/=5-1SV\nVod>Vp -r--\nNA\nV\nI\n1-- ~\n°°\n()'2\n(}4.\n(J.6\nOB\n1.0\n1-2\n1·4\n1-6\n---hox(/1) .\nFig. 5. Thermal noise vs oxide-layer thickness. Drawn curve represents theory.\nthe experimental values of a vs the oxide thickness with constant substrate\ndoping. In this case too, the agreement with the theoretical values (drawn curve)\nis quite good.\nFinally we give a plot of the thermal noise vs the drain voltage (fig. 6). In\n0·5\nf?ogJno\nt\nQ.4- {)O3\n,\n\\\nMOST £13\nNA=3.1015\n1\\\nfIox=0'2p\nVp\"'\" O.6Vg'\n.\\\n.\n\"\n04\n00\n()'2\n().4\n().6\n.\n().8\n1.0\n1·2 1.1;. 1.6\n--(~)\n'9\nFig. 6. Thermal noise vs the drain voltage. Drawn curve represents theory.\nthe same figure we have plotted the theoretical value of a due to eq. (15). I~\nthis case the experimental values deviate somewhat from the theoretical values.\nThese deviations may partly be explained by the fact that the noise resistance\nis difficult to measure below saturation (low gain and large output conductance).\nHowever, this range is not of practical interest.\n514\nF. M. KLAASSEN\nand J. PRINS\nThus, as a general result we may conclude that the thermal noise of a MOS\ntransistor is well described by a theory based on the simple current-voltage\nrelation (eq. (2)).\nAcknowledgement\nOur thanks are due to Dr F. C. Eversteyn and Mr H. L. Peek for preparing\nthe MOS transistors.\nEindhoven, June 1967\nRBFERENCES\n1)\n2)\n3)\n4)\n5)\nC. T. Sah and R. C. Pao, IEEE Trans. EI. Dev. ED-13, 393-409, 1966.\nJ. A. van Nielen and O. W. MemeIink, Phil. Res. Repts 22, 55-71, 1967.\nM. Shoji, IEEE Trans. EI. Dev. ED-13, 520-524, 1966.\nC. T. Sah, S. Y. Wu and F. H. Hielscher, IEEE Trans. EI. Dev. ED-13, 410-414, 1966.\nF. M. Klaassen, to be published.\n```"
]
| [
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.8341231,"math_prob":0.98009634,"size":13744,"snap":"2023-40-2023-50","text_gpt3_token_len":3890,"char_repetition_ratio":0.13340612,"word_repetition_ratio":0.021615008,"special_character_ratio":0.29285505,"punctuation_ratio":0.1365669,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9902753,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-28T01:41:47Z\",\"WARC-Record-ID\":\"<urn:uuid:990e6db2-2db1-4541-8d24-715102d67297>\",\"Content-Length\":\"43348\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:3fcd3c87-a6d2-4c84-ae32-369bf9431ad3>\",\"WARC-Concurrent-To\":\"<urn:uuid:69929b09-7c22-4bbe-a127-240342cd43fb>\",\"WARC-IP-Address\":\"104.21.4.158\",\"WARC-Target-URI\":\"https://studyres.com/doc/24454466/thermal-noise-of-mos-transistors\",\"WARC-Payload-Digest\":\"sha1:LF5DAGG46L6JFVJ5DOJNQN3IG374KKXO\",\"WARC-Block-Digest\":\"sha1:FKG2AI5F4PNJHKKW6XQWGCHP3B3JC6RQ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233510334.9_warc_CC-MAIN-20230927235044-20230928025044-00489.warc.gz\"}"} |
https://www.researchgate.net/scientific-contributions/Vladimir-Vapnik-7305638 | [
"# Vladimir Vapnik's research while affiliated with Columbia University and other places\n\n## Publications (119)\n\nArticle\nThe paper is devoted to two problems: (1) reinforcement of SVM algorithms, and (2) justification of memorization mechanisms for generalization. (1) Current SVM algorithm was designed for the case when the risk for the set of nonnegative slack variables is defined by l1 norm. In this paper, along with that classical l1 norm, we consider risks define...\nArticle\nFull-text available\nThis paper introduces a new learning paradigm, called Learning Using Statistical Invariants (LUSI), which is different from the classical one. In a classical paradigm, the learning machine constructs a classification rule that minimizes the probability of expected error; it is data-driven model of learning. In the LUSI paradigm, in order to constru...\nArticle\nFull-text available\nThe paper considers general machine learning models, where knowledge transfer is positioned as the main method to improve their convergence properties. Previous research was focused on mechanisms of knowledge transfer in the context of SVM framework; the paper shows that this mechanism is applicable to neural network framework as well. The paper de...\nArticle\nThis article describes a method for constructing a special rule (we call it synergy rule) that uses as its input information the outputs (scores) of several monotonic rules which solve the same pattern recognition problem. As an example of scores of such monotonic rules we consider here scores of SVM classifiers. In order to construct the optimal s...\nConference Paper\nThe paper considers several topics on learning with privileged information: (1) general machine learning models, where privileged information is positioned as the main mechanism to improve their convergence properties, (2) existing and novel approaches to leverage that privileged information, (3) algorithmic realization of one of these (namely, kno...\nArticle\nDistillation (Hinton et al., 2015) and privileged information (Vapnik & Izmailov, 2015) are two techniques that enable machines to learn from other machines. This paper unifies these two techniques into generalized distillation, a framework to learn from multiple machines and data representations. We provide theoretical and causal insight about the...\nArticle\nThis paper presents direct settings and rigorous solutions of the main Statistical Inference problems. It shows that rigorous solutions require solving multidimensional Fredholm integral equations of the first kind in the situation where not only the right-hand side of the equation is an approximation, but the operator in the equation is also defin...\nArticle\nThis paper describes a new paradigm of machine learning, in which Intelligent Teacher is involved. During training stage, Intelligent Teacher provides Student with information that contains, along with classification of each example, additional privileged information (for example, explanation) of this example. The paper describes two mechanisms tha...\nConference Paper\nThis paper introduces an advanced setting of machine learning problem in which an Intelligent Teacher is involved. During training stage, Intelligent Teacher provides Student with information that contains, along with classification of each example, additional privileged information (explanation) of this example. The paper describes two mechanisms...\nConference Paper\nThis paper presents direct settings and rigorous solutions of Statistical Inference problems. It shows that rigorous solutions require solving ill-posed Fredholm integral equations of the first kind in the situation where not only the right-hand side of the equation is an approximation, but the operator in the equation is also defined approximately...\nArticle\nWe introduce a constructive setting for the problem of density ratio estimation through the solution of a multidimensional integral equation. In this equation, not only its right hand side is approximately known, but also the integral operator is approximately defined. We show that this ill-posed problem has a rigorous solution and obtain the solut...\nConference Paper\nRadial basis function (RBF) kernels for SVM have been routinely used in a wide range of classification problems, delivering consistently good performance for those problems where the kernel computations are numerically feasible (high-dimensional problems typically use linear kernels). One of the drawbacks of RBF kernels is the necessity of selectin...\nArticle\nWe introduce a general constructive setting of the density ratio estimation problem as a solution of a (multidimensional) integral equation. In this equation, not only its right hand side is known approximately, but also the integral operator is defined approximately. We show that this ill-posed problem has a rigorous solution and obtain the soluti...\nArticle\nWe describe a method for predicting a classification of an object given classifications of the objects in the training set, assuming that the pairs object/classification are generated by an i.i.d. process from a continuous probability distribution. Our method is a modification of Vapnik's support-vector machine; its main novelty is that it gives no...\nConference Paper\nFull-text available\nRecently Vapnik et al. [11, 12, 13] introduced a new learning model, called Learning Using Privileged Information (LUPI). In this model, along with standard training data, the teacher supplies the student with additional (privileged) information. In the optimistic case, the LUPI model can improve the bound for the probability of test error from O(1...\nArticle\nFull-text available\nIn Learning Using Privileged Information (LUPI) paradigm, along with the stan-dard training data in the decision space, a teacher supplies a learner with the priv-ileged information in the correcting space. The goal of the learner is to find a classifier with a low generalization error in the decision space. We consider an empirical risk minimizati...\nConference Paper\nIn Learning Using Privileged Information (LUPI) paradigm, along with the standard training data in the decision space, a teacher supplies a learner with the privileged information in the correcting space. The goal of the learner is to find a classifier with a low generalization error in the decision space. We consider an empirical risk minimization...\nArticle\nIn the Afterword to the second edition of the book \"Estimation of Dependences Based on Empirical Data\" by V. Vapnik, an advanced learning paradigm called Learning Using Hidden Information (LUHI) was introduced. This Afterword also suggested an extension of the SVM method (the so called SVM(gamma)+ method) to implement algorithms which address the L...\nArticle\nFull-text available\nWe compare Karl Popper’s ideas concerning the falsifiability of a theory with similar notions from the part of statistical learning theory known as VC-theory. Popper’s notion of the dimension of a theory is contrasted with the apparently very similar VC-dimension. Having located some divergences, we discuss how best to view Popper’s work from the p...\nConference Paper\nIn this paper we consider a new paradigm of learning: learning using hidden information. The classical paradigm of the supervised learning is to learn a decision rule from labeled data (xi, yi), xi isin X, yi isin {-1, 1}, i = 1, hellip, lscr. In this paper we consider a new setting: given training vectors in space X along with labels and descripti...\nArticle\nThesupport-vector network is a new learning machine for two-group classification problems. The machine conceptually implements the following idea: input vectors are non-linearly mapped to a very high-dimension feature space. In this feature space a linear decision surface is constructed. Special properties of the decision surface ensures high gener...\nConference Paper\nWe focus on distribution-free transductive learning. In this setting the learning algorithm is given a ‘full sample’ of unlabeled points. Then, a training sample is selected uniformly at random from the full sample and the labels of the training points are revealed. The goal is to predict the labels of the remaining unlabeled points as accurately a...\nArticle\nFull-text available\nWe consider a large volume principle for transductive learning that prioritizes the transductive equivalence classes according to the volume they occupy in hypothesis space. We approximate volume maximization using a geometric interpretation of the hypothesis space. The resulting algorithm is defined via a non-convex optimization problem that can s...\nConference Paper\nFull-text available\nWe study classification tasks where one is given a set of labeled examples, and a set of \"non-examples\" of meaningful concepts in the same domain that do not belong to either class (refered to as the universum). We describe an algorithmic approach to leverage universum points and show experimentally that inference based on the labeled data and the...\nChapter\nSupport Vector Machines are used for time series prediction and compared to radial basis function networks. We make use of two different cost functions for Support Vectors: training with (i) an e insensitive loss and (ii) Huber's robust loss function and discuss how to choose the regularization parameters in these models. Two applications are consi...\nArticle\nFull-text available\nWe formulate density estimation as an inverse operator problem. We then use convergence results of empirical distribution functions to true distribution functions to develop an algorithm for multivariate density estimation. The algorithm is based upon a Support Vector Machine (SVM) approach to solving inverse operator problems. The algorithm is imp...\nConference Paper\nFull-text available\nWe describe an algorithm for support vector machines (SVM) that can be parallelized efficiently and scales to very large problems with hundreds of thousands of training vectors. Instead of analyzing the whole training set in one optimization step, the data are split into subsets and optimized separately with multiple SVMs. The partial results are c...\nArticle\nFull-text available\nContents 0 A Roadmap 6 0.1 How to read this Thesis . . . . . . . . . . . . . . . . . . . . . . . 6 0.2 A Short Review of Approximation and Regression Estimation . . 7 0.3 The Reason for Support Vectors . . . . . . . . . . . . . . . . . . . 8 1 Introduction 10 1.1 The Regression Problem . . . . . . . . . . . . . . . . . . . . . . . 10 1.2 A Special...\nArticle\nWe explore methods for incorporating prior knowledge about a problem at hand in Support Vector learning machines. We show that both invariances under group transformations and prior knowledge about locality in images can be incorporated by constructing appropriate kernel functions.\nArticle\nFull-text available\nA new regression technique based on concept of support vectors is introduced. We compare support vector regression with a committee regression technique (bagging) based on regression trees and ridge regression done in feature space. On the basis of these experiments, it is expected that SVR will have advantages in high dimensionality space because...\nArticle\nThe Support Vector (SV) machine is a novel type of learning machine, based on statistical learning theory, which contains polynomial classifiers, neural networks, and radial basis function (RBF) networks as special cases. In the RBF case, the SV algorithm automatically determines centers, weights and threshold such as to minimize an upper bound on...\nArticle\nWe present a novel clustering method using the approach of support vector machines. Data points are mapped by means of a Gaussian kernel to a high dimensional feature space, where we search for the minimal enclosing sphere. This sphere, when mapped back to data space, can separate into several components, each enclosing a separate cluster of points...\nArticle\nDNA micro-arrays now permit scientists to screen thousands of genes simultaneously and determine whether those genes are active, hyperactive or silent in normal or cancerous tissue. Because these new micro-array devices generate bewildering amounts of raw data, new analytical methods must be developed to sort out whether cancer tissues have distinc...\nConference Paper\nWe consider the learning problem of nding a dependency between a general class of objects and another, possibly dierent, general class of objects. The objects can be for example: vectors, images, strings, trees or graphs. Such a task is made possible by employing similarity measures in both input and output spaces using kernel functions, thus embed...\nArticle\nFull-text available\nWe present a novel clustering method using the approach of support vector machines. Data points are mapped by means of a Gaussian kernel to a high dimensional feature space, where we search for the minimal enclosing sphere. This sphere, when mapped back to data space, can separate into several components, each enclosing a separate cluster of points...\nArticle\nFull-text available\nThe problem of automatically tuning multiple parameters for pattern recognition Support Vector Machines (SVMs) is considered. This is done by minimizing some estimates of the generalization error of SVMs using a gradient descent algorithm over the set of parameters. Usual methods for choosing parameters, based on exhaustive search become intractabl...\nArticle\nFull-text available\nWe present a novel method for clustering using the support vector machine approach. Data points are mapped to a high dimensional feature space, where support vectors are used to define a sphere enclosing them. The boundary of the sphere forms in data space a set of closed contours containing the data. Data points enclosed by each contour are define...\nArticle\nFull-text available\nIntroduction Model selection is an important ingredient of many machine learning algorithms, in particular when the sample size in small, in order to strike the right tradeo between overtting and undertting. Previous classical results for linear regression are based on an asymptotical analysis. We present a new penalization method for performing mo...\nArticle\nWe explore methods for incorporating prior knowledge about a problem at hand in Support Vector learning machines. We show that both invariances under group transformations and prior knowledge about locality in images can be incorporated by constructing appropriate kernel functions. 1 INTRODUCTION When we are trying to extract regularities from data...\nArticle\nThis paper compares the performance of several classifier algorithms on a standard database of handwritten digits. We consider not only raw accuracy, but also training time, recognition time, and memory requirements. When available, we report measurements of the fraction of patterns that must be rejected so that the remaining patterns have misclass...\nArticle\nFull-text available\nThis paper compares the performance of several classifier algorithms on a standard database of handwritten digits. We consider not only raw accuracy, but also training time, recognition time, and memory requirements. When available, we report measurements of the fraction of patterns that must be rejected so that the remaining patterns have misclass...\nArticle\nWe report a novel possibility for extracting a small subset of a data base which contains all the information necessary to solve a given classification task: using the Support Vector Algorithm to train three different types of handwritten digit classifiers, we observed that these types of classifiers construct their decision surface from strongly o...\nArticle\nWe introduce an algorithm for estimating the values of a function at a set of test points x `+1 ; : : : ; x `+m given a set of training points (x 1 ; y 1 ); : : : ; (x ` ; y ` ) without estimating (as an intermediate step) the regression function. We demonstrate that this direct (transductive) way for estimating values of the regression (or classic...\nArticle\nNew functionals for parameter (model) selection of Support Vector Machines are introduced based on the concepts of the span of support vectors and rescaling of the feature space. It is shown that using these functionals, one can both predict the best choice of parameters of the model and the relative quality of performance for any value of paramete...\nArticle\nTraditional classification approaches generalize poorly on image classification tasks, because of the high dimensionality of the feature space. This paper shows that Support Vector Machines (SVM) can generalize well on difficult image classification problems where the only features are high dimensional histograms. Heavy-tailed RBF kernels of the fo...\nArticle\nFull-text available\nThis paper compare the performance of classifier algorithms on a stan- dard database of handwritten digits. We consider not only raw accuracy, but also training time, recognition time, and memory requirements. When available, we report measurements of the fraction of patterns that must be rejected so that the remaining patterns have misclassificati...\nArticle\nThis paper compares the performance of classifier algorithms on a standard database of handwritten digits. We consider not raw accuracy, but rejection, training time, recognition time, and memory requirements. \"Comparison of Leaning for Handwritten Digit Recognition\", International Conference on Neural F. and P. Cie Publishers, 1995 Y. Le L. Bottou...\nConference Paper\nThe problem of estimating density, conditional probability, and conditional density is considered as an ill-posed problem of solving integral equations. To solve these equations the support vector method (SVM) is used\nArticle\n. Developed only recently, support vector learning machines achieve high generalization ability by minimizing a bound on the expected test error; however, so far there existed no way of adding knowledge about invariances of a classification problem at hand. We present a method of incorporating prior knowledge about transformation invariances by app...\nArticle\nThe last years have witnessed an increasing interest in Support Vector (SV) machines, which use Mercer kernels for eeciently performing computations in high-dimensional spaces. In pattern recognition, the SV algorithm constructs nonlinear decision functions by training a classiier to perform a linear separation in some high-dimensional space which...\nConference Paper\nFull-text available\nWe introduce a method of feature selection for Support Vector Machines.\nArticle\nFull-text available\nWe introduce a method of feature selection for Support Vector Machines. The method is based upon finding those features which minimize bounds on the leave-one-out error. This search can be efficiently performed via gradient descent. The resulting algorithms are shown to be superior to some standard feature selection algorithms on both toy data and...\nConference Paper\nWe present a novel kernel method for data clustering using a description of the data by support vectors. The kernel reflects a projection of the data points from data space to a high dimensional feature space. Cluster boundaries are defined as spheres in feature space, which represent complex geometric shapes in data space. We utilize this geometri...\nArticle\nFull-text available\nThis paper compares the performance of several classi#er algorithms on a standard database of handwritten digits. We consider not only raw accuracy, but also rejection, training time, recognition time, and memory requirements. 1 COMPARISON OF LEARNING ALGORITHMS FOR HANDWRITTEN DIGIT RECOGNITION Y. LeCun, L. Jackel, L. Bottou, A. Brunot, C. Cortes,...\nArticle\nFull-text available\n> (x); (1.1) 1. `(x) = ( 1; x ? 0 0; otherwise Generic author design sample pages 1999/07/12 15:50 2 Support Vector Density Estimation where instead of knowing the distribution function F (x) we are given the iid (independently and identically distributed) data x 1 ; : : : ; x ` (1.2) generated by F . The problem of density estimation is known to b...\nArticle\nA Support Vector Machine (SVM) algorithm for multivariate density estimation is developed based on regularization principles and bounds on the convergence of empirical distribution functions. The algorithm is compared to Gaussian Mixture Models (GMMs). Our algorithm outperforms GMMs for data drawn from mixtures of gaussians in IR 2 and IR 6 . Our a...\nArticle\nFull-text available\nSupport Vector Machines using ANOVA Decomposition Kernels (SVAD) [Vapng] are a way of imposing a structure on multi-dimensional kernels which are generated as the tensor product of one-dimensional kernels. This gives more accurate control over the capacity of the learning machine (VCdimension) . SVAD uses ideas from ANOVA decomposition methods and...\nArticle\nFull-text available\nthis report we describe how the Support Vector (SV) technique of solving linear operator equations can be applied to the problem of density estimation . We present a new optimization procedure and set of kernels closely related to current SV techniques that guarantee the monotonicity of the approximation. This technique estimates densities with...\nChapter\nContributorsPeter Bartlett, Kristin P. Bennett, Christopher J.C. Burges, Nello Cristianini, Alex Gammerman, Federico Girosi, Simon Haykin, Thorsten Joachims, Linda Kaufman, Jens Kohlmorgen, Ulrich Kreßel, Davide Mattera, Klaus-Robert Müller, Manfred Opper, Edgar E. Osuna, John C. Platt, Gunnar Rätsch, Bernhard Schölkopf, John Shawe-Taylor, Alexande...\nChapter\nContributorsPeter Bartlett, Kristin P. Bennett, Christopher J.C. Burges, Nello Cristianini, Alex Gammerman, Federico Girosi, Simon Haykin, Thorsten Joachims, Linda Kaufman, Jens Kohlmorgen, Ulrich Kreßel, Davide Mattera, Klaus-Robert Müller, Manfred Opper, Edgar E. Osuna, John C. Platt, Gunnar Rätsch, Bernhard Schölkopf, John Shawe-Taylor, Alexande...\nChapter\nContributorsPeter Bartlett, Kristin P. Bennett, Christopher J.C. Burges, Nello Cristianini, Alex Gammerman, Federico Girosi, Simon Haykin, Thorsten Joachims, Linda Kaufman, Jens Kohlmorgen, Ulrich Kreßel, Davide Mattera, Klaus-Robert Müller, Manfred Opper, Edgar E. Osuna, John C. Platt, Gunnar Rätsch, Bernhard Schölkopf, John Shawe-Taylor, Alexande...\nArticle\nFull-text available\nA new regression technique based on Vapnik's concept of support vectors is introduced. We compare support vector regression (SVR) with a committee regression technique (bagging) based on regression trees and ridge regression done in feature space. On the basis of these experiments, it is expected that SVR will have advantages in high dimensionality...\nArticle\nthis report we describe how the Support Vector (SV) technique of solving linear operator equations can be applied to the problem of density estimation . We present a new optimization procedure and set of kernels closely related to current SV techniques that guarantee the monotonicity of the approximation. This technique estimates densities with...\nArticle\nFull-text available\nWe address the problem of determining what size test set guarantees statistically significant results in a character recognition task, as a function of the expected error rate. We provide a statistical analysis showing that if, for example, the expected character error rate is around 1 percent, then, with a test set of at least 10,000 statistically...\nArticle\nFull-text available\nThe support vector (SV) machine is a novel type of learning machine, based on statistical learning theory, which contains polynomial classifiers, neural networks, and radial basis function (RBF) networks as special cases. In the RBF case, the SV algorithm automatically determines centers, weights, and threshold that minimize an upper bound on the e...\nConference Paper\n. Support Vector Machines are used for time series prediction and compared to radial basis function networks. We make use of two different cost functions for Support Vectors: training with (i) an ffl insensitive loss and (ii) Huber's robust loss function and discuss how to choose the regularization parameters in these models. Two applications are c...\nArticle\nLarge VC-dimension classifiers can learn difficult tasks, but are usually impractical because they generalize well only if they are trained with huge quantities of data. In this paper we show that even very high-order polynomial classifiers can be trained with a small amount of training data and yet generalize better than classifiers with a smaller...\nArticle\nWe present a method for discovering informative patterns from data. With this method, large databases can be reduced to only a few representative data entries. Our framework also encompasses methods for cleaning databases containing corrupted data. Both on-line and off-line algorithms are proposed and experimentally checked on databases of handwrit...\nConference Paper\nFull-text available\nTwo view-based object recognition algorithms are compared: (1) a heuristic algorithm based on oriented filters, and (2) a support vector learning machine trained on low-resolution images of the objects. Classification performance is assessed using a high number of images generated by a computer graphics system under precisely controlled conditions....\nConference Paper\nFull-text available\nA new regression technique based on Vapnik's concept of support vectors is introduced. We compare support vector regression (SVR) with a committee regression technique (bagging) based on regression trees and ridge regression done in feature space. On the basis of these experiments, it is expected that SVR will have advantages in high dimensionality...\nArticle\nThesupport-vector network is a new learning machine for two-group classification problems. The machine conceptually implements the following idea: input vectors are non-linearly mapped to a very high-dimension feature space. In this feature space a linear decision surface is constructed. Special properties of the decision surface ensures high gener...\nConference Paper\nFull-text available\nThis paper compares the performance of classifier algorithmson a standard database of handwritten digits. We consider not rawaccuracy, but rejection, training time, recognition time, and memoryrequirements.\"Comparison of Leaning for Handwritten Digit Recognition\", International Conference onNeural F. and P. Cie Publishers, 1995Y. Le L. Bottou, C. J...\nConference Paper\nWe report a novel possibility for extracting asmall subset of a data base which contains allthe information necessary to solve a given classificationtask: using the Support Vector Algorithmto train three different types of handwrittendigit classifiers, we observed that these typesof classifiers construct their decision surface fromstrongly overlapp...\nConference Paper\nIn an optical character recognition problem, we compare (as a function of training set size) the performance of three neural network based ensemble methods (two versions of boosting and a committee of neural networks trained independently) to that of a single network. In boosting, the number of patterns actually used for training is a subset of all...\nConference Paper\nWe compare the performance of three types of neural network-based ensemble techniques to that of a single neural network. The ensemble algorithms are two versions of boosting and committees of neural networks trained independently. For each of the four algorithms, we experimentally determine the test and training error curves in an optical characte..."
]
| [
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.8950961,"math_prob":0.868519,"size":27500,"snap":"2023-14-2023-23","text_gpt3_token_len":5372,"char_repetition_ratio":0.12856415,"word_repetition_ratio":0.35852668,"special_character_ratio":0.19123636,"punctuation_ratio":0.14739648,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9603017,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-03-24T23:32:23Z\",\"WARC-Record-ID\":\"<urn:uuid:dbe211ea-f259-4425-af2b-dfac9ede4bdb>\",\"Content-Length\":\"1050189\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f33a5b52-971c-4b9b-882f-57f31be5efad>\",\"WARC-Concurrent-To\":\"<urn:uuid:8b00a275-1fe6-408b-bef2-993ff7da268b>\",\"WARC-IP-Address\":\"104.17.33.105\",\"WARC-Target-URI\":\"https://www.researchgate.net/scientific-contributions/Vladimir-Vapnik-7305638\",\"WARC-Payload-Digest\":\"sha1:VEQA36CX3UTX4BV23QACZKWMRQORVBCU\",\"WARC-Block-Digest\":\"sha1:Z7XCNTJ5I6ULR7WQ7EX4TJUIO2WZTT3S\",\"WARC-Truncated\":\"length\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-14/CC-MAIN-2023-14_segments_1679296945289.9_warc_CC-MAIN-20230324211121-20230325001121-00418.warc.gz\"}"} |
https://math.stackexchange.com/questions/830219/l-p-metric-on-mathbbrn-and-its-open-balls/830230 | [
"$l_{p}$ metric on $\\mathbb{R}^{n}$ and its open balls\n\nFor $x,y \\in \\mathbb R^n$ let $$d_p(x,y) = \\left(\\sum_{i=1}^n \\def\\abs#1{\\left|#1\\right|}\\abs{x_i - y_i}^p\\right)^{1/p}$$ for $1 \\le p < \\infty$ and $$d_\\infty(x,y) = \\max\\{\\abs{x_i -y_i} \\mid i = 1, \\ldots, n\\}$$ Let $B_p = \\{x\\in \\def\\R{\\mathbb R}\\R^n \\mid d_p(x,0) < 1\\}$, $1 \\le p \\le \\infty$. Which of the following are correct?\n\n1. $B_1$ is open in the $d_\\infty$-metric.\n2. $B_2$ is open in the $d_\\infty$-metric.\n3. $B_1$ is not open in the $d_2$-metric.\n4. $B_2$ is not open in the $d_2$-metric.\n\nHelp me with the geometrical approach to solve this. How to construct those sets and all those ?? and explain the ruled out options in detail\n\n• 1. cannot read indices in the answer. 2. what you mean with ?? – Emanuele Paolini Jun 11 '14 at 9:04\n• I suggest that you type the question yourself in LateX – Vishal Gupta Jun 11 '14 at 9:05\n• Try doing this in $\\mathbb{R}^{2}$ first. See that ball in $d_{1}$ metric is a rhombus, in $d_{2}$ metric, a ball is a circle and in $d_{\\infty}$ metric, it is a square. This is the most geometrical way I can see. Then you can generalize to higher dimensions. – Vishal Gupta Jun 11 '14 at 9:23\n• As Emanuele Paolini pointed out in his answer, all topologies are same and you can see this because you can always get a circle inside a rhombus and vice versa. Simlarly circle inside a square and vice versa and a square inside a rhombus and vice versa. – Vishal Gupta Jun 11 '14 at 9:24\n• @martini Thanks for the edit. – Vishal Gupta Jun 11 '14 at 9:32\n\nProve that for any $p,q$ the ball $B_p$ is contained in some rescaling of the ball $B_q$. As a consequence all the topologies are equal each other hence statements like \"$X$ is open with respect to $d_p$\" do not depend on $p$."
]
| [
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.7138993,"math_prob":0.99930036,"size":644,"snap":"2019-43-2019-47","text_gpt3_token_len":265,"char_repetition_ratio":0.1453125,"word_repetition_ratio":0.0952381,"special_character_ratio":0.42080745,"punctuation_ratio":0.108843535,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99988055,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-10-21T22:51:36Z\",\"WARC-Record-ID\":\"<urn:uuid:452738e4-80c5-4774-aa37-3ea2d912e532>\",\"Content-Length\":\"143301\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9a88c1be-9c6f-4b19-af59-2c4fc92567a9>\",\"WARC-Concurrent-To\":\"<urn:uuid:ab2c2fc4-d740-4594-9ba0-dd45ca6baea9>\",\"WARC-IP-Address\":\"151.101.129.69\",\"WARC-Target-URI\":\"https://math.stackexchange.com/questions/830219/l-p-metric-on-mathbbrn-and-its-open-balls/830230\",\"WARC-Payload-Digest\":\"sha1:TM4CE7OKESSR7CYDQQ43GRZPFVY7YHXL\",\"WARC-Block-Digest\":\"sha1:QALQZJPLBJACW4QBOKHOEHIC34MOH6AS\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-43/CC-MAIN-2019-43_segments_1570987795253.70_warc_CC-MAIN-20191021221245-20191022004745-00228.warc.gz\"}"} |
https://stats.stackexchange.com/questions/339972/gini-coefficients-accounting-for-firm-exits | [
"Gini coefficients - Accounting for firm exits\n\nWhat is the appropriate way to measure the Gini coefficient over time in an industry where a large number of firms are exiting? Should the Gini coefficient be calculated using the initial number of firms, or the number of remaining firms at any point in time?\n\nThis question arises from reading reports about a controversial fisheries policy instrument--individual fishing quotas (IFQs). The implementation of IFQs is commonly associated with a significant drop in the number of fishing vessels. Yet, when many researchers calculate changes in the Gini coefficient on vessel revenue before and after IFQ implementation they do so using revenue per active vessel only. In my view this method ignores the drop in revenue (to zero) for vessels that become inactive as a result of the IFQ program.\n\n_\n\nAn example from the Gulf of Mexico Red Snapper fishery:\n\nNumber of active vessels: before = 482, after =360\n\nGini coefficient among active vessels: before = 0.81, after = 0.79\n\nAverage revenue per active vessel: before = 28,960 after = 58,630\n\n_\n\nIs there a way to use this information to calculate the Gini coefficient for the total number of vessels (active and inactive) before and after IFQ implementation?\n\nAny comment or insight would be greatly appreciated.\n\n-\n\nThe specific paper I'm think about is Performance of federally managed catch share fisheries in the United States by Ayeisha A. Brinson abd Eric M (2016), but similar methods are used for many other US IFQ programs. Thunberg. https://www.sciencedirect.com/science/article/pii/S0165783616300649\n\nThe specific Gini coefficient forumula used:",
null,
"where i = 1 to n; i is the vessel’s rank order in ascending order; x is the annual revenue for vessel i; n is the number of active vessels; u is the mean revenue"
]
| [
null,
"https://i.stack.imgur.com/teKWN.png",
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.9053372,"math_prob":0.7530315,"size":1756,"snap":"2019-26-2019-30","text_gpt3_token_len":385,"char_repetition_ratio":0.14041096,"word_repetition_ratio":0.0,"special_character_ratio":0.22209567,"punctuation_ratio":0.09815951,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9727504,"pos_list":[0,1,2],"im_url_duplicate_count":[null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-06-16T07:20:46Z\",\"WARC-Record-ID\":\"<urn:uuid:514b228a-fd9f-49fb-b243-f50d3ccffc49>\",\"Content-Length\":\"140744\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8dc1d9fd-e554-4f09-bae1-6475d0fcbbfc>\",\"WARC-Concurrent-To\":\"<urn:uuid:56d48099-07bf-4bd8-8108-6621b2680242>\",\"WARC-IP-Address\":\"151.101.193.69\",\"WARC-Target-URI\":\"https://stats.stackexchange.com/questions/339972/gini-coefficients-accounting-for-firm-exits\",\"WARC-Payload-Digest\":\"sha1:M4CCNOBZNXVTDJNO7VY7IFLURRIYONEZ\",\"WARC-Block-Digest\":\"sha1:3GYVJ4IFKD75S5T27V2VNP46YVRA4SHQ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-26/CC-MAIN-2019-26_segments_1560627997801.20_warc_CC-MAIN-20190616062650-20190616084650-00289.warc.gz\"}"} |
https://blog.monkey.codes/tech-interview-cheatsheet-lists-search-sort/ | [
"## monkey codes\n\nRandom bits of knowledge, functional code examples and laughable mistakes from a code monkey working in the real world.\n\nCurious software developer, motorcycle enthusiast, rugby fanatic and biltong connoisseur. My code always works sometimes.\n\n# Tech Interview CheatSheet - Lists, Search & Sort\n\nAn overview of Binary Search algorithms, including Bubble, Merge and Quick Sort.\n\nEvery developer will, or have at some point, encountered this dreadful experience... the intimidating tech interview. In this series of posts I will cover some of the typical topics that arise in tech interviews, starting with searching and sorting algorithms most commonly applied to List based collections.\n\n“Before anything else, preparation is they key to success.”— Alexander Graham Bell\n\n## Arrays\n\nArrays are probably the most common data structure, found in the majority languages used today. Generally 0 based indexes are used (the first element is at index 0).\n\nMainly 2 types exist:\nLinear arrays which are static in size, usually defined during construction.\n\nDynamic arrays that contain extra space reserved for new elements, once it is full elements are copied into a larger array.\n\nArrays are excellent for looking up values by index, which is performed in constant time, $O(1)$, but bad for inserting and deleting elements. The cause for this is that elements need to be copied to make space for new elements or to close the gap caused by deletion. Insert and delete operations are performed in linear time $O(n)$",
null,
"The basic building block of a Linked List is a node. A node contains a single value and a reference to the next node in the list. This dynamic data structure is optimised for insert and delete operations, which happen in $O(1)$\n\nIn contrast to Arrays, index based lookup & searching performance is poor, $O(n)$, since potentially the entire list will have to be traversed to find an element.\n\nIn a Doubly linked list, each node has a reference to the next and previous node. Circularly linked lists have their first and last nodes connected.",
null,
"## Stacks\n\nA Stack provides quick ( $O(1)$ ) access to the head (first) element and is an example of a LIFO (Last In First Out) data structure. Elements are pushed onto the stack and popped off. A Stack can be implemented as a Linked List where the push operation inserts at the head and pop removes from the head.\n\n## Queues\n\nA Queue is an example of a FIFO (First In First Out) data structure. The oldest and newest elements are called the Head and Tail respectively. Elements are Enqueued to the Tail of the queue and Dequeued from the Head. A Queue can be implemented using a Linked List, by keeping track of both the Head and Tail.\n\nA Deque is a double ended Queue, enqueueing or dequeueing can happen on either end.\n\nQueues have the same performance characteristics as Linked Lists, $O(1)$ for enqueue/dequeue but $O(n)$ for searching.\n\n## Searching & Sorting\n\n### Binary Search\n\nGiven an array of sorted numbers, and x (the number being searched for). Recursively look at the element in the middle of the array, repeat on the left side if x is smaller than the middle element or on the right if its bigger. For even number arrays you will have to choose whether to use the lower or upper number as the \"middle\".\n\nExample:\nGiven [1,2,3,4,5,6,7,8] and x = 10:\n\n1. Split the array, using the lower element as the middle for even sized arrays.\n2. Iteration 1: Middle = 4, 10 > 4, search the upper half [5,6,7,8].\n3. Iteration 2: Middle = 6, 10 > 6, search the upper half [7,8].\n4. Iteration 3: Middle = 7, 10 > 7, search .\n5. Iteration 4: 10 > 8, thus the array does not contain the target value x.\n“For interviews, memorize the time efficiency of well know algorithms and learn to spot them”— code monkey\n\n#### Calculating Complexity\n\nWhen faced with an unknown algorithm, one trick is to use a table that tracks the number of iterations for each increase in the input size of the array. Below is a table that applies this technique to the binary search algorithm. Upon closer inspection it roughly looks like the number of iterations increase when the input size doubles:\n\n$$\\begin{array}{|c|c|} \\hline n & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\\\ \\hline \\text{iterations} & 0 & 1 & 2 & 2 & 3 & 3 & 3 & 3 & 4 \\\\ \\hline \\end{array}$$\n\n\\begin{align} n & \\approx 2^\\text{iterations - 1} \\ \\text{iterations}& \\approx log_2 (n) + 1\\ complexity & \\approx O(log_2(n)+1) \\ & \\approx O(log(n))\\ \\end{align}\n\nSimple python implementation (iterative):\n\n### Bubble sort\n\nBubble sort is the simplest of sorting algorithms and probably the one most developers will naturally come up with when they are first faced with the problem. The algorithm basically compares two adjacent elements and swaps them if the order is wrong. This process is repeated until no more swaps happen. Worst case scenario is when the input is in reverse order, that will result in every element being compared to every other element, $O(n^2)$. On the up side, it is an in place sorting algorithm that requires no extra space, $O(1)$.\n\n### Merge sort\n\nMerge sort is an example of a Divide and Conquer algorithm. Recursively split the input array until you have arrays with $\\le 2$ elements. Then merge 2 adjacent arrays by comparing the first elements, since each array is already sorted. Rinse and Repeat.\n\nThe complexity of the algorithm would roughly be $\\text{no_comparisons} \\times \\text{no_iterations}$. The number of comparisons (worst case) is approximately 1 less than the input size. To calculate the number of iterations, the same trick used in Binary Search can be applied, namely map the number of iterations against the input size and try to spot a pattern.\n\n$$\\begin{array}{|c|c|} \\hline & 2^0 & 2^1 & & 2^2 & & & & 2^3 & \\\\ \\hline \\hline n & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\\\ \\hline iterations & 0 & 1 & 2 & 2 & 3 & 3 & 3 & 3 & 4 \\\\ \\hline \\end{array}$$\n\n\\begin{align} n& \\approx 2^\\text{iterations} \\ \\text{iterations}& \\approx log(n) \\ \\text{comparisons}& \\approx n \\ complexity& \\approx O(\\text{comparisons} \\times \\text{iterations}) \\ & \\approx O(n \\ log(n)) \\ \\end{align}\n\n### Quick sort\n\nPick a pivot element at random (convention picks the last number), then move all the numbers less than the pivot to the left of the pivot and all numbers larger to the right. Keep doing this recursively for the lower numbers to the left of the pivot and the higher numbers to the right of the pivot.\n\nQuick Sort is an in place sorting algorithm with a space complexity of $(O(1))$.\n\nWorst case time complexity is when the pivot does not split the array roughly in half. If the pivot belongs at the end, then you will end up comparing it to all the other numbers without swapping. This is compounded if the 2nd number has the same scenario. Thus worst case complexity for Quick Sort is the same as Bubble Sort, $O(n^2)$. Average case is $O(n\\ log(n))$. Quick Sort does offer some opportunity for optimisation, for example the splits can run in parallel.\n\nIn part 2 of this series I will look at hash functions & maps.\n\nhttps://gist.github.com/TSiege/cbb0507082bb18ff7e4b"
]
| [
null,
"https://res.cloudinary.com/monkey-codes/image/upload/v1494831769/algorithms/array_insert_vjgpaz.gif",
null,
"https://res.cloudinary.com/monkey-codes/image/upload/v1494838661/algorithms/linked_list_jncfr9.gif",
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.8599203,"math_prob":0.97251755,"size":6678,"snap":"2023-14-2023-23","text_gpt3_token_len":1685,"char_repetition_ratio":0.12151633,"word_repetition_ratio":0.06717687,"special_character_ratio":0.25965858,"punctuation_ratio":0.0947205,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98168486,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,8,null,6,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-03-28T02:38:23Z\",\"WARC-Record-ID\":\"<urn:uuid:ffd09f79-5b4e-4141-bc0c-e821a3b60b64>\",\"Content-Length\":\"84638\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:62736d3b-1200-40fb-991a-38e92bc60107>\",\"WARC-Concurrent-To\":\"<urn:uuid:0990effa-ac6e-48a5-addd-99f2999de6f9>\",\"WARC-IP-Address\":\"146.75.39.7\",\"WARC-Target-URI\":\"https://blog.monkey.codes/tech-interview-cheatsheet-lists-search-sort/\",\"WARC-Payload-Digest\":\"sha1:Q4353ITCRX4TSYAKATUGI4H5GLBDS4E7\",\"WARC-Block-Digest\":\"sha1:453HXKTJXOIRGL2TC4KEU6MU45W64O3B\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-14/CC-MAIN-2023-14_segments_1679296948756.99_warc_CC-MAIN-20230328011555-20230328041555-00746.warc.gz\"}"} |
https://ask.sagemath.org/question/59374/cauchy-principal-value-integral/ | [
"# Cauchy principal value integral",
null,
"How can I calculate a Cauchy principal Value integral with sagemath\n\nedit retag close merge delete\n\nSort by » oldest newest most voted\n\nI do not know much about the notion of Cauchy principal value integral, but you can reproduce the two examples given on Wikipedia https://en.wikipedia.org/wiki/Cauchy_... :\n\nsage: a = SR.var('a')\nsage: assume(0<2*a<1)\nsage: f = integral(1/x, x, -1, -2*a) + integral(1/x, x, a, 1)\nsage: f.limit(a=0)\nlog(2)\n\n\nand\n\nsage: a = SR.var('a')\nsage: assume(a>0)\nsage: f = integral(2*x/(x^2-1), x, -2*a, a)\nsage: f.limit(a=oo)\n-2*log(2)\n\nmore"
]
| [
null,
"https://www.gravatar.com/avatar/3807813ec3a8536fd55a3897e0ab0a21",
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.6993719,"math_prob":0.9911052,"size":813,"snap":"2021-43-2021-49","text_gpt3_token_len":277,"char_repetition_ratio":0.15203956,"word_repetition_ratio":0.13114753,"special_character_ratio":0.34809348,"punctuation_ratio":0.18592964,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99772036,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-12-07T16:14:22Z\",\"WARC-Record-ID\":\"<urn:uuid:b5b0ce61-9705-4efd-8d29-876b865e4050>\",\"Content-Length\":\"50647\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9c897ff0-1c56-4646-b927-06bbad1858f7>\",\"WARC-Concurrent-To\":\"<urn:uuid:5f3a01c8-d35b-4bb0-8035-b9bca17e1536>\",\"WARC-IP-Address\":\"194.254.163.53\",\"WARC-Target-URI\":\"https://ask.sagemath.org/question/59374/cauchy-principal-value-integral/\",\"WARC-Payload-Digest\":\"sha1:YQRSREOX764SHAH2IVYARSTWZZO4V3K5\",\"WARC-Block-Digest\":\"sha1:WQTU3WOUKXZWDML7TF3GWZMM7JHIBWST\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964363400.19_warc_CC-MAIN-20211207140255-20211207170255-00452.warc.gz\"}"} |
http://justframes.dk/?p=9943 | [
"# Calculate Agreement Between Two Tests\n\nStatistical methods of conformity assessment vary according to the nature of the variables studied and the number of observers between whom the concordance should be assessed. These are summarized in Table 2 and are explained below. If two instruments or techniques are used to measure the same variable on a continuous scale, Bland Altman diagrams can be used to estimate compliance. This diagram is a diagram of the difference between the two measures (Y axis) compared to the average of the two measures (X axis). It therefore offers a graphical representation of the distortion (average difference between the two observers or techniques) with correspondence limits of 95%. The latter are given by the formula: the concordance between the measures refers to the degree of concordance between two (or more) sets of measures. Statistical methods used to verify compliance are used to assess variability between assessors or to decide whether one technique for measuring one variable can replace another. In this article, we look at the statistical concordance levels for different types of data and discuss the differences between these and those that allow correlation to be assessed. Qureshi et al. compared the degree of prostate adenocarnoma assessed by seven pathologists with a standard system (Gleason score). The correspondence between each pathologist and the initial report and between pathologist couples was determined with Cohen`s kappa. That is a useful example. However, we believe that Gleasons Score is an ordinal variable, if weighted kappa would have been a more appropriate choice Dunet V, Klein R, Allenbach G, Renaud J, deKamp R, Prior J.\n\nQuantification of myocardial blood flow by Rb-82 PET/cardiac CT: a detailed reproducibility study between two semi-automatic analysis programs. J Nucl Cardiol. 2015. doi:10.1007/s12350-015-0151-2. Consider a situation in which we would like to evaluate the adequacy between hemoglobin measurements (in g/dl) with a hemoglobinometer on the hospital bed and the formal photometric laboratory technique in ten people [Table 3]. The Bland Altman diagram for this data shows the difference between the two methods for each person [Figure 1]. The mean difference between the values is 1.07 g/dl (with a standard deviation of 0.36 g/dL) and the 95% match limits are 0.35 to 1.79. This means that the hemoglobin level measured by a given person`s photometry can vary from 0.35 g/dl greater than 1.79 g/dl measured by photometry (this is the case for 95% of people; for 5% of individuals, variations could be outside these limits). This obviously means that the two techniques cannot be used as substitutes. It is important that there is no single criterion for acceptable compliance limits; This is a clinical decision that depends on the variables to be measured. Another method of visually assessing the conformity of two tests is to establish a scatter plot of the results of the first test against the results of the second test.\n\nIf the two tests have a good match, we should expect the points to fall on or near the 45° line (i.e. y = x); Deviations from this line would indicate a mismatch. Although the pearson correlation coefficient ρ can be used to assess the strength of a linear relationship between the results of two tests, the Pearson correlation is not an appropriate way to assess conformity: while it is true that test results that match well have a strong pearson correlation, the opposite is not always true, as will be illustrated later…\n\nCategories: Uncategorized"
]
| [
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.93375957,"math_prob":0.9373198,"size":3517,"snap":"2021-31-2021-39","text_gpt3_token_len":746,"char_repetition_ratio":0.12183319,"word_repetition_ratio":0.003552398,"special_character_ratio":0.2052886,"punctuation_ratio":0.085403726,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97210836,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-26T16:47:37Z\",\"WARC-Record-ID\":\"<urn:uuid:178fc7b5-4b14-4199-b147-1052bab80f9a>\",\"Content-Length\":\"15086\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d41102f1-a98b-4104-9d81-86cedbdc6f7f>\",\"WARC-Concurrent-To\":\"<urn:uuid:c64285f9-56a7-42c9-9a05-57c9fb4310cc>\",\"WARC-IP-Address\":\"77.111.240.145\",\"WARC-Target-URI\":\"http://justframes.dk/?p=9943\",\"WARC-Payload-Digest\":\"sha1:I44D5YGATSS4PYXFLDO52QPIIX4ISJF5\",\"WARC-Block-Digest\":\"sha1:23YKGFTOJLDBK3JYBQ37HK64HMSIN2RC\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780057882.56_warc_CC-MAIN-20210926144658-20210926174658-00526.warc.gz\"}"} |
https://de.maplesoft.com/support/help/maple/view.aspx?path=tensor(deprecated)%2Fd2metric | [
"",
null,
"tensor(deprecated)/d2metric - Maple Help\n\ntensor\n\n d2metric\n compute the second partial derivatives of the covariant metric tensor components",
null,
"Calling Sequence d2metric( d1g, coord)",
null,
"Parameters\n\n d1g - first partials of the covariant metric tensor coord - list of names representing the coordinate variables",
null,
"Description\n\nImportant: The tensor package has been deprecated. Use the superseding packages DifferentialGeometry and Physics instead.\n\n • Given the coordinate variables, coord, and the first partials of the covariant metric tensor, d1g, d2metric( d1g, coord ) computes the second partial derivatives of g, which will be a new tensor_type of rank four and which uses the d2met indexing function provided by the tensor package to implement the symmetry in the indices of the second partials.\n • d1g must be a rank 3 tensor_type with character [-1,-1,-1] and must use the cf1 indexing function for its component array. It is recommended that d1g be computed using tensor[d1metric].\n • The extra index due to differentiation is of covariant character, by convention. Thus, the index_char field of the result is [-1, -1, -1, -1].\n • Indexing Function: The d2met indexing function is meant to index the second partials of the covariant metric tensor components. Specifically, it implements the symmetry in the first and second indices (due to the symmetry in the metric components themselves) and the symmetry in the third and fourth indices (due to the symmetry of the mixed partial derivatives) of a four index quantity.\n • Simplification: This routine uses the tensor/d2metric/simp routine for simplification purposes. The simplification routine is applied to each component of result after it is computed. By default, tensor/d2metric/simp is initialized to the tensor/simp routine. It is recommended that the tensor/d2metric/simp routine be customized to suit the needs of the particular problem. For example, if the metric component functions are not fully known but are known to satisfy certain differential equations, the tensor/d2metric/simp routine could be used to make substitutions for the partial derivatives of these functions.\n • For computing the partial derivatives of the components of an arbitrary tensor, use tensor[partial_diff].\n • This function is part of the tensor package, and so can be used in the form d2metric(..) only after performing the command with(tensor) or with(tensor, d2metric). The function can always be accessed in the long form tensor[d2metric](..).",
null,
"Examples\n\nImportant: The tensor package has been deprecated. Use the superseding packages DifferentialGeometry and Physics instead.\n\n > $\\mathrm{with}\\left(\\mathrm{tensor}\\right):$\n\nDefine the coordinate variables and the covariant components of the Schwarzchild metric.\n\n > $\\mathrm{coord}≔\\left[t,r,\\mathrm{θ},\\mathrm{φ}\\right]:$\n > $\\mathrm{g_compts}≔\\mathrm{array}\\left(\\mathrm{symmetric},\\mathrm{sparse},1..4,1..4\\right):$\n > ${\\mathrm{g_compts}}_{1,1}≔1-\\frac{2m}{r}:$${\\mathrm{g_compts}}_{2,2}≔-\\frac{1}{{\\mathrm{g_compts}}_{1,1}}:$\n > ${\\mathrm{g_compts}}_{3,3}≔-{r}^{2}:$${\\mathrm{g_compts}}_{4,4}≔-{r}^{2}{\\mathrm{sin}\\left(\\mathrm{th}\\right)}^{2}:$\n > $g≔\\mathrm{create}\\left(\\left[-1,-1\\right],\\mathrm{eval}\\left(\\mathrm{g_compts}\\right)\\right)$\n ${g}{≔}{table}{}\\left(\\left[{\\mathrm{index_char}}{=}\\left[{-1}{,}{-1}\\right]{,}{\\mathrm{compts}}{=}\\left[\\begin{array}{cccc}{1}{-}\\frac{{2}{}{m}}{{r}}& {0}& {0}& {0}\\\\ {0}& {-}\\frac{{1}}{{1}{-}\\frac{{2}{}{m}}{{r}}}& {0}& {0}\\\\ {0}& {0}& {-}{{r}}^{{2}}& {0}\\\\ {0}& {0}& {0}& {-}{{r}}^{{2}}{}{{\\mathrm{sin}}{}\\left({\\mathrm{th}}\\right)}^{{2}}\\end{array}\\right]\\right]\\right)$ (1)\n\nCompute the first partials of g:\n\n > $\\mathrm{d1g}≔\\mathrm{d1metric}\\left(g,\\mathrm{coord}\\right):$\n\nCompute the second partials of g:\n\n > $\\mathrm{d2g}≔\\mathrm{d2metric}\\left(\\mathrm{d1g},\\mathrm{coord}\\right):$\n\nAll indices are covariant\n\n > $\\mathrm{get_char}\\left(\\mathrm{d2g}\\right)$\n $\\left[{-1}{,}{-1}{,}{-1}{,}{-1}\\right]$ (2)\n > $\\mathrm{compts}≔\\mathrm{get_compts}\\left(\\mathrm{d2g}\\right):$\n\nThe symmetries in the indices are implemented with indexing:\n\n > $\\mathrm{op}\\left(1,\\mathrm{op}\\left(\\mathrm{compts}\\right)\\right)$\n ${\\mathrm{d2met}}$ (3)\n\nExtract only the nonzero entries\n\n > map( proc(x) if compts[op(x)]<>0 then op(x)=compts[op(x)] else NULL end if end proc, [ indices(compts) ] );\n $\\left[\\left({4}{,}{4}{,}{2}{,}{2}\\right){=}{-}{2}{}{{\\mathrm{sin}}{}\\left({\\mathrm{th}}\\right)}^{{2}}{,}\\left({2}{,}{2}{,}{2}{,}{2}\\right){=}\\frac{{4}{}{m}}{{\\left({-}{r}{+}{2}{}{m}\\right)}^{{3}}}{,}\\left({1}{,}{1}{,}{2}{,}{2}\\right){=}{-}\\frac{{4}{}{m}}{{{r}}^{{3}}}{,}\\left({3}{,}{3}{,}{2}{,}{2}\\right){=}{-2}\\right]$ (4)"
]
| [
null,
"https://bat.bing.com/action/0",
null,
"https://de.maplesoft.com/support/help/maple/arrow_down.gif",
null,
"https://de.maplesoft.com/support/help/maple/arrow_down.gif",
null,
"https://de.maplesoft.com/support/help/maple/arrow_down.gif",
null,
"https://de.maplesoft.com/support/help/maple/arrow_down.gif",
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.72309774,"math_prob":0.9994399,"size":3595,"snap":"2023-40-2023-50","text_gpt3_token_len":1015,"char_repetition_ratio":0.15566695,"word_repetition_ratio":0.025477707,"special_character_ratio":0.235605,"punctuation_ratio":0.14809383,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9974961,"pos_list":[0,1,2,3,4,5,6,7,8,9,10],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-10-01T11:13:27Z\",\"WARC-Record-ID\":\"<urn:uuid:ec735454-39cb-4581-aec0-44b5844be436>\",\"Content-Length\":\"155121\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:dcdc1cd8-bd3e-4ce6-9ffa-483e926e9398>\",\"WARC-Concurrent-To\":\"<urn:uuid:787281b3-2d4b-4d49-9e69-874296ceb1fa>\",\"WARC-IP-Address\":\"199.71.183.28\",\"WARC-Target-URI\":\"https://de.maplesoft.com/support/help/maple/view.aspx?path=tensor(deprecated)%2Fd2metric\",\"WARC-Payload-Digest\":\"sha1:BNM2KKYBDNY4IMKVD62BAAKVM7ICXPJE\",\"WARC-Block-Digest\":\"sha1:5NYVNUTFX5VDDU6EWIC7NWADUNABRRHB\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233510888.64_warc_CC-MAIN-20231001105617-20231001135617-00598.warc.gz\"}"} |
https://toph.co/p/set-intersection-and-range | [
"# Practice on Toph\n\nParticipate in exhilarating programming contests, solve unique algorithm and data structure challenges and be a part of an awesome community.\n\n## Set, Intersection and Range\n\nLimits: 1s, 1.0 GB\n\nYou are given N sets of integers (S , S , S , … , S[N]) and Q queries.\n\nEach query has 4 integers X, Y, L, R. You just need to find :\n\n``````A = ( S[X] ) Intersection ( S[Y] )\n[ Intersection means taking the common elements of the two sets ]\n``````\n``````Ans = 0\nFor i = L to R\nif S[i] is equal to A\nAns = Ans + 1\nend if\nend For\n``````\n\nFor each query just print Ans.\n\n#### Input\n\nThe first line contains two integer N ( 1 ≤ N ≤ 105 ) and Q ( 1 ≤ Q ≤ 105 ).\n\nFor the next N lines, first integer denotes the number of elements in the ith set and then the elements of the set is given.\n\nNext there are Q queries with four integers which is described above.\n\nSize of any of the sets are within 105.\nAnd their summation is within 2 * 105.\nThe numbers in a set fits in 32 bit signed integer.\n\n#### Output\n\nFor each query print the corresponding Ans.\n\n#### Samples\n\nInputOutput\n```4 4\n3 1 2 3\n2 2 3\n3 2 3 4\n2 2 3\n1 3 1 4\n1 3 1 2\n1 3 2 4\n1 3 1 1```\n```2\n1\n2\n0\n```\n\nA Set is a collection of unique elements where the elements are always sorted.\n\nDiscussion\nSubmit"
]
| [
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.8582325,"math_prob":0.970933,"size":1166,"snap":"2019-13-2019-22","text_gpt3_token_len":360,"char_repetition_ratio":0.11703958,"word_repetition_ratio":0.0,"special_character_ratio":0.32590052,"punctuation_ratio":0.08646616,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98662,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-03-26T06:59:18Z\",\"WARC-Record-ID\":\"<urn:uuid:054a52fd-23ea-4d7f-a454-7110316d28f4>\",\"Content-Length\":\"9567\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e4635f3b-c0a3-430f-bb2b-41fa8beb009d>\",\"WARC-Concurrent-To\":\"<urn:uuid:56b0be36-feaa-4a38-85b4-f068a3ff430b>\",\"WARC-IP-Address\":\"45.33.70.243\",\"WARC-Target-URI\":\"https://toph.co/p/set-intersection-and-range\",\"WARC-Payload-Digest\":\"sha1:T3YS2HA2OGWKJ46BM7VMXJPWXMKCVJUO\",\"WARC-Block-Digest\":\"sha1:GBVZMMCVUJWGPWD436NFHIE66KV5S2GZ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-13/CC-MAIN-2019-13_segments_1552912204857.82_warc_CC-MAIN-20190326054828-20190326080828-00469.warc.gz\"}"} |
https://doc.ugene.net/wiki/exportword?pageId=65930667 | [
"Date: Wed, 17 Aug 2022 09:28:02 +0700 (NOVT) Message-ID: <[email protected]> Subject: Exported From Confluence MIME-Version: 1.0 Content-Type: multipart/related; boundary=\"----=_Part_100776_173072560.1660703282637\" ------=_Part_100776_173072560.1660703282637 Content-Type: text/html; charset=UTF-8 Content-Transfer-Encoding: quoted-printable Content-Location: file:///C:/exported.html Constraint Element Description\n\n# Constraint Element Description\n\nWhen you describe a constraint element the element_name consis= ts of two parts separated by two hyphens.\n\n=20\n```part1--p=\nart2```\n=20\n\nEach part represents one of the algorithm= s the constraint is imposed on.\n\nIf the algorithm is presented as one elem= ent on a schema (like ORFPattern) the algorithm=E2=80=99s part has format:\n\n=20\n```algorith=\nm_element_name.unit```\n=20\n\nIf the algorithm is presented as two subelements on a schema= (like <= em>RepeatsPrimer) the algorithm=E2=80=99s part h= as format:\n\n=20\n```algorith=\nm_element_name.left```\n=20\n\nor:\n\n=20\n```algorith=\nm_element_name.right```\n=20\n\ndepending on the subelement the constrain= t is imposed on.\n\nAlso you should specify the constraint ty= pe parameter (currently the only available type is distance):\n\n=20\n```type: di=\nstance;```\n=20\n\nAnd specify one of the distance types, for example:\n\n=20\n```distance=\n-type: end-to-start;```\n=20\n\nExample1: The= constraint is imposed on myPattern algorithm elements:\n\n=20\n```myORF.un=\nit--myPattern.unit {\n\ntype: distance;\ndistance-type: start-to-start;\n\n# Other parameters\n}```\n=20\n\nExample2: The= constraint is imposed on myRepeats algorithm = subelement:\n\n=20\n```myORF.un=\nit--myRepeats.left {\n\ntype: distance;\ndistance-type: start-to-end;\n\n# Other parameters\n}```\n=20\n\nThe available constraint elements are described in the Constraint Elements chapter.\n\n------=_Part_100776_173072560.1660703282637--"
]
| [
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.6321265,"math_prob":0.84173816,"size":1621,"snap":"2022-27-2022-33","text_gpt3_token_len":440,"char_repetition_ratio":0.14656772,"word_repetition_ratio":0.041884817,"special_character_ratio":0.30413324,"punctuation_ratio":0.18272425,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.996925,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-08-17T02:28:03Z\",\"WARC-Record-ID\":\"<urn:uuid:6a3cda8e-40fa-4ea8-878d-635861688e19>\",\"Content-Length\":\"11125\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d8b59a38-9cb1-4933-985f-178d60b5ed41>\",\"WARC-Concurrent-To\":\"<urn:uuid:c51d8427-8ab3-44fa-b967-c5da91f0d3b1>\",\"WARC-IP-Address\":\"95.170.150.116\",\"WARC-Target-URI\":\"https://doc.ugene.net/wiki/exportword?pageId=65930667\",\"WARC-Payload-Digest\":\"sha1:V4FCU7CPRETFBXSI6XL5STHTDOOQ3E66\",\"WARC-Block-Digest\":\"sha1:EFRZ6OEGH654X6ILVTFO2UHG5BCKCCDE\",\"WARC-Identified-Payload-Type\":\"message/rfc822\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-33/CC-MAIN-2022-33_segments_1659882572833.78_warc_CC-MAIN-20220817001643-20220817031643-00213.warc.gz\"}"} |
https://www.pip-academy.com/forex/what-is-a-pip-in-forex/intermediate- | [
"top of page\n\n# WHAT IS A \"PIP\" IN FOREX?\n\nAsk price is the lowest price the broker will pay to sell the instrument\n\nBid\n\nBid price is the highest price the broker will pay to purchase the instrument.\n\nEntry Price\n\nPip\n\nA pip is a standardised unit and is the smallest amount by which a currency quote can change. It is usually 0.0001 for U.S.-dollar related currency pairs, and 0.01 for Japanese yen related currency pairs\n\nPipette\n\nA pipette equals 1/10 (one tenth) of a pip and it represents a fraction of 1/100,000 (one in hundred thousand).\n\nA pip is a standardized unit and is the smallest amount by which a currency quote can change. It is usually 0.0001 for U.S.-dollar-related currency pairs and 0.01 for Japanese yen-related currency pairs since they go out to two decimal places.\n\nFor example, if the exchange rate of EUR/USD rises from 1.1343 to 1.1345, that 0.0002 rise in the value equals two pips.\n\nOn the other hand, since the Japanese yen pairs go out to two decimal places, the second decimal 0.01 is considered a pip.\n\nFor example, if the USD/JPY exchange rate rises from 109.74 to 110.74, it is said that USD/JPY has climbed by 100 pips.",
null,
"",
null,
"To illustrate it better, let’s assume that we have a USD/CAD quote of 1.3262. As discussed before, what this quote tells us is that for 1 U.S. dollar, you can receive about 1.3262 Canadian dollars. Therefore, if there were a ten-pip increase in this quote (to 1.3272), the value of the U.S. dollar would rise relative to the Canadian dollar, as now 1 U.S. dollar would allow you to receive slightly more Canadian dollar, which in this example would be equal to 1.3272. To put it differently, now, one Canadian dollar would give you less U.S. dollar (1/1.3272=0.7534 USD).\n\n- 1 USD = 1.3262 CAD\n\n- 1 USD = 1.3272 CAD\n\n- 1 CAD = 0.7534 USD\n\nThe effect that a ten-pip change has on the amount of base currency purchased, or of the pip value itself, depends on the units of base currency purchased.\n\nFor example, suppose a trader buys one standard lot (100,000 units) of U.S. dollar with Canadian dollar or, similarly, buy one lot of USD/CAD. In that case, the price paid will be 132,620 Canadian dollars ([1.3262] x 100,000), assuming the exchange rate for USD/CAD is 1.3262.\n\nIf the exchange rate for this pair experiences a ten-pip increase, and then the trader decides to conduct the same transaction, the price paid now would be 132,720 Canadian dollars ([1.3272] x 100,000) for the same one lot purchase of U.S. dollar.\n\nIn this case, due to a ten-pip rise in the exchange rate of USD/CAD, you have to pay 100 Canadian dollars (132,720-132,620) more to receive the same one lot of the U.S. dollar.\n\nThe pip value on a one-lot purchase of USD/CAD will be Can\\$ 10 per pip (100/10), which equals US\\$ 7.53 per pip.\n\nTherefore, if the trader executed a one-lot buy earlier before the ten pip rise in the exchange rate, he would have saved 100 Canadian dollars. However, if he decided to execute a one-lot buy after the ten-pip rise, he would have paid 100 Canadian dollars more for receiving the same one lot U.S. dollar.\n\nNow, if we assume the trader decides to sell a one-lot USD/CAD, it means that they are selling one lot of U.S. dollars and receiving Canadian dollars. Therefore, assuming the exchange rate remains the same as the example above, 132,620 Canadian dollars will be received for the sell of one lot or 100,000 US dollars.\n\nHowever, if the same scenario happens and the exchange rate jumps by a ten-pip, then the trader decides to sell USD/CAD, he will receive 132,720 Canadian dollars for a one-lot U.S. dollar.\n\nNow, if the trader purchases or sells a mini lot (10,000) instead of a full lot of U.S. dollar, a pip value would have been Cad\\$ 1 or US\\$ 0.75 for a mini lot transaction.\n\nTherefore, the pip value increases depending on the amount of the underlying currency (in this case, U.S. dollar) purchased.\n\nRead the next chapter for more details on pip value calculation for each currency pair and more examples on pip value calculation.\n\nSummary:\n\n1. A pip is a unit of measurement for currency movement and is the fourth decimal place in most currency pairs.\n\n2. For the Japanese yen-related pairs a pip is considered the second decimal place.\n\n3. The pip value depends on the amount of underlying currency that is purchased.\n\nbottom of page"
]
| [
null,
"https://static.wixstatic.com/media/7e82d1_ce67535b52b844e3b8e85a31b1ebd362~mv2.png/v1/fill/w_801,h_340,al_c,q_85,usm_0.66_1.00_0.01,enc_auto/Currency%20Pairs.png",
null,
"data:image/gif;base64,R0lGODlhAQABAIAAAP///wAAACH5BAEAAAAALAAAAAABAAEAAAICRAEAOw==",
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.9251275,"math_prob":0.9214966,"size":4401,"snap":"2022-40-2023-06","text_gpt3_token_len":1152,"char_repetition_ratio":0.14214237,"word_repetition_ratio":0.05154639,"special_character_ratio":0.28198138,"punctuation_ratio":0.14129353,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95824474,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,2,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-02-06T10:11:47Z\",\"WARC-Record-ID\":\"<urn:uuid:f3f38b7f-9d7b-4093-8b61-35d5c71ce54a>\",\"Content-Length\":\"429714\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9c63a15b-b093-4f64-9383-16c05c6778d8>\",\"WARC-Concurrent-To\":\"<urn:uuid:da72a105-6a0b-462f-a756-161c3c338d51>\",\"WARC-IP-Address\":\"34.149.87.45\",\"WARC-Target-URI\":\"https://www.pip-academy.com/forex/what-is-a-pip-in-forex/intermediate-\",\"WARC-Payload-Digest\":\"sha1:HXXUW7FS6B5QXSNCK3DXP6SZJAFGJYKX\",\"WARC-Block-Digest\":\"sha1:G4TJNBVHRNFOPORBHHZ6BKFW77ITBLIS\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764500334.35_warc_CC-MAIN-20230206082428-20230206112428-00325.warc.gz\"}"} |
http://www.ebooklibrary.org/articles/eng/Nonlinearity | [
"",
null,
"#jsDisabledContent { display:none; } My Account | Register | Help",
null,
"Flag as Inappropriate",
null,
"This article will be permanently flagged as inappropriate and made unaccessible to everyone. Are you certain this article is inappropriate? Excessive Violence Sexual Content Political / Social Email this Article Email Address:\n\n# Nonlinearity\n\nArticle Id: WHEBN0015829927\nReproduction Date:\n\n Title: Nonlinearity",
null,
"Author: World Heritage Encyclopedia Language: English Subject: Collection: Publisher: World Heritage Encyclopedia Publication Date:\n\n### Nonlinearity\n\nIn physics and other sciences, a nonlinear system is the opposite of a linear system, that is a system that does not satisfy the superposition principle, which means that the output is not directly proportional to the input.\n\nIn mathematics, a nonlinear system of equations is a set of simultaneous equations in which the unknowns (or the unknown functions in the case of differential equations) appear as variables of a polynomial of degree higher than one or in the argument of a function which is not a polynomial of degree one. In other words, in a nonlinear system of equations, the equation(s) to be solved cannot be written as a linear combination of the unknown variables or functions that appear in it (them). It does not matter if nonlinear known functions appear in the equations. In particular, a differential equation is linear if it is linear in terms of the unknown function and its derivatives, even if nonlinear in terms of the other variables appearing in it.\n\nTypically, the behavior of a nonlinear system is described by a nonlinear system of equations.\n\nNonlinear problems are of interest to engineers, physicists and mathematicians and many other scientists because most systems are inherently nonlinear in nature. As nonlinear equations are difficult to solve, non linear system are commonly approximated by linear equations (linearization). This works well up to some accuracy and some range for the input values, but some interesting phenomena such as chaos and singularities are hidden by linearization. It follows that some aspects of the behavior of a nonlinear system appear commonly to be chaotic, unpredictable or counterintuitive. Although such a chaotic behavior may resemble to a random behavior, it is absolutely not random.\n\nFor example, some aspects of the weather are seen to be chaotic, where simple changes in one part of the system produce complex effects throughout. This nonlinearity is one of the reasons why long term forecast are impossible with the current technology.\n\n## Definition\n\nIn mathematics, a linear function (or map) $f\\left(x\\right)$ is one which satisfies both of the following properties:\n\n• additivity (Superposition), $\\textstyle f\\left(x + y\\right)\\ = f\\left(x\\right)\\ + f\\left(y\\right);$\n• homogeneity, $\\textstyle f\\left(\\alpha x\\right)\\ = \\alpha f\\left(x\\right).$\n\n(Additivity implies homogeneity for any rational α, and, for continuous functions, for any real α. For a complex α, homogeneity does not follow from additivity; for example, an antilinear map is additive but not homogeneous.) The conditions of additivity and homogeneity are often combined in the superposition principle\n\n$f\\left(\\alpha x + \\beta y\\right) = \\alpha f\\left(x\\right) + \\beta f\\left(y\\right) \\,$\n\nAn equation written as\n\n$f\\left(x\\right) = C\\,$\n\nis called linear if $f\\left(x\\right)$ is a linear map (as defined above) and nonlinear otherwise. The equation is called homogeneous if $C = 0$.\n\nThe definition $f\\left(x\\right) = C$ is very general in that $x$ can be any sensible mathematical object (number, vector, function, etc.), and the function $f\\left(x\\right)$ can literally be any mapping, including integration or differentiation with associated constraints (such as boundary values). If $f\\left(x\\right)$ contains differentiation with respect to $x$, the result will be a differential equation.\n\n## Nonlinear algebraic equations\n\nMain article: Algebraic equation\nMain article: Systems of polynomial equations\n\nNonlinear algebraic equations, which are also called polynomial equations, are defined by equating polynomials to zero. For example,\n\n$x^2 + x - 1 = 0\\,.$\n\nFor a single polynomial equation, root-finding algorithms can be used to find solutions to the equation (i.e., sets of values for the variables that satisfy the equation). However, systems of algebraic equations are more complicated; their study is one motivation for the field of algebraic geometry, a difficult branch of modern mathematics. It is even difficult to decide if a given algebraic system has complex solutions (see Hilbert's Nullstellensatz). Nevertheless, in the case of the systems with a finite number of complex solutions, these systems of polynomial equations are now well understood and efficient methods exist for solving them.\n\n## Nonlinear recurrence relations\n\nA nonlinear recurrence relation defines successive terms of a sequence as a nonlinear function of preceding terms. Examples of nonlinear recurrence relations are the logistic map and the relations that define the various Hofstadter sequences. Nonlinear discrete models that represent a wide class of nonlinear recurrence relationships include the NARMAX (Nonlinear Autoregressive Moving Average with eXogenous inputs) model and the related nonlinear system identification and analysis procedures. These approaches can be used to study a wide class of complex nonlinear behaviors in the time, frequency, and spatio-temporal domains.\n\n## Nonlinear differential equations\n\nA system of differential equations is said to be nonlinear if it is not a linear system. Problems involving nonlinear differential equations are extremely diverse, and methods of solution or analysis are problem dependent. Examples of nonlinear differential equations are the Navier–Stokes equations in fluid dynamics and the Lotka–Volterra equations in biology.\n\nOne of the greatest difficulties of nonlinear problems is that it is not generally possible to combine known solutions into new solutions. In linear problems, for example, a family of linearly independent solutions can be used to construct general solutions through the superposition principle. A good example of this is one-dimensional heat transport with Dirichlet boundary conditions, the solution of which can be written as a time-dependent linear combination of sinusoids of differing frequencies; this makes solutions very flexible. It is often possible to find several very specific solutions to nonlinear equations, however the lack of a superposition principle prevents the construction of new solutions.\n\n### Ordinary differential equations\n\nFirst order ordinary differential equations are often exactly solvable by separation of variables, especially for autonomous equations. For example, the nonlinear equation\n\n$\\frac\\left\\{\\operatorname\\left\\{d\\right\\} u\\right\\}\\left\\{\\operatorname\\left\\{d\\right\\} x\\right\\} = -u^2\\,$\n\nwill easily yield u = (x + C)−1 as a general solution. The equation is nonlinear because it may be written as\n\n$\\frac\\left\\{\\operatorname\\left\\{d\\right\\} u\\right\\}\\left\\{\\operatorname\\left\\{d\\right\\} x\\right\\} + u^2=0\\,$\n\nand the left-hand side of the equation is not a linear function of u and its derivatives. Note that if the u2 term were replaced with u, the problem would be linear (the exponential decay problem).\n\nSecond and higher order ordinary differential equations (more generally, systems of nonlinear equations) rarely yield closed form solutions, though implicit solutions and solutions involving nonelementary integrals are encountered.\n\nCommon methods for the qualitative analysis of nonlinear ordinary differential equations include:\n\n### Partial differential equations\n\nThe most common basic approach to studying nonlinear partial differential equations is to change the variables (or otherwise transform the problem) so that the resulting problem is simpler (possibly even linear). Sometimes, the equation may be transformed into one or more ordinary differential equations, as seen in separation of variables, which is always useful whether or not the resulting ordinary differential equation(s) is solvable.\n\nAnother common (though less mathematic) tactic, often seen in fluid and heat mechanics, is to use scale analysis to simplify a general, natural equation in a certain specific boundary value problem. For example, the (very) nonlinear Navier-Stokes equations can be simplified into one linear partial differential equation in the case of transient, laminar, one dimensional flow in a circular pipe; the scale analysis provides conditions under which the flow is laminar and one dimensional and also yields the simplified equation.\n\nOther methods include examining the characteristics and using the methods outlined above for ordinary differential equations.\n\n### Pendula\n\nA classic, extensively studied nonlinear problem is the dynamics of a pendulum under influence of gravity. Using Lagrangian mechanics, it may be shown that the motion of a pendulum can be described by the dimensionless nonlinear equation\n\n$\\frac\\left\\{d^2 \\theta\\right\\}\\left\\{d t^2\\right\\} + \\sin\\left(\\theta\\right) = 0\\,$\n\nwhere gravity points \"downwards\" and $\\theta$ is the angle the pendulum forms with its rest position, as shown in the figure at right. One approach to \"solving\" this equation is to use $d\\theta/dt$ as an integrating factor, which would eventually yield\n\n$\\int \\frac\\left\\{d \\theta\\right\\}\\left\\{\\sqrt\\left\\{C_0 + 2 \\cos\\left(\\theta\\right)\\right\\}\\right\\} = t + C_1\\,$\n\nwhich is an implicit solution involving an elliptic integral. This \"solution\" generally does not have many uses because most of the nature of the solution is hidden in the nonelementary integral (nonelementary even if $C_0 = 0$).\n\nAnother way to approach the problem is to linearize any nonlinearities (the sine function term in this case) at the various points of interest through Taylor expansions. For example, the linearization at $\\theta = 0$, called the small angle approximation, is\n\n$\\frac\\left\\{d^2 \\theta\\right\\}\\left\\{d t^2\\right\\} + \\theta = 0\\,$\n\nsince $\\sin\\left(\\theta\\right) \\approx \\theta$ for $\\theta \\approx 0$. This is a simple harmonic oscillator corresponding to oscillations of the pendulum near the bottom of its path. Another linearization would be at $\\theta = \\pi$, corresponding to the pendulum being straight up:\n\n$\\frac\\left\\{d^2 \\theta\\right\\}\\left\\{d t^2\\right\\} + \\pi - \\theta = 0\\,$\n\nsince $\\sin\\left(\\theta\\right) \\approx \\pi - \\theta$ for $\\theta \\approx \\pi$. The solution to this problem involves hyperbolic sinusoids, and note that unlike the small angle approximation, this approximation is unstable, meaning that $|\\theta|$ will usually grow without limit, though bounded solutions are possible. This corresponds to the difficulty of balancing a pendulum upright, it is literally an unstable state.\n\nOne more interesting linearization is possible around $\\theta = \\pi/2$, around which $\\sin\\left(\\theta\\right) \\approx 1$:\n\n$\\frac\\left\\{d^2 \\theta\\right\\}\\left\\{d t^2\\right\\} + 1 = 0.$\n\nThis corresponds to a free fall problem. A very useful qualitative picture of the pendulum's dynamics may be obtained by piecing together such linearizations, as seen in the figure at right. Other techniques may be used to find (exact) phase portraits and approximate periods.\n\n## Types of nonlinear behaviors\n\n• Classical chaos – the behavior of a system cannot be predicted.\n• Multistability – alternating between two or more exclusive states.\n• Aperiodic oscillations – functions that do not repeat values after some period (otherwise known as chaotic oscillations or chaos).\n• Amplitude death – any oscillations present in the system cease due to some kind of interaction with other system or feedback by the same system.\n• Solitons – self-reinforcing solitary waves"
]
| [
null,
"http://read.images.worldlibrary.org/App_Themes/wel-mem/images/logo.jpg",
null,
"http://read.images.worldlibrary.org/images/SmallBook.gif",
null,
"http://www.ebooklibrary.org/images/delete.jpg",
null,
"http://www.ebooklibrary.org/App_Themes/default/images/icon_new_window.gif",
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.91750574,"math_prob":0.9968216,"size":11913,"snap":"2020-10-2020-16","text_gpt3_token_len":2233,"char_repetition_ratio":0.16877991,"word_repetition_ratio":0.005592841,"special_character_ratio":0.17846051,"punctuation_ratio":0.09888438,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9995919,"pos_list":[0,1,2,3,4,5,6,7,8],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-04-01T17:40:37Z\",\"WARC-Record-ID\":\"<urn:uuid:ecedd2c6-75f7-4ffe-af60-717c26ca0638>\",\"Content-Length\":\"98661\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8f220489-6ee6-4602-8edd-cc71c4dffe7d>\",\"WARC-Concurrent-To\":\"<urn:uuid:aa79d036-7f53-4804-ad11-1990cdcdb756>\",\"WARC-IP-Address\":\"66.27.42.21\",\"WARC-Target-URI\":\"http://www.ebooklibrary.org/articles/eng/Nonlinearity\",\"WARC-Payload-Digest\":\"sha1:BMG64AEGJ5JTRZO3UVFQWUM73BJ6KCFC\",\"WARC-Block-Digest\":\"sha1:6BSKK7DRLGMFJQ4YHDKS4OYU2SJJ6PUS\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-16/CC-MAIN-2020-16_segments_1585370505826.39_warc_CC-MAIN-20200401161832-20200401191832-00027.warc.gz\"}"} |
https://www.colorhexa.com/bdfcc7 | [
"# #bdfcc7 Color Information\n\nIn a RGB color space, hex #bdfcc7 is composed of 74.1% red, 98.8% green and 78% blue. Whereas in a CMYK color space, it is composed of 25% cyan, 0% magenta, 21% yellow and 1.2% black. It has a hue angle of 129.5 degrees, a saturation of 91.3% and a lightness of 86.5%. #bdfcc7 color hex could be obtained by blending #ffffff with #7bf98f. Closest websafe color is: #ccffcc.\n\n• R 74\n• G 99\n• B 78\nRGB color chart\n• C 25\n• M 0\n• Y 21\n• K 1\nCMYK color chart\n\n#bdfcc7 color description : Very soft lime green.\n\n# #bdfcc7 Color Conversion\n\nThe hexadecimal color #bdfcc7 has RGB values of R:189, G:252, B:199 and CMYK values of C:0.25, M:0, Y:0.21, K:0.01. Its decimal value is 12451015.\n\nHex triplet RGB Decimal bdfcc7 `#bdfcc7` 189, 252, 199 `rgb(189,252,199)` 74.1, 98.8, 78 `rgb(74.1%,98.8%,78%)` 25, 0, 21, 1 129.5°, 91.3, 86.5 `hsl(129.5,91.3%,86.5%)` 129.5°, 25, 98.8 ccffcc `#ccffcc`\nCIE-LAB 93.694, -29.825, 19.126 66.103, 84.561, 66.869 0.304, 0.389, 84.561 93.694, 35.43, 147.329 93.694, -31.181, 33.399 91.957, -32.612, 21.256 10111101, 11111100, 11000111\n\n# Color Schemes with #bdfcc7\n\n• #bdfcc7\n``#bdfcc7` `rgb(189,252,199)``\n• #fcbdf2\n``#fcbdf2` `rgb(252,189,242)``\nComplementary Color\n• #d3fcbd\n``#d3fcbd` `rgb(211,252,189)``\n• #bdfcc7\n``#bdfcc7` `rgb(189,252,199)``\n• #bdfce7\n``#bdfce7` `rgb(189,252,231)``\nAnalogous Color\n• #fcbdd3\n``#fcbdd3` `rgb(252,189,211)``\n• #bdfcc7\n``#bdfcc7` `rgb(189,252,199)``\n• #e7bdfc\n``#e7bdfc` `rgb(231,189,252)``\nSplit Complementary Color\n• #fcc7bd\n``#fcc7bd` `rgb(252,199,189)``\n• #bdfcc7\n``#bdfcc7` `rgb(189,252,199)``\n• #c7bdfc\n``#c7bdfc` `rgb(199,189,252)``\n• #f2fcbd\n``#f2fcbd` `rgb(242,252,189)``\n• #bdfcc7\n``#bdfcc7` `rgb(189,252,199)``\n• #c7bdfc\n``#c7bdfc` `rgb(199,189,252)``\n• #fcbdf2\n``#fcbdf2` `rgb(252,189,242)``\n• #74f989\n``#74f989` `rgb(116,249,137)``\n• #8cfa9e\n``#8cfa9e` `rgb(140,250,158)``\n• #a5fbb2\n``#a5fbb2` `rgb(165,251,178)``\n• #bdfcc7\n``#bdfcc7` `rgb(189,252,199)``\n• #d5fddc\n``#d5fddc` `rgb(213,253,220)``\n• #eefef0\n``#eefef0` `rgb(238,254,240)``\n• #ffffff\n``#ffffff` `rgb(255,255,255)``\nMonochromatic Color\n\n# Alternatives to #bdfcc7\n\nBelow, you can see some colors close to #bdfcc7. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #c3fcbd\n``#c3fcbd` `rgb(195,252,189)``\n• #befcbd\n``#befcbd` `rgb(190,252,189)``\n• #bdfcc2\n``#bdfcc2` `rgb(189,252,194)``\n• #bdfcc7\n``#bdfcc7` `rgb(189,252,199)``\n• #bdfccc\n``#bdfccc` `rgb(189,252,204)``\n• #bdfcd2\n``#bdfcd2` `rgb(189,252,210)``\n• #bdfcd7\n``#bdfcd7` `rgb(189,252,215)``\nSimilar Colors\n\n# #bdfcc7 Preview\n\nThis text has a font color of #bdfcc7.\n\n``<span style=\"color:#bdfcc7;\">Text here</span>``\n#bdfcc7 background color\n\nThis paragraph has a background color of #bdfcc7.\n\n``<p style=\"background-color:#bdfcc7;\">Content here</p>``\n#bdfcc7 border color\n\nThis element has a border color of #bdfcc7.\n\n``<div style=\"border:1px solid #bdfcc7;\">Content here</div>``\nCSS codes\n``.text {color:#bdfcc7;}``\n``.background {background-color:#bdfcc7;}``\n``.border {border:1px solid #bdfcc7;}``\n\n# Shades and Tints of #bdfcc7\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, #000902 is the darkest color, while #f5fff7 is the lightest one.\n\n• #000902\n``#000902` `rgb(0,9,2)``\n• #011c05\n``#011c05` `rgb(1,28,5)``\n• #022f09\n``#022f09` `rgb(2,47,9)``\n• #03410d\n``#03410d` `rgb(3,65,13)``\n• #045411\n``#045411` `rgb(4,84,17)``\n• #056714\n``#056714` `rgb(5,103,20)``\n• #067a18\n``#067a18` `rgb(6,122,24)``\n• #068c1c\n``#068c1c` `rgb(6,140,28)``\n• #079f1f\n``#079f1f` `rgb(7,159,31)``\n• #08b223\n``#08b223` `rgb(8,178,35)``\n• #09c527\n``#09c527` `rgb(9,197,39)``\n``#0ad72a` `rgb(10,215,42)``\n• #0bea2e\n``#0bea2e` `rgb(11,234,46)``\n• #14f438\n``#14f438` `rgb(20,244,56)``\n• #27f548\n``#27f548` `rgb(39,245,72)``\n• #3af658\n``#3af658` `rgb(58,246,88)``\n• #4cf767\n``#4cf767` `rgb(76,247,103)``\n• #5ff877\n``#5ff877` `rgb(95,248,119)``\n• #72f987\n``#72f987` `rgb(114,249,135)``\n• #85f997\n``#85f997` `rgb(133,249,151)``\n• #97faa7\n``#97faa7` `rgb(151,250,167)``\n• #aafbb7\n``#aafbb7` `rgb(170,251,183)``\n• #bdfcc7\n``#bdfcc7` `rgb(189,252,199)``\n• #d0fdd7\n``#d0fdd7` `rgb(208,253,215)``\n• #e3fee7\n``#e3fee7` `rgb(227,254,231)``\n• #f5fff7\n``#f5fff7` `rgb(245,255,247)``\nTint Color Variation\n\n# Tones of #bdfcc7\n\nA tone is produced by adding gray to any pure hue. In this case, #dadfdb is the less saturated color, while #baffc5 is the most saturated one.\n\n``#dadfdb` `rgb(218,223,219)``\n• #d8e1d9\n``#d8e1d9` `rgb(216,225,217)``\n• #d5e4d7\n``#d5e4d7` `rgb(213,228,215)``\n• #d2e7d5\n``#d2e7d5` `rgb(210,231,213)``\n• #d0e9d4\n``#d0e9d4` `rgb(208,233,212)``\n• #cdecd2\n``#cdecd2` `rgb(205,236,210)``\n• #caefd0\n``#caefd0` `rgb(202,239,208)``\n• #c8f1ce\n``#c8f1ce` `rgb(200,241,206)``\n• #c5f4cc\n``#c5f4cc` `rgb(197,244,204)``\n• #c2f7cb\n``#c2f7cb` `rgb(194,247,203)``\n• #c0f9c9\n``#c0f9c9` `rgb(192,249,201)``\n• #bdfcc7\n``#bdfcc7` `rgb(189,252,199)``\n• #baffc5\n``#baffc5` `rgb(186,255,197)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #bdfcc7 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":0.5070954,"math_prob":0.6637554,"size":3705,"snap":"2020-24-2020-29","text_gpt3_token_len":1661,"char_repetition_ratio":0.1269927,"word_repetition_ratio":0.011070111,"special_character_ratio":0.51983804,"punctuation_ratio":0.23404256,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9680106,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-06-04T12:30:39Z\",\"WARC-Record-ID\":\"<urn:uuid:b691a2e5-e83e-4e6c-8862-86d45015af9a>\",\"Content-Length\":\"36339\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:41166f71-95f6-4d2d-b34a-13a94ccd7293>\",\"WARC-Concurrent-To\":\"<urn:uuid:6b7b3c1d-adba-43be-83f0-dd7c3e0bc418>\",\"WARC-IP-Address\":\"178.32.117.56\",\"WARC-Target-URI\":\"https://www.colorhexa.com/bdfcc7\",\"WARC-Payload-Digest\":\"sha1:HVYHZGM7ZBODYYMF4EUX47XQWH5TJIK3\",\"WARC-Block-Digest\":\"sha1:NZMNMNBCMPHBWMGSGCMGQBYTCNH25BY5\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-24/CC-MAIN-2020-24_segments_1590347439928.61_warc_CC-MAIN-20200604094848-20200604124848-00277.warc.gz\"}"} |
http://forums.wolfram.com/mathgroup/archive/2011/Oct/msg00196.html | [
"",
null,
"",
null,
"",
null,
"",
null,
"",
null,
"",
null,
"",
null,
"Re: How to simplify a finite sum plus the next element?\n\n• To: mathgroup at smc.vnet.net\n• Subject: [mg121936] Re: How to simplify a finite sum plus the next element?\n• From: \"Dr. Wolfgang Hintze\" <weh at snafu.de>\n• Date: Fri, 7 Oct 2011 04:44:09 -0400 (EDT)\n• Delivered-to: [email protected]\n• References: <j6joce\\$m02\\[email protected]>\n\n```No solution, but it is sometimes a good idea to use FullForm in order\nto see the internal structure of an expression.\nIn our case\n\nFullForm[Sum[x[i], {i, 1, n}] + x[n + 1]]\n\nPlus[Sum[x[i], List[i, 1, n]], x[Plus[1, n]]]\n\nYou can play a little with the internal structure like this:\n\ns = Sum[x[i], {i, 1, n}] + x[n + 1];\nls = List @@ s\n{Sum[x[i], {i, 1, n}], x[1 + n]}\n\nls[[1,1]]\nx[i]\n\nls[[1,2]]\n{i, 1, n}\n\nSum @@ Append[{ls[[1,1]]}, ls[[1,2]] /. n -> n + 1]\nSum[x[i], {i, 1, 1 + n}]\n\nWolfgang\n\n\"Kirill Müller\" <kirillmueller at gmail.com> schrieb im Newsbeitrag\nnews:j6joce\\$m02\\$1 at smc.vnet.net...\n\n> Hello,\n>\n> with Mathematica 8, the following formula does not get simplified as\n> expected:\n>\n> FullSimplify[Sum[x[i], {i, 1, n}] + x[n + 1]]\n>\n> I expect Sum[x[i], {i, 1, n+1}] as output. Instead, the following is\n> output:\n>\n> \\!\\(\n> \\*UnderoverscriptBox[\\(\\[Sum]\\), \\(i = 1\\), \\(n\\)]\\(x[i]\\)\\) +\n> x[1 + n]\n>\n> I have already tried various assumptions for FullSimplify -- no luck.\n> What am I doing wrong here?\n>\n>\n> Cheers\n>\n> Kirill\n>\n\n```\n\n• Prev by Date: Re: average of the consecutive coin tosses\n• Next by Date: Re: average of the consecutive coin tosses\n• Previous by thread: How to simplify a finite sum plus the next element?\n• Next by thread: list equation"
]
| [
null,
"http://forums.wolfram.com/mathgroup/images/head_mathgroup.gif",
null,
"http://forums.wolfram.com/mathgroup/images/head_archive.gif",
null,
"http://forums.wolfram.com/mathgroup/images/numbers/2.gif",
null,
"http://forums.wolfram.com/mathgroup/images/numbers/0.gif",
null,
"http://forums.wolfram.com/mathgroup/images/numbers/1.gif",
null,
"http://forums.wolfram.com/mathgroup/images/numbers/1.gif",
null,
"http://forums.wolfram.com/mathgroup/images/search_archive.gif",
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.6737181,"math_prob":0.9548869,"size":1386,"snap":"2023-40-2023-50","text_gpt3_token_len":507,"char_repetition_ratio":0.1251809,"word_repetition_ratio":0.1,"special_character_ratio":0.4105339,"punctuation_ratio":0.22590362,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9965987,"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\":\"2023-12-07T13:21:02Z\",\"WARC-Record-ID\":\"<urn:uuid:08900f32-3165-4975-b2fd-4cf7b8e38d54>\",\"Content-Length\":\"45027\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ee049771-1ef0-4450-84c3-18763a1d1bfd>\",\"WARC-Concurrent-To\":\"<urn:uuid:05df19f5-bc9f-4a19-b3d8-ec0c40c3046c>\",\"WARC-IP-Address\":\"140.177.9.73\",\"WARC-Target-URI\":\"http://forums.wolfram.com/mathgroup/archive/2011/Oct/msg00196.html\",\"WARC-Payload-Digest\":\"sha1:VEGTO6FOEZGSK3RRX3OD5GE76OREB7UH\",\"WARC-Block-Digest\":\"sha1:PLLWAXN7TMFWKCXR5UBSKX4ZDETTZDFX\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100674.56_warc_CC-MAIN-20231207121942-20231207151942-00308.warc.gz\"}"} |
https://tantalum.academickids.com/encyclopedia/index.php/Student%27s_t-distribution | [
"# Student's t-distribution\n\nTemplate:Probability distribution In probability and statistics, the t-distribution or Student's distribution arises in the problem of estimating the mean of a normally distributed population when the sample size is small. It is the basis of the popular Student's t-tests for the statistical significance of the difference between two sample means, and for confidence intervals for the difference between two population means.\n\nThe derivation of the t-distribution was first published in 1908 by William Sealey Gosset, while he worked at a Guinness brewery in Dublin. He was not allowed to publish under his own name, so the paper was written under the pseudonym Student. The t-test and the associated theory became well-known through the work of R.A. Fisher, who called the distribution \"Student's distribution\".\n\nStudent's distribution arises when (as in nearly all practical statistical work) the population standard deviation is unknown and has to be estimated from the data. Textbook problems treating the standard deviation as if it were known are of two kinds: (1) those in which the sample size is so large that one may treat a data-based estimate of the variance as if it were certain, and (2) those that illustrate mathematical reasoning, in which the problem of estimating the standard deviation is temporarily ignored because that is not the point that the author or instructor is then explaining.\n\n Contents\n\n## Related distributions\n\n• [itex]Y \\sim \\mathrm{F}(\\nu_1 = 1, \\nu_2 = \\nu)[itex] is a F-distribution if [itex]Y = X^2[itex] and [itex]X \\sim \\mathrm{t}(\\nu)[itex] is a Student's t-distribution.\n• [itex]Y \\sim N(0,1)[itex] is a normal distribution as [itex]Y = \\lim_{\\nu \\to \\infty} X[itex] where [itex]X \\sim \\mathrm{t}(\\nu)[itex].\n• [itex]X \\sim \\mathrm{Cauchy}(0,1)[itex] is a Cauchy distribution if [itex]X \\sim \\mathrm{t}(\\nu = 1)[itex].\n\n## How Student's t-distribution arises\n\nSuppose X1, ..., Xn are independent random variables that are normally distributed with expected value μ and variance σ2. Let\n\n[itex]\\overline{X}_n=(X_1+\\cdots+X_n)/n[itex]\n\nbe the \"sample mean\", and\n\n[itex]S_n^2=\\frac{1}{n-1}\\sum_{i=1}^n\\left(X_i-\\overline{X}_n\\right)^2[itex]\n\nbe the \"sample variance\". It is readily shown that the estimate of the mean is represented by Z where\n\n[itex]Z=\\frac{\\overline{X}_n-\\mu}{\\sigma/\\sqrt{n}}[itex]\n\nis normally distributed with mean 0 and variance 1. Gosset studied a related quantity,\n\n[itex]T=\\frac{\\overline{X}_n-\\mu}{S_n/\\sqrt{n}}[itex]\n\nand showed that T has the probability density function\n\n[itex]f(t) = \\frac{\\Gamma((\\nu+1)/2)}{\\sqrt{\\nu\\pi\\,}\\,\\Gamma(\\nu/2)} (1+t^2/\\nu)^{-(\\nu+1)/2}[itex]\n\nwith ν equal to n − 1. The distribution of T is now called the t-distribution. The parameter ν is conventionally called the number of degrees of freedom. The distribution depends on ν, but not μ or σ; the lack of dependence on μ and σ is what makes the t-distribution important in both theory and practice.\n\n## How Student's t-distribution is used\n\nThe interval whose endpoints are\n\n[itex]\\overline{X}_n\\pm A\\frac{S_n}{\\sqrt{n}}[itex]\n\nwhere A is an appropriate percentage-point of the t-distribution, is a confidence interval for μ. Therefore, if we find the mean of a set of observations that we can reasonably expect to have a normal distribution, we can use the t-distribution to examine whether the confidence limits on that mean include some theoretically predicted value - such as the value predicted on a null hypothesis.\n\nIt is this result that is used in the Student's t-tests: since the difference between the means of samples from two normal distributions is itself distributed normally, the t-distribution can be used to examine whether that difference can reasonably be supposed to be zero.\n\nIf the data is normally distributed, the one-sided (1 − a)-upper confidence limit (UCL) of the mean, can be calculated using the following equation:\n\n[itex]UCL _{1-a} = \\overline{X}+{t_{a,n-1}}{S}/{\\sqrt{n}}[itex]\n\nThe resulting UCL will be the the greatest average value that will occur for a given confidence interval and population size.\n\nA number of other statistics can be shown to have t-distributions for samples of moderate size under null hypotheses that are of interest, so that the t-distribution forms the basis for significance tests in other situations as well as when examining the differences between means. For example, the distribution of Spearman's rank correlation coefficient, rho, in the null case (zero correlation) is well approximated by the t distribution for sample sizes above about 20.\n\nSee prediction interval for another example of the use of this distribution.\n\n## Further theory\n\nGosset's result can be stated more generally. (See, for example, Hogg and Craig, Sections 4.4 and 4.8.) Let Z have a normal distribution with mean 0 and variance 1. Let V have a chi-square distribution with ν degrees of freedom. Further suppose that Z and V are independent (see Cochran's theorem). Then the ratio\n\n[itex] \\frac{Z}{\\sqrt{V/\\nu\\ }} [itex]\n\nhas a t-distribution with ν degrees of freedom.\n\nFor a t-distribution with ν degrees of freedom, the expected value is 0, and its variance is ν/(ν − 2) if ν > 2. The skewness is 0 and the kurtosis is 6/(ν − 4) if ν > 4.\n\nThe cumulative distribution function is given by an incomplete beta function,\n\n[itex]\\int_{-\\infty}^t f(u)\\,du = \\left\\{\n\n\\begin{matrix} 1 - \\frac{1}{2} I_x(\\nu/2,1/2) & \\mbox{if}\\quad t > 0, \\\\ \\\\ \\frac{1}{2} I_x(\\nu/2,1/2) & \\mbox{otherwise}, \\end{matrix}\\right.[itex]\n\nwith\n\n[itex]x = \\frac{1}{1+t^2/\\nu}.[itex]\n\nThe t-distribution is related to the F-distribution as follows: the square of a value of t with ν degrees of freedom is distributed as F with 1 and ν degrees of freedom.\n\nThe overall shape of the probability density function of the t-distribution resembles the bell shape of a normally distributed variable with mean 0 and variance 1, except that it is a bit lower and wider. As the number of degrees of freedom grows, the t-distribution approaches the normal distribution with mean 0 and variance 1.\n\nThe following images show the density of the t-distribution for increasing values of ν. The normal distribution is shown as a blue line for comparison. Note that the t-distribution (red line) becomes closer to the normal distribution as ν increases. For ν = 30 the t-distribution is almost the same as the normal distribution.\n\n Missing imageT_distribution_1df.png",
null,
"",
null,
"",
null,
"",
null,
"Missing imageT_distribution_30df.png\n\n## References\n\n• \"Student\" (W.S. Gosset) (1908) The probable error of a mean. Biometrika 6(1):1--25.\n• M. Abramowitz and I. A. Stegun, eds. (1972) Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables. New York: Dover. (See Section 26.7.)\n• R.V. Hogg and A.T. Craig (1978) Introduction to Mathematical Statistics. New York: Macmillan.\n\n• Art and Cultures\n• Countries of the World (http://www.academickids.com/encyclopedia/index.php/Countries)\n• Space and Astronomy"
]
| [
null,
"https://tantalum.academickids.com/encyclopedia/images/thumb/6/6d/240px-T_distribution_2df.png",
null,
"https://tantalum.academickids.com/encyclopedia/images/thumb/6/65/240px-T_distribution_3df.png",
null,
"https://tantalum.academickids.com/encyclopedia/images/thumb/9/9d/240px-T_distribution_5df.png",
null,
"https://tantalum.academickids.com/encyclopedia/images/thumb/8/89/240px-T_distribution_10df.png",
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.86536485,"math_prob":0.9878709,"size":7226,"snap":"2021-43-2021-49","text_gpt3_token_len":1854,"char_repetition_ratio":0.17875935,"word_repetition_ratio":0.019571295,"special_character_ratio":0.25089952,"punctuation_ratio":0.09593023,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99963236,"pos_list":[0,1,2,3,4,5,6,7,8],"im_url_duplicate_count":[null,1,null,1,null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-12-06T20:27:15Z\",\"WARC-Record-ID\":\"<urn:uuid:50b77776-5698-4584-9a5b-6630e2c62098>\",\"Content-Length\":\"37999\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9933681c-0456-49a1-bcf3-d4bd307e5e3c>\",\"WARC-Concurrent-To\":\"<urn:uuid:5322f8c0-0f46-4a51-b9ed-32fed3bf4449>\",\"WARC-IP-Address\":\"108.59.0.28\",\"WARC-Target-URI\":\"https://tantalum.academickids.com/encyclopedia/index.php/Student%27s_t-distribution\",\"WARC-Payload-Digest\":\"sha1:ONA7KI3QTY5XO5XLXB5SRLEHNO7DNVQP\",\"WARC-Block-Digest\":\"sha1:HXROXGSPHMFG5ZOIEOFRJ5C7RDZZWQZX\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964363312.79_warc_CC-MAIN-20211206194128-20211206224128-00314.warc.gz\"}"} |
https://developergang.com/lesson-plan-for-synthetic-division/?amp=1 | [
"# Lesson Plan for Synthetic Division\n\nThe complexities of the mathematical expressions are quite tough to comprehend. One of the challenging concepts of advanced math is synthetic division. Students of math often go through synthetic division to solve the polynomial expressions. Higher degree polynomial roots can be taken for about 3 degrees. The factors of the polynomials are much easier to make.\n\nThe quadratic function and roots are determined by the use of a synthetic division calculator. The function roots are massively valuable for going through the function and making it equal to zero. The quadratic function can be found through different methods. The quadratic function roots are then subjected to the processing of factoring.\n\n## Working of Synthetic Division\n\nBasically, it is known to be a powerful and amazing method to shortcut the long division. Whenever it comes to the long division, consider taking assistance of synthetic division. The long divisions make the calculations quite complex. It utilizes the coefficients of the expression and ensures to save much of the precious time. When you are about to divide any linear factor, you can utilize the synthetic division. It divides the expression or factor in x-a form.\n\nFor instance, you can use it for dividing the linear factor such as x + 4 or x-3 etc. The plain traditional long division is used when the polynomial has not had one in its leading coefficient. Synthetic substitution calculator is one of the most accessible ones. It requires a robust internet connection only. Hence, all the students can avail the services of synthetic division calculators with ease to solve the equations.\n\n## Evaluation of Polynomials\n\nThe synthetic method is the one that eases the determination of polynomial roots. Synthetic division calculator finds out the coefficient and zeros of numerator and denominator, respectively. The evaluation of the division of polynomials is carried out with the use of synthetic division.\n\nThe process of evaluating the polynomial is to divide the polynomials by the synthetic division. The plugging in helps in dealing with the remainder of the synthetic division. The binomial is often used to divide the polynomial function. The practice of the polynomial division by synthetic division will help in solving the factors correctly.\n\n## Online Synthetic Division Calculator\n\nThe synthetic division calculator is one of the most valuable online calculators that intend to figure out the polynomial’s quotient and reminder. The online calculator helps to get the answer to all the polynomial expressions. The synthetic division is a source to find out the functional value. The factoring process helps in finding out the quotient. The immediate solving of the polynomial equations helps the students to excel from others. Indeed, these help in boosting up confidence in algebra. The hesitation to algebra enhances when one is unable to solve the expressions in the right manner. The online calculator polishes the skills of students and guides them about the steps. Hence, the difficulty gets solved fabulously.\n\n## In a Nutshell\n\nA synthetic division calculator minimizes the risk of error that is much more common in the case of manual calculations. Algebra becomes easier when one uses the technological tools for it. Math is such a subject that demands the best use of focus and tremendous practice. The more the questions are solved with the right answers, the more interest will automatically develop in the students. According to the Factor Theorem, when the value of c in the x-c is zero, then it is a factor. Synthetic division calculator is quite easy to access and use. The user has to enter the divisor value in it along with the dividend value.\n\nWe also accepts guest post on Cybersecurity Write For Us category. you can check the page for more information."
]
| [
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.9176971,"math_prob":0.97653234,"size":3809,"snap":"2021-43-2021-49","text_gpt3_token_len":698,"char_repetition_ratio":0.18869908,"word_repetition_ratio":0.006557377,"special_character_ratio":0.17773694,"punctuation_ratio":0.08320951,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99035555,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-12-07T20:33:29Z\",\"WARC-Record-ID\":\"<urn:uuid:40b3deea-66bc-42f9-b7d2-1006ad8e8e80>\",\"Content-Length\":\"110302\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:099299ac-7b0c-4bf9-91d7-d743661be335>\",\"WARC-Concurrent-To\":\"<urn:uuid:aad6c1e3-b9a9-4a8c-8000-cdc99c43b03a>\",\"WARC-IP-Address\":\"162.214.80.52\",\"WARC-Target-URI\":\"https://developergang.com/lesson-plan-for-synthetic-division/?amp=1\",\"WARC-Payload-Digest\":\"sha1:IZTPRYEAV6LAG54WCHMYPLYYIPSJLDWZ\",\"WARC-Block-Digest\":\"sha1:2K7L4VU2SYPGTLTPH7FEXHQTIJG6UTYZ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964363418.83_warc_CC-MAIN-20211207201422-20211207231422-00625.warc.gz\"}"} |
https://7puzzleblog.com/305/ | [
"# DAY/DYDD/GIORNO/NAP 305:",
null,
"The Main Challenge\n\nAnother typical Mathelona challenge taken from our pocket book of number puzzles.\n\nYour task is to make all three lines work out arithmetically by filling the 12 gaps with the following 12 numbers. Can you successfully complete this?\n\n1 1 2 2 2 3 4 4 5 6 7 8\n\n◯ + ◯ = 5 = ◯ – ◯\n◯ + ◯ = 10 = ◯ × ◯\n◯ + ◯ = 2 = ◯ ÷ ◯\n\nFull details of our Mathelona number puzzles can be found by clicking the link.",
null,
"The 7puzzle Challenge\n\nThe playing board of the 7puzzle game is a 7-by-7 grid containing 49 different numbers, ranging from 2 up to 84.\n\nThe 6th & 7th rows contain the following fourteen numbers:\n\n4 5 11 12 18 20 24 27 30 33 49 56 70 77\n\nHow many multiples of 7 are in the list?",
null,
"The Factors Challenge\n\nWhich of the following numbers are factors of 305?\n\n3 5 7 9 11 13 15",
null,
"The Mathematically Possible Challenge\n\nUsing 38 and 10 once each, with + – × ÷ available, which is the ONLY number it is possible to make from the list below?\n\n1 4 9 16 25 36 49 64 81 100\n\n#SquareNumbers",
null,
"The Target Challenge\n\nCan you arrive at 305 by inserting 5, 6, 7, 8 and 9 into the gaps below?\n\n• (◯×◯+◯–◯)×◯ = 305",
null,
"",
null,
"",
null,
""
]
| [
null,
"http://7puzzleblog.com/wp-content/uploads/2018/10/PaulRoundImage2-298x300.jpg",
null,
"http://7puzzleblog.com/wp-content/uploads/2017/06/7-newimage.jpg",
null,
"http://7puzzleblog.com/wp-content/uploads/2017/06/7-newimage.jpg",
null,
"http://7puzzleblog.com/wp-content/uploads/2017/06/7-newimage.jpg",
null,
"http://7puzzleblog.com/wp-content/uploads/2017/06/7-newimage.jpg",
null,
"http://7puzzleblog.com/wp-content/uploads/2018/10/PaulRoundImage2-298x300.jpg",
null,
"http://7puzzleblog.com/wp-content/uploads/2017/06/7-newimage.jpg",
null,
"http://7puzzleblog.com/wp-content/uploads/2018/01/7puzzleblog2-dnp-1-e1515248090297-300x100.jpg",
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.85414183,"math_prob":0.96553063,"size":1107,"snap":"2021-43-2021-49","text_gpt3_token_len":372,"char_repetition_ratio":0.117860384,"word_repetition_ratio":0.008695652,"special_character_ratio":0.33423668,"punctuation_ratio":0.08108108,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9507261,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-25T01:05:50Z\",\"WARC-Record-ID\":\"<urn:uuid:a144d5a0-0d86-4b05-afac-e79687744115>\",\"Content-Length\":\"29509\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4bfdc0a2-ec3f-4674-91fa-ff4e3de119f2>\",\"WARC-Concurrent-To\":\"<urn:uuid:716be341-d026-4597-bad9-b06702cf2a9c>\",\"WARC-IP-Address\":\"185.119.173.216\",\"WARC-Target-URI\":\"https://7puzzleblog.com/305/\",\"WARC-Payload-Digest\":\"sha1:K5ZIBLFXO7WNVSVVUIL4UXV32KWOK64N\",\"WARC-Block-Digest\":\"sha1:7FRANIJJ2DMYXLSJRV3PZXWZ67BT46BC\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323587608.86_warc_CC-MAIN-20211024235512-20211025025512-00348.warc.gz\"}"} |
https://docs.mendix.com/refguide7/add-date-function-calls | [
"Last update:\n\nAdding a time unit to a date. In all examples, the first input can be either a new dateTime (depicted in all examples), a variable of type DateTime, or an attribute of a domain entity of type DateTime.\n\nAdd a number of milliseconds to a date.\n\n### Input parameters\n\n• Initial date Type: DateTime\n• The amount of milliseconds to be added Type: Integer\n\n### Output\n\nA result of type DateTime that corresponds to the initial date plus x milliseconds.\n\nAdd a number of seconds to a date.\n\n### Input parameters\n\n• Initial date Type: DateTime\n• The amount of seconds to be added Type: Integer\n\n### Output\n\nA result of type DateTime that corresponds to the initial date plus x seconds.\n\nAdd a number of minutes to a date.\n\n### Input parameters\n\n• Initial date Type: DateTime\n• The amount of minutes to be added Type: Integer\n\n### Output\n\nA result of type DateTime that corresponds to the initial date plus x minutes.\n\nAdd a number of hours to a date.\n\n### Input parameters\n\n• Initial date Type: DateTime\n• The amount of hours to be added Type: Integer\n\n### Output\n\nA result of type DateTime that corresponds to the initial date plus x hours.\n\nAdd a number of days to a date. `addDays` uses the server’s calendar and `addDaysUTC` uses the UTC calendar.\n\n### Input parameters\n\n• Initial date Type: DateTime\n• The amount of days to be added Type: Integer\n\n### Output\n\nA result of type DateTime that corresponds to the initial date plus x days.\n\nAdd a number of weeks to a date. `addWeeks` uses the server’s calendar and `addWeeksUTC` uses the UTC calendar.\n\n### Input parameters\n\n• Initial date Type: DateTime\n• The amount of weeks to be added Type: Integer\n\n### Output\n\nA result of type DateTime that corresponds to the initial date plus x weeks.\n\nAdd a number of months to a date. `addMonths` uses the server’s calendar and `addMonthsUTC` uses the UTC calendar.\n\n### Input parameters\n\n• Initial date Type: DateTime\n• The amount of months to be added Type: Integer\n\n### Output\n\nA result of type DateTime that corresponds to the initial date plus x months.\n\nAdd a number of years to a date. `addYears` uses the server’s calendar and `addYearsUTC` uses the UTC calendar."
]
| [
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.5291366,"math_prob":0.87326735,"size":3500,"snap":"2020-45-2020-50","text_gpt3_token_len":1004,"char_repetition_ratio":0.17448513,"word_repetition_ratio":0.46921796,"special_character_ratio":0.29371428,"punctuation_ratio":0.15333334,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9759695,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-11-24T07:05:19Z\",\"WARC-Record-ID\":\"<urn:uuid:12327edd-ab97-412e-842b-b5be9ff81cff>\",\"Content-Length\":\"19513\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e9c1f9f4-157b-4f14-9caa-3724142f4ee6>\",\"WARC-Concurrent-To\":\"<urn:uuid:ce8a099c-968c-4154-9d77-a23674b32b82>\",\"WARC-IP-Address\":\"50.17.227.101\",\"WARC-Target-URI\":\"https://docs.mendix.com/refguide7/add-date-function-calls\",\"WARC-Payload-Digest\":\"sha1:QN4NZLXQILVHEPQ4YC3BK7UHJV3IZPQY\",\"WARC-Block-Digest\":\"sha1:MVSELUFTZVLETIYOQT5G6OSJMMZWA6XZ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141171126.6_warc_CC-MAIN-20201124053841-20201124083841-00598.warc.gz\"}"} |
https://nl.mathworks.com/matlabcentral/cody/problems/1895-count-ones/solutions/323757 | [
"Cody\n\n# Problem 1895. Count ones\n\nSolution 323757\n\nSubmitted on 26 Sep 2013 by J.R.! Menzinger\nThis solution is locked. To view this solution, you need to provide a solution of the same size or smaller.\n\n### Test Suite\n\nTest Status Code Input and Output\n1 Pass\n%% x = 1123421; y_correct = 3; assert(isequal(your_fcn_name(x),y_correct))\n\n2 Pass\n%% x = 1111; y_correct = 4; assert(isequal(your_fcn_name(x),y_correct))\n\n3 Pass\n%% x = 345450; y_correct = 0; assert(isequal(your_fcn_name(x),y_correct))\n\n### Community Treasure Hunt\n\nFind the treasures in MATLAB Central and discover how the community can help you!\n\nStart Hunting!"
]
| [
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.69012815,"math_prob":0.9532268,"size":729,"snap":"2020-45-2020-50","text_gpt3_token_len":202,"char_repetition_ratio":0.12689655,"word_repetition_ratio":0.0,"special_character_ratio":0.303155,"punctuation_ratio":0.10769231,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9653944,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-12-04T16:32:09Z\",\"WARC-Record-ID\":\"<urn:uuid:842f8d2d-6e7b-44ba-b49c-9b07b6937572>\",\"Content-Length\":\"78519\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4e918d65-49b8-4c59-8fa6-9ebff6fd4e89>\",\"WARC-Concurrent-To\":\"<urn:uuid:9008b42d-6089-4564-ac3b-656a47c1a12f>\",\"WARC-IP-Address\":\"23.5.129.95\",\"WARC-Target-URI\":\"https://nl.mathworks.com/matlabcentral/cody/problems/1895-count-ones/solutions/323757\",\"WARC-Payload-Digest\":\"sha1:LGKZ7L4TJGTTNZ3WMKM2PU4KWVZLVAPS\",\"WARC-Block-Digest\":\"sha1:ZF5VXKDNYOLNEQFRNZM6A5XSH37XIK3R\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141740670.93_warc_CC-MAIN-20201204162500-20201204192500-00666.warc.gz\"}"} |
https://de.mathworks.com/help/physmod/sps/ref/inductionmachinesquirrelcage.html | [
"Documentation\n\n# Induction Machine Squirrel Cage\n\nSquirrel-cage-rotor induction machine with per-unit or SI parameterization\n\n• Library:\n• Simscape / Electrical / Electromechanical / Asynchronous",
null,
"## Description\n\nThe Induction Machine Squirrel Cage block models a squirrel-cage-rotor induction machine with fundamental parameters expressed in per-unit or in the International System of Units (SI). A squirrel-cage-rotor induction machine is a type of induction machine. All stator connections are accessible on the block. Therefore, you can model soft-start regimes using a switch between wye and delta configurations. If you need access to the rotor windings, use the Induction Machine Wound Rotor block instead.\n\nConnect port ~1 to a three-phase circuit. To connect the stator in delta configuration, connect a Phase Permute block between ports ~1 and ~2. To connect the stator in wye configuration, connect port ~2 to a Grounded Neutral (Three-Phase) or a Floating Neutral (Three-Phase) block.\n\n### Induction Machine Initialization Using Load-Flow Target Values\n\nIf the block is in a network that is compatible with the frequency-time simulation mode, you can perform a load-flow analysis on the network. A load-flow analysis provides steady-state values that you can use to initialize the machine.\n\nFor more information, see Perform a Load-Flow Analysis Using Simscape Electrical and Frequency and Time Simulation Mode (Simscape). For an example that shows how initialize an induction machine using data from a load flow analysis, see Induction Motor Initialization with Loadflow.\n\n### Equations\n\nFor the SI implementation, the block converts the SI values that you enter in the dialog box to per-unit values for simulation. For information on the relationship between SI and per-unit machine parameters, see Per-Unit Conversion for Machine Parameters. For information on per-unit parameterization, see Per-Unit System of Units.\n\nThe induction machine equations are expressed with respect to a synchronous reference frame, defined by\n\n`${\\theta }_{e}\\left(t\\right)=\\underset{0}{\\overset{t}{\\int }}2\\pi {f}_{rated}dt,$`\n\nwhere frated is the value of the Rated electrical frequency parameter.\n\nThe Park transformation maps stator equations to a reference frame that is stationary with respect to the rated electrical frequency. The Park transformation is defined by\n\n`${P}_{s}=\\frac{2}{3}\\left[\\begin{array}{ccc}\\mathrm{cos}{\\theta }_{e}& \\text{cos}\\left({\\theta }_{e}-\\frac{2\\pi }{3}\\right)& \\text{cos}\\left({\\theta }_{e}+\\frac{2\\pi }{3}\\right)\\\\ -\\mathrm{sin}{\\theta }_{e}& -\\text{sin}\\left({\\theta }_{e}-\\frac{2\\pi }{3}\\right)& -\\text{sin}\\left({\\theta }_{e}+\\frac{2\\pi }{3}\\right)\\\\ \\frac{1}{2}& \\frac{1}{2}& \\frac{1}{2}\\end{array}\\right],$`\n\nwhere θe is the electrical angle.\n\nThe Park transformation is used to define the per-unit induction machine equations. The stator voltage equations are defined by\n\n`${v}_{ds}=\\frac{1}{{\\omega }_{base}}\\frac{d{\\psi }_{ds}}{dt}-\\omega {\\psi }_{qs}+{R}_{s}{i}_{ds},$`\n\n`${v}_{qs}=\\frac{1}{{\\omega }_{base}}\\frac{d{\\psi }_{qs}}{dt}+\\omega {\\psi }_{ds}+{R}_{s}{i}_{qs},$`\n\nand\n\n`${v}_{0s}=\\frac{1}{{\\omega }_{base}}\\frac{d{\\psi }_{0s}}{dt}+{R}_{s}{i}_{0s},$`\n\nwhere:\n\n• vds, vqs, and v0s are the d-axis, q-axis, and zero-sequence stator voltages, defined by\n\n$\\left[\\begin{array}{c}{v}_{ds}\\\\ {v}_{qs}\\\\ {v}_{0s}\\end{array}\\right]={P}_{s}\\left[\\begin{array}{c}{v}_{a}\\\\ {v}_{b}\\\\ {v}_{c}\\end{array}\\right].$\n\nva, vb, and vc are the stator voltages across ports ~1 and ~2.\n\n• ωbase is the per-unit base electrical speed.\n\n• ψds, ψqs, and ψ0s are the d-axis, q-axis, and zero-sequence stator flux linkages.\n\n• Rs is the stator resistance.\n\n• ids, iqs, and i0s are the d-axis, q-axis, and zero-sequence stator currents defined by\n\n$\\left[\\begin{array}{c}{i}_{ds}\\\\ {i}_{qs}\\\\ {i}_{0s}\\end{array}\\right]={P}_{s}\\left[\\begin{array}{c}{i}_{a}\\\\ {i}_{b}\\\\ {i}_{c}\\end{array}\\right].$\n\nia, ib, and ic are the stator currents flowing from port ~1 to port ~2.\n\nThe rotor voltage equations are defined by\n\n`${v}_{dr}=\\frac{1}{{\\omega }_{base}}\\frac{d{\\psi }_{dr}}{dt}-\\left(\\omega -{\\omega }_{r}\\right){\\psi }_{qr}+{R}_{rd}{i}_{dr}=0$`\n\nand\n\n`${v}_{qr}=\\frac{1}{{\\omega }_{base}}\\frac{d{\\psi }_{qr}}{dt}+\\left(\\omega -{\\omega }_{r}\\right){\\psi }_{dr}+{R}_{rd}{i}_{qr}=0,$`\n\nwhere:\n\n• vdr and vqr are the d-axis and q-axis rotor voltages.\n\n• ψdr and ψqr are the d-axis and q-axis rotor flux linkages.\n\n• ω is the per-unit synchronous speed. For a synchronous reference frame, the value is 1.\n\n• ωr is the per-unit mechanical rotational speed.\n\n• Rrd is the rotor resistance referred to the stator.\n\n• idr and iqr are the d-axis and q-axis rotor currents.\n\nThe stator flux linkage equations are defined by\n\n`${\\psi }_{ds}={L}_{ss}{i}_{ds}+{L}_{m}{i}_{dr},$`\n\n`${\\psi }_{qs}={L}_{ss}{i}_{qs}+{L}_{m}{i}_{qr},$`\n\nand\n\n`${\\psi }_{0s}={L}_{ss}{i}_{0s},$`\n\nwhere Lss is the stator self-inductance and Lm is the magnetizing inductance.\n\nThe rotor flux linkage equations are defined by\n\n`${\\psi }_{dr}={L}_{rrd}{i}_{dr}+{L}_{m}{i}_{ds}$`\n\nand\n\n`${\\psi }_{qr}={L}_{rrd}{i}_{qr}+{L}_{m}{i}_{qs},$`\n\nwhere Lrrd is the rotor self-inductance referred to the stator.\n\nThe rotor torque is defined by\n\n`$T={\\psi }_{ds}{i}_{qs}-{\\psi }_{qs}{i}_{ds}.$`\n\nThe stator self-inductance Lss, stator leakage inductance Lls, and magnetizing inductance Lm are related by\n\n`${L}_{ss}={L}_{ls}+{L}_{m}.$`\n\nThe rotor self-inductance Lrrd, rotor leakage inductance Llrd, and magnetizing inductance Lm are related by\n\n`${L}_{rrd}={L}_{lrd}+{L}_{m}.$`\n\nWhen a saturation curve is provided, the equations to determine the saturated magnetizing inductance as a function of magnetizing flux are:\n\n`${L}_{m_sat}=f\\left({\\psi }_{m}\\right)$`\n\n`${\\psi }_{m}=\\sqrt{{\\psi }_{dm}^{2}+{\\psi }_{qm}^{2}}$`\n\nFor no saturation, the equation reduces to\n\n`${L}_{m_sat}={L}_{m}$`\n\n### Plotting and Display Options\n\nYou can perform plotting and display actions using the Electrical menu on the block context menu.\n\nRight-click the block and, from the Electrical menu, select an option:\n\n• Display Base Values — Displays the machine per-unit base values in the MATLAB® Command Window.\n\n• Plot Torque Speed (SI) — Plots torque versus speed, both measured in SI units, in a MATLAB figure window using the current machine parameters.\n\n• Plot Torque Speed (pu) — Plots torque versus speed, both measured in per-unit, in a MATLAB figure window using the current machine parameters.\n\n• Plot Open-Circuit Saturation — Plots terminal voltage versus no-load stator current, both in per-unit, in a MATLAB figure window. The plot contains three traces:\n\n• Unsaturated — Stator magnetizing inductance (unsaturated).\n\n• Saturated — Open-circuit lookup table (v versus i) you specify.\n\n• Derived — Open-circuit lookup table derived from the per-unit open-circuit lookup table (v versus i) you specify. This data is used to calculate the saturated magnetizing inductance, Lm_sat, and the saturation factor, Ks, versus magnetic flux linkage, ψm, characteristics.\n\n• Plot Saturation Factor — Plots saturation factor, Ks, versus magnetic flux linkage, ψm, in a MATLAB figure window using the machine parameters. This parameter is derived from other parameters that you specify:\n\n• No-load stator current saturation data, i\n\n• Terminal voltage saturation data, v\n\n• Leakage inductance, Lls\n\n• Plot Saturated Inductance — Plots magnetizing inductance, Lm_sat, versus magnetic flux linkage, ψm, in a MATLAB figure window using the machine parameters. This parameter is derived from other parameters that you specify:\n\n• No-load stator current saturation data, i\n\n• Terminal voltage saturation data, v\n\n• Leakage inductance, Lls\n\nFor the SI implementation, v is in V (phase-phase RMS) and i is in A (rms).\n\n### Variables\n\nUse the Variables settings to specify the priority and initial target values for the block variables before simulation. For more information, see Set Priority and Initial Target for Block Variables (Simscape).\n\nThe type of variables that are visible in the Variables settings depends on the initialization method that you select, in the Main settings, for the Initialization option parameter. To specify target values using:\n\n• Flux variables — Set the Initialization option parameter to `Set targets for flux variables`.\n\n• Data from a load-flow analysis — Set the Initialization option parameter to ```Set targets for load flow variables```.\n\nIf you select `Set targets for load flow variables`, to fully specify the initial condition, you must include an initialization constraint in the form of a high-priority target value. For example, if your induction machine is connected to an Inertia block, the initial condition for the induction machine is completely specified if, in the Variables settings of the Inertia block, the Priority for Rotational velocity is set to `High`. Alternatively, you could set the Priority to `None` for the Inertia block Rotational velocity, and instead set the Priority for the induction machine block Slip, Real power generated, or Mechanical power consumed to `High`.\n\n## Ports\n\n### Input\n\nexpand all\n\nPhysical signal vector port associated with the machine per-unit measurements. The vector elements are:\n\n• pu_torque\n\n• pu_velocity\n\n• pu_vds\n\n• pu_vqs\n\n• pu_v0s\n\n• pu_ids\n\n• pu_iqs\n\n• pu_i0s\n\n### Conserving\n\nexpand all\n\nMechanical rotational conserving port associated with the machine rotor.\n\nMechanical rotational conserving port associated with the machine case.\n\nExpandable three-phase port associated with the stator positive-end connections.\n\nExpandable three-phase port associated with the stator negative-end connections.\n\n## Parameters\n\nexpand all\n\nAll default parameter values are based on a machine delta-winding configuration.\n\n### Main\n\nRated apparent power of the induction machine.\n\nRMS line-line voltage.\n\nNominal electrical frequency corresponding to the rated apparent power.\n\nNumber of machine pole pairs.\n\nUnit system for block parameterization. Choose between `SI`, the international system of units, and `Per unit`, the per-unit system.\n\n#### Dependencies\n\nSelecting:\n\n• `SI` exposes SI parameters in the Impedances and Saturation settings.\n\n• `Per unit` exposes per-unit parameters in the Impedances and Saturation settings.\n\nOption to specify a single or double squirrel cage for the machine.\n\nSetting this parameter to `Double squirrel cage` exposes Impedances parameters for the second cage.\n\nZero-sequence model:\n\n• `Include` — Prioritize model fidelity. An error occurs if you Include zero-sequence terms for simulations that use the Partitioning solver. For more information, see Increase Simulation Speed Using the Partitioning Solver (Simscape).\n\n• `Exclude` — Prioritize simulation speed for desktop simulation or real-time deployment.\n\n#### Dependencies\n\nIf this parameter is set to:\n\n• `Include` and Parameterization unit is set to `SI` — The Stator zero-sequence reactance, X0 parameter in the Impedances settings is visible.\n\n• `Include` and Parameterization unit is set to `Per unit` — The Stator zero-sequence inductance, pu_L0 (pu) parameter in the Impedances settings is visible.\n\n• `Exclude` — The stator zero-sequence parameter in the Impedances settings is not visible.\n\nInitialization method. You can initialize a machine for steady-state simulation using either flux data or data from a load-flow analysis.\n\nIf you select `Set targets for load flow variables`, to fully specify the initial condition, you must include an initialization constraint in the form of a high-priority target value. For example, if your induction machine is connected to an Inertia block, the initial condition for the induction machine is completely specified if, in the Variables settings of the Inertia block, the Priority for Rotational velocity is set to `High`. Alternatively, you could set the Priority to `None` for the Inertia block Rotational velocity, and instead set the Priority for the induction machine block Slip, Real power generated, or Mechanical power consumed to `High`.\n\n#### Dependencies\n\nThe type of variables that are visible in the Variables settings depends on the initialization method that you select, in the Main settings, for the Initialization option parameter. To specify target values using:\n\n• Flux variables — Set the Initialization option parameter to `Set targets for flux variables`.\n\n• Data from a load-flow analysis — Set the Initialization option parameter to ```Set targets for load flow variables```.\n\n### Impedances\n\nFor the Parameterization unit parameter in the Main settings, select `SI` to expose SI parameters or `Per unit` to expose per-unit parameters.\n\nStator resistance.\n\n#### Dependencies\n\nThis parameter is visible only if you set the Parameterization unit parameter in the Main setting to `SI`.\n\nStator leakage reactance.\n\n#### Dependencies\n\nThis parameter is visible only if you set the Parameterization unit parameter in the Main setting to `SI`.\n\nRotor resistance referred to the stator.\n\n#### Dependencies\n\nThis parameter is visible only if you set the Parameterization unit parameter in the Main setting to `SI` and the Squirrel cage parameter in the Main setting to `Single squirrel cage`.\n\nRotor leakage reactance referred to the stator.\n\n#### Dependencies\n\nThis parameter is visible only if you set the Parameterization unit parameter in the Main setting to `SI` and the Squirrel cage parameter in the Main setting to `Single squirrel cage`.\n\nRotor resistance in cage 1.\n\n#### Dependencies\n\nThis parameter is visible only if you set the Parameterization unit parameter in the Main setting to `SI` and the Squirrel cage parameter in the Main setting to `Double squirrel cage`.\n\nRotor leakage reactance in cage 1.\n\n#### Dependencies\n\nThis parameter is visible only if you set the Parameterization unit parameter in the Main setting to `SI` and the Squirrel cage parameter in the Main setting to `Double squirrel cage`.\n\nRotor resistance in cage 2.\n\n#### Dependencies\n\nThis parameter is visible only if you set the Parameterization unit parameter in the Main setting to `SI` and the Squirrel cage parameter in the Main setting to `Double squirrel cage`.\n\nRotor leakage reactance in cage 2.\n\n#### Dependencies\n\nThis parameter is visible only if you set the Parameterization unit parameter in the Main setting to `SI` and the Squirrel cage parameter in the Main setting to `Double squirrel cage`.\n\nMagnetizing reactance.\n\n#### Dependencies\n\nThis parameter is visible only if you set the Parameterization unit parameter in the Main setting to `SI`.\n\nStator zero-sequence reactance.\n\n#### Dependencies\n\nThis parameter is visible only if you set the Parameterization unit parameter in the Main setting to `SI` and the Zero sequence parameter in the Main setting to `Include`.\n\nPer-unit stator resistance.\n\n#### Dependencies\n\nThis parameter is visible only if you set the Parameterization unit parameter in the Main setting to ```Per unit```.\n\nPer-unit stator leakage inductance.\n\n#### Dependencies\n\nThis parameter is visible only if you set the Parameterization unit parameter in the Main setting to ```Per unit```.\n\nPer-unit rotor resistance referred to the stator.\n\n#### Dependencies\n\nThis parameter is visible only if you set the Parameterization unit parameter in the Main setting to ```Per unit``` and the Squirrel cage parameter in the Main setting to `Single squirrel cage`.\n\nPer-unit rotor leakage inductance referred to the stator.\n\n#### Dependencies\n\nThis parameter is visible only if you set the Parameterization unit parameter in the Main setting to ```Per unit``` and the Squirrel cage parameter in the Main setting to `Single squirrel cage`.\n\nPer-unit rotor resistance in cage 1.\n\n#### Dependencies\n\nThis parameter is visible only if you set the Parameterization unit parameter in the Main setting to ```Per unit``` and the Squirrel cage parameter in the Main setting to `Double squirrel cage`.\n\nPer-unit rotor leakage inductance in cage 1.\n\n#### Dependencies\n\nThis parameter is visible only if you set the Parameterization unit parameter in the Main setting to ```Per unit``` and the Squirrel cage parameter in the Main setting to `Double squirrel cage`.\n\nPer-unit rotor resistance in cage 2.\n\n#### Dependencies\n\nThis parameter is visible only if you set the Parameterization unit parameter in the Main setting to ```Per unit``` and the Squirrel cage parameter in the Main setting to `Double squirrel cage`.\n\nPer-unit rotor leakage inductance in cage 2.\n\n#### Dependencies\n\nThis parameter is visible only if you set the Parameterization unit parameter in the Main setting to ```Per unit``` and the Squirrel cage parameter in the Main setting to `Double squirrel cage`.\n\nPer-unit magnetizing inductance, that is, the peak value of stator-rotor mutual inductance.\n\n#### Dependencies\n\nThis parameter is visible only if you set the Parameterization unit parameter in the Main setting to ```Per unit```.\n\nPer-unit stator zero-sequence inductance.\n\n#### Dependencies\n\nThis parameter is visible only if you set the Parameterization unit parameter in the Main setting to ```Per unit``` and the Zero sequence parameter in the Main setting to `Include`.\n\n### Saturation\n\nFor the Parameterization unit parameter in the Main settings, select `SI` to expose SI parameters or ```Per unit``` to expose per-unit parameters.\n\nBlock magnetic saturation representation.\n\n#### Dependencies\n\nIf this parameter is set to ```Open-circuit lookup table (v versus i)```, related parameters are visible.\n\nCurrent i data populates the voltage v versus field current i lookup table. This parameter must contain a vector with at least 10 elements.\n\n#### Dependencies\n\nThis parameter is visible only if the Magnetic saturation representation parameter is set to ```Open-circuit lookup table (v versus i)``` and, in the Main settings, the Parameterization unit parameter is set to `SI`.\n\nTerminal voltage v data populates the voltage v versus current i lookup table. This parameter must contain a vector with at least 10 elements. The number of elements must match the number of elements in the vector for the No-load stator current saturation data, i (rms) parameter.\n\n#### Dependencies\n\nThis parameter is visible only if the Magnetic saturation representation parameter is set to ```Open-circuit lookup table (v versus i)``` and, in the Main settings, the Parameterization unit parameter is set to `SI`.\n\nCurrent i data populates the voltage v versus field current i lookup table. This parameter must contain a vector with at least 10 elements.\n\n#### Dependencies\n\nThis parameter is visible only if the Magnetic saturation representation parameter is set to ```Open-circuit lookup table (v versus i)``` and, in the Main settings, the Parameterization unit parameter is set to ```Per unit```.\n\nTerminal voltage v data populates the voltage v versus current i lookup table. This parameter must contain a vector with at least 10 elements. The number of elements must match the number of elements in the vector for the Per-unit no-load stator current saturation data, i parameter.\n\n#### Dependencies\n\nThis parameter is visible only if the Magnetic saturation representation parameter is set to ```Open-circuit lookup table (v versus i)``` and, in the Main settings, the Parameterization unit parameter is set to ```Per unit```.\n\n Kundur, P. Power System Stability and Control. New York: McGraw Hill, 1993.\n\n Lyshevski, S. E. Electromechanical Systems, Electric Machines and Applied Mechatronics. Boca Raton, FL: CRC Press, 1999.\n\n Ojo, J. O., Consoli, A.,and Lipo, T. A., \"An improved model of saturated induction machines\", IEEE Transactions on Industry Applications. Vol. 26, no. 2, pp. 212-221, 1990."
]
| [
null,
"https://de.mathworks.com/help/physmod/sps/ref/induction_machine_squirrel_cage_ic.png",
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.58728075,"math_prob":0.943177,"size":19869,"snap":"2019-43-2019-47","text_gpt3_token_len":4827,"char_repetition_ratio":0.19280142,"word_repetition_ratio":0.5006515,"special_character_ratio":0.20665358,"punctuation_ratio":0.11054277,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9943178,"pos_list":[0,1,2],"im_url_duplicate_count":[null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-11-14T15:24:29Z\",\"WARC-Record-ID\":\"<urn:uuid:3cfdafcb-a5a6-408c-a5a4-2ac7591daa5a>\",\"Content-Length\":\"184799\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f9a5b385-8a59-43d4-bd76-5540ae9b96ba>\",\"WARC-Concurrent-To\":\"<urn:uuid:fd1027c6-7927-4580-a34f-f1fff22e0af3>\",\"WARC-IP-Address\":\"23.50.228.199\",\"WARC-Target-URI\":\"https://de.mathworks.com/help/physmod/sps/ref/inductionmachinesquirrelcage.html\",\"WARC-Payload-Digest\":\"sha1:N5UCXNENH73EDQO7X3GZNTC5DQ62ONZA\",\"WARC-Block-Digest\":\"sha1:2OQA22QNFBSUDX6EDSIUEXX7CIVRJXLK\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-47/CC-MAIN-2019-47_segments_1573496668525.62_warc_CC-MAIN-20191114131434-20191114155434-00299.warc.gz\"}"} |
https://www.bits.vib.be/software-overview/matlab | [
"### MATLAB",
null,
"MATLAB is a high-level language and interactive environment for numerical computation, visualization, and programming.\n\nIts key features are:\n\n• High-level language for numerical computation, visualization, and application development\n• Interactive environment for iterative exploration, design, and problem solving\n• Mathematical functions for linear algebra, statistics, Fourier analysis, filtering, optimization, numerical integration, and solving ordinary differential equations\n• Built-in graphics for visualizing data and tools for creating custom plots\n• Development tools for improving code quality and maintainability and maximizing performance\n• Tools for building applications with custom graphical interfaces\n• Functions for integrating MATLAB based algorithms with external applications and languages such as C, Java, .NET, and Microsoft Excel\n\n# Get it!\n\nGet MATLAB (VIB scientists only)"
]
| [
null,
"https://www.bits.vib.be/images/images/matlab_logo.gif",
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.83740264,"math_prob":0.6898658,"size":895,"snap":"2020-45-2020-50","text_gpt3_token_len":153,"char_repetition_ratio":0.13019079,"word_repetition_ratio":0.017857144,"special_character_ratio":0.1631285,"punctuation_ratio":0.14503817,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9790762,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-10-26T18:49:16Z\",\"WARC-Record-ID\":\"<urn:uuid:118bbb64-ad9c-46f3-b08e-fc800e8fa6f3>\",\"Content-Length\":\"12646\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8c2d2d4e-e801-4c64-9c54-4e0b0138e76f>\",\"WARC-Concurrent-To\":\"<urn:uuid:c5986de8-9058-4eb7-8acc-84b0fc4ef548>\",\"WARC-IP-Address\":\"185.28.58.152\",\"WARC-Target-URI\":\"https://www.bits.vib.be/software-overview/matlab\",\"WARC-Payload-Digest\":\"sha1:643B25BPUZ6F6AHWO43PQ4KFME34SVEC\",\"WARC-Block-Digest\":\"sha1:K2BE4GVBAMSENLDVVP3YYIKWTHBZCU2H\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-45/CC-MAIN-2020-45_segments_1603107891624.95_warc_CC-MAIN-20201026175019-20201026205019-00601.warc.gz\"}"} |
https://nips.cc/Conferences/2021/ScheduleMultitrack?event=34241 | [
"`\n\nTimezone: »\n\nStatistical Numerical PDE : Fast Rate, Neural Scaling Law and When it’s Optimal\nYiping Lu · Haoxuan Chen · Jianfeng Lu · Lexing Ying · Jose Blanchet\n\nTue Dec 14 06:45 AM -- 07:30 AM (PST) @ None\n\nIn this paper, we study the statistical limits of deep learning techniques for solving elliptic partial differential equations (PDEs) from random samples using the Deep Ritz Method (DRM) and Physics-Informed Neural Networks (PINNs). To simplify the problem, we focus on a prototype elliptic PDE: the Schr\\\"odinger equation on a hypercube with zero Dirichlet boundary condition, which is applied in quantum-mechanical systems. We establish upper and lower bounds for both methods, which improve upon concurrently developed upper bounds for this problem via a fast rate generalization bound. We discover that the current Deep Ritz Method is sub-optimal and propose a modified version of it. We also prove that PINN and the modified version of DRM can achieve minimax optimal bounds over Sobolev spaces. Empirically, following recent work which has shown that the deep model accuracy will improve with growing training sets according to a power law, we supply computational experiments to show similar-behavior of dimension dependent power law for deep PDE solvers."
]
| [
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.6848768,"math_prob":0.87439424,"size":3535,"snap":"2022-27-2022-33","text_gpt3_token_len":861,"char_repetition_ratio":0.113565564,"word_repetition_ratio":0.2956989,"special_character_ratio":0.21442716,"punctuation_ratio":0.060263652,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95882905,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-07-05T06:34:53Z\",\"WARC-Record-ID\":\"<urn:uuid:33d62d60-4e4b-4fad-bec0-333ea5fee119>\",\"Content-Length\":\"44544\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:1bfe9894-27f6-4eff-b1f5-035861b88500>\",\"WARC-Concurrent-To\":\"<urn:uuid:cedee652-48b8-4593-8146-15a1e6720e79>\",\"WARC-IP-Address\":\"198.202.70.65\",\"WARC-Target-URI\":\"https://nips.cc/Conferences/2021/ScheduleMultitrack?event=34241\",\"WARC-Payload-Digest\":\"sha1:CKTXBXWCMTA4MKM7RLDBHWVL575ZR77E\",\"WARC-Block-Digest\":\"sha1:2AGYOMSY5FSSF7YI6UISPEVYYR5IAPER\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656104514861.81_warc_CC-MAIN-20220705053147-20220705083147-00192.warc.gz\"}"} |
https://www.proprofs.com/quiz-school/story.php?title=badm-quiz-1 | [
"15 Questions | Total Attempts: 91",
null,
"",
null,
"Settings",
null,
"",
null,
"All questions are mandatory.\n\n• 1.\nWhich of the following statements about the median is NOT true?\n• A.\n\nIt is equal to the Second Quartile.\n\n• B.\n\nThe median is more affected by extreme values than the mean.\n\n• C.\n\nIt is a measure of central tendency.\n\n• D.\n\nIt is equal to the mode in bell-shaped, symmetrical distributions.\n\n• 2.\nIt is possible that a distribution of data may not actually contain one of the measures of central tendency. Which one might that be?\n• A.\n\nThe Mode\n\n• B.\n\nThe median\n\n• C.\n\nThe Mean\n\n• D.\n\nNone of the above\n\n• 3.\nAs a general rule, a data point is considered to be an outlier if it is\n• A.\n\nMore than three standard deviations away from the mean\n\n• B.\n\nLess than one standard deviation away from the mean\n\n• C.\n\nLess than two standard deviations away from the mean\n\n• D.\n\nMore than one standard deviation away from the mean\n\n• 4.\nIn a right-skewed distribution:\n• A.\n\nThe mean is less than the median.\n\n• B.\n\nThe median equals the mean.\n\n• C.\n\nThe mean is less than the mode.\n\n• D.\n\nThe mean is greater than the median.\n\n• 5.\nApplicants to MBA programs typically must submit graduate-level board scores known as the GMAT. Assuming that the distribution of GMAT scores is symmetrical and bell shaped, then fifty-percent of all scores would be greater than which value?\n• A.\n\nThe mean\n\n• B.\n\nThe median\n\n• C.\n\nThe mode\n\n• D.\n\nThe mean & median\n\n• E.\n\nAll of the above\n\n• 6.\nA survey question which asks whether or not the respondent is “for” or “against” a proposal represents\n• A.\n\nA discrete variable.\n\n• B.\n\nA continuous variable.\n\n• C.\n\nA categorical variable.\n\n• D.\n\nA numerical variable\n\n• 7.\nThe branch of statistics that uses sample data to draw conclusions about an entire population is known as:\n• A.\n\nDescriptive statistics.\n\n• B.\n\nExperimentation.\n\n• C.\n\nPrimary sources.\n\n• D.\n\nInferential statistics\n\n• 8.\nResponses to survey question number one, “How many days did it take from the time you ordered your merchandise to the time you received it?” represents what kind of variable?\n• A.\n\nNumerical\n\n• B.\n\nDiscrete and numerical\n\n• C.\n\nContinuous and numerical\n\n• D.\n\nCategorical\n\n• 9.\nWhat type of distribution is observed when most of the scores cluster around the lower end of the scale?\n• A.\n\nA negatively skewed distribution\n\n• B.\n\nA normal distribution\n\n• C.\n\nA positively skewed distribution\n\n• D.\n\nA related distribution\n\n• E.\n\nA bi-modal distribution\n\n• F.\n\nA flat distribution\n\n• 10.\nWhich two measures use the mean as a baseline and identify the extent to which scores differ from this?\n• A.\n\nVariance and standard deviation\n\n• B.\n\nStandard deviation and range\n\n• C.\n\nSum and variance\n\n• D.\n\nAll of these\n\n• 11.\nA tire manufacturer found a sample of tires created using a new formulation resulted in an average of forty-thousand miles per tire, and concluded that they can safely advertise this finding about the entire population of tires created with the new formula. This decision making process represents the branch of statistics known as descriptive statistics.\n• A.\n\nTrue\n\n• B.\n\nFalse\n\n• 12.\nA business student applying to an MBA program must complete a standardized exam known as the GMAT, which is measured on an interval scale.\n• A.\n\nTrue\n\n• B.\n\nFalse\n\n• 13.\nOne of the advantages of a bar chart is that it clearly shows that the total of all the categories of the bar chart adds to 100%\n• A.\n\nTrue\n\n• B.\n\nFalse\n\n• 14.\nThe coefficient of variation (CV) measures the variability in a data set relative to the size of the median.\n• A.\n\nTrue\n\n• B.\n\nFalse\n\n• 15.\nThe median of the values 3.4, 4.7, 1.9, 7.6, and 6.5 is 3.4\n• A.\n\nTrue\n\n• B.\n\nFalse",
null,
"Back to top"
]
| [
null,
"https://www.proprofs.com/quiz-school/images/story_settings_gear.png",
null,
"https://www.proprofs.com/quiz-school/images/story_settings_gear_color.png",
null,
"https://www.proprofs.com/quiz-school/loader.gif",
null,
"https://media.proprofs.com/images/QM/user_images/misc/rr_1.jpg",
null,
"https://www.proprofs.com/quiz-school/img/top.png",
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.8152288,"math_prob":0.93794876,"size":1224,"snap":"2021-31-2021-39","text_gpt3_token_len":262,"char_repetition_ratio":0.17295082,"word_repetition_ratio":0.14583333,"special_character_ratio":0.1748366,"punctuation_ratio":0.07511737,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9965394,"pos_list":[0,1,2,3,4,5,6,7,8,9,10],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-20T08:31:15Z\",\"WARC-Record-ID\":\"<urn:uuid:d04739c3-ad10-4224-a30c-9ed8c3b3cab9>\",\"Content-Length\":\"261102\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:91552c33-54ea-45fa-84b4-0e871aa95a23>\",\"WARC-Concurrent-To\":\"<urn:uuid:2f9258ed-ec36-48a0-81b8-c85b18de17e5>\",\"WARC-IP-Address\":\"104.26.12.111\",\"WARC-Target-URI\":\"https://www.proprofs.com/quiz-school/story.php?title=badm-quiz-1\",\"WARC-Payload-Digest\":\"sha1:LGAYCDAPX6RVT43ZFOHU7O7P2SXB2OV7\",\"WARC-Block-Digest\":\"sha1:JLBTKQIZD5DN3SET4QASLDHCFHQKZIDA\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780057033.33_warc_CC-MAIN-20210920070754-20210920100754-00622.warc.gz\"}"} |
https://www.php.net/manual/ro/function.pdf-save.php | [
"# PDF_save\n\n(PHP 4, PECL pdflib >= 1.0.0)\n\nPDF_saveSalvează starea grafică\n\n### Descrierea\n\nPDF_save ( resource `\\$p` ) : bool\n\nSalvează starea grafică curentă. Întoarce valoarea `TRUE` în cazul succesului sau `FALSE` în cazul eșecului.",
null,
"add a note\n\n### User Contributed Notes 1 note\n\nbob at nijman dot de\n18 years ago\n``` By playing around with this you'll get a better understanding of what pdf_save and pdf_restore do. <?php //Create & Open PDF-Object \\$pdf = pdf_new(); pdf_open_file(\\$pdf); pdf_set_info(\\$pdf, \"Author\",\"Bob Nijman\"); pdf_set_info(\\$pdf, \"Title\",\"Sponsored by www.nijman.de\"); pdf_set_info(\\$pdf, \"Creator\", \"See Author\"); pdf_set_info(\\$pdf, \"Subject\", \"pdf_restore\"); pdf_begin_page(\\$pdf, 300, 300); /* By encapsulating the changes we make to the coordinate system (pdf_translate() and pdf_rotate()) between pdf_save() and pdf_restore we give these changes only local scope. */ /* SMALL RECTANGLE */ pdf_save(\\$pdf); //move the origin of the coordinate system to (100,100) pdf_translate(\\$pdf, 100, 100); //Rotate the coordinate system by 45 degrees. pdf_rotate(\\$pdf, 45); //Draw a samll rectangle pdf_rect(\\$pdf, 0, 0, 20, 20); pdf_stroke(\\$pdf); //Restore the graphics state to the way it was //before we started to translate and rotate pdf_restore(\\$pdf); /* LARGER RECTANGLE */ pdf_save(\\$pdf); //move the origin of the coordinate system to (10,20) pdf_translate(\\$pdf, 10, 20); //Rotate the coordinate system by 15 degrees. pdf_rotate(\\$pdf, 15); //Draw a larger rectangle pdf_rect(\\$pdf, 0, 0, 40, 40); pdf_stroke(\\$pdf); //Restore the graphics state to the way it was //before we started to translate and rotate pdf_restore(\\$pdf); //close it up pdf_end_page(\\$pdf); pdf_close(\\$pdf); \\$data = pdf_get_buffer(\\$pdf); header('Content-type: application/pdf'); header('Content-disposition: inline; filename=nijman.pdf'); header('Content-length: ' . strlen(\\$data)); echo \\$data; ?> Thanx to: http://www.dynamicwebpages.de/50.tutorials.php?dwp_tutorialID=11 ```",
null,
""
]
| [
null,
"https://www.php.net/images/[email protected]",
null,
"https://www.php.net/images/[email protected]",
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.6056404,"math_prob":0.8329009,"size":1638,"snap":"2019-51-2020-05","text_gpt3_token_len":454,"char_repetition_ratio":0.1621787,"word_repetition_ratio":0.22797927,"special_character_ratio":0.31746033,"punctuation_ratio":0.23934427,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9641184,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-12-06T17:39:26Z\",\"WARC-Record-ID\":\"<urn:uuid:45fa12e5-dfc0-4052-a823-3c7fdf6b1969>\",\"Content-Length\":\"66278\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:99657ccd-657d-45dc-8e35-58772a3a8ce5>\",\"WARC-Concurrent-To\":\"<urn:uuid:05b1927f-3273-43d6-8176-07f09b08640b>\",\"WARC-IP-Address\":\"185.85.0.29\",\"WARC-Target-URI\":\"https://www.php.net/manual/ro/function.pdf-save.php\",\"WARC-Payload-Digest\":\"sha1:EVVYGUGK3BPAKOHPA6EB5HE322KXDPDK\",\"WARC-Block-Digest\":\"sha1:G7I3YUDYMTOPU33CU7SNGTO2L2ZMHH7W\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-51/CC-MAIN-2019-51_segments_1575540490743.16_warc_CC-MAIN-20191206173152-20191206201152-00272.warc.gz\"}"} |
https://redman.xyz/doku.php/gsoc:differentiating_code?do= | [
"Differentiating code\n\nOne step in the spectral upsampling method described by Jakob and Hanika (2019) is computing an error function, which is used to find the model parameters. I wanted to provide derivatives of the error function with respect to its arguments to the optimization algorithm I used, in hopes for faster and more reliable convergence.\n\nHow does one differentiate Python code though? Every program, no matter how complicated, is a series of simple operations and the chain rule can be used. Computer algebra systems (the one I use is http://maxima.sourceforge.net/Maxima) can help with the process, making this a trivial task.\n\nThe error function\n\nThe algorithm is as follows:\n\n• Given a triplet of coefficients (model parameters),",
null,
",",
null,
",",
null,
", compute a reflectance according to the model.",
null,
"",
null,
"• Compute the radiosity of a surface with the reflectance, lit by the given illuminant. This is done by simply multiplying the reflectance by the illuminant's spectral distribution.",
null,
"• Compute the CIE 1931 XYZ tristimulus values describing how the reflected light would be perceived.",
null,
"where",
null,
",",
null,
"and",
null,
"are the cmfs (color matching functions), quantizing human color vision. They are closely related to (but not the same as) spectral sensivities of the three different kinds of cone cells in a human eye. In particular",
null,
"is the luminosity function used in photometry, describing how bright light of different wavelengths appears. The normalization constant",
null,
",",
null,
"assures that",
null,
"for perfectly white surfaces and that arbitrary units can be used for the illuminant and the cmfs.\n\n• Compute the CIE 1976 L*a*b* colorspace coordinates. Start by computing the values of the intermediate lightness function for the three XYZ components.",
null,
"",
null,
"Likewise for the remaining two components.",
null,
"• Compute the Euclidean distance between the coordinates above and the target color (given in the L*a*b* colorspace). This is the CIE 1976 color difference (delta E) formula.",
null,
"Discretization\n\nBecause spectral data is typically discrete, that is measured at some chosen wavelengths (e.g. every 10 nanometers in the visible range), the algorithm above operates on arrays rather than continuous functions. Integrals are approximated simply as the corresponding finite sum:",
null,
"This method turned out to be both fast and accurate enough in this application.\n\nHow does",
null,
"depend on the function inputs? Since this is a scalar function of multiple variables, what we're looking for is the gradient or three derivatives of",
null,
"with respect to each of the inputs. The answer is given by a repeated application of the chain rule.\n\nThe easiest way to approach it is to start from the beginning of the algorithm and compute the three functions",
null,
". (In code the result is a 3-by-n array, where n is the number of discrete wavelengths.) Then compute the derivatives of",
null,
"using the chain rule.",
null,
"The derivatives of",
null,
"have already been computed in the previous step. Doing this step by step will eventually lead to",
null,
",",
null,
"and",
null,
"and ultimately the gradient of",
null,
". Take a look at the full code to see all the formulas.\n\nAutomatic differentation\n\nThere is a clever trick to “automagically” compute derivatives, without having to go step-by-step, differentiating every operation and writing code manually like I did. Dual numbers have the interesting property that, when used correctly, their dual parts are the derivatives of their real parts. They're defined very similarly to complex numbers, but instead of",
null,
",",
null,
", one has",
null,
",",
null,
". (Even more interestingly, this can be extended to higher derivatives as well!). See automatic differentiation on Wikipedia.\n\nThe code I wanted to differentiate was simple enough that I didn't bother looking for and learning a new Python module. I also avoided pulling additional dependencies.",
null,
""
]
| [
null,
"https://redman.xyz/lib/exe/fetch.php",
null,
"https://redman.xyz/lib/exe/fetch.php",
null,
"https://redman.xyz/lib/exe/fetch.php",
null,
"https://redman.xyz/lib/exe/fetch.php",
null,
"https://redman.xyz/lib/exe/fetch.php",
null,
"https://redman.xyz/lib/exe/fetch.php",
null,
"https://redman.xyz/lib/exe/fetch.php",
null,
"https://redman.xyz/lib/exe/fetch.php",
null,
"https://redman.xyz/lib/exe/fetch.php",
null,
"https://redman.xyz/lib/exe/fetch.php",
null,
"https://redman.xyz/lib/exe/fetch.php",
null,
"https://redman.xyz/lib/exe/fetch.php",
null,
"https://redman.xyz/lib/exe/fetch.php",
null,
"https://redman.xyz/lib/exe/fetch.php",
null,
"https://redman.xyz/lib/exe/fetch.php",
null,
"https://redman.xyz/lib/exe/fetch.php",
null,
"https://redman.xyz/lib/exe/fetch.php",
null,
"https://redman.xyz/lib/exe/fetch.php",
null,
"https://redman.xyz/lib/exe/fetch.php",
null,
"https://redman.xyz/lib/exe/fetch.php",
null,
"https://redman.xyz/lib/exe/fetch.php",
null,
"https://redman.xyz/lib/exe/fetch.php",
null,
"https://redman.xyz/lib/exe/fetch.php",
null,
"https://redman.xyz/lib/exe/fetch.php",
null,
"https://redman.xyz/lib/exe/fetch.php",
null,
"https://redman.xyz/lib/exe/fetch.php",
null,
"https://redman.xyz/lib/exe/fetch.php",
null,
"https://redman.xyz/lib/exe/fetch.php",
null,
"https://redman.xyz/lib/exe/fetch.php",
null,
"https://redman.xyz/lib/exe/fetch.php",
null,
"https://redman.xyz/lib/exe/fetch.php",
null,
"https://redman.xyz/lib/exe/fetch.php",
null,
"https://redman.xyz/lib/exe/fetch.php",
null,
"https://redman.xyz/lib/exe/indexer.php",
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.9070311,"math_prob":0.9299773,"size":3747,"snap":"2022-05-2022-21","text_gpt3_token_len":751,"char_repetition_ratio":0.116216935,"word_repetition_ratio":0.0,"special_character_ratio":0.19669068,"punctuation_ratio":0.10730253,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9935209,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50,51,52,53,54,55,56,57,58,59,60,61,62,63,64,65,66,67,68],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-01-27T14:49:43Z\",\"WARC-Record-ID\":\"<urn:uuid:a4e4b0b7-f3a2-4252-9f8a-6cddec04ecce>\",\"Content-Length\":\"18046\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e0de75bd-7b15-4142-9738-80709fd78473>\",\"WARC-Concurrent-To\":\"<urn:uuid:4d7fefe5-b7ae-4753-8f52-ecd8a0af42b6>\",\"WARC-IP-Address\":\"54.38.195.165\",\"WARC-Target-URI\":\"https://redman.xyz/doku.php/gsoc:differentiating_code?do=\",\"WARC-Payload-Digest\":\"sha1:L2BLEBHF2MDPSVRNLLM6B4IDFSN4RVWA\",\"WARC-Block-Digest\":\"sha1:5JZSPMMM63AEOEYDQGLAEDMEASFHJJBA\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-05/CC-MAIN-2022-05_segments_1642320305266.34_warc_CC-MAIN-20220127133107-20220127163107-00309.warc.gz\"}"} |
https://electronics.stackexchange.com/questions/265816/control-system-model-a-simple-flow-valve | [
"Control System: Model a simple flow valve\n\nI was reading a paper about a Controller Design for Temperature Control of Heat Exchanger and I couldn't understand how to model the valve used in the system.\n\nNo need to read the paper as I will wrap up the whole issue:\n\nGiven:\n\n• capacity of control valve is 1.6kg/sec\n• time constant is 3 sec\n• valve input is pressure varying from 3 to 15 psi\n\nThe resulting transfer function is:",
null,
"It is obvious that they considered the valve as a first order transfer function.The gain Kp was calculated by: (1.6kg/sec)/(15-3)psi=1.6/12=0.133. Time constant=3 seconds and that's it.\n\nBut the problem is: shouldn't a step input of 15psi output a signal (1-exponential) with a final value of 1.6kg/sec. But that transfer function won't actually output that. It still needs some kind of a shift. So Am I missing something? Is the model wrong?\n\nI also simulated the transfer function response with MATLAB SIMULINK along with a suggested alternative that probably solve the offset thing:",
null,
"The resulting waveform:",
null,
"zoomed in near time=0seconds:",
null,
"• The yellow waveform is the input step function\n• The blue waveform is the output of the transfer function of the valve according to the paper.\n• The red waveform is the waveform of the transfer function with an offset.\n\n• problem with the blue waveform: wrong final value.\n• problem with the yellow waveform: starts from a negative value (-0.4 kg/sec)\n\nSo what is the correct model?\n\nThe paper is a bit sloppy about transfer functions. If $G_V$ is the transfer from function from control pressure to flow, then it needs an offset. Step response should be measured for a step from 3 to 15 psi."
]
| [
null,
"https://i.stack.imgur.com/Om83q.png",
null,
"https://i.stack.imgur.com/Abl1d.png",
null,
"https://i.stack.imgur.com/QxOOv.png",
null,
"https://i.stack.imgur.com/PcZXU.png",
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.925316,"math_prob":0.9675998,"size":1370,"snap":"2019-26-2019-30","text_gpt3_token_len":318,"char_repetition_ratio":0.14421669,"word_repetition_ratio":0.0,"special_character_ratio":0.23503649,"punctuation_ratio":0.10181818,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9921919,"pos_list":[0,1,2,3,4,5,6,7,8],"im_url_duplicate_count":[null,2,null,3,null,3,null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-06-26T08:22:59Z\",\"WARC-Record-ID\":\"<urn:uuid:ff16d783-80f8-459c-8256-63f9396f8cb8>\",\"Content-Length\":\"143715\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f2f0e72b-8192-4a9e-8ab8-5ec7b139f9a3>\",\"WARC-Concurrent-To\":\"<urn:uuid:86230b50-92e5-42b7-a752-0dc632f006f8>\",\"WARC-IP-Address\":\"151.101.1.69\",\"WARC-Target-URI\":\"https://electronics.stackexchange.com/questions/265816/control-system-model-a-simple-flow-valve\",\"WARC-Payload-Digest\":\"sha1:5PWKGWBIJFSD5VCY6SGWBFMLH5BVWDCC\",\"WARC-Block-Digest\":\"sha1:6B4E5NDYGVPKOYZVCDZZCWVO5S3BLPBJ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-26/CC-MAIN-2019-26_segments_1560628000231.40_warc_CC-MAIN-20190626073946-20190626095946-00231.warc.gz\"}"} |
https://help.altair.com/hwdesktop/hwx/topics/motionview/sdf_curb_to_curb_diameter_and_ackerman_1_and_2_r.htm | [
"# Curb to Curb Diameter and Ackerman 1 and 2\n\nBranch ID: 290\n\nRequest Number: 80000290\n\nTable 1.\n.plt Files .abf, .mrf Files Component Unit Comment\nF1 RESULT(1) Not Used Not Used Not Used\nF2 RESULT(2) Curb to Curb Diameter (Method 1) m The first method of calculating the approximate diameter of the circle the vehicle would move around at the current steer angles assuming no rear wheel steer.\nWhen the Vehicle is Steering Right\n\nouter = abs((left toe-in)/rtod)\n\ninner = abs((right toe-in)/rtod)\n\nctoc diam = 2*sqrt(((turn radius)+track/2)^2+ (wheel base)^2)/1000\n\nWhen the Vehicle is Steering Left\n\ninner = -(left toe-in)/rtod\n\nctoc diam = 2*sqrt(((turn radius)+track/2)^2+ (wheel base)^2)/1000\n\nF3 RESULT(3) Percent Ackerman (Method 1) % The first method of calculating percent Ackerman. Ackerman is used to quantify the amount of scrubbing occurring on the front of the vehicle during a turn assuming no rear wheel steer. A vehicle whose front wheels are parallel (the same steer angle) has 0% Ackerman. Conversely a vehicle is said to be 100% Ackerman when the lines through the left and right spindle axes intersect in the top view at a lateral line passing through the rear wheel centers. The following equation is used to calculate percent Ackerman with Method 1. See F2/RESULT(2) above for intermediate term calculations.\nWhen the Vehicle is Steering Right\n\npercent Ackerman = 100*((wheel base)/tan(outer)-(turn radius))/ (right spindle at kingpin y)\n\nWhen the Vehicle is Steering Left\n\npercent Ackerman = 100*((wheel base)/tan(outer)-(turn radius))/-(left spindle at kingpin y)\n\nF4 RESULT(4) Ackerman Deviation (Method 1) deg The first method of calculating Ackerman deviation. Ackerman deviation, like percent Ackerman, is used to quantify the amount of scrubbing occurring on the front of the vehicle during a turn assuming no rear wheel steer. Zero Ackerman deviation means that lines through the left and right spindle axes intersect in the top view at a lateral line passing through the rear wheel centers. The following equation is used to calculate Ackerman deviation with Method 1. See F2/RESULT(2) above for intermediate term calculations.\n\ndeviation = (outer-alpha)*rtod\n\nF5 RESULT(5) Not Used Not Used Not Used\nF6 RESULT(6) Curb to Curb Diameter (Method 2) m The second method of calculating the approximate diameter of the circle the vehicle would move around at the current steer angles assuming no rear wheel steer.\nWhen the Vehicle is Steering Right\n\nouter = abs((left toe-in)/rtod)\n\ninner = abs((right toe-in)/rtod)\n\nctoc diam = 2*sqrt(((turn radius)+track/2)^2+ (wheel base)^2)/1000\n\nWhen the Vehicle is Steering Left\n\ninner = -(left toe-in)/rtod\n\nctoc diam = 2*sqrt(((turn radius)+track/2)^2+ (wheel base)^2)/1000\n\nF7 RESULT(7) Percent Ackerman (Method 2) %\n\nThe second method of calculating percent Ackerman. See F3/RESULT(3) above for an explanation of Percent Ackerman. See F6/RESULT(6) above for intermediate term calculations."
]
| [
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.810347,"math_prob":0.9711863,"size":2534,"snap":"2023-40-2023-50","text_gpt3_token_len":671,"char_repetition_ratio":0.16166008,"word_repetition_ratio":0.45118734,"special_character_ratio":0.23993686,"punctuation_ratio":0.05922551,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99456644,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-09T08:01:55Z\",\"WARC-Record-ID\":\"<urn:uuid:c6c81b91-8f3e-4ce1-9ff4-c97e57cb22d5>\",\"Content-Length\":\"134687\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a1745809-acb1-4dd9-93dc-cb70aef06668>\",\"WARC-Concurrent-To\":\"<urn:uuid:6c5a346b-fdc7-4d14-8879-5b81fc71a76b>\",\"WARC-IP-Address\":\"173.225.177.121\",\"WARC-Target-URI\":\"https://help.altair.com/hwdesktop/hwx/topics/motionview/sdf_curb_to_curb_diameter_and_ackerman_1_and_2_r.htm\",\"WARC-Payload-Digest\":\"sha1:HPBQPA3B5EI3DMKVZUEMQJDHEYJ6IJEE\",\"WARC-Block-Digest\":\"sha1:ANRRMYZD5TSLOTLGSXL2TWZEXFCZIVPE\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100873.6_warc_CC-MAIN-20231209071722-20231209101722-00132.warc.gz\"}"} |
https://www.semantic-mediawiki.org/wiki/Help:Statistics | [
"# Help:Statistics\n\nSeveral result formats exist to deliver descriptive statistics from ask queries.\n\n Min minimum value (smallest observation) Lower Quartile first quartile (Q1 = 25th percentile) Median middle value (Q2 = 50th percentile, separating the greater and lesser halves of a data set) Upper Quartile second quartile (Q3 = 75th percentile) Max maximum value (largest observation) Average arithmetic mean (Sum of values of a data set divided by number of values) Mode mode (most frequent value in a data set) Variance variance (squared deviation from the mean) Standard deviation standard deviation (square root of the variance) Interquartile range interquartile range (difference between upper and lower quartiles) Interquartile mean interquartile mean (truncated mean of the interquartile range) Sum sum of values\n\nThere can be differences in calculation whether or not the query returns all values (total population) or just a set of values (sample), e. g. Samplevariance and Samplestandarddeviation."
]
| [
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.64354795,"math_prob":0.9820011,"size":1120,"snap":"2021-43-2021-49","text_gpt3_token_len":282,"char_repetition_ratio":0.140681,"word_repetition_ratio":0.0,"special_character_ratio":0.22410715,"punctuation_ratio":0.08571429,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9988129,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-21T02:01:20Z\",\"WARC-Record-ID\":\"<urn:uuid:5752a405-518e-4b41-973e-2692e7e2a595>\",\"Content-Length\":\"28588\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d8ce1160-5c3b-4e54-98e7-8f2083d83f23>\",\"WARC-Concurrent-To\":\"<urn:uuid:ddd2c4e6-fb07-45f7-809b-de0a2880902e>\",\"WARC-IP-Address\":\"116.203.61.64\",\"WARC-Target-URI\":\"https://www.semantic-mediawiki.org/wiki/Help:Statistics\",\"WARC-Payload-Digest\":\"sha1:DJUAPM3NBXR4E7IAXOUYMYWJPBJIFL76\",\"WARC-Block-Digest\":\"sha1:6QVCSHJXJOJ5H6SZUSGXL3LAWNC23P2S\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323585380.70_warc_CC-MAIN-20211021005314-20211021035314-00082.warc.gz\"}"} |
https://blog.eduguru.in/tutorial/c-tutorial-tutorial/infographics/conditional-operators-c-program-example | [
"",
null,
"# conditional operators c program example\n\n## conditional operators c program example",
null,
"conditional operators c program example\n\nWhat would be the output of the following programs:\n\nmain( )\n{\nint k, num = 30 ;\nk = ( num > 5 ? ( num <= 10 ? 100 : 200 ) : 500 ) ;\nprintf ( “\\n%d”, num ) ;\n}"
]
| [
null,
"https://blog.eduguru.in/wp-content/uploads/2018/07/c-program-for-egg-game.jpg",
null,
"https://blog.eduguru.in/wp-content/uploads/2018/09/conditional-operators-c-program-example.png",
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.84826136,"math_prob":0.9860375,"size":560,"snap":"2021-04-2021-17","text_gpt3_token_len":129,"char_repetition_ratio":0.14028777,"word_repetition_ratio":0.01923077,"special_character_ratio":0.28214285,"punctuation_ratio":0.11881188,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95141435,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,2,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-04-17T15:43:01Z\",\"WARC-Record-ID\":\"<urn:uuid:156b817a-71d4-43af-be7d-91a7b93bc23f>\",\"Content-Length\":\"43244\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:deffe9c6-9096-439b-ba22-4c119d0532af>\",\"WARC-Concurrent-To\":\"<urn:uuid:743b96d0-0805-4e1d-8ec0-79af37162645>\",\"WARC-IP-Address\":\"103.110.7.83\",\"WARC-Target-URI\":\"https://blog.eduguru.in/tutorial/c-tutorial-tutorial/infographics/conditional-operators-c-program-example\",\"WARC-Payload-Digest\":\"sha1:J5L2CYYZSAW5ZR33VD2I5WMPI6EFD4TJ\",\"WARC-Block-Digest\":\"sha1:4E2LUYIUXW6BMYNYFPQ63LFAYFEVKNO7\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-17/CC-MAIN-2021-17_segments_1618038460648.48_warc_CC-MAIN-20210417132441-20210417162441-00498.warc.gz\"}"} |
http://digitalforexnfashionstyles.com/cat3/560-relationship-between-currency-and-interest-rates.html | [
"## How do changes in national interest rates affect a\n\nWeekly Trading Lesson: The Relationship Between if there is indeed any relationship between the US interest rate environment interest rates in.\n\n## What Is the Relationship Between Interest Rates Currency\n\nRelationship between Inflation and Interest nonofficial currency rates. They studied relationship between the relationship between.\n\n## The relationship between exchange rates, interest rates\n\nThe NonMonotonic Relationship between Interest Rates and term interest rates to defend against currency relationship between interest rates and.\n\n## 6 Factors That Influence Exchange Rates Investopedia\n\nThe relationship between exchange rates, interest then arbitrageurs could prot by immediately changing currency The relationship between interest rates.\n\n## What is the relationship between interest rates, inflation",
null,
"## Weekly Trading Lesson: The Relationship Between Interest",
null,
"## What is the relationship between interest rates and\n\nHow interest rate affects currency. of a given country's currency. The higher interest rates that can be earned tend Relationship between interest rate and. This paper revisits the relationship between interest rates and exchange rates an active position to defend the currency. By raising interest rates high.\nRelationships among Inflation, Interest Rates and FIN 439 Ch 8: Relationships among Inflation, Interest Rates Relationship between the interest. relationship between interest rate differentials between countries and the impact on their exchange rates. Although interest rate differentials are only one.",
null,
"## What is the relationship between bond prices and currency",
null,
"## Rising Rates and the US Dollar WisdomTree\n\nThe NonMonotonic Relationship between Interest Rates and relationship between nominal interest rates increase in interest rates depreciates the currency. Currency Tools; Futures Contracts The incredible goldinterest rate from an econometric formula based on the last decades relationship between gold and. What Is the Relationship Between Treasury Notes and Mortgage Rates? That's because investors who want a steady and safe return compare interest rates of all.\nRELATIONSHIP BETWEEN INTEREST RATES AND EXCHANGE RATES Relationship between interest rates and exchange rates Relationship between interest rates and exchange rates\nThe Relationship Between Interest Rates And Exchange Rates Finance Essay. Published: 23rd March, 2015 Last Edited: 23rd March, 2015. This essay has been submitted by.",
null,
"## Relationship Between Real Exchange Rates And Interest",
null,
"## The Relationship Between Interest Rates And - UKEssays",
null,
"## INFLATION, INTEREST RATE, AND EXCHANGE RATE: WHAT IS THE",
null,
"Sep 29, 2011Difference Between Exchange Rate and Interest Rate. There is a close relationship between interest rates and Difference Between Macroeconomics. and therefore from one currency to another. Higher interest rates in The Effect of Interest Rates on Relationship of high interest rates and inflation.\nINFLATION, INTEREST RATE, AND EXCHANGE RATE: between interest and inflation rates. The IFE theory suggests that currency a relationship between interest rate.\nThe LongRun Relationship Between Real Exchange Rates and the real exchange ratereal interest rate relationship. REAL EXCHANGE RATES AND REAL INTEREST.",
null,
"",
null,
"",
null,
"",
null,
"Relationship between currency and interest rates\n\n## How do changes in national interest rates affect a\n\nWeekly Trading Lesson: The Relationship Between if there is indeed any relationship between the US interest rate environment interest rates in.\n\n## What Is the Relationship Between Interest Rates Currency\n\nRelationship between Inflation and Interest nonofficial currency rates. They studied relationship between the relationship between.\n\n## The relationship between exchange rates, interest rates\n\nThe NonMonotonic Relationship between Interest Rates and term interest rates to defend against currency relationship between interest rates and.\n\n## 6 Factors That Influence Exchange Rates Investopedia\n\nThe relationship between exchange rates, interest then arbitrageurs could prot by immediately changing currency The relationship between interest rates.\n\n## What is the relationship between interest rates, inflation",
null,
"## Weekly Trading Lesson: The Relationship Between Interest",
null,
"## What is the relationship between interest rates and\n\nHow interest rate affects currency. of a given country's currency. The higher interest rates that can be earned tend Relationship between interest rate and. This paper revisits the relationship between interest rates and exchange rates an active position to defend the currency. By raising interest rates high. Relationships among Inflation, Interest Rates and FIN 439 Ch 8: Relationships among Inflation, Interest Rates Relationship between the interest. relationship between interest rate differentials between countries and the impact on their exchange rates. Although interest rate differentials are only one.",
null,
"## What is the relationship between bond prices and currency",
null,
"## Rising Rates and the US Dollar WisdomTree\n\nThe NonMonotonic Relationship between Interest Rates and relationship between nominal interest rates increase in interest rates depreciates the currency. Currency Tools; Futures Contracts The incredible goldinterest rate from an econometric formula based on the last decades relationship between gold and. What Is the Relationship Between Treasury Notes and Mortgage Rates? That's because investors who want a steady and safe return compare interest rates of all. RELATIONSHIP BETWEEN INTEREST RATES AND EXCHANGE RATES Relationship between interest rates and exchange rates Relationship between interest rates and exchange rates\nThe Relationship Between Interest Rates And Exchange Rates Finance Essay. Published: 23rd March, 2015 Last Edited: 23rd March, 2015. This essay has been submitted by.",
null,
"## Relationship Between Real Exchange Rates And Interest",
null,
"## The Relationship Between Interest Rates And - UKEssays",
null,
"## INFLATION, INTEREST RATE, AND EXCHANGE RATE: WHAT IS THE",
null,
"Sep 29, 2011Difference Between Exchange Rate and Interest Rate. There is a close relationship between interest rates and Difference Between Macroeconomics. and therefore from one currency to another. Higher interest rates in The Effect of Interest Rates on Relationship of high interest rates and inflation.\nINFLATION, INTEREST RATE, AND EXCHANGE RATE: between interest and inflation rates. The IFE theory suggests that currency a relationship between interest rate.\nThe LongRun Relationship Between Real Exchange Rates and the real exchange ratereal interest rate relationship. REAL EXCHANGE RATES AND REAL INTEREST.",
null,
"",
null,
"",
null,
"",
null,
""
]
| [
null,
"https://www.ifund.com.hk/ifund/share/images/iClass/04-01E%281%29.png",
null,
"http://www.docslide.com/assets/images/large/2011/08/22612-relationships-between-inflation-interest-rates-and-exchange-rates-p1.gif",
null,
"http://3.bp.blogspot.com/-KM7z8VN9uio/U6m9X6VVbkI/AAAAAAAAChQ/4v2IPKjdtpM/s1600/4.png",
null,
"http://www.smarterwithmoney.in/Content/Article_Image1111120121756.jpg",
null,
"http://zuberbuehler-associates.ch/blog/wp-content/uploads/2014/08/relationship-between-spot-rates-and-forward-rates.jpg",
null,
"http://cdn.economicsdiscussion.net/wp-content/uploads/2015/11/clip_image002156.jpg",
null,
"http://2.bp.blogspot.com/-IG-FL_68Ygo/VIl_UGvcAhI/AAAAAAAAECo/mD2_dYXTlRk/s1600/Inflation%2Band%2BInterest.png",
null,
"https://i.ytimg.com/vi/I7FDx4DPapw/maxresdefault.jpg",
null,
"http://echelonim.com/wp-content/uploads/2015/10/bond-vs-stock.jpg",
null,
"https://www.fidelity.com/bin-public/060_www_fidelity_com/images/LC/70_2_1.png",
null,
"http://voxeu.org/sites/default/files/image/FromMay2014/hameedfig3.png",
null,
"http://www.sgmoneymatters.com/wp-content/uploads/2016/01/bond-price-interest-rate-relationship.jpg",
null,
"https://www.ifund.com.hk/ifund/share/images/iClass/04-01E%281%29.png",
null,
"http://www.docslide.com/assets/images/large/2011/08/22612-relationships-between-inflation-interest-rates-and-exchange-rates-p1.gif",
null,
"http://3.bp.blogspot.com/-KM7z8VN9uio/U6m9X6VVbkI/AAAAAAAAChQ/4v2IPKjdtpM/s1600/4.png",
null,
"http://www.smarterwithmoney.in/Content/Article_Image1111120121756.jpg",
null,
"http://zuberbuehler-associates.ch/blog/wp-content/uploads/2014/08/relationship-between-spot-rates-and-forward-rates.jpg",
null,
"http://cdn.economicsdiscussion.net/wp-content/uploads/2015/11/clip_image002156.jpg",
null,
"http://2.bp.blogspot.com/-IG-FL_68Ygo/VIl_UGvcAhI/AAAAAAAAECo/mD2_dYXTlRk/s1600/Inflation%2Band%2BInterest.png",
null,
"https://i.ytimg.com/vi/I7FDx4DPapw/maxresdefault.jpg",
null,
"http://echelonim.com/wp-content/uploads/2015/10/bond-vs-stock.jpg",
null,
"https://www.fidelity.com/bin-public/060_www_fidelity_com/images/LC/70_2_1.png",
null,
"http://voxeu.org/sites/default/files/image/FromMay2014/hameedfig3.png",
null,
"http://www.sgmoneymatters.com/wp-content/uploads/2016/01/bond-price-interest-rate-relationship.jpg",
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.9396496,"math_prob":0.9062339,"size":2601,"snap":"2019-26-2019-30","text_gpt3_token_len":456,"char_repetition_ratio":0.31882942,"word_repetition_ratio":0.050561797,"special_character_ratio":0.1633987,"punctuation_ratio":0.104477614,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98473567,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48],"im_url_duplicate_count":[null,4,null,8,null,4,null,10,null,6,null,4,null,4,null,9,null,null,null,null,null,4,null,6,null,4,null,8,null,4,null,10,null,6,null,4,null,4,null,9,null,null,null,null,null,4,null,6,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-07-17T02:46:10Z\",\"WARC-Record-ID\":\"<urn:uuid:c2581dc5-329d-4924-b406-3a6b298b4899>\",\"Content-Length\":\"35018\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d6626e5c-6744-4154-9f60-bc1132805c50>\",\"WARC-Concurrent-To\":\"<urn:uuid:b11bc5dd-dac8-40df-ad71-ce8f107868a5>\",\"WARC-IP-Address\":\"5.196.127.17\",\"WARC-Target-URI\":\"http://digitalforexnfashionstyles.com/cat3/560-relationship-between-currency-and-interest-rates.html\",\"WARC-Payload-Digest\":\"sha1:SV6J2MWDFXR64BQP4UDB6HK6PQUN23UR\",\"WARC-Block-Digest\":\"sha1:XBRGVEROTDHEO6HGMWBIOBNEJIU64HUM\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-30/CC-MAIN-2019-30_segments_1563195525009.36_warc_CC-MAIN-20190717021428-20190717043428-00316.warc.gz\"}"} |
http://www.numbersaplenty.com/150666322 | [
"Search a number\nBaseRepresentation\nbin10001111101011…\n…11110001010010\n3101111111122100011\n420332233301102\n5302032310242\n622541145134\n73506433262\noct1076576122\n9344448304\n10150666322\n1178057924\n124255b1aa\n13252a3291\n141601d7a2\n15d361d17\nhex8fafc52\n\n150666322 has 4 divisors (see below), whose sum is σ = 225999486. Its totient is φ = 75333160.\n\nThe previous prime is 150666319. The next prime is 150666337. The reversal of 150666322 is 223666051.\n\nIt is a semiprime because it is the product of two primes.\n\nIt can be written as a sum of positive squares in only one way, i.e., 85396081 + 65270241 = 9241^2 + 8079^2 .\n\nIt is a Smith number, since the sum of its digits (31) coincides with the sum of the digits of its prime factors. Since it is squarefree, it is also a hoax number.\n\nIt is an unprimeable number.\n\nIt is a polite number, since it can be written as a sum of consecutive naturals, namely, 37666579 + ... + 37666582.\n\nAlmost surely, 2150666322 is an apocalyptic number.\n\n150666322 is a deficient number, since it is larger than the sum of its proper divisors (75333164).\n\n150666322 is an equidigital number, since it uses as much as digits as its factorization.\n\n150666322 is an evil number, because the sum of its binary digits is even.\n\nThe sum of its prime factors is 75333163.\n\nThe product of its (nonzero) digits is 12960, while the sum is 31.\n\nThe square root of 150666322 is about 12274.6210532138. The cubic root of 150666322 is about 532.1148700453.\n\nThe spelling of 150666322 in words is \"one hundred fifty million, six hundred sixty-six thousand, three hundred twenty-two\".\n\nDivisors: 1 2 75333161 150666322"
]
| [
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.8658896,"math_prob":0.9955506,"size":1712,"snap":"2020-34-2020-40","text_gpt3_token_len":529,"char_repetition_ratio":0.15398127,"word_repetition_ratio":0.028169014,"special_character_ratio":0.4661215,"punctuation_ratio":0.13981763,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99741584,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-08-08T17:47:07Z\",\"WARC-Record-ID\":\"<urn:uuid:5e47cabf-dcf4-431a-8f66-1001bfb336e2>\",\"Content-Length\":\"8285\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:debfe04a-d253-453a-bc1c-93e490fa911e>\",\"WARC-Concurrent-To\":\"<urn:uuid:67161eae-c688-45e7-b092-e26ac537d85b>\",\"WARC-IP-Address\":\"62.149.142.170\",\"WARC-Target-URI\":\"http://www.numbersaplenty.com/150666322\",\"WARC-Payload-Digest\":\"sha1:TI4GODZY6YAM5I7GLQJXZ6MVN4Q5K4X3\",\"WARC-Block-Digest\":\"sha1:Y5R6D2AE4YQBXXROLXMGGAYPQO3UMI6I\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-34/CC-MAIN-2020-34_segments_1596439738015.38_warc_CC-MAIN-20200808165417-20200808195417-00502.warc.gz\"}"} |
https://universalium.enacademic.com/133004/integration | [
"\n\n# integration\n\n/in'ti gray\"sheuhn/, n.\n1. an act or instance of combining into an integral whole.\n2. an act or instance of integrating a racial, religious, or ethnic group.\n3. an act or instance of integrating an organization, place of business, school, etc.\n4. Math. the operation of finding the integral of a function or equation, esp. solving a differential equation.\n5. behavior, as of an individual, that is in harmony with the environment.\n6. Psychol. the organization of the constituent elements of the personality into a coordinated, harmonious whole.\n[1610-20; INTEGRATE + -ION; cf. L integratio renewal]\nSyn. 1. combination, blending, fusing.\n\n* * *\n\nIn calculus, the process of finding a function whose derivative is a given function.\n\nThe term, sometimes used interchangeably with \"antidifferentiation,\" is indicated symbolically with the integral sign ∫. (The differential dx usually follows to indicate x as the variable.) The basic rules of integration are: (1) ∫(f + g)dx = ∫fdx + ∫gdx (where f and g are functions of the variable x), (2) ∫kfdx = kfdx (k is a constant), and (3)",
null,
"(C is a constant). Note that any constant value may be added onto an indefinite integral without changing its derivative. Thus, the indefinite integral of 2x is x2 + C, where C can be any real number. A definite integral is an indefinite integral evaluated over an interval. The result is not affected by the choice for the value of C. See also differentiation.\n\n* * *\n\nin mathematics, technique of finding a function g(x) the derivative of which, Dg(x), is equal to a given function f(x). This is indicated by the integral sign “∫,” as in ∫f(x), usually called the indefinite integral of the function. (The symbol dx is usually added, which merely identifies x as the variable.) The definite integral, written\n\nwith a and b called the limits of integration, is equal to g(b) − g(a), where Dg(x) = f(x).\n\nSome antiderivatives can be calculated by merely recalling which function has a given derivative, but the techniques of integration mostly involve classifying the functions according to which types of manipulations will change the function into a form the antiderivative of which can be more easily recognized. For example, if one is familiar with derivatives, the function 1/(x + 1) can be easily recognized as the derivative of loge(x + 1). The antiderivative of (x2 + x + 1)/(x + 1) cannot be so easily recognized, but if written as x(x + 1)/(x + 1) + 1/(x + 1) = x + 1/(x + 1), it then can be recognized as the derivative of x2/2 + loge(x + 1). One useful aid for integration is the theorem known as integration by parts. In symbols, the rule is ∫fDg = fg − ∫gDf. That is, if a function is the product of two other functions, f and one that can be recognized as the derivative of some function g, then the original problem can be solved if one can integrate the product gDf. For example, if f = x, and Dg = cos x, then ∫x·cos x = x·sin x − ∫sin x = x·sin x − cos x + C. Integrals are used to evaluate such quantities as area, volume, work, and, in general, any quantity that can be interpreted as the area under a curve.\n\n* * *\n\nUniversalium. 2010.\n\n### Look at other dictionaries:\n\n• Integration — Integration … Deutsch Wörterbuch\n\n• intégration — [ ɛ̃tegrasjɔ̃ ] n. f. • 1700; « rétablissement » 1309; lat. integratio → intégrer 1 ♦ Math. Opération (inverse de la différentiation) par laquelle on détermine la grandeur limite de la somme de quantités infinitésimales en nombre indéfiniment… … Encyclopédie Universelle\n\n• integration — in‧te‧gra‧tion [ˌɪntˈgreɪʆn] noun [uncountable] 1. the combining of two or more organizations, activities etc so that they become more effective, make better profits etc: • The important achievement of the European Union is economic integration … Financial and business terms\n\n• Integration — is a process of combining or accumulating. It may specifically refer to:*The bringing together of two or more economies, cultures, religions (usually called syncretism), etc.;in sociology and economy *Social integration *Racial integration,… … Wikipedia\n\n• Integration — Intégration Cette page d’homonymie répertorie les différents sujets et articles partageant un même nom … Wikipédia en Français\n\n• integration — in·te·gra·tion n: the act or process or an instance of integrating: as a: a writing that embodies a complete and final agreement between parties b: incorporation as equals into society or an organization of individuals of different groups (as… … Law dictionary\n\n• Integration — In te*gra tion ([i^]n t[ e]*gr[=a] sh[u^]n), n. [L. integratio a renewing, restoring: cf. F. int[ e]gration.] [1913 Webster] 1. The act or process of making whole or entire. [1913 Webster] 2. (Math.) The operation of finding the primitive… … The Collaborative International Dictionary of English\n\n• integration — (n.) 1610s, from Fr. intégration and directly from L. integrationem (nom. integratio) renewal, restoration, noun of action from pp. stem of integrare (see INTEGRATE (Cf. integrate)). Anti discrimination sense is recorded from 1940 in a S.African… … Etymology dictionary\n\n• integration — integration. См. интеграция. (Источник: «Англо русский толковый словарь генетических терминов». Арефьев В.А., Лисовенко Л.А., Москва: Изд во ВНИРО, 1995 г.) … Молекулярная биология и генетика. Толковый словарь.\n\n• integration — articulation, concatenation (see under INTEGRATE vb) Analogous words: unification, consolidation, concentration (see corresponding verbs at COMPACT): integrity, union, *unity, solidarity … New Dictionary of Synonyms\n\n• integration — ► NOUN 1) the action or process of integrating. 2) the intermixing of peoples or groups previously segregated. DERIVATIVES integrationist noun … English terms dictionary"
]
| [
null,
"https://universalium.enacademic.com/pictures/universalium/09HMK27M2P_B5BE9F26.jpg",
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.77777195,"math_prob":0.9850748,"size":5730,"snap":"2020-45-2020-50","text_gpt3_token_len":1515,"char_repetition_ratio":0.1526371,"word_repetition_ratio":0.01748634,"special_character_ratio":0.23961605,"punctuation_ratio":0.15592904,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9748431,"pos_list":[0,1,2],"im_url_duplicate_count":[null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-10-31T10:17:30Z\",\"WARC-Record-ID\":\"<urn:uuid:fd69b038-e6a1-46e3-a4d0-075fd877737a>\",\"Content-Length\":\"59843\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:90f8d150-cc59-4031-b66b-6390db6dee4f>\",\"WARC-Concurrent-To\":\"<urn:uuid:f926310f-92d3-4e04-ba55-bd33e23d632a>\",\"WARC-IP-Address\":\"18.232.245.187\",\"WARC-Target-URI\":\"https://universalium.enacademic.com/133004/integration\",\"WARC-Payload-Digest\":\"sha1:MUROUAGOWGJBXPINJ5YP4BYK2XNISJN4\",\"WARC-Block-Digest\":\"sha1:ARHVKAY75F6CRPLDVYJU73XXBNNYZXXN\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-45/CC-MAIN-2020-45_segments_1603107917390.91_warc_CC-MAIN-20201031092246-20201031122246-00364.warc.gz\"}"} |
http://tonyohagan.co.uk/academic/abs/uncert.html | [
"",
null,
"Tony O'Hagan - Academic pages - Abstracts",
null,
"## Bayesian Inference for the Uncertainty Distribution of Computer Model Outputs\n\nJeremy Oakley and Anthony O'Hagan\n\nUniversity of Sheffield\n\nPublication details: Biometrika 89, 769-784, 2002.\n\n### Abstract\n\nWe consider a problem of inference for the output of a computationally expensive computer model. We suppose that the model is to be used in a context where the values of one or more inputs are uncertain, so that the input configuration is a random variable. We require to make inference about the induced distribution of the output. This distribution is called the uncertainty distribution, and the general problem is known to users of computer models as uncertainty analysis. Specifically, we develop Bayesian inference for the distribution and density functions of the model output. Modelling the output, as a function of its inputs, as a Gaussian process, we derive expressions for the posterior mean and variance of the distribution and density functions, based on data comprising observed outputs at a sample of input configurations. We show that direct computation of these expressions may encounter numerical difficulties. We develop an alternative approach based on simulating approximate realisations from the posterior distribution of the output function. Our methods are illustrated using a model describing the effective dose received by individuals on ingesting radioactive iodine.\n\nKeywords: Computer experiment; Gaussian process; Uncertainty analysis."
]
| [
null,
"http://tonyohagan.co.uk/Images/tuoslogo_rgb_lo.gif",
null,
"http://tonyohagan.co.uk/academic/abs/none",
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.8778278,"math_prob":0.7957794,"size":1374,"snap":"2023-40-2023-50","text_gpt3_token_len":247,"char_repetition_ratio":0.12773722,"word_repetition_ratio":0.0,"special_character_ratio":0.17467248,"punctuation_ratio":0.10526316,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9540281,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,8,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-05T17:31:46Z\",\"WARC-Record-ID\":\"<urn:uuid:15597ca2-c717-40a5-8a7d-9cd3d645e5af>\",\"Content-Length\":\"2625\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:1ea15ea2-7f2f-464f-8f18-bba3cce5225f>\",\"WARC-Concurrent-To\":\"<urn:uuid:64051a10-453c-4d2e-903e-71ff324cbc03>\",\"WARC-IP-Address\":\"188.64.184.30\",\"WARC-Target-URI\":\"http://tonyohagan.co.uk/academic/abs/uncert.html\",\"WARC-Payload-Digest\":\"sha1:57QC3INVQHSV4NERRS4BBMSEQYYD2KUL\",\"WARC-Block-Digest\":\"sha1:EYY4EPFSS3Q24W3J3GCWDQG3UI2GUNJZ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100555.27_warc_CC-MAIN-20231205172745-20231205202745-00853.warc.gz\"}"} |
https://www.jshero.net/en/koans/subfunction.html | [
"The browser storage localStorage is not available. Either your browser does not support it or you have disabled it or the maximum memory size is exceeded. Without localStorage your solutions will not be stored.\n\n# Functions call functions\n\nFunctions can call other functions:\n``````function gross(net) {\nlet vat = 20;\nlet tax = net * vat / 100;\nreturn net + tax;\n}\n\n`addWithVat` takes two net amounts and returns the sum of the gross amounts. `gross` calculates the gross amount from a net amount. Thereby a value added tax of 20% is assumed. Instead of calculating the gross amount itself, `addWithVat` use the function `brutto`. This has two advantages: The code is easier to read and the gross calculation is performed centrally in one place.\nWrite a function `sum` that takes an array of numbers and returns the sum of these numbers. Write a function `mean` that takes an array of numbers and returns the average of these numbers. The `mean` function should use the `sum` function."
]
| [
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.78853667,"math_prob":0.98108524,"size":1079,"snap":"2020-34-2020-40","text_gpt3_token_len":247,"char_repetition_ratio":0.15255813,"word_repetition_ratio":0.06629834,"special_character_ratio":0.21594068,"punctuation_ratio":0.10606061,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99053353,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-09-27T17:52:39Z\",\"WARC-Record-ID\":\"<urn:uuid:b799ab0b-bbd5-4a85-8f0f-42e441308991>\",\"Content-Length\":\"5250\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7dc32f7d-5acf-47f2-9a89-cf58d8766450>\",\"WARC-Concurrent-To\":\"<urn:uuid:cede5989-210b-45ed-ad7d-2982be6a9ab1>\",\"WARC-IP-Address\":\"185.26.156.172\",\"WARC-Target-URI\":\"https://www.jshero.net/en/koans/subfunction.html\",\"WARC-Payload-Digest\":\"sha1:YILXTO6SOORGUBF6FJYTAGNRSUVMYOOX\",\"WARC-Block-Digest\":\"sha1:KPZYAT4VE3E7IRK24OPH7PMDEPL3H6MI\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-40/CC-MAIN-2020-40_segments_1600400283990.75_warc_CC-MAIN-20200927152349-20200927182349-00606.warc.gz\"}"} |
https://probabilityhowto.com/cauchy-distribution/ | [
"# Cauchy Distribution\n\n< Back to Probability Distribution List < Cauchy distribution\n\n## What is the Cauchy distribution?\n\nThe Cauchy distribution (also called the Lorentz distribution, Cauchy–Lorentz distributionLorentz(ian) function, or Breit–Wigner distribution) is a family of continuous probably distributions that resemble normal distributions with taller peaks. However, their fat tails decay much more slowly than the normal distribution.\n\nThe Cauchy distribution, named after 18th century mathematician Augustin-Louis Cauchy, is well known for the fact that it’s expected value and other moments do not exist. This is one reason why this ill-behaved distribution is “…best known as a pathological case.” It is not stable under convolution and it doesn’t belong to any of the common parametric distribution families. In addition, the following are undefined or do not exist:\n\nDespite its limitations, the Cauchy’s properties — such as robustness — make it useful in some areas of study.\n\n## Cauchy Distribution Functions\n\nProbability Density Function\n\nThe general formula for the probability density function (PDF) is :\n\nWhere\n\nWhen t = 0, s = 1, the equation reduces to the standard Cauchy distribution:\n\nSupport (range) for the PDF is on (-∞, ∞)\n\nCumulative Distribution Function (CDF):\n\nThe Cauchy percent point function is\n\nThe Cauchy hazard function is\n\nThe Cauchy cumulative hazard function is\n\nThe Cauchy survival function is defined as\n\nInverse survival function:\n\nQuantile Function\n\n## Applications\n\nThe Cauchy distribution is often used as an example of a “pathological” distribution — a distribution that is ill-behaved. However, it does have a few practical applications. For example:\n\n• Robustness studies. The Cauchy is robust because it has heavy tails, which means is has a high probability of producing extreme values — thus making it a good choice for robustness studies.\n• Modeling a ratio of two normal random variables.\n• Modeling polar and non-polar liquids in porous glasses .\n• In quantum mechanics, it models the distribution of energy of an unstable state ."
]
| [
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.8455457,"math_prob":0.90053004,"size":2760,"snap":"2023-40-2023-50","text_gpt3_token_len":661,"char_repetition_ratio":0.16328011,"word_repetition_ratio":0.0049875313,"special_character_ratio":0.22463769,"punctuation_ratio":0.14226805,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9775318,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-04T12:26:49Z\",\"WARC-Record-ID\":\"<urn:uuid:517096d4-7a60-4448-ad91-7ab0400d7a4c>\",\"Content-Length\":\"135756\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:290ebeb5-2314-4647-ab97-cf82211a1bb8>\",\"WARC-Concurrent-To\":\"<urn:uuid:8343409b-a2ba-48eb-8834-e983d95c599e>\",\"WARC-IP-Address\":\"160.153.0.5\",\"WARC-Target-URI\":\"https://probabilityhowto.com/cauchy-distribution/\",\"WARC-Payload-Digest\":\"sha1:EGM35CYDCJG6TZ3CZJGCLYSNOPKEDALP\",\"WARC-Block-Digest\":\"sha1:4BYCXE6M2LQJ2WOATA5DQDG72GTJKV27\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100529.8_warc_CC-MAIN-20231204115419-20231204145419-00473.warc.gz\"}"} |
https://www.toppersbulletin.com/net-profit-ratio/ | [
"Home » Finance » Ratio Analysis » Net Profit Ratio\n\n# Net Profit Ratio\n\nNet Profit Ratio is the ratio of net profit after tax to net sales. It explains per rupee profit generating capacity of sales. This ratio is a measure of the overall profitability net profit is arrived at after taking into account both the operating and non-operating items of incomes and expenses. This ratio is very useful to the proprietors and prospective investors because it reveals the overall profitability of the concern. It is calculated as follows:",
null,
"For example if net profit after tax is Rs.80,000 and net sales are Rs.5,00,000, the net profit ratio will be calculated as follow:",
null,
"It is better the higher ratio because it gives idea of improved efficiency of the concern.\n\nNEXT – Profitability Indicator Ratios: Return on Capital Employed"
]
| [
null,
"data:image/svg+xml,%3Csvg%20xmlns='http://www.w3.org/2000/svg'%20viewBox='0%200%20468%2073'%3E%3C/svg%3E",
null,
"data:image/svg+xml,%3Csvg%20xmlns='http://www.w3.org/2000/svg'%20viewBox='0%200%20422%2069'%3E%3C/svg%3E",
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.8735671,"math_prob":0.9622288,"size":1143,"snap":"2023-40-2023-50","text_gpt3_token_len":238,"char_repetition_ratio":0.21510096,"word_repetition_ratio":0.046783626,"special_character_ratio":0.19422573,"punctuation_ratio":0.104477614,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9576747,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-23T16:46:00Z\",\"WARC-Record-ID\":\"<urn:uuid:23bba265-b2e3-42d6-8f22-9042787d735c>\",\"Content-Length\":\"73706\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8c1fe746-e62b-402d-aa2b-025f0d8332cf>\",\"WARC-Concurrent-To\":\"<urn:uuid:87588ce8-4d58-4a7c-a51f-b56223ae59dd>\",\"WARC-IP-Address\":\"104.21.13.20\",\"WARC-Target-URI\":\"https://www.toppersbulletin.com/net-profit-ratio/\",\"WARC-Payload-Digest\":\"sha1:2ULM7QQIY24KO3YI2BDZZNOTW7X3JTZT\",\"WARC-Block-Digest\":\"sha1:OYY36MSDD2VPVCQWVZYT2K7HZ5CSG6DJ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233506528.19_warc_CC-MAIN-20230923162848-20230923192848-00584.warc.gz\"}"} |
https://calculator.academy/charge-to-current-calculator/ | [
"Enter the total charge (C) and the change in time (s) into the calculator to determine the Current From Charge.\n\n## Current From Charge Formula\n\nThe following formula is used to calculate the Current From Charge.\n\nI = Q / dT\n\n• Where I is the Current From Charge (amps)\n• Q is the total charge (C)\n• dT is the change in time (s)\n\n## How to Calculate Current From Charge?\n\nThe following two example problems outline how to calculate the Current From Charge.\n\nExample Problem #1:\n\n1. First, determine the total charge (C). In this example, the total charge (C) is measured to be 140.\n2. Next, determine the change in time (s). For this problem, the change in time (s) is calculated to be 23.\n3. Finally, calculate the Current From Charge using the formula above:\n\nI = Q / dT\n\nInserting the values from above and solving the equation with the imputed values gives:\n\nI = 140 / 23 = 6.086 (amps)\n\nExample Problem #2:\n\nUsing the same process as example problem 1, we first define the variables outlined by the formula. In this case, the values are:\n\ntotal charge (C) = 290\n\nchange in time (s) = 50\n\nEntering these values into the formula or calculator above gives us::\n\nI = 290 / 50 = 5.8 (amps)"
]
| [
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.85861045,"math_prob":0.9981678,"size":1294,"snap":"2022-05-2022-21","text_gpt3_token_len":317,"char_repetition_ratio":0.17829457,"word_repetition_ratio":0.03846154,"special_character_ratio":0.26352397,"punctuation_ratio":0.10288066,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.999997,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-22T08:35:31Z\",\"WARC-Record-ID\":\"<urn:uuid:df71cca3-188f-479e-b23a-2d0d4ff7a61a>\",\"Content-Length\":\"129998\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:2aa9c055-d20e-40e9-99fe-ccbf5415afac>\",\"WARC-Concurrent-To\":\"<urn:uuid:a306cdcb-6c8d-4a47-b8f7-815f835fd31c>\",\"WARC-IP-Address\":\"172.67.69.208\",\"WARC-Target-URI\":\"https://calculator.academy/charge-to-current-calculator/\",\"WARC-Payload-Digest\":\"sha1:5MA63IKAH3FB3B4IQJURQAFY55TIH3PC\",\"WARC-Block-Digest\":\"sha1:HQH37QNTBVBVDZBXDZGUM2IM22YCYHEB\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652662545090.44_warc_CC-MAIN-20220522063657-20220522093657-00737.warc.gz\"}"} |
https://freeyourmindtravel.com/get-visa/how-far-does-a-car-travel-in-5-seconds.html | [
"How far does a car travel in 5 seconds?\n\nContents\n\nHow far does the car travel in 5 seconds?\n\nIn 5 seconds the car will travel a distance of 5 × 25 = 125 meters.\n\nHow far can a car travel in a second?\n\nAnswer: A car travels 0.017 miles in 1 second.\n\nHow far does a car go in 10 seconds?\n\nSo, 10 seconds= 10/3600= 1/360 hr. Distance = speed *time = 36*(1/360)=1/10km.\n\nHow far do you travel in 2 seconds?\n\nIn 2 seconds, it will fall d = 16(22) = 64 feet.\n\nHow far does a car travel in 2 seconds at 60 mph?\n\n316800 ft ÷ 3600 sec = 88 ft. per sec. So at 60 mph he is traveling 88 ft. per second.\n\nHow far does a car travel in 1 second at 70 mph?\n\n70 mph is just over 100 feet per second (102.7 ft/sec).\n\nHow far can a car travel in 3 seconds?\n\nThe Three-Second Rule Chart\n\nThree-Second Rule 3 seconds\nSpeed Distance Traveled Good\n25 m.p.h. 37 ft. per second 111 ft.\n35 m.p.h. 52 ft. per second 166 ft.\n45 m.p.h. 66 ft. per second 198 ft.\n\nHow far do you travel in 1 second at 60 kph?\n\nFor a fit and alert driver, in good driving conditions, at 60 km/h it takes about 3/4 of a second or 12 metres of travel for you to realise that you have to brake and at least another 20 metres to stop. At higher speeds, the distance will be even greater.\n\nHow far would a car travel in 4 seconds if it is moving with a speed of 45 km h?\n\nThus, the car would travel a distance of 50 metres in 4 seconds.\n\nHow fast would you have to go to go a mile a second?\n\nPace (Various Sports)\n\nmile per second to minutes per kilometer 0.01036\nmile per second to minutes per mile 0.01667\nmile per second to seconds per mile 1\nmile per second to time per mile (HH:MM:SS) 00:00:01\nmile per second to seconds per 100 yards 0.05682\n\nHow do you find the distance?\n\nTo solve for distance use the formula for distance d = st, or distance equals speed times time. Rate and speed are similar since they both represent some distance per unit time like miles per hour or kilometers per hour. If rate r is the same as speed s, r = s = d/t.\n\nHow far does a car travel in 1 second at 65 mph?\n\nAt 65 mph, keeping one second of following distance means the vehicle is traveling 100 feet behind the vehicle in front. It takes an automobile a minimum of 150 feet to stop. With a 100 foot gap plus the auto’s stopping distance of 150 feet, an impact would occur within 250 feet.\n\nIMPORTANT: Can I work in India with green card?"
]
| [
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.8973102,"math_prob":0.98605037,"size":2297,"snap":"2022-05-2022-21","text_gpt3_token_len":657,"char_repetition_ratio":0.20061055,"word_repetition_ratio":0.08008658,"special_character_ratio":0.3286896,"punctuation_ratio":0.12,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98507595,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-01-17T07:47:03Z\",\"WARC-Record-ID\":\"<urn:uuid:3dc8f5ce-7400-4650-98d2-f74aa479efbc>\",\"Content-Length\":\"72099\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:da422262-7476-4687-9a83-fad462317c71>\",\"WARC-Concurrent-To\":\"<urn:uuid:e4d73bd1-3fe8-43ab-b2d5-d17e6fdc2dfd>\",\"WARC-IP-Address\":\"82.146.60.64\",\"WARC-Target-URI\":\"https://freeyourmindtravel.com/get-visa/how-far-does-a-car-travel-in-5-seconds.html\",\"WARC-Payload-Digest\":\"sha1:ZKVO3G52I2TYEFBWD7TEWG3TT6VVT7UQ\",\"WARC-Block-Digest\":\"sha1:KDMJ5ZQZDFO7VYSPDJKM44USV6B6ECLS\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-05/CC-MAIN-2022-05_segments_1642320300343.4_warc_CC-MAIN-20220117061125-20220117091125-00247.warc.gz\"}"} |
https://www.worldwideconverter.com/length/feet/to/fermis | [
"",
null,
"Convert Feet (ft) to Fermis (fm) | ft in fm\nYou are: Home > Length > Feet to Fermis\n\n# Let's convert Feet (ft) to Fermis (fm)\n\nThis quick and easy calculator will convert Feet (ft) to Fermis (fm) and show formula, brief history on the units and quick maths for the conversion.\n\n## Quick Reference for Converting Feet to Fermis\n\nFormula\nfm = ft x 3.048E+14\nQuick Rough Maths\nTo get the Fermis, multiply the number of Feet by 304.8 trillion\nFeet (ft) in 1 Fermi\nThere are 0 Feet in 1 Fermi\nFermis (fm) in 1 Foot\nThere are 3.048E+14 Fermis in 1 Foot\n\n## Unit Information\n\nFoot\n/fʊt/\nSymbol: ft\nUnit System: Imperial\n\n### What is the Foot?\n\nThe foot is a unit of length in the imperial unit system and uses the symbol ft.\n\nOne foot is exactly equal to 12 inches. It is also exactly equal to 0.3048m. One yard is comprised of three feet.\n\nThe foot is just behind the metre in terms of widespread use due to its previous popularity. The US is the only developed country that still uses the foot in preference to the metre.\n\nThe UK still uses feet to express human height more than metres.\n\nFermi\n/ˈfəːmi/\nSymbol: fm\nUnit System: Non-SI metric\n\n### What is the Fermi?\n\nThe fermis is a unit of length in the non-SI metric system and uses the symbol fm.\n\nOne fermi is equal to 1×10−15m or 3.9370×10−14 in.\n\nIt is used to measure nuclear distances and was named after the Italian physicist Enrico Fermi (1901–1954). It has now been replaced with the SI-derived term femtometre."
]
| [
null,
"https://www.facebook.com/tr",
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.95719147,"math_prob":0.9077571,"size":746,"snap":"2023-40-2023-50","text_gpt3_token_len":184,"char_repetition_ratio":0.1212938,"word_repetition_ratio":0.05970149,"special_character_ratio":0.25201073,"punctuation_ratio":0.08125,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98026514,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-09T04:18:17Z\",\"WARC-Record-ID\":\"<urn:uuid:a91db3bb-2feb-499d-9875-15ac1e762de0>\",\"Content-Length\":\"29037\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8dee9389-c99a-4e17-9b59-f662baf3afeb>\",\"WARC-Concurrent-To\":\"<urn:uuid:44f907ab-fdc0-4616-a215-bf55658410e8>\",\"WARC-IP-Address\":\"77.68.116.144\",\"WARC-Target-URI\":\"https://www.worldwideconverter.com/length/feet/to/fermis\",\"WARC-Payload-Digest\":\"sha1:NDVRQ4L6IOFULPAC3E6U4F5CAWUJUQDJ\",\"WARC-Block-Digest\":\"sha1:D6VP6UALS24GQKWXL4WIGI72XWB5URUV\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100800.25_warc_CC-MAIN-20231209040008-20231209070008-00802.warc.gz\"}"} |
https://cs.stackexchange.com/questions?sort=active | [
"All Questions\n\n31,082 questions\n3 views\n\nForcing an edge to be in S-T min-cut\n\nGiven a Flow-Network N=(G,c,s,t) and an edge e=(u,v), i am trying to build an algorithm that finds a minimum (S,T) cut in the given network, that includes e. So, I tried couple of steps, first, I ...\n3 views\n\nMistake in a proof of termination phase of Simplex algorithm in CLRS?\n\nThere is a pseudo-code for Simplex algorithm in CLRS: The proof consists from three-part loop invariant: Proof We use the following three-part loop invariant: At the start of each iteration ...\n6 views\n\nWhere does fact constellation schema differ from star and snowflake schema?\n\nIn the 3 main (for my level) schemas for data warehouse modelling are the star, snowflake and the fact constellation schemas. I understand that the star schema is a purely denormalized schema as the ...\n27 views\n\nHow to give a context-sensitive grammar for words like $a^nba^nba^nb$?\n\ni am struggling on this problem since days: $L = \\{a^nba^nba^nb \\mid n \\in \\Bbb N\\}$ i have to give for this language a context-sensitive grammar.\n21 views\n\nAssigning $m$ balls to $n$ buckets - recursive algorithm\n\nI came across the following problem and the answer to that problem: Given $m$ balls and $n$ bins, find out how many ways to assign the balls to the bins. Notice the bins have no order: for example, ...\n13 views\n\ncounting the integer solutions to 1000a + 100b + 10c + d where 0 < x,y < 20 [on hold]\n\nI am trying to figure out the integer solutions to 1000a + 100b + 10c + d where 0 < x,y < 20. I can print out the solutions by replacing n = n + 1 with 'print(a,b,c,d)' and removing the 'print(n)...\n28 views\n\nGiven price and number of pages of each book, What is the maximum number of pages you can buy?\n\nYou are in a book shop which sells n different books. You know the price and number of pages of each book. You have decided that the total price of your purchases will be at most x. What is the ...\n20 views\n\nAlgorithm for deducing values\n\nI have a group of logical conditions and need to deduce values that would NOT satisfy them. For example: City != New York && Location = Museum ...\n11 views\n\nAre the languages recognized by one-counter machines equivalent to deterministic context free language?\n\nWhen learning automata theory, I notice that John Hopcroft mentioned In fact, a PDA In fact the languages of one counter machines are accepted by deterministic PDA's although the proof is ...\n9 views\n\nOverlay two Voronoi Diagrams and calculate membership and areas of intersecting polygons\n\nI would like to generate a composite diagram of two Voronoi diagrams. I'm currently researching the cgal library for options, but I'm not sure if my precise application is covered. Basically, I have ...\n20 views\n\nHow to prove that extended euclidean algorithm has time complexity log(max(m,n)) ?'\n\nI tried to search on internet and also thought by myself but was unsuccessful. Intuitively i think it should be O(max(m,n)). can someone give easy explanation since i am beginner in algorithms.\n40 views\n\nIs there any relationship between time complexity and space complexity of an algorithm?\n\nFor example: If algorithm A takes an input of size n, and has a time complexity of O(a^n) and a space complexity of O(1) Is there a way to increase the space complexity to something like O(n^2) that ...\n26 views\n\nAcceptance problem for CFGs is not regular\n\nLet $ACFG$ be the language of all encodings $(C,x)$ where $C$ is a context free grammar that generates a language containing $x$, i.e. $ACFG$ is the acceptance problem for context free grammars. It ...\n36 views\n\nIntersection of two deterministic parity automata\n\nGiven two deterministic parity automata $A_1=(Q_1,\\Sigma,\\delta,q_{01},c_1)$ and $A_2=(Q_2,\\Sigma,\\delta,q_{02},c_2)$ with the finite set of states $Q_i$, the finite alphabet $\\Sigma_i$, the ...\n17 views\n\nI need advices and guidance! [on hold]\n\nI have a lot of questions and a lot of doubts of what should i do with my life. The thing is, i finished IT college in Bosnia and Herzegovina ( 3 year bachelors degree). I didn’t master any IT skills ...\n49 views\n\nDSPACE(f(n)) closed under complement\n\nI think you can create the complementary language that is an element of DSPACE($f(n)$), where $f(n) \\geq \\log(n)$ by adding a step to the algorithm that reverses the answer. By that the function $f(n)$...\n25 views\n\nLongest simple circuit and P=NP relation\n\nGiven the following function: $$\\:f\\left(G,v\\right)\\:=\\:size\\:of\\:the\\:longest\\:simple\\:circuit\\:in\\:a\\:directed\\:graph\\:G\\:that\\:contains\\:v$$ Output: Function returns a natural number or 0, which ...\n35 views\n\nHelp: Context Free Grammar [on hold]\n\nConstruct the CFG given the following language: $$\\{a^i \\; b^j \\; c^k \\;|\\; i = j \\; or \\; j = k \\}$$\n36 views\n\n511 views\n\nDecisional problems vs Optimization problems : NP-COMPLETE vs NP-HARD\n\nI'm studying some of computation theory and i encounter a big question mark. I have an optimization problem and i have to proof that is NP-HARD. I know that my problem can be reduced to another np-...\n175 views\n\nProving there is no polynomial algorithm for independent set\n\nI need some guidance in an assignment I'm doing. I'm at complete loss, he says the the MAXIMUM INDEPENDENT SET problem is NP-hard and then asks me to prove that there is no polynomial time for the ...\n98 views\n\nWhy does coNP⊆NP∖P imply that the polynomial hierarchy collapses?\n\nI was looking for some information on 1-in-3 SAT and came across this paper, last updated 9 days ago, which claims that the Polynomial Time Hierarchy collapses \"to the level above P=NP\". That's quite ...\n38 views\n\nDealing with test condition '=' for a while loop when determining a bound function/loop variant\n\nThe following is the definition of what a bound function for a while loop must satisfy: The bound function is an integer-valued, total function of some of the inputs, variables and global data that ...\n38 views\n\nBinary search uneven split number of queries?\n\nFor even split binary search (repeated halving) number of queries is log with base 2. According to Skiena's Algorithm Design Manual, if the split in binary search is by ratio 1/3:2/3 instead of 1/2:1/...\n26 views\n\nGiven a randomised algorithm for a problem, is there a corresponding deterministic algorithm?\n\nDoes every randomised algorithm have a corresponding deterministic algorithm solving the same problem? I'm going to guess that the deterministic algorithm will take more time, while the randomised ...\n23k views\n\nDoes a byte contain 8 bits, or 9?\n\nI read in this assembly programming tutorial that 8 bits are used for data while 1 bit is for parity, which is then used for detecting parity error (caused by hardware fault or electrical disturbance)....\n30 views\n\nWhy do P-class problems not require a witness like NP?\n\nIf I am given a problem in P, to claim that I have solved it, I would need to give proof, right? For example, to show a number is not prime, if I use some deterministic algorithm, I can return the ...\n21 views\n\nUsing the decision problem, PATH in order to solve the optimization problem, SHORTEST-PATH in polynomial time\n\nSo, if I were using a black-box decision algorithm, PATH in which I could say, \"does a path of weight k exist in this graph from ...\n59 views\n\nHow much can we trust mathematical software when working with large numbers, and how much memory it needs to work with these numbers?\n\nFor example, I want to evaluate the expression: $3^{3^{{3}^{3}}}$ so I used wolframalpha.com (it's free, and I don't own any software), which returned the scientific notation of the number above, ...\n51 views\n\nHeuristics vs meta-heuristics vs hyper-heuristics?\n\nThe wikipedia page on meta-heuristics states that they are \"heuristics designed to find, generate, or select a heuristic\". The wikipedia page on hyper-heuristics states that they are \"heuristic ...\n25 views\n\nHow do i make a percentage based loot system C# [on hold]\n\nso i have been trying to figure this one out on my own for a little while now for my game i am creating, but i am not able to get my head around it. I basically would like too know how to create a ..."
]
| [
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.9041082,"math_prob":0.9544554,"size":14030,"snap":"2019-26-2019-30","text_gpt3_token_len":3650,"char_repetition_ratio":0.13859974,"word_repetition_ratio":0.017312448,"special_character_ratio":0.26072702,"punctuation_ratio":0.13090278,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99665934,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-06-20T22:07:44Z\",\"WARC-Record-ID\":\"<urn:uuid:4232a939-5454-425c-b088-b841d16dda80>\",\"Content-Length\":\"228410\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4a3e0966-4bfb-4588-9f60-6de737c6eb15>\",\"WARC-Concurrent-To\":\"<urn:uuid:2a0020a7-a2e2-4a94-8bc1-f9b1956e58d9>\",\"WARC-IP-Address\":\"151.101.1.69\",\"WARC-Target-URI\":\"https://cs.stackexchange.com/questions?sort=active\",\"WARC-Payload-Digest\":\"sha1:OQPXAN7MPQHQDAWZF33OCJ5WYSR2PSFF\",\"WARC-Block-Digest\":\"sha1:SUOZYB4DEOL2QYDY36MG2NFI2Y4CNOLB\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-26/CC-MAIN-2019-26_segments_1560627999273.79_warc_CC-MAIN-20190620210153-20190620232153-00420.warc.gz\"}"} |
https://www.brightstorm.com/tag/depression/ | [
"# depression 5 videos\n\n• #### Angle of Elevation and Depression\n\n##### GeometryBasic Trigonometry\n\nHow to define an angle of elevation or an angle of depression.\n\ntrigonometry angle of elevation angle of depression\n• #### Angle of Elevation and Depression\n\n##### TrigonometryBasic Trigonometry\n\nHow to define an angle of elevation or an angle of depression.\n\ntrigonometry angle of elevation angle of depression\n• #### Angle of Elevation and Depression\n\n##### PrecalculusBasic Trigonometry\n\nHow to define an angle of elevation or an angle of depression.\n\ntrigonometry angle of elevation angle of depression\n• #### Freezing Point Depression\n\n##### ChemistryChemical Solutions\n\nUnderstanding the colligative property of solutions that decreases the freezing point.\n\nfreezing point depression elevation pressure vapor"
]
| [
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.63829666,"math_prob":0.6440173,"size":756,"snap":"2023-14-2023-23","text_gpt3_token_len":151,"char_repetition_ratio":0.2712766,"word_repetition_ratio":0.6057692,"special_character_ratio":0.15476191,"punctuation_ratio":0.036036037,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9914286,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-03-21T20:10:35Z\",\"WARC-Record-ID\":\"<urn:uuid:df6f9a96-be93-4ec8-a623-9b64d70c53f8>\",\"Content-Length\":\"83121\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:67eb7e73-be5e-4432-a20f-fd6611821b27>\",\"WARC-Concurrent-To\":\"<urn:uuid:1d22accc-8c01-4837-98c9-c21930d847a6>\",\"WARC-IP-Address\":\"34.207.32.175\",\"WARC-Target-URI\":\"https://www.brightstorm.com/tag/depression/\",\"WARC-Payload-Digest\":\"sha1:Z3RFAPWZRDF4PW23XNTWEKJ3TAIKBUDU\",\"WARC-Block-Digest\":\"sha1:LCCEBGAT4I4SM3YCUXKOFHJRGQOZZRYG\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-14/CC-MAIN-2023-14_segments_1679296943746.73_warc_CC-MAIN-20230321193811-20230321223811-00416.warc.gz\"}"} |
https://math.stackexchange.com/questions/850813/branches-of-the-complex-logarithm/850819 | [
"# Branches of the complex logarithm\n\nProblem statement\n\nLet $\\Omega \\subset \\mathbb C^*$ be open, we call branch of logarithm of $z$ in $\\Omega$ to every continuous function $f:\\Omega \\to \\mathbb C$ such that $e^{g(z)}=z$ for all $z \\in \\Omega$.\n\n$(i)$ Prove that every branch of logarithm is injective and holomorphic on $\\Omega$. Let $f_1,f_2$ be two branches of logarithm on $\\Omega$. Show that if $\\Omega$ is connected and there is $z_0 \\in \\Omega$: $f_1(z_0)=f_2(z_0) \\implies f_1=f_2$ on $\\Omega$.\n\n$(ii)$ Prove that if there exists a branch of logarithm on $\\Omega$, then $S^1 \\not \\subset \\Omega$.\n\nMy attempt at a solution\n\nFor $(i)$, I suppose I must find a continuous function $f$ continuous on $\\Omega$ and such that $e^{f(z)}=z$. I know that for $z=a+ib \\in \\mathbb C$, we define $$e^z=e^{a+ib}=e^ae^{ib}=e^a(\\cos(b)+i\\sin(b))$$ Also, if $e^z=w=|w|\\dfrac{w}{|w|}$, then it is clear that $e^a=|w|$ and $e^{ib}=\\dfrac{w}{|w|}$.\n\nSo, if I define the function $f(z)=log(|z|)+iarg(z)$, then $e^{f(z)}=e^{log(|z|)}e^{iarg(z)}=|z|(\\cos(arg(z))+i\\sin(arg(z)))=z$.\n\nAs $f$ is sum of two continous functions, then $f$ is continuous. Now, is $f$ is well defined? I ask this because for $z \\in \\mathbb C$, it is the same to describe $z$ by $arg(z)$ or $arg(z)+i2k\\pi$.\n\nI don't know how to prove the other part other part of $(i)$ and I also don't know what it means the notation $\\mathbb C^*$.\n\nAs for $(ii)$ I don't know how to prove that statement but what's more important is that I intuitively don't get why can't $S^1$ be in $\\Omega$ for existence of a branch.\n\n• $\\mathbb C^*$ means $\\mathbb C -(0,0)$. Intuitively, IMHO, the complications with the log have to see , not with the Real part, but with its complex part, which is the argument.Once you wind around the origin, your argument cannot be continuous, since it must jump from $0$ to $2\\pi$ upon winding around the origin. The condition of winding around the origin is equivalent to containing a copy of $S^1$. – gary Jun 28 '14 at 23:40\n\nAs you correctly observe, defining a branch of log in $\\Omega$ comes down to defining arg continuously over all of $\\Omega$. For any particular $0 \\ne z \\in \\mathbb{C}$, you have many choices for arg. They are obtained by taking any one choice and then adding or subtracting a multiple of $2\\pi$. For example, if $z=1$ then the choices are the integer multiples of $2\\pi$.\n\nFor (i), the idea is that if $\\Omega$ is connected, once you choose a value for arg for any single point $z_0 \\in \\Omega$, the rest of the values are forced. Thus if $f_1$ and $f_2$ agree at a single point, they agree on all of $\\Omega$.\n\nFormally, if arg is continuous in the neighborhood of a point $z_0 \\in \\Omega$, then there is a neighborhood $N$ of $z_0$ where the image $\\arg N$ is contained in an open interval of length < $2\\pi$. Thus arg is uniquely determined by its value at $z_0$. Now for any other point $z \\in \\Omega$, take a path from $z_0$ to $z$. We can find neighborhoods $N_w$ as above for any point $w$ along the path. By compactness of the path, we can find finitely many overlapping $N_w$ that cover the path, so arg is uniquely determined along the entire path to $z$. See picture at: http://en.wikipedia.org/wiki/Monodromy_theorem\n\nFor (ii), observe that if $S^1 \\subset \\Omega$ then when you wind around $S^1$ once counterclockwise, whatever value of arg you started with will increase by $2\\pi$ by the time you come back around. Thus you cannot make a consistent choice of arg in $\\Omega$.\n\n• I have some doubts: how can I formally prove that connectedness determines the arg? On the other hand, I've proposed the function $f(z)=log(|z|)+iarg(z)$ I know that $log(x)$ is differentiable for $x>0$, but why can I affirm that $f$ is holomorphic? – user100106 Jun 29 '14 at 18:05\n• @user100106 See some details for (i) edited in. – Ted Jun 29 '14 at 18:41\n• @Ted Would u be able to have me with this releated about branch points question as well? Thanks math.stackexchange.com/questions/1905092/… – ys wong Aug 27 '16 at 15:21\n\nThe way that branches of logz make more sense to me is that they are local inverses of $e^z$; $e^z$ is not globally-invertible, since it is infinite-to-1 , because it is periodic, but , e.g., by the inverse function theorem, since $d/dz(e^z)=e^z \\neq 0$, $e^z$ does have local inverses, which are called branches. Each branch is a rectangle of infinite width, and with height $2\\pi^-$ , i.e., the height is of the form $[y, y+2\\pi)$, which is made to prevent $e^z$ from realizing a period within the region, i.e., since $e^z=e^{z+2\\pi}$ we want an injection, so we want to avoid having $z-$ values that are $2\\pi$ appart.\n\nFor 2), a standard result is that a branch of logz exists in $\\Omega$ when $\\Omega$ is simply-connected, and it does not wind around the origin. For one thing, logz is defined as $\\int_{\\gamma} \\frac{dz}{z}$ , where $\\gamma \\in \\Omega$ , and for the integral to be well-defined, we need independence of path, for which we need a simply-connected region. If logz was globally defined, then the integral about a closed curve would be zero. But the integral of $dz/z$ about $S^1$ is known to be $2\\pi i$; this will happen whenever you wind around the origin, which is equivalent to having a copy of $S^1$ in your region $\\Omega$.\n\nFor1), you can just use the fact that in a branch , as you said, you have $e^{logz}=z=log(e^z)$, then, by invertibility, you must have a bijection and, in particular, an injection.\n\n• There are two things where I still have doubts: in part (i) I still don't know how to prove that if $\\Omega$ is connected and there is $z_0 \\in \\Omega$ such that $f_1(z_0)=f_2(z_0) \\implies f_1=f_2$ on $\\Omega$. For $(ii)$, I understand and like your argument but here the hypothesis is that $\\Omega \\subset \\mathbb C^*$, so I cannot use that $\\Omega$ is simple connected, just the fact that $\\Omega \\subset \\mathbb C^*$ and that $\\Omega$ is open. – user100106 Jun 29 '14 at 6:10\n• For i), if I understood you correctly, you have that branches are local inverses, so these are bijections of $\\mathbb R \\times [y,y+2\\pi)$ with $\\mathbb C^*$, so that they are $1-1$. So if $f_1(z)=f_2(z)$, these must both be the same branch of logz. for ii), I assume $S^1$:={$z:|z|=1$}, so that if $\\Omega$ contains a copy of $S^1$ , then $\\Omega$ must either not be simply-connected, or $\\Omega$ must wind around the origin. I hope I understood (and answered) your questions correctly. – gary Jun 29 '14 at 6:19\n• You are assuming what you are trying to prove. Assuming $e^{log(z)}=log(e^z)$ is assuming they are inverses (i.e. one to one) before you have proven that they are one to one. Same thing with the inverse function theorem. You are assuming that the logarithms are differentiable when that is what the OP is asking to prove. – Bobby Ocean Jun 29 '14 at 6:39\n• By definition/construction a branch logz of the log is a local inverse of $e^z$, so that $e^{logz}=log(e^z)=z$ within the branch. I don't know where differentiability comes up, but $e^z$ is differentiable, with $d/dz(e^z)=e^z \\neq 0$ , so that the inverse function theorem guarantees a local inverse; that local inverse is the branch $logz$. – gary Jun 29 '14 at 6:44\n• @gary I agree with what you are saying except that what you are saying is not proof. The OP was not told that every branch of the lograithm is a local inverse of $e^z$''. I agree that is true, but I would think that is what the OP is suppose to prove (not just state it is so). Likewise, the inverse function theorem states that the local inverse of a non-zero analytic function is differentiable (I agree), but that is not what was being asked. The OP is suppose to prove that statement for the branchs of the logarithm, not just state it is so. Unless I am misunderstanding the OP. – Bobby Ocean Jun 29 '14 at 6:53\n\nProof of (ii), Suppose there exists a continuous branch of argument on $$\\Omega$$, with $$S^1 \\subset \\Omega$$, say $$f$$.\n\nConsider the function $$g(\\theta)=\\frac{f(e^{i\\theta})+f(e^{-i\\theta})}{2\\pi}\\hspace{5mm} \\theta \\in [0,2\\pi)$$ Then $$g$$ is continuous on $$[0,2\\pi)$$ (connected set) and $$g$$ takes only integer values. So, $$g$$ is constant function. So, $$g(0)=g(\\pi) \\implies f(1)=f(-1)$$, which is a contradiction, as $$f(1)\\in \\{2n\\pi: n \\in \\mathbb{Z}\\}$$ and $$f(-1)\\in \\{2n\\pi-\\pi: n \\in \\mathbb{Z}\\}$$.\n\nSo there does not exist a continuous branch of argument on $$\\Omega$$, with $$S^1 \\in \\Omega$$, and hence there does not exist a holomorphic branch of logarithm on $$\\Omega$$"
]
| [
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.8443694,"math_prob":0.9997323,"size":1521,"snap":"2021-04-2021-17","text_gpt3_token_len":536,"char_repetition_ratio":0.12722479,"word_repetition_ratio":0.017021276,"special_character_ratio":0.3530572,"punctuation_ratio":0.08083832,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99998415,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-01-20T13:15:16Z\",\"WARC-Record-ID\":\"<urn:uuid:772e5d94-1cf5-4d29-b776-2923f5ea85fd>\",\"Content-Length\":\"179391\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:3e96bf10-3965-4a7a-9054-a7b24e7b129f>\",\"WARC-Concurrent-To\":\"<urn:uuid:d13a8664-debb-4af1-9a69-2ec02270545b>\",\"WARC-IP-Address\":\"151.101.65.69\",\"WARC-Target-URI\":\"https://math.stackexchange.com/questions/850813/branches-of-the-complex-logarithm/850819\",\"WARC-Payload-Digest\":\"sha1:IXLXSPS43BBCH5TPCNAFVRPFQVOEX7ZZ\",\"WARC-Block-Digest\":\"sha1:ESQUBGRE4XJGABV2CFT3FS3GKLXQ2KQJ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-04/CC-MAIN-2021-04_segments_1610703520883.15_warc_CC-MAIN-20210120120242-20210120150242-00616.warc.gz\"}"} |
https://nyuscholars.nyu.edu/en/publications/filtering-nonlinear-dynamical-systems-with-linear-stochastic-mode | [
"# Filtering nonlinear dynamical systems with linear stochastic models\n\nJ. Harlim, A. J. Majda\n\nResearch output: Contribution to journalArticlepeer-review\n\n## Abstract\n\nAn important emerging scientific issue is the real time filtering through observations of noisy signals for nonlinear dynamical systems as well as the statistical accuracy of spatio-temporal discretizations for filtering such systems. From the practical standpoint, the demand for operationally practical filtering methods escalates as the model resolution is significantly increased. For example, in numerical weather forecasting the current generation of global circulation models with resolution of 35 km has a total of billions of state variables. Numerous ensemble based Kalman filters (Evensen 2003 Ocean Dyn. 53 343-67; Bishop et al 2001 Mon. Weather Rev. 129 420-36; Anderson 2001 Mon. Weather Rev. 129 2884-903; Szunyogh et al 2005 Tellus A 57 528-45; Hunt et al 2007 Physica D 230 112-26) show promising results in addressing this issue; however, all these methods are very sensitive to model resolution, observation frequency, and the nature of the turbulent signals when a practical limited ensemble size (typically less than 100) is used. In this paper, we implement a radical filtering approach to a relatively low (40) dimensional toy model, the L-96 model (Lorenz 1996 Proc. on Predictability (ECMWF, 4-8 September 1995) pp 1-18) in various chaotic regimes in order to address the 'curse of ensemble size' for complex nonlinear systems. Practically, our approach has several desirable features such as extremely high computational efficiency, filter robustness towards variations of ensemble size (we found that the filter is reasonably stable even with a single realization) which makes it feasible for high dimensional problems, and it is independent of any tunable parameters such as the variance inflation coefficient in an ensemble Kalman filter. This radical filtering strategy decouples the problem of filtering a spatially extended nonlinear deterministic system to filtering a Fourier diagonal system of parametrized linear stochastic differential equations (Majda and Grote 2007 Proc. Natl Acad. Sci. 104 1124-9; Castronovo et al 2008 J. Comput. Phys. 227 3678-714); for the linear stochastically forced partial differential equations with constant coefficients such as in (Castronovo et al 2008 J. Comput. Phys. 227 3678-714), this Fourier diagonal decoupling is a natural approach provided that the system noise is chosen to be independent in the Fourier space; for a nonlinear problem, however, there is a strong mixing and correlations between different Fourier modes. Our strategy is to radically assume for the purposes of filtering that different Fourier modes are uncorrelated. In particular, we introduce physical model errors by replacing the nonlinearity in the original model with a suitable Ornstein-Uhlenbeck process. We show that even with this 'poor-man's' stochastic model, when the appropriate parametrization strategy is guided by mathematical offline test criteria, it is able to produce reasonably skilful filtered solutions. In the highly turbulent regime with infrequent observation time, this approach is at least as good as trusting the observations while the ensemble Kalman filter implemented in a perfect model scenario diverges. Since these Fourier diagonal linear filters have large model error compared with the nonlinear dynamics, an essential part of the study below is the interplay between this error and the mathematical criteria for a given linear filter in order to produce skilful filtered solutions through the radical strategy.\n\nOriginal language English (US) 1281-1306 26 Nonlinearity 21 6 https://doi.org/10.1088/0951-7715/21/6/008 Published - Jun 1 2008\n\n## ASJC Scopus subject areas\n\n• Statistical and Nonlinear Physics\n• Mathematical Physics\n• Physics and Astronomy(all)\n• Applied Mathematics"
]
| [
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.8797195,"math_prob":0.73704356,"size":3810,"snap":"2021-21-2021-25","text_gpt3_token_len":810,"char_repetition_ratio":0.11403048,"word_repetition_ratio":0.014260249,"special_character_ratio":0.21548556,"punctuation_ratio":0.08557845,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9515921,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-06-19T18:13:35Z\",\"WARC-Record-ID\":\"<urn:uuid:fc425de5-1e9f-4633-adbc-8fb0c5fd15fc>\",\"Content-Length\":\"59442\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:28c9115e-4f1b-4e41-966d-43c62a1ec57b>\",\"WARC-Concurrent-To\":\"<urn:uuid:97195c82-62f3-4688-961c-5353c8fafa3c>\",\"WARC-IP-Address\":\"3.90.122.189\",\"WARC-Target-URI\":\"https://nyuscholars.nyu.edu/en/publications/filtering-nonlinear-dynamical-systems-with-linear-stochastic-mode\",\"WARC-Payload-Digest\":\"sha1:4OTVUZZA3D4JR3Z65B6XZV2HVUBLZXDV\",\"WARC-Block-Digest\":\"sha1:CBQKR2GV6YFKFSCJJTSIEZG5MSMTKIOF\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-25/CC-MAIN-2021-25_segments_1623487649688.44_warc_CC-MAIN-20210619172612-20210619202612-00118.warc.gz\"}"} |
http://self.gutenberg.org/articles/eng/Pentagon | [
"",
null,
"#jsDisabledContent { display:none; } My Account | Register | Help",
null,
"Flag as Inappropriate",
null,
"This article will be permanently flagged as inappropriate and made unaccessible to everyone. Are you certain this article is inappropriate? Excessive Violence Sexual Content Political / Social Email this Article Email Address:\n\n# Pentagon\n\nArticle Id: WHEBN0020741014\nReproduction Date:\n\n Title: Pentagon",
null,
"Author: World Heritage Encyclopedia Language: English Subject: Collection: Publisher: World Heritage Encyclopedia Publication Date:\n\n### Pentagon\n\nRegular pentagon",
null,
"A regular pentagon\nType Regular polygon\nEdges and vertices 5\nSchläfli symbol {5}\nCoxeter diagram\nSymmetry group Dihedral (D5), order 2×5\nInternal angle (degrees) 108°\nDual polygon self\nProperties convex, cyclic, equilateral, isogonal, isotoxal\n\nIn geometry, a pentagon (from the Greek pente and gonia, meaning five and angle) is any five-sided polygon. A pentagon may be simple or self-intersecting. The sum of the internal angles in a simple pentagon is 540°. A pentagram is an example of a self-intersecting pentagon.\n\n## Contents\n\n• Regular pentagons 1\n• Derivation of the area formula 1.1\n• Derivation of the diagonal length formula 1.2\n• Chords from the circumscribing circle to the vertices 1.3\n• Construction of a regular pentagon 2\n• Richmond's method 2.1\n• Carlyle circles 2.2\n• Using trigonometry and the Pythagorean Theorem 2.3\n• The construction 2.3.1\n• † Proof that cos 36° = \\tfrac{1+\\sqrt{5}}{4} 2.3.2\n• Another method 2.4\n• Euclid's method 2.5\n• Simply using a protractor (not a classical construction) 2.6\n• Physical methods 2.7\n• Cyclic pentagons 3\n• General convex pentagons 4\n• Graphs 5\n• Examples of pentagons 6\n• Plants 6.1\n• Animals 6.2\n• Artificial 6.3\n• Pentagons in tiling 7\n• Pentagons in polyhedra 8\n• In-line notes and references 10\n\n## Regular pentagons\n\nIn a regular pentagon, all sides are equal in length and each interior angle is 108°. A regular pentagon has five lines of reflectional symmetry, and rotational symmetry of order 5 (through 72°, 144°, 216° and 288°). Its Schläfli symbol is {5}. The diagonals of a regular pentagon are in golden ratio to its sides.\n\nThe area of a regular convex pentagon with side length t is given by\n\nA = \\frac{4} = \\frac{5t^2 \\tan(54^\\circ)}{4} \\approx 1.720477401 t^2.\n\nA pentagram or pentangle is a regular star pentagon. Its Schläfli symbol is {5/2}. Its sides form the diagonals of a regular convex pentagon – in this arrangement the sides of the two pentagons are in the golden ratio.\n\nWhen a regular pentagon is inscribed in a circle with radius R, its edge length t is given by the expression\n\nt = R\\ {\\sqrt { \\frac {5-\\sqrt{5}}{2}} } = 2R\\sin 36^\\circ = 2R\\sin\\frac{\\pi}{5} \\approx 1.17557050458 R.\n\n### Derivation of the area formula\n\nThe area of any regular polygon is:\n\nA = \\frac{1}{2}Pa\n\nwhere P is the perimeter of the polygon, a is the apothem. One can then substitute the respective values for P and a, which makes the formula:\n\nA = \\frac{1}{2} \\times \\frac{5t}{1} \\times \\frac{t\\tan(54^\\circ)}{2}\n\nwith t as the given side length. Then we can then rearrange the formula as:\n\nA = \\frac{1}{2} \\times \\frac{5t^2\\tan(54^\\circ)}{2}\n\nand then, we combine the two terms to get the final formula, which is:\n\nA = \\frac{5t^2\\tan(54^\\circ)}{4}.\n\n### Derivation of the diagonal length formula\n\nThe diagonals of a regular pentagon (here represented by d) can be calculated based upon the golden ratio φ and the known side t (see discussion of the pentagon in Golden ratio):\n\n\\frac {d}{t} = \\varphi = \\frac {1+ \\sqrt {5} }{2} \\ ,\n\nAccordingly:\n\nd = t \\times \\varphi \\ = R\\ {\\sqrt { \\frac {5+\\sqrt{5}}{2}} } = 2R\\cos 18^\\circ = 2R\\cos\\frac{\\pi}{10} \\approx 1.90211303259 R.\n\n### Chords from the circumscribing circle to the vertices\n\nIf a regular pentagon with successive vertices A, B, C, D, E is inscribed in a circle, and if P is any point on that circle between points B and C, then PA + PD = PB + PC + PE.\n\n## Construction of a regular pentagon\n\nA variety of methods are known for constructing a regular pentagon. Some are discussed below.\n\n### Richmond's method\n\nOne method to construct a regular pentagon in a given circle is described by Richmond and further discussed in Cromwell's \"Polyhedra.\"\n\nThe top panel shows the construction used in Richmond's method to create the side of the inscribed pentagon. The circle defining the pentagon has unit radius. Its center is located at point C and a midpoint M is marked halfway along its radius. This point is joined to the periphery vertically above the center at point D. Angle CMD is bisected, and the bisector intersects the vertical axis at point Q. A horizontal line through Q intersects the circle at point P, and chord PD is the required side of the inscribed pentagon.\n\nTo determine the length of this side, the two right triangles DCM and QCM are depicted below the circle. Using Pythagoras' theorem and two sides, the hypotenuse of the larger triangle is found as \\scriptstyle \\sqrt{5}/2. Side h of the smaller triangle then is found using the half-angle formula:\n\n\\tan ( \\phi/2) = \\frac{1-\\cos(\\phi)}{\\sin (\\phi)} \\ ,\n\nwhere cosine and sine of ϕ are known from the larger triangle. The result is:\n\nh = \\frac{\\sqrt 5 - 1}{4} \\ .\n\nWith this side known, attention turns to the lower diagram to find the side s of the regular pentagon. First, side a of the right-hand triangle is found using Pythagoras' theorem again:\n\na^2 = 1-h^2 \\ ; \\ a = \\frac{1}{2}\\sqrt { \\frac {5+\\sqrt 5}{2}} \\ .\n\nThen s is found using Pythagoras' theorem and the left-hand triangle as:\n\ns^2 = (1-h)^2 + a^2 = (1-h)^2 + 1-h^2 = 1-2h+h^2 + 1-h^2 = 2-2h=2-2\\left(\\frac{\\sqrt 5 - 1}{4}\\right) \\\n=\\frac {5-\\sqrt 5}{2} \\ .\n\nThe side s is therefore:\n\ns = \\sqrt{ \\frac {5-\\sqrt 5}{2}} \\ ,\n\na well established result. Consequently, this construction of the pentagon is valid.\n\n### Carlyle circles\n\nSee main article: Carlyle circle\n\nThe Carlyle circle was invented as a geometric method to find the roots of a quadratic equation. This methodology leads to a procedure for constructing a regular pentagon. The steps are as follows:\n\n1. Draw a circle in which to inscribe the pentagon and mark the center point O.\n2. Draw a horizontal line through the center of the circle. Mark one intersection with the circle as point B.\n3. Construct a vertical line through the center. Mark one intersection with the circle as point A.\n4. Construct the point M as the midpoint of O and B.\n5. Draw a circle centered at M through the point A. Mark its intersection with the horizontal line (inside the original circle) as the point W and its intersection outside the circle as the point V.\n6. Draw a circle of radius OA and center W. It intersects the original circle at two of the vertices of the pentagon.\n7. Draw a circle of radius OA and center V. It intersects the original circle at two of the vertices of the pentagon.\n8. The fifth vertex is the rightmost intersection of the horizontal line with the original circle.\n\nSteps 6-8 are equivalent to the following version, shown in the animation:\n\n6a. Construct point F as the midpoint of O and W.\n7a. Construct a vertical line through F. It intersects the original circle at two of the vertices of the pentagon. The third vertex is the rightmost intersection of the horizontal line with the original circle.\n8a. Construct the other two vertices using the compass and the length of the vertex found in step 7a.\n\n### Using trigonometry and the Pythagorean Theorem",
null,
"Using trigonometry and the Pythagorean Theorem to construct a regular pentagon.\n\n#### The construction\n\n1. We first note that a regular pentagon can be divided into 10 congruent triangles as shown in the Observation. Also, cos 36° = \\tfrac{1+\\sqrt{5}}{4}.\n2. In Step 1, we use four units (shown in blue) and a right angle to construct a segment of length 1+5, specifically by creating a 1-2-5 right triangle and then extending the hypotenuse of 5 by a length of 1. We then bisect that segment – and then bisect again – to create a segment of length \\tfrac{1+\\sqrt{5}}{4} (shown in red.)\n3. In Step 2, we construct two concentric circles centered at O with radii of length 1 and length \\tfrac{1+\\sqrt{5}}{4}. We then place P arbitrarily on the smaller circle, as shown. Constructing a line perpendicular to OP passing through P, we construct the first side of the pentagon by using the points created at the intersection of the tangent and the unit circle. Copying that length four times along the outer edge of the unit circles gives us our regular pentagon.\n\n#### † Proof that cos 36° = \\tfrac{1+\\sqrt{5}}{4}\n\n0 = \\cos 90\n= \\cos (72+18)\n= \\cos 72\\cos 18 - \\sin 72\\sin 18 (using the angle addition formula for cosine)\n= (2\\cos ^2 36 -1)\\sqrt{\\tfrac{1+\\cos 36}{2}}-2\\sin 36\\cos 36\\sqrt{\\tfrac{1-\\cos 36}{2}} (using double and half angle formulas)\nLet u = cos 36. First, note that 0 < u < 1 (which will help us simplify as we work.) Now,\n\\begin{align} 0 & {} = (2u^2 -1)\\sqrt{\\tfrac{1+u}{2}}-2\\sqrt{1-u^2}\\cdot u\\sqrt{\\tfrac{1-u}{2}} \\\\ 2\\sqrt{1-u^2}\\cdot u\\sqrt{\\tfrac{1-u}{2}} & {} = (2u^2 -1)\\sqrt{\\tfrac{1+u}{2}} \\\\ 2\\sqrt{1+u}\\sqrt{1-u}\\cdot u\\sqrt{1-u} & {} = (2u^2 -1)\\sqrt{1+u} \\\\ 2u(1-u) & {} = 2u^2-1 \\\\ 2u-2u^2 & {} = 2u^2-1 \\\\ 0 & {} = 4u^2-2u-1 \\\\ u & {} = \\tfrac{2+\\sqrt{(-2)^2-4(4)(-1)}}{2(4)} \\\\ u & {} = \\tfrac{2+\\sqrt{20}}{8} \\\\ u & {} = \\tfrac{1+\\sqrt{5}}{4} \\end{align}\n\n### Another method\n\nAnother method is this:\n\n1. Draw a circle in which to inscribe the pentagon and mark the center point O. (This is the green circle in the diagram to the right).\n2. Choose a point A on the circle that will serve as one vertex of the pentagon. Draw a line through O and A.\n3. Construct a line perpendicular to the line OA passing through O. Mark its intersection with one side of the circle as the point B.\n4. Construct the point C as the midpoint of the line OB.\n5. Draw a circle centered at C through the point A. Mark its intersection with the line OB (inside the original circle) as the point D.\n6. Draw a circle centered at A through the point D. Mark its intersections with the original (green) circle as the points E and F.\n7. Draw a circle centered at E through the point A. Mark its other intersection with the original circle as the point G.\n8. Draw a circle centered at F through the point A. Mark its other intersection with the original circle as the point H.\n9. Construct the regular pentagon AEGHF.\n\n### Euclid's method\n\nA regular pentagon is constructible using a compass and straightedge, either by inscribing one in a given circle or constructing one on a given edge. This process was described by Euclid in his Elements circa 300 BC.\n\n### Simply using a protractor (not a classical construction)\n\nA direct method using degrees follows:\n\n1. Draw a circle and choose a point to be the pentagon's (e.g. top center)\n2. Choose a point A on the circle that will serve as one vertex of the pentagon. Draw a line through O and A.\n3. Draw a guideline through it and the circle's center\n4. Draw lines at 54° (from the guideline) intersecting the pentagon's point\n5. Where those intersect the circle, draw lines at 18° (from parallels to the guideline)\n6. Join where they intersect the circle\n\nAfter forming a regular convex pentagon, if one joins the non-adjacent corners (drawing the diagonals of the pentagon), one obtains a pentagram, with a smaller regular pentagon in the center. Or if one extends the sides until the non-adjacent sides meet, one obtains a larger pentagram. The accuracy of this method depends on the accuracy of the protractor used to measure the angles.\n\n### Physical methods\n\n• A regular pentagon may be created from just a strip of paper by tying an overhand knot into the strip and carefully flattening the knot by pulling the ends of the paper strip. Folding one of the ends back over the pentagon will reveal a pentagram when backlit.\n• Construct a regular hexagon on stiff paper or card. Crease along the three diameters between opposite vertices. Cut from one vertex to the center to make an equilateral triangular flap. Fix this flap underneath its neighbor to make a pentagonal pyramid. The base of the pyramid is a regular pentagon.\n\n## Cyclic pentagons\n\nA cyclic pentagon is one for which a circle called the circumcircle goes through all five vertices. The regular pentagon is an example of a cyclic pentagon. The area of a cyclic pentagon, whether regular or not, can be expressed as one fourth the square root of one of the roots of a septic equation whose coefficients are functions of the sides of the pentagon.\n\nThere exist cyclic pentagons with rational sides and rational area; these are called Robbins pentagons. In a Robbins pentagon, either all diagonals are rational or all are irrational, and it is conjectured that all the diagonals must be rational.\n\n## General convex pentagons\n\nFor all convex pentagons, the sum of the squares of the diagonals is less than 3 times the sum of the squares of the sides.:p.75,#1854\n\n## Graphs\n\nThe K5 complete graph is often drawn as a regular pentagon with all 10 edges connected. This graph also represents an orthographic projection of the 5 vertices and 10 edges of the 5-cell. The rectified 5-cell, with vertices at the mid-edges of the 5-cell is projected inside a pentagon.",
null,
"5-cell (4D)",
null,
"Rectified 5-cell (4D)\n\n## Pentagons in tiling",
null,
"The best known packing of equal-sized regular pentagons on a plane is a double lattice structure which covers 92.131% of the plane.\n\nA pentagon cannot appear in any tiling made by regular polygons. To prove a pentagon cannot form a regular tiling (one in which all faces are congruent), observe that 360 / 108 = 313, which is not a whole number. More difficult is proving a pentagon cannot be in any edge-to-edge tiling made by regular polygons:\n\nThere are no combinations of regular polygons with 4 or more meeting at a vertex that contain a pentagon. For combinations with 3, if 3 polygons meet at a vertex and one has an odd number of sides, the other 2 must be congruent. The reason for this is that the polygons that touch the edges of the pentagon must alternate around the pentagon, which is impossible because of the pentagon's odd number of sides. For the pentagon, this results in a polygon whose angles are all (360 − 108) / 2 = 126°. To find the number of sides this polygon has, the result is 360 / (180 − 126) = 623, which is not a whole number. Therefore, a pentagon cannot appear in any tiling made by regular polygons.\n\n## Pentagons in polyhedra",
null,
"",
null,
"",
null,
"",
null,
"",
null,
"Dodecahedron Pyritohedron Pentagonal icositetrahedron Pentagonal hexecontahedron Truncated_trapezohedron\n\n## In-line notes and references\n\n1. ^ \"pentagon, adj. and n.\" OED Online. Oxford University Press, June 2014. Web. 17 August 2014.\n2. ^ Herbert W Richmond (1893). \"Pentagon\".\n3. ^ Peter R. Cromwell. Polyhedra. p. 63.\n4. ^ This result agrees with Herbert Edwin Hawkes, William Arthur Luby, Frank Charles Touton (1920). \"Exercise 175\". Plane geometry. Ginn & Co. p. 302.\n5. ^ Eric W. Weisstein (2003). CRC concise encyclopedia of mathematics (2nd ed.). CRC Press. p. 329.\n6. ^ Duane W DeTemple (1991). \"Carlyle Circles and the Lemoine Simplicity of Polygon Constructions\". The American Mathematical Monthly 98 (2): 97–108. JSTOR link\n7. ^ George Edward Martin (1998). Geometric constructions. Springer. p. 6.\n8. ^ Weisstein, Eric W. \"Cyclic Pentagon.\" From MathWorld--A Wolfram Web Resource. \n9. ^ Robbins, D. P. \"Areas of Polygons Inscribed in a Circle.\" Discr. Comput. Geom. 12, 223-236, 1994.\n10. ^ Robbins, D. P. \"Areas of Polygons Inscribed in a Circle.\" Amer. Math. Monthly 102, 523-530, 1995.\n11. ^ *Buchholz, Ralph H.; MacDougall, James A. (2008), \"Cyclic polygons with rational sides and area\", .\n12. ^ Inequalities proposed in “Crux Mathematicorum”, ."
]
| [
null,
"http://read.images.worldlibrary.org/App_Themes/pg/images/logo.jpg",
null,
"http://read.images.worldlibrary.org/images/SmallBook.gif",
null,
"http://self.gutenberg.org/images/delete.jpg",
null,
"http://self.gutenberg.org/App_Themes/default/images/icon_new_window.gif",
null,
"http://images.worldlibrary.net/articles/eng/File:Regular_polygon_5_annotated.svg",
null,
"http://images.worldlibrary.net/articles/eng/File:Regular_pentagon_with_trig.gif",
null,
"http://images.worldlibrary.net/articles/eng/File:4-simplex_t0.svg",
null,
"http://images.worldlibrary.net/articles/eng/File:4-simplex_t1.svg",
null,
"http://images.worldlibrary.net/articles/eng/File:2-d_pentagon_packing_dual.svg",
null,
"http://images.worldlibrary.net/articles/eng/File:Dodecahedron.jpg",
null,
"http://images.worldlibrary.net/articles/eng/File:Pyritohedron.png",
null,
"http://images.worldlibrary.net/articles/eng/File:Pentagonalicositetrahedronccw.jpg",
null,
"http://images.worldlibrary.net/articles/eng/File:Pentagonalhexecontahedronccw.jpg",
null,
"http://images.worldlibrary.net/articles/eng/File:Pentagonal_truncated_trapezohedron.png",
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.8523513,"math_prob":0.9883215,"size":16460,"snap":"2020-24-2020-29","text_gpt3_token_len":4463,"char_repetition_ratio":0.1713053,"word_repetition_ratio":0.107282415,"special_character_ratio":0.27430135,"punctuation_ratio":0.107209444,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9988505,"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,null,null,null,null,4,null,4,null,8,null,5,null,4,null,4,null,10,null,7,null,10,null,4,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-06-06T05:26:54Z\",\"WARC-Record-ID\":\"<urn:uuid:278d4946-9ac6-4bb3-a2d5-4fb87ebf7dda>\",\"Content-Length\":\"127823\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:58834ded-f4f0-4117-8d0f-bef84bf73d81>\",\"WARC-Concurrent-To\":\"<urn:uuid:eaf623df-3835-46c5-becb-57bb58461d98>\",\"WARC-IP-Address\":\"66.27.42.21\",\"WARC-Target-URI\":\"http://self.gutenberg.org/articles/eng/Pentagon\",\"WARC-Payload-Digest\":\"sha1:RC4NXJHC6LCT33LF3KANASP4ITNFNCRY\",\"WARC-Block-Digest\":\"sha1:EKHASMWJESCHWT36ZPMQRG5XMHLRYO45\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-24/CC-MAIN-2020-24_segments_1590348509972.80_warc_CC-MAIN-20200606031557-20200606061557-00160.warc.gz\"}"} |
https://www.trashrite.com/blogs/news/43cm-in-inches-converting-centimeters-to-inches | [
"## The Conversion Puzzle Unraveled\n\nHave you ever wondered how to convert centimeters to inches? If you've come across the measurement 43cm and need to know its equivalent in inches, you've come to the right place. In this comprehensive guide, we will delve into the world of measurement conversion, focusing specifically on converting 43cm to inches. Whether you're a student, a DIY enthusiast, or simply curious about measurements, understanding how to convert centimeters to inches is a valuable skill. Let's get started!\n\n## The Basics: What is 43cm in Inches?\n\nBefore we dive into the conversion process, let's clarify the basic measurements we'll be working with. Centimeters (cm) and inches (in) are both units of length, but they differ in their magnitude. The centimeter is part of the metric system, commonly used in many countries around the world, while the inch is primarily used in the United States and a few other countries. Now, let's tackle the main question at hand: what is 43cm in inches?\n\n43cm in inches is approximately 16.93 inches.\n\n## How to Convert 43cm to Inches: Step-by-Step Guide\n\nConverting centimeters to inches involves a simple mathematical calculation. By following these steps, you'll be able to convert 43cm to inches accurately:\n\n1. Write down the given measurement: Start by writing down the value you want to convert, which in this case is 43cm.\n2. Use the conversion factor: Since 1 inch is equal to 2.54 centimeters, we can use this conversion factor to find the equivalent inches. Multiply the given measurement (43cm) by the conversion factor (2.54) to obtain the value in inches.\n3. Perform the calculation: Multiply 43cm by 2.54 to get the result.\n• 43cm * 2.54 = 109.22cm\n4. Round the answer: To make the conversion more practical, round the value to the nearest hundredth, which yields 109.22cm ≈ 109.22 inches.\n5. Write the final answer: The final result is approximately 16.93 inches.\n\nRemember, this method can be used to convert any measurement from centimeters to inches. Simply substitute the given value in step 1 and follow the remaining steps accordingly.\n\n## FAQs about Converting 43cm to Inches\n\n1. Can I use an online converter to convert 43cm to inches? Yes, there are numerous online conversion tools available that can quickly provide you with the conversion result. Just make sure to choose a reputable and reliable website or app.\n\n2. Why do we use inches and centimeters for measuring length? Inches are commonly used in the United States and a few other countries, while centimeters are part of the metric system used in many other nations. The choice of measurement unit often depends on cultural conventions and regional preferences.\n\n3. Can I use a ruler to measure 43cm? Yes, you can use a ruler to measure 43cm. Simply align the starting point of the ruler with the object's edge and count the centimeters until you reach 43cm.\n\n4. Are there any other common conversions between inches and centimeters? Yes, there are several common conversions between inches and centimeters, such as 1 inch = 2.54 cm, 12 inches = 30.48 cm, and 100 cm = 39.37 inches.\n\n5. How can I convert inches to centimeters? To convert inches to centimeters, you can use the conversion factor of 2.54 cm per inch. Multiply the number of inches by 2.54 to obtain the equivalent centimeters.\n\n6. Are there any shortcuts for converting centimeters to inches mentally? While there is no universal shortcut, some people find it helpful to memorize common conversions, such as 2.5cm ≈ 1 inch, to estimate the approximate values quickly.\n\n## Conclusion\n\nIn conclusion, converting 43cm to inches is a straightforward process. By using the conversion factor of 2.54 and following a few simple steps, you can accurately determine that 43cm is approximately equal to 16.93 inches. Remember, understanding how to convert between centimeters and inches can be beneficial in various contexts, from academic pursuits to practical everyday tasks. So the next time you encounter a measurement in centimeters, you'll have the knowledge to convert it to inches effortlessly.\n\nBy Raied Muheisen 0 comment"
]
| [
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.9025756,"math_prob":0.95267755,"size":4098,"snap":"2023-40-2023-50","text_gpt3_token_len":907,"char_repetition_ratio":0.17855398,"word_repetition_ratio":0.026706232,"special_character_ratio":0.2249878,"punctuation_ratio":0.1285347,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9928744,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-09T05:20:33Z\",\"WARC-Record-ID\":\"<urn:uuid:26e7e580-beaf-4767-81e0-7c783e3bd382>\",\"Content-Length\":\"282767\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4282b61f-0de8-4abc-b8f4-6ebd653565ad>\",\"WARC-Concurrent-To\":\"<urn:uuid:b8fef9fc-3128-427a-aa4d-86c0e3459940>\",\"WARC-IP-Address\":\"23.227.38.74\",\"WARC-Target-URI\":\"https://www.trashrite.com/blogs/news/43cm-in-inches-converting-centimeters-to-inches\",\"WARC-Payload-Digest\":\"sha1:EEMLHQ4SYHEISTRBF4W67JKYUTLK2GBJ\",\"WARC-Block-Digest\":\"sha1:3HSRKQH4JPLGT57LGMKB5LUYMJOXTTDB\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100800.25_warc_CC-MAIN-20231209040008-20231209070008-00608.warc.gz\"}"} |
https://www.colorhexa.com/0f4625 | [
"# #0f4625 Color Information\n\nIn a RGB color space, hex #0f4625 is composed of 5.9% red, 27.5% green and 14.5% blue. Whereas in a CMYK color space, it is composed of 78.6% cyan, 0% magenta, 47.1% yellow and 72.5% black. It has a hue angle of 144 degrees, a saturation of 64.7% and a lightness of 16.7%. #0f4625 color hex could be obtained by blending #1e8c4a with #000000. Closest websafe color is: #003333.\n\n• R 6\n• G 27\n• B 15\nRGB color chart\n• C 79\n• M 0\n• Y 47\n• K 73\nCMYK color chart\n\n#0f4625 color description : Very dark cyan - lime green.\n\n# #0f4625 Color Conversion\n\nThe hexadecimal color #0f4625 has RGB values of R:15, G:70, B:37 and CMYK values of C:0.79, M:0, Y:0.47, K:0.73. Its decimal value is 1000997.\n\nHex triplet RGB Decimal 0f4625 `#0f4625` 15, 70, 37 `rgb(15,70,37)` 5.9, 27.5, 14.5 `rgb(5.9%,27.5%,14.5%)` 79, 0, 47, 73 144°, 64.7, 16.7 `hsl(144,64.7%,16.7%)` 144°, 78.6, 27.5 003333 `#003333`\nCIE-LAB 25.609, -26.395, 14.913 2.721, 4.615, 2.498 0.277, 0.469, 4.615 25.609, 30.317, 150.533 25.609, -20.254, 18.152 21.483, -14.988, 8.145 00001111, 01000110, 00100101\n\n# Color Schemes with #0f4625\n\n• #0f4625\n``#0f4625` `rgb(15,70,37)``\n• #460f30\n``#460f30` `rgb(70,15,48)``\nComplementary Color\n• #15460f\n``#15460f` `rgb(21,70,15)``\n• #0f4625\n``#0f4625` `rgb(15,70,37)``\n• #0f4641\n``#0f4641` `rgb(15,70,65)``\nAnalogous Color\n• #460f15\n``#460f15` `rgb(70,15,21)``\n• #0f4625\n``#0f4625` `rgb(15,70,37)``\n• #410f46\n``#410f46` `rgb(65,15,70)``\nSplit Complementary Color\n• #46250f\n``#46250f` `rgb(70,37,15)``\n• #0f4625\n``#0f4625` `rgb(15,70,37)``\n• #250f46\n``#250f46` `rgb(37,15,70)``\n• #30460f\n``#30460f` `rgb(48,70,15)``\n• #0f4625\n``#0f4625` `rgb(15,70,37)``\n• #250f46\n``#250f46` `rgb(37,15,70)``\n• #460f30\n``#460f30` `rgb(70,15,48)``\n• #020704\n``#020704` `rgb(2,7,4)``\n• #061c0f\n``#061c0f` `rgb(6,28,15)``\n• #0b311a\n``#0b311a` `rgb(11,49,26)``\n• #0f4625\n``#0f4625` `rgb(15,70,37)``\n• #145b30\n``#145b30` `rgb(20,91,48)``\n• #18703b\n``#18703b` `rgb(24,112,59)``\n• #1d8546\n``#1d8546` `rgb(29,133,70)``\nMonochromatic Color\n\n# Alternatives to #0f4625\n\nBelow, you can see some colors close to #0f4625. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #0f4617\n``#0f4617` `rgb(15,70,23)``\n• #0f461c\n``#0f461c` `rgb(15,70,28)``\n• #0f4620\n``#0f4620` `rgb(15,70,32)``\n• #0f4625\n``#0f4625` `rgb(15,70,37)``\n• #0f462a\n``#0f462a` `rgb(15,70,42)``\n• #0f462e\n``#0f462e` `rgb(15,70,46)``\n• #0f4633\n``#0f4633` `rgb(15,70,51)``\nSimilar Colors\n\n# #0f4625 Preview\n\nThis text has a font color of #0f4625.\n\n``<span style=\"color:#0f4625;\">Text here</span>``\n#0f4625 background color\n\nThis paragraph has a background color of #0f4625.\n\n``<p style=\"background-color:#0f4625;\">Content here</p>``\n#0f4625 border color\n\nThis element has a border color of #0f4625.\n\n``<div style=\"border:1px solid #0f4625;\">Content here</div>``\nCSS codes\n``.text {color:#0f4625;}``\n``.background {background-color:#0f4625;}``\n``.border {border:1px solid #0f4625;}``\n\n# Shades and Tints of #0f4625\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, #010503 is the darkest color, while #f4fdf8 is the lightest one.\n\n• #010503\n``#010503` `rgb(1,5,3)``\n• #05160b\n``#05160b` `rgb(5,22,11)``\n• #082614\n``#082614` `rgb(8,38,20)``\n• #0c361c\n``#0c361c` `rgb(12,54,28)``\n• #0f4625\n``#0f4625` `rgb(15,70,37)``\n• #12562e\n``#12562e` `rgb(18,86,46)``\n• #166636\n``#166636` `rgb(22,102,54)``\n• #19763f\n``#19763f` `rgb(25,118,63)``\n• #1d8747\n``#1d8747` `rgb(29,135,71)``\n• #209750\n``#209750` `rgb(32,151,80)``\n• #24a758\n``#24a758` `rgb(36,167,88)``\n• #27b761\n``#27b761` `rgb(39,183,97)``\n• #2bc769\n``#2bc769` `rgb(43,199,105)``\n• #32d373\n``#32d373` `rgb(50,211,115)``\n• #43d77e\n``#43d77e` `rgb(67,215,126)``\n• #53da89\n``#53da89` `rgb(83,218,137)``\n• #63de94\n``#63de94` `rgb(99,222,148)``\n• #73e19f\n``#73e19f` `rgb(115,225,159)``\n• #83e4aa\n``#83e4aa` `rgb(131,228,170)``\n• #93e8b5\n``#93e8b5` `rgb(147,232,181)``\n• #a3ebc0\n``#a3ebc0` `rgb(163,235,192)``\n• #b4efcb\n``#b4efcb` `rgb(180,239,203)``\n• #c4f2d6\n``#c4f2d6` `rgb(196,242,214)``\n• #d4f6e1\n``#d4f6e1` `rgb(212,246,225)``\n• #e4f9ed\n``#e4f9ed` `rgb(228,249,237)``\n• #f4fdf8\n``#f4fdf8` `rgb(244,253,248)``\nTint Color Variation\n\n# Tones of #0f4625\n\nA tone is produced by adding gray to any pure hue. In this case, #292c2a is the less saturated color, while #025322 is the most saturated one.\n\n• #292c2a\n``#292c2a` `rgb(41,44,42)``\n• #262f2a\n``#262f2a` `rgb(38,47,42)``\n• #233229\n``#233229` `rgb(35,50,41)``\n• #1f3628\n``#1f3628` `rgb(31,54,40)``\n• #1c3928\n``#1c3928` `rgb(28,57,40)``\n• #193c27\n``#193c27` `rgb(25,60,39)``\n• #163f26\n``#163f26` `rgb(22,63,38)``\n• #124326\n``#124326` `rgb(18,67,38)``\n• #0f4625\n``#0f4625` `rgb(15,70,37)``\n• #0c4924\n``#0c4924` `rgb(12,73,36)``\n• #084d24\n``#084d24` `rgb(8,77,36)``\n• #055023\n``#055023` `rgb(5,80,35)``\n• #025322\n``#025322` `rgb(2,83,34)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #0f4625 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":0.5224647,"math_prob":0.7787683,"size":3673,"snap":"2019-43-2019-47","text_gpt3_token_len":1647,"char_repetition_ratio":0.12482966,"word_repetition_ratio":0.011029412,"special_character_ratio":0.57473457,"punctuation_ratio":0.23496659,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98887897,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-11-14T03:39:39Z\",\"WARC-Record-ID\":\"<urn:uuid:97befec0-940c-4641-afff-26f6c0a7e631>\",\"Content-Length\":\"36218\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:0b98c684-66cd-4f80-88b2-2443d13154ad>\",\"WARC-Concurrent-To\":\"<urn:uuid:3a05199e-6ca7-46bc-a021-f5c09adea5a9>\",\"WARC-IP-Address\":\"178.32.117.56\",\"WARC-Target-URI\":\"https://www.colorhexa.com/0f4625\",\"WARC-Payload-Digest\":\"sha1:J27NVWHDYEJ4XOIKKIISJQ53JBJCWKJW\",\"WARC-Block-Digest\":\"sha1:WPP3RNH7DWFG3KO4RJKGE3ET4YCWROJ6\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-47/CC-MAIN-2019-47_segments_1573496667945.28_warc_CC-MAIN-20191114030315-20191114054315-00453.warc.gz\"}"} |
https://www.proofwiki.org/wiki/Partition_of_Integer_into_Odd_Parts | [
"# Partition of Integer into Odd Parts\n\n## Examples\n\n### Partition of $5$ into Odd Parts\n\nThe integer $5$ can be partitioned into odd parts as follows:\n\n$5$\n$3 + 1 + 1$\n$1 + 1 + 1 + 1 + 1$"
]
| [
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.810914,"math_prob":0.9986342,"size":889,"snap":"2022-05-2022-21","text_gpt3_token_len":244,"char_repetition_ratio":0.108474575,"word_repetition_ratio":0.32911393,"special_character_ratio":0.26771653,"punctuation_ratio":0.0872093,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99750507,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-25T21:54:11Z\",\"WARC-Record-ID\":\"<urn:uuid:f19ec20b-654e-4ed3-913e-9d1a1ec3bc33>\",\"Content-Length\":\"39219\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:42118165-e650-4d20-b04a-1452e8d3bc6b>\",\"WARC-Concurrent-To\":\"<urn:uuid:0a54a2d9-1266-4b01-a83c-257a04b42607>\",\"WARC-IP-Address\":\"172.67.198.93\",\"WARC-Target-URI\":\"https://www.proofwiki.org/wiki/Partition_of_Integer_into_Odd_Parts\",\"WARC-Payload-Digest\":\"sha1:WDJGIQONQL4NKN22JCKXRCA427D5MIR2\",\"WARC-Block-Digest\":\"sha1:ENCHVYS5O2VLSJHTYSXRJCAWPH33PGSU\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652662594414.79_warc_CC-MAIN-20220525213545-20220526003545-00754.warc.gz\"}"} |
https://www.benfordonline.net/fullreference/501 | [
"## View Complete Reference\n\n### Sarkar, BP (1973)\n\n#### An Observation on the Significant Digits of Binomial Coefficients and Factorials\n\nSankhya - The Indian Journal of Statistics Series B 35(3), 363-364.\n\nISSN/ISBN: 0581-5738 DOI: Not available at this time.\n\nAbstract: The distribution of the first significant digit has been the subject of a number of papers, both empirical and theoretical. Because binomial coefficients and factorials may (in theory) be generated by a multiplicative process one would anticipate that in table of either, the fraction of entries beginning with a first significant digit of A would be approximately log10(A+1)/A. This borne out by the tables in the paper.\n\nBibtex:\n```@article{, title={An observation on the significant digits of binomial coefficients and factorials}, author={Sarkar, PB}, journal={Sankhy{\\=a} B}, volume={35}, pages={363--364}, year={1973}, ISSN={0581-5738}, }```\n\nReference Type: Journal Article\n\nSubject Area(s): General Interest"
]
| [
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.78021586,"math_prob":0.4209634,"size":838,"snap":"2021-43-2021-49","text_gpt3_token_len":205,"char_repetition_ratio":0.09232614,"word_repetition_ratio":0.0,"special_character_ratio":0.2673031,"punctuation_ratio":0.1474359,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95171666,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-27T08:56:50Z\",\"WARC-Record-ID\":\"<urn:uuid:63b8b118-63fd-4950-a235-afca55aa4dfc>\",\"Content-Length\":\"3996\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4ca36f30-1a92-48c4-bfb1-28df8a41ea8c>\",\"WARC-Concurrent-To\":\"<urn:uuid:c812c5d1-eaa3-4a0b-9647-83bb5f61ae4e>\",\"WARC-IP-Address\":\"192.124.249.34\",\"WARC-Target-URI\":\"https://www.benfordonline.net/fullreference/501\",\"WARC-Payload-Digest\":\"sha1:BATDKEFCX35ITXCLVABNS3CLONMYTZG7\",\"WARC-Block-Digest\":\"sha1:ZTCWQG4BA4CRHGD5WK4IB6GG2YEGZTB4\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323588113.25_warc_CC-MAIN-20211027084718-20211027114718-00389.warc.gz\"}"} |
https://x80.org/synchron2016/ | [
"## Towards Certified Digital Audio Processing\n\n### Emilio Jesús Gallego Arias (joint work with Pierre Jouvelot)\n\n#### MINES ParisTech, PSL Research University, France",
null,
"##### December 5th 2016 - Synchron 2016 - Bamberg\n\nFEEVER ANR Project\n\nReal Time Signal Processing\n$\\cap$ Programming Language Theory\n$\\cap$ Theorem Proving\n\nKey Goal: Arbitrary Formal Proofs\n\n• Linearity:\n• A set of programs correspond to an LTI system.\n• Simulation:\n• Different executions/implementations denote the same program. Including compilation!\n• Approximation:\n• A program approximates another program or mathematical system. Quite free-form! Think of relating a program to its Z-transform.\n\n## Basic Audio Objects\n\n###### Recursive 2nd order Filter",
null,
"Image Credits: Julius Orion Smith III\n\n###### Wave-guide Resonator",
null,
"Image Credits: Julius Orion Smith III\n\n##### Programs correspond to synchronous data-flow.\n\nJulius Orion Smith III audio-focused book's series:\n\n• Mathematics of the Discrete Fourier Transform (DFT)\n• [Partially handled + missing inf. sum.]\n• Introduction to Digital Filters\n• [Work of this talk]\n• Physical Audio Signal Processing\n• [Minor bits in this talk + missing ODE]\n• Spectral Audio Signal Processing\n• [Future work]\n##### Main Relevant Points:\n• Free mix of mathematics and computation.\n• Linear algebra is pervasive.\n• Proofs of uneven difficulty, not constructive-friendly.\n\n## Where to Look?\n\n### VeriDrone Project[Ricketts, Machela, Lerner, ...]\n\n[EMSOFT 2016] traces+LTL, continuity, Lyapunov.\n\n### Lustre Certified Compiler[Bourke et al.]\n\nWork in progress, denotational semantics?\n\n### FRP/Arrows[Krishnaswami, Elliot, Hughes, Hudak, Nilsson, ...]\n\nNice functional PL techniques; too abstract for real-time sample-based DSP.\n\n### Synchronous Languages [Berry, Caspi, Halbwachs, Pouzet, Guatto, ...]\n\nGuatto's PhD Most related to our approach. Suited to linear algebra/DSP?\nData-intensive vs control-intensive require quite different control techniques. [Berry, 2000]\n\n### Isabelle/HOL[Akbarpour, Tahar et al.]\n\nFixed systems, numerical issues.\n\n## The Plan\n\n##### Formal Proof and Semantics\n$$\\forall p, ⟦p⟧_A = ⟦T(p)⟧_B$$\n\nMain question, what is ⟦ ⟧?\n\n##### Borrow Techniques from Functional Programming to Attack the Problem:\n• Define a functional, dataflow-like language.\n• Untyped operational semantics!\n• Classify well-behaved programs wrt the enviroment.\n• Logical Relations\n• ?\n• Profit!\nReplace stratification by domain theory/types by operational step-indexing.\n\n## What now: Pre-Wagner\n\nSimple λ-calculus with feedback, pre, and stable types.\n\n#### Types and Syntax:\n\n$$\\newcommand{\\ip}{\\langle{#1},{#2}\\rangle} \\newcommand{\\inferrule}{\\frac{#1}{#2}} \\newcommand{\\fpo}{\\varphi} \\newcommand{\\fpw}{\\psi} \\newcommand{\\fpt}{\\theta} \\newcommand{\\vm}{\\vec{m}} \\newcommand{\\vn}{\\vec{n}} \\newcommand{\\vo}{\\vec{o}} \\newcommand{\\vx}{\\vec{x}} \\newcommand{\\ve}{\\vec{e}} \\newcommand{\\vv}{\\vec{v}} \\newcommand{\\rn}{{~\\mathsf{#1}}} \\newcommand{\\fpre}{{#1}_{\\{#2\\}}} \\newcommand{\\fjud}{\\{ #1 \\} ~~ {#2} ~~ \\{ #3 \\}} \\begin{array}{lrll} T & := & R \\mid I_n \\mid R[n] & \\text{Samples, Ordinals, Arrays} \\\\ & ∣& R@r & \\text{rated stream} \\\\ & ∣& T ⊗ T & \\text{product} \\\\ & ∣& R → T & \\text{simple function} \\\\ & ∣& R@r ⇒_m T & \\text{stream processor} \\\\ \\end{array}$$ $$\\newcommand{\\R}{\\mathbf{R}} \\newcommand{\\feed}{\\text{feed}~{#1} = {#2}} \\newcommand{\\let}{\\text{let}~{#1}={#2}~\\text{in}~{#3}} \\newcommand{\\feedin}{\\text{feed}~{#1} = {#2}~\\text{in}~{#3}} \\begin{array}{lrll} e & := & x \\mid c \\mid p & \\text{Var, Const, Prim} \\\\ & \\mid & \\pi_n~e \\mid (e, e) & \\text{Products} \\\\ & \\mid & λ x.~e \\mid (e_f~e_a ) & \\text{Regular functions} \\\\ & \\mid & Λ x.~e \\mid \\{e_1, …, e_n\\} & \\text{Stream functions} \\\\ & \\mid & e_{\\{j\\}} \\mid \\feed{\\vec{x}}{\\vec{e}} & \\text{Previous, Feedback} \\end{array}$$\n\n## Pre-Wagner: Op. Semantics\n\nAssume causal feedback. Big-step, time-indexed relation. Every program reduces to a value at time step $n$. Similar to synchronous data-flow.\n\n$$\\newcommand{\\ev}{{#1} \\downarrow_{#2} {#3}} \\begin{gather} \\inferrule { } {\\ev{x}{n}{x}} \\qquad \\inferrule {\\ev{e_1}{n}{v_1} \\quad … \\quad \\ev{e_k}{n}{v_k} \\quad P(v_1, …, v_1) ↓ v } {\\ev{P(e_1,…,e_k)}{n}{v}} \\\\[1em] \\inferrule {\\ev{eₐ}{n}{vₐ} \\quad \\ev{e_f[x/vₐ]}{n}{v}} {\\ev{(λ~x.e_f)~eₐ}{n}{v}} \\qquad \\inferrule {\\ev{eₐ}{n}{vₐ} \\quad \\ev{e_f[x/vₐ]}{n}{v}} {\\ev{(Λ~x.e_f)~eₐ}{n}{v}} \\\\[1.5em] \\inferrule { \\ev{e_1}{n}{v_1} ~\\ldots~ \\ev{e_k}{n}{v_k} } { \\ev{ \\{e_1, \\ldots, e_k \\} }{n}{\\{v_1, \\ldots, v_k \\} } } \\qquad \\inferrule { } {\\ev{e_{\\{k+1\\}}}{0}{0_e}} \\qquad \\inferrule {\\ev{e_{\\{k\\}}}{n}{v}} {\\ev{e_{\\{k+1\\}}}{n+1}{v}} \\\\[2em] \\inferrule {\\ev{\\ve[\\vx/0_\\ve]}{0}{\\vv}} {\\ev{\\feed{\\vx}{\\ve}}{0}{\\vv}} \\qquad \\inferrule { \\ev{\\feed{\\vx}{\\ve}}{n}{\\vv} \\quad \\ev{\\ve[\\vx/\\vv]}{n+1}{\\vv} } { \\ev{\\feed{\\vx}{\\ve}}{n+1}{\\vec{v}} } \\end{gather}$$\n\n## Examples: DF-I Filter",
null,
"$$\\mathsf{df1} ≡ λ~(a:R[k_a])~(b : R[k_b]). Λ x.~\\feed{y}{ \\{ b \\cdot x + a \\cdot y} \\}$$\n\nNote size polymorphism of the dot operator:\n\n$$x \\cdot y ≡ \\sum_i^k x[i] ⬝ \\fpre{y}{i}$$\n\n## Examples: DF-II",
null,
"$$\\mathsf{df2} ≡ Λ x.~ \\feedin{v}{ \\{ x + a \\cdot v \\}}{ \\{ b \\cdot v \\} }$$\n\nCompare with df1:\n\n$$\\mathsf{df1} ≡ Λ x.~\\feed{y}{ \\{ b \\cdot x + a \\cdot y} \\}$$\n\n## Examples: Wave-Guide OSC",
null,
"$$\\begin{array}{rcl} \\text{feed}~x & = & C \\cdot (G \\cdot x' + y') - y' \\\\ y & = & C \\cdot (G \\cdot x' + y') + G \\cdot x' \\end{array}$$\n\nwe write $x' \\equiv \\fpre{x}{1}, y' \\equiv \\fpre{y}{1}$. With different sugar:\n\n$$\\feed{z}{ \\{ (C \\cdot (G \\cdot z'_1 + z'_2) - z'_2, C \\cdot (G \\cdot z'_1 + z'_2) + G \\cdot z'_1) \\} }$$\n\n## Towards a New Semantics Or some critics of \"pre\"\n\nA) Compare a typical FIR:\n\n$$x \\cdot y ≡ \\sum_i x[i] ⬝ \\fpre{y}{i} \\qquad \\text{vs} \\qquad x \\cdot y ≡ \\sum_i x[i] ⬝ y[i]$$\n• Model where pre behaves like array indexing?\n\nB) Evaluation strategy and caching?\n\n$$\\begin{array}{ll} \\text{collect} &: R ⊗ \\dots ⊗ R ⇒_0 R \\\\ \\text{avg} &: R ⇒_{100} R \\\\ \\text{process} &: R ⊗ \\dots ⊗ R ⇒_? R \\\\ \\text{process} &= avg ∘ collect \\end{array}$$\n\nWhat should be the type of $\\text{process}$ ? A caching of 100? Ups!\n\nC) How can we drop the causality requirement? \"pre\" doesn't need it!\n\n## Val-Wagner\n\nAssume CBV, only variables can be time-addressed:\n\n#### Types and Syntax:\n\n$$\\begin{array}{lrll} T & := & R \\mid I_n \\mid R[n] & \\text{Samples, Ordinals, Arrays} \\\\ & ∣& R@r & \\text{rated stream} \\\\ & ∣& T ⊗ T & \\text{product} \\\\ & ∣& R → T & \\text{simple function} \\\\ & ∣& R@r ⇒_m T & \\text{stream processor} \\\\ \\end{array}$$ $$\\begin{array}{lrll} e & := & \\fpre{x}{k} \\mid c \\mid p & \\text{Var, Const, Prim} \\\\ & \\mid & \\pi_n~e \\mid (e, e) & \\text{Products} \\\\ & \\mid & λ x.~e \\mid (e_f~e_a ) & \\text{Regular functions} \\\\ & \\mid & Λ x.~e \\mid \\{e_1, …, e_n\\} & \\text{Stream functions} \\\\ & \\mid & \\feed{\\vx}{\\ve} & \\text{Feedback} \\end{array}$$\n\n## Where is my Pre?\n\nVariables are arrays with history; we recover \"pre\" by \"eta\":\n\n$$\\fpre{e}{k} ≡ (Λ x.~\\fpre{x}{k})~e$$\n\nUntyped Operational Semantics using a stack of arrays ${\\cal E}$:\n\n$$\\inferrule{ }{ ⟨{\\cal E} ∥ \\fpre{x}{k}⟩ ↓_n {\\cal E}(x)[k] } \\\\[1.1em] \\inferrule{ ⟨{\\cal E} ∥ f ⟩ ↓_n λ x. v_f \\quad ⟨{\\cal E} ∥ e ⟩ ↓_{(0 ≤ i ≤ n)} {v_e}_i \\quad ⟨[{v_e}_0, …, {v_e}_n] ∷ {\\cal E} ∥ v_f ⟩ ↓_n v }{ ⟨{\\cal E} ∥ f~e ⟩ ↓_n v } \\\\[1.1em] \\inferrule{ ⟨{\\cal E} ∥ \\feed{\\vx}{\\ve} ⟩ ↓_{(0 ≤ i ≤ n)} {\\vec{v_e}}_i \\quad ⟨[{\\vec{v_e}}_0, …, {\\vec{v_e}}_n, \\vec{0}] ∷ {\\cal E} ∥ \\ve ⟩ ↓_{n+1} \\vv }{ ⟨{\\cal E} ∥ \\feed{\\vx}{\\ve} ⟩ ↓_{n+1} v }$$\n\nFeedback is interpreted now by \"sequential\" execution! We allow it to update the environment. We can alternatively replace $ε$ for $\\vec{0}$.\n\n## Interaction and Typing\n\nEnvironment ${\\cal E}$ interacts well with $e$ if it contains enough history.\n\n$${\\cal E}~ \\bot_n~ e ↔ ∃ v, ⟨ {\\cal E} ∥ e ⟩ ↓_n v$$\n\n#### Type System:\n\nCo-effects keep track of history-access. Array subtyping and polarity (optional):\n\n$$\\begin{gather} \\inferrule { k \\leq n } { (Γ_\\vm, x :_n R@r ⊢ \\fpre{x}{k} : R[r] } \\rn{Var} \\qquad \\inferrule { (Γ_\\vm, x :_n R@r ⊢ e : T } { Γ_\\vm ⊢ Λ x.~e : R@r ⇒_n T } \\rn{Sp_I} \\qquad \\inferrule { \\Gamma_\\vm, x :_0 R[r(n+1)] ⊢ e : T} { \\Gamma_\\vm, x :_n R@r ⊢ e : T}\\rn{Sub} \\end{gather}$$\n\n#### Main Lemma:\n\nIf $Γ_\\vm ⊢ e : T$ and $Γ_\\vm \\vdash {\\cal E}$ and $e$ is causal (definition of your choice!) then $$∀ n. {\\cal E}~ \\bot_n~ e \\qquad \\qquad \\qquad (\\equiv {\\cal E}~ \\bot~ e)$$\n\nSemantics for types $⟦T⟧$: sets closed under $⊥$.\n\n## Case Study I: Linearity\n\nWe characterize the set of linear Wagner programs, a first step for the Z-transform theory of linear programs.\n\nLin-Wagner is obtained by dropping constants and non-linear operations. What is the right definition of linearity? We use a logical relation:\n\n$$\\begin{array}{lcl} \\mathsf{linR_{n}}^{\\vec{R}}(e) &≡& \\forall {\\cal E}_1~{\\cal E}_2.~ ⟨{\\cal E}_1 ∥ e⟩_n - ⟨{\\cal E}_2 ∥ e⟩_n = ⟨{\\cal E}_1 - {\\cal E}_2 ∥ e⟩_n \\\\ \\mathsf{linR_{n}}^{\\vec{R} ⇒ \\vec{R}}(f) &≡& \\forall {\\cal E}_1 ~{\\cal E}_2 ~r_1 ~r_2.\\\\ && \\quad ⟨{\\cal E}_1 ∥ f~r_1⟩_n - ⟨{\\cal E}_2 ∥ f~r_2⟩_n = ⟨{\\cal E}_1 - {\\cal E}_2 ∥ f~(r_1 - r_2)⟩_n \\end{array}$$\n\nThe relation is simple, but works at higher-types. Every Lin-Wagner program satisfies the relation! The proof proceeds, first, by induction on the typing derivation, then by induction on the number of steps.\n\n## Case Study I: Linearity\n\nWhat about multi-linearity? A brief comment:\n\nCompare: $$\\begin{array}{l} λ x. Λ y . x \\cdot y : R → S ⇒ T \\\\ Λ x. Λ y . x \\cdot y : R ⇒ S ⇒ T \\end{array}$$\n\n## Case Study II: The SSSM Abstract Machine\n\nLin/Val-Wagner semantics compute the full history of the argument at each function call. We introduce a heap and define a monadic semantics for types:\n\nThe binary heap follows the structure of expressions.\n\n## Case Study II: The SSSM Abstract Machine\n\nEvaluation as usual. Operations to \"focus\" the heap, add a new value, etc...\n\nIdea: put/get respect initial buffer size.\n\n## Case Study II: The SSSM Abstract Machine\n\nA well-formed heap for that program contains the proper amount of buffering for $e$, plus the correct values for the past. The main lemma is then:\n\n$${\\cal E} ≈_n ({\\cal D, H}) → ⟨ {\\cal E} ∥ e ⟩_{n+1} = \\mathsf{exprV~{\\cal D}~e}~H$$\n\n## Case Study III: Frequency Domain\n\n#### Quick Review:\n\n• Frequency Response = impulse response output.\n• A filter is stable = its Frequency Response is summable.\n• Stability implied by the existence of the DTFT.\n• The DFT may approximate the DTFT.\n• The DTFT is the evaluation of the Z-transform for the unit circle.\n\n#### Current state:\n\n• Previous linearity theorem and a notion of DFT in Coq are prerequisites.\n• Work in progress, Z-semantics defined propositionaly.\n• Requires a representation theorem to interpret types.\n\n## Conclusions\n\n• Long path, technique is powerful. Applications to other areas, extensions (references, polymorphism).\n• Multirate/clocks: Very simple reclocking primitives, enough for our needs. Better support under study.\n• Some other limitations: false cycle circuits.\n• Implementation in Ocaml with a core type system. Interesting design space, in particular wrt to closures.\n• Coq formalization: working but more experience is needed to decide on some implementation details.\n• Connection to State Space analysis, control theory, MIMO systems.\n• Floating Point Interpretation. Numerical issues are pervasive in this domain, but trying to get there.\n• Verification of program and transform optimization. [Püschel]\n\n## The fun starts !!\n\nWe embed Val-Wagner into Coq using a lightweight dependenly typed syntax. We focus on a reduced subset, leaving higher order, rates, and stable values for later. Recall that: $R ⇒ R ⇒ R ≈ R ⊗ R → R$. Thus:\n\n## Program Interpreation:\n\nThe lightly-typed syntax allows to overcome typical termination issues in our embedding and provide an \"executable\" semantics inside Coq, we want something like:\n\n$$⟦ \\_ ⟧_E : ∀~\\mathsf{τ}~({\\cal E} : \\mathsf{env})~(\\mathsf{e} : \\mathsf{expr~τ})~(\\mathsf{n} : ℕ), ⟦ \\mathsf{τ} ⟧_T$$\n\nWe take a simple choice and use $\\mathsf{env} = \\mathsf{seq}~(\\mathsf{seq}~ℝ)$. For type interpretation we use:\n\n$$\\begin{array}{lcl} ⟦ ⊗~n ⟧_T &= & \\mathsf{cV[ℝ]_n} \\\\ ⟦ n_a ⇒ n_r ⟧_T &= & \\mathsf{seq~cV[ℝ]_{n_a} → cV[ℝ]_{n_r}} \\\\ \\end{array}$$\n\nThus functions have access to all history. We also provide an initial value for every type. Many more refinements are possible!\n\n## Step-Indexing\n\nThe definition of step-indexing in Coq is tricky, interpretation proceeds by induction on time and expressions. We must use a stronger induction principle for Coq to be happy:\n\n## Geometric Signal Theory\n\n#### The DFT:\n\n$$X(\\omega_k) = \\ip{x^T, \\omega_k} ~~\\text{where}~~ \\omega_k = \\omega^{k*0},\\ldots,\\omega^{k*(N-1)}$$\n\nMost properties are an immediate consequency of matrix multiplication lemmas in math-comp.\n\n##### Primitive roots:\n\nThe constructive algebraic numbers in mathcomp provides us with a primitive root of the unity such that $ω^N = 1$.\n\n## In an Ideal World...\n\n#### ... we'd have everything we need in our library.\n\n• We are not so far from it.\n• Took the definitions from classfun.v and proved:\n\nEnergy theorems are easy corollaries. Compact development for now, see the library.\n\n## Transfer Function:\n\nWe now want to relate our programs to their transfer function. The first step is to define the Z-interpreation for types:\n\nWe'd like a theorem relating the summability of:\n\nto the evaluation of the transfer function. We can establish:\n\n## Z-Interpretation of Programs\n\nIt is work in progress; we tried to build an effective procedure, it is difficult.\n\n## First Try: Mini-Faust [LAC2015]\n\n• Faust [Orlarey 2002]: DSL for DSP arrow-like, point-free combinators.\n• Coq semantics: step-indexing, math-comp.\n\n### Problems with the approach:\n\n• DSP not friendly to point-free combinators (matrices).\n• PL methods awkward for DSP experts.\n• Doesn't work well for complex examples.\n• Key point for proofs: Semantics of programs.\n\n## Executing Equations:\n\nThis system is simple but convenient due to its natural, computable, and canonical embedding, what can be done in the case of Lustre-like equations?\n\n1. Define a topological type $\\mathsf{ts}$ and sort function $\\mathsf{topo : eqn \\to option~ ts}$, from sets of equations to schedules.\n2. Define an interpreter for $\\mathsf{ts}$.\n3. [Optional]: If the program is well-clocked, then $\\mathsf{topo}$ succeeds.\n4. Prove a): Prove topo is sound (the output respects the original eqn).\n5. Prove b): All possible schedules return the same value in the interpreter. (invariance under permutation)\n6. The rest of the proof should be \"free\" of scheduling choices. The intrepreter should expose the internal state from the begining.\n\n## Handling Clocked Streams\n\n##### [The beginning of it all]\n\nPort of the Lucid paper to Coq/Ssreflect; replace dependently typed streams by sequences plus a decidable well-clocked relation:\n\nDecidability provides \"inversion views\" (also called small inversion sometimes) which are very convenient.\n\n## Example: Filter Stability\n\n$$\\fjud{\\fpo}{f}{\\fpw} \\iff \\forall t, \\fpo(i(t)) \\Rightarrow \\fpw(f(i)(t))$$\n\nIn PL terms: $\\fjud{x \\in [a,b]}{\\mathsf{*(1-c) : + \\sim *(c)}}{x \\in [a,b]}$",
null,
"In DSP terms: The impulse response decays to 0 as time goes to infinity.\n\n• Higher-order filters may make PL methods impractical.\n• State seen as matrices/arrays.\n\n## Context of the ANR project:\n\nPractice of real time DSP still far from convenient.\n\n##### Starting point, Faust:\n• Faust [Orlarey 2002], a functional PL for audio programming.\n• Abstracts away low-level complexity, efficient execution.\n• A success, good number of users, high-interest topic. (CCRMA workshops, Kadenze course, etc...)\n##### Faust's Future:\n• Extend Faust to multirate processing.\n• Reasoning about audio programs is not easy.\n\n/"
]
| [
null,
"https://x80.org/synchron2016/img/sleak.gif",
null,
"https://x80.org/synchron2016/img/df1.png",
null,
"https://x80.org/synchron2016/img/wgr.png",
null,
"https://x80.org/synchron2016/img/df1.png",
null,
"https://x80.org/synchron2016/img/df2.png",
null,
"https://x80.org/synchron2016/img/wgr.png",
null,
"https://x80.org/synchron2016/img/fstab.gif",
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.61091834,"math_prob":0.9898415,"size":15598,"snap":"2023-40-2023-50","text_gpt3_token_len":5329,"char_repetition_ratio":0.10978582,"word_repetition_ratio":0.11415525,"special_character_ratio":0.32786256,"punctuation_ratio":0.13517241,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99891233,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14],"im_url_duplicate_count":[null,4,null,8,null,8,null,8,null,4,null,8,null,4,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-29T03:56:28Z\",\"WARC-Record-ID\":\"<urn:uuid:1770b622-6692-4b4a-a9f4-165a91ab4162>\",\"Content-Length\":\"37087\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8a812cb0-a803-4340-acd5-b16a87ec4786>\",\"WARC-Concurrent-To\":\"<urn:uuid:1dc668f9-2228-4afc-a4c2-5405e4b2aa5f>\",\"WARC-IP-Address\":\"188.226.171.132\",\"WARC-Target-URI\":\"https://x80.org/synchron2016/\",\"WARC-Payload-Digest\":\"sha1:ZWA7VFTHOYEIFXEUVQZ2RHY2BV5N65HT\",\"WARC-Block-Digest\":\"sha1:HFRRC6MH6OU4FYRHPQEE5OZBHYLZAQI5\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233510481.79_warc_CC-MAIN-20230929022639-20230929052639-00287.warc.gz\"}"} |
https://www.mankier.com/3/cungtr.f | [
"# cungtr.f man page\n\ncungtr.f\n\n## Synopsis\n\n### Functions/Subroutines\n\nsubroutine cungtr (UPLO, N, A, LDA, TAU, WORK, LWORK, INFO)\nCUNGTR\n\n## Function/Subroutine Documentation\n\n### subroutine cungtr (character UPLO, integer N, complex, dimension( lda, * ) A, integer LDA, complex, dimension( * ) TAU, complex, dimension( * ) WORK, integer LWORK, integer INFO)\n\nCUNGTR\n\nPurpose:\n\n``` CUNGTR generates a complex unitary matrix Q which is defined as the\nproduct of n-1 elementary reflectors of order N, as returned by\nCHETRD:\n\nif UPLO = 'U', Q = H(n-1) . . . H(2) H(1),\n\nif UPLO = 'L', Q = H(1) H(2) . . . H(n-1).```\nParameters:\n\nUPLO\n\n``` UPLO is CHARACTER*1\n= 'U': Upper triangle of A contains elementary reflectors\nfrom CHETRD;\n= 'L': Lower triangle of A contains elementary reflectors\nfrom CHETRD.```\n\nN\n\n``` N is INTEGER\nThe order of the matrix Q. N >= 0.```\n\nA\n\n``` A is COMPLEX array, dimension (LDA,N)\nOn entry, the vectors which define the elementary reflectors,\nas returned by CHETRD.\nOn exit, the N-by-N unitary matrix Q.```\n\nLDA\n\n``` LDA is INTEGER\nThe leading dimension of the array A. LDA >= N.```\n\nTAU\n\n``` TAU is COMPLEX array, dimension (N-1)\nTAU(i) must contain the scalar factor of the elementary\nreflector H(i), as returned by CHETRD.```\n\nWORK\n\n``` WORK is COMPLEX array, dimension (MAX(1,LWORK))\nOn exit, if INFO = 0, WORK(1) returns the optimal LWORK.```\n\nLWORK\n\n``` LWORK is INTEGER\nThe dimension of the array WORK. LWORK >= N-1.\nFor optimum performance LWORK >= (N-1)*NB, where NB is\nthe optimal blocksize.\n\nIf LWORK = -1, then a workspace query is assumed; the routine\nonly calculates the optimal size of the WORK array, returns\nthis value as the first entry of the WORK array, and no error\nmessage related to LWORK is issued by XERBLA.```\n\nINFO\n\n``` INFO is INTEGER\n= 0: successful exit\n< 0: if INFO = -i, the i-th argument had an illegal value```\nAuthor:\n\nUniv. of Tennessee\n\nUniv. of California Berkeley\n\nNAG Ltd.\n\nDate:\n\nDecember 2016\n\nDefinition at line 125 of file cungtr.f.\n\n## Author\n\nGenerated automatically by Doxygen for LAPACK from the source code.\n\n## Referenced By\n\nThe man page cungtr(3) is an alias of cungtr.f(3).\n\nTue Nov 14 2017 Version 3.8.0 LAPACK"
]
| [
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.6759465,"math_prob":0.9725021,"size":2391,"snap":"2019-43-2019-47","text_gpt3_token_len":723,"char_repetition_ratio":0.11562631,"word_repetition_ratio":0.29135802,"special_character_ratio":0.26390633,"punctuation_ratio":0.18359375,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9885294,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-11-13T21:57:11Z\",\"WARC-Record-ID\":\"<urn:uuid:259f6c40-7416-4082-a30a-47890bc90658>\",\"Content-Length\":\"8045\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:18f9b7c1-df56-4f2d-a4ba-ac955745cb33>\",\"WARC-Concurrent-To\":\"<urn:uuid:bd89dbe5-9767-4b7f-af6e-61810e4f2b22>\",\"WARC-IP-Address\":\"104.31.89.183\",\"WARC-Target-URI\":\"https://www.mankier.com/3/cungtr.f\",\"WARC-Payload-Digest\":\"sha1:KPDYWUQCQB6QMJXDA6O2ZYN7NKTGEW26\",\"WARC-Block-Digest\":\"sha1:W77GWN6TMY4BNNTXIBIKCQPJ4SD5XWKH\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-47/CC-MAIN-2019-47_segments_1573496667442.36_warc_CC-MAIN-20191113215021-20191114003021-00011.warc.gz\"}"} |
http://lrp2.steinkuehler.net/Packages/man/fdformat.8.man.htm | [
"",
null,
"# fdformat.8\n\n```\n\n```\n\n## NAME\n\n``` fdformat - Low-level formats a floppy disk\n\n```\n\n## SYNOPSIS\n\n``` fdformat [ -n ] device\n\n```\n\n## DESCRIPTION\n\n``` fdformat does a low level format on a floppy disk. device\nis usually one of the following (for floppy devices, the\nmajor = 2, and the minor is shown for informational purş\nposes only):\n\n/dev/fd0d360 (minor = 4)\n/dev/fd0h1200 (minor = 8)\n/dev/fd0D360 (minor = 12)\n/dev/fd0H360 (minor = 12)\n/dev/fd0D720 (minor = 16)\n/dev/fd0H720 (minor = 16)\n/dev/fd0h360 (minor = 20)\n/dev/fd0h720 (minor = 24)\n/dev/fd0H1440 (minor = 28)\n\n/dev/fd1d360 (minor = 5)\n/dev/fd1h1200 (minor = 9)\n/dev/fd1D360 (minor = 13)\n/dev/fd1H360 (minor = 13)\n/dev/fd1D720 (minor = 17)\n/dev/fd1H720 (minor = 17)\n/dev/fd1h360 (minor = 21)\n/dev/fd1h720 (minor = 25)\n/dev/fd1H1440 (minor = 29)\n\nThe generic floppy devices, /dev/fd0 and /dev/fd1, will\nfail to work with fdformat when a non-standard format is\nbeing used, or if the format has not been autodetected\nearlier. In this case, use setfdprm(8) to load the disk\nparameters.\n\n```\n\n## OPTIONS\n\n``` -n No verify. This option will disable the verificaş\ntion that is performed after the format.\n\n```\n\n## SEE ALSO\n\n``` fd(4), setfdprm(8), mkfs(8), emkfs(8)\n\n```\n\n## AUTHOR\n\n``` Werner Almesberger ([email protected])\n\n```\n\nMan(1) output converted with man2html"
]
| [
null,
"http://lrp2.steinkuehler.net/local/images/logo.jpg",
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.71802044,"math_prob":0.6962183,"size":1240,"snap":"2019-35-2019-39","text_gpt3_token_len":453,"char_repetition_ratio":0.27831715,"word_repetition_ratio":0.0,"special_character_ratio":0.36612904,"punctuation_ratio":0.08888889,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97011393,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-08-25T23:21:09Z\",\"WARC-Record-ID\":\"<urn:uuid:afe764cc-092b-4370-85eb-7d50d7e0084f>\",\"Content-Length\":\"3733\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9b9b32bc-a20b-416d-b241-07447a01b7fd>\",\"WARC-Concurrent-To\":\"<urn:uuid:a2af1360-7918-474f-93f3-ade3ce6e95b3>\",\"WARC-IP-Address\":\"139.146.185.30\",\"WARC-Target-URI\":\"http://lrp2.steinkuehler.net/Packages/man/fdformat.8.man.htm\",\"WARC-Payload-Digest\":\"sha1:6RP3TYYCYAT5XMSFIC76OZRHWWBPPD4X\",\"WARC-Block-Digest\":\"sha1:7BYUZYU3XH7DPRDJ4V3WJNKKP7WP4AQL\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-35/CC-MAIN-2019-35_segments_1566027330907.46_warc_CC-MAIN-20190825215958-20190826001958-00109.warc.gz\"}"} |
https://nl.mathworks.com/matlabcentral/answers/483175-how-can-i-make-a-step-response-graph-for-open-loop-ziegler-nichols-method | [
"# How can I make a step response graph for open Loop Ziegler-Nichols Method?\n\n15 views (last 30 days)\nMilon Chowdhury on 2 Oct 2019\nCommented: David K. on 3 Oct 2019\nHI,\nGreetings. I am very new in MATLAB. I want to draw a step response graph for Open Loop Ziegler-Nichols Method. the equation is attached here. Actually, I can't understand, how to code the numerator and denominator. Thanks in advance.",
null,
"David K. on 2 Oct 2019\nWhen trying to do transfer functions in matlab one very nice way to do it is with\ns = tf('s');\nNow, s indicates it is a transfer function you can actually just type the rest of your equations as you would want.\nk = k; d = d; tau = tau; % These are just saying that you should define these yourself\nG = k*exp(-d*s)/(tau*s+1); % Create the transfer function\nstep(G); % Plot the transfer function\n\nMilon Chowdhury on 3 Oct 2019\nThanks for your reply. I have drawn a nice step response graph using your code. I am facing another problem. In Simulink code, how can I mention/put this equation?",
null,
"David K. on 3 Oct 2019\nGenerally if you have a followup question you should put it in a comment on my answer, your question, an edit to your question, or a new question instead of an answer.\nYou can do it by putting a transfer function block with\nnum = [km]\nden = [tau 1]\nin series after a Transport Delay Block with the delay set to d. This is because an exponent in laplace domain results in a time delay of d seconds."
]
| [
null,
"https://www.mathworks.com/matlabcentral/answers/uploaded_files/240777/image.jpeg",
null,
"https://www.mathworks.com/matlabcentral/answers/uploaded_files/240953/Transfer%20function.jpeg",
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.90960234,"math_prob":0.57165337,"size":1521,"snap":"2023-40-2023-50","text_gpt3_token_len":389,"char_repetition_ratio":0.106789716,"word_repetition_ratio":0.021428572,"special_character_ratio":0.25180802,"punctuation_ratio":0.0952381,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97207636,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,3,null,3,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-10-02T07:30:51Z\",\"WARC-Record-ID\":\"<urn:uuid:2c82b79a-7bda-4993-bac1-9e33468dc139>\",\"Content-Length\":\"125072\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e71912bc-2e35-4293-a32e-29420caa8f65>\",\"WARC-Concurrent-To\":\"<urn:uuid:b2139f1c-e526-4f39-8709-22229cd81527>\",\"WARC-IP-Address\":\"104.86.80.92\",\"WARC-Target-URI\":\"https://nl.mathworks.com/matlabcentral/answers/483175-how-can-i-make-a-step-response-graph-for-open-loop-ziegler-nichols-method\",\"WARC-Payload-Digest\":\"sha1:RZXDHK5B5FMM3SGDHTZNEDR4QXHP3FL6\",\"WARC-Block-Digest\":\"sha1:V4SG75BJ5PXQP33GSO4VI3YFOLL2AGWG\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233510983.45_warc_CC-MAIN-20231002064957-20231002094957-00060.warc.gz\"}"} |
https://www.physicsforums.com/threads/bisection-method-and-numerical-integration.867060/ | [
"# Bisection method and numerical integration\n\n• MATLAB\n\n## Main Question or Discussion Point\n\nIn Matlab I am trying to use the composite Simpson's rule to find $x_l$ so that\n\n$$170=\\int^{x_l}_0 \\sqrt{1+(y')^2} dx = \\int^{x_l}_0 \\sqrt{1+\\left( \\frac{x^2}{68000} \\right)^2} dx$$\n\nFor convenience this can be written as\n\n$$I(x) = 170 - \\int^x_0 \\sqrt{1 + (\\frac{x^2}{68000})} dx$$\n\nThe limits of integration would be from $0$ to $x=170$. Now to find the zero of this function I want to employ the bisection method while using Simpsons rule to evaluate the integral involved in evaluating $I(x)$ at each step.\n\nHere's my code so far:\n\nCode:\na=0; b=170;\n\nfor x=[0:]\n\nf = sqrt(1+((x.^2)./68000).^2);\n\n%Simpson's rule\nif numel(f)>1\nn=numel(f)-1; h=(b-a)/n;\nI= h/3*(f(1)+2*sum(f(3:2:end-2))+4*sum(f(2:2:end))+f(end));\nelse\nh=(b-a)/n; xi=a:h:b;\nI= h/3*(f(xi(1))+2*sum(f(xi(3:2:end-2)))+4*sum(f(xi(2:2:end)))+f(xi(end)));\nend\nf = 170 - I;\n\ntol=1e-6;\nwhile abs(b-a) > tol\nx = (a+b)/2;\ny = ff(x);\n\nif y == 0\nrv = x;\nbreak\nend\n\nif ff(a)*y < 0\nb = x;\nelse\na = x;\nend\nend\n\nrv=(a+b)/2;\nBut I get lots of errors and the code does not run. I think it is because I need to get the Simpson's rule to calculate new value for the function at each midpoint of the bisection method. I'm not really sure how to fix this.",
null,
"Alternatively, if bisection does not work with Simpson's method, I would appreciate it if anyone could show me how to exactly incorporate built in root finders into my code.\n\nAny help would be greatly appreicated.\n\nLast edited:\n\nRelated MATLAB, Maple, Mathematica, LaTeX News on Phys.org\nDr Transport\nGold Member\nI'd use regula-falsi for the root finding part of this problem. If memory serves me correctly, MATLAB has a simpson's rule built in.....\n\n•",
null,
"roam\nThe question is confusing because first you say you are trying to find the upper limit of integration $x_l$ (such that the value of the integral is 170), but then later you say that the upper limit of integration is 170.\n\nIf you are trying to find the upper integration limit such that the value of the definite integral is 170, the following will work. quadl is a numerical integrator that comes with MATLAB, but you could also use quadgk or integral to the same effect. Also, fzero is the built-in root finder.\n\nCode:\n>> f = @(xl) quadl(@(x)sqrt(1+(x.^2/68000).^2),0,xl);\n>> R = fzero(@(x) 170-f(x),4)\nR =\n167.2340\n170.0000\n•",
null,
""
]
| [
null,
"data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7",
null,
"data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7",
null,
"data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7",
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.780627,"math_prob":0.99731266,"size":2152,"snap":"2020-10-2020-16","text_gpt3_token_len":722,"char_repetition_ratio":0.095437616,"word_repetition_ratio":0.5648415,"special_character_ratio":0.375,"punctuation_ratio":0.103238866,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99997044,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-03-30T23:21:09Z\",\"WARC-Record-ID\":\"<urn:uuid:4e859f68-3820-465d-b80b-1059a7dcd422>\",\"Content-Length\":\"82009\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b6e01925-97b7-4043-a9e2-1ebbde59e263>\",\"WARC-Concurrent-To\":\"<urn:uuid:4c0e93e2-1f9b-441c-8e6f-4bc44e7d453a>\",\"WARC-IP-Address\":\"23.111.143.85\",\"WARC-Target-URI\":\"https://www.physicsforums.com/threads/bisection-method-and-numerical-integration.867060/\",\"WARC-Payload-Digest\":\"sha1:3DGXMAJQGT4DZHFSXXFFM6V5BMSX5NJA\",\"WARC-Block-Digest\":\"sha1:DK7NIOZ4R3BNMSXECW6BFEPTZADBBVCT\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-16/CC-MAIN-2020-16_segments_1585370497309.31_warc_CC-MAIN-20200330212722-20200331002722-00524.warc.gz\"}"} |
https://geokush.com/2020/02/17/regression-analysis-in-biological-anthropology/?shared=email&msg=fail | [
"# Analysis Working Papers: Regression Analysis in Biological Anthropology\n\nWhat is Regression?\n\nRegression analysis comprises several related analytical tools that are useful to researchers in a variety of disciplines, including biological anthropology. Different styles of regression are available for different types of dependent variables (see Table 1 for a non-exhaustive list) and can accommodate a wide range of contexts that are challenging for many standard statistical methods, including controlling for any number of potential confounding variables, adjusting for non-independent sampling units, and using cases that have missing data.",
null,
"Table 1. Some types of regression.\n\nMost types of regression analysis can be used to: (a) test hypotheses about the association between variables, (b) model the relationship between variables and provide measures of the strength of model fit and proportion of variation in a dependent variable explained by the independent variables, and (c) build equations that can be used for prediction. Regression analysis can accommodate any number of independent variables, ranging from the simplest cases where you are interested in the effect of a single independent variable on the dependent variable, to the case of multiple regression where you are interested in two or more independent variables. The simplest case is where the independent variables are continuous, but other types of independent variables can be accommodated. Table 2 introduces some resources for issues in the application of regression analysis.",
null,
"Table 2. Resources for issues in the application of regression analysis.\n\nReferences\n\nAgresti, A. (2013). Categorical data analysis (3E). NY: Wiley.\n\nAllison, P.D. (2001). Missing data. Thousand Oaks, CA: Sage Publications.\n\nAllison, P.D. (2014). Event history and survival analysis. Thousand Oaks, CA: Sage Publications.\n\nBurnham, K. P., & Anderson, D. R. (2004). Multimodel inference: understanding AIC and BIC in model selection. Sociological Methods & Research, 33, 261-304.\n\nCoxe, S., West, S. G., & Aiken, L. S. (2009). The analysis of count data: A gentle introduction to Poisson regression and its alternatives. Journal of Personality Assessment, 91(2), 121-136.\n\nGelman, A., & Hill, J. (2007). Data analysis using regression and multilevel/hierarchical models. NY: Cambridge University Press.\n\nGraham, M. H. (2003). Confronting multicollinearity in ecological multiple regression. Ecology, 84(11), 2809-15.\n\nGraubard, B. I., & Korn, E. L. (1994). Regression analysis with clustered data. Statistics in Medicine, 13, 509-522.\n\nHardy, M. A. (1993). Regression with dummy variables. Newbury Park, CA: Sage Publications.\n\nHarrison, X. A., Donaldson, L., Correa-Cano, et al. (2018). A brief introduction to mixed effects modelling and multi-model inference in ecology. PeerJ, 6, e4794.\n\nHilba, J.M. (2009). Logistic regression models. Boca Raton, FL: CRC Press.\n\nHosmer, D.W. (2008). Applied survival analysis: Regression modeling of time-to-event data. NJ: Wiley.\n\nHosmer, D.W., Lemeshow, S., & Sturdivant, R.X. (2013). Applied logistic regression (3E). NY: John Wiley.\n\nKlein, J.P. (2016) Handbook of survival analysis. London: CRC Press.\n\nJaccard, J. (2001). Interaction effects in logistic regression. Newbury Park, CA: Sage Publications.\n\nJaccard, J., Turrisi, R., & Wan, C. K. (1990). Interaction effects in multiple regression. Newbury Park, CA: Sage Publications.\n\nJohnson, J., & Omland, K. S. (2004). Model selection in ecology and evolution. Trends in Ecology & Evolution, 19(2), 101-108.\n\nMckinnon, A. (2010). The use and reporting of multiple imputation in medical research — a review. Journal of Internal Medicine, 268, 586-593.\n\nMcNamee, R. (2005). Regression modelling and other methods to control confounding. Occupational and Environmental Medicine, 62, 500-506.\n\nMontgomery, D.C. (2013). Introduction to linear regression analysis (5E). Somerset: John Wiley and Sons.\n\nO’Brien, R. (2007). A caution regarding rules of thumb for variance inflation factors. Quality & Quantity, 41, 673.\n\nPalmer, P. B., & O’Connell, D. G. (2009). Regression analysis for prediction: Understanding the process. Cardiopulmonary Physical Therapy Journal, 20(3), 23-26.\n\nWeisberg, S. (2014). Applied linear regression. Hoboken, NJ: Wiley.\n\nWestfall, J., & Yarkoni, T. (2016). Statistically controlling for confounding constructs is harder than you think. PLoS ONE, 11(3), e0152719.\n\nWritten by:\nDr Geoff Kushnick ([email protected])\nThe Australian National University, Canberra\n\nDownload a PDF copy of this post by clicking here.\n\nLeave a comment\n\nFiled under Analysis Working Papers"
]
| [
null,
"https://hbeanu.files.wordpress.com/2020/02/table-1.jpg",
null,
"https://hbeanu.files.wordpress.com/2020/02/table-2.jpg",
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.7409533,"math_prob":0.6138913,"size":4477,"snap":"2021-04-2021-17","text_gpt3_token_len":1104,"char_repetition_ratio":0.13950369,"word_repetition_ratio":0.014218009,"special_character_ratio":0.26468617,"punctuation_ratio":0.27631578,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97593987,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,10,null,10,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-04-14T02:15:47Z\",\"WARC-Record-ID\":\"<urn:uuid:e47dd269-d9c2-4913-bf0d-db44f850f629>\",\"Content-Length\":\"69995\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:0059482e-42a7-4141-8861-2311ad353bbf>\",\"WARC-Concurrent-To\":\"<urn:uuid:4f9ddfc4-2b0b-4d87-94eb-70c04d2245bf>\",\"WARC-IP-Address\":\"192.0.78.25\",\"WARC-Target-URI\":\"https://geokush.com/2020/02/17/regression-analysis-in-biological-anthropology/?shared=email&msg=fail\",\"WARC-Payload-Digest\":\"sha1:B6AAROQCQUBPQKI3NTRSS5YPLDTYJHIJ\",\"WARC-Block-Digest\":\"sha1:X2JFFXJLOARKRP7IBHBDUYS4YC7TEEWA\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-17/CC-MAIN-2021-17_segments_1618038076454.41_warc_CC-MAIN-20210414004149-20210414034149-00606.warc.gz\"}"} |
https://www.wyzant.com/resources/lessons/math/calculus/derivative_proofs/power_rule | [
"# Derivative Proof of Power Rule",
null,
"This proof requires a lot of work if you are not familiar with implicit differentiation, which is basically differentiating a variable in terms of x. Some may try to prove the power rule by repeatedly using product rule. Though it is not a \"proper proof,\" it can still be good practice using mathematical induction. A common proof that is used is using the Binomial Theorem:",
null,
"The limit definition for xn would be as follows",
null,
"Using the Binomial Theorem, we get",
null,
"Subtract the xn",
null,
"Factor out an h",
null,
"All of the terms with an h will go to 0, and then we are left with",
null,
"",
null,
"## Implicit Differentiation Proof of Power Rule\n\nIf we don't want to get messy with the Binomial Theorem, we can simply use implicit differentiation, which is basically treating y as f(x) and using Chain rule.\n\nLet",
null,
"Take the natural log of both sides",
null,
"Take the derivative with respect to x",
null,
"Notice that we took the derivative of lny and used chain rule as well to take the derivative of the inside function y.\n\nMultiply both sides by y",
null,
"Substitute xc back in for y",
null,
"",
null,
"Sign up for free to access more calculus resources like . Wyzant Resources features blogs, videos, lessons, and more about calculus and over 250 other subjects. Stop struggling and start learning today with thousands of free resources!\nif (isMyPost) { }"
]
| [
null,
"https://dj1hlxw0wr920.cloudfront.net/userfiles/wyzfiles/18e0fbaf-d421-456a-b327-daf1b42e01f7.gif",
null,
"https://dj1hlxw0wr920.cloudfront.net/userfiles/wyzfiles/16d36ae2-3f4e-4670-96b2-41cc38913c2e.gif",
null,
"https://dj1hlxw0wr920.cloudfront.net/userfiles/wyzfiles/dea1009a-fdcc-4d25-83b3-02c25e36e089.gif",
null,
"https://dj1hlxw0wr920.cloudfront.net/userfiles/wyzfiles/158f3e27-4a2c-4c4f-ab50-1733f9daf352.gif",
null,
"https://dj1hlxw0wr920.cloudfront.net/userfiles/wyzfiles/0e4004eb-b1d3-48fd-98eb-062a76977090.gif",
null,
"https://dj1hlxw0wr920.cloudfront.net/userfiles/wyzfiles/252d9d78-612b-4e65-a304-607b8a8ae718.gif",
null,
"https://dj1hlxw0wr920.cloudfront.net/userfiles/wyzfiles/abb9e6d4-386e-4f89-b4dc-83532a4a153c.gif",
null,
"https://dj1hlxw0wr920.cloudfront.net/userfiles/wyzfiles/6c90c7e7-d9b9-4ed5-a333-b6b6de96b960.gif",
null,
"https://dj1hlxw0wr920.cloudfront.net/userfiles/wyzfiles/581f79b2-b9aa-4243-a8b1-a5a75de4c90e.gif",
null,
"https://dj1hlxw0wr920.cloudfront.net/userfiles/wyzfiles/c0f166fe-b135-400c-ba1a-a29f72c13995.gif",
null,
"https://dj1hlxw0wr920.cloudfront.net/userfiles/wyzfiles/bcb40b26-8314-4d57-9e92-29f17f46f89f.gif",
null,
"https://dj1hlxw0wr920.cloudfront.net/userfiles/wyzfiles/6e8b5796-b829-4138-ac3d-15e740436b95.gif",
null,
"https://dj1hlxw0wr920.cloudfront.net/userfiles/wyzfiles/4d688bb8-4ea1-4e11-9bbc-ec5e578c7093.gif",
null,
"https://dj1hlxw0wr920.cloudfront.net/userfiles/wyzfiles/6c90c7e7-d9b9-4ed5-a333-b6b6de96b960.gif",
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.944967,"math_prob":0.93142146,"size":1042,"snap":"2021-04-2021-17","text_gpt3_token_len":232,"char_repetition_ratio":0.111753374,"word_repetition_ratio":0.010638298,"special_character_ratio":0.20057581,"punctuation_ratio":0.058536585,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9954004,"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,3,null,2,null,1,null,1,null,1,null,1,null,1,null,null,null,1,null,1,null,1,null,1,null,1,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-04-18T11:23:18Z\",\"WARC-Record-ID\":\"<urn:uuid:dd80964e-392d-4545-88a3-a46706146e9b>\",\"Content-Length\":\"54648\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:96733404-768b-4c2d-9302-7193adbfd16c>\",\"WARC-Concurrent-To\":\"<urn:uuid:98b4190b-838e-416c-8b71-82862565ad40>\",\"WARC-IP-Address\":\"34.117.195.90\",\"WARC-Target-URI\":\"https://www.wyzant.com/resources/lessons/math/calculus/derivative_proofs/power_rule\",\"WARC-Payload-Digest\":\"sha1:L73HGHVXIEWRWGN73MKPISGNN5SQHFM6\",\"WARC-Block-Digest\":\"sha1:YCPZ5C7DAL6STKYH5S32LPUGTQKAW2XS\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-17/CC-MAIN-2021-17_segments_1618038476606.60_warc_CC-MAIN-20210418103545-20210418133545-00416.warc.gz\"}"} |
https://metanumbers.com/71998 | [
"71998 (number)\n\n71,998 (seventy-one thousand nine hundred ninety-eight) is an even five-digits composite number following 71997 and preceding 71999. In scientific notation, it is written as 7.1998 × 104. The sum of its digits is 34. It has a total of 2 prime factors and 4 positive divisors. There are 35,998 positive integers (up to 71998) that are relatively prime to 71998.\n\nBasic properties\n\n• Is Prime? No\n• Number parity Even\n• Number length 5\n• Sum of Digits 34\n• Digital Root 7\n\nName\n\nShort name 71 thousand 998 seventy-one thousand nine hundred ninety-eight\n\nNotation\n\nScientific notation 7.1998 × 104 71.998 × 103\n\nPrime Factorization of 71998\n\nPrime Factorization 2 × 35999\n\nComposite number\nDistinct Factors Total Factors Radical ω(n) 2 Total number of distinct prime factors Ω(n) 2 Total number of prime factors rad(n) 71998 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0\n\nThe prime factorization of 71,998 is 2 × 35999. Since it has a total of 2 prime factors, 71,998 is a composite number.\n\nDivisors of 71998\n\n1, 2, 35999, 71998\n\n4 divisors\n\n Even divisors 2 2 1 1\nTotal Divisors Sum of Divisors Aliquot Sum τ(n) 4 Total number of the positive divisors of n σ(n) 108000 Sum of all the positive divisors of n s(n) 36002 Sum of the proper positive divisors of n A(n) 27000 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 268.324 Returns the nth root of the product of n divisors H(n) 2.66659 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors\n\nThe number 71,998 can be divided by 4 positive divisors (out of which 2 are even, and 2 are odd). The sum of these divisors (counting 71,998) is 108,000, the average is 27,000.\n\nOther Arithmetic Functions (n = 71998)\n\n1 φ(n) n\nEuler Totient Carmichael Lambda Prime Pi φ(n) 35998 Total number of positive integers not greater than n that are coprime to n λ(n) 35998 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 7121 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares\n\nThere are 35,998 positive integers (less than 71,998) that are coprime with 71,998. And there are approximately 7,121 prime numbers less than or equal to 71,998.\n\nDivisibility of 71998\n\n m n mod m 2 3 4 5 6 7 8 9 0 1 2 3 4 3 6 7\n\nThe number 71,998 is divisible by 2.\n\nClassification of 71998\n\n• Arithmetic\n• Semiprime\n• Deficient\n\nExpressible via specific sums\n\n• Polite\n• Non-hypotenuse\n\n• Square Free\n\nBase conversion (71998)\n\nBase System Value\n2 Binary 10001100100111110\n3 Ternary 10122202121\n4 Quaternary 101210332\n5 Quinary 4300443\n6 Senary 1313154\n8 Octal 214476\n10 Decimal 71998\n12 Duodecimal 357ba\n20 Vigesimal 8jji\n36 Base36 1jjy\n\nBasic calculations (n = 71998)\n\nMultiplication\n\nn×y\n n×2 143996 215994 287992 359990\n\nDivision\n\nn÷y\n n÷2 35999 23999.3 17999.5 14399.6\n\nExponentiation\n\nny\n n2 5183712004 373216896863992 26870870140413696016 1934648908369505285759968\n\nNth Root\n\ny√n\n 2√n 268.324 41.6013 16.3806 9.36406\n\n71998 as geometric shapes\n\nCircle\n\n Diameter 143996 452377 1.62851e+10\n\nSphere\n\n Volume 1.56333e+15 6.51404e+10 452377\n\nSquare\n\nLength = n\n Perimeter 287992 5.18371e+09 101821\n\nCube\n\nLength = n\n Surface area 3.11023e+10 3.73217e+14 124704\n\nEquilateral Triangle\n\nLength = n\n Perimeter 215994 2.24461e+09 62352.1\n\nTriangular Pyramid\n\nLength = n\n Surface area 8.97845e+09 4.3984e+13 58786.1"
]
| [
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.61857224,"math_prob":0.9894765,"size":4565,"snap":"2022-05-2022-21","text_gpt3_token_len":1616,"char_repetition_ratio":0.11839509,"word_repetition_ratio":0.028315946,"special_character_ratio":0.4543264,"punctuation_ratio":0.07483871,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9990355,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-01-25T20:04:56Z\",\"WARC-Record-ID\":\"<urn:uuid:9f73ef6a-8ee3-481d-a52a-ff7605e7e471>\",\"Content-Length\":\"39613\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c489c242-bea0-48cd-a85d-9e3422654afa>\",\"WARC-Concurrent-To\":\"<urn:uuid:df903f0d-6c87-41c8-856e-332b557788be>\",\"WARC-IP-Address\":\"46.105.53.190\",\"WARC-Target-URI\":\"https://metanumbers.com/71998\",\"WARC-Payload-Digest\":\"sha1:NUSBDJJO4NBRSYRM4FR4WXDW3FKEANYG\",\"WARC-Block-Digest\":\"sha1:3DROMBWPOUGDDDZQYRQFPZL6IWYZ3SUB\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-05/CC-MAIN-2022-05_segments_1642320304872.21_warc_CC-MAIN-20220125190255-20220125220255-00631.warc.gz\"}"} |
https://invasivecardiology.wordpress.com/2013/12/08/going-with-the-flow-in-vitro-assessment-of-ffr-in-2-stenoses/ | [
"## Going With the Flow: In Vitro Assessment of FFR in >2 Stenoses\n\nBy Naritatsu Saito\n\nAs many physicians know, a simple fractional flow reserve (FFR) measurement does not predict the true functional severity of an individual stenosis in multiple sequential coronary stenoses because of the complex fluid dynamic interaction between the stenoses. Theoretical equations to predict the true FFR of individual stenosis in a tandem lesion were developed more than 10 years ago, and the equations have been cited repeatedly in many articles. However, the application of the equations is limited in a tandem lesion. We mathematically derived two novel equations applicable to multiple sequential coronary stenoses with more than 3 stenoses. One is the equation which predicts the true FFR of each stenosis (equation A), and the other predicts the true FFR after releasing a given stenosis in multiple sequential stenoses (equation B).",
null,
"One may consider that both equation A and equation B are difficult to apply in daily clinical practice since both equations require wedge pressure measurement during the maximum hyperemia. However, both equations are important when considering the theoretical background of the FFR assessment in a sequential lesion. Equation A suggests that the true FFR of individual stenosis is always smaller than the apparent FFR value. The severity of individual stenosis is always underestimated in multiple sequential stenoses. Equation B suggests that the true FFR of entire stenoses after releasing a given stenosis is always smaller than the apparent FFR value obtained by a simple addition of baseline FFR of the entire stenoses plus ∆FFR across the target lesion. A better understanding of the background theory helps to improve the performance of daily practice.\n\nRead the entire article at:\n\nA selection of articles on Fractional Flow Reserve available at http://www.invasivecardiology.com"
]
| [
null,
"https://invasivecardiology.files.wordpress.com/2013/12/13-00242-blog-formula-art.jpg",
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.85703343,"math_prob":0.98765755,"size":2975,"snap":"2019-13-2019-22","text_gpt3_token_len":708,"char_repetition_ratio":0.1268933,"word_repetition_ratio":0.045454547,"special_character_ratio":0.21243697,"punctuation_ratio":0.14102565,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9888337,"pos_list":[0,1,2],"im_url_duplicate_count":[null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-03-23T02:12:43Z\",\"WARC-Record-ID\":\"<urn:uuid:75cbc0bd-2547-48ac-bc15-23644e9d1a89>\",\"Content-Length\":\"60621\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:eeb5a3ef-80c8-471e-8b43-d6ea42845f0e>\",\"WARC-Concurrent-To\":\"<urn:uuid:2f3b6f4a-0d39-4021-89f5-b579a9b55811>\",\"WARC-IP-Address\":\"192.0.78.12\",\"WARC-Target-URI\":\"https://invasivecardiology.wordpress.com/2013/12/08/going-with-the-flow-in-vitro-assessment-of-ffr-in-2-stenoses/\",\"WARC-Payload-Digest\":\"sha1:K3CAYYYI7KJKY5BCLBO5ZZYBCLHSELX2\",\"WARC-Block-Digest\":\"sha1:IR5XIE4G6LSEXXI62GYXYKDDLUE5K5ZH\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-13/CC-MAIN-2019-13_segments_1552912202711.3_warc_CC-MAIN-20190323020538-20190323042538-00423.warc.gz\"}"} |
https://networkx.org/documentation/networkx-2.3/reference/algorithms/generated/networkx.algorithms.connectivity.edge_kcomponents.k_edge_components.html | [
"Warning\n\nThis documents an unmaintained version of NetworkX. Please upgrade to a maintained version and see the current NetworkX documentation.\n\n# networkx.algorithms.connectivity.edge_kcomponents.k_edge_components¶\n\nk_edge_components(G, k)[source]\n\nGenerates nodes in each maximal k-edge-connected component in G.\n\nParameters: G (NetworkX graph) k (Integer) – Desired edge connectivity k_edge_components – will have k-edge-connectivity in the graph G. a generator of k-edge-ccs. Each set of returned nodes\n\nlocal_edge_connectivity()\n\nk_edge_subgraphs()\nsimilar to this function, but the subgraph defined by the nodes must also have k-edge-connectivity.\nk_components()\nsimilar to this function, but uses node-connectivity instead of edge-connectivity\nRaises: NetworkXNotImplemented: – If the input graph is a multigraph. ValueError: – If k is less than 1\n\nNotes\n\nAttempts to use the most efficient implementation available based on k. If k=1, this is simply simply connected components for directed graphs and connected components for undirected graphs. If k=2 on an efficient bridge connected component algorithm from _ is run based on the chain decomposition. Otherwise, the algorithm from _ is used.\n\nExample\n\n>>> import itertools as it\n>>> from networkx.utils import pairwise\n>>> paths = [\n... (1, 2, 4, 3, 1, 4),\n... (5, 6, 7, 8, 5, 7, 8, 6),\n... ]\n>>> G = nx.Graph()"
]
| [
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.7677426,"math_prob":0.8888441,"size":1758,"snap":"2022-27-2022-33","text_gpt3_token_len":475,"char_repetition_ratio":0.15108323,"word_repetition_ratio":0.01724138,"special_character_ratio":0.29522184,"punctuation_ratio":0.22740525,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9740601,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-08-09T07:43:34Z\",\"WARC-Record-ID\":\"<urn:uuid:06bd2379-0c30-409a-be26-1ed24d1422cd>\",\"Content-Length\":\"24791\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:964f74e4-ee09-441b-9cae-63b3a78d71c2>\",\"WARC-Concurrent-To\":\"<urn:uuid:0daf5d9d-90b9-4ea2-a2d6-3aad7c77f915>\",\"WARC-IP-Address\":\"185.199.110.153\",\"WARC-Target-URI\":\"https://networkx.org/documentation/networkx-2.3/reference/algorithms/generated/networkx.algorithms.connectivity.edge_kcomponents.k_edge_components.html\",\"WARC-Payload-Digest\":\"sha1:KO6BS57K7TRWMG6S2OPQDIWZ7BZIHHPW\",\"WARC-Block-Digest\":\"sha1:7QRSMHYGJH5BXWA7EMIITCIOCQ4WWLH7\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-33/CC-MAIN-2022-33_segments_1659882570913.16_warc_CC-MAIN-20220809064307-20220809094307-00435.warc.gz\"}"} |
https://annals.math.princeton.edu/2010/171-3/p05 | [
"# The Atiyah-Singer index formula for subelliptic operators on contact manifolds. Part I\n\n### Abstract\n\nThe Atiyah-Singer index theorem gives a topological formula for the index of an elliptic differential operator. The topological index depends on a cohomology class that is constructed from the principal symbol of the operator. On contact manifolds, the important Fredholm operators are not elliptic, but hypoelliptic. Their symbolic calculus is noncommutative, and is closely related to analysis on the Heisenberg group. For a hypoelliptic differential operator in the Heisenberg calculus on a contact manifold we construct a symbol class in the $K$-theory of a noncommutative $C^*$-algebra that is associated to the algebra of symbols. There is a canonical map from this analytic $K$-theory group to the ordinary cohomology of the manifold, which gives a de Rham class to which the Atiyah-Singer formula can be applied. We prove that the index formula holds for these hypoelliptic operators. Our methods derive from Connes’ tangent groupoid proof of the index theorem."
]
| [
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.89389384,"math_prob":0.8823352,"size":978,"snap":"2023-40-2023-50","text_gpt3_token_len":218,"char_repetition_ratio":0.1201232,"word_repetition_ratio":0.0,"special_character_ratio":0.17791411,"punctuation_ratio":0.072289154,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9930311,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-11-29T22:58:41Z\",\"WARC-Record-ID\":\"<urn:uuid:09c8ab54-2e0b-4c16-9f1b-dcced0bfdb88>\",\"Content-Length\":\"23234\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:30abdf7f-bc22-4173-9f7b-db5f46f155a5>\",\"WARC-Concurrent-To\":\"<urn:uuid:35b38ca1-f8c1-445f-a196-fc698aaf6815>\",\"WARC-IP-Address\":\"128.112.19.56\",\"WARC-Target-URI\":\"https://annals.math.princeton.edu/2010/171-3/p05\",\"WARC-Payload-Digest\":\"sha1:74AEIAX2YLPNMRWLTXY44N3KITZOLMA5\",\"WARC-Block-Digest\":\"sha1:XKPM3O54R7SYLBGSZ3HNRHMK4W54D256\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100146.5_warc_CC-MAIN-20231129204528-20231129234528-00745.warc.gz\"}"} |
https://www.arxiv-vanity.com/papers/0711.0336/ | [
"# QCD Thermodynamics\n\nZoltan Fodor\nJohn von Neumann Institute for Computing (NIC), DESY, D-15738, Zeuthen / FZJ, D-52425, Juelich, Germany\nDepartment of Physics, University of Wuppertal, Gauss Strasse 20, D-42119, Wuppertal, Germany\nInstitute for Theoretical Physics, Eotvos University, Pazmany 1, H-1117 Budapest, Hungary Email:\n###### Abstract:\n\nRecent results on QCD thermodynamics are presented. The nature of the T0 transition is determined, which turns out to be an analytic cross-over. The absolute scale for this transition is calculated. The temperature dependent static potential is given. The results were obtained by using a Symanzik improved gauge and stout-link improved fermionic action. In order to approach the continuum limit four different sets of lattice spacings were used with temporal extensions =4, 6, 8 and 10 (they correspond to lattice spacings 0.3, 0.2, 0.15 and 0.12 fm). A new technique is presented, which –in contrast to earlier methods– enables one to determine the equation of state at very large temperatures.\n\n## 1 Introduction\n\nThe QCD transition at non-vanishing temperatures () plays an important role in the physics of the early Universe and of heavy ion collisions (most recently at RHIC at BNL; LHC at CERN and FAIR at GSI will be the next generation of machines). The main goal of the present summary is to present results of the Budapest-Wuppertal group on the QCD transition at vanishing chemical potential (=0), which is of direct relevance for the early universe ( is negligible there) and for present and future heavy ion collisions (at RHIC 40 MeV, which is far less than the typical hadronic scale). Since these results were obtained at four different sets of lattice spacings and a careful continuum extrapolation was performed, we consider them as full results. In addition, a new technique will be introduced, which closes the gap between lattice QCD and the perturbative approach for bulk thermodynamical quantities.",
null,
"Figure 1: The phase diagram of QCD on the hypothetical light quark mass versus strange quark mass plane. Thick lines correspond to second order phase transitions, the purple regions represent first order phase transitions and the yellow region represents an analytic cross-over.\n\nThe standard picture for the QCD phase diagram on the light quark mass () versus strange quark mass () plane is shown by Figure 1. It contains two regions at small and at large quark masses, for which the QCD transition is of first order. Between them one finds a cross-over region, for which the QCD transition is an analytic one. The first order transition regions and the cross-over region are separated by lines, which correspond to second order phase transitions.\n\nWhen we analyze the nature and/or the absolute scale of the QCD transition for the physically relevant case two ingredients are quite important.\n\nFirst of all, one should use physical quark masses. As Figure 1 shows the nature of the transition depends on the quark mass, thus for small or large quark masses it is a first order phase transition, whereas for intermediate quark masses it is an analytic cross over. Though in the chirally broken phase chiral perturbation theory provides a controlled technique to gain information for the quark mass dependence, it can not be applied for the QCD transition (which deals with the restoration of the chiral symmetry). In principle, the behaviour of different quantities in the critical region (in the vicinity of the second order phase transition line) might give some guidance. However, a priori it is not known how large this region is. Thus, the only consistent way to eliminate uncertainties related to non-physical quark masses is to use physical quark masses (which is, of course, quite CPU demanding).",
null,
"Figure 2: The ratio f(cont)/f(Nt) as a function of 1/N2t (left panel). f(cont) is the continuum extrapolated free energy of the staggered fermionic gas in the non-interactive, infinitely high temperature limit. f(Nt) is the value obtained on a lattice with Nt temporal extension. The black line shows our choice (stout improvement, only next-neighbour terms in the action), whereas the red and blue lines represent the Naik and p4 actions, respectively. Masses and taste symmetry violation for different approaches in the literature (right panel). The smallest, physical quark mass and the smallest taste symmetry violation was reached by our works (black dot, [8, 9]). Somewhat larger taste symmetry violation and about three times larger quark masses were reached by the MILC analysis on QCD thermodynamics (red dot, ). Even larger taste symmetry violation and about four times the physical quark masses are the characteristics of the Bielefeld-Brookhaven-Columbia-Riken result on Tc (blue dot, ).\n\nSecondly, the nature of the QCD transition is known to suffer from discretization errors [1, 2, 3]. The three flavour theory with standard action on =4 lattices predicts a critical pseudoscalar mass of about 300 MeV. This point separates the first order and cross-over regions of Figure 1. If we took another discretization, with another discretization error, e.g. the p4 action and =4, the critical pseudoscalar mass turns out to be around 70 MeV (similar effect is observed if one used stout smearing improvement and/or =6). Since the physical pseudoscalar mass (135 MeV) is just between these two values, the discretization errors in the first case would lead to a first order transition, whereas in the second case to a cross-over. The only way to resolve this inconclusive situation is to carry out a careful continuum limit analysis.\n\nSince the nature of the transition influences the absolute scale () of the transition –its value, mass dependence, uniqueness etc.– the above comments are valid for the determination of , too.\n\nTherefore, we have to answer the question: what happens for physical quark masses, in the continuum, at what ? In order to get a reliable answer we used physical quark masses on =4,6,8 and 10 lattices, which correspond to approximately 0.3, 0.2, 0.15 and 0.12 fm lattice spacings, respectively.\n\nIn the presentation of the results of the Bielefeld-Brookhaven-Columbia-Riken Collaboration published results from =4 and 6 were shown (and some unpublished figures for =8, which were obtained within the HotQCD Collaboration). Since the CPU requirements for thermodynamics increase as our =10 simulations need about 50 times more CPU than =6. Do we have 50 times more resources for QCD thermodynamics than our competitors? Of course not (it is almost the other way around). Instead, reaching =10 is a fine balance. It is partly related to the choice of our action (which will be discussed in the next section), partly to the arrangements of the financial resources. For instance, as increases, one needs more and more statistics. Thus the thermalization can be done only once on a relatively expensive, scalable machine, such as Blue-Gene/L, whereas a large fraction of the non-vanishing T simulations can be done on more cost effective devices such as personal computer graphics cards. A 2 years old model can accommodate =6 lattices, on a one-and-a-half year old model you can put =8 lattices and the one year old model can work with quite large =10 lattices. It costs a few hundred dollars and can provide upto 30–60 Gflops sustained QCD performance. They are not easy to code, adding two numbers needs 3 pages, but recently more efficient programming environments were introduced. Clearly, this type of hardware provides a very advantageous price–performance ratio for lattice QCD.\n\n## 2 The choice of the action",
null,
"Figure 3: The volume dependence of the susceptibility peaks for pure SU(3) gauge theory (Polyakov-loop susceptibility, left panel) and for full QCD (chiral susceptibility on Nt=4 and 6 lattices, middle and right panels, respectively).\n\nThe first step is to choose an action, which respects all the needs of a thermodynamic analysis. T=0 simulations are needed to set the scale and for renormalization. T¿0 simulations are needed to map the behaviour of the system at non-vanishing temperatures. The action should maintain a balance between these two needs, leading to approximately the same uncertainties for both sectors (otherwise a large fraction of the CPU-power is used just for “over-killing” one of the two sectors). We used Symanzik improved gauge and stout improved 2+1 flavour staggered fermions (due to the stout improvement we have only next-neighbour terms in the fermionic part of the action). The simulations were done along the line of constant physics. The parameters were tuned with a quite high precision, thus at all lattice spacings the and ratios were set to their experimental values with an accuracy better than 2%.",
null,
"Figure 4: Continuum extrapolated susceptibilities T4/(m2Δχ) as a function of 1/(T3cV). For true phase transitions the infinite volume extrapolation should be consistent with zero, whereas for an analytic crossover the infinite volume extrapolation gives a non-vanishing value. The continuum-extrapolated susceptibilities show no phase-transition-like volume dependence, though the volume changes by a factor of five. The V→∞ extrapolated value is 22(2) which is 11σ away from zero. For illustration, we fit the expected asymptotic behaviour for first-order and O(4) (second order) phase transitions shown by dotted and dashed lines, which results in chance probabilities of 10−19 (7×10−13), respectively.\n\nThe choice of the action has advantages and disadvantages. As we will see the advantages are probably more important than the disadvantages. The left panel of Figure 2 shows the continuum free energy devided by its value at a given . A related plot is usually shown by the Bielefeld-Brookhaven-Columbia-Riken collaboration as a function of . Since in staggered QCD most lattice corrections scale with , which is proportional to 1/, it is instructive to show this ratio as a function of 1/, for our action, for Naik and for p4. Extrapolations from 4 and 6 always overshoot or undershoot. Clearly, the Naik and p4 actions reach the continuum value much faster than our choice, but the scaling appears quite early even for actions with next-neighbour interactions. Extrapolations from and +2 with our action are approximately as good as with the p4 action (which was taylored to be optimal for this quantity, namely for the free energy at infinitely large temperatures). In practice, it means that our choice with =8,10 gives approximately 2% error for the free energy. In a balanced analysis you do not need more, because the corresponding lattice spacings 0.15 and 0.12 fm are most probably not fine enough to set the scale unambiguously with the same accuracy. (E.g. the asqtad action at 10, which corresponds about a=0.12 fm lattice spacing, still has 10% scale difference between & .) Since the p4 action is almost 20 times more expensive than the stout action, it is not worth to pay this price and improve one part of the calculation, which hinders you to reach a reasonable accuracy in another part of the calculation.\n\nThis is the balance one should remember in thermodynamics. Indeed, taste symmetry violation should be suppressed for many reasons (setting the scale at , restoring chiral symmetry at etc.) As it was argued above it is more important to improve on this sector of the calculation than on the infinitely high temperature behaviour. The left panel of Figure 2 shows the splitting between the Goldstone and the first non-Goldstone pions for the Bielefeld-Brookhaven- Columbia-Riken Collaboration (which is far beyond their kaon mass), for MILC and for us. Our small splitting is partly related to the stout improvement and partly to the cost issues, since smaller lattice spacings could be used, which resulted in smaller splitting. In addition, the cost issues allowed us to use in our finite T simulations physical quark masses instead of much larger masses.\n\n## 3 The nature of the QCD transition\n\nThe next topic to be discussed is the nature of the QCD transition. Physical quark masses were used and a continuum extrapolation was carried out by using four different lattice spacings. The details of the calculations can be found in . In order to determine the nature of the transition one should apply finite size scaling techniques for the chiral susceptibility . This quantity shows a pronounced peak as a function of the temperature. For a first order phase transition, such as in the pure gauge theory, the peak of the analogous Polyakov susceptibility gets more and more singular as we increase the volume (V). The width scales with 1/V the height scales with volume (see left panel of Figure 3). A second order transition shows a similar singular behaviour with critical indices. For an analytic transition (what we call a cross-over) the peak width and height saturates to a constant value. That is what we observe in full QCD on =4 and 6 lattices (middle and right panels of Figure 3). We see an order of magnitude difference between the volumes, but a volume independent scaling. It is a clear indication for a cross-over. These results were obtained with physical quark masses for two sets of lattice spacings. Note, however, that for a final conclusion the important question remains: do we get the same volume independent scaling in the continuum; or we have the unlucky case what we had in the Introduction for 3 flavour QCD (namely the discretization errors changed the nature of the transition for the physical pseudoscalar mass case)?\n\nWe carried out a finite size scaling analyses with the continuum extrapolated height of the renormalized susceptibility. The renormalization of the chiral susceptibility can be done by taking the second derivative of the free energy density () with respect to the renormalized mass (). We apply the usual definition: =\n. This quantity has a correct continuum limit. The subtraction term is obtained at =0, for which simulations are carried out on lattices with , spatial and temporal extensions (otherwise at the same parameters of the action). The bare light quark mass () is related to by the mass renormalization constant =. Note that falls out of the combination /=/. Thus, also has a continuum limit (for its maximum values for different , and in the continuum limit we use the shorthand notation ).",
null,
"Figure 6: Temperature dependence of the renormalized chiral susceptibility (m2Δχ¯ψψ/T4), the strange quark number susceptibility (χs/T2) and the renormalized Polyakov-loop (PR) in the transition region. The different symbols show the results for Nt=4,6,8 and 10 lattice spacings (filled and empty boxes for Nt=4 and 6, filled and open circles for Nt=8 and 10). The vertical bands indicate the corresponding transition temperatures and their uncertainties coming from the T≠0 analyses. This error is given by the number in the first parenthesis, whereas the error of the overall scale determination is indicated by the number in the second parenthesis. The orange bands show our continuum limit estimates for the three renormalized quantities as a function of the temperature with their uncertainties.\n\nIn order to carry out the finite volume scaling in the continuum limit we took three different physical volumes (see Figure 4). The inverses of the volumes are shown in units of . For these 3 physical volumes we calculated the dimensionless combination at 4 different lattice spacings: 0.3fm was always off, otherwise the continuum extrapolations could be carried out, which are shown on Figure 4. Our result is consistent with an approximately constant behaviour, despite the fact that we had a factor of 5 difference in the volume. The chance probabilities, that statistical fluctuations changed the dominant behaviour of the volume dependence are negligible. As a conclusion we can say that the staggered QCD transition at mu=0 is a cross-over.\n\n## 4 The transition temperature\n\nAn analytic cross-over, like the QCD transition has no unique . A particularly nice example for that is the water-vapor transition (c.f. Figure 5). Up to about 650 K the transition is a first order one, which ends at a second order critical point. For a first or second order phase transition the different observables (such as density or heat capacity) have their singularity (a jump or an infinitly high peak) at the same pressure. However, at even higher temperatures the transition is an analytic cross-over, for which the most singular points are different. The blue curve shows the peak of the heat capacity and the red one the inflection point of the density. Clearly, these transition temperatures are different, which is a characteristic feature of an analytic transition (cross-over).",
null,
"Figure 7: Difference between the the Tc values obtained by the Polyakov loop and by the chiral condensate as a function of a2.\n\nIn QCD we will study the chiral and the quark number susceptibilities and the Polyakov loop. Usually they give different values, but there is nothing wrong with it. As it was illustrated by the water-vapor transition it is a physical ambiguity, related to the analytic behaviour of the transition. There is another, non-physical, ambiguity. If we used different observables (such as the string tension, , the rho mass or the kaon decay constant), particularly at large lattice spacings we obtain different overall scales. They lead to different values. This ambiguity disappeares in the continuum limit. According to our experiences, at finite lattice spacings, the best choice is the kaon decay constant . It is known experimentally (in contrast to the string tension or ), thus no intermediate calculation with unknown systematics is involved. Furthermore it can be measured on the lattice quite precisely.\n\nFigure 6 shows the results for the chiral susceptibility, for the quark number susceptibility and for the Polyakov loop. Red, blue, green, and purple indicate =4,6,8 and 10 lattices. =4 is always off, the rest scales nicely. The shaded regions indicate the continuum estimates. There is a surprising several sigma effect. The remnant of the chiral transition happens at a quite different temperature than that of the deconfining transition. It is quite a robust statement, since the Polyakov transition region is quite off the -peak, and the -peak is quite far from the inflection point of the Polyakov loop. This quite large differece is also related to the fact that the transition is fairly broad. The widths are around 30-40 MeV.\n\nDue to the broadness of the transition the normalization prescription changes , too. It is easy to imagine why, just multiply a Gaussian by and the peak is shifted. That means using gives about 10 MeV higher than our definition, for which a normalization was applied. (Note, that for the unrenormalized a normalization is natural, whereas for the renormalized the natural normalization is done by . This kind of naturalness manifests itself as possibly small errors of the observable.)\n\nFigure 7 shows the difference between the values obtained by the Polyakov loop and by the chiral condensate as a function of the lattice spacing squared. The blue band indicates the difference for the chiral susceptibility peak position for the and normalization. Thus using the normalization no difference can be seen for =4 and 6, a slight difference is observed for =8 and a reliable continuum extrapolation needs =6,8 and 10.\n\nOur result on and that of the MILC Collaboration (=169(12)(4) MeV ) are consistent within the (quite sizable) errorbars.",
null,
"Figure 8: Possible contributions to the 40 MeV difference between the results of Refs. and .",
null,
"Figure 9: Continuum extrapolations based on Nt=4 and 6 (left panel: inconsistent continuum limit) and using Nt=6,8 and 10 (right panel: consistent continuum limit).\n\nThe ambiguity related to the inconsistent continuum limit is clearly unphysical, and it is resolved as we take smaller and smaller lattice spacings (c.f. Figure 9). The differences between the values for different observables are physical, it is a consequence of the cross-over nature of the QCD transition. There is another phenomenon, namely the volume dependence of , which is also physical. Recently A. Bazavov and B. Berg studied the volume dependence of in the pure SU(3) theory (for which the transition is a first order phase transition). They applied the usual periodic boundary condition (which approximates the thermodynamic limit in a very effective way) and also a disorder wall boundary condition. The difference between the values can be as large as 30 MeV (c.f. Figure 10). Clearly, a detailed study of this phenomenon is needed for dynamical QCD.",
null,
"Figure 10: Transition temperature for the pure SU(3) theory as a function of the spatial size for Nt=4 and 6. The lower, almost horizontal lines represent the results with periodic boundary condition, whereas the upper curves show the results with disorder wall boundary condition (Figure from Ref. ).\n\n## 5 The renormalized static potential in the continuum limit\n\nOur aim is to compute the free energy of a static quark-antiquark pair. There are several measurements on this quantity in the literature (for recent publications cf. [12, 13, 14, 15]). Here we go beyond these computations (for details see ), we use physical quark masses and perform a careful continuum limit extrapolation with the necessary renormalization procedure.\n\nThe quark-antiquark free energy can be expressed as correlators of Polyakov loops:\n\n e−F¯qq(r)/T∼∑x⟨TrP(x)TrP†(x+r)⟩, (1)\n\nwhere r is a vector in the spatial direction, is the temperature and runs over all the spatial lattice sites. is the Polyakov loop.\n\nIn full dynamical QCD at large distances it is favorable to generate a quark-antiquark pair from the vacuum, which then screens the color field between the two Polyakov loops . From this point (the string breaking scale) the lowest energy level will be insensitive of the position of the heavy quarks, resulting in a constant free energy.\n\nAt finite temperature the above picture persists, but we can also have general expectations about the temperature dependence. Physically we expect that in a thermal vacuum it is easier to generate a quark-antiquark pair than at , since there are thermally excited particles around which can scatter on the gluonic string between the static quark-antiquark pair. The gluonic string can more easily break into a dynamical quark-antiquark pair.\n\nWhen we approach the continuum limit, the value of the un-renormalized free energy diverges. This is because in a single Polyakov loop the self-energy is divergent. We expect:\n\n ⟨TrP(x)⟩∣∣∣div=e−C(a)Nta=e−C(a)/T, (2)\n\nwhere in the continuum limit. At finite the specific value of has no physical meaning, since it depends on how we define the “divergent part” of the self-energy (renormalization scheme). In the literature there are several ways to fix this constant [14, 9]. A possible way of fixing is to take a physical observable based on , and require that it should be independent of . In fact, the most useful quantity in our calculation was the constant value of the free energy after the string breaking/screening, at a fixed temperature (its value was set to MeV in the calculation). The value of MeV was motivated by the fact that it is already in the deconfined phase where the statistical errors of the free energy are much smaller than in the confined phase. For the calculation we used the gauge configurations of Ref . Having determined the renormalized free energy for all lattice spacings, we could take the continuum limit by using the and 10 free energies, and extrapolating in . In the studied region and 10 results almost completely coincide. Therefore a safe extrapolation to is possible. We estimate the systematic error of this extrapolation by comparing the results coming from extrapolation and extrapolation. The result for the renormalized free energy at different temperatures, including both the statistical and the systematic errors, can be seen on Figure 11.\n\n## 6 The equation of state at high temperatures\n\nWhile available lattice results for the equation of state (both for pure gauge theory and for full QCD) end at around , standard perturbation theory converges only at extremely high temperatures (at , the different perturbative orders can not even tell the sign of the deviation from the Stefan-Boltzmann limit). Until recently no link between the two most systematic methods of QCD, namely perturbation theory and lattice\nQCD, existed for bulk thermodynamical quantities.\n\nIt is easy to understand the difficulties for lattice calculations beyond these (e.g. ) temperatures. The correctly renormalized pressure is obtained by subtracting the vacuum term, thus =-, where = and is the partition function. Since the partition function itself is difficult to determine, usually the integral method is used, which e.g. for the plaquette action is given by\n\n pren(T)=p(T)−p(T=0)=∫dβ(⟨Pl⟩T−⟨Pl⟩0) (3)\n\nAs an example imagine a calculation at on =8 lattices. A back of an envelope estimate can convince you that it corresponds to a lattice spacings of about 0.0075 fm or lattice sizes of about 1000 in full QCD. This is clearly out of reach in the near future.\n\nIn the perturbative approach the infrared properties of the non-abelian gauge theories at non-vanishing temperatures are responsible for the bad convergence.",
null,
"Figure 12: The pressure, normalized to its Stefan-Boltzmann value, as a function of the temperature obtained by our new technique. Results with small discretization errors (Nt=8, blue circles), seem to fit to the improved perturbation theory prediction, and also reproduce results obtained by the standard method at lower temperatures. At the highest temperature, 3⋅107⋅Tc, the pressure (within its statistical uncertainty) is consistent with the Stefan-Boltzmann limit.\n\nTo create a link between these two systematic approaches, we suggested a method to obtain lattice results for the pressure at temperatures, which were previously unreachable. It becomes possible to compare the data with perturbation theory formulae. The results are obtained using two different techniques: the first one is a new way to renormalize the pressure, the second one is a direct method to measure the pressure. For illustration, the techniques and the results for pure SU(3) gauge theory are presented. The extension to full QCD is straightforward.\n\nThe basic idea of the first technique is the observation, that the the divergences that are removed by renormalization are independent of the temperature. Thus the subtraction of the vacuum term, which is needed for renormalization, can be done at non-vanishing temperatures (actually even at very high temperatures). Thus, one can build up as a sum of differences =[-]+[-]+ In fact, we usually measure the dimensionless pressure, which can be obtained by including the factors. This leads to increasing powers of 1/ in the successive terms, thus only a few of them are needed. Applying this scheme, we may reach arbitrarily high temperatures using lattices with only and temporal extents. It is worth mentioning, that a similar formula can be constructed for the case of the normalized interaction measure .\n\nNote, that independently of the renormalization procedure, a problem emerges within the integral method framework. Since strictly speaking, the pressure is only exactly zero at , in principle we would have to carry out the integration starting from zero temperature. Thus, there is an uncertainty of setting the lower point of the integral. This problem is solved by our direct method (c.f. ).\n\nLet us consider the first bracketed term in .\n\n [p(T)−p(T/2)]=1NtN3slogZ(Nt)−12NtN3slogZ(2Nt)=12NtN3slog(Z(Nt)2Z(2Nt)). (4)\n\nThis can be determined by using (action with one boundary in the temporal direction), (action with two boundaries in the temporal direction) and an interpolating partition function = Using , one obtains\n\n [p(T)−p(T/2)]∼log(Z(Nt)2Z(2Nt))=log(¯Z(1)¯Z(0))=∫10dαdlog¯Z(α)dα=∫10dα⟨S1b−S2b⟩ (5)\n\nSo we can calculate the pressure itself at any given temperature, without carrying out simulations at lower temperatures, and then performing an integral. (Note, however, that huge cancellations appear also within this framework).\n\nThe method discussed above gives us the possibility to measure the pressure at very high temperatures. This was carried out using lattices with temporal extension , and . The temperature interval ranged from upto . Our results are shown on figure 12. A comparison can be done with the standard method, namely the result of for smaller temperatures. Here we present the results which are closest to the continuum limit, thus . These results nicely follow the perturbative predictions ( for a recent paper see and references therein).\n\n## 7 Conclusion\n\nQCD thermodynamics results from the Budapest-Wuppertal group were summarized. The necessary balance between and simulations was discussed in detail. As a consequence, Symanzik improved gauge and stout-link improved staggered fermionic lattice action was used in the simulations with an exact simulation algorithm. Physical masses were taken both for the light quarks and for the strange quark. The parameters were tuned with a quite high precision, thus at all lattice spacings the and ratios were set to their experimental values with an accuracy better than 2%. Four sets of lattice spacings on lattices with =4,6,8 and 10 temporal extensions were used (they correspond to lattice spacing 0.3, 0.2, 0.15 and 0.12 fm) to carry out the continuum extrapolation. It turned out that only =6,8 and 10 can be used for a controlled extrapolation, =4 is out of the scaling region.\n\nThe nature of the T0 transition was determined. The renormalized chiral susceptibility was extrapolated to vanishing lattice spacing for three physical volumes, the smallest and largest of which differ by a factor of five. This ensures that a true transition should result in a dramatic increase of the susceptibilities. No such behaviour is observed: the finite-size scaling analysis showed that the finite-temperature QCD transition in the hot early Universe was not a real phase transition, but an analytic crossover (involving a rapid change, as opposed to a jump, as the temperature varied). As such, it will be difficult to find experimental evidence of this transition from astronomical observations. Since for present day heavy ion experiments the baryonic chemical potential is also very small, the above results apply for them, too.\n\nThe absolute scale for the T0 transition was calculated. Since the QCD transition is a non-singular cross-over there is no unique . This well-known phenomenon was illustrated on the water-vapor phase diagram. Different observables led to different numerical values in the continuum and thermodynamic limit also in QCD. Three observables were used to determine the corresponding transition temperatures. The peak of the renormalized chiral susceptibility predicted =151(3)(3) MeV, whereas based on the strange quark number susceptibility resulted in 24(4) MeV larger value. Another quantity, which is related to the deconfining phase transition in the large quark mass limit is the Polyakov loop. Its behavior predicted a 25(4) MeV larger transition temperature, than that of the chiral susceptibility. Another consequence of the cross-over are the non-vanishing widths of the peaks even in the thermodynamic limit, which were also determined. For the chiral susceptibility, strange quark number susceptibility and Polyakov-loop we obtained widths of 28(5)(1) MeV, 42(4)(1) MeV and 38(5)(1) MeV, respectively.\n\nThe temperature dependent static potential was given. The same action and the same set of configurations were used as for the determination of . Since results for different lattice spacings were available a careful renormalization program was carried out.\n\nThese features, numbers and functions are attempted to be the full result for the 0 transition, though other lattice fermion formulations —e.g. Wilson fermions (for ongoing projects see e.g. [21, 22]) or chiral fermions (for an early dynamical overlap test see , for the domain wall approach a recent presentation can be found in Ref.)— are needed to cross-check the findings with staggered fermions.\n\nA new technique is presented, which –in contrast to earlier methods– enables one to determine the equation of state at very large temperatures. The method is based on the observation, that the divergences, which are removed by renormalization, are independent of the temperature. Thus the subtraction of the vacuum term, which is needed for renormalization, can be done at non-vanishing temperatures (actually even at very high temperatures). A direct method was also suggested, which does not need an integration over the temperature. Results for the pure SU(3) theory are presented upto temperatures.\n\n## Acknowledgment\n\nPartial support of grants of DFG F0 502/1, EU I3HP, OTKA AT049652 and OTKA K68108 is acknowledged. The author thanks R. Hoffmann, S.D. Katz and K.K. Szabo for careful reading of the manuscript."
]
| [
null,
"https://media.arxiv-vanity.com/render-output/3685609/x1.png",
null,
"https://media.arxiv-vanity.com/render-output/3685609/x2.png",
null,
"https://media.arxiv-vanity.com/render-output/3685609/x4.png",
null,
"https://media.arxiv-vanity.com/render-output/3685609/x5.png",
null,
"https://media.arxiv-vanity.com/render-output/3685609/x7.png",
null,
"https://media.arxiv-vanity.com/render-output/3685609/x8.png",
null,
"https://media.arxiv-vanity.com/render-output/3685609/x9.png",
null,
"https://media.arxiv-vanity.com/render-output/3685609/x10.png",
null,
"https://media.arxiv-vanity.com/render-output/3685609/x11.png",
null,
"https://media.arxiv-vanity.com/render-output/3685609/x13.png",
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.9163746,"math_prob":0.9650784,"size":34460,"snap":"2021-04-2021-17","text_gpt3_token_len":7922,"char_repetition_ratio":0.15239726,"word_repetition_ratio":0.047283884,"special_character_ratio":0.22504354,"punctuation_ratio":0.12484625,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97369796,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20],"im_url_duplicate_count":[null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-01-28T15:24:08Z\",\"WARC-Record-ID\":\"<urn:uuid:8bc5b88b-7c69-4238-99b6-b8b4bc218821>\",\"Content-Length\":\"367145\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b3357857-238d-4c1d-b8a8-76811455d9d1>\",\"WARC-Concurrent-To\":\"<urn:uuid:156b5a0e-f2cd-4610-b8a0-097ceefd4fcb>\",\"WARC-IP-Address\":\"172.67.158.169\",\"WARC-Target-URI\":\"https://www.arxiv-vanity.com/papers/0711.0336/\",\"WARC-Payload-Digest\":\"sha1:3KXISQFIDU6WQ6T7AUK4SXOOUJN4LZGJ\",\"WARC-Block-Digest\":\"sha1:DUOG6G6IVIMLSVCEKTM25URGUXXLLDOF\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-04/CC-MAIN-2021-04_segments_1610704847953.98_warc_CC-MAIN-20210128134124-20210128164124-00249.warc.gz\"}"} |
https://oeis.org/A054899 | [
"This site is supported by donations to The OEIS Foundation.",
null,
"Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)\n A054899 a(n) = Sum_{k>0} floor(n/10^k). 58\n 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 11, 11, 11 (list; graph; refs; listen; history; text; internal format)\n OFFSET 0,21 COMMENTS The old definition of this sequence was \"Highest power of 10 dividing n!\", but that is wrong (see A027868). For example, the highest power of 10 dividing 5!=120 is 1; however, a(5)=0. - Hieronymus Fischer, Jun 18 2007 Highest power of 10 dividing the quotient of multifactorials product{k>=1, M(10^k,10^k*floor(n/10^k))}/product{k>=1, M(10^(k-1),10^(k-1)*floor(n/10^k))} where M(r,s) is the r-multifactorial (r>=1) of s which is defined by M(r,s)=s*M(r,s-r) for s>0, M(r,s)=1 for s<=0. This is because that quotient of multifactorials evaluates to the product 10^floor(n/10)*10^floor(n/100)*... - Hieronymus Fischer, Jun 14 2007 Partial sums of A122840. - Hieronymus Fischer, Jun 06 2012 LINKS Hieronymus Fischer, Table of n, a(n) for n = 0..10000 Eric Weisstein's World of Mathematics, Multifactorial. FORMULA floor[n/10] + floor[n/100] + floor[n/1000] + floor[n/10000] + .... a(n) = (n - A007953(n))/9. From Hieronymus Fischer, Jun 14 2007, Jun 25 2007, and Aug 13 2007: (Start) a(n) = sum {k>0, floor(n/10^k)}. a(n) = sum{10<=k<=n, sum{j|k,j>=10, floor(log_10(j))-floor(log_10(j-1))}}. G.f.: g(x) = sum{k>0, x^(10^k)/(1-x^(10^k))}/(1-x). G.f. expressed in terms of Lambert series: g(x) = L[b(k)](x)/(1-x) where L[b(k)](x) = sum{k>=0, b(k)*x^k/(1-x^k)} is a Lambert series with b(k)=1, if k>1 is a power of 10, else b(k)=0. G.f.: g(x) = sum{k>0, c(k)*x^k}/(1-x), where c(k)=sum{j>1,j|k, floor(log_10(j))-floor(log_10(j-1))}. a(n) = sum_{0<=k<=floor(log_10(n))} ds_10(floor(n/10^k))*10^k - n where ds_10(x) = digital sum of x in base 10. a(n) = sum_{0<=k<=floor(log_10(n))} A007953(floor(n/10^k))*10^k - n. Recurrence: a(n) = floor(n/10) + a(floor(n/10)). a(10*n) = n + a(n). a(n*10^m) = n*(10^m-1)/9 + a(n). a(k*10^m) = k*(10^m-1)/9, for 0<=k<10, m>=0. Asymptotic behavior: a(n) = n/9 + O(log(n)), a(n+1) - a(n) = O(log(n)), which follows from the inequalities below. a(n) <= (n - 1)/9; equality holds for powers of 10. a(n) >= n/9 - 1 - floor(log_10(n)); equality holds for n=10^m-1, m>0. lim inf (n/9 - a(n)) = 1/9, for n-->oo. lim sup (n/9 - log_10(n) - a(n)) = 0, for n-->oo. lim sup (a(n+1) - a(n) - log_10(n)) = 0, for n-->oo. (End) EXAMPLE a(11)=1 a(111)=12. a(1111)=123. a(11111)=1234. a(111111)=12345. a(1111111)=123456. a(11111111)=1234567. a(111111111)=12345678. a(1111111111)=123456789. MATHEMATICA Table[t = 0; p = 10; While[s = Floor[n/p]; t = t + s; s > 0, p *= 10]; t, {n, 0, 100} ] PROG (PARI) a(n)=my(s); while(n\\=10, s+=n); s \\\\ Charles R Greathouse IV, Jul 19 2011 CROSSREFS Cf. A011371 and A054861 for analogs involving powers of 2 and 3. Different from the highest power of 10 dividing n! (see A027868 for reference). Cf. A027868, A067080, A098844, A132027, A122840. Sequence in context: A132272 A179051 A324160 * A061217 A102684 A156821 Adjacent sequences: A054896 A054897 A054898 * A054900 A054901 A054902 KEYWORD nonn AUTHOR Henry Bottomley, May 23 2000 EXTENSIONS An incorrect g.f. was deleted by N. J. A. Sloane, Sep 13 2009 Examples added by Hieronymus Fischer, Jun 06 2012 STATUS approved\n\nLookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam\nContribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recent\nThe OEIS Community | Maintained by The OEIS Foundation Inc.\n\nLast modified November 22 13:47 EST 2019. Contains 329393 sequences. (Running on oeis4.)"
]
| [
null,
"https://oeis.org/banner2021.jpg",
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.58178467,"math_prob":0.99880034,"size":3554,"snap":"2019-43-2019-47","text_gpt3_token_len":1644,"char_repetition_ratio":0.1371831,"word_repetition_ratio":0.11456628,"special_character_ratio":0.5655599,"punctuation_ratio":0.23980099,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99973994,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-11-22T19:17:13Z\",\"WARC-Record-ID\":\"<urn:uuid:46daf13f-d861-4ead-9d92-fcf0205aa69a>\",\"Content-Length\":\"25056\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e3b78a15-d4a2-4b58-abb8-03b3f8ab89b3>\",\"WARC-Concurrent-To\":\"<urn:uuid:4009d1e3-2070-46b3-809c-03b77828ab8b>\",\"WARC-IP-Address\":\"104.239.138.29\",\"WARC-Target-URI\":\"https://oeis.org/A054899\",\"WARC-Payload-Digest\":\"sha1:2B2AH2ZXWJFWKS7VIAGG5QADAATOBTE5\",\"WARC-Block-Digest\":\"sha1:NQWTXHITPOGHCRQ4MNXZ6NEEFQM6C73B\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-47/CC-MAIN-2019-47_segments_1573496671411.14_warc_CC-MAIN-20191122171140-20191122200140-00097.warc.gz\"}"} |
https://vdocuments.site/chapter-5-solid-state-diffusion.html | [
"# chapter 5 solid state diffusion",
null,
"",
null,
"",
null,
"",
null,
"",
null,
"Post on 03-Jan-2017\n\n223 views\n\nCategory:\n\n## Documents\n\n0 download\n\nEmbed Size (px)\n\nTRANSCRIPT\n\n• 1\n\n18.03.05 10:36\n\nCHAPTER 5\n\nSOLID STATE DIFFUSION INTRODUCTION Numerous chemical reactions or micro-structural changes in solids take place through solid state diffusion, i.e. the movement and transport of atoms in solid phases. The diffusion takes place because of the presence of defects in solids. Point defects, e.g. vacancies and interstitial ions, are responsible for lattice diffusion. Diffusion also takes place along line and surface defects which include grain boundaries, dislocations, inner and outer surfaces, etc. As diffusion along linear, planar and surface defects is generally faster than in the lattice, they are also termed high diffusivity or easy diffusion paths. Another frequently used term is short circuit diffusion. The relative contribution of the different types of diffusion in oxides and other inorganic compounds are functions of the temperature, partial pressures or activities of the constituents of the compounds, the microstructure, grain size, porosity etc. Grain boundary and dislocation diffusion generally have smaller activation energies than lattice diffusion and as a result become increasingly important the lower the temperature in solids with a given microstructure. In the literature on diffusion and diffusion-controlled reactions or processes one encounters many different terms that describe the diffusional behaviour under different experimental conditions: tracer and self-diffusion of atoms and ions, diffusion of defects, chemical diffusion, ambipolar diffusion, a.o. Many of these are used for treating diffusion in compounds, and in the following chapters these phenomena and terms will be described in more detail. Here we will start out with a few simple phenomenological descriptions, and for simplicity we look only at diffusion of neutral, independent particles. PHENOMENOLOGICAL DESCRIPTIONS OF DIFFUSION It is a well known phenomenon that heat flows from hot to cold regions. Such a flow of heat in a temperature gradient is described by Fourier's law\n\n• 2\n\nJq = - cond\n\ndTdx (5.1)\n\nJq is the heat flux, i.e. the flow of heat per unit area of the plane through which the heat\n\ntraverses per second, dTdx the temperature gradient, and cond the thermal conductivity. It may\n\nbe noted that the minus sign reflects that the heat flows from high to low temperatures; in the direction of the heat flow the temperature gradient is - dTdx .\n\nFick's first law The expression for the flow of particles from high to low concentrations is analogous to that for the flow of heat and is given by the expression\n\nJparticles = - D dcdx\n\n(5.2)\n\nJparticles is the particle flux, dcdx\n\nthe concentration gradient of the particles and D the\n\ndiffusion coefficient. As in the equation for the heat flux, the minus sign reflects that the particles flow from high to low concentration of particles. This relation is called Fick's first law after A. Fick who first formulated this relation.\n\nFig.5.1. Schematic illustration of Fick's first law. The particle concentration gradient is the \"driving force\" of the diffusion\n\nJparticles represents the number of particles (or moles) crossing a unit area (cm2 or m2) per unit time (second). If the concentration of particles is expressed in number of particles per cm3 and the distance x in cm, the diffusion coefficient has the dimension cm2s-1. In SI units the concentration is expressed in number of particles per m3 and the diffusion coefficient has\n\nx\n\nCon\n\nc.\n\n-dc/dxGradient:\n\n• 3\n\nthe dimension m2s-1. The particle flux and particle gradient is illustrated schematically in Fig.5.1. Diffusion: potential gradients as the driving force. In Fick's first law in Eq.5.2 the particle gradient, dcdx , may be considered to be an expression for the \"driving force\" for the\n\nparticle flux (Eq.5.2). The larger the concentration gradient, the larger the particle flux. But to be more accurate the driving force for the diffusion really constitutes the chemical potential gradient of the particles that diffuse (provided that no other forces act on the particles). Correspondingly, the driving force for the transport of electrical charges in a homogeneous solid is the electrical potential gradient. In the following is given a brief derivation of Fick's first law using the chemical potential gradient as the driving force. Let us consider the homogeneous transport of particles of type \"i\" across a plane under a driving force F. The particle flux through a plane is given by the product of the volume concentration ci of the particles at the plane and the average migration or drift velocity vi of the particles ji (particles/(cm2sec)) = ci (particles/cm3) .vi (cm/sec) (5.3) For uncorrelated movements the drift velocity of a particle is proportional to the driving force F exerted on the particle: vi = BiF (5.4) The proportionality factor Bi is termed the mobility (\"Beweglichkeit\") of the particles and is defined as the average drift velocity per unit driving force. The driving force is, in turn, given by the negative value of the potential gradient normal to the cross-sectional area in the plane F = - dPdx (5.5)\n\nwhere P is a potential. The negative sign is - as above - due to the fact that the transport takes place from higher to lower values of P. In the direction of transport the potential gradient is -dP/dx This is illustrated in Fig.5.2.\n\n• 4\n\nFig.5.2 Schematic illustration of Fick's first law. The potential gradient is the driving force of the diffusion.\n\nWhen one combines Eqs.5.3 - 5.5 the particle flux becomes ji = civi = -ci Bi\n\ndPdx (5.6)\n\nIf the particles move in a chemical potential gradient, the potential P equals the chemical potential of particles of type i: P = i. Equation 5.6 then takes the form\n\nji = -ci Bi didx (5.7)\n\nThe chemical potential i is related to the chemical activity ai of species i through i =\n\noi + kT lnai (5.8)\n\nIf ideal conditions can be assumed, the activity can be set equal to the concentration of i, ci, and the chemical potential gradient is then given by\n\ndidx = kT\n\ndlncidx =\n\nkTci\n\ndcidx (5.9)\n\nWhen one substitutes the expression for the chemical potential gradient in Eq.5.7, the particle flux becomes\n\nji = - ciBididx = - BikT\n\ndcidx (5.10)\n\nBikT is termed the diffusion coefficient Di of particles i: Di = BikT (5.11)\n\nx\n\nPot\n\nentia\n\nl-dP/dxGradient:\n\n• 5\n\nBy combination with Eq.5.11, Eq.5.10 takes the form of Fick's first law as given in Eq.5.2:\n\nji = - Di dcidx (5.2)\n\nThus this expression for Fick's first law presupposes that ideal conditions prevail and thus that the concentration ci of particles i can be substituted for their activity ai. If this is not so, Fick's first law should be expressed in terms of the chemical potential gradient, e.g.\n\nji = - ciDikT\n\ndidx (5.12)\n\nThis expression for the particle flux will be used later on, e.g. in considering transport of ions in an electrochemical potential gradient as for instance in the derivation of the rate constant in Carl Wagner's treatment of the parabolic oxidation of metals. It is very important to be aware that Eqs. 5.2 and 5.12 are fully valid only for ideal cases of diffusion of neutral particles in chemical potential gradients only. If the particles are charged, we need to take into account both the electrical potential gradent and requirements to the combination of fluxes to maintain a given total current and electroneutrality. These cases will be treated later on. If the activity differs greatly from the concentration and thus that the chemical potential gradient differs greatly from the particle gradient, it is in principle possible to have diffusion along the chemical potential gradient but against the particle concentration gradient. This phenomenon has been termed \"uphill diffusion\". In the following ideal conditions will be assumed and concentrations will be used for activities of atoms, ions and various types of defects. But it should be recalled that this always represents an approximation. Simplified model for one-dimensional diffusion Following Eq.5.2 the diffusion coefficient represents the proportionality constant between the particle flux and the concentration gradient. But the diffusion of atoms in solids represents the sum of a large number of particles or atoms that each makes a large number of jumps . We will therefore attempt to describe the diffusion coefficient in terms of the number of jumps per unit time (the jump frequency) and the distance that each particle or atom moves in each jump.\n\n• 6\n\nFor this purpose let us consider a simplified one-dimensional model where particles jump\n\nbetween parallel planes separated by a distance s as illustrated in Fig.5.3. The two\n\nneighbouring planes under consideration are termed plane 1 and 2. The number of particles\n\nper unit area in plane 1 and 2 is termed n1 and n2, respectively, and let us further assume that\n\nn1>n2. Consider further that the particles in plane 1 and 2 may jump from one plane to a\n\nneighbouring plane at a jump frequency . The particles in plane 1 have an equal probability\n\nof jumping to plane 2 and to the neighbouring plane in the opposite direction. The total\n\nnumber of particles jumping out of a unit area of plane 1 per unit time is equal to the product\n\nof number of particles per unit area times the jump frequency: n1. As the particles may jump\n\nin opposite directions, the number of particles jumping from plane 1\n\nRecommended"
]
| [
null,
"data:image/png;base64,R0lGODlhAQABAAD/ACwAAAAAAQABAAACADs=",
null,
"data:image/png;base64,R0lGODlhAQABAAD/ACwAAAAAAQABAAACADs=",
null,
"data:image/png;base64,R0lGODlhAQABAAD/ACwAAAAAAQABAAACADs=",
null,
"data:image/png;base64,R0lGODlhAQABAAD/ACwAAAAAAQABAAACADs=",
null,
"data:image/png;base64,R0lGODlhAQABAAD/ACwAAAAAAQABAAACADs=",
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.89089924,"math_prob":0.9532121,"size":9431,"snap":"2021-21-2021-25","text_gpt3_token_len":2068,"char_repetition_ratio":0.18489446,"word_repetition_ratio":0.018099548,"special_character_ratio":0.20824939,"punctuation_ratio":0.09823964,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9864266,"pos_list":[0,1,2,3,4,5,6,7,8,9,10],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-06-19T22:46:11Z\",\"WARC-Record-ID\":\"<urn:uuid:6c47023a-036c-4f5a-9508-d578c878013a>\",\"Content-Length\":\"92759\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d0315cbc-5c23-40ea-b57b-fd5cd0db35d2>\",\"WARC-Concurrent-To\":\"<urn:uuid:78fe365d-8db3-4827-ac92-8777bca08733>\",\"WARC-IP-Address\":\"5.196.140.102\",\"WARC-Target-URI\":\"https://vdocuments.site/chapter-5-solid-state-diffusion.html\",\"WARC-Payload-Digest\":\"sha1:RWA43JDQJTQO2TQKGGQA7MSGSRHZG5ME\",\"WARC-Block-Digest\":\"sha1:54Z55FNYCODFLRA3M4NSVSJWDPUUFMFA\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-25/CC-MAIN-2021-25_segments_1623487649731.59_warc_CC-MAIN-20210619203250-20210619233250-00150.warc.gz\"}"} |
https://ncna17.kattis.com/problems/stoichiometry | [
"",
null,
"2017 North Central NA Regional Contest",
null,
"#### Start\n\n2017-10-28 16:05 UTC\n\n## 2017 North Central NA Regional Contest\n\n#### End\n\n2017-10-28 21:05 UTC\nThe end is near!\nContest is over.\nNot yet started.\nContest is starting in -749 days 14:56:47\n\n5:00:00\n\n0:00:00\n\n# Problem AStoichiometry",
null,
"Image by Gerd Altmann.\n\nYou have landed a lucrative contract with Amalgamated Chemical Manufacturing (ACM), to help their chemists with stoichiometry. Stoichiometry is the calculation of reactants and products in chemical reactions, based on the law of conservation of mass, which states that the total mass of the reactants equals the total mass of the products. The relations among quantities of reactants and products typically form a ratio of positive integers. If the amounts of the separate reactants are known, then the amount of the product can be calculated, and vice-versa. The relationship of reactants to products can be described using a soichiometric equation such as:\n\n\\begin{equation} \\rm {C H_4 + 2 O_2 \\rightarrow C O_2 + 2 H_2 O}, \\end{equation}\n\nwhich can be read as: “One molecule of $\\rm C H_4$ and two molecules of $\\rm O_2$ yield one molecule of $\\rm C O_2$ and two molecules of $\\rm H_2 O$.” The total number of atoms of each element on the left hand side of the stoichiometric equation must match the number of atoms of that element on right hand side. Your task is to write a program that, given an equation of the form:\n\n\\begin{equation} \\rm {\\_ H_2 O + \\_ C O_2 \\rightarrow \\_ O_2 + \\_ C_6 H_{12} O_6}, \\label{exampleeq} \\end{equation}\n\nwill fill in the blanks to produce a balanced equation. For example, the above equation could be balanced as follows:\n\n\\begin{equation} \\rm {6H_2O + 6CO_2 \\rightarrow 6O_2 + 1C_6H_{12}O_6}. \\end{equation}\n\n## Input\n\nAn equation is input in the form of a sequence of $M$ $(1 < M \\le 20)$ lines, one for each molecule in the formula (e.g., $\\rm {H_2 O}$ or $\\rm {CO_2}$). Each line $m$ ($1\\le m \\le M$) has the following fields:\n\n\\begin{equation*} sign_ m\\; \\; N_ m\\; \\; element_{m,1}\\; \\; count_{m,1}\\; \\; \\ldots \\; \\; element_{m,{N_ m}}\\; \\; count_{m,{N_ m}} \\end{equation*}\n\nwhere $sign_ m$ is either +1 or -1, with a sign of +1 indicating that this molecule appears on the left of the equation, and -1 indicating that it appears on the right. $N_ m$, where $0 < N_ m < 20$, is the number of $element$/$count$ pairs following on the line. Each $element_{m,n}$, where $1 \\le n \\le N_ m$, is an element name consisting of one or two upper or lowercase letters, and each $count_{m,n}$ is a positive integer, $1 \\leq count_{m,n} \\leq 12$. For example, the element/count pair “Fe 2” indicates that the molecule contains two atoms of the element Fe (iron). There will be no more than 10 unique elements in a single equation.\n\nNote that an element may be repeated in a given line of input, as in\n+1 6 C 1 H 5 C 1 O 1 O 1 H 1\nwhich specifies that at least one molecule of $\\rm CH_5COOH$ appears on the left side of the equation. Note that $\\rm CH_5COOH$ can be written as $\\rm C_2H_6O_2$.\n\nInput ends with a line of the form\n0 0\n\n## Output\n\nThe program must output the numbers $C_1,\\ldots ,C_ M$ $(0<C_ i\\le 1\\, 000)$, in order, to fill in the blanks in the equation. Each number, $C_ m\\mid 1\\le m \\le M$, must be the minimum number for which the equation is balanced (i.e. there is no common factor that could reduce all of the $C_ m$ coefficients). You may assume that every input test case has exactly one valid solution meeting these criteria.\n\nSample Input 1 Sample Output 1\n+1 2 H 2 O 1\n+1 2 C 1 O 2\n-1 1 O 2\n-1 3 C 6 H 12 O 6\n0 0\n\n6 6 6 1\n\nSample Input 2 Sample Output 2\n+1 5 Be 2 C 1 O 3 O 2 H 2\n+1 3 Ac 1 O 1 H 1\n-1 4 Be 4 O 1 Ac 6 O 6\n-1 2 H 2 O 1\n-1 2 C 1 O 2\n0 0\n\n2 6 1 5 2"
]
| [
null,
"https://ncna17.kattis.com/images/site-logo",
null,
"https://ncna17.kattis.com/images/sessions/ncna17",
null,
"https://ncna17.kattis.com/problems/stoichiometry/file/statement/en/img-0001.png",
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.8008076,"math_prob":0.99933755,"size":3392,"snap":"2019-43-2019-47","text_gpt3_token_len":1099,"char_repetition_ratio":0.13252656,"word_repetition_ratio":0.01540832,"special_character_ratio":0.3304835,"punctuation_ratio":0.10013351,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99951017,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,null,null,null,null,3,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-11-17T07:01:46Z\",\"WARC-Record-ID\":\"<urn:uuid:e0f0526f-4bf4-437f-b98f-b43fed26d684>\",\"Content-Length\":\"21952\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ac04f824-d361-499e-a1e5-96aa9572b8aa>\",\"WARC-Concurrent-To\":\"<urn:uuid:d57be5db-5407-4e39-938e-75755d6fbe69>\",\"WARC-IP-Address\":\"104.26.12.234\",\"WARC-Target-URI\":\"https://ncna17.kattis.com/problems/stoichiometry\",\"WARC-Payload-Digest\":\"sha1:EGJGOPXOH5WFAPKSQ4JNLPNY62Q3OOSC\",\"WARC-Block-Digest\":\"sha1:KUHJAAPDLJB7YJFSC2DOVRQ54XPVEFGK\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-47/CC-MAIN-2019-47_segments_1573496668896.47_warc_CC-MAIN-20191117064703-20191117092703-00110.warc.gz\"}"} |
https://www.hindawi.com/journals/jdm/2013/721051/ | [
"/ / Article\n\nResearch Article | Open Access\n\nVolume 2013 |Article ID 721051 | https://doi.org/10.1155/2013/721051\n\nS. Arumugam, I. Sahul Hamid, V. M. Abraham, \"Decomposition of Graphs into Paths and Cycles\", Journal of Discrete Mathematics, vol. 2013, Article ID 721051, 6 pages, 2013. https://doi.org/10.1155/2013/721051\n\n# Decomposition of Graphs into Paths and Cycles\n\nRevised12 Mar 2013\nAccepted12 Mar 2013\nPublished14 Apr 2013\n\n#### Abstract\n\nA decomposition of a graph is a collection of edge-disjoint subgraphs of such that every edge of belongs to exactly one . If each is a path or a cycle in , then is called a path decomposition of . If each is a path in , then is called an acyclic path decomposition of . The minimum cardinality of a path decomposition (acyclic path decomposition) of is called the path decomposition number (acyclic path decomposition number) of and is denoted by () (()). In this paper we initiate a study of the parameter and determine the value of for some standard graphs. Further, we obtain some bounds for and characterize graphs attaining the bounds. We also prove that the difference between the parameters and can be made arbitrarily large.\n\n#### 1. Introduction\n\nGraph decomposition problems rank among the most prominent areas of research in graph theory and combinatorics and further it has numerous applications in various fields such as networking, block designs, and bioinformatics.\n\nA decomposition of a graph is a collection of edge-disjoint subgraphs of such that every edge of belongs to exactly one . Various types of path decompositions and corresponding parameters have been studied by several authors by imposing conditions on the paths in the decomposition. It is obvious that every graph admits a decomposition in which each subgraph is either a path or a cycle. In this connection, Erds asked what is the minimum number of paths into which every connected graph on vertices can be decomposed and Gallai conjectured that this number is at most as stated below.\n\nGallai's Conjecture (see ). If is a connected graph on vertices, then can be decomposed into paths.\n\nA good number of research articles have been published in which Gallai's is the focus of study and still this conjecture remains unsettled for more than 30 years. Towards a proof of the conjecture, Lovasz made the first significant contribution by proving the following theorem.\n\nTheorem 1 (see ). A graph on vertices (not necessarily connected) can be decomposed into paths and cycles.\n\nGallai's conjecture and Theorem 1 motivate the following definition.\n\nDefinition 2. Let be a decomposition of a graph . If each is either a path or a cycle, then is called a path decomposition of . If each is a path, then is called an acyclic path decomposition of . The minimum cardinality of a path decomposition of is called the path decomposition number. Similarly the acyclic path decomposition number of is defined and is denoted by .\n\nThe parameter was introduced by Harary and further studied extensively by Harary and Schwenk , Péroche , Stanton et al. [4, 5], and Arumugam and Suseela . This paper initiates a study of the parameter path decomposition number.\n\n#### 2. Basic Terminologies and Results\n\nBy a graph we mean a finite, connected, and undirected graph without loops or multiple edges. For graph theoretic terminology we refer to Chartrand and Lesniak . The order and size of a graph are denoted by and , respectively.\n\nIf is a path in a graph , then the vertices are called internal vertices of and and are called external vertices of . For a cycle , the vertex is called the initial vertex and all other vertices are internal vertices. Two paths and of a graph are said to be internally disjoint if no vertex of is an internal vertex of both and . If and are two paths in , then the walk obtained by concatenating and at is denoted by , that is, . Further when , the path is denoted by .\n\nA caterpillar is a tree in which removal of all the pendant vertices results in a path. For vertices and in a connected graph , the detour distance is the length of a longest path in . The detour diameter of is defined to be . Two graphs are said to be homeomorphic if both can be obtained from the same graph by a sequence of subdivisions of edges. The length of a longest cycle in a graph is called the circumference of .\n\nWe need the following theorems.\n\nTheorem 3 (see ). For any tree , , where is the number of vertices of odd degree.\n\nTheorem 4 (see ). Let be a unicyclic graph with cycle . Let denote the number of vertices of degree greater than two on . Let denote the number of vertices of odd degree. Then\n\nTheorem 5 (see ). For the complete graph , one has .\n\nTheorem 6 (see ). For the complete bipartite graph with , one has where is the conventional Kronecker delta.\n\n#### 3. Bounds for\n\nIn this section we obtain some bounds for the path decomposition number and characterize the graphs attaining some of the bounds obtained. Let us first establish an upper and lower bound involving the order of the graph and the number of vertices of odd degree. For this we define the term of a graph . Given a path decomposition for a graph , let , where denotes the number of internal vertices of . Now, define , where the maximum is taken over all path decompositions of .\n\nTheorem 7. Let be a graph on vertices. Then , where is the number of vertices of odd degree in .\n\nProof. Let be the collection of all path decompositions of and let .\nThen Hence, so that . Also, it is not difficult to see that and hence , which implies that . The upper bound follows immediately from Theorem 1.\n\nRemark 8. If is a graph with vertices of odd degree, then if and only if there exists a path decomposition of such that every vertex is an internal vertex of members in .\n\nThe following examples show that there are infinite families of graphs attaining the bounds given in the above theorem and consequently prove that these bounds are sharp. That is, the family of all trees and an infinite class of unicyclic graphs achieve the lower bound, whereas some common classes of graphs such as complete graphs, wheels, and complete bipartite graphs of odd size attain the upper bound. In this connection we need to note down the inequality that which follows from the fact that every acyclic path decomposition of a graph is also a path decomposition.\n\nExample 9. (i) For any tree , .\n(ii) If is a unicyclic graph with cycle and if denotes the number of vertices on with degree greater than two, then\n\nProof. (i) Since every path decomposition of is an acyclic path decomposition of , we have and hence the result follows from Theorem 3.\n(ii) If , then so that . If , then it follows from Theorem 4 that and hence . Further, for any path decomposition of all the odd vertices and at least one vertex on are external vertices of some paths in so that . Thus . If , by Theorem 4, we have and hence . Thus as .\n\nExample 10. (i) For the complete graph , we have .\n(ii) For the complete bipartite graph with\n(iii) If denotes the wheel on vertices, then .\n\nProof. (i) Let be any path decomposition of . Since every member of covers at most edges, we have so that . Further, it follows from Theorem 1 that . Thus .\n(ii) Assume first that . Now, if is odd, then every vertex is of odd degree and hence . If is even, then can be decomposed into hamiltonian cycles and hence . Further, for any path decomposition of , every member of covers at most edges so that and hence . Thus .\nSuppose . If is odd, then every vertex is of odd degree and hence . If is even, then . Further, for any path decomposition of , every member in covers at most edges and hence . Thus .\n(iii) Since the number of vertices of odd degree in is or accordingly as is odd or even, it follows from Theorem 7 that . Also by Theorem 1, we have and hence .\n\nWe now proceed to obtain some lower and upper bounds for involving detour diameter, circumference, and maximum degree of a graph along with the characterization of graphs attaining the bounds.\n\nTheorem 11. For any graph , , where is the detour diameter of . Further, equality holds if and only if is a caterpillar with .\n\nProof. Let be a path in of length . Then is a path decomposition of and . Hence, .\nSuppose . Let be a path of length in . If there is a vertex not on with , let be a longest path which contains and edge-disjoint from . Thus, , where is the set of edges not covered by the paths and is a path decomposition of and , which is a contradiction. Thus every vertex not on is a pendent vertex and hence is a caterpillar.\nNow, if there exists a vertex on with , let and be two vertices not on which are adjacent to . Then , where is the set of edges not covered by the paths and is a path decomposition of and , which is a contradiction. Hence, , for all .\nThe converse is obvious.\n\nTheorem 12. For any graph , , where is the circumference of . Further, equality holds if and only if is either a cycle or a cycle with exactly one chord or a unicyclic graph with and every vertex not on the cycle is a pendant vertex.\n\nProof. Let be a cycle of length . Then is a path decomposition of and . Hence .\nSuppose . Let be a cycle of length in . We consider the following two cases.\nCase 1 . We claim that has at most one chord. Suppose has two chords say and . If and , then , where is the set of edges not covered by and the path is a path decomposition of and , which is a contradiction. Suppose and . Let and denote the two -sections of the cycle and without loss of generality we assume that contains one of the -sections of . Now, let denote the path consisting of the edge , followed by the path and the edge . Then , where denotes the set of edges not covered by the paths and is a path decomposition of and , which is a contradiction. Hence, has at most one chord. Thus is a cycle or a cycle with exactly one chord.\nCase 2 . Suppose that there exists a vertex not on with . Let be a longest path which contains and edge-disjoint from . Thus , where is the set of edges not covered by and is a path decomposition of and , which is a contradiction. Thus every vertex not on is a pendant vertex.\nIf there exists a vertex on with , let be a longest path containing and edge-disjoint from . Then , where is the set of edges not covered by and is a path decomposition of with , which is a contradiction. Hence every vertex on has degree less than or equal to .\nNow, we claim that the cycle has no chord. Suppose is a chord of . Let be a pendant vertex not on . Then there exists a vertex on different from the vertices and such that and are adjacent. Let and denote the two -sections of and without loss of generality we assume that lies on . Let be the path consisting of the edge followed by the path and the edge . Now, , where is the set of edges not covered by the paths and is a path decomposition of and , which is a contradiction. Hence, has no chords.\nThus is a unicyclic graph with and every vertex not on the cycle is a pendant vertex. The converse is obvious.\n\nRemark 13. Obviously, for any vertex of a graph , at most two of the edges incident with can be covered by a member of a path decomposition of . As this is true in particular for a vertex of maximum degree it follows that . This bound is sharp as a complete graph of odd order attains this. In fact, every graph which is hamiltonian cycle decomposable attains the bound. Further, for any tree , if and only if is homeomorphic to the star .\n\nThe following theorem characterizes the family of unicyclic graphs achieving the bound given in Remark 13.\n\nTheorem 14. Let be a unicyclic graph with cycle and . Then if and only if every vertex not on has degree 1 or 2, at most two vertices on have degree and has exactly one vertex with degree if .\n\nProof. Let be a unicyclic graph with . Let be a minimum path decomposition of . Let be a vertex in with . Then every member of passes through .\nWe claim that lies on the cycle . This is obvious if . If , let and be the paths in covering the edges of the cycle . Since both and pass through ; it follows that lies on . Thus every vertex of degree lies on and consequently every vertex not on has degree 1 or 2.\nFurther if , it follows that contains exactly one vertex of degree and if , then contains at most two vertices of degree .\nThe converse is obvious.\n\nIn the following theorem we characterize graphs of order with and having minimum (maximum) number of edges.\n\nTheorem 15. Let be a graph of order with . Then . Further, if and only if is homeomorphic to the star . Also if and only if is hamiltonian cycle decomposable or is the complete graph with even.\n\nProof. Let be a graph with . Then is connected and hence .\nNow, suppose . Then is a tree. As it follows that is homeomorphic to the star . Conversely, if is homeomorphic to , then .\nNow, suppose . Then is -regular. If is even, then for any minimum path decomposition of , we have and each path in must cover edges so that is hamiltonian cycle decomposable. Conversely, if is hamiltonian cycle decomposable, then . If is odd, then and hence . Hence, and is even. The converse follows from Example 10(i).\n\n#### 4. Relation between the Parameters and\n\nIn this section we establish some relation between the parameters and .\n\nTheorem 16. For any graph , one has . Further, if and only if every component of is a cycle.\n\nProof. The first inequality is already seen. For the other inequality, consider a path decomposition of and spilt each cycle member of into two paths and thus an acyclic path decomposition of with cardinality at most is obtained, which proves the required inequality. Now suppose . Let be a minimum path decomposition of . Then every member of is a cycle. Further, all the cycles in are vertex-disjoint and hence every component of is a cycle. The converse is obvious.\n\nRemark 17. The first part of the inequalities given in the above theorem is strict in the sense that they both are equal for infinitely many graphs. For example, one has the following.(i)For even integers , we have , which follows from Theorem 5 and Example 10(i). (ii)Example 9 and Theorem 4 together show that the parameters and are equal for all trees and also for the class of all unicyclic graphs other than the cycles.\n\nThough the upper bound suggested by Gallai's conjecture for and the upper bound given by Theorem 1 for differ by at most 1, the difference between the actual values of these parameters can be made arbitrarily large, which we proceed to prove. The following lemma is useful in this regard.\n\nLemma 18. Let be a connected graph which can be decomposed into a cycle and a path . Then can be decomposed into two paths if and only if has a vertex of degree 2 which lies on or on the -section of where is the smallest positive integer such that and and is the largest positive integer such that and .\n\nProof. Suppose can be decomposed into two paths and . Since and are the only two vertices of odd degree in , we may assume that is the origin of and is the origin of . Let be the terminus of . Since the degree of is even, it follows that is not an internal vertex of so that is the terminus of as well, and the degree of is 2. Clearly cannot lie on the -section or the -section of .\nConversely, suppose has a vertex of degree two satisfying the conditions stated in the lemma. We consider two cases.\nCase 1 . Since is connected, . If , then trivially can be decomposed into two paths. Hence, we assume that . If there exists a -section of which is internally disjoint from , let be the path consisting of the -section of followed by the -section of which is not internally disjoint from and let be the path consisting of the -section of which is internally disjoint from followed by the -section of . Then is a decomposition of . The proof is similar if there exists a -section of which is internally disjoint from .\nNow, suppose both the -sections of and both the -sections of are not internally disjoint from . Let be the largest positive integer such that and . Then the -section of and one of the -sections of , say , are internally disjoint. If has length greater than 1, then any internal vertex of has degree 2 and the -section of is internally disjoint from and hence the proof is complete. If the -section of has length greater than 1, let be the path consisting of the -section of followed by the -section of which is different from followed by the edge and let be the path consisting of the -section of followed by followed by -section of . Then is a decomposition of .\nCase 2 (no vertex on has degree 2). In this case lies on the -section of . Let and let and be integers with such that and no internal vertex of the -section of lies on . Without loss of generality we assume that there exists a -section of , say , which contains and and let be the other -section of . Let be the path consisting of the -section of , followed by and the -section of . Let be the path consisting of the -section of followed by and the -section of . Then is a decomposition of .\n\nCorollary 19. If is a graph admitting a minimum path decomposition having exactly one cycle , then . Further, if and only if there exists a path in with and the induced subgraph satisfies the conditions stated in Lemma 18.\n\nTheorem 20. Given any positive integer , there exists a graph such that .\n\nProof. Let be the graph consisting of vertex-disjoint cycles and a - path , where , for . Clearly . Further, . We now prove that by induction on .\nIf , then it follows from Lemma 18 that . Assume that , for all graphs as described above with cycles. Let be such a graph with cycles. Let be an acyclic path decomposition of . Let be the path in containing the edge . Let , , and . Let be the paths in which decompose the graph and let be the paths in which decompose the graph . Clearly . We now claim that . Let be the end vertex of which lies in . Then the -section of together with the paths decompose the graph . Hence, by induction hypothesis so that . Hence, so that . Thus and so .\n\nTheorem 21. Let be a connected graph. Then if and only if is the graph consisting of vertex-disjoint cycles and a path satisfying the following conditions. (i)Every vertex of the cycle , , lies on the path . (ii)If and denote, respectively, the smallest and the largest positive integer such that , then every edge of the -section of is a chord of the cycle .\n\nProof. Suppose . Let be a minimum path decomposition of . Then consists of exactly one path, say and vertex-disjoint cycles, say . Let . Since is connected, for all . Now, if there exists a vertex of degree 2 in which lies either on any of the cycles or on the -section of then it follows from Lemma 18 that , which is a contradiction. Hence, (i) and (ii) hold.\nNow, the graph given in the theorem is nothing but the graph described in Corollary 19 with and thus the converse follows.\n\nTheorem 22. Given two positive integers and with , there exists a graph such that and .\n\nProof. If , let be the graph with components, each component being a cycle. If , let be the graph given in Theorem 21. If , let be the graph consisting of exactly one cut vertex and blocks, each block being a cycle. In all these cases, and .\nNow, suppose . Let , where . Let be the connected graph consisting of vertex-disjoint cycles and a path satisfying the following conditions: (i) if and is a proper nonempty subset of if ; (ii)if and denote, respectively, the smallest and largest positive integer such that , , then each edge of the -section of is a chord of the cycle .\nClearly and it follows from Lemma 18 that .\n\nWe conclude the paper by posing the following problem for further researchs. (i)Characterize graphs for which , where is the number of vertices of odd degree in . (ii)Characterize graphs for which . (iii)Characterize graphs for which . (iv)Characterize graphs for which .\n\n1. L. Lovasz, “On covering of graphs,” in Theory of Graphs, P. Erdös and G. Katona, Eds., Procedure Collage, pp. 231–236, Academic Press, Tihany, Hungary, 1968. View at: Google Scholar\n2. F. Harary and A. J. Schwenk, “Evolution of the path number of a graph: covering and packing in graphs II,” in Graph Theory and Computing, C. Road, Ed., pp. 39–45, Academic Press, New York, NY, USA, 1972.\n3. B. Péroche, “The path-numbers of some multipartite graphs,” Annals of Discrete Mathematics, vol. 9, pp. 195–197, 1980.\n4. R. G. Stanton, D. D. Cowan, and L. O. James, “Some results on path numbers,” in Proceedings of the Louisiana Conference on Combinatorics, Graph Theory and computing, pp. 112–1135, 1970. View at: Google Scholar\n5. R. G. Stanton, L. O. James, and D. D. Cowan, “Tripartite path numbers,” in Graph Theory and Computing, R. C. Read, Ed., pp. 285–294, Academic Press, New York, NY, USA, 1972.\n6. S. Arumugam and J. S. Suseela, “Acyclic graphoidal covers and path partitions in a graph,” Discrete Mathematics, vol. 190, no. 1–3, pp. 67–77, 1998.\n7. G. Chartrand and L. Lesniak, Graphs & Digraphs, CRC Press, Boca Raton, Fla, USA, 4th edition, 2005. View at: Zentralblatt MATH | MathSciNet\n\n#### More related articles\n\nWe are committed to sharing findings related to COVID-19 as quickly as possible. We will be providing unlimited waivers of publication charges for accepted research articles as well as case reports and case series related to COVID-19. Review articles are excluded from this waiver policy. Sign up here as a reviewer to help fast-track new submissions."
]
| [
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.9336201,"math_prob":0.9744535,"size":21340,"snap":"2020-45-2020-50","text_gpt3_token_len":4943,"char_repetition_ratio":0.1875703,"word_repetition_ratio":0.2175,"special_character_ratio":0.23580131,"punctuation_ratio":0.14315021,"nsfw_num_words":2,"has_unicode_error":false,"math_prob_llama3":0.99959284,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-12-05T06:24:53Z\",\"WARC-Record-ID\":\"<urn:uuid:f1fa4067-fe49-4b63-9583-d0d7f2cde8dd>\",\"Content-Length\":\"1049254\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e19e953c-6962-4d97-939f-7fe9a4d1e1a6>\",\"WARC-Concurrent-To\":\"<urn:uuid:bcd2c8a3-b05c-42db-90e4-48104605c41b>\",\"WARC-IP-Address\":\"13.32.204.76\",\"WARC-Target-URI\":\"https://www.hindawi.com/journals/jdm/2013/721051/\",\"WARC-Payload-Digest\":\"sha1:7VY3JUERS3BKJ6AVBB4RHEOOS6IB6MW4\",\"WARC-Block-Digest\":\"sha1:V22GKTQONJ7MJ2DCP4PCBVMFXTFYWJVT\",\"WARC-Truncated\":\"length\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141746320.91_warc_CC-MAIN-20201205044004-20201205074004-00130.warc.gz\"}"} |
https://ouncetocup.com/1.5-oz-to-cups/ | [
"# 1.5 Ounce to Cups (1.5 oz to cups conversion)\n\nAre you cooking your favorite dish? The detailed chart in the recipe includes the calculation of 1.5 ounces to cups conversion? Do not worry; check this conversion tool to find how many 1.5 ounces equal to cups in a minute. This 1.5 oz to cup converter gives an exact measurement for any recipe you prepare.\n\nOunce Value:\n\nOunces\n\nCup Value:\n\nCups\n\n1.5 Ounce = 0.1875 Cup\n(1.5 oz = 0.1875 cup)\n\nTry our auto 1.5 ounce to cup calculator (Without Convert Button), Just change the first field value and you got final value.",
null,
"## How many cups is 1.5 oz?\n\nWe know that the volume value of 1.5 oz is equal to 0.1875 cups. If you want to convert 1.5 fluid oz to an equal number of cups, just divide the volume value by 8. Hence, 1.5 Ounce is equal to 0.1875 cup.\n\nThe Answer is: 1.5 US Fluid Ounces = 0.1875 US Cup\n\n1.5 oz = 0.1875 cup\n\nMany of them try to search or find an answer for what is 1.5 ounces in cups? So, we’ll start with 1.5 fl oz to cups conversion to know how big is 1.5 oz.\n\n## How To Calculate 1.5 fluid oz to cups?\n\nTo calculate 1.5 fluid ounces to an equal number of cups, simply follow the steps below.\n\nFluid Ounces to Cups formula is:\n\nCup = Fluid Ounce / 8\n\nAssume that we are finding out how many cups were found in 1.5 fl oz of water, multiply it either by 0.125 or divide the value by 8 to get the result.\n\nApplying to Formula: Cup = 1.5 oz / 8 = 0.1875 cup.\n\n## How to convert 1.5 oz to cups?\n\n• To convert 1.5 fluid ounces to cups,\n• Simply multiply the 1.5 fluid ounces by 0.125, or simply divide them by 8.\n• Applying to the formula, cups = 1.5 ounces * 0.125 [1.5x0.125] (or) cups = 1.5 ounces/8 = 1.5/8 = 0.1875.\n• Hence, 1.5 ounces is equal to 0.1875 cups.\n\n## Some quick table references for fluid ounce to cup conversions:\n\nOunce [oz] Cup [c]\n1 oz0.125 cup\n2 oz0.25 cup\n3 oz0.375 cup\n4 oz0.5 cup\n5 oz0.625 cup\n6 oz0.75 cup\n7 oz0.875 cup\n8 oz1 cup\n9 oz1.125 cup\n10 oz1.25 cup\n11 oz1.375 cup\n12 oz1.5 cup\n\n## Reverse Calculation: How many ounces are in 1.5 cups?\n\n• To convert 1.5 cups to ounces,\n• Simply multiply the 1.5 cups by 8.\n• Then, applying the formula, ounce = 1.5 cups * 8 [1.5x8 = 12].\n• Hence, 1.5 cup is equal to 12 ounces.\n\n### Related Converter:\n\nFormula: Ounce to Cup\n\nCup = Ounce / 8\n\nApplying to Formula,\n\nCup = 1.5/8 = 0.1875\n\n1.5 Ounce = 0.1875 Cup"
]
| [
null,
"https://ouncetocup.com/images/1.5-oz-to-cups.jpg",
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.8127335,"math_prob":0.9990534,"size":2070,"snap":"2023-40-2023-50","text_gpt3_token_len":731,"char_repetition_ratio":0.18151017,"word_repetition_ratio":0.023752969,"special_character_ratio":0.38115942,"punctuation_ratio":0.1797153,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99947065,"pos_list":[0,1,2],"im_url_duplicate_count":[null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-27T08:28:45Z\",\"WARC-Record-ID\":\"<urn:uuid:6f37b090-1a8b-4044-a328-fd311392a0ad>\",\"Content-Length\":\"16527\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:71b74773-3c4e-4b1c-90de-c7a3dce33219>\",\"WARC-Concurrent-To\":\"<urn:uuid:1e3d24e8-e835-4fd4-a690-ab9b01c45be6>\",\"WARC-IP-Address\":\"172.67.201.37\",\"WARC-Target-URI\":\"https://ouncetocup.com/1.5-oz-to-cups/\",\"WARC-Payload-Digest\":\"sha1:GK7XZ3O7HVKA2HMSMTUZCQAOX6URQABC\",\"WARC-Block-Digest\":\"sha1:ACA6D665AJZNTDSVUDOAGRL7NBME3IZN\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233510284.49_warc_CC-MAIN-20230927071345-20230927101345-00610.warc.gz\"}"} |
https://uk.mathworks.com/matlabcentral/cody/problems/1861-binary/solutions/863149 | [
"Cody\n\n# Problem 1861. Binary\n\nSolution 863149\n\nSubmitted on 3 Apr 2016 by HH\nThis solution is locked. To view this solution, you need to provide a solution of the same size or smaller.\n\n### Test Suite\n\nTest Status Code Input and Output\n1 Pass\nn = 1; y_correct = ; assert(isequal(binary(n),y_correct))\n\ny = 1\n\n2 Pass\nn = 7; y_correct = [1 1 1]; assert(isequal(binary(n),y_correct))\n\ny = 1 1 1\n\n3 Pass\nn = 10; y_correct = [1 0 1 0]; assert(isequal(binary(n),y_correct))\n\ny = 1 0 1 0"
]
| [
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.5012624,"math_prob":0.9946066,"size":568,"snap":"2020-24-2020-29","text_gpt3_token_len":185,"char_repetition_ratio":0.14361702,"word_repetition_ratio":0.0,"special_character_ratio":0.36619717,"punctuation_ratio":0.102564104,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9940924,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-07-13T16:57:21Z\",\"WARC-Record-ID\":\"<urn:uuid:fb28010f-bdd4-42e0-a94a-d633aa95b813>\",\"Content-Length\":\"71906\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:96382667-4e5c-4690-801e-2da3c0e11a90>\",\"WARC-Concurrent-To\":\"<urn:uuid:d6d45914-a9d6-4148-a106-b85a5a1904fa>\",\"WARC-IP-Address\":\"23.66.56.59\",\"WARC-Target-URI\":\"https://uk.mathworks.com/matlabcentral/cody/problems/1861-binary/solutions/863149\",\"WARC-Payload-Digest\":\"sha1:O32KJG75NZFZATCWV5SZWTN5YUG4D76J\",\"WARC-Block-Digest\":\"sha1:VQSCMHTN65SN3IV4IS62HOQDAUS4Z52Y\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-29/CC-MAIN-2020-29_segments_1593657146247.90_warc_CC-MAIN-20200713162746-20200713192746-00095.warc.gz\"}"} |
https://www.manualslib.com/manual/1582109/Hp-12c-Platinum.html?page=260 | [
"# Straight-line Depreciation; Sum-of-the-years-digits Depreciation; Declining-balance Depreciation - HP 12C Platinum User Manual\n\nFinancial calculator.\n\n260 Appendix E: Formulas Used\n\n## Straight-Line Depreciation\n\nKeyboard function:\nSBV\nSAL\nDPN\n=\nJ\nL\nProgram for partial first year:\nSBV\nSAL\nDPN\n=\n1\nL\nSBV\nSAL\nDPN\n=\nJ\nL\nDPN\n= RDV\nL + 1\nL\n\n### Sum-of-the-Years-Digits Depreciation\n\n(\nW\n+\nSOYD\n=\nk\nwhere W = integer part of k\nF = fractional part of k.\n(i.e., for k = 12.25 years, W = 12 and F = 0.25).\nKeyboard function:\n(\n) 1\nL\nj\n+\nDPN\n=\nJ\nSOYD\nL\nProgram for partial year:\nL\nDPN\n=\n1\nSOYD\nDPN\n=\nj\nSOYD\nL\n=\nwhere\n\n### Declining-Balance Depreciation\n\nKeyboard function:\nDPN\nRBV\n=\nj\nj\n1\nProgram for partial first year:\nFACT\nDPN\nSBV\n=\n1\n100\nfor j = 1, 2, ..., L\nY\n1\n12\nfor j = 2, 3, ..., L\n)( 1\nW\n2\nF\n)\n+\n2\n(\nSBV\nSAL\n)\nY\n1\n(\n⋅ ⎟\n⋅ ⎟\nSBV\nSAL\n12\nj\n2\n+\n⋅ ⎟ ⎟\n(\nSBV\nD\n1\nY\n1\n12\nFACT\nfor j = 1, 2, ..., L\n100\nL\nY\n1\nL\n12\n)\nSAL\n)\nfor j ≠ 1",
null,
"",
null,
""
]
| [
null,
"https://www.manualslib.com/public/img/loading.gif",
null,
"https://www.manualslib.com/code.php",
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.54937285,"math_prob":0.97160584,"size":888,"snap":"2019-51-2020-05","text_gpt3_token_len":535,"char_repetition_ratio":0.12330317,"word_repetition_ratio":0.06276151,"special_character_ratio":0.4358108,"punctuation_ratio":0.16831683,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9970263,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-12-06T15:08:38Z\",\"WARC-Record-ID\":\"<urn:uuid:ab2b177e-88bc-45ae-881a-0a30695322bf>\",\"Content-Length\":\"322039\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:3b03ddab-8e85-43b8-9169-e0a504936d2d>\",\"WARC-Concurrent-To\":\"<urn:uuid:104870f3-5496-4b33-9800-525c353abde7>\",\"WARC-IP-Address\":\"213.174.130.220\",\"WARC-Target-URI\":\"https://www.manualslib.com/manual/1582109/Hp-12c-Platinum.html?page=260\",\"WARC-Payload-Digest\":\"sha1:4RKHUFPNOTEGX4SX25EQX5VPQ2Z4RIS7\",\"WARC-Block-Digest\":\"sha1:QEPURAZSHLB3X5JYELGQRE3VENHFAOMP\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-51/CC-MAIN-2019-51_segments_1575540488870.33_warc_CC-MAIN-20191206145958-20191206173958-00033.warc.gz\"}"} |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.